{"id": 0, "contents": "CHAPTER 1 <br> Structure and Bonding - \nFIGURE 1.1 The enzyme HMG-CoA reductase, shown here as a so-called ribbon model, catalyzes a crucial step in the body's synthesis of cholesterol. Understanding how this enzyme functions has led to the development of drugs credited with saving millions of lives. (credit: image from the RCSB PDB (rcsb.org) of PBD ID 1HW9 (E.S. Istvan, J. Deisenhofer) (2001) Structural mechanism for statin inhibition of HMGCoA reductase Science 292: 1160-1164/RCSB PDB, CC BY 1.0)"}
{"id": 1, "contents": "CHAPTER CONTENTS - \n1.1 Atomic Structure: The Nucleus\n1.2 Atomic Structure: Orbitals\n1.3 Atomic Structure: Electron Configurations\n1.4 Development of Chemical Bonding Theory\n1.5 Describing Chemical Bonds: Valence Bond Theory\n$1.6 s p^{3}$ Hybrid Orbitals and the Structure of Methane\n$1.7 s p^{3}$ Hybrid Orbitals and the Structure of Ethane\n$1.8 s p^{2}$ Hybrid Orbitals and the Structure of Ethylene\n1.9 sp Hybrid Orbitals and the Structure of Acetylene\n1.10 Hybridization of Nitrogen, Oxygen, Phosphorus, and Sulfur\n1.11 Describing Chemical Bonds: Molecular Orbital Theory\n1.12 Drawing Chemical Structures\n\nWHY THIS CHAPTER? We'll ease into the study of organic chemistry by first reviewing some ideas about atoms, bonds, and molecular geometry that you may recall from your general chemistry course. Much of the material in this chapter and the next is likely to be familiar to you, but it's nevertheless a good idea to make sure you understand it before moving on.\n\nWhat is organic chemistry, and why should you study it? The answers to these questions are all around you.\n\nEvery living organism is made of organic chemicals. The proteins that make up your hair, skin, and muscles; the DNA that controls your genetic heritage; the foods that nourish you; and the medicines that heal you are all organic chemicals. Anyone with a curiosity about life and living things, and anyone who wants to be a part of the remarkable advances taking place in medicine and the biological sciences, must first understand organic chemistry. Look at the following drawings for instance, which show the chemical structures of some molecules whose names might be familiar to you. Although the drawings may appear unintelligible at this point, don't worry. They'll make perfectly good sense before long, and you'll soon be drawing similar structures for any substance you're interested in.\n\n\nOxycodone (OxyContin)\n\n\nCholesterol\n\n\nBenzylpenicillin"}
{"id": 2, "contents": "CHAPTER CONTENTS - \nOxycodone (OxyContin)\n\n\nCholesterol\n\n\nBenzylpenicillin\n\nHistorically, the term organic chemistry dates to the mid-1700s, when it was used to mean the chemistry of substances found in living organisms. Little was known about chemistry at that time, and the behavior of the \"organic\" substances isolated from plants and animals seemed different from that of the \"inorganic\" substances found in minerals. Organic compounds were generally low-melting solids and were usually more difficult to isolate, purify, and work with than high-melting inorganic compounds.\n\nBy the mid-1800s, however, it was clear that there was no fundamental difference between organic and inorganic compounds. The only distinguishing characteristic of organic compounds is that all contain the element carbon.\n\nOrganic chemistry, then, is the study of carbon compounds. But why is carbon special? Why, of the more than 197 million presently known chemical compounds, do almost all of them contain carbon? The answers to these questions come from carbon's electronic structure and its consequent position in the periodic table (FIGURE 1.2). As a group 4A element, carbon can share four valence electrons and form four strong covalent bonds. Furthermore, carbon atoms can bond to one another, forming long chains and rings. Carbon, alone of all elements, is able to form an immense diversity of compounds, from the simple methane, with one carbon atom, to the staggeringly complex DNA, which can have more than 100 million carbons."}
{"id": 3, "contents": "CHAPTER CONTENTS - \n| Grou 1A |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | 8A |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| H | 2A |  |  |  |  |  |  |  |  |  |  | 3A | 4A | 5A | 6A | 7A | He |\n| Li | Be |  |  |  |  |  |  |  |  |  |  | B | C | N | 0 | F | Ne |\n| Na | Mg |  |  |  |  |  |  |  |  |  |  | Al | Si | P | S | Cl | Ar |\n| K | Ca | Sc | Ti | V | Cr | Mn | Fe | Co | Ni | Cu | Zn | Ga | Ge | As | Se | Br | Kr |\n| Rb | Sr | Y | Zr | Nb | Mo | Tc | Ru | Rh | Pd | Ag | Cd | In | Sn | Sb | Te | I | Xe |\n| Cs | Ba | La | Hf | Ta | W | Re | Os | Ir | Pt | Au | Hg | TI | Pb | Bi | Po | At | Rn |\n| Fr | Ra | Ac |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |\n\nFIGURE 1.2 Carbon, hydrogen, and other elements commonly found in organic compounds are shown in the colors typically used to represent them.\n\nNot all carbon compounds are derived from living organisms, however. Modern chemists have developed a remarkably sophisticated ability to design and synthesize new organic compounds in the laboratory-medicines, dyes, polymers, and a host of other substances. Organic chemistry touches the lives of\neveryone. Its study can be a fascinating undertaking."}
{"id": 4, "contents": "CHAPTER CONTENTS - 1.1 Atomic Structure: The Nucleus\nAs you might remember from your general chemistry course, an atom consists of a dense, positively charged nucleus surrounded at a relatively large distance by negatively charged electrons (FIGURE 1.3). The nucleus consists of subatomic particles called neutrons, which are electrically neutral, and protons, which are positively charged. Because an atom is neutral overall, the number of positive protons in the nucleus and the number of negative electrons surrounding the nucleus are the same.\n\n\nFIGURE 1.3 A schematic view of an atom. The dense, positively charged nucleus contains most of the atom's mass and is surrounded by negatively charged electrons. The three-dimensional view on the right shows calculated electron-density surfaces. Electron density increases steadily toward the nucleus and is 40 times greater at the blue solid surface than at the gray mesh surface.\n\nAlthough extremely small-about $10^{-14}$ to $10^{-15}$ meter ( m ) in diameter-the nucleus nevertheless contains essentially all the mass of the atom. Electrons have negligible mass and circulate around the nucleus at a distance of approximately $10^{-10} \\mathrm{~m}$. Thus, the diameter of a typical atom is about $2 \\times 10^{-10} \\mathrm{~m}$, or 200 picometers (pm), where $1 \\mathrm{pm}=10^{-12} \\mathrm{~m}$. To give you an idea of how small this is, a thin pencil line is about 3 million carbon atoms wide. Although most chemists throughout the world use the International System (SI) of units and describe small distances in picometers, many organic chemists and biochemists in the United States still use the unit angstrom ( $\\AA$ ) to express atomic distances, where $1 \\AA=100 \\mathrm{pm}=10^{-10} \\mathrm{~m}$. As you probably did in your general chemistry course, however, we'll stay with SI units in this book."}
{"id": 5, "contents": "CHAPTER CONTENTS - 1.1 Atomic Structure: The Nucleus\nA specific atom is described by its atomic number ( $Z$ ), which gives the number of protons (or electrons) it contains, and its mass number ( $\\boldsymbol{A}$, which gives the total number of protons and neutrons in its nucleus. All the atoms of a given element have the same atomic number: 1 for hydrogen, 6 for carbon, 15 for phosphorus, and so on; but they can have different mass numbers depending on how many neutrons they contain. Atoms with the same atomic number but different mass numbers are called isotopes. The element carbon, for instance, has three isotopes that occur naturally, with mass numbers of 12,13 , and 14. Carbon- 12 has a natural abundance of $98.89 \\%$, carbon-13 has a natural abundance of $1.11 \\%$, and carbon-14 has only a negligible natural abundance.\n\nThe weighted-average of an element's naturally occurring isotopes is called atomic weight and is given in unified atomic mass units (u) or daltons (Da) where 1 u or 1 Da is defined as one twelfth the mass of one atom of carbon-12. Thus, the atomic weight is 1.008 u for hydrogen, 12.011 u for carbon, 30.974 u for phosphorus, and so on. Atomic weights of all elements are given in the periodic table in Appendix $D$."}
{"id": 6, "contents": "CHAPTER CONTENTS - 1.2 Atomic Structure: Orbitals\nHow are the electrons distributed in an atom? You might recall from your general chemistry course that, according to the quantum mechanical model, the behavior of a specific electron in an atom can be described by a mathematical expression called a wave equation-the same type of expression used to describe the motion of waves in a fluid. The solution to a wave equation is called a wave function, or orbital, and is denoted by the lowercase Greek letter psi ( $\\psi$ ).\nWhen the square of the wave function, $\\psi^{2}$, is plotted in three-dimensional space, an orbital describes the volume of space around a nucleus that an electron is most likely to occupy. You might therefore think of an orbital as looking like a photograph of the electron taken at a slow shutter speed. In such a photo, the orbital would appear as a blurry cloud, indicating the region of space where the electron has been. This electron cloud doesn't have a sharp boundary, but for practical purposes we can set the limits by saying that an orbital represents the space where an electron spends $90 \\%$ to $95 \\%$ of its time.\n\nWhat do orbitals look like? There are four different kinds of orbitals, denoted $s, p, d$, and $f$, each with a different shape. Of the four, we'll be concerned primarily with $s$ and $p$ orbitals because these are the most common in\norganic and biological chemistry. An $s$ orbital has a spherical shape, with the nucleus at its center; a $p$ orbital has a dumbbell shape with two parts, or lobes; and four of the five $d$ orbitals have a cloverleaf shape with four lobes, as shown in FIGURE 1.4. The fifth $d$ orbital is shaped like an elongated dumbbell with a doughnut around its middle.\n\n\nAn $s$ orbital\n\n\nA p orbital\n\n\nAd orbital\n\nFIGURE 1.4 Representations of $\\boldsymbol{s}, \\boldsymbol{p}$, and $\\boldsymbol{d}$ orbitals. An $s$ orbital is spherical, a $p$ orbital is dumbbell-shaped, and four of the five $d$ orbitals are cloverleaf-shaped. Different lobes of $p$ orbitals are often drawn for convenience as teardrops, but their actual shape is more like that of a doorknob, as indicated."}
{"id": 7, "contents": "CHAPTER CONTENTS - 1.2 Atomic Structure: Orbitals\nThe orbitals in an atom are organized into different layers around the nucleus called electron shells, which are centered around the nucleus and have successively larger size and energy. Different shells contain different numbers and kinds of orbitals, and each orbital within a shell can be occupied by two electrons. The first shell contains only a single $s$ orbital, denoted $1 s$, and thus holds only 2 electrons. The second shell contains one $2 s$ orbital and three $2 p$ orbitals and thus holds a total of 8 electrons. The third shell contains a $3 s$ orbital, three $3 p$ orbitals, and five $3 d$ orbitals, for a total capacity of 18 electrons. These orbital groupings and their energy levels are shown in FIGURE 1.5.\n\n\nFIGURE 1.5 Energy levels of electrons in an atom. The first shell holds a maximum of 2 electrons in one $1 s$ orbital; the second shell holds a maximum of 8 electrons in one $2 s$ and three $2 p$ orbitals; the third shell holds a maximum of 18 electrons in one $3 s$, three $3 p$, and $3 d$ orbitals; and so on. The two electrons in each orbital are represented by five up and down arrows, $\\uparrow \\downarrow$. Although not shown, the energy level of the $4 s$ orbital falls between $3 p$ and $3 d$.\n\nThe three different $p$ orbitals within a given shell are oriented in space along mutually perpendicular directions, denoted $p_{\\mathrm{x}}, p_{\\mathrm{y}}$, and $p_{\\mathrm{z}}$. As shown in FIGURE 1.6, the two lobes of each $p$ orbital are separated by a region of zero electron density called a node. Furthermore, the two orbital regions separated by the node have different algebraic signs, + and -, in the wave function, as represented by the different colors in FIGURE 1.4 and FIGURE 1.6. As we'll see in Section 1.11, these algebraic signs for different orbital lobes have important consequences with respect to chemical bonding and chemical reactivity.\n\n\nA $2 p_{\\mathrm{x}}$ orbital\n\n\nA $2 p_{\\mathrm{y}}$ orbital"}
{"id": 8, "contents": "CHAPTER CONTENTS - 1.2 Atomic Structure: Orbitals\nA $2 p_{\\mathrm{x}}$ orbital\n\n\nA $2 p_{\\mathrm{y}}$ orbital\n\n\nA $\\mathbf{2} p_{z}$ orbital\n\nFIGURE 1.6 Shapes of the $2 p$ orbitals. Each of the three mutually perpendicular, dumbbell-shaped orbitals has two lobes separated by a node. The two lobes have different algebraic signs in the corresponding wave function, as indicated by the different colors."}
{"id": 9, "contents": "CHAPTER CONTENTS - 1.3 Atomic Structure: Electron Configurations\nThe lowest-energy arrangement, or ground-state electron configuration, of an atom is a list of the orbitals occupied by its electrons. We can predict this arrangement by following three rules."}
{"id": 10, "contents": "RULE 1 - \nThe lowest-energy orbitals fill up first, $1 s \\rightarrow 2 s \\rightarrow 2 p \\rightarrow 3 s \\rightarrow 3 p \\rightarrow 4 s \\rightarrow 3 d$, according to the following graphic, a statement called the Aufbau principle. Note that the $4 s$ orbital lies between the $3 p$ and $3 d$ orbitals in energy.\n\n\nRULE 2\nElectrons act in some ways as if they were spinning around an axis, somewhat as the earth spins. This spin can have two orientations, denoted as up ( $\\uparrow$ ) and down ( $\\downarrow$ ). Only two electrons can occupy an orbital, and they must have opposite spins, a statement called the Pauli exclusion principle.\n\nRULE 3\nIf two or more empty orbitals of equal energy are available, one electron occupies each with spins parallel until all orbitals are half-full, a statement called Hund's rule.\n\nSome examples of how these rules apply are shown in TABLE 1.1. Hydrogen, for instance, has only one electron, which must occupy the lowest-energy orbital. Thus, hydrogen has a $1 s$ ground-state configuration. Carbon has six electrons and the ground-state configuration $1 s^{2} 2 s^{2} 2 p_{x}{ }^{1} 2 p_{y}{ }^{1}$, and so forth. Note that a superscript is used to represent the number of electrons in a particular orbital.\n\n| TABLE 1.1 <br> Elements | Ground-State Electron Configurations of Some |  |  |  |\n| :--- | :--- | :--- | :--- | :--- |\n| Element | Atomic number | Configuration |  |  |\n| Hydrogen | 1 | $1 s$ | $\\uparrow$ |  |\n| Carbon | 6 | $2 p$ | $\\uparrow \\uparrow$ | $\\uparrow$ |\n\nPROBLEM What is the ground-state electron configuration of each of the following elements:\n1-1 (a) Oxygen (b) Nitrogen (c) Sulfur\n\nPROBLEM How many electrons does each of the following biological trace elements have in its outermost $\\mathbf{1 - 2}$ electron shell?\n(a) Magnesium\n(b) Cobalt\n(c) Selenium"}
{"id": 11, "contents": "RULE 1 - 1.4 Development of Chemical Bonding Theory\nBy the mid-1800s, the new science of chemistry was developing rapidly, especially in Europe, and chemists had begun to probe the forces holding compounds together. In 1858, the German chemist August Kekul\u00e9 and the Scottish chemist Archibald Couper independently proposed that, in all organic compounds, carbon is tetravalent-it always forms four bonds when it joins other elements to form stable compounds. Furthermore, said Kekul\u00e9, carbon atoms can bond to one another to form extended chains of linked atoms. In 1865, Kekul\u00e9 provided another major advance when he suggested that carbon chains can double back on themselves to form rings of atoms.\n\nAlthough Kekule and Couper were correct in describing the tetravalent nature of carbon, chemistry was still viewed in a two-dimensional way until 1874. In that year, the Dutch chemist Jacobus van't Hoff and French chemist Joseph Le Bel added a third dimension to our ideas about organic compounds when they proposed that the four bonds of carbon are not oriented randomly but have specific spatial directions. Van't Hoff went even further and suggested that the four atoms to which carbon is bonded sit at the corners of a regular tetrahedron, with carbon in the center.\n\nA representation of a tetrahedral carbon atom is shown in FIGURE 1.7. Note the conventions used to show threedimensionality: solid lines represent bonds in the plane of the page, the heavy wedged line represents a bond coming out of the page toward the viewer, and the dashed line represents a bond receding back behind the page, away from the viewer. Get used to them; these representations will be used throughout the text.\n\n\nFIGURE 1.7 A representation of van't Hoff's tetrahedral carbon atom. The solid lines represent bonds in the plane of the paper, the heavy wedged line represents a bond coming out of the plane of the page toward the viewer, and the dashed line represents a bond going back behind the plane of the page away from the viewer."}
{"id": 12, "contents": "RULE 1 - 1.4 Development of Chemical Bonding Theory\nWhy, though, do atoms bond together, and how can chemical bonds be described electronically? The why question is relatively easy to answer: atoms bond together because the compound that results is more stable and lower in energy than the separate atoms. Energy-usually as heat-is always released and flows out of the chemical system when a bond forms. Conversely, energy is added to the chemical system when a bond breaks. Making bonds always releases energy, and breaking bonds always absorbs energy. The how question is more difficult. To answer it, we need to know more about the electronic properties of atoms.\n\nWe know through observation that eight electrons (an electron octet) in an atom's outermost shell, or valence shell, impart special stability to the noble-gas elements in group 8A of the periodic table: $\\mathrm{Ne}(2+8)$; $\\mathrm{Ar}(2+8+$ 8 ); $\\mathrm{Kr}(2+8+18+8)$. We also know that the chemistry of the main-group elements on the left and right sides of the periodic table is governed by their tendency to take on the electron configuration of the nearest noble gas. The alkali metals such as sodium in group 1A, for example, achieve a noble-gas configuration by losing the single $s$ electron from their valence shell to form a cation, while the halogens such as chlorine in group 7A achieve a noble-gas configuration by gaining a $p$ electron to fill their valence shell and form an anion. The resultant ions are held together in compounds like $\\mathrm{Na}^{+} \\mathrm{Cl}^{-}$by the electrical attraction of unlike charges that we call an ionic bond."}
{"id": 13, "contents": "RULE 1 - 1.4 Development of Chemical Bonding Theory\nBut how do elements closer to the middle of the periodic table form bonds? Look at methane, $\\mathrm{CH}_{4}$, the main constituent of natural gas, for example. The bonding in methane is not ionic because it would take too much energy for carbon ( $1 s^{2} 2 s^{2} 2 p^{2}$ ) either to gain or lose four electrons to achieve a noble-gas configuration. Instead, carbon bonds to other atoms, not by gaining or losing electrons, but by sharing them. Such a sharedelectron bond, first proposed in 1916 by the American chemist G. N. Lewis, is called a covalent bond. The neutral collection of atoms held together by covalent bonds is called a molecule. Ionic compounds such as sodium chloride, however, are not called molecules.\n\nA simple way of indicating the covalent bonds in molecules is to use what are called Lewis structures, or electron-dot structures, in which the valence-shell electrons of an atom are represented as dots. Thus, hydrogen has one dot representing its $1 s$ electron, carbon has four dots ( $2 s^{2} 2 p^{2}$ ), oxygen has six dots ( $2 s^{2} 2 p^{4}$ ), and so on. A stable molecule results whenever a noble-gas configuration of eight dots (an octet) is achieved for all main-group atoms or two dots for hydrogen. Even simpler than Lewis structures is the use of Kekul\u00e9 structures, or line-bond structures, in which the two-electron covalent bonds are indicated as lines drawn between atoms."}
{"id": 14, "contents": "RULE 1 - 1.4 Development of Chemical Bonding Theory\n| Electron-dot structures (Lewis structures) | <smiles></smiles> | $\\begin{gathered} H: \\ddot{\\mathrm{N}}: \\mathrm{H} \\\\ \\ddot{\\mathrm{H}} \\end{gathered}$ | $H: \\ddot{O}: H$ | <smiles></smiles> |\n| :---: | :---: | :---: | :---: | :---: |\n| Line-bond structures (Kekul\u00e9 structures) | <smiles>C</smiles> | <smiles>[NH]</smiles> | $\\mathrm{H}-\\ddot{\\mathrm{O}}-\\mathrm{H}$ | <smiles>CO</smiles> |\n|  | Methane $\\left(\\mathrm{CH}_{4}\\right)$ | Ammonia $\\left(\\mathrm{NH}_{3}\\right)$ | Water $\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)$ | Methanol $\\left(\\mathrm{CH}_{3} \\mathrm{OH}\\right)$ |\n\nThe number of covalent bonds an atom forms depends on how many additional valence electrons it needs to reach a noble-gas configuration. Hydrogen has one valence electron (1s) and needs only one more to reach the helium configuration ( $1 s^{2}$ ), so it forms one bond. Carbon has four valence electrons ( $2 s^{2} 2 p^{2}$ ) and needs four more to reach the neon configuration $\\left(2 s^{2} 2 p^{6}\\right)$, so it forms four bonds. Nitrogen has five valence electrons ( $2 s^{2}$ $2 p^{3}$ ), needs three more, and forms three bonds; oxygen has six valence electrons ( $2 s^{2} 2 p^{4}$ ), needs two more, and forms two bonds; and the halogens have seven valence electrons, need one more, and form one bond."}
{"id": 15, "contents": "RULE 1 - 1.4 Development of Chemical Bonding Theory\nValence electrons that are not used for bonding remain as dots in structures and are called lone-pair electrons, or nonbonding electrons. The nitrogen atom in ammonia, $\\mathrm{NH}_{3}$, for instance, shares six valence electrons in three covalent bonds and has its remaining two valence electrons as two dots in a nonbonding lone pair. As a time-saving shorthand, nonbonding electrons are often omitted when drawing line-bond structures, but you still have to keep them in mind since they're often crucial in chemical reactions.\n\n\nAmmonia"}
{"id": 16, "contents": "Predicting the Number of Bonds Formed by Atoms in Molecules - \nHow many hydrogen atoms does phosphorus bond to in forming phosphine, $\\mathrm{PH}_{\\text {? }}$ ?"}
{"id": 17, "contents": "Strategy - \nIdentify the periodic group of phosphorus, and find from that how many electrons (bonds) are needed to make an octet."}
{"id": 18, "contents": "Solution - \nPhosphorus is in group 5A of the periodic table and has five valence electrons. It thus needs to share three more electrons to make an octet and therefore bonds to three hydrogen atoms, giving $\\mathrm{PH}_{3}$."}
{"id": 19, "contents": "Drawing Electron-Dot and Line-Bond Structures - \nDraw both electron-dot and line-bond structures for chloromethane, $\\mathrm{CH}_{3} \\mathrm{Cl}$."}
{"id": 20, "contents": "Strategy - \nRemember that a covalent bond-that is, a pair of shared electrons-is represented as a line between atoms."}
{"id": 21, "contents": "Solution - \nHydrogen has one valence electron, carbon has four valence electrons, and chlorine has seven valence electrons. Thus, chloromethane is represented as\n\n\nChloromethane\n\nPROBLEM Draw a molecule of chloroform, $\\mathrm{CHCl}_{3}$, using solid, wedged, and dashed lines to show its tetrahedral 1-3 geometry.\n\nPROBLEM Convert the following representation of ethane, $\\mathrm{C}_{2} \\mathrm{H}_{6}$, into a conventional drawing that uses solid, 1-4 wedged, and dashed lines to indicate tetrahedral geometry around each carbon (black = C, gray = H)."}
{"id": 22, "contents": "Ethane - \nPROBLEM What are likely formulas for the following substances?\n1-5 (a) $\\mathrm{CCl}_{\\text {? }}$\n(b) AlH ?\n(c) $\\mathrm{CH}_{?} \\mathrm{Cl}_{2}$\n(d) SiF\n(e) $\\mathrm{CH}_{3} \\mathrm{NH}_{?}$\n\nPROBLEM Write line-bond structures for the following substances, showing all nonbonding electrons:\n1-6 (a) $\\mathrm{CHCl}_{3}$, chloroform (b) $\\mathrm{H}_{2} \\mathrm{~S}$, hydrogen sulfide (c) $\\mathrm{CH}_{3} \\mathrm{NH}_{2}$, methylamine\n(d) $\\mathrm{CH}_{3} \\mathrm{Li}$, methyllithium\n\nPROBLEM Why can't an organic molecule have the formula $\\mathrm{C}_{2} \\mathrm{H}_{7}$ ?"}
{"id": 23, "contents": "1-7 - 1.5 Describing Chemical Bonds: Valence Bond Theory\nHow does electron sharing lead to bonding between atoms? Two models have been developed to describe covalent bonding: valence bond theory and molecular orbital theory. Each model has its strengths and weaknesses, and chemists tend to use them interchangeably depending on the circumstances. Valence bond theory is the more easily visualized of the two, so most of the descriptions we'll use in this book derive from that approach.\n\nAccording to valence bond (VB) theory, a covalent bond forms when two atoms approach each other closely and a singly occupied orbital on one atom overlaps a singly occupied orbital on the other atom. The electrons are now paired in the overlapping orbitals and are attracted to the nuclei of both atoms, thus bonding the atoms together. In the $\\mathrm{H}_{2}$ molecule, for instance, the $\\mathrm{H}-\\mathrm{H}$ bond results from the overlap of two singly occupied hydrogen $1 s$ orbitals.\n\n\nThe overlapping orbitals in the $\\mathrm{H}_{2}$ molecule have the elongated egg shape we might get by pressing two spheres together. If a plane were to pass through the middle of the bond, the intersection of the plane and the overlapping orbitals would be a circle. In other words, the $\\mathrm{H}-\\mathrm{H}$ bond is cylindrically symmetrical, as shown in FIGURE 1.8. Such bonds, which are formed by the head-on overlap of two atomic orbitals along a line drawn between the nuclei, are called sigma ( $\\boldsymbol{\\sigma}$ ) bonds.\n\n\nFIGURE 1.8 The cylindrical symmetry of the $\\mathbf{H - H} \\boldsymbol{\\sigma}$ bond in an $\\mathbf{H}_{\\mathbf{2}}$ molecule. The intersection of a plane cutting through the $\\boldsymbol{\\sigma}$ bond is a circle."}
{"id": 24, "contents": "1-7 - 1.5 Describing Chemical Bonds: Valence Bond Theory\nDuring the bond-forming reaction $2 \\mathrm{H} \\cdot \\longrightarrow \\mathrm{H} 2,436 \\mathrm{~kJ} / \\mathrm{mol}(104 \\mathrm{kcal} / \\mathrm{mol})$ of energy is released. Because the product $\\mathrm{H}_{2}$ molecule has $436 \\mathrm{~kJ} / \\mathrm{mol}$ less energy than the starting $2 \\mathrm{H} \\cdot$ atoms, the product is more stable than the reactant and we say that the $\\mathrm{H}-\\mathrm{H}$ bond has a bond strength of $436 \\mathrm{~kJ} / \\mathrm{mol}$. In other words, we would have to put $436 \\mathrm{~kJ} / \\mathrm{mol}$ of energy into the $\\mathrm{H}-\\mathrm{H}$ bond to break the $\\mathrm{H}_{2}$ molecule apart into two H atoms (FIGURE 1.9). For convenience, we'll generally give energies in both kilocalories (kcal) and the SI unit kilojoules (kJ): $1 \\mathrm{~kJ}=0.2390$ kcal; $1 \\mathrm{kcal}=4.184 \\mathrm{~kJ}$.\n\n\nFIGURE 1.9 Relative energy levels of two $\\mathbf{H}$ atoms and the $\\mathbf{H}_{\\mathbf{2}}$ molecule. The $\\mathrm{H}_{2}$ molecule has $436 \\mathrm{~kJ} / \\mathrm{mol}(104 \\mathrm{kcal} / \\mathrm{mol})$ less energy than the two separate H atoms, so $436 \\mathrm{~kJ} / \\mathrm{mol}$ of energy is released when the $\\mathrm{H}-\\mathrm{H}$ bond forms. Conversely, $436 \\mathrm{~kJ} / \\mathrm{mol}$ is absorbed when the $\\mathrm{H}-\\mathrm{H}$ bond breaks."}
{"id": 25, "contents": "1-7 - 1.5 Describing Chemical Bonds: Valence Bond Theory\nHow close are the two nuclei in the $\\mathrm{H}_{2}$ molecule? If they are too close, they will repel each other because both are positively charged. Yet if they're too far apart, they won't be able to share the bonding electrons. Thus, there is an optimum distance between nuclei that leads to maximum stability (FIGURE 1.10). Called the bond length, this distance is 74 pm in the $\\mathrm{H}-\\mathrm{H}$ molecule. Every covalent bond has both a characteristic bond strength and bond length.\n\n\nFIGURE 1.10 A plot of energy versus internuclear distance for two hydrogen atoms. The distance between nuclei at the minimum energy point is the bond length."}
{"id": 26, "contents": "$1.6 s p^{3}$ Hybrid Orbitals and the Structure of Methane - \nThe bonding in the hydrogen molecule is fairly straightforward, but the situation is more complicated in organic molecules with tetravalent carbon atoms. Take methane, $\\mathrm{CH}_{4}$, for instance. As we've seen, carbon has four valence electrons $\\left(2 s^{2} 2 p^{2}\\right)$ and forms four bonds. Because carbon uses two kinds of orbitals for bonding, $2 s$ and $2 p$, we might expect methane to have two kinds of $\\mathrm{C}-\\mathrm{H}$ bonds. In fact, though, all four $\\mathrm{C}-\\mathrm{H}$ bonds in methane are identical and are spatially oriented toward the corners of a regular tetrahedron, as shown previously in FIGURE 1.7. How can we explain this?\n\nAn answer was provided in 1931 by Linus Pauling, who showed mathematically how an $s$ orbital and three $p$ orbitals on an atom can combine, or hybridize, to form four equivalent atomic orbitals with tetrahedral orientation. Shown in FIGURE 1.11, these tetrahedrally oriented orbitals are called $\\boldsymbol{s p}^{\\mathbf{3}}$ hybrid orbitals. Note that the superscript 3 in the name $s p^{3}$ tells how many of each type of atomic orbital combine to form the hybrid, not how many electrons occupy it.\n\n\nFIGURE 1.11 Four $s p^{3}$ hybrid orbitals, oriented toward the corners of a regular tetrahedron, are formed by the combination of an $s$ orbital and three $\\boldsymbol{p}$ orbitals (red/blue). The $s p^{3}$ hybrids have two lobes and are unsymmetrical about the nucleus, giving them a directionality and allowing them to form strong bonds to other atoms."}
{"id": 27, "contents": "$1.6 s p^{3}$ Hybrid Orbitals and the Structure of Methane - \nThe concept of hybridization explains how carbon forms four equivalent tetrahedral bonds but not why it does so. The shape of the hybrid orbital suggests the answer to why. When an $s$ orbital hybridizes with three $p$ orbitals, the resultant $s p^{3}$ hybrid orbitals are unsymmetrical about the nucleus. One of the two lobes is larger than the other and can therefore overlap more effectively with an orbital from another atom to form a bond. As a result,\n$s p^{3}$ hybrid orbitals form stronger bonds than do unhybridized $s$ or $p$ orbitals.\nThe asymmetry of $s p^{3}$ orbitals arises because, as noted previously, the two lobes of a $p$ orbital have different algebraic signs, + and - in the wave function. Thus, when a $p$ orbital hybridizes with an $s$ orbital, the positive $p$ lobe adds to the $s$ orbital but the negative $p$ lobe subtracts from the $s$ orbital. The resultant hybrid orbital is therefore unsymmetrical about the nucleus and is strongly oriented in one direction.\n\nWhen each of the four identical $s p^{3}$ hybrid orbitals of a carbon atom overlaps with the $1 s$ orbital of a hydrogen atom, four identical $\\mathrm{C}-\\mathrm{H}$ bonds are formed and methane results. Each $\\mathrm{C}-\\mathrm{H}$ bond in methane has a strength of $439 \\mathrm{~kJ} / \\mathrm{mol}(105 \\mathrm{kcal} / \\mathrm{mol})$ and a length of 109 pm . Because the four bonds have a specific geometry, we also can define a property called the bond angle. The angle formed by each $\\mathrm{H}-\\mathrm{C}-\\mathrm{H}$ is $109.5^{\\circ}$, the so-called tetrahedral angle. Methane thus has the structure shown in FIGURE 1.12.\n\n\nFIGURE 1.12 The structure of methane, showing its $109 . \\mathbf{5}^{\\circ}$ bond angles."}
{"id": 28, "contents": "$1.7 s p^{3}$ Hybrid Orbitals and the Structure of Ethane - \nThe same kind of orbital hybridization that accounts for the methane structure also accounts for the bonding together of carbon atoms into chains and rings to make possible many millions of organic compounds. Ethane, $\\mathrm{C}_{2} \\mathrm{H}_{6}$, is the simplest molecule containing a carbon-carbon bond.\n\n\nWe can picture the ethane molecule by imagining that the two carbon atoms bond to each other by head-on sigma ( $\\sigma$ ) overlap of an $s p^{3}$ hybrid orbital from each (FIGURE 1.13). The remaining three $s p^{3}$ hybrid orbitals on each carbon overlap with the 1 s orbitals of three hydrogens to form the six $\\mathrm{C}-\\mathrm{H}$ bonds. The $\\mathrm{C}-\\mathrm{H}$ bonds in ethane are similar to those in methane, although a bit weaker: $421 \\mathrm{~kJ} / \\mathrm{mol}$ ( $101 \\mathrm{kcal} / \\mathrm{mol}$ ) for ethane versus $439 \\mathrm{~kJ} / \\mathrm{mol}$ for methane. The C-C bond is 153 pm in length and has a strength of $377 \\mathrm{~kJ} / \\mathrm{mol}(90 \\mathrm{kcal} / \\mathrm{mol})$. All the bond angles of ethane are near, although not exactly at, the tetrahedral value of $109.5^{\\circ}$.\n\n\nFIGURE 1.13 The structure of ethane. The carbon-carbon bond is formed by $\\sigma$ overlap of two $s p^{3}$ hybrid orbitals. For clarity, the smaller\nlobes of the $s p^{3}$ hybrid orbitals are not shown.\nPROBLEM Draw a line-bond structure for propane, $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$. Predict the value of each bond angle, and 1-8 indicate the overall shape of the molecule.\n\nPROBLEM Convert the following molecular model of hexane, a component of gasoline, into a line-bond 1-9 structure (black $=\\mathrm{C}$, gray $=\\mathrm{H}$ ).\n\n\nHexane"}
{"id": 29, "contents": "$1.8 s p^{2}$ Hybrid Orbitals and the Structure of Ethylene - \nThe bonds we've seen in methane and ethane are called single bonds because they result from the sharing of one electron pair between bonded atoms. It was recognized nearly 150 years ago, however, that carbon atoms can also form double bonds by sharing two electron pairs between atoms or triple bonds by sharing three electron pairs. Ethylene, for instance, has the structure $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}_{2}$ and contains a carbon-carbon double bond, while acetylene has the structure $\\mathrm{HC} \\equiv \\mathrm{CH}$ and contains a carbon-carbon triple bond.\n\nHow are multiple bonds described by valence bond theory? When we discussed $s p^{3}$ hybrid orbitals in Section 1.6, we said that the four valence-shell atomic orbitals of carbon combine to form four equivalent $s p^{3}$ hybrids. Imagine instead that the $2 s$ orbital combines with only $t w o$ of the three available $2 p$ orbitals. Three $\\boldsymbol{s p}^{\\mathbf{2}}$ hybrid orbitals result, and one $2 p$ orbital remains unchanged. Like $s p^{3}$ hybrids, $s p^{2}$ hybrid orbitals are unsymmetrical about the nucleus and are strongly oriented in a specific direction so they can form strong bonds. The three $s p^{2}$ orbitals lie in a plane at angles of $120^{\\circ}$ to one another, with the remaining $p$ orbital perpendicular to the $s p^{2}$ plane, as shown in FIGURE 1.14.\n\n\nSide view\n\n\nTop view\n\nFIGURE $1.14 \\boldsymbol{s p} \\boldsymbol{p}^{2}$ Hybridization. The three equivalent $\\boldsymbol{s p ^ { 2 }}$ hybrid orbitals lie in a plane at angles of $120^{\\circ}$ to one another, and a single unhybridized $\\boldsymbol{p}$ orbital (red/blue) is perpendicular to the $s p^{2}$ plane."}
{"id": 30, "contents": "$1.8 s p^{2}$ Hybrid Orbitals and the Structure of Ethylene - \nWhen two carbons with $s p^{2}$ hybridization approach each other, they form a strong $\\sigma$ bond by $s p^{2}-s p^{2}$ head-on overlap. At the same time, the unhybridized $p$ orbitals interact by sideways overlap to form what is called a pi ( $\\boldsymbol{\\pi}$ ) bond. The combination of an $s p^{2}-s p^{2} \\sigma$ bond and a $2 p-2 p \\pi$ bond results in the sharing of four electrons and the formation of a carbon-carbon double bond (FIGURE 1.15). Note that the electrons in the $\\sigma$ bond occupy the region centered between nuclei, while the electrons in the $\\pi$ bond occupy regions above and below a line drawn between nuclei.\n\nTo complete the structure of ethylene, four hydrogen atoms form $\\sigma$ bonds with the remaining four $s p^{2}$ orbitals. Ethylene thus has a planar structure, with $\\mathrm{H}-\\mathrm{C}-\\mathrm{H}$ and $\\mathrm{H}-\\mathrm{C}-\\mathrm{C}$ bond angles of approximately $120^{\\circ}$. (The actual values are $117.4^{\\circ}$ for the $\\mathrm{H}-\\mathrm{C}-\\mathrm{H}$ bond angle and $121.3^{\\circ}$ for the $\\mathrm{H}-\\mathrm{C}-\\mathrm{C}$ bond angle.) Each $\\mathrm{C}-\\mathrm{H}$ bond has a length of 108.7 pm and a strength of $464 \\mathrm{~kJ} / \\mathrm{mol}(111 \\mathrm{kcal} / \\mathrm{mol})$.\n\n\nFIGURE 1.15 The structure of ethylene. One part of the double bond in ethylene results from $\\sigma$ (head-on) overlap of $s p^{2}$ hybrid orbitals, and the other part results from $\\pi$ (sideways) overlap of unhybridized $\\boldsymbol{p}$ orbitals (red/blue). The $\\pi$ bond has regions of electron density above and below a line drawn between nuclei."}
{"id": 31, "contents": "$1.8 s p^{2}$ Hybrid Orbitals and the Structure of Ethylene - \nAs you might expect, the carbon-carbon double bond in ethylene is both shorter and stronger than the single bond in ethane because it has four electrons bonding the nuclei together rather than two. Ethylene has a $\\mathrm{C}=\\mathrm{C}$ bond length of 134 pm and a strength of $728 \\mathrm{~kJ} / \\mathrm{mol}$ ( $174 \\mathrm{kcal} / \\mathrm{mol}$ ) versus a C-C length of 153 pm and a strength of $377 \\mathrm{~kJ} / \\mathrm{mol}$ for ethane. The carbon-carbon double bond is less than twice as strong as a single bond because the sideways overlap in the $\\pi$ part of the double bond is not as great as the head-on overlap in the $\\sigma$ part."}
{"id": 32, "contents": "Drawing Electron-Dot and Line-Bond Structures - \nCommonly used in biology as a tissue preservative, formaldehyde, $\\mathrm{CH}_{2} \\mathrm{O}$, contains a carbon-oxygen double bond. Draw electron-dot and line-bond structures of formaldehyde, and indicate the hybridization of the carbon orbitals."}
{"id": 33, "contents": "Strategy - \nWe know that hydrogen forms one covalent bond, carbon forms four, and oxygen forms two. Trial and error, combined with intuition, is needed to fit the atoms together."}
{"id": 34, "contents": "Solution - \nThere is only one way that two hydrogens, one carbon, and one oxygen can combine:\n\n\nLike the carbon atoms in ethylene, the carbon atom in formaldehyde is in a double bond and its orbitals are therefore $s p^{2}$-hybridized.\n\nPROBLEM Draw a line-bond structure for propene, $\\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{CH}_{2}$. Indicate the hybridization of the orbitals on 1-10 each carbon, and predict the value of each bond angle.\n\nPROBLEM Draw a line-bond structure for 1,3-butadiene, $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}-\\mathrm{CH}=\\mathrm{CH}_{2}$. Indicate the hybridization of the 1-11 orbitals on each carbon, and predict the value of each bond angle.\n\nPROBLEM A molecular model of aspirin (acetylsalicylic acid) is shown. Identify the hybridization of the\n1-12 orbitals on each carbon atom in aspirin, and tell which atoms have lone pairs of electrons (black = C, red $=0$, gray $=H$ )."}
{"id": 35, "contents": "1.9 sp Hybrid Orbitals and the Structure of Acetylene - \nIn addition to forming single and double bonds by sharing two and four electrons, respectively, carbon can also form a triple bond by sharing six electrons. To account for the triple bond in a molecule such as acetylene, $\\mathrm{H}-\\mathrm{C} \\equiv \\mathrm{C}-\\mathrm{H}$, we need a third kind of hybrid orbital, an $\\boldsymbol{s p}$ hybrid. Imagine that, instead of combining with two or three $p$ orbitals, a carbon $2 s$ orbital hybridizes with only a single $p$ orbital. Two sphybrid orbitals result, and two $p$ orbitals remain unchanged. The two $s p$ orbitals are oriented $180^{\\circ}$ apart on the right-left ( $x$ ) axis, while the p orbitals are perpendicular on the up-down $(y)$ axis and the in-out $(z)$ axis, as shown in FIGURE 1.16.\n\n\nOne $s p$ hybrid Another $s p$ hybrid\nFIGURE 1.16 sp Hybridization. The two $s p$ hybrid orbitals are oriented $180^{\\circ}$ away from each other, perpendicular to the two remaining $\\boldsymbol{p}$ orbitals (red/blue).\n\nWhen two $s p$-hybridized carbon atoms approach each other, $s p$ hybrid orbitals on each carbon overlap headon to form a strong $s p-s p \\sigma$ bond. At the same time, the $p_{\\mathrm{Z}}$ orbitals from each carbon form a $p_{\\mathrm{Z}}-p_{\\mathrm{Z}} \\pi$ bond by sideways overlap, and the $p_{\\mathrm{y}}$ orbitals overlap similarly to form a $p_{\\mathrm{y}}-p_{\\mathrm{y}} \\pi$ bond. The net effect is the sharing of six electrons and formation of a carbon-carbon triple bond. Each of the two remaining $s p$ hybrid orbitals forms a $\\sigma$ bond with hydrogen to complete the acetylene molecule (FIGURE 1.17)."}
{"id": 36, "contents": "1.9 sp Hybrid Orbitals and the Structure of Acetylene - \nFIGURE 1.17 The structure of acetylene. The two carbon atoms are joined by one $s p-s p \\sigma$ bond and two $p-p \\pi$ bonds.\nAs suggested by $s p$ hybridization, acetylene is a linear molecule with $\\mathrm{H}-\\mathrm{C}-\\mathrm{C}$ bond angles of $180^{\\circ}$. The $\\mathrm{C}-\\mathrm{H}$ bonds have a length of 106 pm and a strength of $558 \\mathrm{~kJ} / \\mathrm{mol}(133 \\mathrm{kcal} / \\mathrm{mol})$. The $\\mathrm{C}-\\mathrm{C}$ bond length in acetylene is 120 pm , and its strength is about $965 \\mathrm{~kJ} / \\mathrm{mol}(231 \\mathrm{kcal} / \\mathrm{mol})$, making it the shortest and strongest of any carbon-carbon bond. A comparison of $s p, s p^{2}$, and $s p^{3}$ hybridization is given in TABLE 1.2."}
{"id": 37, "contents": "1.9 sp Hybrid Orbitals and the Structure of Acetylene - \n| Molecule | Bond | Bond strength |  | Bond length (pm) |\n| :---: | :---: | :---: | :---: | :---: |\n|  |  | (kJ/mol) | ( $\\mathrm{kcal} / \\mathrm{mol}$ ) |  |\n| Methane, $\\mathrm{CH}_{4}$ | $\\left(s p^{3}\\right) \\mathrm{C}-\\mathrm{H}$ | 439 | 105 | 109 |\n| Ethane, $\\mathrm{CH}_{3} \\mathrm{CH}_{3}$ | $\\left(s p^{3}\\right) \\mathrm{C}-\\mathrm{C}\\left(s p^{3}\\right)$ | 377 | 90 | 153 |\n|  | $\\left(s p^{3}\\right) \\mathrm{C}-\\mathrm{H}$ | 421 | 101 | 109 |\n| Ethylene, $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}_{2}$ | $\\left(s p^{2}\\right) \\mathrm{C}=\\mathrm{C}\\left(s p^{2}\\right)$ | 728 | 174 | 134 |\n|  | $\\left(s p^{2}\\right) \\mathrm{C}-\\mathrm{H}$ | 464 | 111 | 109 |\n| Acetylene, $\\mathrm{HC} \\equiv \\mathrm{CH}$ | $(s p) \\mathrm{C} \\equiv \\mathrm{C}(s p)$ | 965 | 231 | 120 |\n|  | (sp) $\\mathrm{C}-\\mathrm{H}$ | 558 | 133 | 106 |\n\nPROBLEM Draw a line-bond structure for propyne, $\\mathrm{CH}_{3} \\mathrm{C} \\equiv \\mathrm{CH}$. Indicate the hybridization of the orbitals on\n1-13 each carbon, and predict a value for each bond angle."}
{"id": 38, "contents": "1.9 sp Hybrid Orbitals and the Structure of Acetylene - 1.10 Hybridization of Nitrogen, Oxygen, Phosphorus, and Sulfur\nThe valence-bond concept of orbital hybridization described in the previous four sections is not limited to carbon. Covalent bonds formed by other elements can also be described using hybrid orbitals. Look, for instance, at the nitrogen atom in methylamine $\\left(\\mathrm{CH}_{3} \\mathrm{NH}_{2}\\right)$, an organic derivative of ammonia ( $\\mathrm{NH}_{3}$ ) and the substance responsible for the odor of rotting fish.\n\nThe experimentally measured $\\mathrm{H}-\\mathrm{N}-\\mathrm{H}$ bond angle in methylamine is $107.1^{\\circ}$, and the $\\mathrm{C}-\\mathrm{N}-\\mathrm{H}$ bond angle is\n$110.3^{\\circ}$, both of which are close to the $109.5^{\\circ}$ tetrahedral angle found in methane. We therefore assume that nitrogen forms four $s p^{3}$-hybridized orbitals, just as carbon does. One of the four $s p^{3}$ orbitals is occupied by two nonbonding electrons (a lone pair), and the other three hybrid orbitals have one electron each. Overlap of these three half-filled nitrogen orbitals with half-filled orbitals from other atoms ( C or H ) gives methylamine. Note that the unshared lone pair of electrons in the fourth $s p^{3}$ hybrid orbital of nitrogen occupies as much space as an $\\mathrm{N}-\\mathrm{H}$ bond does and is very important to the chemistry of methylamine and other nitrogen-containing organic molecules.\n\n\nMethylamine\nLike the carbon atom in methane and the nitrogen atom in methylamine, the oxygen atom in methanol (methyl alcohol) and many other organic molecules can be described as $s p^{3}$-hybridized. The $\\mathrm{C}-\\mathrm{O}-\\mathrm{H}$ bond angle in methanol is $108.5^{\\circ}$, very close to the $109.5^{\\circ}$ tetrahedral angle. Two of the four $s p^{3}$ hybrid orbitals on oxygen are occupied by nonbonding electron lone pairs, and two are used to form bonds."}
{"id": 39, "contents": "1.9 sp Hybrid Orbitals and the Structure of Acetylene - 1.10 Hybridization of Nitrogen, Oxygen, Phosphorus, and Sulfur\nIn the periodic table, phosphorus and sulfur are the third-row analogs of nitrogen and oxygen, and the bonding in both can be described using hybrid orbitals. Because of their positions in the third row, however, both phosphorus and sulfur can expand their outer-shell octets and form more than the typical number of covalent bonds. Phosphorus, for instance, often forms five covalent bonds, and sulfur often forms four.\n\nPhosphorus is most commonly encountered in biological molecules in compounds called organophosphates, which contain a phosphorus atom bonded to four oxygens, with one of the oxygens also bonded to carbon. Methyl phosphate, $\\mathrm{CH}_{3} \\mathrm{OPO}_{3}{ }^{2-}$, is the simplest example. The $\\mathrm{O}-\\mathrm{P}-\\mathrm{O}$ bond angle in such compounds is typically in the range $110^{\\circ}$ to $112^{\\circ}$, implying $s p^{3}$ hybridization for phosphorus orbitals.\n\n\nMethyl phosphate (an organophosphate)\nSulfur is most commonly encountered in biological molecules either in compounds called thiols, which have a sulfur atom bonded to one hydrogen and one carbon, $\\mathrm{C}-\\mathrm{S}-\\mathrm{H}$ or in sulfides, which have a sulfur atom bonded to two carbons, $\\mathrm{C}-\\mathrm{S}-\\mathrm{C}$. Produced by some bacteria, methanethiol $\\left(\\mathrm{CH}_{3} \\mathrm{SH}\\right)$ is the simplest example of a thiol, and dimethyl sulfide, $\\mathrm{H}_{3} \\mathrm{C}-\\mathrm{S}-\\mathrm{CH}_{3}$, is the simplest example of a sulfide. Both can be described by approximate $s p^{3}$ hybridization around sulfur, although both have significant deviation from the $109.5^{\\circ}$ tetrahedral angle.\n\n\nMethanethiol\n\n\nDimethyl sulfide"}
{"id": 40, "contents": "1.9 sp Hybrid Orbitals and the Structure of Acetylene - 1.10 Hybridization of Nitrogen, Oxygen, Phosphorus, and Sulfur\nMethanethiol\n\n\nDimethyl sulfide\n\nPROBLEM Identify all nonbonding lone pairs of electrons in the following molecules, and tell what geometry\n1-14 you expect for each of the indicated atoms.\n(a) The oxygen atom in dimethyl ether, $\\mathrm{CH}_{3}-\\mathrm{O}-\\mathrm{CH}_{3}$ (b) The nitrogen atom in trimethylamine,\n\n(c) The phosphorus atom in phosphine, $\\mathrm{PH}_{3}$\n(d) The sulfur atom in the amino acid methionine,"}
{"id": 41, "contents": "1.9 sp Hybrid Orbitals and the Structure of Acetylene - 1.11 Describing Chemical Bonds: Molecular Orbital Theory\nWe said in Section 1.5 that chemists use two models for describing covalent bonds: valence bond theory and molecular orbital theory. Having now seen the valence bond approach, which uses hybrid atomic orbitals to account for geometry and assumes the overlap of atomic orbitals to account for electron sharing, let's look briefly at the molecular orbital approach to bonding. We'll return to this topic in Chapters 14, 15, and 30 for a more in-depth discussion.\n\nMolecular orbital (MO) theory describes covalent bond formation as arising from a mathematical combination of atomic orbitals (wave functions) on different atoms to form molecular orbitals, so called because they belong to the entire molecule rather than to an individual atom. Just as an atomic orbital, whether unhybridized or hybridized, describes a region of space around an atom where an electron is likely to be found, so a molecular orbital describes a region of space in a molecule where electrons are most likely to be found.\n\nLike an atomic orbital, a molecular orbital has a specific size, shape, and energy. In the $\\mathrm{H}_{2}$ molecule, for example, two singly occupied $1 s$ atomic orbitals combine to form two molecular orbitals. There are two ways for the orbital combination to occur-an additive way and a subtractive way. The additive combination leads to formation of a molecular orbital that is lower in energy and roughly egg-shaped, while the subtractive combination leads to a molecular orbital that is higher in energy and has a node between nuclei (FIGURE 1.18). Note that the additive combination is a single, egg-shaped, molecular orbital; it is not the same as the two overlapping 1 s atomic orbitals of the valence bond description. Similarly, the subtractive combination is a single molecular orbital with the shape of an elongated dumbbell.\n\n\nFIGURE 1.18 Molecular orbitals of $\\mathbf{H}_{\\mathbf{2}}$. Combination of two hydrogen $1 s$ atomic orbitals leads to two $\\mathrm{H}_{2}$ molecular orbitals. The lowerenergy, bonding MO is filled, and the higher-energy, antibonding MO is unfilled."}
{"id": 42, "contents": "1.9 sp Hybrid Orbitals and the Structure of Acetylene - 1.11 Describing Chemical Bonds: Molecular Orbital Theory\nThe additive combination is lower in energy than the two hydrogen $1 s$ atomic orbitals and is called a bonding MO because electrons in this MO spend most of their time in the region between the two nuclei, thereby bonding the atoms together. The subtractive combination is higher in energy than the two hydrogen $1 s$ orbitals and is called an antibonding MO because any electrons it contains can't occupy the central region between the nuclei, where there is a node, and thus can't contribute to bonding. The two nuclei therefore repel each other.\n\nJust as bonding and antibonding $\\sigma$ molecular orbitals result from the head-on combination of two $s$ atomic orbitals in $\\mathrm{H}_{2}$, so bonding and antibonding $\\pi$ molecular orbitals result from the sideways combination of two $p$ atomic orbitals in ethylene. As shown in FIGURE 1.19, the lower-energy, $\\pi$ bonding MO has no node between nuclei and results from the combination of $p$ orbital lobes with the same algebraic sign. The higher-energy, $\\pi$ antibonding MO has a node between nuclei and results from the combination of lobes with opposite algebraic signs. Only the bonding MO is occupied; the higher-energy, antibonding MO is vacant. We'll see in Chapters 14, 15 , and 30 that molecular orbital theory is particularly useful for describing $\\pi$ bonds in compounds that have more than one double bond.\n\n\nFIGURE 1.19 A molecular orbital description of the C-C $\\boldsymbol{\\pi}$ bond in ethylene. The lower-energy, $\\boldsymbol{\\pi}$ bonding MO results from an additive combination of $p$ orbital lobes with the same algebraic sign and is filled. The higher-energy, $\\boldsymbol{\\pi}$ antibonding MO results from a subtractive combination of $p$ orbital lobes with opposite algebraic signs and is unfilled."}
{"id": 43, "contents": "1.9 sp Hybrid Orbitals and the Structure of Acetylene - 1.12 Drawing Chemical Structures\nLet's cover just one more point before ending this introductory chapter. In the structures we've been drawing until now, a line between atoms has represented the two electrons in a covalent bond. Drawing every bond and every atom is tedious, however, so chemists have devised several shorthand ways for writing structures. In condensed structures, carbon-hydrogen and carbon-carbon single bonds aren't shown; instead, they're understood. If a carbon has three hydrogens bonded to it, we write $\\mathrm{CH}_{3}$; if a carbon has two hydrogens bonded to it, we write $\\mathrm{CH}_{2}$; and so on. The compound called 2-methylbutane, for example, is written as follows:\n\n\n2-Methylbutane\nNote that the horizontal bonds between carbons aren't shown in condensed structures-the $\\mathrm{CH}_{3}, \\mathrm{CH}_{2}$, and CH units are simply placed next to each other-but vertical carbon-carbon bonds like that of the first of the condensed structures drawn above is shown for clarity. Notice also in the second of the condensed structures that the two $\\mathrm{CH}_{3}$ units attached to the CH carbon are grouped together as $\\left(\\mathrm{CH}_{3}\\right)_{2}$.\n\nEven simpler than condensed structures are skeletal structures such as those shown in TABLE 1.3. The rules for drawing skeletal structures are straightforward."}
{"id": 44, "contents": "RULE 1 - \nCarbon atoms aren't usually shown. Instead, a carbon atom is assumed to be at each intersection of two lines (bonds) and at the end of each line. Occasionally, a carbon atom might be indicated for emphasis or clarity."}
{"id": 45, "contents": "RULE 2 - \nHydrogen atoms bonded to carbon aren't shown. Because carbon always has a valence of 4, we mentally supply the correct number of hydrogen atoms for each carbon."}
{"id": 46, "contents": "RULE 3 - \nAtoms other than carbon and hydrogen are shown.\nOne further comment: Although such groupings as $-\\mathrm{CH}_{3},-\\mathrm{OH}$, and $-\\mathrm{NH}_{2}$ are usually written with the C , O , or N atom first and the H atom second, the order of writing is sometimes inverted to $\\mathrm{H}_{3} \\mathrm{C}-, \\mathrm{HO}-$, and $\\mathrm{H}_{2} \\mathrm{~N}$ - if needed to make the bonding connections clearer. Larger units such as $-\\mathrm{CH}_{2} \\mathrm{CH}_{3}$ are not inverted, though; we don't write $\\mathrm{H}_{3} \\mathrm{CH}_{2} \\mathrm{C}$ - because it would be confusing. There are, however, no well-defined rules that cover all cases; it's largely a matter of preference.\n\n| TABLE 1.3 Line-bond and Skeletal Structures for Some Compounds |\n| :--- |\n| Compound |\n| Isoprene, $\\mathrm{C}_{5} \\mathrm{H}_{8}$ |\n| Line-bond structure |\n| Shethylcyclohexane, $\\mathrm{C}_{7} \\mathrm{H}_{14}$ |"}
{"id": 47, "contents": "Interpreting a Line-Bond Structure - \nCarvone, a substance responsible for the odor of spearmint, has the following structure. Tell how many hydrogens are bonded to each carbon, and give the molecular formula of carvone.\n\n\nCarvone"}
{"id": 48, "contents": "Strategy - \nThe end of a line represents a carbon atom with 3 hydrogens, $\\mathrm{CH}_{3}$; a two-way intersection is a carbon atom with 2 hydrogens, $\\mathrm{CH}_{2}$; a three-way intersection is a carbon atom with 1 hydrogen, CH ; and a four-way intersection is a carbon atom with no attached hydrogens.\n\nSolution\n\n\nCarvone ( $\\mathrm{C}_{\\mathbf{1 0}} \\mathrm{H}_{\\mathbf{1 4}} \\mathrm{O}$ )\n\nPROBLEM How many hydrogens are bonded to each carbon in the following compounds, and what is the 1-15 molecular formula of each substance?\n(a)\n\n\nAdrenaline\nEstrone (a hormone)\n\nPROBLEM Propose skeletal structures for compounds that satisfy the following molecular formulas: There is\n1-16 more than one possibility in each case.\n(a) $\\mathrm{C}_{5} \\mathrm{H}_{12}$\n(b) $\\mathrm{C}_{2} \\mathrm{H}_{7} \\mathrm{~N}$\n(c) $\\mathrm{C}_{3} \\mathrm{H}_{6} \\mathrm{O}$\n(d) $\\mathrm{C}_{4} \\mathrm{H}_{9} \\mathrm{Cl}$\n\nPROBLEM The following molecular model is a representation of para-aminobenzoic acid (PABA), the active\n1-17 ingredient in many sunscreens. Indicate the positions of the multiple bonds, and draw a skeletal structure (black $=\\mathrm{C}$, red $=\\mathrm{O}$, blue $=\\mathrm{N}$, gray $=\\mathrm{H}$ ).\n\npara-Aminobenzoic acid\n(PABA)"}
{"id": 49, "contents": "Organic Foods: Risk versus Benefit - \nContrary to what you may hear in supermarkets or on television, all foods are organic-that is, complex mixtures of organic molecules. Even so, when applied to food, the word organic has come to mean an absence of synthetic chemicals, typically pesticides, antibiotics, and preservatives. How concerned should we be about traces of pesticides in the food we eat? Or toxins in the water we drink? Or pollutants in the air we breathe?\n\nLife is not risk-free-we all take many risks each day without even thinking about it. We decide to ride a bike rather than drive, even though there is a ten times greater likelihood per mile of dying in a bicycling accident than in a car. We decide to walk down stairs rather than take an elevator, even though 32,000 people die from falls each year in the United States. Some of us decide to smoke cigarettes, even though it increases our chance of getting cancer by $50 \\%$. But what about risks from chemicals like pesticides?\n\n\nFIGURE 1.20 How dangerous is the pesticide being sprayed on this crop? (credit: \"NRCSAR83001(265)\" by USDA Natural Resources Conservation Service/Wikimedia Commons, Public Domain)\n\nOne thing is certain: without pesticides, whether they target weeds (herbicides), insects (insecticides), or molds and fungi (fungicides), crop production would drop significantly, food prices would increase, and famines would occur in less developed parts of the world. Take the herbicide atrazine, for instance. In the United States alone, approximately 100 million pounds of atrazine are used each year to kill weeds in corn, sorghum, and sugarcane fields, greatly improving the yields of these crops. Nevertheless, the use of atrazine continues to be a concern because traces persist in the environment. Indeed, heavy atrazine exposure can pose health risks to humans and some animals. Because of these risks, the United States Environmental Protection Agency (EPA) has decided not to ban its use because doing so would result in lower crop yields and increased food costs, and because there is no suitable alternative herbicide available."}
{"id": 50, "contents": "Organic Foods: Risk versus Benefit - \nAtrazine\nHow can the potential hazards from a chemical like atrazine be determined? Risk evaluation of chemicals is carried out by exposing test animals, usually mice or rats, to the chemical and then monitoring the animals for signs of harm. To limit the expense and time needed, the amounts administered are typically hundreds or thousands of times greater than those a person might normally encounter. The results obtained in animal tests are then distilled into a single number called an $\\mathrm{LD}_{50}$, the amount of substance per kilogram of body weight that\nis a lethal dose for $50 \\%$ of the test animals. For atrazine, the $\\mathrm{LD}_{50}$ value is between 1 and $4 \\mathrm{~g} / \\mathrm{kg}$ depending on the animal species. Aspirin, for comparison, has an $\\mathrm{LD}_{50}$ of $1.1 \\mathrm{~g} / \\mathrm{kg}$, and ethanol (ethyl alcohol) has an $\\mathrm{LD}_{50}$ of $10.6 \\mathrm{~g} / \\mathrm{kg}$.\n\nTABLE 1.4 lists the $\\mathrm{LD}_{50}$ for some other familiar substances. The lower the value, the more toxic the substance. Note, though, that $\\mathrm{LD}_{50}$ values only pertain to the effects of heavy exposure for a relatively short time. They say nothing about the risks of long-term exposure, such as whether the substance can cause cancer or interfere with development in the unborn.\n\n| TABLE 1.4 Some LD 50 |  |  |  |\n| :--- | :--- | :--- | :---: |\n| Substance | $\\mathrm{LD}_{50}(\\mathrm{~g} / \\mathrm{kg})$ | Substance | $\\mathrm{LD}_{50}(\\mathrm{~g} / \\mathrm{kg})$ |\n| Strychnine 0.005 Chloroform 1.2 <br> Arsenic trioxide 0.015 Iron(II) sulfate 1.5 <br> DDT 0.115 Ethyl alcohol 10.6 <br> Aspirin 1.1 Sodium cyclamate 17 |  |  |  |"}
{"id": 51, "contents": "Organic Foods: Risk versus Benefit - \nSo, should we still use atrazine? All decisions involve tradeoffs, and the answer is rarely obvious. Does the benefit of increased food production outweigh possible health risks of a pesticide? Do the beneficial effects of a new drug outweigh a potentially dangerous side effect in a small number of users? Different people will have different opinions, but an honest evaluation of facts is surely the best way to start. As of June 2022, atrazine was still approved for continued use in the United States because the EPA believes that the benefits of increased food production outweigh possible health risks. At the same time, atrazine is little used, though not banned, in the European Union."}
{"id": 52, "contents": "Key Terms - \n- antibonding MO\n- atomic number ( $Z$ )\n- Aufbau principle\n- bond angle\n- bond length\n- bond strength\n- bonding MO\n- condensed structure\n- covalent bond\n- electron shell\n- electron-dot structure\n- ground-state electron configuration\n- Hund's rule\n- ionic bond\n- isotope\n- Kekul\u00e9 structure\n- Lewis structure\n- line-bond structure\n- lone-pair electrons\n- mass number ( $A$ )\n- molecular orbital (MO) theory\n- molecule\n- node\n- nonbonding electron\n- orbital\n- organic chemistry\n- Pauli exclusion principle\n- pi ( $\\pi$ ) bond\n- sigma ( $\\sigma$ ) bond\n- skeletal structure\n- sp hybrid orbital\n- $s p^{2}$ hybrid orbital\n- $s p^{3}$ hybrid orbital\n- valence bond (VB) theory\n- valence shell"}
{"id": 53, "contents": "Summary - \nThe purpose of this chapter has been to get you up to speed-to review some ideas about atoms, bonds, and molecular geometry. As we've seen, organic chemistry is the study of carbon compounds. Although a division into organic and inorganic chemistry occurred historically, there is no scientific reason for the division.\n\nAn atom consists of a positively charged nucleus surrounded by one or more negatively charged electrons. The electronic structure of an atom can be described by a quantum mechanical wave equation, in which electrons are considered to occupy orbitals around the nucleus. Different orbitals have different energy levels and different shapes. For example, $s$ orbitals are spherical and $p$ orbitals are dumbbell-shaped. The groundstate electron configuration of an atom can be found by assigning electrons to the proper orbitals, beginning with the lowest-energy ones.\n\nA covalent bond is formed when an electron pair is shared between atoms. According to valence bond (VB) theory, electron sharing occurs by the overlap of two atomic orbitals. According to molecular orbital (MO) theory, bonds result from the mathematical combination of atomic orbitals to give molecular orbitals, which belong to the entire molecule. Bonds that have a circular cross-section and are formed by head-on interaction are called sigma ( $\\boldsymbol{\\sigma}$ ) bonds; bonds formed by sideways interaction of $p$ orbitals are called $\\mathbf{p i}$ ( $\\boldsymbol{\\pi}$ ) bonds.\n\nIn the valence bond description, carbon uses hybrid orbitals to form bonds in organic molecules. When forming only single bonds with tetrahedral geometry, carbon uses four equivalent $\\boldsymbol{s p}^{\\mathbf{3}}$ hybrid orbitals. When forming a double bond with planar geometry, carbon uses three equivalent $\\boldsymbol{s p}^{\\mathbf{2}}$ hybrid orbitals and one unhybridized $p$ orbital. When forming a triple bond with linear geometry, carbon uses two equivalent $\\boldsymbol{s p h}$ hybrid orbitals and two unhybridized $p$ orbitals. Other atoms such as nitrogen, phosphorus, oxygen, and sulfur also use hybrid orbitals to form strong, oriented bonds."}
{"id": 54, "contents": "Summary - \nOrganic molecules are usually drawn using either condensed structures or skeletal structures. In condensed structures, carbon-carbon and carbon-hydrogen bonds aren't shown. In skeletal structures, only the bonds and not the atoms are shown. A carbon atom is assumed to be at the ends and at the junctions of lines (bonds), and the correct number of hydrogens is supplied mentally."}
{"id": 55, "contents": "WHY YOU SHOULD WORK PROBLEMS - \nThere's no surer way to learn organic chemistry than by working problems. Although careful reading and rereading of this text are important, reading alone isn't enough. You must also be able to use the information you've read and be able to apply your knowledge in new situations. Working problems gives you practice at doing this.\n\nEach chapter in this book provides many problems of different sorts. The in-chapter problems are placed for immediate reinforcement of ideas just learned, while end-of-chapter problems provide additional practice and come in several forms. They often begin with a short section called \"Visualizing Chemistry,\" which helps you see the microscopic world of molecules and provides practice for working in three dimensions. After the visualizations are many further problems, which are organized by topic. Early problems are primarily of the drill type, providing an opportunity for you to practice your command of the fundamentals. Later problems tend to be more thought-provoking, and some are real challenges.\n\nAs you study organic chemistry, take the time to work the problems. Do the ones you can, and ask for help on the ones you can't. If you're stumped by a particular problem, check the accompanying Study Guide and Student Solutions Manual for an explanation that should help clarify the difficulty. Working problems takes effort, but the payoff in knowledge and understanding is immense."}
{"id": 56, "contents": "Visualizing Chemistry - \nPROBLEM Convert each of the following molecular models into a skeletal structure, and give the formula of\n1-18 each. Only the connections between atoms are shown; multiple bonds are not indicated (black = C, red $=0$, blue $=\\mathrm{N}$, gray $=\\mathrm{H}$ ) .\n(a)\n\n\nConiine (the toxic substance in poison hemlock)\n(b)\n\n\nAlanine (an amino acid)\n\nPROBLEM The following model is a representation of citric acid, the key substance in the so-called citric acid\n1-19 cycle, by which food molecules are metabolized in the body. Only the connections between atoms are shown; multiple bonds are not indicated. Complete the structure by indicating the positions of multiple bonds and lone-pair electrons (black = C, red = O, gray $=\\mathrm{H}$ ).\n\n\nPROBLEM The following model is a representation of acetaminophen, a pain reliever sold in drugstores\n1-20 under a variety of names, including Tylenol. Identify the hybridization of each carbon atom in acetaminophen, and tell which atoms have lone pairs of electrons (black = C, red = O, blue = N, gray $=\\mathrm{H})$.\n\n\nPROBLEM The following model is a representation of aspartame, $\\mathrm{C}_{14} \\mathrm{H}_{18} \\mathrm{~N}_{2} \\mathrm{O}_{5}$, known commercially under\n1-21 many names, including NutraSweet. Only the connections between atoms are shown; multiple bonds are not indicated. Complete the structure for aspartame, and indicate the positions of multiple bonds (black $=\\mathrm{C}$, red $=\\mathrm{O}$, blue $=\\mathrm{N}$, gray $=\\mathrm{H}$ )."}
{"id": 57, "contents": "Electron Configurations - \nPROBLEM How many valence electrons does each of the following dietary trace elements have?\n1-22\n(a) Zinc (b) Iodine (c) Silicon (d) Iron\n\nPROBLEM Give the ground-state electron configuration for each of the following elements:\n1-23\n(a) Potassium (b) Arsenic\n(c) Aluminum\n(d) Germanium"}
{"id": 58, "contents": "Electron-Dot and Line-Bond Structures - \nPROBLEM What are likely formulas for the following molecules?\n1-24 (a) $\\mathrm{NH}_{?} \\mathrm{OH}$\n(b) $\\mathrm{AlCl}_{\\text {? }}$\n(c) $\\mathrm{CF}_{2} \\mathrm{Cl}_{?}$\n(d) $\\mathrm{CH}_{?} \\mathrm{O}$\n\nPROBLEM Why can't molecules with the following formulas exist?\n$\\mathbf{1 - 2 5}$ (a) $\\mathrm{CH}_{5}$ (b) $\\mathrm{C}_{2} \\mathrm{H}_{6} \\mathrm{~N}$ (c) $\\mathrm{C}_{3} \\mathrm{H}_{5} \\mathrm{Br}_{2}$\nPROBLEM Draw an electron-dot structure for acetonitrile, $\\mathrm{C}_{2} \\mathrm{H}_{3} \\mathrm{~N}$, which contains a carbon-nitrogen triple\n1-26 bond. How many electrons does the nitrogen atom have in its outer shell? How many are bonding, and how many are nonbonding?\n\nPROBLEM Draw a line-bond structure for vinyl chloride, $\\mathrm{C}_{2} \\mathrm{H}_{3} \\mathrm{Cl}$, the starting material from which PVC\n1-27 poly(vinyl chloride) plastic is made.\nPROBLEM Fill in any nonbonding valence electrons that are missing from the following structures:\n1-28\n(a)\n\nDimethyl disulfide\n(b)\n\nAcetamide\n(c)\n\nAcetate ion\n\nPROBLEM Convert the following line-bond structures into molecular formulas:\n1-29\n(a)\n\n(b)\n\n(c)\n\nVitamin C (ascorbic acid)\nNicotine\n(d)\n\nGlucose"}
{"id": 59, "contents": "Electron-Dot and Line-Bond Structures - \nAcetate ion\n\nPROBLEM Convert the following line-bond structures into molecular formulas:\n1-29\n(a)\n\n(b)\n\n(c)\n\nVitamin C (ascorbic acid)\nNicotine\n(d)\n\nGlucose\n\nPROBLEM Convert the following molecular formulas into line-bond structures that are consistent with valence\n1-30 rules:\n(a) $\\mathrm{C}_{3} \\mathrm{H}_{8}$\n(b) $\\mathrm{CH}_{5} \\mathrm{~N}$\n(c) $\\mathrm{C}_{2} \\mathrm{H}_{6} \\mathrm{O}$ (2 possibilities)\n(d) $\\mathrm{C}_{3} \\mathrm{H}_{7} \\mathrm{Br}$ (2 possibilities)\n(e) $\\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}$ (3 possibilities) (f) $\\mathrm{C}_{3} \\mathrm{H}_{9} \\mathrm{~N}$ (4 possibilities)\n\nPROBLEM Draw a three-dimensional representation of the oxygen-bearing carbon atom in ethanol,\n1-31 $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$, using the standard convention of solid, wedged, and dashed lines."}
{"id": 60, "contents": "Electron-Dot and Line-Bond Structures - \nPROBLEM Oxaloacetic acid, an important intermediate in food metabolism, has the formula $\\mathrm{C}_{4} \\mathrm{H}_{4} \\mathrm{O}_{5}$ and\n1-32 contains three $\\mathrm{C}=\\mathrm{O}$ bonds and two $\\mathrm{O}-\\mathrm{H}$ bonds. Propose two possible structures.\nPROBLEM Draw structures for the following molecules, showing lone pairs:\n1-33 (a) Acrylonitrile, $\\mathrm{C}_{3} \\mathrm{H}_{3} \\mathrm{~N}$, which contains a carbon-carbon double bond and a carbon-nitrogen triple bond\n(b) Ethyl methyl ether, $\\mathrm{C}_{3} \\mathrm{H}_{8} \\mathrm{O}$, which contains an oxygen atom bonded to two carbons\n(c) Butane, $\\mathrm{C}_{4} \\mathrm{H}_{10}$, which contains a chain of four carbon atoms\n(d) Cyclohexene, $\\mathrm{C}_{6} \\mathrm{H}_{10}$, which contains a ring of six carbon atoms and one carbon-carbon double bond\n\nHybridization\nPROBLEM What is the hybridization of each carbon atom in acetonitrile (Problem 1-26)?\n1-34\nPROBLEM What kind of hybridization do you expect for each carbon atom in the following molecules?\n$1-35$\n(a) Propane, $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$\n(b) 2-Methylpropene,\n\n(c) But-1-en-3-yne, $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}-\\mathrm{C} \\equiv \\mathrm{CH}$\n(d) Acetic acid,\n\n\nPROBLEM What is the shape of benzene, and what hybridization do you expect for each carbon?\n1-36\n\n\nPROBLEM What bond angle do you expect for each of the indicated atoms, and what kind of hybridization do\n1-37 you expect for the central atom in each molecule?\n(a)\n\n(b)\n\nPyridine\n(c)\n\nLactic acid\n(in sour milk)"}
{"id": 61, "contents": "Electron-Dot and Line-Bond Structures - \n(b)\n\nPyridine\n(c)\n\nLactic acid\n(in sour milk)\n\nPROBLEM Propose structures for molecules that meet the following descriptions:\n1-38 (a) Contains two $s p^{2}$-hybridized carbons and two $s p^{3}$-hybridized carbons\n(b) Contains only four carbons, all of which are $s p^{2}$-hybridized\n(c) Contains two $s p$-hybridized carbons and two $s p^{2}$-hybridized carbons\n\nPROBLEM What kind of hybridization do you expect for each carbon atom in the following molecules:\n1-39\n(a)\n\n(b)\n\n\nPROBLEM Pyridoxal phosphate, a close relative of vitamin $\\mathrm{B}_{6}$, is involved in a large number of metabolic\n1-40 reactions. What is the hybridization and the bond angle for each nonterminal atom?\n\n\nPyridoxal phosphate\n\nSkeletal Structures\nPROBLEM Convert the following structures into skeletal drawings:\n1-41\n(a)\n\nIndole\n(b)\n\n(c)\n\n(d)\n\nBenzoquinone\n\nPROBLEM How many hydrogens are bonded to each carbon atom in the following substances, and what is the\n1-42 molecular formula of each?\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Quetiapine, marketed as Seroquel, is a heavily prescribed antipsychotic drug used in the treatment\n1-43 of schizophrenia and bipolar disorder. Convert the following representation into a skeletal structure, and give the molecular formula of quetiapine.\n\n\nPROBLEM How many hydrogens are bonded to each carbon atom in (a) the antiinfluenza agent oseltamivir, 1-44 marketed as Tamiflu, and (b) the platelet aggregation inhibitor clopidogrel, marketed as Plavix? Give the molecular formula of each.\n(a)\n\nOseltamivir\n(Tamiflu)\n(b)\n\nClopidogrel\n(Plavix)"}
{"id": 62, "contents": "General Problems - \nPROBLEM Why do you suppose no one has ever been able to make cyclopentyne as a stable molecule?\n1-45\n\n\nCyclopentyne\n\nPROBLEM Allene, $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{C}=\\mathrm{CH}_{2}$, has two adjacent double bonds. Draw a picture showing the orbitals involved\n1-46 in the $\\sigma$ and $\\pi$ bonds of allene. Is the central carbon atom $s p^{2}$ - or $s p$-hybridized? What about the hybridization of the terminal carbons? What shape do you predict for allene?\n\nPROBLEM Allene (see Problem 1-46) is structurally related to carbon dioxide, $\\mathrm{CO}_{2}$. Draw a picture showing the\n1-47 orbitals involved in the $\\sigma$ and $\\pi$ bonds of $\\mathrm{CO}_{2}$, and identify the likely hybridization of carbon.\nPROBLEM Complete the electron-dot structure of caffeine, showing all lone-pair electrons, and identify the\n1-48 hybridization of the indicated atoms.\n\n\nPROBLEM Most stable organic species have tetravalent carbon atoms, but species with trivalent carbon atoms\n1-49 also exist. Carbocations are one such class of compounds."}
{"id": 63, "contents": "A carbocation - \n(a) How many valence electrons does the positively charged carbon atom have?\n(b) What hybridization do you expect this carbon atom to have?\n(c) What geometry is the carbocation likely to have?\n\nPROBLEM A carbanion is a species that contains a negatively charged, trivalent carbon.\n1-50\n\n\nA carbanion\n(a) What is the electronic relationship between a carbanion and a trivalent nitrogen compound such as $\\mathrm{NH}_{3}$ ?\n(b) How many valence electrons does the negatively charged carbon atom have?\n(c) What hybridization do you expect this carbon atom to have?\n(d) What geometry is the carbanion likely to have?\n\nPROBLEM Divalent carbon species called carbenes are capable of fleeting existence. For example, methylene,\n1-51 $: \\mathrm{CH}_{2}$, is the simplest carbene. The two unshared electrons in methylene can be either paired in a single orbital or unpaired in different orbitals. Predict the type of hybridization you expect carbon to adopt in singlet (spin-paired) methylene and triplet (spin-unpaired) methylene. Draw a picture of each, and identify the valence orbitals on carbon.\n\nPROBLEM Two different substances have the formula $\\mathrm{C}_{4} \\mathrm{H}_{10}$. Draw both, and tell how they differ.\n1-52\nPROBLEM Two different substances have the formula $\\mathrm{C}_{3} \\mathrm{H}_{6}$. Draw both, and tell how they differ.\n1-53\nPROBLEM Two different substances have the formula $\\mathrm{C}_{2} \\mathrm{H}_{6} \\mathrm{O}$. Draw both, and tell how they differ.\n1-54\nPROBLEM Three different substances contain a carbon-carbon double bond and have the formula $\\mathrm{C}_{4} \\mathrm{H}_{8}$. Draw\n1-55 them, and tell how they differ.\nPROBLEM Among the most common over-the-counter drugs you might find in a medicine cabinet are mild\n1-56 pain relievers such ibuprofen (Advil, Motrin), naproxen (Aleve), and acetaminophen (Tylenol).\n\n\nIbuprofen\n\n\nNaproxen"}
{"id": 64, "contents": "A carbocation - \nIbuprofen\n\n\nNaproxen\n\n\nAcetaminophen\n(a) How many $s p^{3}$-hybridized carbons does each molecule have?\n(b) How many $s p^{2}$-hybridized carbons does each molecule have?\n(c) What similarities can you see in their structures?"}
{"id": 65, "contents": "CHAPTER 2 - \nPolar Covalent Bonds; Acids and Bases\n\n\nFIGURE 2.1 The opium poppy is the source of morphine, one of the first \"vegetable alkali,\" or alkaloids, to be isolated. (credit: modification of work \"Papaver somniferum\" by Liz West/Flickr, CC BY 2.0)"}
{"id": 66, "contents": "CHAPTER CONTENTS - \n2.1 Polar Covalent Bonds and Electronegativity\n2.2 Polar Covalent Bonds and Dipole Moments\n2.3 Formal Charges\n2.4 Resonance\n2.5 Rules for Resonance Forms\n2.6 Drawing Resonance Forms\n2.7 Acids and Bases: The Br\u00f8nsted-Lowry Definition\n2.8 Acid and Base Strength\n2.9 Predicting Acid-Base Reactions from $\\mathrm{p} K_{\\mathrm{a}}$ Values\n2.10 Organic Acids and Organic Bases\n2.11 Acids and Bases: The Lewis Definition\n2.12 Noncovalent Interactions between Molecules\n\nWHY THIS CHAPTER? Understanding organic chemistry means knowing not just what happens but also why and how it happens at the molecular level. In this chapter, we'll look at some of the ways that chemists describe and account for chemical reactivity, thereby providing a foundation to understand the specific reactions discussed in subsequent chapters. Topics such as bond polarity, the acid-base behavior of molecules, and hydrogen-bonding are a particularly important part of that foundation.\n\nWe saw in the previous chapter how covalent bonds between atoms are described, and we looked at the valence\nbond model, which uses hybrid orbitals to account for the observed shapes of organic molecules. Before going on to a systematic study of organic chemistry, however, we still need to review a few fundamental topics. In particular, we need to look more closely at how electrons are distributed in covalent bonds and at some of the consequences that arise when the electrons in a bond are not shared equally between atoms."}
{"id": 67, "contents": "CHAPTER CONTENTS - 2.1 Polar Covalent Bonds and Electronegativity\nUp to this point, we've treated chemical bonds as either ionic or covalent. The bond in sodium chloride, for instance, is ionic. Sodium transfers an electron to chlorine to produce $\\mathrm{Na}^{+}$and $\\mathrm{Cl}^{-}$ions, which are held together in the solid by electrostatic attractions between unlike charges. The $\\mathrm{C}-\\mathrm{C}$ bond in ethane, however, is covalent. The two bonding electrons are shared equally by the two equivalent carbon atoms, resulting in a symmetrical electron distribution in the bond. Most bonds, however, are neither fully ionic nor fully covalent but are somewhere between the two extremes. Such bonds are called polar covalent bonds, meaning that the bonding electrons are attracted more strongly by one atom than the other so that the electron distribution between atoms is not symmetrical (FIGURE 2.2).\n\n\nFIGURE 2.2 The continuum in bonding from covalent to ionic is a result of an unequal distribution of bonding electrons between atoms. The symbol $\\delta$ (lowercase Greek letter delta) means partial charge, either partial positive ( $\\delta+$ ) for the electron-poor atom or partial negative ( $\\delta-$ ) for the electron-rich atom.\n\nBond polarity is due to differences in electronegativity (EN), the intrinsic ability of an atom to attract the shared electrons in a covalent bond. As shown in FIGURE 2.3, electronegativities are based on an arbitrary scale, with fluorine the most electronegative ( $\\mathrm{EN}=4.0$ ) and cesium the least ( $\\mathrm{EN}=0.7$ ). Metals on the left side of the periodic table attract electrons weakly and have lower electronegativities, while oxygen, nitrogen, and halogens on the right side of the periodic table attract electrons strongly and have higher electronegativities. Carbon, the most important element in organic compounds, has an intermediate electronegativity value of 2.5 ."}
{"id": 68, "contents": "CHAPTER CONTENTS - 2.1 Polar Covalent Bonds and Electronegativity\n| $\\begin{gathered} \\mathrm{H} \\\\ 2.1 \\end{gathered}$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | He |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $\\begin{aligned} & \\mathrm{Li} \\\\ & 1.0 \\end{aligned}$ | $\\begin{aligned} & \\hline \\mathrm{Be} \\\\ & 1.6 \\end{aligned}$ |  |  |  |  |  |  |  |  |  |  | $\\begin{gathered} \\text { B } \\\\ 2.0 \\end{gathered}$ | $\\begin{gathered} C \\\\ C \\\\ 2.5 \\end{gathered}$ | $\\begin{gathered} \\mathrm{N} \\\\ 3.0 \\end{gathered}$ | $\\begin{gathered} 0 \\\\ 3.5 \\end{gathered}$ | $\\begin{gathered} \\text { F } \\\\ 4.0 \\end{gathered}$ | Ne |"}
{"id": 69, "contents": "CHAPTER CONTENTS - 2.1 Polar Covalent Bonds and Electronegativity\n| $\\begin{aligned} & \\mathrm{Na} \\\\ & 0.9 \\end{aligned}$ | $\\begin{aligned} & \\hline \\mathrm{Mg} \\\\ & 1.2 \\end{aligned}$ |  |  |  |  |  |  |  |  |  |  | $\\begin{aligned} & \\mathrm{Al} \\\\ & 1.5 \\end{aligned}$ | $\\begin{gathered} \\mathrm{Si} \\\\ 1.8 \\end{gathered}$ | $\\begin{gathered} \\hline P \\\\ 2.1 \\end{gathered}$ | $\\begin{gathered} \\mathrm{S} \\\\ 2.5 \\end{gathered}$ | $\\begin{gathered} \\hline \\mathrm{Cl} \\\\ 3.0 \\end{gathered}$ | Ar |"}
{"id": 70, "contents": "CHAPTER CONTENTS - 2.1 Polar Covalent Bonds and Electronegativity\n| $\\begin{gathered} \\mathrm{K} \\\\ 0.8 \\end{gathered}$ | $\\begin{aligned} & \\hline \\mathrm{Ca} \\\\ & 1.0 \\end{aligned}$ | $\\begin{aligned} & \\mathrm{Sc} \\\\ & 1.3 \\end{aligned}$ | Ti 1.5 | $\\begin{gathered} \\hline \\mathrm{V} \\\\ 1.6 \\end{gathered}$ | $\\begin{gathered} \\mathrm{Cr} \\\\ 1.6 \\end{gathered}$ | $\\begin{aligned} & \\mathrm{Mn} \\\\ & 1.5 \\end{aligned}$ | $\\begin{aligned} & \\hline \\mathrm{Fe} \\\\ & 1.8 \\end{aligned}$ | Co 1.9 | $\\begin{gathered} \\hline \\mathrm{Ni} \\\\ 1.9 \\end{gathered}$ | Cu 1.9 | Zn 1.6 | $\\begin{aligned} & \\hline \\mathrm{Ga} \\\\ & 1.6 \\end{aligned}$ | $\\begin{aligned} & \\hline \\mathrm{Ge} \\\\ & 1.8 \\end{aligned}$ | As 2.0 | $\\begin{aligned} & \\hline \\mathrm{Se} \\\\ & 2.4 \\end{aligned}$ | $\\begin{aligned} & \\hline \\mathrm{Br} \\\\ & 2.8 \\end{aligned}$ | Kr |"}
{"id": 71, "contents": "CHAPTER CONTENTS - 2.1 Polar Covalent Bonds and Electronegativity\n| $\\begin{aligned} & \\hline \\mathrm{Rb} \\\\ & 0.8 \\end{aligned}$ | $\\begin{gathered} \\hline \\mathrm{Sr} \\\\ 1.0 \\end{gathered}$ | Y 1.2 | $\\begin{aligned} & \\mathrm{Zr} \\\\ & 1.4 \\end{aligned}$ | $\\begin{aligned} & \\hline \\mathrm{Nb} \\\\ & 1.6 \\end{aligned}$ | $\\begin{aligned} & \\hline \\text { Mo } \\\\ & 1.8 \\end{aligned}$ | $\\begin{aligned} & \\hline \\text { Tc } \\\\ & 1.9 \\end{aligned}$ | $\\begin{aligned} & \\mathrm{Ru} \\\\ & 2.2 \\end{aligned}$ | Rh 2.2 | $\\begin{aligned} & \\hline \\mathrm{Pd} \\\\ & 2.2 \\end{aligned}$ | Ag 1.9 | Cd 1.7 | $\\begin{aligned} & \\hline \\text { In } \\\\ & 1.7 \\end{aligned}$ | Sn | Sb | $\\begin{aligned} & \\hline \\mathrm{Te} \\\\ & 2.1 \\end{aligned}$ | $\\begin{gathered} \\mathrm{I} \\\\ 2.5 \\end{gathered}$ | Xe |"}
{"id": 72, "contents": "CHAPTER CONTENTS - 2.1 Polar Covalent Bonds and Electronegativity\n| $\\begin{gathered} \\hline \\mathrm{Cs} \\\\ 0.7 \\end{gathered}$ | $\\begin{aligned} & \\mathrm{Ba} \\\\ & 0.9 \\end{aligned}$ | La 1.0 | Hf 1.3 | Ta 1.5 | W | Re 1.9 | $\\begin{aligned} & \\hline \\text { Os } \\\\ & 2.2 \\end{aligned}$ | Ir 2.2 | Pt 2.2 | Au 2.4 | Hg 1.9 | $\\begin{gathered} \\hline \\mathrm{Tl} \\\\ 1.8 \\end{gathered}$ | $\\mathrm{Pb}$ | $\\begin{gathered} \\mathrm{Bi} \\\\ 1.9 \\end{gathered}$ | $\\begin{aligned} & \\hline \\text { Po } \\\\ & 2.0 \\end{aligned}$ | $\\begin{gathered} \\hline \\text { At } \\\\ 2.1 \\end{gathered}$ | Rn |"}
{"id": 73, "contents": "CHAPTER CONTENTS - 2.1 Polar Covalent Bonds and Electronegativity\nFIGURE 2.3 Electronegativity values and trends. Electronegativity generally increases from left to right across the periodic table and decreases from top to bottom. The values are on an arbitrary scale, with $\\mathrm{F}=4.0$ and $\\mathrm{Cs}=0.7$. Elements in red are the most electronegative, those in yellow are medium, and those in green are the least electronegative.\n\nAs a rough guide, bonds between atoms whose electronegativities differ by less than 0.5 are nonpolar covalent, bonds between atoms whose electronegativities differ by 0.5 to 2 are polar covalent, and bonds between atoms whose electronegativities differ by more than 2 are largely ionic. Carbon-hydrogen bonds, for example, are relatively nonpolar because carbon ( $\\mathrm{EN}=2.5$ ) and hydrogen ( $\\mathrm{EN}=2.1$ ) have similar electronegativities. Bonds between carbon and more electronegative elements such as oxygen ( $\\mathrm{EN}=3.5$ ) and nitrogen ( $\\mathrm{EN}=3.0$ ), by contrast, are polarized so that the bonding electrons are drawn away from carbon toward the electronegative atom. This leaves carbon with a partial positive charge, $\\delta$-, and the electronegative atom with a partial negative charge, $\\delta$-. An example is the $\\mathrm{C}-\\mathrm{O}$ bond in methanol, $\\mathrm{CH}_{3} \\mathrm{OH}$ (FIGURE 2.4a). Bonds between carbon and less\nelectronegative elements are polarized so that carbon bears a partial negative charge and the other atom bears a partial positive charge. An example is the $\\mathrm{C}-\\mathrm{Li}$ bond in methyllithium, $\\mathrm{CH}_{3} \\mathrm{Li}$ (FIGURE 2.4b).\n(a)\n\n(b)\n\n\n\nMethyllithium"}
{"id": 74, "contents": "CHAPTER CONTENTS - 2.1 Polar Covalent Bonds and Electronegativity\n(b)\n\n\n\nMethyllithium\n\nFIGURE 2.4 Polar covalent bonds. (a) Methanol, $\\mathrm{CH}_{3} \\mathrm{OH}$, has a polar covalent $\\mathrm{C}-\\mathrm{O}$ bond, and (b) methyllithium, $\\mathrm{CH}_{3} \\mathrm{Li}$, has a polar covalent C-Li bond. The computer-generated representations, called electrostatic potential maps, use color to show calculated charge distributions, ranging from red (electron-rich; $\\delta$-) to blue (electron-poor; $\\delta+$ ).\n\nNote in the representations of methanol and methyllithium in FIGURE 2.4 that a crossed arrow $\\longrightarrow$ is used to indicate the direction of bond polarity. By convention, electrons are displaced in the direction of the arrow. The tail of the arrow (which looks like a plus sign) is electron-poor ( $\\delta+$ ), and the head of the arrow is electron-rich ( $\\delta$-).\n\nNote also in FIGURE 2.4 that calculated charge distributions in molecules can be displayed visually with what are called electrostatic potential maps, which use color to indicate electron-rich (red; $\\delta$-) and electron-poor (blue; $\\delta+$ ) regions. In methanol, oxygen carries a partial negative charge and is colored red, while the carbon and hydrogen atoms carry partial positive charges and are colored blue-green. In methyllithium, lithium carries a partial positive charge (blue), while carbon and the hydrogen atoms carry partial negative charges (red). Electrostatic potential maps are useful because they show at a glance the electron-rich and electron-poor atoms in molecules. We'll make frequent use of these maps throughout the text and will see many examples of how electronic structure correlates with chemical reactivity.\n\nWhen speaking of an atom's ability to polarize a bond, we often use the term inductive effect. An inductive effect is simply the shifting of electrons in a $\\sigma$ bond in response to the electronegativity of nearby atoms. Metals, such as lithium and magnesium, inductively donate electrons, whereas reactive nonmetals, such as oxygen and nitrogen, inductively withdraw electrons. Inductive effects play a major role in understanding chemical reactivity, and we'll use them many times throughout this text to explain a variety of chemical observations."}
{"id": 75, "contents": "CHAPTER CONTENTS - 2.1 Polar Covalent Bonds and Electronegativity\nPROBLEM Which element in each of the following pairs is more electronegative?\n2-1 (a) Li or H\n(b) B or Br\n(c) Cl or I (d) C or H\n\nPROBLEM Use the $\\delta+/ \\delta$ - convention to indicate the direction of expected polarity for each of the bonds 2-2 indicated.\n(a) $\\mathrm{H}_{3} \\mathrm{C}-\\mathrm{Cl}$\n(b) $\\mathrm{H}_{3} \\mathrm{C}-\\mathrm{NH}_{2}$\n(c) $\\mathrm{H}_{2} \\mathrm{~N}-\\mathrm{H}$\n(d) $\\mathrm{H}_{3} \\mathrm{C}-\\mathrm{SH}$\n(e) $\\mathrm{H}_{3} \\mathrm{C}-\\mathrm{MgBr}$ (f) $\\quad \\mathrm{H}_{3} \\mathrm{C}-\\mathrm{F}$\n\nPROBLEM Use the electronegativity values shown in Figure 2.3 to rank the following bonds from least polar to 2-3 most polar: $\\mathrm{H}_{3} \\mathrm{C}-\\mathrm{Li}, \\mathrm{H}_{3} \\mathrm{C}-\\mathrm{K}, \\mathrm{H}_{3} \\mathrm{C}-\\mathrm{F}, \\mathrm{H}_{3} \\mathrm{C}-\\mathrm{MgBr}, \\mathrm{H}_{3} \\mathrm{C}-\\mathrm{OH}$\n\nPROBLEM Look at the following electrostatic potential map of methylamine, a substance responsible for the $\\mathbf{2 - 4}$ odor of rotting fish, and tell the direction of polarization of the $\\mathrm{C}-\\mathrm{N}$ bond:\n\n\nMethylamine"}
{"id": 76, "contents": "CHAPTER CONTENTS - 2.2 Polar Covalent Bonds and Dipole Moments\nJust as individual bonds are often polar, molecules as a whole are often polar as well. Molecular polarity results from the vector summation of all individual bond polarities and lone-pair contributions in the molecule. As a practical matter, strongly polar substances are often soluble in polar solvents like water, whereas less polar substances are insoluble in water.\n\nNet polarity is measured by a quantity called the dipole moment and can be thought of in the following way: assume that there is a center of mass of all positive charges (nuclei) in a molecule and a center of mass of all negative charges (electrons). If these two centers don't coincide, then the molecule has a net polarity.\n\nThe dipole moment, $\\mu$ (lowercase Greek letter mu), is defined as the magnitude of the charge $Q$ at either end of the molecular dipole times the distance $r$ between the charges, $\\mu=Q \\times r$. Dipole moments are expressed in debyes $(\\mathrm{D})$, where $1 \\mathrm{D}=3.336 \\times 10^{-30}$ coulomb meters ( $\\mathrm{C} \\cdot \\mathrm{m}$ ) in SI units. For example, the unit charge on an electron is $1.60 \\times 10^{-19} \\mathrm{C}$. Thus, if one positive charge and one negative charge are separated by 100 pm (a bit less than the length of a typical covalent bond), the dipole moment is $1.60 \\times 10^{-29} \\mathrm{C} \\cdot \\mathrm{m}$, or 4.80 D .\n\n$$\n\\begin{aligned}\n& \\mu=Q \\times r \\\\\n& \\mu=\\left(1.60 \\times 10^{-19} \\mathrm{C}\\right)\\left(100 \\times 10^{-12} \\mathrm{~m}\\right)\\left(\\frac{1 \\mathrm{D}}{3.336 \\times 10^{-30} \\mathrm{C} \\cdot \\mathrm{~m}}\\right)=4.80 \\mathrm{D}\n\\end{aligned}\n$$"}
{"id": 77, "contents": "CHAPTER CONTENTS - 2.2 Polar Covalent Bonds and Dipole Moments\nDipole moments for some common substances are given in TABLE 2.1. Of the compounds shown in the table, sodium chloride has the largest dipole moment $(9.00 \\mathrm{D})$ because it is ionic. Even small molecules like water ( $\\mu=1.85 \\mathrm{D}$ ), methanol ( $\\mathrm{CH}_{3} \\mathrm{OH} ; \\mu=1.70 \\mathrm{D}$ ), and ammonia ( $\\mu=1.47 \\mathrm{D}$ ), have substantial dipole moments, however, both because they contain strongly electronegative atoms (oxygen and nitrogen) and because all three molecules have lone-pair electrons. The lone-pair electrons on oxygen and nitrogen stick out into space away from the positively charged nuclei, giving rise to a considerable charge separation and making a large contribution to the dipole moment.\n\nTABLE 2.1 Dipole Moments of Some Compounds\n\n| Compound | Dipole moment (D) |  | Compound |\n| :--- | :--- | :--- | :--- |\n| NaCl | 9.00 | $\\mathrm{NH}_{3}$ | 1.47 |\n| $\\mathrm{CH}_{2} \\mathrm{O}$ | 2.33 | $\\mathrm{CH}_{3} \\mathrm{NH}_{2}$ | 1.31 |\n| $\\mathrm{CH}_{3} \\mathrm{Cl}$ | 1.87 | $\\mathrm{CO}_{2}$ | 0 |\n| $\\mathrm{H}_{2} \\mathrm{O}$ | 1.85 | $\\mathrm{CH}_{4}$ | 0 |\n| $\\mathrm{CH}_{3} \\mathrm{OH}$ | 1.70 | $\\mathrm{CH}_{3} \\mathrm{CH}_{3}$ | 0 |\n\nTABLE 2.1 Dipole Moments of Some Compounds"}
{"id": 78, "contents": "CHAPTER CONTENTS - 2.2 Polar Covalent Bonds and Dipole Moments\nTABLE 2.1 Dipole Moments of Some Compounds\n\n| Compound | Dipole moment (D) | Compound | Dipole moment (D) |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$ | 1.70 | <smiles>c1ccccc1</smiles> <br> Benzene | 0 |\n| $\\mathrm{CH}_{3} \\mathrm{SH}$ | 1.52 |  |  |\n| <smiles>C1=C[C+]2CC(O2)[Te]1</smiles><smiles>H1([H])C@H[C@H]2C=C[C@H]1C2</smiles><smiles>C[I-]NH+[I-]</smiles>Water Methanol Ammonia <br> $(\\mu=1.85 \\mathrm{D})$ $(\\mu=1.70 \\mathrm{D})$ $(\\mu=1.47 \\mathrm{D})$ |  |  |  |\n|  |  |  |  |\n\nIn contrast with water, methanol, and ammonia, molecules such as carbon dioxide, methane, ethane, and benzene have zero dipole moments. Because of the symmetrical structures of these molecules, the individual bond polarities and lone-pair contributions exactly cancel.\n\n$$\n\\mathrm{O}=\\mathrm{C}=\\mathrm{O}\n$$\n\nCarbon dioxide ( $\\mu=0$ )\n\n\nMethane\n( $\\mu=0$ )\n\n\nEthane\n( $\\mu=0$ )\n\n\nBenzene ( $\\mu=0$ )"}
{"id": 79, "contents": "Predicting the Direction of a Dipole Moment - \nMake a three-dimensional drawing of methylamine, $\\mathrm{CH}_{3} \\mathrm{NH}_{2}$, and show the direction of its dipole moment $(\\mu=$ 1.31)."}
{"id": 80, "contents": "Strategy - \nLook for any lone-pair electrons, and identify any atom with an electronegativity substantially different from that of carbon. (Usually, this means O, N, F, Cl, or Br.) Electron density will be displaced in the general direction of the electronegative atoms and the lone pairs."}
{"id": 81, "contents": "Solution - \nMethylamine contains an electronegative nitrogen atom with a lone pair of electrons. The dipole moment thus points generally from $-\\mathrm{CH}_{3}$ toward the lone pair.\n\n\nMethylamine\n( $\\mu=1.31$ )\n\nPROBLEM Ethylene glycol, $\\mathrm{HOCH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}$, may look nonpolar when drawn, but an internal hydrogen bond 2-5 between the two - OH groups results in a dipole moment. Explain.\n\nPROBLEM Make three-dimensional drawings of the following molecules, and predict whether each has a 2-6 dipole moment. If you expect a dipole moment, show its direction.\n(a) $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}_{2}$\n(b) $\\mathrm{CHCl}_{3}$\n(c) $\\mathrm{CH}_{2} \\mathrm{Cl}_{2}$\n(d) $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CCl}_{2}$"}
{"id": 82, "contents": "Solution - 2.3 Formal Charges\nClosely related to the ideas of bond polarity and dipole moment is the assignment of formal charges to specific atoms within a molecule, particularly atoms that have an apparently \"abnormal\" number of bonds. Look at dimethyl sulfoxide $\\left(\\mathrm{CH}_{3} \\mathrm{SOCH}_{3}\\right)$, for instance, a solvent commonly used for preserving biological cell lines at low temperature. The sulfur atom in dimethyl sulfoxide has three bonds rather than the usual two and has a formal positive charge. The oxygen atom, by contrast, has one bond rather than the usual two and has a formal negative charge. Note that an electrostatic potential map of dimethyl sulfoxide shows the oxygen as negative (red) and the sulfur as relatively positive (blue), in accordance with the formal charges."}
{"id": 83, "contents": "Dimethyl sulfoxide - \nFormal charges, as the name suggests, are a formalism and don't imply the presence of actual ionic charges in a molecule. Instead, they're a device for electron \"bookkeeping\" and can be thought of in the following way: A typical covalent bond is formed when each atom donates one electron. Although the bonding electrons are shared by both atoms, each atom can still be considered to \"own\" one electron for bookkeeping purposes. In methane, for instance, the carbon atom owns one electron in each of the four $\\mathrm{C}-\\mathrm{H}$ bonds. Because a neutral, isolated carbon atom has four valence electrons, and because the carbon atom in methane still owns four, the methane carbon atom is neutral and has no formal charge.\n\n\nThe same is true for the nitrogen atom in ammonia, which has three covalent $\\mathrm{N}-\\mathrm{H}$ bonds and two nonbonding electrons (a lone pair). Atomic nitrogen has five valence electrons, and the ammonia nitrogen also has five-one in each of three shared $\\mathrm{N}-\\mathrm{H}$ bonds plus two in the lone pair. Thus, the nitrogen atom in ammonia has no formal charge.\n\n\nThe situation is different in dimethyl sulfoxide. Atomic sulfur has six valence electrons, but the dimethyl sulfoxide sulfur owns only five-one in each of the two $\\mathrm{S}-\\mathrm{C}$ single bonds, one in the $\\mathrm{S}-\\mathrm{O}$ single bond, and two in a lone pair. Thus, the sulfur atom has formally lost an electron and therefore has a positive formal charge. A similar calculation for the oxygen atom shows that it has formally gained an electron and has a negative charge. Atomic oxygen has six valence electrons, but the oxygen in dimethyl sulfoxide has seven-one in the $O-S$ bond and two in each of three lone pairs. Thus, the oxygen has formally gained an electron and has a negative formal charge.\n\n\nTo express the calculations in a general way, the formal charge on an atom is equal to the number of valence electrons in a neutral, isolated atom minus the number of electrons owned by that bonded atom in a molecule. The number of electrons in the bonded atom, in turn, is equal to half the number of bonding electrons plus the nonbonding, lone-pair electrons."}
{"id": 84, "contents": "Dimethyl sulfoxide - \n$$\n\\begin{aligned}\n\\text { Formal charge } & =\\left(\\begin{array}{c}\n\\text { Number of } \\\\\n\\text { valence electrons } \\\\\n\\text { in free atom }\n\\end{array}\\right)-\\left(\\begin{array}{c}\n\\text { Number of } \\\\\n\\text { valence electrons } \\\\\n\\text { in bonded atom }\n\\end{array}\\right) \\\\\n& =\\left(\\begin{array}{c}\n\\text { Number of } \\\\\n\\text { valence electrons } \\\\\n\\text { in free atom }\n\\end{array}\\right)-\\left(\\begin{array}{c}\n\\text { Number of } \\\\\n\\text { bonding electrons } \\\\\n2\n\\end{array} \\begin{array}{c}\n\\text { Number of } \\\\\n\\text { nonbonding } \\\\\n\\text { electrons }\n\\end{array}\\right)\n\\end{aligned}\n$$\n\nA summary of commonly encountered formal charges and the bonding situations in which they occur is given in TABLE 2.2. Although only a bookkeeping device, formal charges often give clues about chemical reactivity, so it's helpful to be able to identify and calculate them correctly.\n\nTABLE 2.2 A Summary of Common Formal Charges"}
{"id": 85, "contents": "Dimethyl sulfoxide - \nTABLE 2.2 A Summary of Common Formal Charges\n\n| Atom | C |  |  | N |  | 0 |  | S |  | P |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| Structure | <smiles>CC(C)(C)C</smiles> | <smiles>CC(C)(C)C</smiles> | <smiles>CC(C)(C)C</smiles> |  | <smiles>CN+C</smiles> | <smiles></smiles> | - $\\ddot{0}$ | $-\\ddot{S}{ }^{+}$ | - |  |\n| Valence electrons | 4 | 4 | 4 | 5 | 5 | 6 | 6 | 6 | 6 | 5 |\n| Number of bonds | 3 | 3 | 3 | 4 | 2 | 3 | 1 | 3 | 1 | 4 |\n| Number of nonbonding electrons | 1 | 0 | 2 | 0 | 4 | 2 | 6 | 2 | 6 | 0 |\n| Formal charge | 0 | +1 | -1 | +1 | -1 | +1 | -1 | +1 | -1 | +1 |\n\nPROBLEM Calculate formal charges for the nonhydrogen atoms in the following molecules:\n2-7 (a)\nDiazomethane, $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{N}=\\ddot{\\mathrm{N}}$ :\n(b) Acetonitrile oxide, $\\mathrm{H}_{3} \\mathrm{C}-\\mathrm{C} \\equiv \\mathrm{N}-\\ddot{\\mathrm{O}}$ :\n(c) Methyl isocyanide, $\\mathrm{H}_{3} \\mathrm{C}-\\mathrm{N} \\equiv \\mathrm{C}$ :\n\nPROBLEM Organic phosphate groups occur commonly in biological molecules. Calculate formal charges on\nthe four O atoms in the methyl phosphate dianion."}
{"id": 86, "contents": "Dimethyl sulfoxide - 2.4 Resonance\nMost substances can be represented unambiguously by the Kekul\u00e9 line-bond structures we've been using up to this point, but an interesting problem sometimes arises. Look at the acetate ion, for instance. When we draw a line-bond structure for acetate, we need to show a double bond to one oxygen and a single bond to the other. But which oxygen is which? Should we draw a double bond to the \"top\" oxygen and a single bond to the \"bottom\" oxygen, or vice versa?\n\n\nAlthough the two oxygen atoms in the acetate ion appear different in line-bond structures, experiments show that they are equivalent. Both carbon-oxygen bonds, for instance, are 127 pm in length, midway between the length of a typical $\\mathrm{C}-\\mathrm{O}$ single bond ( 135 pm ) and a typical $\\mathrm{C}=\\mathrm{O}$ double bond ( 120 pm ). In other words, neither of the two structures for acetate is correct by itself. The true structure is intermediate between the two, and an electrostatic potential map shows that both oxygen atoms share the negative charge and have equal electron densities (red).\n\n\nThe two individual line-bond structures for acetate ion are called resonance forms, and their special resonance relationship is indicated by the double-headed arrow between them. The only difference between the two resonance forms is the placement of the $\\pi$ and nonbonding valence electrons. The atoms themselves occupy exactly the same place in both resonance forms, the connections between atoms are the same, and the threedimensional shapes of the resonance forms are the same.\n\nA good way to think about resonance forms is to realize that a substance like the acetate ion is the same as any other. Acetate doesn't jump back and forth between two resonance forms, spending part of the time looking like one and part of the time looking like the other. Rather, acetate has a single unchanging structure that we say is a resonance hybrid of the two individual forms and has characteristics of both. The only \"problem\" with acetate is that we can't draw it accurately using a familiar line-bond structure-line-bond structures just don't work for resonance hybrids. The difficulty, however, is with the representation of acetate on paper, not with acetate itself."}
{"id": 87, "contents": "Dimethyl sulfoxide - 2.4 Resonance\nResonance is a very useful concept that we'll return to on numerous occasions throughout the rest of this book. We'll see in Chapter 15, for instance, that the six carbon-carbon bonds in aromatic compounds, such as benzene, are equivalent and that benzene is best represented as a hybrid of two resonance forms. Although each individual resonance form seems to imply that benzene has alternating single and double bonds, neither form is correct by itself. The true benzene structure is a hybrid of the two individual forms, and all six carbon-carbon bonds are equivalent. This symmetrical distribution of electrons around the molecule is evident in an\nelectrostatic potential map.\n\n\n\nBenzene (two resonance forms)"}
{"id": 88, "contents": "Dimethyl sulfoxide - 2.5 Rules for Resonance Forms\nWhen first dealing with resonance forms, it's useful to have a set of guidelines that describe how to draw and interpret them. The following rules should be helpful:"}
{"id": 89, "contents": "RULE 1 - \nIndividual resonance forms are imaginary, not real. The real structure is a composite, or resonance hybrid, of the different forms. Species such as the acetate ion and benzene are no different from any other. They have single, unchanging structures, and they don't switch back and forth between resonance forms. The only difference between these and other substances is in the way they are represented in drawings."}
{"id": 90, "contents": "RULE 2 - \nResonance forms differ only in the placement of their $\\boldsymbol{\\pi}$ or nonbonding electrons. Neither the position nor the hybridization of any atom changes from one resonance form to another. In the acetate ion, for instance, the carbon atom is $s p^{2}$-hybridized and the oxygen atoms remain in exactly the same place in both resonance forms. Only the positions of the $\\pi$ electrons in the $\\mathrm{C}=\\mathrm{O}$ bond and the lone-pair electrons on oxygen differ from one form to another. This movement of electrons from one resonance structure to another can be indicated with curved arrows. A curved arrow always indicates the movement of electrons, not the movement of atoms. An arrow shows that a pair of electrons moves from the atom or bond at the tail of the arrow to the atom or bond at the head of the arrow.\n\n\nThe situation with benzene is similar to that with acetate. The $\\pi$ electrons in the double bonds move, as shown with curved arrows, but the carbon and hydrogen atoms remain in place.\n\n\nRULE 3\nDifferent resonance forms of a substance don't have to be equivalent. As an example, we'll see in Chapter 22 that a compound such as acetone, which contains a $\\mathrm{C}=\\mathrm{O}$ bond, can be converted into its anion by reaction with a strong base. The resultant anion has two resonance forms. One form contains a carbon-oxygen double bond and has a negative charge on carbon; the other contains a carbon-carbon double bond and has a negative charge on oxygen. Even though the two resonance forms aren't equivalent, both contribute to the overall\nresonance hybrid.\n\n\nWhen two resonance forms are nonequivalent, the actual structure of the resonance hybrid resembles the more stable form. Thus, we might expect the true structure of the acetone anion to be more like that of the form that places the negative charge on the electronegative oxygen atom rather than on the carbon.\n\nRULE 4\nResonance forms obey normal rules of valency. A resonance form is like any other structure: the octet rule still applies to second-row, main-group atoms. For example, one of the following structures for the acetate ion is not a valid resonance form because the carbon atom has five bonds and ten valence electrons:"}
{"id": 91, "contents": "RULE 2 - \nRULE 5\nThe resonance hybrid is more stable than any individual resonance form. In other words, resonance leads to stability. Generally speaking, the larger the number of resonance forms a substance has, the more stable the substance is, because its electrons are spread out over a larger part of the molecule and are closer to more nuclei. We'll see in Chapter 15, for instance, that a benzene ring is more stable because of resonance than might otherwise be expected."}
{"id": 92, "contents": "RULE 2 - 2.6 Drawing Resonance Forms\nLook back at the resonance forms of the acetate ion and the acetone anion shown in the previous section. The pattern seen there is a common one that leads to a useful technique for drawing resonance forms. In general, any three-atom grouping with a $p$ orbital on each atom has two resonance forms:\n\n\nThe atoms $\\mathrm{X}, \\mathrm{Y}$, and Z in the general structure might be $\\mathrm{C}, \\mathrm{N}, \\mathrm{O}, \\mathrm{P}, \\mathrm{S}$, or others, and the asterisk (*) might mean that the $p$ orbital on atom $Z$ is vacant, that it contains a single electron, or that it contains a lone pair of electrons. The two resonance forms differ simply by an exchange in position of the multiple bond and the asterisk from one end of the three-atom grouping to the other.\n\nBy learning to recognize such three-atom groupings within larger structures, resonance forms can be systematically generated. Look, for instance, at the anion produced when $\\mathrm{H}^{+}$is removed from 2,4-pentanedione by reaction with a base. How many resonance structures does the resultant anion have?\n\n\nThe 2,4-pentanedione anion has a lone pair of electrons and a formal negative charge on the central carbon atom, next to a $\\mathrm{C}=\\mathrm{O}$ bond on the left. The $\\mathrm{O}=\\mathrm{C}-\\mathrm{C}:^{-}$grouping is a typical one for which two resonance structures can be drawn.\n\n\nJust as there is a $\\mathrm{C}=\\mathrm{O}$ bond to the left of the lone pair, there is a second $\\mathrm{C}=\\mathrm{O}$ bond to the right. Thus, we can draw a total of three resonance structures for the 2,4-pentanedione anion."}
{"id": 93, "contents": "Drawing Resonance Forms for an Anion - \nDraw three resonance structures for the carbonate ion, $\\mathrm{CO}_{3}{ }^{2-}$."}
{"id": 94, "contents": "Strategy - \nLook for three-atom groupings that contain a multiple bond next to an atom with a $p$ orbital. Then exchange the positions of the multiple bond and the electrons in the $p$ orbital. In the carbonate ion, each singly bonded oxygen atom with three lone pairs and a negative charge is adjacent to the $\\mathrm{C}=\\mathrm{O}$ double bond, giving the grouping \u00d6: $=C-$ \u00d6:-"}
{"id": 95, "contents": "Solution - \nExchanging the position of the double bond and an electron lone pair in each grouping generates three resonance structures."}
{"id": 96, "contents": "Drawing Resonance Forms for a Radical - \nDraw three resonance forms for the pentadienyl radical, where a radical is a substance that contains a single, unpaired electron in one of its orbitals, denoted by a dot $(\\cdot)$."}
{"id": 97, "contents": "Strategy - \nFind the three-atom groupings that contain a multiple bond next to an atom with a $p$ orbital."}
{"id": 98, "contents": "Solution - \nThe unpaired electron is on a carbon atom next to a $\\mathrm{C}=\\mathrm{C}$ bond, giving a typical three-atom grouping that has two resonance forms.\n\n\nIn the second resonance form, the unpaired electron is next to another double bond, giving another three-atom grouping and leading to another resonance form.\nThree-atom grouping\n\n\n\nThus, the three resonance forms for the pentadienyl radical are:\n\n\nPROBLEM Which of the following pairs of structures represent resonance forms, and which do not? Explain. 2-9\n(a)\n\n(b)\nand\n\n\nPROBLEM Draw the indicated number of resonance forms for each of the following substances:\n2-10 (a) The methyl phosphate anion, $\\mathrm{CH}_{3} \\mathrm{OPO}_{3}{ }^{2-}$ (3)\n(b) The nitrate anion, $\\mathrm{NO}_{3}{ }^{-}$(3)\n(c) The allyl cation, $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}-\\mathrm{CH}_{2}{ }^{+}$(2)\n(d) The benzoate anion (2)"}
{"id": 99, "contents": "Solution - 2.7 Acids and Bases: The Br\u00f8nsted-Lowry Definition\nPerhaps the most important of all concepts related to electronegativity and polarity is that of acidity and\nbasicity. We'll soon see, in fact, that the acid-base behavior of organic molecules explains much of their chemistry. You may recall from a course in general chemistry that two definitions of acidity are frequently used: the Br\u00f8nsted-Lowry definition and the Lewis definition. We'll look at the Br\u00f8nsted-Lowry definition in this and the following three sections and then discuss the Lewis definition in Section 2.11.\n\nA Br\u00f8nsted-Lowry acid is a substance that donates a hydrogen ion, $\\mathrm{H}^{+}$, and a Br\u00f8nsted-Lowry base is a substance that accepts a hydrogen ion. (The name proton is often used as a synonym for $\\mathrm{H}^{+}$because loss of the valence electron from a neutral hydrogen atom leaves only the hydrogen nucleus-a proton.) When gaseous hydrogen chloride dissolves in water, for example, a polar HCl molecule acts as an acid and donates a proton, while a water molecule acts as a base and accepts the proton, yielding chloride anion ( $\\mathrm{Cl}^{-}$) and hydronium cation $\\left(\\mathrm{H}_{3} \\mathrm{O}^{+}\\right)$. This and other acid-base reactions are reversible, so we'll write them with double, forward-andbackward arrows.\n\n\n\nAcid\n\nBase\n\n\nConjugate base\n\n\nConjugate acid\n\nChloride ion, the product that results when the acid HCl loses a proton, is called the conjugate base of the acid, and hydronium ion, the product that results when the base $\\mathrm{H}_{2} \\mathrm{O}$ gains a proton, is called the conjugate acid of the base. Other common mineral acids such as $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ and $\\mathrm{HNO}_{3}$ behave similarly, as do organic acids such as acetic acid, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$.\n\nIn a general sense,"}
{"id": 100, "contents": "Solution - 2.7 Acids and Bases: The Br\u00f8nsted-Lowry Definition\nIn a general sense,\n\n$$\n\\begin{array}{cc}\n\\mathrm{H}-\\mathrm{A} & +\\mathrm{B} \\\\\n\\text { Acid } & \\text { Base }\n\\end{array}\n$$\n\nFor example:\n\n\nNotice that water can act either as an acid or as a base, depending on the circumstances. In its reaction with HCl , water is a base that accepts a proton to give the hydronium ion, $\\mathrm{H}_{3} \\mathrm{O}^{+}$. In its reaction with ammonia $\\left(\\mathrm{NH}_{3}\\right)$, however, water is an acid that donates a proton to give ammonium ion $\\left(\\mathrm{NH}_{4}^{+}\\right)$and hydroxide ion, $\\mathrm{HO}^{-}$.\n\nPROBLEM Nitric acid $\\left(\\mathrm{HNO}_{3}\\right)$ reacts with ammonia $\\left(\\mathrm{NH}_{3}\\right)$ to yield ammonium nitrate. Write the reaction, and\n2-11 identify the acid, the base, the conjugate acid product, and the conjugate base product."}
{"id": 101, "contents": "Solution - 2.8 Acid and Base Strength\nDifferent acids differ in their ability to donate $\\mathrm{H}^{+}$. Stronger acids, such as HCl , react almost completely with water, whereas weaker acids, such as acetic acid $\\left(\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right)$, react only slightly. The exact strength of a given acid HA in water solution is described using the acidity constant ( $\\boldsymbol{K}_{\\mathbf{a}}$ ) for the acid-dissociation equilibrium. Recall from general chemistry that the concentration of solvent is ignored in the equilibrium expression and that brackets [ ] around a substance refer to the concentration of the enclosed species in moles per liter.\n\n$$\n\\begin{gathered}\n\\mathrm{HA}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathrm{~A}^{-}+\\mathrm{H}_{3} \\mathrm{O}^{+} \\\\\nK_{\\mathrm{a}}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]}\n\\end{gathered}\n$$\n\nStronger acids have their equilibria toward the right and thus have larger acidity constants, whereas weaker acids have their equilibria toward the left and have smaller acidity constants. The range of $K_{\\mathrm{a}}$ values for different acids is enormous, running from about $10^{15}$ for the strongest acids to about $10^{-60}$ for the weakest. Common inorganic acids such as $\\mathrm{H}_{2} \\mathrm{SO}_{4}, \\mathrm{HNO}_{3}$, and HCl have $K_{\\mathrm{a}}$ 's in the range of $10^{2}$ to $10^{9}$, while organic acids generally have $K_{\\mathrm{a}}$ 's in the range of $10^{-5}$ to $10^{-15}$. As you gain experience, you'll develop a rough feeling for which acids are \"strong\" and which are \"weak\" (always remembering that the terms are relative, not absolute)."}
{"id": 102, "contents": "Solution - 2.8 Acid and Base Strength\nAcid strengths are normally expressed using $\\mathrm{p} K_{\\mathrm{a}}$ values rather than $K_{\\mathrm{a}}$ values, where the $\\mathbf{p} \\boldsymbol{K}_{\\mathbf{a}}$ is the negative common logarithm of the $K_{\\mathrm{a}}$ :\n\n$$\n\\mathrm{p} K_{\\mathrm{a}}=-\\log K_{\\mathrm{a}}\n$$\n\nA stronger acid (larger $K_{\\mathrm{a}}$ ) has a smaller $\\mathrm{p} K_{\\mathrm{a}}$, and a weaker acid (smaller $K_{\\mathrm{a}}$ ) has a larger $\\mathrm{p} K_{\\mathrm{a}}$. TABLE 2.3 lists the $\\mathrm{p} K_{\\mathrm{a}}$ 's of some common acids in order of their strength, and a more comprehensive table is given in Appendix B .\n\nNotice that the $\\mathrm{p} K_{\\mathrm{a}}$ value shown in TABLE 2.3 for water is 15.74 , which results from the following calculation. Because water is both the acid and the solvent, the equilibrium expression is"}
{"id": 103, "contents": "Solution - 2.8 Acid and Base Strength\n$$\n\\begin{gathered}\n\\begin{array}{c}\n\\mathrm{H}_{2} \\mathrm{O}\n\\end{array}+\\underset{\\text { (acid) }}{\\mathrm{H}_{2} \\mathrm{O}} \\rightleftharpoons \\mathrm{OH}^{-}+\\mathrm{H}_{3} \\mathrm{O}^{+} \\\\\nK_{\\mathrm{a}}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]}=\\frac{\\begin{array}{c}\n{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]}\n\\end{array}}{\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]}=\\frac{\\left[1.0 \\times 10^{-7}\\right]\\left[1.0 \\times 10^{-7}\\right]}{[55.4]}=1.8 \\times 10^{-16} \\\\\n\\mathrm{p} K_{\\mathrm{a}}=15.74\n\\end{gathered}\n$$\n\nThe numerator in this expression is the so-called ion-product constant for water, $K_{\\mathrm{w}}=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]=1.00 \\times$ $10^{-14}$, and the denominator is the molar concentration of pure water, $\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]=55.4 \\mathrm{M}$ at $25^{\\circ} \\mathrm{C}$. The calculation is artificial in that the concentration of \"solvent\" water is ignored while the concentration of \"acid\" water is not, but it is nevertheless useful for making a comparison of water with other weak acids on a similar footing."}
{"id": 104, "contents": "Solution - 2.8 Acid and Base Strength\nNotice also in TABLE 2.3 that there is an inverse relationship between the acid strength of an acid and the base strength of its conjugate base. A strong acid has a weak conjugate base, and a weak acid has a strong conjugate base. To understand this inverse relationship, think about what is happening to the acidic hydrogen in an acid-base reaction. A strong acid is one that loses $\\mathrm{H}^{+}$easily, meaning that its conjugate base holds the $\\mathrm{H}^{+}$ weakly and is therefore a weak base. A weak acid is one that loses $\\mathrm{H}^{+}$with difficulty, meaning that its conjugate base holds the proton tightly and is therefore a strong base. The fact that HCl is a strong acid, for example, means that $\\mathrm{Cl}^{-}$does not hold $\\mathrm{H}^{+}$tightly and is thus a weak base. Water, on the other hand, is a weak acid, meaning that $\\mathrm{OH}^{-}$holds $\\mathrm{H}^{+}$tightly and is a strong base."}
{"id": 105, "contents": "Solution - 2.8 Acid and Base Strength\n|  | Acid | Name | $\\mathrm{p} K_{\\mathrm{a}}$ | Conjugate base | Name |  |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| Weaker acid | $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$ | Ethanol | 16.00 | $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{O}^{-}$ | Ethoxide ion | Stronger base |\n|  | $\\mathrm{H}_{2} \\mathrm{O}$ | Water | 15.74 | $\\mathrm{HO}^{-}$ | Hydroxide ion |  |\n|  | HCN | Hydrocyanic acid | 9.31 | $\\mathrm{CN}^{-}$ | Cyanide ion |  |\n|  | $\\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{-}$ | Dihydrogen phosphate ion | 7.21 | $\\mathrm{HPO}_{4}{ }^{2-}$ | Hydrogen phosphate ion |  |\n|  | $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$ | Acetic acid | 4.76 | $\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}$ | Acetate ion |  |\n|  | $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ | Phosphoric acid | 2.16 | $\\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{-}$ | Dihydrogen phosphate ion |  |\n|  | $\\mathrm{HNO}_{3}$ | Nitric acid | -1.3 | $\\mathrm{NO}_{3}{ }^{-}$ | Nitrate ion |  |\n| acid | HCl | Hydrochloric acid | -7.0 | $\\mathrm{Cl}^{-}$ | Chloride ion | base |\n\nPROBLEM The amino acid phenylalanine has $\\mathrm{p} K_{\\mathrm{a}}=1.83$, and tryptophan has $\\mathrm{p} K_{\\mathrm{a}}=2.83$. Which is the stronger $\\mathbf{2 - 1 2}$ acid?"}
{"id": 106, "contents": "Solution - 2.8 Acid and Base Strength\nPhenylalanine\n( $\\mathrm{p} K_{\\mathrm{a}}=1.83$ )\nTryptophan\n$\\left(\\mathrm{p} K_{\\mathrm{a}}=2.83\\right)$\nPROBLEM Amide ion, $\\mathrm{H}_{2} \\mathrm{~N}^{-}$, is a much stronger base than hydroxide ion, $\\mathrm{HO}^{-}$. Which is the stronger acid, $\\mathrm{NH}_{3}$\n2-13 or $\\mathrm{H}_{2} \\mathrm{O}$ ? Explain."}
{"id": 107, "contents": "Solution - 2.9 Predicting Acid-Base Reactions from $\\mathrm{p} K_{\\mathrm{a}}$ Values\nCompilations of $\\mathrm{p} K_{\\mathrm{a}}$ values like those in TABLE 2.3 and Appendix B are useful for predicting whether a given acid-base reaction will take place, because $\\mathrm{H}^{+}$will always go from the stronger acid to the stronger base. That is, an acid will donate a proton to the conjugate base of a weaker acid, and the conjugate base of a weaker acid will remove a proton from a stronger acid. Since water ( $\\mathrm{p} K_{\\mathrm{a}}=15.74$ ) is a weaker acid than acetic acid ( $\\mathrm{p} K_{\\mathrm{a}}=4.76$ ), for example, hydroxide ion holds a proton more tightly than acetate ion does. Hydroxide ion will therefore react to a large extent with acetic acid, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$, to yield acetate ion and $\\mathrm{H}_{2} \\mathrm{O}$.\n\n\n\nAnother way to predict acid-base reactivity is to remember that the product conjugate acid in an acid-base reaction must be weaker and less reactive than the starting acid and the product conjugate base must be weaker and less reactive than the starting base. In the reaction of acetic acid with hydroxide ion, for example, the product conjugate acid $\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)$ is weaker than the starting acid $\\left(\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right)$, and the product conjugate base $\\left(\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}\\right)$is weaker than the starting base $\\left(\\mathrm{OH}^{-}\\right)$."}
{"id": 108, "contents": "Predicting Acid Strengths from $\\mathbf{p} K_{\\mathbf{a}}$ Values - \nWater has $\\mathrm{p} K_{\\mathrm{a}}=15.74$, and acetylene has $\\mathrm{p} K_{\\mathrm{a}}=25$. Which is the stronger acid? Does hydroxide ion react to a significant extent with acetylene?\n\n\nAcetylene"}
{"id": 109, "contents": "Strategy - \nIn comparing two acids, the one with the lower $\\mathrm{p} K_{\\mathrm{a}}$ is stronger. Thus, water is a stronger acid than acetylene and gives up $\\mathrm{H}^{+}$more easily."}
{"id": 110, "contents": "Solution - \nBecause water is a stronger acid and gives up $\\mathrm{H}^{+}$more easily than acetylene, the $\\mathrm{HO}^{-}$ion must have less affinity for $\\mathrm{H}^{+}$than the $\\mathrm{HC} \\equiv \\mathrm{C}$ : $^{-}$ion. In other words, the anion of acetylene is a stronger base than hydroxide ion, and the reaction will not proceed significantly as written."}
{"id": 111, "contents": "Calculating $K_{\\mathrm{a}}$ from $\\mathrm{p} K_{\\mathrm{a}}$ - \nAccording to the data in TABLE 2.3, acetic acid has $\\mathrm{p} K_{\\mathrm{a}}=4.76$. What is its $K_{\\mathrm{a}}$ ?"}
{"id": 112, "contents": "Strategy - \nSince $\\mathrm{p} K_{\\mathrm{a}}$ is the negative logarithm of $K_{\\mathrm{a}}$, it's necessary to use a calculator with an ANTILOG or INV LOG function. Enter the value of the $\\mathrm{p} K_{\\mathrm{a}}(4.76)$, change the sign $(-4.76)$, and then find the antilog $\\left(1.74 \\times 10^{-5}\\right)$."}
{"id": 113, "contents": "Solution - \n$K_{\\mathrm{a}}=1.74 \\times 10^{-5}$.\n\nPROBLEM Will either of the following reactions take place to a significant extent as written, according to the 2-14 data in Table 2.3?\n\n\nPROBLEM Ammonia, $\\mathrm{NH}_{3}$, has $\\mathrm{p} K_{\\mathrm{a}} \\approx 36$, and acetone has $\\mathrm{p} K_{\\mathrm{a}} \\approx 19$. Will the following reaction take place to a 2-15 significant extent?\n\n\nAcetone\nPROBLEM What is the $K_{\\mathrm{a}}$ of HCN if its $\\mathrm{p} K_{\\mathrm{a}}=9.31$ ?\n2-16"}
{"id": 114, "contents": "Solution - 2.10 Organic Acids and Organic Bases\nMany of the reactions we'll be seeing in future chapters, including practically all biological reactions, involve organic acids and organic bases. Although it's too early to go into the details of these processes now, you might keep the following generalities in mind:"}
{"id": 115, "contents": "Organic Acids - \nOrganic acids are characterized by the presence of a positively polarized hydrogen atom (blue in electrostatic potential maps) and are of two main kinds: acids such as methanol and acetic acid that contain a hydrogen atom bonded to an electronegative oxygen atom $(\\mathrm{O}-\\mathrm{H})$ and those such as acetone (Section 2.5) that contain a hydrogen atom bonded to a carbon atom next to a $\\mathrm{C}=\\mathrm{O}$ bond $(\\mathrm{O}=\\mathrm{C}-\\mathrm{C}-\\mathrm{H})$.\n\n\n\nMethanol\n$\\left(\\mathrm{p} K_{a}=15.54\\right)$\n\n\nAcetic acid\n$\\left(\\mathrm{p} K_{\\mathrm{a}}=4.76\\right)$\n\n\nAcetone\n$\\left(p K_{a}=19.3\\right)$\n\nMethanol contains an $\\mathrm{O}-\\mathrm{H}$ bond and is a weak acid, while acetic acid also contains an $\\mathrm{O}-\\mathrm{H}$ bond and is a somewhat stronger acid. In both cases, acidity is due to the fact that the conjugate base resulting from loss of $\\mathrm{H}^{+}$ is stabilized by having its negative charge on a strongly electronegative oxygen atom. In addition, the conjugate base of acetic acid is stabilized by resonance (Section 2.4 and Section 2.5).\n\n\nAnion is stabilized both by having negative charge on a highly electronegative atom and by resonance.\n\nThe acidity of acetone and other compounds with $\\mathrm{C}=\\mathrm{O}$ bonds is due to the fact that the conjugate base resulting from loss of $\\mathrm{H}^{+}$is stabilized by resonance. In addition, one of the resonance forms stabilizes the negative charge by placing it on an electronegative oxygen atom.\n\n\nAnion is stabilized both by resonance and by having negative charge on a highly electronegative atom.\n\nElectrostatic potential maps of the conjugate bases from methanol, acetic acid, and acetone are shown in FIGURE 2.5. As you might expect, all three show a substantial amount of negative charge (red) on oxygen.\n(a)"}
{"id": 116, "contents": "Organic Acids - \n$\\mathrm{CH}_{3} \\mathrm{O}^{-}$\n(b)\n\n\n(c)\n\n\n\nFIGURE 2.5 Electrostatic potential maps of the conjugate bases of (a) methanol, (b) acetic acid, and (c) acetone. The electronegative oxygen atoms stabilize the negative charge in all three.\n\nCompounds called carboxylic acids, which contain the $-\\mathrm{CO}_{2} \\mathrm{H}$ grouping, occur abundantly in all living organisms and are involved in almost all metabolic pathways. Acetic acid, pyruvic acid, and citric acid are examples. You might note that at the typical pH of 7.3 found within cells, carboxylic acids are usually dissociated and exist as their carboxylate anions, $-\\mathrm{CO}_{2}{ }^{-}$.\n\n\nAcetic acid\n\n\nPyruvic acid\n\n\nCitric acid"}
{"id": 117, "contents": "Organic Bases - \nOrganic bases are characterized by the presence of an atom (reddish in electrostatic potential maps) with a lone pair of electrons that can bond to $\\mathrm{H}^{+}$. Nitrogen-containing compounds such as methylamine are the most common organic bases and are involved in almost all metabolic pathways, but oxygen-containing compounds can also act as bases when reacting with a sufficiently strong acid. Note that some oxygen-containing compounds can act both as acids and as bases depending on the circumstances, just as water can. Methanol and acetone, for instance, act as acids when they donate a proton but as bases when their oxygen atom accepts a proton.\n\n\n\nMethylamine\n\n\n\nMethanol\n\n\n\nAcetone\n\nWe'll see in Chapter 26 that substances called amino acids, so-named because they are both amines $\\left(-\\mathrm{NH}_{2}\\right)$ and carboxylic acids $\\left(-\\mathrm{CO}_{2} \\mathrm{H}\\right)$, are the building blocks from which the proteins in all living organisms are made. Twenty different amino acids go into making up proteins-alanine is an example. Interestingly, alanine and other amino acids exist primarily in a doubly charged form called a zwitterion rather than in the uncharged form. The zwitterion form arises because amino acids have both acidic and basic sites within the same molecule and therefore undergo an internal acid-base reaction.\n\n\nAlanine\n(uncharged form)\n\n\nAlanine\n(zwitterion form)"}
{"id": 118, "contents": "Organic Bases - 2.11 Acids and Bases: The Lewis Definition\nThe Lewis definition of acids and bases is more encompassing than the Br\u00f8nsted-Lowry definition because it's not limited to substances that donate or accept just protons. A Lewis acid is a substance that accepts an electron pair, and a Lewis base is a substance that donates an electron pair. The donated electron pair is shared between the acid and the base in a covalent bond."}
{"id": 119, "contents": "Lewis Acids and the Curved Arrow Formalism - \nThe fact that a Lewis acid is able to accept an electron pair means that it must have either a vacant, low-energy orbital or a polar bond to hydrogen so that it can donate $\\mathrm{H}^{+}$(which has an empty $1 s$ orbital). Thus, the Lewis definition of acidity includes many species in addition to $\\mathrm{H}^{+}$. For example, various metal cations, such as $\\mathrm{Mg}^{2+}$, are Lewis acids because they accept a pair of electrons when they form a bond to a base. We'll also see in later chapters that certain metabolic reactions begin with an acid-base reaction between $\\mathrm{Mg}^{2+}$ as a Lewis acid and an organic diphosphate or triphosphate ion as the Lewis base.\n\n\nLewis acid\nLewis base\nAcid-base complex (an organodiphosphate ion)\nIn the same way, compounds of group 3A elements, such as $\\mathrm{BF}_{3}$ and $\\mathrm{AlCl}_{3}$, are Lewis acids because they have\nunfilled valence orbitals and can accept electron pairs from Lewis bases, as shown in FIGURE 2.6. Similarly, many transition-metal compounds, such as $\\mathrm{TiCl}_{4}, \\mathrm{FeCl}_{3}, \\mathrm{ZnCl}_{2}$, and $\\mathrm{SnCl}_{4}$, are Lewis acids.\n\n\n\n| Boron <br> trifluoride <br> (Lewis acid) | Dimethyl <br> ether <br> (Lewis base) |\n| :---: | :---: |\n\nAcid-base complex\n\nFIGURE 2.6 The reaction of boron trifluoride, a Lewis acid, with dimethyl ether, a Lewis base. The Lewis acid accepts a pair of electrons, and the Lewis base donates a pair of nonbonding electrons. Note how the movement of electrons from the Lewis base to the Lewis acid is indicated by a curved arrow. Note also how, in electrostatic potential maps, the boron becomes more negative (red) after reaction because it has gained electrons and the oxygen atom becomes more positive (blue) because it has donated electrons."}
{"id": 120, "contents": "Lewis Acids and the Curved Arrow Formalism - \nLook closely at the acid-base reaction in FIGURE 2.6, and notice how it's shown. Dimethyl ether, the Lewis base, donates an electron pair to a vacant valence orbital of the boron atom in $\\mathrm{BF}_{3}$, a Lewis acid. The direction of electron-pair flow from base to acid is shown using a curved arrow, just as the direction of electron flow from one resonance structure to another was shown using curved arrows in Section 2.5. We'll use this curved-arrow notation throughout the remainder of this text to indicate electron flow during reactions, so get used to seeing it.\n\nSome further examples of Lewis acids follow:"}
{"id": 121, "contents": "Lewis Bases - \nThe Lewis definition of a base-a compound with a pair of nonbonding electrons that it can use to bond to a Lewis acid-is similar to the Br\u00f8nsted-Lowry definition. Thus, $\\mathrm{H}_{2} \\mathrm{O}$, with its two pairs of nonbonding electrons\non oxygen, acts as a Lewis base by donating an electron pair to an $\\mathrm{H}^{+}$in forming the hydronium ion, $\\mathrm{H}_{3} \\mathrm{O}^{+}$.\n\n\nIn a more general sense, most oxygen- and nitrogen-containing organic compounds can act as Lewis bases because they have lone pairs of electrons. A divalent oxygen compound has two lone pairs of electrons, and a trivalent nitrogen compound has one lone pair. Note in the following examples that some compounds can act as both acids and bases, just as water can. Alcohols and carboxylic acids, for instance, act as acids when they donate an $\\mathrm{H}^{+}$but as bases when their oxygen atom accepts an $\\mathrm{H}^{+}$.\n\n\nNote also that some Lewis bases, such as carboxylic acids, esters, and amides, have more than one atom with a lone pair of electrons and can therefore react at more than one site. Acetic acid, for example, can be protonated either on the doubly bonded oxygen atom or on the singly bonded oxygen atom. Reaction normally occurs only once in such instances, and the more stable of the two possible protonation products is formed. For acetic acid, protonation by reaction with sulfuric acid occurs on the doubly bonded oxygen because that product is stabilized by two resonance forms.\n\n\nAcetic acid (base)"}
{"id": 122, "contents": "Using Curved Arrows to Show Electron Flow - \nUsing curved arrows, show how acetaldehyde, $\\mathrm{CH}_{3} \\mathrm{CHO}$, can act as a Lewis base."}
{"id": 123, "contents": "Strategy - \nA Lewis base donates an electron pair to a Lewis acid. We therefore need to locate the electron lone pairs on acetaldehyde and use a curved arrow to show the movement of a pair toward the H atom of the acid."}
{"id": 124, "contents": "Solution - \nAcetaldehyde\n\nPROBLEM Using curved arrows, show how the species in part (a) can act as Lewis bases in their reactions with 2-17 HCl , and show how the species in part (b) can act as Lewis acids in their reaction with $\\mathrm{OH}^{-}$.\n(a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}, \\mathrm{HN}\\left(\\mathrm{CH}_{3}\\right)_{2}, \\mathrm{P}\\left(\\mathrm{CH}_{3}\\right)_{3}$\n(b) $\\mathrm{H}_{3} \\mathrm{C}^{+}, \\mathrm{B}\\left(\\mathrm{CH}_{3}\\right)_{3}, \\mathrm{MgBr}_{2}$\n\nPROBLEM Imidazole, which forms part of amino acid histidine, can act as both an acid and a base.\n2-18\n\n\nImidazole\n\n\nHistidine\n(a) Look at the electrostatic potential map of imidazole, and identify the most acidic hydrogen atom and the most basic nitrogen atom.\n(b) Draw structures for the resonance forms of the products that result when imidazole is protonated by an acid and deprotonated by a base."}
{"id": 125, "contents": "Solution - 2.12 Noncovalent Interactions between Molecules\nWhen thinking about chemical reactivity, chemists usually focus their attention on bonds, the covalent interactions between atoms within molecules. Also important, however, particularly in large biomolecules like proteins and nucleic acids, are a variety of interactions between molecules that strongly affect molecular properties. Collectively called either intermolecular forces, van der Waals forces, or noncovalent interactions, they are of several different types: dipole-dipole forces, dispersion forces, and hydrogen bonds.\n\nDipole-dipole forces occur between polar molecules as a result of electrostatic interactions among dipoles. The forces can be either attractive or repulsive depending on the orientation of the molecules-attractive when unlike charges are together and repulsive when like charges are together. The attractive geometry is lower in energy and therefore predominates (FIGURE 2.7).\n\n\nFIGURE 2.7 Dipole-dipole forces cause polar molecules (a) to attract one another when they orient with unlike charges together, but (b) to repel one another when they orient with like charges together.\n\nDispersion forces occur between all neighboring molecules and arise because the electron distribution within molecules is constantly changing. Although uniform on a time-averaged basis, the electron distribution even in nonpolar molecules is likely to be nonuniform at any given instant. One side of a molecule may, by chance, have a slight excess of electrons relative to the opposite side, giving the molecule a temporary dipole. This temporary dipole in one molecule causes a nearby molecule to adopt a temporarily opposite dipole, resulting in a tiny attraction between the two (FIGURE 2.8). Temporary molecular dipoles have only a fleeting existence and are constantly changing, but their cumulative effect is often strong enough to hold molecules close together so that a substance is a liquid or solid rather than a gas.\n\n\nFIGURE 2.8 Attractive dispersion forces in nonpolar molecules are caused by temporary dipoles, as shown in these models of pentane, $\\mathrm{C}_{5} \\mathrm{H}_{12}$."}
{"id": 126, "contents": "Solution - 2.12 Noncovalent Interactions between Molecules\nFIGURE 2.8 Attractive dispersion forces in nonpolar molecules are caused by temporary dipoles, as shown in these models of pentane, $\\mathrm{C}_{5} \\mathrm{H}_{12}$.\n\nPerhaps the most important noncovalent interaction in biological molecules is the hydrogen bond, an attractive interaction between a hydrogen atom bonded to an electronegative O or N atom and an unshared electron pair on another O or N atom. In essence, a hydrogen bond is a very strong dipole-dipole interaction involving polarized $\\mathrm{O}-\\mathrm{H}$ or $\\mathrm{N}-\\mathrm{H}$ bonds. Electrostatic potential maps of water and ammonia clearly show the positively polarized hydrogens (blue) and the negatively polarized oxygens and nitrogens (red).\n\n\n\n\nHydrogen bonding has enormous consequences for living organisms. Hydrogen bonds cause water to be a liquid rather than a gas at ordinary temperatures, they hold enzymes in the shapes necessary for catalyzing biological reactions, and they cause strands of deoxyribonucleic acid (DNA) to pair up and coil into the double helix that stores genetic information."}
{"id": 127, "contents": "A deoxyribonucleic acid segment - \nOne further point before leaving the subject of noncovalent interactions: biochemists frequently use the term hydrophilic, meaning \"water-loving,\" to describe a substance that is attracted to water and the term hydrophobic, meaning \"water-fearing,\" to describe a substance that is not strongly attracted to water.\n\nHydrophilic substances, such as table sugar, often have a number of -OH groups in their structure so they can form hydrogen bonds and dissolve in water, whereas hydrophobic substances, such as vegetable oil, do not have groups that form hydrogen bonds and do not dissolve in water.\n\n\nPROBLEM Of the two vitamins A and C, one is hydrophilic and water-soluble while the other is hydrophobic 2-19 and fat-soluble. Which is which?\n\n\nVitamin A\n(retinol)\n\n\nVitamin C (ascorbic acid)"}
{"id": 128, "contents": "Alkaloids: From Cocaine to Dental Anesthetics - \nJust as ammonia $\\left(\\mathrm{NH}_{3}\\right)$ is a weak base, there are a large number of nitrogen-containing organic compounds called amines that are also weak bases. In the early days of organic chemistry, basic amines derived from natural sources were known as vegetable alkali, but they are now called alkaloids. More than 20,000 alkaloids are known. Their study provided much of the impetus for the growth of organic chemistry in the nineteenth century and remains today an active and fascinating area of research.\n\n\nFIGURE 2.9 The coca bush Erythroxylon coca, native to upland rain forest areas of Colombia, Ecuador, Peru, Bolivia, and western Brazil, is the source of the alkaloid cocaine. (credit: \"Erythroxylum coca\" by Danna Guevara/Wikimedia Commons, CC BY 4.0)\n\nMany alkaloids have pronounced biological properties, and approximately $50 \\%$ of pharmaceutical agents used today are derived from naturally occurring amines. As just three examples, morphine, an analgesic agent (painkiller), is obtained from the opium poppy Papaver somniferum. Ephedrine, a bronchodilator, decongestant, and appetite suppressant, is obtained from Ephedra sinica, an evergreen shrub native to Mongolia and northeastern China. Cocaine, both an anesthetic and a stimulant, is obtained from the coca bush Erythroxylon coca, endemic to the upland rain forest areas of central South America. (And yes, there really was a small amount of cocaine in the original Coca-Cola recipe, although it was removed in 1906.)\n\n\nMorphine\n\n\nEphedrine\n\n\nCocaine"}
{"id": 129, "contents": "Alkaloids: From Cocaine to Dental Anesthetics - \nMorphine\n\n\nEphedrine\n\n\nCocaine\n\nCocaine itself is rarely used medically because it is too addictive, but its anesthetic properties provoked a long search for related but nonaddictive compounds. This search ultimately resulted in the synthesis of the \"caine\" anesthetics that are commonly used today in dental and surgical anesthesia. Procaine, the first such compound, was synthesized in 1898 and marketed under the name Novocain. It was rapidly adopted and remains in use today as a topical anesthetic. Other related compounds with different activity profiles followed: Lidocaine, marketed as Xylocaine, was introduced in 1943, and mepivacaine (Carbocaine) in the early 1960s. More recently, bupivacaine (Marcaine) and prilocaine (Citanest) have gained popularity. Both are quick-acting, but the effects of bupivacaine last for 3 to 6 hours while those of prilocaine fade after 45 minutes. Notice some structural similarity of all the caines to cocaine itself.\n\n\nProcaine\n(Novocain)\n\n\nMepivacaine (Carbocaine)\n\n\nBupivacaine (Marcaine)\n\n\nLidocaine\n(Xylocaine)\n\n\nPrilocaine\n(Citanest)\nAn estimate from the U.S. National Academy of Sciences is that less than $1 \\%$ of all living species have been characterized. Thus, alkaloid chemistry remains an active area of research, and innumerable substances with potentially useful properties have yet to be discovered. Undoubtedly even the caine anesthetics will become obsolete at some point, perhaps supplanted by newly discovered alkaloids."}
{"id": 130, "contents": "Key Terms - \n- acidity constant $\\left(K_{\\mathrm{a}}\\right)$\n- alkaloid\n- Br\u00f8nsted-Lowry acid\n- Br\u00f8nsted-Lowry base\n- conjugate acid\n- conjugate base\n- dipole moment ( $\\mu$ )\n- dispersion force\n- electronegativity (EN)\n- electrostatic potential map\n- formal charge\n- hydrogen bond\n- hydrophilic\n- hydrophobic\n- inductive effect\n- intermolecular force\n- Lewis acid\n- Lewis base\n- noncovalent interaction\n- $\\mathrm{p} K_{\\mathrm{a}}$\n- polar covalent bond\n- resonance form\n- resonance hybrid\n- van der Waals force"}
{"id": 131, "contents": "Summary - \nUnderstanding organic chemistry means knowing not just what happens but also why and how it happens at the molecular level. In this chapter, we've reviewed some of the ways that chemists describe and account for chemical reactivity, thereby providing a foundation for understanding the specific reactions that will be discussed in subsequent chapters.\n\nOrganic molecules often have polar covalent bonds as a result of unsymmetrical electron sharing caused by differences in the electronegativity of atoms. A carbon-oxygen bond is polar, for example, because oxygen attracts the shared electrons more strongly than carbon does. Carbon-hydrogen bonds are relatively nonpolar. Many molecules as a whole are also polar, owing to the presence of individual polar bonds and electron lone pairs. The polarity of a molecule is measured by its dipole moment, $\\boldsymbol{\\mu}$.\n\nPlus (+) and minus (-) signs are often used to indicate the presence of formal charges on atoms in molecules. Assigning formal charges to specific atoms is a bookkeeping technique that makes it possible to keep track of the valence electrons around an atom and offers some clues about chemical reactivity."}
{"id": 132, "contents": "Summary - \nSome substances, such as acetate ion and benzene, can't be represented by a single line-bond structure and must be considered as a resonance hybrid of two or more structures, none of which would be correct by themselves. The only difference between two resonance forms is in the location of their $\\pi$ and nonbonding electrons. The nuclei remain in the same places in both structures, and the hybridization of the atoms remains\nthe same.\nAcidity and basicity are closely related to the ideas of polarity and electronegativity. A Br\u00f8nsted-Lowry acid is a compound that can donate a proton (hydrogen ion, $\\mathrm{H}^{+}$), and a Br\u00f8nsted-Lowry base is a compound that can accept a proton. The strength of a Br\u00f8nsted-Lowry acid or base is expressed by its acidity constant, $\\boldsymbol{K}_{\\mathbf{a}}$, or by the negative logarithm of the acidity constant, $\\mathbf{p} \\boldsymbol{K}_{\\mathbf{a}}$. The larger the $\\mathrm{p} K_{\\mathrm{a}}$, the weaker the acid. More useful is the Lewis definition of acids and bases. A Lewis acid is a compound that has a low-energy empty orbital that can accept an electron pair; $\\mathrm{Mg}^{2+}, \\mathrm{BF}_{3}, \\mathrm{AlCl}_{3}$, and $\\mathrm{H}^{+}$are examples. A Lewis base is a compound that can donate an unshared electron pair; $\\mathrm{NH}_{3}$ and $\\mathrm{H}_{2} \\mathrm{O}$ are examples. Most organic molecules that contain oxygen and nitrogen can act as Lewis bases toward sufficiently strong acids.\n\nA variety of noncovalent interactions have a significant effect on the properties of large biomolecules. Hydrogen bonding-the attractive interaction between a positively polarized hydrogen atom bonded to an oxygen or nitrogen atom with an unshared electron pair on another O or N atom, is particularly important in giving proteins and nucleic acids their shapes."}
{"id": 133, "contents": "Visualizing Chemistry - \nPROBLEM Fill in the multiple bonds in the following model of naphthalene, $\\mathrm{C}_{10} \\mathrm{H}_{8}$ (black $=\\mathrm{C}$, gray $=\\mathrm{H}$ ). How 2-20 many resonance structures does naphthalene have? Draw them.\n\n\nPROBLEM The following model is a representation of ibuprofen, a common over-the-counter pain reliever.\n2-21 Indicate the positions of the multiple bonds, and draw a skeletal structure (black $=\\mathrm{C}$, red = O, gray $=\\mathrm{H})$.\n\n\nPROBLEM cis-1,2-Dichloroethylene and trans-1,2-dichloroethylene are isomers, compounds with the same\n$\\mathbf{2 - 2 2}$ formula but different chemical structures. Look at the following electrostatic potential maps, and tell whether either compound has a dipole moment.\n\ncis-1,2-Dichloroethylene\n\ntrans-1,2-Dichloroethylene\n\nPROBLEM The following molecular models are representations of (a) adenine and (b) cytosine, constituents of\n2-23 DNA (deoxyribonucleic acid). Indicate the positions of multiple bonds and lone pairs for both, and draw skeletal structures (black = C, red = O, blue = N, gray = H).\n(a)\n\n\nAdenine\n(b)\n\n\nCytosine"}
{"id": 134, "contents": "Mechanism Problems - \nPROBLEM Predict the product(s) of the following acid/base reactions. Draw curved arrows to show the 2-24 formation and breaking of bonds.\n(a)\n\n(b)\n\n\n(c)\n\n\nPROBLEM Use curved arrows to draw the protonated form of the following Lewis bases.\n2-25 (a)\n\n\n(b)\n\n\n(c)\n\n(d)\n\n\nPROBLEM Use the curved-arrow formalism to show how the electrons flow in the resonance form on the left to 2-26 give the one on the right.\n\n(a)\n\n(b)\n\nc)\n\n\nPROBLEM Double bonds can also act like Lewis bases, sharing their electrons with Lewis acids. Use curved\n2-27 arrows to show how each of the following double bonds will react with HCl and draw the resulting\ncarbocation.\n(a) $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}_{2}$\n(b)\n\n(c)"}
{"id": 135, "contents": "Electronegativity and Dipole Moments - \nPROBLEM Identify the most electronegative element in each of the following molecules:\n2-28 (a) $\\mathrm{CH}_{2} \\mathrm{FCl}$\n(b) $\\mathrm{FCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{Br}$\n(c) $\\mathrm{HOCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}$\n(d) $\\mathrm{CH}_{3} \\mathrm{OCH}_{2} \\mathrm{Li}$\n\nPROBLEM Use the electronegativity table given in Figure 2.3 to predict which bond in each of the following\n2-29 pairs is more polar, and indicate the direction of bond polarity for each compound.\n(a) $\\mathrm{H}_{3} \\mathrm{C}-\\mathrm{Cl}$ or $\\mathrm{Cl}-\\mathrm{Cl}$\n(b) $\\mathrm{H}_{3} \\mathrm{C}-\\mathrm{H}$ or $\\mathrm{H}-\\mathrm{Cl}$\n(c) $\\mathrm{HO}-\\mathrm{CH}_{3}$ or $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{Si}-\\mathrm{CH}_{3}$\n(d) $\\mathrm{H}_{3} \\mathrm{C}-\\mathrm{Li}$ or $\\mathrm{Li}-\\mathrm{OH}$\n\nPROBLEM Which of the following molecules has a dipole moment? Indicate the expected direction of each.\n2-30 (a)\n\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM (a) The $\\mathrm{H}-\\mathrm{Cl}$ bond length is 136 pm . What would the dipole moment of HCl be if the molecule\n2-31 were $100 \\%$ ionic, $\\mathrm{H}^{+} \\mathrm{Cl}^{-}$?\n(b) The actual dipole moment of HCl is 1.08 D . What is the percent ionic character of the $\\mathrm{H}-\\mathrm{Cl}$ bond?"}
{"id": 136, "contents": "Electronegativity and Dipole Moments - \nPROBLEM Phosgene, $\\mathrm{Cl}_{2} \\mathrm{C}=\\mathrm{O}$, has a smaller dipole moment than formaldehyde, $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{O}$, even though it\n2-32 contains electronegative chlorine atoms in place of hydrogen. Explain.\nPROBLEM Fluoromethane $\\left(\\mathrm{CH}_{3} \\mathrm{~F}, \\mu=1.81 \\mathrm{D}\\right)$ has a smaller dipole moment than chloromethane $\\left(\\mathrm{CH}_{3} \\mathrm{Cl}, \\mu=\\right.$\n2-33 1.87 D ) even though fluorine is more electronegative than chlorine. Explain.\nPROBLEM Methanethiol, $\\mathrm{CH}_{3} \\mathrm{SH}$, has a substantial dipole moment ( $\\mu=1.52$ ) even though carbon and sulfur\n2-34 have identical electronegativities. Explain."}
{"id": 137, "contents": "Formal Charges - \nPROBLEM Calculate the formal charges on the atoms shown in red.\n2-35 (a) ${ }_{\\left(\\mathrm{CH}_{3}\\right)_{2}} \\mathrm{OO}_{\\mathrm{BF}}^{3}$\n(b) $\\mathrm{H}_{2} \\ddot{\\mathrm{C}}-\\mathrm{N} \\equiv \\mathrm{N}$ :\n(c) $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{N}=\\ddot{\\mathrm{N}}$ :\n(d) $: \\ddot{0}=\\ddot{0}-\\ddot{0}:$\n\n(f)\n\n\nPROBLEM Assign formal charges to the atoms in each of the following molecules:\n2-36 (a)\n\n(b) $\\mathrm{H}_{3} \\mathrm{C}-\\ddot{\\mathrm{N}}-\\mathrm{N} \\equiv \\mathrm{N}$ :\n(c) $\\mathrm{H}_{3} \\mathrm{C}-\\ddot{\\mathrm{N}}=\\mathrm{N}=\\ddot{\\mathrm{N}}$ :"}
{"id": 138, "contents": "Resonance - \nPROBLEM Which of the following pairs of structures represent resonance forms?\n2-37\n(a)\n\nand\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM Draw as many resonance structures as you can for the following species:\n2-38 (a)\n\n\n(b)\n\n(c)\n\n(e)\n\n\nPROBLEM 1,3-Cyclobutadiene is a rectangular molecule with two shorter double bonds and two longer single 2-39 bonds. Why do the following structures not represent resonance forms?\n\n\nAcids and Bases\nPROBLEM Alcohols can act either as weak acids or as weak bases, just as water can. Show the reaction of\n2-40 methanol, $\\mathrm{CH}_{3} \\mathrm{OH}$, with a strong acid such as HCl and with a strong base such as $\\mathrm{Na}^{+}{ }^{-} \\mathrm{NH}_{2}$\nPROBLEM The $\\mathrm{O}-\\mathrm{H}$ hydrogen in acetic acid is more acidic than any of the $\\mathrm{C}-\\mathrm{H}$ hydrogens. Explain this result\n2-41 using resonance structures.\n\n\nAcetic acid\n\nPROBLEM Draw electron-dot structures for the following molecules, indicating any unshared electron pairs.\n2-42 Which of the compounds are likely to act as Lewis acids and which as Lewis bases?\n(a) $\\mathrm{AlBr}_{3}$\n(b) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{NH}_{2}$\n(c) $\\mathrm{BH}_{3}$\n(d) HF\n(e) $\\mathrm{CH}_{3} \\mathrm{SCH}_{3}$\n(f) $\\mathrm{TiCl}_{4}$\n\nPROBLEM Write the products of the following acid-base reactions:\n2-43 (a) $\\mathrm{CH}_{3} \\mathrm{OH}+\\mathrm{H}_{2} \\mathrm{SO}_{4} \\rightleftarrows$ ? (b) $\\mathrm{CH}_{3} \\mathrm{OH}+\\mathrm{NaNH}_{2} \\rightleftarrows$ ?\n(c) $\\mathrm{CH}_{3} \\mathrm{NH}_{3}^{+} \\mathrm{Cl}^{-}+\\mathrm{NaOH} \\rightleftarrows$ ?\n\nPROBLEM Rank the following substances in order of increasing acidity:\n2-44"}
{"id": 139, "contents": "Resonance - \nPROBLEM Rank the following substances in order of increasing acidity:\n2-44\n\n\nAcetone ( $\\mathrm{p} K_{\\mathrm{a}}=19.3$ )\n\n\n2,4-Pentanedione\n$\\left(\\mathrm{p} K_{\\mathrm{a}}=9\\right)$\n\n\nPhenol\n$\\left(p K_{a}=9.9\\right)$\n\n\nAcetic acid\n$\\left(p K_{a}=4.76\\right)$\n\nPROBLEM Which, if any, of the substances in Problem 2-44 is a strong enough acid to react almost completely\n2-45 with NaOH ? (The $\\mathrm{p} K_{\\mathrm{a}}$ of $\\mathrm{H}_{2} \\mathrm{O}$ is 15.74.)\nPROBLEM The ammonium ion $\\left(\\mathrm{NH}_{4}{ }^{+}, \\mathrm{p} K_{\\mathrm{a}}=9.25\\right)$ has a lower $\\mathrm{p} K_{\\mathrm{a}}$ than the methylammonium ion $\\left(\\mathrm{CH}_{3} \\mathrm{NH}_{3}{ }^{+}\\right.$,\n2-46 $\\mathrm{p} K_{\\mathrm{a}}=10.66$ ). Which is the stronger base, ammonia $\\left(\\mathrm{NH}_{3}\\right)$ or methylamine $\\left(\\mathrm{CH}_{3} \\mathrm{NH}_{2}\\right)$ ? Explain.\nPROBLEM Is tert-butoxide anion a strong enough base to react significantly with water? In other words,\n2-47 can a solution of potassium tert-butoxide be prepared in water? The $\\mathrm{p} K_{\\mathrm{a}}$ of tert-butyl alcohol is approximately 18."}
{"id": 140, "contents": "Potassium tert-butoxide - \nPROBLEM Predict the structure of the product formed in the reaction of the organic base pyridine with the\n2-48 organic acid acetic acid, and use curved arrows to indicate the direction of electron flow.\n\n\nPyridine Acetic acid\nPROBLEM Calculate $K_{\\mathrm{a}}$ values from the following $\\mathrm{p} K_{\\mathrm{a}}$ 's:\n2-49 (a) Acetone, $\\mathrm{p} K_{\\mathrm{a}}=19.3$ (b) Formic acid, $\\mathrm{p} K_{\\mathrm{a}}=3.75$\nPROBLEM Calculate $\\mathrm{p} K_{\\mathrm{a}}$ values from the following $K_{\\mathrm{a}}$ 's:\n2-50 (a) Nitromethane, $K_{\\mathrm{a}}=5.0 \\times 10^{-11}$\n(b) Acrylic acid, $K_{\\mathrm{a}}=5.6 \\times 10^{-5}$\n\nPROBLEM What is the pH of a 0.050 M solution of formic acid, $\\mathrm{p} K_{\\mathrm{a}}=3.75$ ?\n2-51\nPROBLEM Sodium bicarbonate, $\\mathrm{NaHCO}_{3}$, is the sodium salt of carbonic acid $\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right), \\mathrm{p} K_{\\mathrm{a}}=6.37$. Which of the\n2-52 substances shown in Problem 2-44 will react significantly with sodium bicarbonate?"}
{"id": 141, "contents": "General Problems - \nPROBLEM Maleic acid has a dipole moment, but the closely related fumaric acid, a substance involved in the\n2-53 citric acid cycle by which food molecules are metabolized, does not. Explain.\n\n\nMaleic acid\n\n\nFumaric acid\n\nPROBLEM Assume that you have two unlabeled bottles, one of which contains phenol ( $\\mathrm{p} K_{\\mathrm{a}}=9.9$ ) and one of\n2-54 which contains acetic acid ( $\\mathrm{p} K_{\\mathrm{a}}=4.76$ ). In light of your answer to Problem 2-52, suggest a simple way to determine what is in each bottle.\n\nPROBLEM Identify the acids and bases in the following reactions:\n2-55 (a)\n\n\n(c)\n\n(d)\n\n\nPROBLEM Which of the following pairs represent resonance structures?\n2-56\n(a)\n\n(b)\n\nand\n\n(c)\n\nand\n\n(d)\n\n\n\nPROBLEM Draw as many resonance structures as you can for the following species, adding appropriate formal\n2-57 charges to each:\n(a) Nitromethane,\n\n(b)\nOzone,\n\n(c) Diazomethane, $\\mathrm{H}_{2} \\mathrm{C}=\\stackrel{+}{\\mathrm{N}}=\\stackrel{\\overline{\\mathrm{N}}}{ }$ :\n\nPROBLEM Carbocations, which contain a trivalent, positively charged carbon atom, react with water to give\n2-58 alcohols:\n\n\nHow can you account for the fact that the following carbocation gives a mixture of two alcohols on reaction with water?\n\n\nPROBLEM We'll see in the next chapter that organic molecules can be classified according to the functional\n2-59 groups they contain, where a functional group is a collection of atoms with a characteristic chemical reactivity. Use the electronegativity values given in Figure 2.3 to predict the direction of polarization of the following functional groups.\n\n\n\nKetone\n(b)\n\n\nAlcohol\n(c)\n\n\nAmide\n(d) $-\\mathrm{C} \\equiv \\mathrm{N}$\n\nNitrile"}
{"id": 142, "contents": "General Problems - \nKetone\n(b)\n\n\nAlcohol\n(c)\n\n\nAmide\n(d) $-\\mathrm{C} \\equiv \\mathrm{N}$\n\nNitrile\n\nPROBLEM The azide functional group, which occurs in azidobenzene, contains three adjacent nitrogen atoms.\n2-60 One resonance structure for azidobenzene is shown. Draw three additional resonance structures, and assign appropriate formal charges to the atoms in all four.\n\n\nAzidobenzene\n\nPROBLEM Phenol, $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{OH}$, is a stronger acid than methanol, $\\mathrm{CH}_{3} \\mathrm{OH}$, even though both contain an $\\mathrm{O}-\\mathrm{H}$ bond.\n2-61 Draw the structures of the anions resulting from loss of $\\mathrm{H}^{+}$from phenol and methanol, and use resonance structures to explain the difference in acidity.\n\n\nPhenol $\\left(\\mathrm{p} K_{\\mathrm{a}}=9.89\\right) \\quad$ Methanol $\\left(\\mathrm{p} K_{\\mathrm{a}}=15.54\\right)$\nPROBLEM Thiamin diphosphate (TPP), a derivative of vitamin $B_{1}$ required for glucose metabolism, is a weak\n2-62 acid that can be deprotonated by a base. Assign formal charges to the appropriate atoms in both TPP and its deprotonation product.\n\n\nPROBLEM Which of the following compounds or ions have a dipole moment?\n2-63\n(a) Carbonate ion $\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$\n(b)\n\n(c) $\\oplus$\n$\\mathrm{C}\\left(\\mathrm{CH}_{3}\\right)_{3}$\n\nPROBLEM Use the $\\mathrm{p} K_{\\mathrm{a}}$ table in Appendix B to determine in which direction the equilibrium is favored.\n2-64 (a)\n\n(b)\n\n(c)\n\n$$\n\\overline{\\mathrm{C}} \\mathrm{H}_{3}+\\mathrm{CH}_{3} \\mathrm{NO}_{2} \\rightleftarrows \\mathrm{CH}_{4}+\\overline{\\mathrm{C}} \\mathrm{H}_{2} \\mathrm{NO}_{2}\n$$"}
{"id": 143, "contents": "General Problems - \nPROBLEM Which intermolecular force is predominantly responsible for each observation below?\n$\\mathbf{2 - 6 5}$ (a) $\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{29} \\mathrm{CH}_{3}$, a component found in paraffin wax, is a solid at room temperature while $\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{6} \\mathrm{CH}_{3}$ is a liquid.\n(b) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}$ has a higher boiling point than $\\mathrm{CH}_{4}$.\n(c) $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$, which is found in vinegar, will dissolve in water but not in oil. Assume that oil is $\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{4} \\mathrm{CH}_{3}$.\n$702 \\cdot$ Additional Problems"}
{"id": 144, "contents": "CHAPTER 3 - \nOrganic Compounds: Alkanes and Their Stereochemistry\n\n\nFIGURE 3.1 The bristlecone pine is the oldest living organism on Earth. The waxy coating on its needles contains a mixture of organic compounds called alkanes, the subject of this chapter. (credit: modification of work \"Gnarly Bristlecone Pine\" by Rick Goldwaser/Flickr, CC BY 2.0)"}
{"id": 145, "contents": "CHAPTER CONTENTS - 3.1 Functional Groups\n3.2 Alkanes and Alkane Isomers\n3.3 Alkyl Groups\n3.4 Naming Alkanes\n3.5 Properties of Alkanes\n3.6 Conformations of Ethane\n3.7 Conformations of Other Alkanes\n\nWHY THIS CHAPTER? The group of organic compounds called alkanes are simple and relatively unreactive, but they nevertheless provide a useful vehicle for introducing some important general ideas. In this chapter, we'll use alkanes to introduce the basic approach to naming organic compounds and to take an initial look at some of the three-dimensional aspects of molecules, a topic of particular importance in understanding biological organic chemistry.\n\nAccording to Chemical Abstracts, the publication that abstracts and indexes the chemical literature, there are more than 195 million known organic compounds. Each of these compounds has its own physical properties, such as melting point and boiling point, and each has its own chemical reactivity.\n\nChemists have learned through years of experience that organic compounds can be classified into families according to their structural features and that the members of a given family have similar chemical behavior. Instead of 195 million compounds with random reactivity, there are a few dozen families of organic compounds whose chemistry is reasonably predictable. We'll study the chemistry of specific families throughout much of this book, beginning in this chapter with a look at the simplest family, the alkanes."}
{"id": 146, "contents": "CHAPTER CONTENTS - 3.1 Functional Groups\nThe structural features that make it possible to classify compounds into families are called functional groups. A functional group is a group of atoms within a molecule that has a characteristic chemical behavior. Chemically, a given functional group behaves in nearly the same way in every molecule it's a part of. For example, compare ethylene, a plant hormone that causes fruit to ripen, with menthene, a much more complicated molecule found in peppermint oil. Both substances contain a carbon-carbon double-bond functional group, and both therefore react with $\\mathrm{Br}_{2}$ in the same way to give a product in which a Br atom has added to each of the double-bond carbons (FIGURE 3.2). This example is typical: the chemistry of every organic molecule, regardless of size and complexity, is determined by the functional groups it contains.\n\n\n\nEthylene\n\n\n\nMenthene\n\n\n\nFIGURE 3.2 The reactions of ethylene and menthene with bromine. In both molecules, the carbon-carbon double-bond functional group has a similar polarity pattern, so both molecules react with $\\mathrm{Br}_{2}$ in the same way. The size and complexity of the molecules are not important.\n\nLook at TABLE 3.1, which lists many of the common functional groups and gives simple examples of their occurrence. Some functional groups have only carbon-carbon double or triple bonds; others have halogen atoms; and still others contain oxygen, nitrogen, or sulfur. Much of the chemistry you'll be studying is the chemistry of these functional groups."}
{"id": 147, "contents": "Functional Groups with Carbon-Carbon Multiple Bonds - \nAlkenes, alkynes, and arenes (aromatic compounds) all contain carbon-carbon multiple bonds. Alkenes have a double bond, alkynes have a triple bond, and arenes have alternating double and single bonds in a sixmembered ring of carbon atoms. They look different, but because of their structural similarities, they also have chemical similarities.\n\n\nTABLE 3.1 Structures of Some Common Functional Groups"}
{"id": 148, "contents": "Functional Groups with Carbon-Carbon Multiple Bonds - \n| Name | Structure* | Name ending | Example |\n| :---: | :---: | :---: | :---: |\n| Amine | <smiles>CN(C)C(C)(C)C</smiles> | -amine | $\\mathrm{CH}_{3} \\mathrm{NH}_{2}$ <br> Methylamine |\n| Imine (Schiff base) | <smiles>CCC(=NC)C(C)(C)C</smiles> | None | <smiles>C=NC</smiles> <br> Acetone imine |\n| Nitrile | -C $\\equiv \\mathrm{N}$ | -nitrile | $\\mathrm{CH}_{3} \\mathrm{C} \\equiv \\mathrm{~N}$ <br> Ethanenitrile |\n| Thiol | <smiles>CC(C)(C)S</smiles> | -thiol | $\\mathrm{CH}_{3} \\mathrm{SH}$ <br> Methanethiol |\n| Sulfide | <smiles></smiles> | sulfide | $\\mathrm{CH}_{3} \\mathrm{SCH}_{3}$ <br> Dimethyl sulfide |\n| Disulfide | <smiles>CSC(C)(C)C</smiles> | disulfide | $\\mathrm{CH}_{3} \\mathrm{SSCH}_{3}$ <br> Dimethyl disulfide |\n| Sulfoxide | <smiles>CCS+C(C)(C)C</smiles> | sulfoxide | <smiles>C[SeH]=[OH+]</smiles> <br> Dimethyl sulfoxide |\n| Aldehyde | <smiles>CC=O</smiles> | -al | Ethanal |\n| Ketone | <smiles>CCC(=O)C(C)(C)C</smiles> | -one | <smiles>C=C=O</smiles> <br> Propanone |\n| Carboxylic acid | <smiles>CC(C)(C)C(=O)O</smiles> | -oic acid | <smiles>C[14CH2]O</smiles> <br> Ethanoic acid |"}
{"id": 149, "contents": "Functional Groups with Carbon-Carbon Multiple Bonds - \n| Ester | <smiles>CC(C)(C)C(=O)O</smiles> | -oate | <smiles>CC(C)=O</smiles> <br> Methyl ethanoate |\n| Thioester | <smiles>CC(C)(C)C(=O)S</smiles> | -thioate | <smiles>C[AsH]=O</smiles> <br> Methyl ethanethioate |"}
{"id": 150, "contents": "Functional Groups with Carbon-Carbon Multiple Bonds - \nTABLE 3.1 Structures of Some Common Functional Groups\n\n| Name | Structure* | Name ending | Example |\n| :---: | :---: | :---: | :---: |\n| Amide | <smiles>CN(C)C(=O)C(C)(C)C</smiles> | -amide | <smiles>CC(N)=O</smiles> <br> Ethanamide |\n| Acid chloride | <smiles>CC(C)(C)C(=O)Cl</smiles> | -oyl chloride | <smiles>CC(=O)Cl</smiles> <br> Ethanoyl chloride |\n| Carboxylic acid anhydride | <smiles>CC(C)(C)C(=O)OC(=O)C(C)(C)C</smiles> | -oic anhydride | <smiles>COC(C)=O</smiles> <br> Ethanoic anhydride |\n\n*The bonds whose connections aren't specified are assumed to be attached to carbon or hydrogen atoms in the rest of the molecule."}
{"id": 151, "contents": "Functional Groups with Carbon Singly Bonded to an Electronegative Atom - \nAlkyl halides (haloalkanes), alcohols, ethers, alkyl phosphates, amines, thiols, sulfides, and disulfides all have a carbon atom singly bonded to an electronegative atom-halogen, oxygen, nitrogen, or sulfur. Alkyl halides have a carbon atom bonded to halogen ( -X ), alcohols have a carbon atom bonded to the oxygen of a hydroxyl group ( -OH ), ethers have two carbon atoms bonded to the same oxygen, organophosphates have a carbon atom bonded to the oxygen of a phosphate group $\\left(-\\mathrm{OPO}_{3}{ }^{2-}\\right)$, amines have a carbon atom bonded to a nitrogen, thiols have a carbon atom bonded to the sulfur of an -SH group, sulfides have two carbon atoms bonded to the same sulfur, and disulfides have carbon atoms bonded to two sulfurs that are joined together. In all cases, the bonds are polar, with the carbon atom bearing a partial positive charge ( $\\delta+$ ) and the electronegative atom bearing a partial negative charge ( $\\delta-$ ).\n\n\n\nAlkyl halide (haloalkane)\n\n\n\nAmine\n\n\n\nAlcohol\n\n\n\nThiol\n\n\n\nEther\n\n\n\nSulfide\n\n\n\nPhosphate\n\n\n\nDisulfide"}
{"id": 152, "contents": "Functional Groups with a Carbon-Oxygen Double Bond (Carbonyl Groups) - \nThe carbonyl group, $\\mathrm{C}=\\mathrm{O}$ (pronounced car-bo-neel) is common to many of the families listed in TABLE 3.1. Carbonyl groups are present in a majority of organic compounds and in practically all biological molecules. These compounds therefore behave similarly in many respects but differ depending on the identity of the other atoms bonded to the carbonyl-group carbon. Aldehydes have at least one hydrogen bonded to the $\\mathrm{C}=\\mathrm{O}$, ketones have two carbons bonded to the $\\mathrm{C}=\\mathrm{O}$, carboxylic acids have an -OH group bonded to the $\\mathrm{C}=\\mathrm{O}$, esters have an ether-like oxygen bonded to the $\\mathrm{C}=\\mathrm{O}$, thioesters have a sulfide-like sulfur bonded to the $\\mathrm{C}=\\mathrm{O}$, amides have an amine-like nitrogen bonded to the $\\mathrm{C}=\\mathrm{O}$, acid chlorides have a chlorine bonded to the $\\mathrm{C}=\\mathrm{O}$, and so on. In all these functional groups, the carbonyl carbon atom bears a partial positive charge ( $\\delta+$ ), and the oxygen bears a partial negative charge ( $\\delta$-).\n\n\nAcetone-a typical carbonyl compound\n\n\nAldehyde\n\n\nKetone\n\n\nCarboxylic acid\n\n\nEster\n\n\nThioester\n\n\nAmide\n\n\nAcid chloride\n\nPROBLEM Use Table 3.1 to identify the functional groups in each of the following molecules:\n3-1 (a)\nMethionine, an amino acid:\n(b) Ibuprofen, a pain reliever:\n\n\n\n(c) Capsaicin, the pungent substance in chili peppers:\n\n\nPROBLEM Propose structures for simple molecules that contain the following functional groups:\n3-2 (a) Alcohol (b) Aromatic ring\n(c) Carboxylic acid (d) Amine\n(e) Both ketone and amine (f) Two double bonds"}
{"id": 153, "contents": "Functional Groups with a Carbon-Oxygen Double Bond (Carbonyl Groups) - \nPROBLEM Identify the functional groups in the following model of arecoline, a veterinary drug used to control 3-3 worms in animals. Convert the drawing into a line-bond structure and a molecular formula (red = O , blue $=\\mathrm{N}$, black $=\\mathrm{C}$, gray $=\\mathrm{H}$ ) ."}
{"id": 154, "contents": "Functional Groups with a Carbon-Oxygen Double Bond (Carbonyl Groups) - 3.2 Alkanes and Alkane Isomers\nBefore beginning a systematic study of the different functional groups, let's look first at the simplest family of molecules to develop some general ideas that apply to all families. We saw in Section 1.7 that the carbon-carbon single bond in ethane results from $\\sigma$ (head-on) overlap of carbon $s p^{3}$ hybrid orbitals. If we imagine joining three, four, five, or even more carbon atoms by $\\mathrm{C}-\\mathrm{C}$ single bonds, we can generate the large family of molecules called alkanes.\n\n\nMethane\n\n\nEthane\n\n\nPropane\n\n\nButane\n\nAlkanes are often described as saturated hydrocarbons: hydrocarbons because they contain only carbon and hydrogen; saturated because they have only $\\mathrm{C}-\\mathrm{C}$ and $\\mathrm{C}-\\mathrm{H}$ single bonds and thus contain the maximum possible number of hydrogens per carbon. They have the general formula $\\mathrm{C}_{n} \\mathrm{H}_{2 n+2}$, where $n$ is an integer. Alkanes are also occasionally called aliphatic compounds, a name derived from the Greek aleiphas, meaning \"fat.\" We'll see in Section 27.1 that many animal fats contain long carbon chains similar to alkanes.\n\n\nA typical animal fat\nThink about the ways that carbon and hydrogen might combine to make alkanes. With one carbon and four hydrogens, only one structure is possible: methane, $\\mathrm{CH}_{4}$. Similarly, there is only one combination of two carbons with six hydrogens (ethane, $\\mathrm{CH}_{3} \\mathrm{CH}_{3}$ ) and only one combination of three carbons with eight hydrogens (propane, $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$ ). When larger numbers of carbons and hydrogens combine, however, more than one structure is possible. For example, there are two substances with the formula $\\mathrm{C}_{4} \\mathrm{H}_{10}$ : the four carbons can all be in a row (butane), or they can branch (isobutane). Similarly, there are three $\\mathrm{C}_{5} \\mathrm{H}_{12}$ molecules, and so on for larger alkanes."}
{"id": 155, "contents": "Functional Groups with a Carbon-Oxygen Double Bond (Carbonyl Groups) - 3.2 Alkanes and Alkane Isomers\nCompounds like butane and pentane, whose carbons are all connected in a row, are called straight-chain alkanes, or normal alkanes. Compounds like 2-methylpropane (isobutane), 2-methylbutane, and 2,2-dimethylpropane, whose carbon chains branch, are called branched-chain alkanes.\n\nCompounds like the two $\\mathrm{C}_{4} \\mathrm{H}_{10}$ molecules and the three $\\mathrm{C}_{5} \\mathrm{H}_{12}$ molecules, which have the same formula but different structures, are called Isomers, from the Greek isos + meros, meaning \"made of the same parts.\" Isomers have the same numbers and kinds of atoms but differ in the way the atoms are arranged. Compounds like butane and isobutane, whose atoms are connected differently, are called constitutional isomers. We'll see shortly that other kinds of isomers are also possible, even among compounds whose atoms are connected in the same order. As TABLE 3.2 shows, the number of possible alkane isomers increases dramatically with the number of carbon atoms.\n\n| TABLE 3.2 Number of Alkane Isomers |  |  |  |\n| :--- | :--- | :--- | :--- |\n| Formula | Number of isomers | Formula | Number of isomers |\n| $\\mathrm{C}_{6} \\mathrm{H}_{14}$ | 5 | $\\mathrm{C}_{10} \\mathrm{H}_{22}$ | 75 |\n| $\\mathrm{C}_{7} \\mathrm{H}_{16}$ | 9 | $\\mathrm{C}_{15} \\mathrm{H}_{32}$ | 4347 |\n| $\\mathrm{C}_{8} \\mathrm{H}_{18}$ | 18 | $\\mathrm{C}_{20} \\mathrm{H}_{42}$ | 366,319 |\n| $\\mathrm{C}_{9} \\mathrm{H}_{20}$ | 35 | $\\mathrm{C}_{30} \\mathrm{H}_{62}$ | $4,111,846,763$ |"}
{"id": 156, "contents": "Functional Groups with a Carbon-Oxygen Double Bond (Carbonyl Groups) - 3.2 Alkanes and Alkane Isomers\nConstitutional isomerism is not limited to alkanes-it occurs widely throughout organic chemistry. Constitutional isomers may have different carbon skeletons (as in isobutane and butane), different functional groups (as in ethanol and dimethyl ether), or different locations of a functional group along the chain (as in isopropylamine and propylamine). Regardless of the reason for the isomerism, constitutional isomers are always different compounds with different properties but with the same formula."}
{"id": 157, "contents": "Different carbon skeletons $\\mathrm{C}_{4} \\mathrm{H}_{10}$ - \nDifferent functional groups\n$\\mathrm{C}_{2} \\mathrm{H}_{6} \\mathrm{O}$\n\nDifferent position of functional groups $\\mathrm{C}_{3} \\mathrm{H}_{9} \\mathrm{~N}$\n\n\n2-Methylpropane\n(isobutane)\n$\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$\nEthanol\n\n\nIsopropylamine\n\n$$\n\\text { and } \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}\n$$\n\nButane\n$\\mathrm{CH}_{3} \\mathrm{OCH}_{3}$\nDimethyl ether\nand\n\n\nPropylamine\n\nA given alkane can be drawn in many ways. For example, the straight-chain, four-carbon alkane called butane can be represented by any of the structures shown in FIGURE 3.3. These structures don't imply any particular three-dimensional geometry for butane; they indicate only the connections among atoms. In practice, as noted in Section 1.12, chemists rarely draw all the bonds in a molecule and usually refer to butane by the condensed structure, $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$ or $\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{2} \\mathrm{CH}_{3}$. Still more simply, butane can be represented as $n-\\mathrm{C}_{4} \\mathrm{H}_{10}$, where $n$ denotes normal (straight-chain) butane.\n\n\n\n$$\n\\mathrm{CH}_{3}-\\mathrm{CH}_{2}-\\mathrm{CH}_{2}-\\mathrm{CH}_{3} \\quad \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3} \\quad \\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{2} \\mathrm{CH}_{3}\n$$"}
{"id": 158, "contents": "Different carbon skeletons $\\mathrm{C}_{4} \\mathrm{H}_{10}$ - \nFIGURE 3.3 Some representations of butane, $\\mathbf{C}_{\\mathbf{4}} \\mathbf{H}_{\\mathbf{1 0}}$. The molecule is the same regardless of how it's drawn. These structures imply only that butane has a continuous chain of four carbon atoms; they do not imply any specific geometry.\n\nStraight-chain alkanes are named according to the number of carbon atoms they contain, as shown in TABLE 3.3. With the exception of the first four compounds-methane, ethane, propane, and butane-whose names have historical roots, the alkanes are named based on Greek numbers. The suffix -ane is added to the end of each name to indicate that the molecule identified is an alkane. Thus, pentane is the five-carbon alkane, hexane is the six-carbon alkane, and so on. We'll soon see that these alkane names form the basis for naming all other organic compounds, so at least the first ten should be memorized."}
{"id": 159, "contents": "TABLE 3.3 Names of Straight-Chain Alkanes - \n| Number of carbons <br> $(n)$ | Name | Formula <br> $\\left(\\mathrm{C}_{n} \\mathrm{H}_{2 n+2}\\right)$ | Number of carbons <br> $(n)$ | Name | Formula <br> $\\left(\\mathrm{C}_{n} \\mathrm{H}_{2 n+2}\\right)$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n| 1 | Methane | $\\mathrm{CH}_{4}$ | 9 | Nonane | $\\mathrm{C}_{9} \\mathrm{H}_{20}$ |\n| 2 | Ethane | $\\mathrm{C}_{2} \\mathrm{H}_{6}$ | 10 | Decane | $\\mathrm{C}_{10} \\mathrm{H}_{22}$ |\n| 3 | Propane | $\\mathrm{C}_{3} \\mathrm{H}_{8}$ | 11 | Undecane | $\\mathrm{C}_{11} \\mathrm{H}_{24}$ |\n| 4 | Butane | $\\mathrm{C}_{4} \\mathrm{H}_{10}$ | 12 | Dodecane | $\\mathrm{C}_{12} \\mathrm{H}_{26}$ |\n| 5 | Pentane | $\\mathrm{C}_{5} \\mathrm{H}_{12}$ | 13 | Tridecane | $\\mathrm{C}_{13} \\mathrm{H}_{28}$ |\n\nTABLE 3.3 Names of Straight-Chain Alkanes"}
{"id": 160, "contents": "TABLE 3.3 Names of Straight-Chain Alkanes - \nTABLE 3.3 Names of Straight-Chain Alkanes\n\n| Number of carbons <br> $(n)$ | Name | Formula <br> $\\left(\\mathrm{C}_{n} \\mathrm{H}_{2 n+2}\\right)$ | Number of carbons <br> $(n)$ | Name | Formula <br> $\\left(\\mathrm{C}_{n} \\mathrm{H}_{2 n+2}\\right)$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n| 6 | Hexane | $\\mathrm{C}_{6} \\mathrm{H}_{14}$ | 20 | Icosane | $\\mathrm{C}_{20} \\mathrm{H}_{42}$ |\n| 7 | Heptane | $\\mathrm{C}_{7} \\mathrm{H}_{16}$ | 30 | Triacontane | $\\mathrm{C}_{30} \\mathrm{H}_{62}$ |\n| 8 | Octane | $\\mathrm{C}_{8} \\mathrm{H}_{18}$ |  |  |  |"}
{"id": 161, "contents": "Drawing the Structures of Isomers - \nPropose structures for two isomers with the formula $\\mathrm{C}_{2} \\mathrm{H}_{7} \\mathrm{~N}$."}
{"id": 162, "contents": "Strategy - \nWe know that carbon forms four bonds, nitrogen forms three, and hydrogen forms one. Write down the carbon atoms first, and then use trial and error plus intuition to put the pieces together."}
{"id": 163, "contents": "Solution - \nThere are two isomeric structures. One has the connection $\\mathrm{C}-\\mathrm{C}-\\mathrm{N}$, and the other has the connection $\\mathrm{C}-\\mathrm{N}-\\mathrm{C}$.\n\n\nPROBLEM Draw structures of the five isomers of $\\mathrm{C}_{6} \\mathrm{H}_{14}$.\n3-4\nPROBLEM Propose structures that meet the following descriptions:\n3-5 (a) Two isomeric esters with the formula $\\mathrm{C}_{5} \\mathrm{H}_{10} \\mathrm{O}_{2}$ :\n(b) Two isomeric nitriles with the formula $\\mathrm{C}_{4} \\mathrm{H}_{7} \\mathrm{~N}$\n(c) Two isomeric disulfides with the formula $\\mathrm{C}_{4} \\mathrm{H}_{10} \\mathrm{~S}_{2}$\n\nPROBLEM How many isomers are there with the following descriptions?\n3-6 (a) Alcohols with the formula $\\mathrm{C}_{3} \\mathrm{H}_{8} \\mathrm{O}$ (b) Bromoalkanes with the formula $\\mathrm{C}_{4} \\mathrm{H}_{9} \\mathrm{Br}$\n(c) Thioesters with the formula $\\mathrm{C}_{4} \\mathrm{H}_{8} \\mathrm{OS}$"}
{"id": 164, "contents": "Solution - 3.3 Alkyl Groups\nIf you imagine removing a hydrogen atom from an alkane, the partial structure that remains is called an alkyl group. Alkyl groups are not stable compounds themselves, they are simply parts of larger compounds and are named by replacing the -ane ending of the parent alkane with an $-y l$ ending. For example, removal of a hydrogen from methane, $\\mathrm{CH}_{4}$, generates a methyl group, $-\\mathrm{CH}_{3}$, and removal of a hydrogen from ethane, $\\mathrm{CH}_{3} \\mathrm{CH}_{3}$, generates an ethyl group, $-\\mathrm{CH}_{2} \\mathrm{CH}_{3}$. Similarly, removal of a hydrogen atom from the end carbon of any straight-chain alkane gives the series of straight-chain alkyl groups shown in TABLE 3.4. Combining an alkyl group with any of the functional groups listed earlier makes it possible to generate and name many thousands of compounds. For example:\n\n\n\nMethane\n\n\n\nA methyl group\n\n\n\nMethyl alcohol (methanol)\n\n\nMethylamine\nTABLE 3.4 Some Straight-Chain Alkyl Groups"}
{"id": 165, "contents": "Solution - 3.3 Alkyl Groups\nMethane\n\n\n\nA methyl group\n\n\n\nMethyl alcohol (methanol)\n\n\nMethylamine\nTABLE 3.4 Some Straight-Chain Alkyl Groups\n\n| Alkane |  | Name | Alkyl group |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{CH}_{4}$ | Methane | $-\\mathrm{CH}_{3}$ | Mame (abbreviation) |\n| $\\mathrm{CH}_{3} \\mathrm{CH}_{3}$ | Ethane (Me) | $-\\mathrm{CH}_{2} \\mathrm{CH}_{3}$ | Ethyl (Et) |\n| $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$ | Propane | $-\\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$ | Propyl (Pr) |\n| $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$ | Butane | $-\\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$ | Butyl (Bu) |\n| $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$ | Pentane | $-\\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$ | Pentyl, or amyl |\n\nJust as straight-chain alkyl groups are generated by removing a hydrogen from an end carbon, branched alkyl groups are generated by removing a hydrogen atom from an internal carbon. Two 3-carbon alkyl groups and four 4-carbon alkyl groups are possible (FIGURE 3.4).\n\n\n$\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\frac{?}{2}$\nPropyl\n\n\n\nIsopropyl\n\n$\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$\nButane"}
{"id": 166, "contents": "Solution - 3.3 Alkyl Groups\nIsopropyl\n\n$\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$\nButane\n\n\n$\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2}$ \u4e4b\nButyl\n\n\n\nIsobutyl\n\n~\n\nsec-Butyl\n\n\ntert-Butyl\n\nFIGURE 3.4 Alkyl groups generated from straight-chain alkanes.\nOne further comment about naming alkyl groups: the prefixes sec- (for secondary) and tert- (for tertiary) used for the $\\mathrm{C}_{4}$ alkyl groups in FIGURE 3.4 refer to the number of other carbon atoms attached to the branching carbon atom. There are four possibilities: primary $\\left(1^{\\circ}\\right)$, secondary $\\left(2^{\\circ}\\right)$, tertiary $\\left(3^{\\circ}\\right)$, and quaternary $\\left(4^{\\circ}\\right)$.\n\n\nPrimary carbon ( $1^{\\circ}$ )\nis bonded to one other carbon.\n\n\nSecondary carbon ( $\\mathbf{2}^{\\circ}$ ) is bonded to two other carbons.\n\n\nTertiary carbon ( $3^{\\circ}$ ) is bonded to three other carbons."}
{"id": 167, "contents": "Quaternary carbon (40) <br> is bonded to four - \nother carbons.The symbol $\\mathbf{R}$ is used here and throughout organic chemistry to represent a generalized organic group. The $R$ group can be methyl, ethyl, propyl, or any of a multitude of others. You might think of $\\mathbf{R}$ as representing the Rest of the molecule, which isn't specified.\n\nThe terms primary, secondary, tertiary, and quaternary are routinely used in organic chemistry, and their meanings need to become second nature. For example, if we were to say, \"Citric acid is a tertiary alcohol,\" we would mean that it has an alcohol functional group $(-\\mathrm{OH})$ bonded to a carbon atom that is itself bonded to three other carbons.\n\n\nGeneral class of tertiary alcohols, $\\mathrm{R}_{3} \\mathrm{COH}$\n\n\nCitric acid-a specific tertiary alcohol\n\nIn addition to speaking of carbon atoms as being primary, secondary, or tertiary, we speak of hydrogens in the same way. Primary hydrogen atoms are attached to primary carbons $\\left(\\mathrm{RCH}_{3}\\right)$, secondary hydrogens are attached to secondary carbons $\\left(\\mathrm{R}_{2} \\mathrm{CH}_{2}\\right)$, and tertiary hydrogens are attached to tertiary carbons $\\left(\\mathrm{R}_{3} \\mathrm{CH}\\right)$. There is, however, no such thing as a quaternary hydrogen. (Why not?)\n\n\nPROBLEM Draw the eight 5-carbon alkyl groups (pentyl isomers). 3-7\n\nPROBLEM Identify the carbon atoms in the following molecules as primary, secondary, tertiary, or quaternary:\n3-8 (a)\n\n(b)\n\n(c)\n\n\nPROBLEM Identify the hydrogen atoms on the compounds shown in Problem 3-8 as primary, secondary, or 3-9 tertiary.\n\nPROBLEM Draw structures of alkanes that meet the following descriptions:\n3-10 (a) An alkane with two tertiary carbons (b) An alkane that contains an isopropyl group\n(c) An alkane that has one quaternary and one secondary carbon"}
{"id": 168, "contents": "Quaternary carbon (40) <br> is bonded to four - 3.4 Naming Alkanes\nIn earlier times, when relatively few pure organic chemicals were known, new compounds were named at the whim of their discoverer. Thus, urea $\\left(\\mathrm{CH}_{4} \\mathrm{~N}_{2} \\mathrm{O}\\right)$ is a crystalline substance isolated from urine; morphine $\\left(\\mathrm{C}_{17} \\mathrm{H}_{19} \\mathrm{NO}_{3}\\right)$ is an analgesic (painkiller) named after Morpheus, the Greek god of dreams; and acetic acid, the primary organic constituent of vinegar, is named from the Latin word for vinegar, acetum.\n\nAs the science of organic chemistry slowly grew in the 19th century, so too did the number of known compounds and the need for a systematic method of naming them. The system of naming (nomenclature) we'll use in this book is that devised by the International Union of Pure and Applied Chemistry (IUPAC, usually spoken as eye-you-pac).\n\nA chemical name typically has four parts in the IUPAC system: parent, prefix, locant, and suffix. The parent name identifies the main part of the molecule and tells how many carbon atoms are in that part. Prefixes identify the various substituent groups attached to the parent. Locants give the positions of the attached substituents. And the suffix identifies the primary functional group attached to the parent.\n\n\nAs we cover new functional groups in later chapters, the applicable IUPAC rules of nomenclature will be given. In addition, Appendix $A$ at the back of this book gives an overall view of organic nomenclature and shows how compounds that contain more than one functional group are named. (If preferred, you can study that appendix now.) For the present, let's see how to name branched-chain alkanes and learn some general rules that are applicable to all compounds.\n\nAll but the most complex branched-chain alkanes can be named by following four steps. For a very few compounds, a fifth step is needed."}
{"id": 169, "contents": "Identify the parent hydrocarbon. - \n(a) Find the longest continuous chain of carbon atoms in the molecule, and use the name of that chain as the parent name. The longest chain may not always be apparent from the manner of writing; you may have to \"turn corners.\"\n\n(b) If two different chains of equal length are present, choose the one with the larger number of branch points as the parent.\n\n\nNamed as a hexane with two substituents\n\n\nNOT\nas a hexane with one substituent\n\nSTEP 2"}
{"id": 170, "contents": "Number the atoms in the longest chain. - \n(a) Beginning at the end nearer the first branch point, number each carbon atom in the parent chain.\n\n\nNOT\n\n\nThe first branch occurs at C3 in the proper system of numbering, not at C4.\n(b) If there is branching an equal distance away from both ends of the parent chain, begin numbering at the end nearer the second branch point.\n\n\nNOT\n\n\nSTEP 3\nIdentify and number the substituents.\n(a) Assign a number to each substituent to locate its point of attachment to the parent chain.\n\n(b) If there are two substituents on the same carbon, give both the same number. There must be as many numbers in the name as there are substituents.\n\n\nNamed as a hexane\n\nSubstituents:\nOn C2, $\\mathrm{CH}_{3}$\nOn C4, CH3\nOn C4, $\\mathrm{CH}_{2} \\mathrm{CH}_{3}$\n\n$$\n\\begin{aligned}\n& \\text { (2-methyl) } \\\\\n& \\text { (4-methyl) } \\\\\n& \\text { (4-ethyl) }\n\\end{aligned}\n$$"}
{"id": 171, "contents": "Write the name as a single word. - \nUse hyphens to separate the different prefixes, and use commas to separate numbers. If two or more different substituents are present, cite them in alphabetical order. If two or more identical substituents are present on the parent chain, use one of the multiplier prefixes di-, tri-, tetra-, and so forth, but don't use these prefixes for alphabetizing. Full names for some of the examples we have been using are as follows:\n\n\nSTEP 5"}
{"id": 172, "contents": "Name a branched substituent as though it were itself a compound. - \nIn some particularly complex cases, a fifth step is necessary. It occasionally happens that a substituent on the main chain is itself branched. In the following case, for instance, the substituent at C 6 is a three-carbon chain with a methyl group. To name the compound fully, the branched substituent must first be named.\n\n\nNamed as a 2,3,6trisubstituted decane\n\n\nA 2-methylpropyl substituent\n\nNumber the branched substituent beginning at the point of its attachment to the main chain, and identify it-in this case, a 2-methylpropyl group. The substituent is treated as a whole and is alphabetized according to the first letter of its complete name, including any numerical prefix. It is set off in parentheses when naming the entire molecule.\n\n\n2,3-Dimethyl-6-(2-methylpropyl)decane\nAs a further example:\n\n\n5-(1,2-Dimethylpropyl)-2-methylnonane\n\n\nA 1,2-dimethylpropyl group\n\nFor historical reasons, some of the simpler branched-chain alkyl groups also have nonsystematic, common names, as noted earlier.\n\n\n\n\nNeopentyl\ntert-Pentyl, also called\ntert-amyl $(t-a m y l)$\nIsopentyl, also called"}
{"id": 173, "contents": "5-Carbon alkyl groups - \nThe common names of these simple alkyl groups are so well entrenched in the chemical literature that IUPAC rules make allowance for them. Thus, the following compound is properly named either 4-(1-methylethyl)heptane or 4-isopropylheptane. There's no choice but to memorize these common names; fortunately, there are only a few of them.\n\n\n4-(1-Methylethyl)heptane or 4-Isopropylheptane\nWhen writing an alkane name, the nonhyphenated prefix iso- is considered part of the alkyl-group name for alphabetizing purposes, but the hyphenated and italicized prefixes sec- and tert- are not. Thus, isopropyl and isobutyl are listed alphabetically under $i$, but sec-butyl and tert-butyl are listed under $b$."}
{"id": 174, "contents": "Naming Alkanes - \nWhat is the IUPAC name for the following alkane?"}
{"id": 175, "contents": "Strategy - \nFind the longest continuous carbon chain in the molecule, and use that as the parent name. This molecule has a chain of eight carbons-octane-with two methyl substituents. (You have to turn corners to see it.) Numbering from the end nearer the first methyl substituent indicates that the methyls are at C2 and C6."}
{"id": 176, "contents": "Converting a Chemical Name into a Structure - \nDraw the structure of 3-isopropyl-2-methylhexane."}
{"id": 177, "contents": "Strategy - \nThis is the reverse of Worked Example 3.2 and uses a reverse strategy. Look at the parent name (hexane), and draw its carbon structure.\n\n$$\n\\mathrm{C}-\\mathrm{C}-\\mathrm{C}-\\mathrm{C}-\\mathrm{C}-\\mathrm{C} \\text { Hexane }\n$$\n\nNext, find the substituents (3-isopropyl and 2-methyl), and place them on the proper carbons.\n\n\nFinally, add hydrogens to complete the structure.\nSolution\n\n\n3-Isopropyl-2-methylhexane\n\nPROBLEM Give IUPAC names for the following compounds:\n3-11 (a) The three isomers of $\\mathrm{C}_{5} \\mathrm{H}_{12}$\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM Draw structures corresponding to the following IUPAC names:\n3-12 (a) 3,4-Dimethylnonane (b) 3-Ethyl-4,4-dimethylheptane (c) 2,2-Dimethyl-4-propyloctane (d) 2,2,4-Trimethylpentane\n\nPROBLEM Name the eight 5-carbon alkyl groups you drew in Problem 3-7.\n\nPROBLEM Give the IUPAC name for the following hydrocarbon, and convert the drawing into a skeletal $\\mathbf{3 - 1 4}$ structure."}
{"id": 178, "contents": "Strategy - 3.5 Properties of Alkanes\nAlkanes are sometimes referred to as paraffins, a word derived from the Latin parum affinis, meaning \"little affinity.\" This term aptly describes their behavior, for alkanes show little chemical affinity for other substances and are chemically inert to most laboratory reagents. They are also relatively inert biologically and are not often involved in the chemistry of living organisms. Alkanes do, however, react with oxygen, halogens, and a few other substances under appropriate conditions.\n\nReaction with oxygen occurs during combustion in an engine or furnace when an alkane is used as a fuel. Carbon dioxide and water are formed as products, and a large amount of heat is released. For example, methane (natural gas) reacts with oxygen according to the equation\n\n$$\n\\mathrm{CH}_{4}+2 \\mathrm{O}_{2} \\longrightarrow \\mathrm{CO}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}+890 \\mathrm{~kJ} / \\mathrm{mol}(213 \\mathrm{kcal} / \\mathrm{mol})\n$$\n\nThe reaction of an alkane with $\\mathrm{Cl}_{2}$ occurs when a mixture of the two is irradiated with ultraviolet light (denoted $h v$, where $v$ is the Greek letter $n u$ ). Depending on the time allowed and the relative amounts of the two reactants, a sequential substitution of the alkane hydrogen atoms by chlorine occurs, leading to a mixture of chlorinated products. Methane, for instance, reacts with $\\mathrm{Cl}_{2}$ to yield a mixture of $\\mathrm{CH}_{3} \\mathrm{Cl}_{1}, \\mathrm{CH}_{2} \\mathrm{Cl}_{2}, \\mathrm{CHCl}_{3}$, and $\\mathrm{CCl}_{4}$. We'll look at this reaction in more detail in Section 6.6.\n\n\nAlkanes show regular increases in both boiling point and melting point as molecular weight increases (FIGURE 3.5), an effect due to the presence of weak dispersion forces between molecules (Section 2.12). Only when sufficient energy is applied to overcome these forces does the solid melt or liquid boil. As you might expect, dispersion forces increase as molecular size increases, accounting for the higher melting and boiling points of larger alkanes."}
{"id": 179, "contents": "Strategy - 3.5 Properties of Alkanes\nFIGURE 3.5 A plot of melting and boiling points versus number of carbon atoms for the C1-C14 straight-chain alkanes. There is a regular increase with molecular size.\n\nAnother effect seen in alkanes is that increased branching lowers an alkane's boiling point. Thus, pentane has no branches and boils at $36.1^{\\circ} \\mathrm{C}$, isopentane (2-methylbutane) has one branch and boils at $27.85{ }^{\\circ} \\mathrm{C}$, and neopentane (2,2-dimethylpropane) has two branches and boils at $9.5^{\\circ} \\mathrm{C}$. Similarly, octane boils at $125.7^{\\circ} \\mathrm{C}$, whereas isooctane ( $2,2,4$-trimethylpentane) boils at $99.3^{\\circ} \\mathrm{C}$. Branched-chain alkanes are lower-boiling because they are more nearly spherical than straight-chain alkanes, have smaller surface areas, and consequently have smaller dispersion forces."}
{"id": 180, "contents": "Strategy - 3.6 Conformations of Ethane\nUp until now, we've viewed molecules primarily in a two-dimensional way and have given little thought to any consequences that might arise from the spatial arrangement of atoms in molecules. Now it's time to add a third dimension to our study. Stereochemistry is the branch of chemistry concerned with the three-dimensional aspects of molecules. We'll see on many occasions in future chapters that the exact three-dimensional structure of a molecule is often crucial to determining its properties and biological behavior.\n\nWe know from Section 1.5 that $\\sigma$ bonds are cylindrically symmetrical. In other words, the intersection of a plane cutting through a carbon-carbon single-bond orbital looks like a circle. Because of this cylindrical symmetry, rotation is possible around carbon-carbon bonds in open-chain molecules. In ethane, for instance, rotation around the $\\mathrm{C}-\\mathrm{C}$ bond occurs freely, constantly changing the spatial relationships between the hydrogens on one carbon and those on the other (FIGURE 3.6).\n\n\nFIGURE 3.6 Rotation occurs around the carbon-carbon single bond in ethane because of $\\boldsymbol{\\sigma}$ bond cylindrical symmetry.\nThe different arrangements of atoms that result from bond rotation are called conformations, and molecules that have different arrangements are called conformational isomers, or conformers. Unlike constitutional isomers, however, different conformers often can't be isolated because they interconvert too rapidly.\n\nConformational isomers are represented in two ways, as shown in FIGURE 3.7. A sawhorse representation views the carbon-carbon bond from an oblique angle and indicates spatial orientation by showing all $\\mathrm{C}-\\mathrm{H}$ bonds. A Newman projection views the carbon-carbon bond directly end-on and represents the two carbon atoms by a circle. Bonds attached to the front carbon are represented by lines to the center of the circle, and bonds attached to the rear carbon are represented by lines to the edge of the circle.\n\n\n\nSawhorse representation\n\n\n\nNewman\nprojection"}
{"id": 181, "contents": "Strategy - 3.6 Conformations of Ethane\nSawhorse representation\n\n\n\nNewman\nprojection\n\nFIGURE 3.7 A sawhorse representation and a Newman projection of ethane. The sawhorse representation views the molecule from an oblique angle, while the Newman projection views the molecule end-on. Note that the molecular model of the Newman projection appears at first to have six atoms attached to a single carbon. Actually, the front carbon, with three attached green atoms, is directly in front of the rear carbon, with three attached red atoms.\n\nDespite what we've just said, we actually don't observe perfectly free rotation in ethane. Experiments show that there is a small $(12 \\mathrm{~kJ} / \\mathrm{mol} ; 2.9 \\mathrm{kcal} / \\mathrm{mol})$ barrier to rotation and that some conformations are more stable than others. The lowest-energy, most stable conformation is the one in which all six $\\mathrm{C}-\\mathrm{H}$ bonds are as far away from one another as possible-staggered when viewed end-on in a Newman projection. The highestenergy, least stable conformation is the one in which the six $\\mathrm{C}-\\mathrm{H}$ bonds are as close as possible-eclipsed in a Newman projection. At any given instant, about $99 \\%$ of ethane molecules have an approximately staggered conformation and only about $1 \\%$ are near the eclipsed conformation."}
{"id": 182, "contents": "Strategy - 3.6 Conformations of Ethane\nThe extra $12 \\mathrm{~kJ} / \\mathrm{mol}$ of energy present in the eclipsed conformation of ethane is called torsional strain. Its cause has been the subject of controversy, but the major factor is an interaction between $\\mathrm{C}-\\mathrm{H}$ bonding orbitals on one carbon and antibonding orbitals on the adjacent carbon, which stabilizes the staggered conformation relative to the eclipsed one. Because a total strain of $12 \\mathrm{~kJ} / \\mathrm{mol}$ arises from three equal hydrogen-hydrogen eclipsing interactions, we can assign a value of approximately $4.0 \\mathrm{~kJ} / \\mathrm{mol}(1.0 \\mathrm{kcal} / \\mathrm{mol})$ to each single interaction. The barrier to rotation that results can be represented on a graph of potential energy versus degree of rotation, in which the angle between $\\mathrm{C}-\\mathrm{H}$ bonds on the front and back carbons as viewed end-on (the dihedral angle) goes full circle from 0 to $360^{\\circ}$. Energy minima occur at staggered conformations, and energy maxima occur at eclipsed conformations, as shown in FIGURE 3.8.\n\n\nFIGURE 3.8 A graph of potential energy versus bond rotation in ethane. The staggered conformations are $12 \\mathrm{~kJ} / \\mathrm{mol}$ lower in energy than the eclipsed conformations."}
{"id": 183, "contents": "Strategy - 3.7 Conformations of Other Alkanes\nPropane, the next-higher member in the alkane series, also has a torsional barrier that results in hindered rotation around the carbon-carbon bonds. The barrier is slightly higher in propane than in ethane-a total of 14 $\\mathrm{kJ} / \\mathrm{mol}(3.4 \\mathrm{kcal} / \\mathrm{mol})$ versus $12 \\mathrm{~kJ} / \\mathrm{mol}$.\n\nThe eclipsed conformation of propane has three interactions-two ethane-type hydrogen-hydrogen interactions and one additional hydrogen-methyl interaction. Since each eclipsing $\\mathrm{H} \\longleftrightarrow \\mathrm{H}$ interaction is the same as that in ethane and thus has an energy \"cost\" of $4.0 \\mathrm{~kJ} / \\mathrm{mol}$, we can assign a value of $14-(2 \\times 4.0)=6.0$ $\\mathrm{kJ} / \\mathrm{mol}(1.4 \\mathrm{kcal} / \\mathrm{mol})$ to the eclipsing $\\mathrm{H} \\longleftrightarrow \\mathrm{CH}_{3}$ interaction (FIGURE 3.9).\n\n\nFIGURE 3.9 Newman projections of propane showing staggered and eclipsed conformations. The staggered conformer is lower in energy by $14 \\mathrm{~kJ} / \\mathrm{mol}$."}
{"id": 184, "contents": "Strategy - 3.7 Conformations of Other Alkanes\nFIGURE 3.9 Newman projections of propane showing staggered and eclipsed conformations. The staggered conformer is lower in energy by $14 \\mathrm{~kJ} / \\mathrm{mol}$.\n\nThe conformational situation becomes more complex for larger alkanes because not all staggered conformations have the same energy and not all eclipsed conformations have the same energy. In butane, for instance, the lowest-energy arrangement, called the anti conformation, is the one in which the two methyl groups are as far apart as possible $-180^{\\circ}$ away from each other. As rotation around the $\\mathrm{C} 2-\\mathrm{C} 3$ bond occurs, an eclipsed conformation is reached where there are two $\\mathrm{CH}_{3} \\longleftrightarrow \\mathrm{H}$ interactions and one $\\mathrm{H} \\leftrightarrows \\mathrm{H}$ interaction. Using the energy values derived previously from ethane and propane, this eclipsed conformation is more strained than the anti conformation by $2 \\times 6.0 \\mathrm{~kJ} / \\mathrm{mol}+4.0 \\mathrm{~kJ} / \\mathrm{mol}$ (two $\\mathrm{CH}_{3} \\longleftrightarrow \\mathrm{H}$ interactions plus one $\\mathrm{H} \\longleftrightarrow \\mathrm{H}$ interaction), for a total of $16 \\mathrm{~kJ} / \\mathrm{mol}(3.8 \\mathrm{kcal} / \\mathrm{mol})$.\n\n\n\nButane-anti conformation ( $0 \\mathrm{~kJ} / \\mathrm{mol}$ )"}
{"id": 185, "contents": "Strategy - 3.7 Conformations of Other Alkanes\nButane-anti conformation ( $0 \\mathrm{~kJ} / \\mathrm{mol}$ )\n\nButane-eclipsed\nconformation\n$(16 \\mathrm{~kJ} / \\mathrm{mol})$\nAs bond rotation continues, an energy minimum is reached at the staggered conformation where the methyl groups are $60^{\\circ}$ apart. Called the gauche conformation, it lies $3.8 \\mathrm{~kJ} / \\mathrm{mol}(0.9 \\mathrm{kcal} / \\mathrm{mol})$ higher in energy than the anti conformation even though it has no eclipsing interactions. This energy difference occurs because the hydrogen atoms of the methyl groups are near one another in the gauche conformation, resulting in what is called steric strain. Steric strain is the repulsive interaction that occurs when atoms are forced closer together than their atomic radii allow. It's the result of trying to force two atoms to occupy the same space.\n\n\nAs the dihedral angle between the methyl groups approaches zero, an energy maximum is reached at a second eclipsed conformation. Because the methyl groups are forced even closer together than in the gauche conformation, both torsional strain and steric strain are present. A total strain energy of $19 \\mathrm{~kJ} / \\mathrm{mol}(4.5 \\mathrm{kcal} /$ mol ) has been estimated for this conformation, making it possible to calculate a value of $11 \\mathrm{~kJ} / \\mathrm{mol}(2.6 \\mathrm{kcal} /$ mol ) for the $\\mathrm{CH}_{3} \\longleftrightarrow \\mathrm{CH}_{3}$ eclipsing interaction: total strain of $19 \\mathrm{~kJ} / \\mathrm{mol}$ minus the strain of two $\\mathrm{H} \\longleftrightarrow \\mathrm{H}$ eclipsing interactions ( $2 \\times 4.0 \\mathrm{kcal} / \\mathrm{mol}$ ) equals $11 \\mathrm{~kJ} / \\mathrm{mol}$."}
{"id": 186, "contents": "Strategy - 3.7 Conformations of Other Alkanes\nAfter $0^{\\circ}$, the rotation becomes a mirror image of what we've already seen: another gauche conformation is reached, another eclipsed conformation, and finally a return to the anti conformation. A plot of potential energy versus rotation about the $\\mathrm{C} 2-\\mathrm{C} 3$ bond is shown in FIGURE 3.10.\n\n\nFIGURE 3.10 A plot of potential energy versus rotation for the C2-C3 bond in butane. The energy maximum occurs when the two methyl groups eclipse each other, and the energy minimum occurs when the two methyl groups are $180^{\\circ}$ apart (anti).\n\nThe notion of assigning definite energy values to specific interactions within a molecule is very useful, and we'll return to it in the next chapter. A summary of what we've seen thus far is given in TABLE 3.5.\n\nThe same principles just developed for butane apply to pentane, hexane, and all higher alkanes. The most favorable conformation for any alkane has the carbon-carbon bonds in staggered arrangements, with large substituents arranged anti to one another. A generalized alkane structure is shown in FIGURE 3.11.\n\n| Interaction | Cause | Energy cost |  |\n| :---: | :---: | :---: | :---: |\n|  |  | ( $\\mathrm{kJ} / \\mathrm{mol}$ ) | ( $\\mathrm{kcal} / \\mathrm{mol}$ ) |\n| $\\mathrm{H} \\leftrightarrow \\mathrm{H}$ eclipsed | Torsional strain | 4.0 | 1.0 |\n| $\\mathrm{H} \\leftrightarrow \\mathrm{CH}_{3}$ eclipsed | Mostly torsional strain | 6.0 | 1.4 |\n| $\\mathrm{CH}_{3} \\longleftrightarrow \\mathrm{CH}_{3}$ eclipsed | Torsional and steric strain | 11.0 | 2.6 |\n| $\\mathrm{CH}_{3} \\longleftrightarrow \\mathrm{CH}_{3}$ gauche | Steric strain | 3.8 | 0.9 |"}
{"id": 187, "contents": "Strategy - 3.7 Conformations of Other Alkanes\nFIGURE 3.11 The most stable alkane conformation is the one in which all substituents are staggered and the carbon-carbon bonds are arranged anti, as shown in this model of decane.\nOne final point: saying that one particular conformer is \"more stable\" than another doesn't mean the molecule adopts and maintains only the more stable conformation. At room temperature, rotations around $\\sigma$ bonds occur so rapidly that all conformers are in equilibrium. At any given instant, however, a larger percentage of molecules will be found in a more stable conformation than in a less stable one."}
{"id": 188, "contents": "Newman Projections - \nSight along the C1-C2 bond of 1-chloropropane, and draw Newman projections of the most stable and least stable conformations."}
{"id": 189, "contents": "Strategy - \nThe most stable conformation of a substituted alkane is generally a staggered one in which large groups have an anti relationship. The least stable conformation is generally an eclipsed one in which large groups are as close as possible."}
{"id": 190, "contents": "Solution - \nMost stable (staggered)\n\n\nLeast stable (eclipsed)\n\nPROBLEM Make a graph of potential energy versus angle of bond rotation for propane, and assign values to the 3-15 energy maxima.\n\nPROBLEM Sight along the C2-C1 bond of 2-methylpropane (isobutane).\n3-16 (a) Draw a Newman projection of the most stable conformation.\n(b) Draw a Newman projection of the least stable conformation.\n(c) Make a graph of energy versus angle of rotation around the $\\mathrm{C} 2-\\mathrm{C} 1$ bond.\n(d) Assign relative values to the maxima and minima in your graph, given that an $\\mathrm{H} \\longleftrightarrow \\mathrm{H}$ eclipsing interaction costs $4.0 \\mathrm{~kJ} / \\mathrm{mol}$ and an $\\mathrm{H} \\longleftrightarrow \\mathrm{CH}_{3}$ eclipsing interaction costs $6.0 \\mathrm{~kJ} / \\mathrm{mol}$.\n\nPROBLEM Sight along the C2-C3 bond of 2,3-dimethylbutane, and draw a Newman projection of the most\n3-17 stable conformation.\nPROBLEM Draw a Newman projection along the C2-C3 bond of the following conformation of 3-18 2,3-dimethylbutane, and calculate a total strain energy:"}
{"id": 191, "contents": "Gasoline - \nBritish Foreign Minister Ernest Bevin once said that \"The Kingdom of Heaven runs on righteousness, but the Kingdom of Earth runs on alkanes.\" (Actually, he said \"runs on oil\" not \"runs on alkanes,\" but they're essentially the same.) By far, the major sources of alkanes are the world's natural gas and petroleum deposits. Laid down eons ago, these deposits are thought to be derived primarily from the decomposition of tiny singlecelled marine organisms called foraminifera. Natural gas consists chiefly of methane but also contains ethane, propane, and butane. Petroleum is a complex mixture of hydrocarbons that must be separated into fractions and then further refined before it can be used.\n\n\nFIGURE 3.12 Gasoline is a finite resource. It won't be around forever. (credit: \"The first oil well\" (https://www.loc.gov/item/ 2010649522/) by Unknown/Library of Congress)\n\nThe petroleum era began in August 1859, when the world's first oil well was drilled by Edwin Drake near Titusville, Pennsylvania. The petroleum was distilled into fractions according to boiling point, but it was highboiling kerosene, or lamp oil, rather than gasoline that was primarily sought. Literacy was becoming widespread at the time, and people wanted better light for reading than was available from candles. Gasoline was too volatile for use in lamps and was initially considered a waste by-product. The world has changed greatly since those early days, however, and it is now gasoline rather than lamp oil that is prized.\n\nPetroleum refining begins by fractional distillation of crude oil into three principal cuts according to boiling point (bp): straight-run gasoline (bp 30-200 ${ }^{\\circ} \\mathrm{C}$ ), kerosene (bp $175-300^{\\circ} \\mathrm{C}$ ), and heating oil, or diesel fuel (bp $275-400{ }^{\\circ} \\mathrm{C}$ ). Further distillation under reduced pressure then yields lubricating oils and waxes and leaves a tarry residue of asphalt. The distillation of crude oil is only the first step in gasoline production, however. Straight-run gasoline turns out to be a poor fuel in automobiles because of engine knock, an uncontrolled combustion that can occur in a hot engine causing potentially serious damage."}
{"id": 192, "contents": "Gasoline - \nThe octane number of a fuel is the measure by which its antiknock properties are judged. It was recognized long ago that straight-chain hydrocarbons are far more prone to inducing engine knock than highly branched compounds. Heptane, a particularly bad fuel, is assigned a base value of 0 octane number, and 2,2,4-trimethylpentane, commonly known as isooctane, has a rating of 100 .\n\n\nHeptane\n(octane number $=\\mathbf{0}$ )\n\n\n2,2,4-Trimethylpentane\n(octane number $=100$ )\n\nBecause straight-run gasoline burns so poorly in engines, petroleum chemists have devised numerous methods for producing higher-quality fuels. One of these methods, catalytic cracking, involves taking the high-boiling kerosene cut ( $\\mathrm{C}_{11}-\\mathrm{C}_{14}$ ) and \"cracking\" it into smaller branched molecules suitable for use in gasoline. Another process, called reforming, is used to convert $\\mathrm{C}_{6}-\\mathrm{C}_{8}$ alkanes to aromatic compounds such as benzene and toluene, which have substantially higher octane numbers than alkanes. The final product that goes in your tank has an approximate composition of $15 \\% \\mathrm{C}_{4}-\\mathrm{C}_{8}$ straight-chain alkanes, $25 \\%$ to $40 \\% \\mathrm{C}_{4}-\\mathrm{C}_{10}$ branched-chain alkanes, $10 \\%$ cyclic alkanes, $10 \\%$ straight-chain and cyclic alkenes, and $25 \\%$ arenes (aromatics)."}
{"id": 193, "contents": "Key Terms - \n```\n- alcohol\nester\n- aldehyde\n- ether\n- aliphatic\n- functional group\n- alkane\n- gauche conformation\n- alkene\n- hydrocarbon\n- alkyl group\n- isomer\n- alkyl halide\n- ketone\n- alkyne\n- Newman projection\n- amide\n- nitrile\n- amine\n- R group\n- anti conformation\n- saturated\n- arene\n- sawhorse representation\n- branched-chain alkane\n- staggered conformation\n- carbonyl group\n- stereochemistry\n- carboxylic acid\n- steric strain\n- conformation\n- straight-chain alkane\n- conformational isomer\n- substituent\n- conformer\n- sulfide\n- constitutional isomer\n- thiol\n- eclipsed conformation\n- torsional strain\n```"}
{"id": 194, "contents": "Summary - \nAlkanes are relatively unreactive and rarely involved in chemical reactions, but they nevertheless provide a useful vehicle for introducing some important general ideas. In this chapter, we've used alkanes to introduce the basic approach to naming organic compounds and to take an initial look at some of the three-dimensional aspects of molecules.\n\nA functional group is a group of atoms within a larger molecule that has a characteristic chemical reactivity. Because functional groups behave in approximately the same way in all molecules where they occur, the chemical reactions of an organic molecule are largely determined by its functional groups.\n\n[^0]structures are called isomers. More specifically, compounds such as butane and isobutane, which differ in their connections between atoms, are called constitutional isomers.\n\nCarbon-carbon single bonds in alkanes are formed by $\\sigma$ overlap of carbon $s p^{3}$ hybrid orbitals. Rotation is possible around $\\sigma$ bonds because of their cylindrical symmetry, and alkanes therefore exist in a large number of rapidly interconverting conformations. Newman projections make it possible to visualize the spatial consequences of bond rotation by sighting directly along a carbon-carbon bond axis. Not all alkane conformations are equally stable. The staggered conformation of ethane is $12 \\mathrm{~kJ} / \\mathrm{mol}(2.9 \\mathrm{kcal} / \\mathrm{mol}) \\mathrm{more}$ stable than the eclipsed conformation because of torsional strain. In general, any alkane is most stable when all its bonds are staggered."}
{"id": 195, "contents": "Visualizing Chemistry - \nPROBLEM Identify the functional groups in the following substances, and convert each drawing into a 3-19 molecular formula (red $=0$, blue $=\\mathrm{N}$ ).\n(a)\n\nPhenylalanine\n(b)\n\n\nLidocaine\nPROBLEM Give IUPAC names for the following alkanes, and convert each drawing into a skeletal structure.\n\n3-20 (a)\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM Draw a Newman projection along the C2-C3 bond of the following conformation of 2-butanol.\n3-21"}
{"id": 196, "contents": "Functional Groups - \nPROBLEM Locate and identify the functional groups in the following molecules.\n\n3-22 (a)\n\n(e)\n\n\n(f)\n\n\n\n(c)\n(d)\n\n\n\n(d)\n\n\n\n\nPROBLEM Propose structures that meet the following descriptions:\n3-23 (a) A ketone with five carbons (b) A four-carbon amide (c) A five-carbon ester\n(d) An aromatic aldehyde\n(e) A keto ester\n(f) An amino alcohol\n\nPROBLEM Propose structures for the following:\n3-24 (a) A ketone, $\\mathrm{C}_{4} \\mathrm{H}_{8} \\mathrm{O}$ (b) A nitrile, $\\mathrm{C}_{5} \\mathrm{H}_{9} \\mathrm{~N}$ (c) A dialdehyde, $\\mathrm{C}_{4} \\mathrm{H}_{6} \\mathrm{O}_{2}$\n(d) A bromoalkene, $\\mathrm{C}_{6} \\mathrm{H}_{11} \\mathrm{Br}$ (e) An alkane, $\\mathrm{C}_{6} \\mathrm{H}_{14}$ (f) cyclic saturated hydrocarbon, $\\mathrm{C}_{6} \\mathrm{H}_{12}$\n(g) A diene (dialkene), $\\mathrm{C}_{5} \\mathrm{H}_{8}$\n(h) A keto alkene, $\\mathrm{C}_{5} \\mathrm{H}_{8} \\mathrm{O}$\n\nPROBLEM Predict the hybridization of the carbon atom in each of the following functional groups:\n3-25 (a) Ketone\n(b) Nitrile\n(c) Carboxylic acid"}
{"id": 197, "contents": "Functional Groups - \nPROBLEM Predict the hybridization of the carbon atom in each of the following functional groups:\n3-25 (a) Ketone\n(b) Nitrile\n(c) Carboxylic acid\n\nPROBLEM Draw the structures of the following molecules:\n3-26 (a) Biacetyl, $\\mathrm{C}_{4} \\mathrm{H}_{6} \\mathrm{O}_{2}$, a substance with the aroma of butter; it contains no rings or carbon-carbon multiple bonds.\n(b) Ethylenimine, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{~N}$, a substance used in the synthesis of melamine polymers; it contains no multiple bonds.\n(c) Glycerol, $\\mathrm{C}_{3} \\mathrm{H}_{8} \\mathrm{O}_{3}$, a substance isolated from fat and used in cosmetics; it has an-OH group on each carbon."}
{"id": 198, "contents": "Isomers - \nPROBLEM Draw structures that meet the following descriptions (there are many possibilities):\n3-27 (a) Three isomers with the formula $\\mathrm{C}_{8} \\mathrm{H}_{18}$ (b) Two isomers with the formula $\\mathrm{C}_{4} \\mathrm{H}_{8} \\mathrm{O}_{2}$\nPROBLEM Draw structures of the nine isomers of $\\mathrm{C}_{7} \\mathrm{H}_{16}$.\n3-28\nPROBLEM In each of the following sets, which structures represent the same compound and which represent 3-29 different compounds?\n(a)\n\n\n\n(b)\n\n\n\n(c)\n\n\n\n\nPROBLEM Seven constitutional isomers have the formula $\\mathrm{C}_{4} \\mathrm{H}_{10}$ O. Draw as many as you can.\n3-30\nPROBLEM Draw as many compounds as you can that fit the following descriptions:\n3-31 (a) Alcohols with formula $\\mathrm{C}_{4} \\mathrm{H}_{10} \\mathrm{O}$\n(b) Amines with formula $\\mathrm{C}_{5} \\mathrm{H}_{13} \\mathrm{~N}$\n(c) Ketones with formula $\\mathrm{C}_{5} \\mathrm{H}_{10} \\mathrm{O}$\n(d) Aldehydes with formula $\\mathrm{C}_{5} \\mathrm{H}_{10} \\mathrm{O}$\n(e) Esters with formula $\\mathrm{C}_{4} \\mathrm{H}_{8} \\mathrm{O}_{2}$ (f) Ethers with formula $\\mathrm{C}_{4} \\mathrm{H}_{10} \\mathrm{O}$\n\nPROBLEM Draw compounds that contain the following:\n3-32 (a) A primary alcohol (b) A tertiary nitrile (c) A secondary thiol\n(d) Both primary and secondary alcohols (e) An isopropyl group (f) A quaternary carbon"}
{"id": 199, "contents": "Isomers - \nNaming Compounds\nPROBLEM Draw and name all monobromo derivatives of pentane, $\\mathrm{C}_{5} \\mathrm{H}_{11} \\mathrm{Br}$.\n3-33\nPROBLEM Draw and name all monochloro derivatives of 2,5-dimethylhexane, $\\mathrm{C}_{8} \\mathrm{H}_{17} \\mathrm{Cl}$.\n3-34\nPROBLEM Draw structures for the following:\n3-35 (a) 2-Methylheptane (b) 4-Ethyl-2,2-dimethylhexane (c) 4-Ethyl-3,4-dimethyloctane\n(d) 2,4,4-Trimethylheptane (e) 3,3-Diethyl-2,5-dimethylnonane\n(f) 4-Isopropyl-3-methylheptane\n\nPROBLEM Draw a compound that:\n3-36 (a) Has only primary and tertiary carbons (b) Has no secondary or tertiary carbons\n(c) Has no secondary or tertiary carbons\n\nPROBLEM Draw a compound that:\n3-37 (a) Has nine primary hydrogens (b) Has only primary hydrogens\nPROBLEM Give IUPAC names for the following compounds:\n3-38 (a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n\nPROBLEM Name the five isomers of $\\mathrm{C}_{6} \\mathrm{H}_{14}$.\n3-39\nPROBLEM Explain why each of the following names is incorrect:\n3-40 (a) 2,2-Dimethyl-6-ethylheptane\n(b) 4-Ethyl-5,5-dimethylpentane\n(c) 3-Ethyl-4,4-dimethylhexane (d) 5,5,6-Trimethyloctane\n(e) 2-Isopropyl-4-methylheptane\n\nPROBLEM Propose structures and give IUPAC names for the following:\n3-41 (a) A diethyldimethylhexane (b) A (3-methylbutyl)-substituted alkane"}
{"id": 200, "contents": "Conformations - \nPROBLEM Consider 2-methylbutane (isopentane). Sighting along the C2-C3 bond:\n3-42 (a) Draw a Newman projection of the most stable conformation.\n(b) Draw a Newman projection of the least stable conformation.\n(c) If a $\\mathrm{CH}_{3} \\longleftrightarrow \\mathrm{CH}_{3}$ eclipsing interaction costs $11 \\mathrm{~kJ} / \\mathrm{mol}(2.5 \\mathrm{kcal} / \\mathrm{mol})$ and a $\\mathrm{CH}_{3} \\longleftrightarrow \\mathrm{CH}_{3}$ gauche interaction costs $3.8 \\mathrm{~kJ} / \\mathrm{mol}(0.9 \\mathrm{kcal} / \\mathrm{mol})$, make a quantitative plot of energy versus rotation about the $\\mathrm{C} 2-\\mathrm{C} 3$ bond.\n\nPROBLEM What are the relative energies of the three possible staggered conformations around the C2-C3\n3-43 bond in 2,3-dimethylbutane? (See Problem 3-42.)\nPROBLEM Construct a qualitative potential-energy diagram for rotation about the $\\mathrm{C}-\\mathrm{C}$ bond of 3-44 1,2-dibromoethane. Which conformation would you expect to be most stable? Label the anti and gauche conformations of 1,2-dibromoethane.\n\nPROBLEM Which conformation of 1,2-dibromoethane (Problem 3-44) would you expect to have the largest\n3-45 dipole moment? The observed dipole moment of 1,2-dibromoethane is $\\mu=1.0 \\mathrm{D}$. What does this tell you about the actual conformation of the molecule?\n\nPROBLEM Draw the most stable conformation of pentane, using wedges and dashes to represent bonds\n3-46 coming out of the paper and going behind the paper, respectively.\nPROBLEM Draw the most stable conformation of 1,4-dichlorobutane, using wedges and dashes to represent\n3-47 bonds coming out of the paper and going behind the paper, respectively."}
{"id": 201, "contents": "General Problems - \nPROBLEM For each of the following compounds, draw an isomer that has the same functional groups.\n\n3-48 (a)\n\n(b)\n\n(c) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{C} \\equiv \\mathrm{N}$\n(d)\n\n(e) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CHO}$\n(f)\n\n\nPROBLEM Malic acid, $\\mathrm{C}_{4} \\mathrm{H}_{6} \\mathrm{O}_{5}$, has been isolated from apples. Because this compound reacts with 2 molar 3-49 equivalents of base, it is a dicarboxylic acid.\n(a) Draw at least five possible structures.\n(b) If malic acid is a secondary alcohol, what is its structure?\n\nPROBLEM Formaldehyde, $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{O}$, is known to all biologists because of its usefulness as a tissue preservative.\n3-50 When pure, formaldehyde trimerizes to give trioxane, $\\mathrm{C}_{3} \\mathrm{H}_{6} \\mathrm{O}_{3}$, which, surprisingly enough, has no carbonyl groups. Only one monobromo derivative $\\left(\\mathrm{C}_{3} \\mathrm{H}_{5} \\mathrm{BrO}_{3}\\right)$ of trioxane is possible. Propose a structure for trioxane.\n\nPROBLEM The barrier to rotation about the $\\mathrm{C}-\\mathrm{C}$ bond in bromoethane is $15 \\mathrm{~kJ} / \\mathrm{mol}(3.6 \\mathrm{kcal} / \\mathrm{mol})$.\n3-51 (a) What energy value can you assign to an $\\mathrm{H} \\leftrightarrow \\mathrm{Br}$ eclipsing interaction?\n(b) Construct a quantitative diagram of potential energy versus bond rotation for bromoethane."}
{"id": 202, "contents": "General Problems - \nPROBLEM Increased substitution around a bond leads to increased strain. Take the four substituted butanes\n3-52 listed below, for example. For each compound, sight along the C2-C3 bond and draw Newman projections of the most stable and least stable conformations. Use the data in Table 3.5 to assign strain-energy values to each conformation. Which of the eight conformations is most strained? Which is least strained?\n(a) 2-Methylbutane\n(b) 2,2-Dimethylbutane\n(c) 2,3-Dimethylbutane\n(d) 2,2,3-Trimethylbutane\n\nPROBLEM The cholesterol-lowering agents called statins, such as simvastatin (Zocor) and pravastatin\n3-53 (Pravachol), are among the most widely prescribed drugs in the world, with annual sales estimated at approximately $\\$ 25$ billion. Identify the functional groups in both, and tell how the two substances differ.\n\n\nSimvastatin (Zocor)\n\n\nPravastatin\n(Pravachol)\n\nPROBLEM In the next chapter we'll look at cycloalkanes-saturated cyclic hydrocarbons-and we'll see that\n3-54 the molecules generally adopt puckered, nonplanar conformations. Cyclohexane, for instance, has a puckered shape like a lounge chair rather than a flat shape. Why?\n\n\nNonplanar cyclohexane\n\n\nPlanar cyclohexane\n\nPROBLEM We'll see in the next chapter that there are two isomeric substances, both named\n3-55 1,2-dimethylcyclohexane. Explain."}
{"id": 203, "contents": "1,2-Dimethylcyclohexane - \n104 3 \u2022 Additional Problems"}
{"id": 204, "contents": "CHAPTER 4 - \nOrganic Compounds: Cycloalkanes and Their Stereochemistry\n\n\nFIGURE 4.1 The musk gland of the male Himalayan musk deer secretes a substance once used in perfumery that contains cycloalkanes of 14 to $\\mathbf{1 8}$ carbons. (credit: modification of work \"Siberian musk deer in the tiaga\" by ErikAdamsson/Wikimedia Commons, CCO 1.0)"}
{"id": 205, "contents": "CHAPTER CONTENTS - \n4.1 Naming Cycloalkanes\n4.2 Cis-Trans Isomerism in Cycloalkanes\n4.3 Stability of Cycloalkanes: Ring Strain\n4.4 Conformations of Cycloalkanes\n4.5 Conformations of Cyclohexane\n4.6 Axial and Equatorial Bonds in Cyclohexane\n4.7 Conformations of Monosubstituted Cyclohexanes\n4.8 Conformations of Disubstituted Cyclohexanes\n4.9 Conformations of Polycyclic Molecules\n\nWHY THIS CHAPTER? We'll see numerous instances in future chapters where the chemistry of a given functional group is affected by being in a ring rather than an open chain. Because cyclic molecules are encountered in most pharmaceuticals and in all classes of biomolecules, including proteins, lipids, carbohydrates, and nucleic acids, it's important to understand the behavior of cyclic structures.\n\nAlthough we've only discussed open-chain compounds up to now, most organic compounds contain rings of carbon atoms. Chrysanthemic acid, for instance, whose esters occur naturally as the active insecticidal constituents of chrysanthemum flowers, contains a three-membered (cyclopropane) ring."}
{"id": 206, "contents": "Chrysanthemic acid - \nProstaglandins, potent hormones that control an extraordinary variety of physiological functions in humans, contain a five-membered (cyclopentane) ring.\n\n\nProstaglandin $\\mathrm{E}_{1}$\n\nSteroids, such as cortisone, contain four rings joined together-three six-membered (cyclohexane) and one fivemembered. We'll discuss steroids and their properties in more detail in Sections 27.6 and 27.7."}
{"id": 207, "contents": "Cortisone - 4.1 Naming Cycloalkanes\nSaturated cyclic hydrocarbons are called cycloalkanes, or alicyclic compounds (aliphatic cyclic). Because cycloalkanes consist of rings of $-\\mathrm{CH}_{2}$-units, they have the general formula $\\left(\\mathrm{CH}_{2}\\right)_{n}$, or $\\mathrm{C}_{n} \\mathrm{H}_{2 n}$, and can be represented by polygons in skeletal drawings.\n\n\nSubstituted cycloalkanes are named by rules similar to those we saw in (Section 3.4) for open-chain alkanes. For most compounds, there are only two steps."}
{"id": 208, "contents": "Find the parent. - \nCount the number of carbon atoms in the ring and the number in the largest substituent. If the number of carbon atoms in the ring is equal to or greater than the number in the substituent, the compound is named as an alkyl-substituted cycloalkane. If the number of carbon atoms in the largest substituent is greater than the number in the ring, the compound is named as a cycloalkyl-substituted alkane. For example:\n\n\nMethylcyclopentane\n\n\n3 carbons 4 carbons"}
{"id": 209, "contents": "Number the substituents, and write the name. - \nFor an alkyl- or halo-substituted cycloalkane, choose a point of attachment as carbon 1 and number the substituents on the ring so that the second substituent has as low a number as possible. If ambiguity still exists, number so that the third or fourth substituent has as low a number as possible, until a point of difference is found.\n\n\n\n\n\nNOT\n\n\n\n\n1-Ethyl-2,6-dimethylcycloheptane\n\nHigher\n\n\n3-Ethyl-1,4-dimethylcycloheptane\n$\\uparrow_{\\text {Higher }}^{\\uparrow}$\n(a) When two or more different alkyl groups are present that could potentially take the same numbers, number them by alphabetical priority, ignoring numerical prefixes such as di- and tri-.\n\n\n1-Ethyl-2-methylcyclopentane\n\nNOT\n\n\n2-Ethyl-1-methylcyclopentane\n(b) If halogens are present, treat them just like alkyl groups.\n\n\n1-Bromo-2-methylcyclobutane\n\n\n2-Bromo-1-methylcyclobutane\n\nSome additional examples follow:\n\n\n1-Bromo-3-ethyl-5-methylcyclohexane\n\n(1-Methylpropyl)cyclobutane or sec-butylcyclobutane\n\n\n1-Chloro-3-ethyl-2-methylcyclopentane\n\nPROBLEM Give IUPAC names for the following cycloalkanes:\n4-1 (a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n\nPROBLEM Draw structures corresponding to the following IUPAC names:\n4-2 (a) 1,1-Dimethylcyclooctane\n(b) 3-Cyclobutylhexane\n(c) 1,2-Dichlorocyclopentane\n(d) 1,3-Dibromo-5-methylcyclohexane\n\nPROBLEM Name the following cycloalkane:"}
{"id": 210, "contents": "4-3 - 4.2 Cis-Trans Isomerism in Cycloalkanes\nIn many respects, the chemistry of cycloalkanes is like that of open-chain alkanes: both are nonpolar and fairly inert. There are, however, some important differences. One difference is that cycloalkanes are less flexible than open-chain alkanes. In contrast with the relatively free rotation around single bonds in open-chain alkanes (Section 3.6 and Section 3.7 ), there is much less freedom in cycloalkanes. Cyclopropane, for example, must be a rigid, planar molecule because three points (the carbon atoms) define a plane. No bond rotation can take place around a cyclopropane carbon-carbon bond without breaking open the ring (FIGURE 4.2).\n(a)\n\n\n\n(b)\n\n\n\nFIGURE 4.2 Bond rotation in ethane and cyclopropane. (a) Rotation occurs around the carbon-carbon bond in ethane, but (b) no rotation is possible around the carbon-carbon bonds in cyclopropane without breaking open the ring.\n\nLarger cycloalkanes have increasing rotational freedom, and very large rings ( $\\mathrm{C}_{25}$ and up) are so floppy that they are nearly indistinguishable from open-chain alkanes. The common ring sizes ( $\\mathrm{C}_{3}-\\mathrm{C}_{7}$ ), however, are severely restricted in their molecular motions.\n\nBecause of their cyclic structures, cycloalkanes have two faces when viewed edge-on, a \"top\" face and a \"bottom\" face. As a result, isomerism is possible in substituted cycloalkanes. For example, there are two different 1,2-dimethylcyclopropane isomers, one with the two methyl groups on the same face of the ring and one with the methyl groups on opposite faces (FIGURE 4.3). Both isomers are stable compounds, and neither can be converted into the other without breaking and reforming chemical bonds.\n\ncis-1,2-Dimethylcyclopropane\n\ntrans-1,2-Dimethylcyclopropane"}
{"id": 211, "contents": "4-3 - 4.2 Cis-Trans Isomerism in Cycloalkanes\ncis-1,2-Dimethylcyclopropane\n\ntrans-1,2-Dimethylcyclopropane\n\nFIGURE 4.3 There are two different 1,2-dimethylcyclopropane isomers, one with the methyl groups on the same face of the ring (cis) and the other with the methyl groups on opposite faces of the ring (trans). The two isomers do not interconvert.\nUnlike the constitutional isomers butane and isobutane, which have their atoms connected in a different order (Section 3.2), the two 1,2-dimethylcyclopropanes have the same order of connections but differ in the spatial orientation of the atoms. Such compounds, with atoms connected in the same order but differing in threedimensional orientation, are called stereochemical isomers, or stereoisomers. As we saw in Section 3.6, the term stereochemistry is used generally to refer to the three-dimensional aspects of structure and reactivity."}
{"id": 212, "contents": "Constitutional isomers (different connections between atoms) - \nand $\\mathrm{CH}_{3}-\\mathrm{CH}_{2}-\\mathrm{CH}_{2}-\\mathrm{CH}_{3}$\n\nStereoisomers (same connections but different threedimensional geometry)\n\nand\n\n\nThe 1,2-dimethylcyclopropanes are members of a subclass of stereoisomers called cis-trans isomers. The prefixes cis- (Latin \"on the same side\") and trans- (Latin \"across\") are used to distinguish between them. Cis-trans isomerism is a common occurrence in substituted cycloalkanes and in many cyclic biological molecules.\n\ncis-1,3-Dimethylcyclobutane\n\ntrans-1-Bromo-3-ethylcyclopentane"}
{"id": 213, "contents": "Naming Cycloalkanes - \nName the following substances, including the cis- or trans- prefix:\n(a)\n\n(b)"}
{"id": 214, "contents": "Strategy - \nIn these views, the ring is roughly in the plane of the page, a wedged bond protrudes out of the page, and a dashed bond recedes into the page. Two substituents are cis if they are both out of or both into the page, and they are trans if one is out of and one is into the page."}
{"id": 215, "contents": "Solution - \n(a) trans-1,3-Dimethylcyclopentane\n(b) cis-1,2-Dichlorocyclohexane\n\nPROBLEM Name the following substances, including the cis- or trans- prefix:\n4-4 (a)\n\n(b)\n\n\nPROBLEM Draw the structures of the following molecules:\n4-5 (a) trans-1-Bromo-3-methylcyclohexane (b) cis-1,2-Dimethylcyclobutane\n(c) trans-1-tert-Butyl-2-ethylcyclohexane\n\nPROBLEM Prostaglandin $\\mathrm{F}_{2 \\alpha}$, a hormone that causes uterine contraction during childbirth, has the following\n4-6 structure. Are the two hydroxyl groups $(-\\mathrm{OH})$ on the cyclopentane ring cis or trans to each other? What about the two carbon chains attached to the ring?\n\n\nProstaglandin $\\mathrm{F}_{2 \\alpha}$\n\nPROBLEM Name the following substances, including the cis- or trans- prefix (red-brown $=\\mathrm{Br}$ ):\n4-7 (a)\n\n(b)"}
{"id": 216, "contents": "Solution - 4.3 Stability of Cycloalkanes: Ring Strain\nChemists in the late 1800s knew that cyclic molecules existed, but the limitations on ring size were unclear. Although numerous compounds containing five-membered and six-membered rings were known, smaller and larger ring sizes had not been prepared, despite many attempts.\n\nA theoretical interpretation of this observation was proposed in 1885 by Adolf von Baeyer, who suggested that small and large rings might be unstable due to angle strain-the strain induced in a molecule when bond angles are forced to deviate from the ideal $109^{\\circ}$ tetrahedral value. Baeyer based his suggestion on the simple geometric notion that a three-membered ring (cyclopropane) should be an equilateral triangle with bond angles of $60^{\\circ}$ rather than $109^{\\circ}$, a four-membered ring (cyclobutane) should be a square with bond angles of $90^{\\circ}$, a fivemembered ring should be a regular pentagon with bond angles of $108^{\\circ}$, and so on. Continuing this argument, large rings should be strained by having bond angles that are much greater than $109^{\\circ}$.\n\n\nWhat are the facts? To measure the amount of strain in a compound, we have to measure the total energy of the compound and then subtract the energy of a strain-free reference compound. The difference between the two values should represent the amount of extra energy in the molecule due to strain. The simplest experimental way to do this for a cycloalkane is to measure its heat of combustion, the amount of heat released when the compound burns completely with oxygen. The more energy (strain) the compound contains, the more energy (heat) is released by combustion.\n\n$$\n\\left(\\mathrm{CH}_{2}\\right)_{n}+3 n / 2 \\mathrm{O}_{2} \\longrightarrow n \\mathrm{CO}_{2}+n \\mathrm{H}_{2} \\mathrm{O}+\\text { Heat (add) }\n$$"}
{"id": 217, "contents": "Solution - 4.3 Stability of Cycloalkanes: Ring Strain\nBecause the heat of combustion of a cycloalkane depends on size, we need to look at heats of combustion per $\\mathrm{CH}_{2}$ unit. Subtracting a reference value derived from a strain-free acyclic alkane and then multiplying by the number of $\\mathrm{CH}_{2}$ units in the ring gives the overall strain energy. FIGURE 4.4 shows the results.\n\n\nFIGURE 4.4 Cycloalkane strain energies, as calculated by taking the difference between cycloalkane heat of combustion per $\\mathbf{C H}_{2}$ and acyclic alkane heat of combustion per $\\mathbf{C H}_{2}$, and multiplying by the number of $\\mathbf{C H}_{2}$ units in a ring. Small and medium rings are strained, but cyclohexane rings and very large rings are strain-free.\n\nThe data in FIGURE 4.4 show that Baeyer's theory is only partially correct. Cyclopropane and cyclobutane are indeed strained, just as predicted, but cyclopentane is more strained than predicted, and cyclohexane is strainfree. Cycloalkanes of intermediate size have only modest strain, and rings of 14 carbons or more are strain-free. Why is Baeyer's theory wrong?\n\nBaeyer's theory is wrong for the simple reason that he assumed all cycloalkanes to be flat. In fact, as we'll see in the next section, most cycloalkanes are not flat; instead, they adopt puckered three-dimensional conformations that allow bond angles to be nearly tetrahedral. As a result, angle strain occurs only in three- and fourmembered rings, which have little flexibility. For most ring sizes, particularly the medium-ring ( $\\mathrm{C}_{7}-\\mathrm{C}_{11}$ ) cycloalkanes, torsional strain caused by $\\mathrm{H} \\leftrightarrows \\mathrm{H}$ eclipsing interactions at adjacent carbons (Section 3.6) and steric strain caused by the repulsion between nonbonded atoms that approach too closely (Section 3.7) are the most important factors. Thus, three kinds of strain contribute to the overall energy of a cycloalkane."}
{"id": 218, "contents": "Solution - 4.3 Stability of Cycloalkanes: Ring Strain\n- Angle strain-the strain due to expansion or compression of bond angles\n- Torsional strain-the strain due to eclipsing of bonds between neighboring atoms\n- Steric strain-the strain due to repulsive interactions when atoms approach each other too closely\n\nPROBLEM Each $\\mathrm{H} \\longleftrightarrow \\mathrm{H}$ eclipsing interaction in ethane costs about $4.0 \\mathrm{~kJ} / \\mathrm{mol}$. How many such interactions $\\mathbf{4 - 8}$ are present in cyclopropane? What fraction of the overall $115 \\mathrm{~kJ} / \\mathrm{mol}(27.5 \\mathrm{kcal} / \\mathrm{mol})$ strain energy of cyclopropane is due to torsional strain?\n\nPROBLEM cis-1,2-Dimethylcyclopropane has more strain than trans-1,2-dimethylcyclopropane. How can you 4-9 account for this difference? Which of the two compounds is more stable?"}
{"id": 219, "contents": "Cyclopropane - \nCyclopropane is the most strained of all rings, primarily due to the angle strain caused by its $60^{\\circ} \\mathrm{C}-\\mathrm{C}-\\mathrm{C}$ bond angles. In addition, cyclopropane has considerable torsional strain because the $\\mathrm{C}-\\mathrm{H}$ bonds on neighboring carbon atoms are eclipsed (FIGURE 4.5).\n\n\nFIGURE 4.5 The structure of cyclopropane, showing the eclipsing of neighboring C-H bonds that gives rise to torsional strain. Part (b) is a Newman projection along a $\\mathrm{C}-\\mathrm{C}$ bond.\n\nHow can the hybrid-orbital model of bonding account for the large distortion of bond angles from the normal $109^{\\circ}$ tetrahedral value to $60^{\\circ}$ in cyclopropane? The answer is that cyclopropane has bent bonds. In an unstrained alkane, maximum bonding is achieved when two atoms have their overlapping orbitals pointing directly toward each other. In cyclopropane, though, the orbitals can't point directly toward each other; instead, they overlap at a slight angle. The result is that cyclopropane bonds are weaker and more reactive than typical alkane bonds- $255 \\mathrm{~kJ} / \\mathrm{mol}$ ( $61 \\mathrm{kcal} / \\mathrm{mol}$ ) for a C-C bond in cyclopropane versus $370 \\mathrm{~kJ} / \\mathrm{mol}(88 \\mathrm{kcal} / \\mathrm{mol}$ ) for a $\\mathrm{C}-\\mathrm{C}$ bond in open-chain propane.\n\n\nTypical alkane $\\mathrm{C}-\\mathrm{C}$ bonds\n\n\nTypical bent cyclopropane $\\mathrm{C}-\\mathrm{C}$ bonds"}
{"id": 220, "contents": "Cyclobutane - \nCyclobutane has less angle strain than cyclopropane but has more torsional strain because of its larger number of ring hydrogens. As a result, the total strain for the two compounds is nearly the same $-110 \\mathrm{~kJ} / \\mathrm{mol}(26.4 \\mathrm{kcal} /$ $\\mathrm{mol})$ for cyclobutane versus $115 \\mathrm{~kJ} / \\mathrm{mol}(27.5 \\mathrm{kcal} / \\mathrm{mol})$ for cyclopropane. Cyclobutane is not quite flat but is slightly bent so that one carbon atom lies about $25^{\\circ}$ above the plane of the other three (FIGURE 4.6). The effect of this slight bend is to increase angle strain but to decrease torsional strain, until a minimum-energy balance between the two opposing effects is achieved.\n(a)\n\n(b)\n\n(c) Not quite\n\n\nFIGURE 4.6 The conformation of cyclobutane. Part (c) is a Newman projection along a $\\mathrm{C}-\\mathrm{C}$ bond, showing that neighboring $\\mathrm{C}-\\mathrm{H}$ bonds are\nnot quite eclipsed."}
{"id": 221, "contents": "Cyclopentane - \nCyclopentane was predicted by Baeyer to be nearly strain-free, but it actually has a total strain energy of 26 $\\mathrm{kJ} / \\mathrm{mol}(6.2 \\mathrm{kcal} / \\mathrm{mol})$. Although planar cyclopentane has practically no angle strain, it has a large torsional strain. Cyclopentane therefore twists to adopt a puckered, nonplanar conformation that strikes a balance between increased angle strain and decreased torsional strain. Four of the cyclopentane carbon atoms are in approximately the same plane, with the fifth carbon atom bent out of the plane. Most of the hydrogens are nearly staggered with respect to their neighbors (FIGURE 4.7).\n(a)\n\n(b)\n\nObserver\n(c)\n\n\n\nFIGURE 4.7 The conformation of cyclopentane. Carbons $1,2,3$, and 4 are nearly coplanar, but carbon 5 is out of the plane. Part (c) is a Newman projection along the C1-C2 bond, showing that neighboring C-H bonds are nearly staggered.\n\nPROBLEM How many $\\mathrm{H} \\longleftrightarrow \\mathrm{H}$ eclipsing interactions would be present if cyclopentane were planar? Assuming\n4-10 an energy cost of $4.0 \\mathrm{~kJ} / \\mathrm{mol}$ for each eclipsing interaction, how much torsional strain would planar cyclopentane have? Since the measured total strain of cyclopentane is $26 \\mathrm{~kJ} / \\mathrm{mol}$, how much of the torsional strain is relieved by puckering?\n\nPROBLEM Two conformations of cis-1,3-dimethylcyclobutane are shown. What is the difference between\n4-11 them, and which do you think is likely to be more stable?\n(a)\n\n(b)"}
{"id": 222, "contents": "Cyclopentane - 4.5 Conformations of Cyclohexane\nSubstituted cyclohexanes are the most common cycloalkanes and occur widely in nature. A large number of compounds, including steroids and many pharmaceutical agents, have cyclohexane rings. The flavoring agent menthol, for instance, has three substituents on a six-membered ring.\n\n\nMenthol\n\nCyclohexane adopts a strain-free, three-dimensional shape that is called a chair conformation because of its similarity to a lounge chair, with a back, seat, and footrest (FIGURE 4.8). Chair cyclohexane has neither angle strain nor torsional strain-all C-C-C bond angles are near the $109^{\\circ}$ tetrahedral value, and all neighboring $\\mathrm{C}-\\mathrm{H}$ bonds are staggered.\n(a)\n\n(b)\n\nObserver\n(c)\n\n\nFIGURE 4.8 The strain-free chair conformation of cyclohexane. All C-C-C bond angles are $111.5^{\\circ}$, close to the ideal $109^{\\circ}$ tetrahedral angle, and all neighboring $\\mathrm{C}-\\mathrm{H}$ bonds are staggered.\n\nThe easiest way to visualize chair cyclohexane is to build a molecular model if you have access to a model kit, or alternatively to explore with one of the many computer-based modeling programs you may have access to.\n\nThe chair conformation of cyclohexane can be drawn in three steps."}
{"id": 223, "contents": "STEP 1 - \nDraw two parallel lines, slanted downward and slightly offset from each other. This means that four of the cyclohexane carbons lie in a plane."}
{"id": 224, "contents": "STEP 2 - \nPlace the topmost carbon atom above and to the right of the plane of the other four, and connect the bonds."}
{"id": 225, "contents": "STEP 3 - \nPlace the bottommost carbon atom below and to the left of the plane of the middle four, and connect the bonds. Note that the bonds to the bottommost carbon atom are parallel to the bonds to the topmost carbon.\n\nWhen viewing cyclohexane, it's helpful to remember that the lower bond is in front and the upper bond is in back. If this convention isn't defined, it can appear that the reverse is true. For clarity, all cyclohexane rings drawn in this book will have the front (lower) bond heavily shaded to indicate nearness to the viewer.\n\n\nIn addition to the chair conformation of cyclohexane, there is an alternative conformation of cyclohexane that bears a slight resemblance to a twisted boat. Called the twist-boat conformation, it is nearly free of angle strain. It does, however, have both steric strain and torsional strain and is about $23 \\mathrm{~kJ} / \\mathrm{mol}(5.5 \\mathrm{kcal} / \\mathrm{mol})$ higher in\nenergy than the chair conformation. As a result, molecules adopt the twist-boat geometry only rarely.\n\n\n\nTwist-boat cyclohexane ( $23 \\mathrm{~kJ} / \\mathrm{mol}$ strain)"}
{"id": 226, "contents": "STEP 3 - 4.6 Axial and Equatorial Bonds in Cyclohexane\nThe chair conformation of cyclohexane has many consequences. We'll see in Section 11.9, for instance, that the chemical behavior of many substituted cyclohexanes is influenced by their conformation. In addition, we'll see in Section 25.5 that simple carbohydrates, such as glucose, adopt a conformation based on the cyclohexane chair and that their chemistry is directly affected as a result.\n\n\nCyclohexane (chair conformation)\n\n\nGlucose\n(chair conformation)\n\nAnother trait of the chair conformation is that there are two kinds of positions for substituents on the cyclohexane ring: axial positions and equatorial positions (as shown in FIGURE 4.9). The six axial positions are parallel to the ring axis, while the six equatorial positions are in the rough plane of the ring, around the ring equator.\n\n\nFIGURE 4.9 Axial and equatorial positions in chair cyclohexane. The six axial hydrogens are parallel to the ring axis, and the six equatorial hydrogens are in a band around the ring equator.\n\nAs shown in FIGURE 4.9, each carbon atom in chair cyclohexane has one axial and one equatorial hydrogen. Furthermore, each side of the ring has three axial and three equatorial hydrogens in an alternating arrangement. For example, if the top side of the ring has axial hydrogens on carbons 1, 3, and 5, then it has equatorial hydrogens on carbons 2,4 , and 6 . The reverse is true for the bottom side: carbons 1,3 , and 5 have equatorial hydrogens, but carbons 2, 4, and 6 have axial hydrogens (FIGURE 4.10).\n\n\nFIGURE 4.10 Alternating axial and equatorial positions in chair cyclohexane, looking directly down the ring axis. Each carbon atom has one axial and one equatorial position, and each face has alternating axial and equatorial positions.\n\nNote that we haven't used the words cis and trans in this discussion of cyclohexane conformation. Two hydrogens on the same side of the ring are always cis, regardless of whether they're axial or equatorial and regardless of whether they're adjacent. Similarly, two hydrogens on opposite sides of the ring are always trans."}
{"id": 227, "contents": "STEP 3 - 4.6 Axial and Equatorial Bonds in Cyclohexane\nAxial and equatorial bonds can be drawn following the procedure shown in FIGURE 4.11. If possible, look at a molecular model as you practice.\n\nAxial bonds: The six axial bonds, one on each carbon, are parallel and alternate up-down.\n\n\nEquatorial bonds: The six equatorial bonds, one on each carbon, come in three sets of two parallel lines. Each set is also parallel to two ring bonds. Equatorial bonds\n\n\nalternate between sides around the ring.\n\n\nFIGURE 4.11 A procedure for drawing axial and equatorial bonds in chair cyclohexane.\nBecause chair cyclohexane has two kinds of positions-axial and equatorial-we might expect to find two isomeric forms of a monosubstituted cyclohexane. In fact, we don't. There is only one methylcyclohexane, one bromocyclohexane, one cyclohexanol (hydroxycyclohexane), and so on, because cyclohexane rings are conformationally mobile at room temperature. Different chair conformations readily interconvert, exchanging axial and equatorial positions. This interconversion, called a ring-flip, is shown in FIGURE 4.12.\n\n\nFIGURE 4.12 A ring-flip in chair cyclohexane interconverts axial and equatorial positions. What is axial in the starting structure becomes equatorial in the ring-flipped structure, and what is equatorial in the starting structure is axial after ring-flip.\n\nAs shown in FIGURE 4.12, a chair cyclohexane can be ring-flipped by keeping the middle four carbon atoms in place while folding the two end carbons in opposite directions. In so doing, an axial substituent in one chair form becomes an equatorial substituent in the ring-flipped chair form and vice versa. For example, axial bromocyclohexane becomes equatorial bromocyclohexane after a ring-flip. Since the energy barrier to chair-chair interconversion is only about $45 \\mathrm{~kJ} / \\mathrm{mol}(10.8 \\mathrm{kcal} / \\mathrm{mol})$, the process is rapid at room temperature and we see what appears to be a single structure rather than distinct axial and equatorial isomers.\n\n\n\nAxial bromocyclohexane\n\n\nRing-flip"}
{"id": 228, "contents": "STEP 3 - 4.6 Axial and Equatorial Bonds in Cyclohexane\nAxial bromocyclohexane\n\n\nRing-flip\n\n\nEquatorial bromocyclohexane"}
{"id": 229, "contents": "Drawing the Chair Conformation of a Substituted Cyclohexane - \nDraw 1,1-dimethylcyclohexane in a chair conformation, indicating which methyl group in your drawing is axial and which is equatorial."}
{"id": 230, "contents": "Strategy - \nDraw a chair cyclohexane ring using the procedure in FIGURE 4.11, and then put two methyl groups on the same carbon. The methyl group in the rough plane of the ring is equatorial, and the one directly above or below the ring is axial."}
{"id": 231, "contents": "Solution - \nPROBLEM Draw two different chair conformations of cyclohexanol (hydroxycyclohexane), showing all 4-12 hydrogen atoms. Identify each position as axial or equatorial.\n\nPROBLEM Draw two different chair conformations of trans-1,4-dimethylcyclohexane, and label all positions 4-13 as axial or equatorial.\n\nPROBLEM Identify each of the colored positions-red, blue, and green-as axial or equatorial. Then carry out a 4-14 ring-flip, and show the new positions occupied by each color."}
{"id": 232, "contents": "Solution - 4.7 Conformations of Monosubstituted Cyclohexanes\nEven though cyclohexane rings flip rapidly between chair conformations at room temperature, the two conformations of a monosubstituted cyclohexane aren't equally stable. In methylcyclohexane, for instance, the equatorial conformation is more stable than the axial conformation by $7.6 \\mathrm{~kJ} / \\mathrm{mol}(1.8 \\mathrm{kcal} / \\mathrm{mol})$. The same is true of other monosubstituted cyclohexanes: a substituent is almost always more stable in an equatorial position than in an axial position.\n\nYou might recall from your general chemistry course that it's possible to calculate the percentages of two isomers at equilibrium using the equation $\\Delta E=-R T \\ln K$, where $\\Delta E$ is the energy difference between isomers, $R$ is the gas constant [8.315 J/(K\u2022mol)], $T$ is the Kelvin temperature, and $K$ is the equilibrium constant between isomers. For example, an energy difference of $7.6 \\mathrm{~kJ} / \\mathrm{mol}$ means that about $95 \\%$ of methylcyclohexane molecules have an equatorial methyl group at any given instant while only $5 \\%$ have an axial methyl group. FIGURE 4.13 plots the relationship between energy and isomer percentages.\n\n\nFIGURE 4.13 A plot of the percentages of two isomers at equilibrium versus the energy difference between them. The curves are calculated using the equation $\\Delta E=-R T \\ln K$.\n\nThe energy difference between axial and equatorial conformations is due to steric strain caused by $\\mathbf{1 , 3} \\mathbf{3}$ diaxial interactions. The axial methyl group on C1 is too close to the axial hydrogens three carbons away on C3 and C5, resulting in $7.6 \\mathrm{~kJ} / \\mathrm{mol}$ of steric strain (FIGURE 4.14).\n\n\nFIGURE 4.14 Interconversion of axial and equatorial methylcyclohexane, represented in several formats. The equatorial conformation is more stable than the axial conformation by $7.6 \\mathrm{~kJ} / \\mathrm{mol}$."}
{"id": 233, "contents": "Solution - 4.7 Conformations of Monosubstituted Cyclohexanes\nThe 1,3-diaxial steric strain in substituted methylcyclohexane is already familiar-we saw it previously as the steric strain between methyl groups in gauche in Section 3.7. Gauche butane is less stable than anti butane by $3.8 \\mathrm{~kJ} / \\mathrm{mol}(0.9 \\mathrm{kcal} / \\mathrm{mol})$ because of steric interference between hydrogen atoms on the two methyl groups. Comparing a four-carbon fragment of axial methylcyclohexane with gauche butane shows that the steric interaction is the same in both (FIGURE 4.15). Because axial methylcyclohexane has two such interactions, it has $2 \\times 3.8=7.6 \\mathrm{~kJ} / \\mathrm{mol}$ of steric strain. Equatorial methylcyclohexane has no such interactions and is therefore more stable.\n\n\nFIGURE 4.15 The origin of 1,3-diaxial interactions in methylcyclohexane. The steric strain between an axial methyl group and an axial hydrogen atom three carbons away is identical to the steric strain in gauche butane. (To display clearly the diaxial interactions in methylcyclohexane, two of the equatorial hydrogens are not shown.)\n\nThe exact amount of 1,3-diaxial steric strain in a given substituted cyclohexane depends on the nature and size of the substituent, as indicated in TABLE 4.1. Not surprisingly, the amount of steric strain increases through the series $\\mathrm{H}_{3} \\mathrm{C}-<\\mathrm{CH}_{3} \\mathrm{CH}_{2}-<\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CH}-\\ll\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{C}-$, paralleling the increasing size of the alkyl groups. Note that the values given in TABLE 4.1 refer to 1,3-diaxial interactions of the substituent with a single hydrogen atom. These values must be doubled to arrive at the amount of strain in a monosubstituted cyclohexane.\n\nTABLE 4.1 Steric Strain in Monosubstituted Cyclohexanes"}
{"id": 234, "contents": "Solution - 4.7 Conformations of Monosubstituted Cyclohexanes\nTABLE 4.1 Steric Strain in Monosubstituted Cyclohexanes\n\n| 1,3-Diaxial strain <br> (kJ/mol) <br> (kcal/mol) |  |  |  |\n| :--- | :--- | :--- | :--- |\n| Y |  |  |  |\n| F | 0.5 | 0.12 |  |\n\nTABLE 4.1 Steric Strain in Monosubstituted Cyclohexanes\n\n| 1,3-Diaxial strain |  |  |  |\n| :---: | :---: | :---: | :---: |\n| Y | ( $\\mathrm{kJ} / \\mathrm{mol}$ ) | ( $\\mathrm{kcal} / \\mathrm{mol}$ ) | $\\sim$ |\n| $\\mathrm{Cl}, \\mathrm{Br}$ | 1.0 | 0.25 |  |\n| OH | 2.1 | 0.5 |  |\n| $\\mathrm{CH}_{3}$ | 3.8 | 0.9 |  |\n| $\\mathrm{CH}_{2} \\mathrm{CH}_{3}$ | 4.0 | 0.95 |  |\n| $\\mathrm{CH}\\left(\\mathrm{CH}_{3}\\right)_{2}$ | 4.6 | 1.1 |  |\n| $\\mathrm{C}\\left(\\mathrm{CH}_{3}\\right)_{3}$ | 11.4 | 2.7 |  |\n| $\\mathrm{C}_{6} \\mathrm{H}_{5}$ | 6.3 | 1.5 |  |\n| $\\mathrm{CO}_{2} \\mathrm{H}$ | 2.9 | 0.7 |  |\n| CN | 0.4 | 0.1 |  |\n\nPROBLEM What is the energy difference between the axial and equatorial conformations of cyclohexanol 4-15 (hydroxycyclohexane)?\n\nPROBLEM Why do you suppose an axial cyano (-CN) substituent causes practically no 1,3-diaxial steric strain 4-16 ( $0.4 \\mathrm{~kJ} / \\mathrm{mol}$ )?"}
{"id": 235, "contents": "Solution - 4.7 Conformations of Monosubstituted Cyclohexanes\nPROBLEM Look back at Figure 4.13 and estimate the percentages of axial and equatorial conformations 4-17 present at equilibrium in bromocyclohexane."}
{"id": 236, "contents": "Solution - 4.8 Conformations of Disubstituted Cyclohexanes\nMonosubstituted cyclohexanes are always more stable with their substituent in an equatorial position, but the situation with disubstituted cyclohexanes is more complex because the steric effects of both substituents must be taken into account. All steric interactions for both possible chair conformations must be analyzed before deciding which conformation is favored.\n\nLet's look at 1,2-dimethylcyclohexane as an example. There are two isomers, cis-1,2-dimethylcyclohexane and trans-1,2-dimethylcyclohexane, which must be considered separately. In the cis isomer, both methyl groups are on the same face of the ring and the compound can exist in either of the two chair conformations shown in FIGURE 4.16. (It may be easier for you to see whether a compound is cis- or trans-disubstituted by first drawing the ring as a flat representation and then converting it to a chair conformation.)\ncis-1,2-Dimethylcyclohexane\nOne gauche\ninteraction ( $3.8 \\mathrm{~kJ} / \\mathrm{mol}$ ) Two $\\mathrm{CH}_{3} \\leftrightarrow \\mathrm{H}$ diaxial interactions ( $7.6 \\mathrm{~kJ} / \\mathrm{mol}$ )\nTotal strain: $\\mathbf{3 . 8}+\\mathbf{7 . 6}=\\mathbf{1 1 . 4} \\mathbf{~ k J} / \\mathbf{m o l}$\n\n\nOne gauche interaction ( $3.8 \\mathrm{~kJ} / \\mathrm{mol}$ ) Two $\\mathrm{CH}_{3} \\leftrightarrow \\mathrm{H}$ diaxial interactions ( $7.6 \\mathrm{~kJ} / \\mathrm{mol}$ )\nTotal strain: $3.8+7.6=11.4 \\mathrm{~kJ} / \\mathbf{m o l}$"}
{"id": 237, "contents": "Solution - 4.8 Conformations of Disubstituted Cyclohexanes\nFIGURE 4.16 Conformations of cis-1,2-dimethylcyclohexane. The two chair conformations are equal in energy because each has one axial methyl group and one equatorial methyl group.\nBoth chair conformations of cis-1,2-dimethylcyclohexane have one axial methyl group and one equatorial methyl group. The top conformation in FIGURE 4.16 has an axial methyl group at C2, which has 1,3-diaxial interactions with hydrogens on C4 and C6. The ring-flipped conformation has an axial methyl group at C1, which has 1,3-diaxial interactions with hydrogens on C3 and C5. In addition, both conformations have gauche butane interactions between the two methyl groups. The two conformations are equal in energy, with a total steric strain of $3 \\times 3.8 \\mathrm{~kJ} / \\mathrm{mol}=11.4 \\mathrm{~kJ} / \\mathrm{mol}(2.7 \\mathrm{kcal} / \\mathrm{mol})$."}
{"id": 238, "contents": "Solution - 4.8 Conformations of Disubstituted Cyclohexanes\nIn trans-1,2-dimethylcyclohexane, the two methyl groups are on opposite sides of the ring and the compound can exist in either of the two chair conformations shown in FIGURE 4.17. The situation here is quite different from that of the cis isomer. The top conformation in FIGURE 4.17 has both methyl groups equatorial with only a gauche butane interaction between them ( $3.8 \\mathrm{~kJ} / \\mathrm{mol}$ ) but no 1,3-diaxial interactions. The ring-flipped conformation, however, has both methyl groups axial. The axial methyl group at C1 interacts with axial hydrogens at C3 and C5, and the axial methyl group at C2 interacts with axial hydrogens at C4 and C6. These four 1,3-diaxial interactions produce a steric strain of $4 \\times 3.8 \\mathrm{~kJ} / \\mathrm{mol}=15.2 \\mathrm{~kJ} / \\mathrm{mol}$ and make the diaxial conformation $15.2-3.8=11.4 \\mathrm{~kJ} / \\mathrm{mol}$ less favorable than the diequatorial conformation. We therefore predict that trans-1,2-dimethylcyclohexane will exist almost exclusively in the diequatorial conformation."}
{"id": 239, "contents": "trans-1,2-Dimethylcyclohexane - \nOne gauche interaction ( $3.8 \\mathrm{~kJ} / \\mathrm{mol}$ )\n\n\nFour $\\mathrm{CH}_{3} \\leftrightarrow \\mathrm{H}$ diaxial interactions ( $15.2 \\mathrm{~kJ} / \\mathrm{mol}$ )\n\n\nFIGURE 4.17 Conformations of trans-1,2-dimethylcyclohexane. The conformation with both methyl groups equatorial (top) is favored by $11.4 \\mathrm{~kJ} / \\mathrm{mol}(2.7 \\mathrm{kcal} / \\mathrm{mol})$ over the conformation with both methyl groups axial (bottom).\nThe same kind of conformational analysis just carried out for cis- and trans-1,2-dimethylcyclohexane can be done for any substituted cyclohexane, such as cis-1-tert-butyl-4-chlorocyclohexane (see Worked Example 4.3).\n\nAs you might imagine, though, the situation becomes more complex as the number of substituents increases. For instance, compare glucose with mannose, a carbohydrate present in seaweed. Which do you think is more strained? In glucose, all substituents on the six-membered ring are equatorial, while in mannose, one of the -OH groups is axial, making it more strained.\n\n\n\nGlucose\n\n\n\nMannose\n\nA summary of the various axial and equatorial relationships among substituent groups in the different possible cis and trans substitution patterns for disubstituted cyclohexanes is given in TABLE 4.2.\n\nTABLE 4.2 Axial and Equatorial Relationships in Cis- and Trans-Disubstituted Cyclohexanes"}
{"id": 240, "contents": "trans-1,2-Dimethylcyclohexane - \nTABLE 4.2 Axial and Equatorial Relationships in Cis- and Trans-Disubstituted Cyclohexanes\n\n| Cis/trans substitution pattern |  | Axial/equatorial relationships |  |\n| :--- | :--- | :--- | :--- |\n| 1,2-Cis disubstituted | a,e | or | e,a |\n| 1,2-Trans disubstituted | a,a | or | e,e |\n| 1,3-Cis disubstituted | a,a | or | e,e |\n| 1,3-Trans disubstituted | a,e | or | e,a |\n| 1,4-Cis disubstituted | a,e | or | e,a |\n| 1,4-Trans disubstituted | a,a | or | e,e |"}
{"id": 241, "contents": "Drawing the Most Stable Conformation of a Substituted Cyclohexane - \nDraw the more stable chair conformation of cis-1-tert-butyl-4-chlorocyclohexane. By how much is it favored?"}
{"id": 242, "contents": "Strategy - \nDraw the two possible chair conformations, and calculate the strain energy in each. Remember that equatorial substituents cause less strain than axial substituents."}
{"id": 243, "contents": "Solution - \nFirst draw the two chair conformations of the molecule:\n\n\nIn the conformation on the left, the tert-butyl group is equatorial and the chlorine is axial. In the conformation\non the right, the tert-butyl group is axial and the chlorine is equatorial. These conformations aren't of equal energy because an axial tert-butyl substituent and an axial chloro substituent produce different amounts of steric strain. TABLE 4.1 shows that the 1,3-diaxial interaction between a hydrogen and a tert-butyl group costs $11.4 \\mathrm{~kJ} / \\mathrm{mol}(2.7 \\mathrm{kcal} / \\mathrm{mol})$, whereas the interaction between a hydrogen and a chlorine costs only $1.0 \\mathrm{~kJ} /$ $\\mathrm{mol}(0.25 \\mathrm{kcal} / \\mathrm{mol})$. An axial tert-butyl group therefore produces $(2 \\times 11.4 \\mathrm{~kJ} / \\mathrm{mol})-(2 \\times 1.0 \\mathrm{~kJ} / \\mathrm{mol})=20.8$ $\\mathrm{kJ} / \\mathrm{mol}(4.9 \\mathrm{kcal} / \\mathrm{mol})$ more steric strain than an axial chlorine, and the compound preferentially adopts the conformation with the chlorine axial and the tert-butyl equatorial.\n\nPROBLEM Draw the more stable chair conformation of the following molecules, and estimate the amount of\n4-18 strain in each:\n(a) trans-1-Chloro-3-methylcyclohexane\n(b) cis-1-Ethyl-2-methylcyclohexane\n(c) cis-1-Bromo-4-ethylcyclohexane\n(d) cis-1-tert-Butyl-4-ethylcyclohexane\n\nPROBLEM Identify each substituent in the following compound as axial or equatorial, and tell whether the 4-19 conformation shown is the more stable or less stable chair form (green $=\\mathrm{Cl}$ ):"}
{"id": 244, "contents": "Solution - 4.9 Conformations of Polycyclic Molecules\nThe final point we'll consider about cycloalkane stereochemistry is to see what happens when two or more cycloalkane rings are fused together along a common bond to construct a polycyclic molecule-for example, decalin."}
{"id": 245, "contents": "Decalin-two fused cyclohexane rings - \nDecalin consists of two cyclohexane rings joined to share two carbon atoms (the bridgehead carbons, C1 and C6) and a common bond. Decalin can exist in either of two isomeric forms, depending on whether the rings are trans fused or cis fused. In cis-decalin, the hydrogen atoms at the bridgehead carbons are on the same side of the rings; in trans-decalin, the bridgehead hydrogens are on opposite sides. FIGURE 4.18 shows how both compounds can be represented using chair cyclohexane conformations. Note that the two decalin isomers are not interconvertible by ring-flips or other rotations. They are cis-trans stereoisomers and have the same relationship to each other that cis- and trans-1,2-dimethylcyclohexane have.\n\ncis-Decalin\n\n\ntrans-Decalin\n\nFIGURE 4.18 Representations of cis- and trans-decalin. Hydrogen atoms at the bridgehead carbons are on the same face of the rings in the cis isomer but on opposite faces in the trans isomer.\n\nPolycyclic compounds are common in nature, and many valuable substances have fused-ring structures. For example, steroids, such as testosterone, the primary sex hormone in males, have three six-membered rings and one five-membered ring fused together. Although steroids look complicated compared with cyclohexane or decalin, the same principles that apply to the conformational analysis of simple cyclohexane rings apply equally well (and often better) to steroids."}
{"id": 246, "contents": "Testosterone (a steroid) - \nAnother common ring system is the norbornane, or bicyclo[2.2.1]heptane, structure. Like decalin, norbornane is a bicycloalkane, so called because two rings would have to be broken open to generate an acyclic structure. Its systematic name, bicyclo[2.2.1]heptane, reflects the fact that the molecule has seven carbons, is bicyclic, and has three \"bridges\" of 2, 2, and 1 carbon atoms connecting the two bridgehead carbons.\n\n\nNorbornane (bicyclo[2.2.1]heptane)\nNorbornane has a conformationally locked boat cyclohexane ring (Section 4.5) in which carbons 1 and 4 are joined by an additional $\\mathrm{CH}_{2}$ group. In drawing this structure, a break in the rear bond indicates that the vertical bond crosses in front of it. Making a molecular model is particularly helpful when trying to see the three-\ndimensionality of norbornane.\nSubstituted norbornanes, such as camphor, are found widely in nature, and many have been important historically in developing organic structural theories.\n\n\nPROBLEM Which isomer is more stable, cis-decalin or trans-decalin (Figure 4.18)? Explain. 4-20\n\nPROBLEM Look at the following structure of estrone, the primary sex hormone in females, and tell whether 4-21 each of the two indicated (red) ring-fusions is cis or trans.\n\n\n\nEstrone"}
{"id": 247, "contents": "Molecular Mechanics - \nAll the structural models in this book are computer-drawn. To make sure they accurately represent bond angles, bond lengths, torsional interactions, and steric interactions, the most stable geometry of each molecule has been calculated on a desktop computer using a commercially available molecular mechanics program based on work by Norman Allinger at the University of Georgia.\n\nThe idea behind molecular mechanics is to begin with a rough geometry for a molecule and then calculate a total strain energy for that starting geometry, using mathematical equations that assign values to specific kinds of molecular interactions. Bond angles that are too large or too small cause angle strain; bond lengths that are too short or too long cause stretching or compressing strain; unfavorable eclipsing interactions around single bonds cause torsional strain; and nonbonded atoms that approach each other too closely cause steric, or van der Waals, strain.\n\n$$\nE_{\\text {total }}=E_{\\mathrm{bond} \\text { stretching }}+E_{\\text {angle strain }}+E_{\\text {torsional strain }}+E_{\\text {van der Waals }}\n$$\n\n\n\nFIGURE 4.19 Computer programs make it possible to accurately represent molecular geometry. (credit: \"Molecular geometry\" by Jane Whitney/Flickr, CC BY 4.0)\n\nAfter calculating a total strain energy for the starting geometry, the program automatically changes the geometry slightly in an attempt to lower strain-perhaps by lengthening a bond that is too short or decreasing an angle that is too large. Strain is recalculated for the new geometry, more changes are made, and more calculations are done. After dozens or hundreds of iterations, the calculation ultimately converges on a minimum energy that corresponds to the most favorable, least strained conformation of the molecule.\n\nSimilar calculations have proven to be particularly useful in pharmaceutical research, where a complementary fit between a drug molecule and a receptor molecule in the body is often the key to designing new pharmaceutical agents (FIGURE 4.20).\n\n\nFIGURE 4.20 The structure of Tamiflu (oseltamivir), an antiviral agent active against type A influenza, along with a molecular model of its minimum-energy conformation as calculated by molecular mechanics."}
{"id": 248, "contents": "Key Terms - \n- alicyclic\n- angle strain\n- axial position (cyclohexane)\n- boat cyclohexane\n- Bridgehead atom\n- chair conformation\n- cis-trans isomers\n- conformational analysis\n- cycloalkanes"}
{"id": 249, "contents": "- 1,3-diaxial interaction - \n- Equatorial position (cyclohexane)\n- polycyclic molecule\n- ring-flip (cyclohexane)\n- stereoisomer\n- steric strain\n- torsional strain\n- twist-boat conformation"}
{"id": 250, "contents": "Summary - \nCyclic molecules are so commonly encountered throughout organic and biological chemistry that it's important to understand the consequences of their cyclic structures. Thus, we've taken a close look at cyclic structures in this chapter.\n\nCycloalkanes are saturated cyclic hydrocarbons with the general formula $\\mathrm{C}_{n} \\mathrm{H}_{2 n}$. In contrast to open-chain alkanes, where nearly free rotation occurs around $\\mathrm{C}-\\mathrm{C}$ bonds, rotation is greatly reduced in cycloalkanes. Disubstituted cycloalkanes can therefore exist as cis-trans isomers. The cis isomer has both substituents on the same side of the ring; the trans isomer has substituents on opposite sides. Cis-trans isomers are just one kind of stereoisomer-compounds that have the same connections between atoms but different threedimensional arrangements.\n\nNot all cycloalkanes are equally stable. Three kinds of strain contribute to the overall energy of a cycloalkane: (1) angle strain is the resistance of a bond angle to compression or expansion from the normal $109^{\\circ}$ tetrahedral value, (2) torsional strain is the energy cost of having neighboring $\\mathrm{C}-\\mathrm{H}$ bonds eclipsed rather than staggered, and (3) steric strain is the repulsive interaction that arises when two groups attempt to occupy the same space.\n\nCyclopropane ( $115 \\mathrm{~kJ} / \\mathrm{mol}$ strain) and cyclobutane ( $110.4 \\mathrm{~kJ} / \\mathrm{mol}$ strain) have both angle strain and torsional strain. Cyclopentane is free of angle strain but has a substantial torsional strain due to its large number of eclipsing interactions. Both cyclobutane and cyclopentane pucker slightly away from planarity to relieve torsional strain."}
{"id": 251, "contents": "Summary - \nCyclohexane is strain-free because it adopts a puckered chair conformation, in which all bond angles are near $109^{\\circ}$ and all neighboring $\\mathrm{C}-\\mathrm{H}$ bonds are staggered. Chair cyclohexane has two kinds of positions: axial and equatorial. Axial positions are oriented up and down, parallel to the ring axis, while equatorial positions lie in a belt around the equator of the ring. Each carbon atom has one axial and one equatorial position.\n\nChair cyclohexanes are conformationally mobile and can undergo a ring-flip, which interconverts axial and equatorial positions. Substituents on the ring are more stable in the equatorial position because axial substituents cause 1,3-diaxial interactions. The amount of 1,3-diaxial steric strain caused by an axial substituent depends on its size."}
{"id": 252, "contents": "Visualizing Chemistry - \nPROBLEM Name the following cycloalkanes:\n\n4-22 (a)\n\n(b)\n\n\nPROBLEM Name the following compound, identify each substituent as axial or equatorial, and tell whether the\n4-23 conformation shown is the more stable or less stable chair form (green $=\\mathrm{Cl}$ ):\n\n\nPROBLEM A trisubstituted cyclohexane with three substituents-red, green, and blue-undergoes a ring-flip\n4-24 to its alternate chair conformation. Identify each substituent as axial or equatorial, and show the positions occupied by the three substituents in the ring-flipped form.\n\n\nPROBLEM The following cyclohexane derivative has three substituents-red, green, and blue. Identify each\n4-25 substituent as axial or equatorial, and identify each pair of relationships (red-blue, red-green, and blue-green) as cis or trans.\n\n\nPROBLEM Glucose exists in two forms having a 36:64 ratio at equilibrium. Draw a skeletal structure of each,\n4-26 describe the difference between them, and tell which of the two you think is more stable (red = 0).\n\n\nCycloalkane Isomers\nPROBLEM Draw the five cycloalkanes with the formula $\\mathrm{C}_{5} \\mathrm{H}_{10}$.\n4-27\nPROBLEM Give IUPAC names for the following compounds.\n4-28 (a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n\nPROBLEM Draw a stereoisomer of trans-1,3-dimethylcyclobutane.\n4-29\nPROBLEM Tell whether the following pairs of compounds are identical, constitutional isomers, stereoisomers, 4-30 or unrelated.\n(a) cis-1,3-Dibromocyclohexane and trans-1,4-dibromocyclohexane\n(b) 2,3-Dimethylhexane and 2,3,3-trimethylpentane\n(c)\nand\n\n\nPROBLEM Draw three isomers of trans-1,2-dichlorocyclobutane, and label them as either constitutional 4-31 isomers or stereoisomers."}
{"id": 253, "contents": "Visualizing Chemistry - \nPROBLEM Draw three isomers of trans-1,2-dichlorocyclobutane, and label them as either constitutional 4-31 isomers or stereoisomers.\n\nPROBLEM Identify each pair of relationships among the -OH groups in glucose (red-blue, red-green, 4-32 red-black, blue-green, blue-black, green-black) as cis or trans.\n\n\nGlucose\n\nPROBLEM Draw 1,3,5-trimethylcyclohexane using a hexagon to represent the ring. How many cis-trans 4-33 stereoisomers are possible?"}
{"id": 254, "contents": "Cycloalkane Conformation and Stability - \nPROBLEM Hydrocortisone, a naturally occurring hormone produced in the adrenal glands, is often used to\n4-34 treat inflammation, severe allergies, and numerous other conditions. Is the indicated -OH group axial or equatorial?\n\n\nHydrocortisone\n\nPROBLEM A 1,2-cis disubstituted cyclohexane, such as cis-1,2-dichlorocyclohexane, must have one group 4-35 axial and one group equatorial. Explain.\n\nPROBLEM A 1,2-trans disubstituted cyclohexane must have either both groups axial or both groups equatorial. 4-36 Explain.\n\nPROBLEM Why is a 1,3-cis disubstituted cyclohexane more stable than its trans isomer? 4-37\n\nPROBLEM Which is more stable, a 1,4-trans disubstituted cyclohexane or its cis isomer?\n4-38\nPROBLEM cis-1,2-Dimethylcyclobutane is less stable than its trans isomer, but cis-1,3-dimethylcyclobutane is\n4-39 more stable than its trans isomer. Draw the most stable conformations of both, and explain.\nPROBLEM From the data in Figure 4.13 and Table 4.1, estimate the percentages of molecules that have their\n4-40 substituents in an axial orientation for the following compounds:\n(a) Isopropylcyclohexane\n(b) Fluorocyclohexane\n(c) Cyclohexanecarbonitrile, $\\mathrm{C}_{6} \\mathrm{H}_{11} \\mathrm{CN}$"}
{"id": 255, "contents": "Cycloalkane Conformation and Stability - \nPROBLEM Assume that you have a variety of cyclohexanes substituted in the positions indicated. Identify\n4-41 the substituents as either axial or equatorial. For example, a 1,2-cis relationship means that one substituent must be axial and one equatorial, whereas a 1,2-trans relationship means that both substituents are axial or both are equatorial.\n(a) 1,3-Trans disubstituted (b) 1,4-Cis disubstituted (c) 1,3-Cis disubstituted\n(d) 1,5-Trans disubstituted\n(e) 1,5-Cis disubstituted (f) 1,6-Trans disubstituted\n\nCyclohexane Conformational Analysis\nPROBLEM Draw the two chair conformations of cis-1-chloro-2-methylcyclohexane. Which is more stable, and\n4-42 by how much?\nPROBLEM Draw the two chair conformations of trans-1-chloro-2-methylcyclohexane. Which is more stable? 4-43\n\nPROBLEM Galactose, a sugar related to glucose, contains a six-membered ring in which all the substituents\n4-44 except the -OH group, indicated below in red, are equatorial. Draw galactose in its more stable chair conformation.\n\n\nGalactose\n\nPROBLEM Draw the two chair conformations of menthol, and tell which is more stable.\n4-45"}
{"id": 256, "contents": "Menthol - \nPROBLEM There are four cis-trans isomers of menthol (Problem 4-45), including the one shown. Draw the\n4-46 other three.\nPROBLEM The diaxial conformation of cis-1,3-dimethylcyclohexane is approximately $23 \\mathrm{~kJ} / \\mathrm{mol}(5.4 \\mathrm{kcal} / \\mathrm{mol})$\n4-47 less stable than the diequatorial conformation. Draw the two possible chair conformations, and suggest a reason for the large energy difference.\n\nPROBLEM Approximately how much steric strain does the 1,3-diaxial interaction between the two methyl\n4-48 groups introduce into the diaxial conformation of cis-1,3-dimethylcyclohexane? (See Problem 4-47.)\n\nPROBLEM In light of your answer to Problem 4-48, draw the two chair conformations of\n4-49 1,1,3-trimethylcyclohexane and estimate the amount of strain energy in each. Which conformation is favored?\n\nPROBLEM One of the two chair structures of cis-1-chloro-3-methylcyclohexane is more stable than the other\n4-50 by $15.5 \\mathrm{~kJ} / \\mathrm{mol}(3.7 \\mathrm{kcal} / \\mathrm{mol})$. Which is it? What is the energy cost of a 1,3 -diaxial interaction between a chlorine and a methyl group?"}
{"id": 257, "contents": "General Problems - \nPROBLEM We saw in Problem 4-20 that cis-decalin is less stable than trans-decalin. Assume that the\n4-51 1,3-diaxial interactions in cis-decalin are similar to those in axial methylcyclohexane [that is, one $\\mathrm{CH} 2 \\leftrightarrow \\mathrm{H}$ interaction costs $3.8 \\mathrm{~kJ} / \\mathrm{mol}(0.9 \\mathrm{kcal} / \\mathrm{mol})]$, and calculate the magnitude of the energy difference between cis- and trans-decalin.\n\nPROBLEM Using molecular models as well as structural drawings, explain why trans-decalin is rigid and\n4-52 cannot ring-flip whereas cis-decalin can easily ring-flip.\nPROBLEM trans-Decalin is more stable than its cis isomer, but cis-bicyclo[4.1.0]heptane is more stable than its\n4-53 trans isomer. Explain.\n\ntrans-Decalin\n\ncis-Bicyclo[4.1.0]heptane\n\nPROBLEM As mentioned in Problem 3-53, the statin drugs, such as simvastatin (Zocor), pravastatin (Pravachol),\n4-54 and atorvastatin (Lipitor) are the most widely prescribed drugs in the world.\n\n(a) Are the two indicated bonds on simvastatin cis or trans?\n(b) What are the cis/trans relationships among the three indicated bonds on pravastatin?\n(c) Why can't the three indicated bonds on atorvastatin be identified as cis or trans?\n\nPROBLEM myo-Inositol, one of the isomers of 1,2,3,4,5,6-hexahydroxycyclohexane, acts as a growth factor in\n4-55 both animals and microorganisms. Draw the most stable chair conformation of myo-inositol.\n\nmyo-Inositol"}
{"id": 258, "contents": "General Problems - \nmyo-Inositol\n\nPROBLEM How many cis-trans stereoisomers of myo-inositol (Problem 4-55) are there? Draw the structure of\n4-56 the most stable isomer.\nPROBLEM Julius Bredt, discoverer of the structure of camphor, proposed in 1935 that bicycloalkenes such as\n4-57 1-norbornene, which have a double bond to a bridgehead carbon, are too strained to exist. Explain. (Making a molecular model will be helpful.\n\n\n1-Norbornene\n\nPROBLEM Tell whether each of the following substituents on a steroid is axial or equatorial. (A substituent that\n4-58 is \"up\" is on the top side of the molecule as drawn, and a substituent that is \"down\" is on the bottom side.\n(a) Substituent up at C3 (b) Substituent down at C7\n(c) Substituent down at C11\n\n\nPROBLEM Amantadine is an antiviral agent that is active against influenza type A infection. Draw a three-\n4-59 dimensional representation of amantadine, showing the chair cyclohexane rings.\n\n\nAmantadine\n\nPROBLEM There are two different isomers named trans-1,2-dimethylcyclopentane. Similarly, you have two\n4-60 different appendages called hands. What is the relationship between them? (We'll explore this kind of isomerism in the next chapter.)\n\nand\n\n\nPROBLEM Ketones react with alcohols to yield products called ketals. Why does the all-cis isomer of\n4-61 4-tert-butyl-1,3-cyclohexanediol react readily with acetone and an acid catalyst to form a ketal, but other stereoisomers do not react? In formulating your answer, draw the more stable chair conformations of all four stereoisomers and the product ketal for each one."}
{"id": 259, "contents": "General Problems - \nA ketal\nPROBLEM Alcohols undergo an oxidation reaction to yield carbonyl compounds when treatment with $\\mathrm{CrO}_{3}$.\n4-62 For example, 2-tert-butylcyclohexanol gives 2-tert-butylcyclohexanone. If axial -OH groups are generally more reactive than their equatorial isomers, which do you think reacts faster, the cis isomer of 2-tert-butylcyclohexanol or the trans isomer? Explain."}
{"id": 260, "contents": "CHAPTER 5 <br> Stereochemistry at Tetrahedral Centers - \nFIGURE 5.1 Like the mountain whose image is reflected in a lake, many organic molecules also have mirror-image counterparts. (credit: modification of work \"Crystal Lake sunrise reflection\" by Sandy Horvath-Dori/Wikimedia Commons, CC BY 2.0)"}
{"id": 261, "contents": "CHAPTER CONTENTS - 5.1 Enantiomers and the Tetrahedral Carbon\n5.2 The Reason for Handedness in Molecules: Chirality"}
{"id": 262, "contents": "CHAPTER CONTENTS - 5.3 Optical Activity\n5.4 Pasteur's Discovery of Enantiomers\n5.5 Sequence Rules for Specifying Configuration\n5.6 Diastereomers\n5.7 Meso Compounds"}
{"id": 263, "contents": "CHAPTER CONTENTS - 5.9 A Review of Isomerism\n5.10 Chirality at Nitrogen, Phosphorus, and Sulfur"}
{"id": 264, "contents": "CHAPTER CONTENTS - 5.11 Prochirality\n5.12 Chirality in Nature and Chiral Environments\n\nWHY THIS CHAPTER? Understanding the causes and consequences of molecular handedness is crucial to understanding organic and biological chemistry. The subject can be a bit complex at first, but the material covered in this chapter nevertheless forms the basis for much of the remainder of the book.\n\nAre you right-handed or left-handed? You may not spend much time thinking about it, but handedness plays a surprisingly large role in your daily activities. Many musical instruments, such as oboes and clarinets, have a handedness to them; the last available softball glove always fits the wrong hand. The reason for these difficulties is that our hands aren't identical; rather, they're mirror images. When you hold a left hand up to a mirror, the image you see looks like a right hand. Try it.\n\n\nLeft hand\n\n\nRight hand\n\nHandedness is also important in organic and biological chemistry, where it arises primarily as a consequence of the tetrahedral stereochemistry of $s p^{3}$-hybridized carbon atoms. Many drugs and almost all the molecules in our bodies-amino acids, carbohydrates, nucleic acids, and many more-have a handedness. Furthermore, molecular handedness enables the precise interactions between enzymes and their substrates that are involved in the hundreds of thousands of chemical reactions on which life is based."}
{"id": 265, "contents": "CHAPTER CONTENTS - 5.1 Enantiomers and the Tetrahedral Carbon\nWhat causes molecular handedness? Look at generalized molecules of the type $\\mathrm{CH}_{3} \\mathrm{X}, \\mathrm{CH}_{2} \\mathrm{XY}$, and CHXYZ shown in FIGURE 5.2. On the left are three molecules, and on the right are their images reflected in a mirror. The $\\mathrm{CH}_{3} \\mathrm{X}$ and $\\mathrm{CH}_{2} \\mathrm{XY}$ molecules are identical to their mirror images and thus are not handed. If you make a molecular model of each molecule and its mirror image, you find that you can superimpose one on the other so that all atoms coincide. The CHXYZ molecule, by contrast, is not identical to its mirror image. You can't superimpose a model of this molecule on a model of its mirror image for the same reason that you can't superimpose a left hand on a right hand: they simply aren't the same.\n\n\nFIGURE 5.2 Tetrahedral carbon atoms and their mirror images. Molecules of the type $\\mathrm{CH}_{3} \\mathrm{X}$ and $\\mathrm{CH}_{2} \\mathrm{XY}$ are identical to their mirror images, but a molecule of the type CHXYZ is not. $\\mathrm{A} C H X Y Z$ molecule is related to its mirror image in the same way a right hand is related to a left hand."}
{"id": 266, "contents": "CHAPTER CONTENTS - 5.1 Enantiomers and the Tetrahedral Carbon\nA molecule that is not identical to its mirror image is a kind of stereoisomer (Section 4.2) called an enantiomer (e-nan-tee-oh-mer, from the Greek enantio, meaning \"opposite\"). Enantiomers are related to each other as a right hand is related to a left hand and result whenever a tetrahedral carbon is bonded to four different substituents (one need not be H). For example, lactic acid (2-hydroxypropanoic acid) exists as a pair of enantiomers because there are four different groups ( $-\\mathrm{H},-\\mathrm{OH},-\\mathrm{CH}_{3},-\\mathrm{CO}_{2} \\mathrm{H}$ ) bonded to the central carbon atom. The enantiomers are called ( + )-lactic acid and ( - )-lactic acid. Both are found in sour milk, but only the ( + ) enantiomer occurs in muscle tissue."}
{"id": 267, "contents": "Lactic acid: a molecule of the general formula CHXYZ - \n(+)-Lactic acid\n\n\n\nNo matter how hard you try, you can't superimpose a molecule of (+)-lactic acid on a molecule of (-)-lactic acid. If any two groups match up, say -H and $-\\mathrm{CO}_{2} \\mathrm{H}$, the remaining two groups don't match (FIGURE 5.3).\n\n\n\nFIGURE 5.3 Attempts at superimposing the mirror-image forms of lactic acid. (a) When the -H and -OH substituents match up, the $-\\mathrm{CO}_{2} \\mathrm{H}$ and $-\\mathrm{CH}_{3}$ substituents don't; (b) when $-\\mathrm{CO}_{2} \\mathrm{H}$ and $-\\mathrm{CH}_{3}$ match up, -H and -OH don't. Regardless of how the molecules are oriented, they aren't identical."}
{"id": 268, "contents": "Lactic acid: a molecule of the general formula CHXYZ - 5.2 The Reason for Handedness in Molecules: Chirality\nA molecule that is not identical to its mirror image is said to be chiral (ky-ral, from the Greek cheir, meaning \"hand\"). You can't take a chiral molecule and its enantiomer and place one on the other so that all atoms coincide.\n\nHow can you predict whether a given molecule is or is not chiral? A molecule is not chiral if it has a plane of symmetry. A plane of symmetry is a plane that cuts through the middle of a molecule (or any object) in such a way that one half of the molecule or object is a mirror image of the other half. A coffee mug, for example, has a plane of symmetry. If you were to cut the coffee mug in half from top to bottom, one half would be a mirror image of the other half. A hand, however, does not have a plane of symmetry. One \"half\" of a hand is not a mirror image of the other half (FIGURE 5.4).\n\n\nFIGURE 5.4 The meaning of symmetry plane. (a) An object like the coffee mug has a symmetry plane cutting through it so that right and left halves are mirror images. (b) An object like a hand has no symmetry plane; the right \"half\" of a hand is not a mirror image of the left half.\n\nA molecule that has a plane of symmetry in any conformation must be identical to its mirror image and must be nonchiral, or achiral. Thus, propanoic acid, $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$, has a plane of symmetry when lined up as shown in FIGURE 5.5 and is achiral, while lactic acid, $\\mathrm{CH}_{3} \\mathrm{CH}(\\mathrm{OH}) \\mathrm{CO}_{2} \\mathrm{H}$, has no plane of symmetry in any conformation\nand is chiral.\n\n\n\n\n\nFIGURE 5.5 The achiral propanoic acid molecule versus the chiral lactic acid molecule. Propanoic acid has a plane of symmetry that makes one side of the molecule a mirror image of the other. Lactic acid has no such symmetry plane."}
{"id": 269, "contents": "Lactic acid: a molecule of the general formula CHXYZ - 5.2 The Reason for Handedness in Molecules: Chirality\nThe most common, although not the only, cause of chirality in organic molecules is the presence of a tetrahedral carbon atom bonded to four different groups-for example, the central carbon atom in lactic acid. Such carbons are referred to as chirality centers, although other terms such as stereocenter, asymmetric center, and stereogenic center have also been used. Note that chirality is a property of the entire molecule, whereas a chirality center is the cause of chirality.\n\nDetecting a chirality center in a complex molecule takes practice because it's not always immediately apparent whether four different groups are bonded to a given carbon. The differences don't necessarily appear right next to the chirality center. For example, 5 -bromodecane is a chiral molecule because four different groups are bonded to C 5 , the chirality center (marked with an asterisk). A butyl substituent is similar to a pentyl substituent, but it isn't identical. The difference isn't apparent until looking four carbon atoms away from the chirality center, but there's still a difference.\n\n\nAs other possible examples, look at methylcyclohexane and 2-methylcyclohexanone. Methylcyclohexane is achiral because no carbon atom in the molecule is bonded to four different groups. You can immediately eliminate all $-\\mathrm{CH}_{2}$ - carbons and the $-\\mathrm{CH}_{3}$ carbon from consideration, but what about C 1 on the ring? The C 1 carbon atom is bonded to $\\mathrm{a}-\\mathrm{CH}_{3}$ group, to an -H atom, and to C 2 and C 6 of the ring. Carbons 2 and 6 are equivalent, however, as are carbons 3 and 5 . Thus, the C6-C5-C4 \"substituent\" is equivalent to the C2-C3-C4 substituent, and methylcyclohexane is achiral. Another way of reaching the same conclusion is to realize that methylcyclohexane has a symmetry plane, which passes through the methyl group and through C1 and C4 of the ring."}
{"id": 270, "contents": "Lactic acid: a molecule of the general formula CHXYZ - 5.2 The Reason for Handedness in Molecules: Chirality\nThe situation is different for 2-methylcyclohexanone. 2-Methylcyclohexanone has no symmetry plane and is chiral because its C 2 is bonded to four different groups: $-\\mathrm{CH}_{3}$ group, an -H atom, $\\mathrm{a}-\\mathrm{COCH}_{2}-$ ring bond (C1), and a $-\\mathrm{CH}_{2} \\mathrm{CH}_{2}-$ ring bond (C3).\n\n\nSeveral more examples of chiral molecules are shown below. Check for yourself that the labeled carbons are chirality centers. You might note that carbons in $-\\mathrm{CH}_{2}-,-\\mathrm{CH}_{3}, \\mathrm{C}=\\mathrm{O}, \\mathrm{C}=\\mathrm{C}$, and $\\mathrm{C} \\equiv \\mathrm{C}$ groups can't be chirality centers. (Why not?)\n\n\nCarvone (spearmint oil)\n\n\nNootkatone (grapefruit oil)"}
{"id": 271, "contents": "Drawing the Three-Dimensional Structure of a Chiral Molecule - \nDraw the structure of a chiral alcohol."}
{"id": 272, "contents": "Strategy - \nAn alcohol is a compound that contains the -OH functional group. To make an alcohol chiral, we need to have four different groups bonded to a single carbon atom, say $-\\mathrm{H},-\\mathrm{OH},-\\mathrm{CH}_{3}$, and $-\\mathrm{CH}_{2} \\mathrm{CH}_{3}$\n\nSolution\n\n\n2-Butanol\n(chiral)\n\nPROBLEM Which of the following objects are chiral?\n5-1\n(a) Soda can\n(b) Screwdriver\n(c) Screw\n(d) Shoe\n\nPROBLEM Which of the following molecules are chiral? Identify the chirality center(s) in each.\n(a)\n\nConiine (poison hemlock)\n\n\nMenthol (flavoring agent)\n(c)\n\n\nDextromethorphan (cough suppressant)\n\nPROBLEM Alanine, an amino acid found in proteins, is chiral. Draw the two enantiomers of alanine using the 5-3 standard convention of solid, wedged, and dashed lines.\n\n\nAlanine\nPROBLEM Identify the chirality centers in the following molecules (gray $=\\mathrm{H}$, black $=\\mathrm{C}$, red $=\\mathrm{O}$, green $=\\mathrm{Cl}$, 5-4 yellow-green = F):\n\n\nThreose\n(a sugar)\n(b)\n\n\nEnflurane (an anesthetic)"}
{"id": 273, "contents": "Strategy - 5.3 Optical Activity\nThe study of chirality originated in the early 19th century during investigations by the French physicist JeanBaptiste Biot into the nature of plane-polarized light. A beam of ordinary light consists of electromagnetic waves that oscillate in an infinite number of planes at right angles to its direction of travel. When a beam of ordinary light passes through a device called a polarizer, however, only the light waves oscillating in a single plane pass through and the light is said to be plane-polarized. Light waves in all other planes are blocked out.\n\nBiot made the remarkable observation that when a beam of plane-polarized light passes through a solution of certain organic molecules, such as sugar or camphor, the plane of polarization is rotated through an angle, $\\alpha$. Not all organic substances exhibit this property, but those that do are said to be optically active.\n\nThe angle of rotation can be measured with an instrument called a polarimeter, represented in FIGURE 5.6. A solution of optically active organic molecules is placed in a sample tube, plane-polarized light is passed through the tube, and rotation of the polarization plane occurs. The light then goes through a second polarizer called the analyzer. By rotating the analyzer until the light passes through it, we can find the new plane of polarization and can tell to what extent rotation has occurred.\n\n\nFIGURE 5.6 Schematic representation of a polarimeter. Plane-polarized light passes through a solution of optically active molecules, which rotate the plane of polarization.\n\nIn addition to determining the extent of rotation, we can also find the direction. From the vantage point of the observer looking directly at the analyzer, some optically active molecules rotate polarized light to the left (counterclockwise) and are said to be levorotatory, whereas others rotate polarized light to the right (clockwise) and are said to be dextrorotatory. By convention, rotation to the left is given a minus sign (-) and rotation to the right is given a plus sign (+). (-)-Morphine, for example, is levorotatory, and (+)-sucrose is dextrorotatory."}
{"id": 274, "contents": "Strategy - 5.3 Optical Activity\nThe extent of rotation observed in a polarimetry experiment depends on the number of optically active molecules encountered by the light beam. This number, in turn, depends on sample concentration and sample pathlength. If the concentration of the sample is doubled, the observed rotation doubles. If the concentration is kept constant but the length of the sample tube is doubled, the observed rotation doubles. It also happens that the angle of rotation depends on the wavelength of the light used.\n\nTo express optical rotations in a meaningful way so that comparisons can be made, we have to choose standard conditions. The specific rotation, $[\\boldsymbol{\\alpha}]_{\\mathbf{D}}$, of a compound is defined as the observed rotation when light of 589.6 nanometer ( $\\mathrm{nm} ; 1 \\mathrm{~nm}=10^{-9} \\mathrm{~m}$ ) wavelength is used with a sample pathlength 1 of 1 decimeter ( $\\mathrm{dm} ; 1 \\mathrm{dm}=10$ cm ) and a sample concentration $c$ of $1 \\mathrm{~g} / \\mathrm{cm}^{3}$. (Light of 589.6 nm , the so-called sodium D line, is the yellow light emitted from common sodium street lamps.)\n\n$$\n[\\alpha]_{\\mathrm{D}}=\\frac{\\text { Observed rotation (degrees) }}{\\text { Pathlength, } l(\\mathrm{dm}) \\times \\text { Concentration, } c\\left(\\mathrm{~g} / \\mathrm{cm}^{3}\\right)}=\\frac{\\alpha}{l \\times c}\n$$"}
{"id": 275, "contents": "Strategy - 5.3 Optical Activity\nWhen optical rotations are expressed in this standard way, the specific rotation, $[\\alpha]_{\\mathrm{D}}$, is a physical constant characteristic of a given optically active compound. For example, (+)-lactic acid has $[\\alpha]_{D}=+3.82$, and (-)-lactic acid has $[\\alpha]_{D}=-3.82$. That is, the two enantiomers rotate plane-polarized light to exactly the same extent but in opposite directions. Note that the units of specific rotation are [ $\\left.\\left(\\mathrm{deg} \\cdot \\mathrm{cm}^{2}\\right) / \\mathrm{g}\\right]$ but the values are usually expressed without units. Some additional examples are listed in TABLE 5.1.\n\n| TABLE 5.1 Specific Rotations of Some Organic Molecules |  |  |  |\n| :--- | :--- | :--- | :---: |\n| Compound | $[\\boldsymbol{\\alpha}]_{\\mathrm{D}}$ | Compound | $[\\boldsymbol{\\alpha}]_{\\mathrm{D}}$ |\n| Penicillin V | +233 | Cholesterol | -31.5 |\n| Sucrose | +66.47 | Morphine | -132 |\n| Camphor | +44.26 | Cocaine | -16 |\n| Chloroform | 0 | Acetic acid | 0 |"}
{"id": 276, "contents": "Calculating an Optical Rotation - \nA 1.20 g sample of cocaine, $[\\alpha]_{D}=-16$, was dissolved in 7.50 mL of chloroform and placed in a sample tube having a pathlength of 5.00 cm . What was the observed rotation?\n\n\nCocaine"}
{"id": 277, "contents": "Strategy - \nSince $[\\alpha]_{\\mathrm{D}}=\\frac{\\alpha}{l \\times c}$\nThen $\\alpha=l \\times c \\times[\\alpha]_{\\mathrm{D}}$\nwhere $[\\alpha]_{\\mathrm{D}}=-16 ; 1=5.00 \\mathrm{~cm}=0.500 \\mathrm{dm} ; c=1.20 \\mathrm{~g} / 7.50 \\mathrm{~cm}^{3}=0.160 \\mathrm{~g} / \\mathrm{cm}^{3}$"}
{"id": 278, "contents": "Solution - \n$\\alpha=(-16)(0.500)(0.160)=-1.3^{\\circ}$.\n\nPROBLEM Is cocaine (Worked Example 5.2) dextrorotatory or levorotatory?\n5-5\nPROBLEM A 1.50 g sample of coniine, the toxic extract of poison hemlock, was dissolved in 10.0 mL of ethanol\n5-6 and placed in a sample cell with a 5.00 cm pathlength. The observed rotation at the sodium D line was $+1.21^{\\circ}$. Calculate $[\\alpha]_{D}$ for coniine."}
{"id": 279, "contents": "Solution - 5.4 Pasteur's Discovery of Enantiomers\nLittle was done to build on Biot's discovery of optical activity until 1848, when Louis Pasteur began work on a study of crystalline tartaric acid salts derived from wine. On crystallizing a concentrated solution of sodium ammonium tartrate below $28^{\\circ} \\mathrm{C}$, Pasteur made the surprising observation that two distinct kinds of crystals were obtained. Furthermore, the two kinds of crystals were nonsuperimposable mirror images and were related in the same way that a right hand is related to a left hand.\n\nWorking carefully with tweezers, Pasteur was able to separate the crystals into two piles, one of \"right-handed\" crystals and one of \"left-handed\" crystals, like those shown in FIGURE 5.7. Although the original sample, a 50 : 50 mixture of right and left, was optically inactive, solutions of the crystals from each of the sorted piles were optically active and their specific rotations were equal in magnitude but opposite in sign.\n\n\n\nSodium ammonium tartrate\nFIGURE 5.7 Drawings of sodium ammonium tartrate crystals taken from Pasteur's original sketches. One of the crystals is dextrorotatory\nin solution, and the other is levorotatory.\nPasteur was far ahead of his time. Although the structural theory of Kekul\u00e9 had not yet been proposed, Pasteur explained his results by speaking of the molecules themselves, saying, \"There is no doubt that [in the dextro tartaric acid] there exists an asymmetric arrangement having a nonsuperimposable image. It is no less certain that the atoms of the levo acid have precisely the inverse asymmetric arrangement.\" Pasteur's vision was extraordinary, for it was not until 25 years later that his ideas regarding asymmetric carbon atoms were confirmed.\n\nToday, we would describe Pasteur's work by saying that he had discovered enantiomers. Enantiomers, also called optical isomers, have identical physical properties, such as melting point and boiling point, but differ in the direction in which their solutions rotate plane-polarized light."}
{"id": 280, "contents": "Solution - 5.5 Sequence Rules for Specifying Configuration\nStructural drawings provide a visual representation of stereochemistry, but a written method for indicating the three-dimensional arrangement, or configuration, of substituents at a chirality center is also needed. The method used a set of sequence rules to rank the four groups attached to the chirality center and then looks at the handedness with which those groups are attached. Called the Cahn-Ingold-Prelog rules after the chemists who proposed them, the sequence rules are as follows:"}
{"id": 281, "contents": "RULE 1 - \nLook at the four atoms directly attached to the chirality center, and rank them according to atomic number. The atom with the highest atomic number has the highest ranking (first), and the atom with the lowest atomic number (usually hydrogen) has the lowest ranking (fourth). When different isotopes of the same element are compared, such as deuterium $\\left({ }^{2} \\mathrm{H}\\right)$ and protium $\\left({ }^{1} \\mathrm{H}\\right)$, the heavier isotope ranks higher than the lighter isotope. Thus, atoms commonly found in organic compounds have the following order.\n\n| Atomic number | 35 | 17 | 16 | 15 | 8 | 7 | 6 | $(2)$ | $(1)$ |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Higher rank | $\\mathrm{Br}>\\mathrm{Cl}>\\mathrm{S}$ | $>\\mathrm{P}$ | $>\\mathrm{O}$ | $>\\mathrm{N}$ | $>\\mathrm{C}$ | $>{ }^{2} \\mathrm{H}>{ }^{1} \\mathrm{H} \\quad$ Lower rank |  |  |  |\n\nRULE 2\nIf a decision can't be reached by ranking the first atoms in the substituent, look at the second, third, or fourth atoms away from the chirality center until the first difference is found. $\\mathrm{A}-\\mathrm{CH}_{2} \\mathrm{CH}_{3}$ substituent and a $-\\mathrm{CH}_{3}$ substituent are equivalent by rule 1 because both have carbon as the first atom. By rule 2, however, ethyl ranks higher than methyl because ethyl has a carbon as its highest second atom, while methyl has only hydrogen as its second atom. Look at the following pairs of examples to see how the rule works:\n\n\n\nHigher\n\n\nLower\n\n\nLower\n\n\nHigher\n\n\nLower\n\n\nHigher"}
{"id": 282, "contents": "RULE 1 - \nHigher\n\n\nLower\n\n\nLower\n\n\nHigher\n\n\nLower\n\n\nHigher\n\nRULE 3\nMultiple-bonded atoms are equivalent to the same number of single-bonded atoms. For example, an aldehyde substituent $(-\\mathrm{CH}=\\mathrm{O})$, which has a carbon atom doubly bonded to one oxygen, is equivalent to a substituent having a carbon atom singly bonded to two oxygens:\n\n\nAs further examples, the following pairs are equivalent:\n\n\nHaving ranked the four groups attached to a chiral carbon, we describe the stereochemical configuration around the carbon by orienting the molecule so that the group with the lowest ranking (4) points directly away from us. We then look at the three remaining substituents, which now appear to radiate toward us like the spokes on a steering wheel (FIGURE 5.8). If a curved arrow drawn from the highest to second-highest to thirdhighest ranked substituent $(1 \\rightarrow 2 \\rightarrow 3)$ is clockwise, we say that the chirality center has the $\\boldsymbol{R}$ configuration ( $S$ for the Latin rectus, meaning \"right\"). If an arrow from $1 \\rightarrow 2 \\rightarrow 3$ is counterclockwise, the chirality center has the $\\boldsymbol{S}$ configuration (Latin sinister, meaning \"left\"). To remember these assignments, think of a car's steering wheel when making a Right (clockwise) turn.\n\n\nFIGURE 5.8 Assigning $\\boldsymbol{R}$ and $\\boldsymbol{S}$ configurations to chirality centers. When the molecule is oriented so that the lowest-ranked group (4) is toward the rear, the remaining three groups radiate toward the viewer like the spokes of a steering wheel. If the direction of travel $1 \\rightarrow 2 \\rightarrow 3$ is clockwise (right turn), the center has the $R$ configuration. If the direction of travel $1 \\rightarrow 2 \\rightarrow 3$ is counterclockwise (left turn), the center is $S$."}
{"id": 283, "contents": "RULE 1 - \nLook at (-)-lactic acid in FIGURE 5.9 for an example of how to assign configuration. Sequence rule 1 says that -OH is ranked 1 and -H is ranked 4 , but it doesn't allow us to distinguish between $-\\mathrm{CH}_{3}$ and $-\\mathrm{CO}_{2} \\mathrm{H}$ because both groups have carbon as their first atom. Sequence rule 2, however, says that $-\\mathrm{CO}_{2} \\mathrm{H}$ ranks higher than $-\\mathrm{CH}_{3}$ because O (the highest second atom in $-\\mathrm{CO}_{2} \\mathrm{H}$ ) outranks H (the highest second atom in $-\\mathrm{CH}_{3}$ ). Now, turn the\nmolecule so that the fourth-ranked group ( -H ) is oriented toward the rear, away from the observer. Since a curved arrow from $1(-\\mathrm{OH})$ to $2\\left(-\\mathrm{CO}_{2} \\mathrm{H}\\right)$ to $3\\left(-\\mathrm{CH}_{3}\\right)$ is clockwise (right turn of the steering wheel), ( - )-lactic acid has the $R$ configuration. Applying the same procedure to (+)-lactic acid leads to the opposite assignment.\n(a)\n\n$R$ configuration\n(-)-Lactic acid\n\n(b)\n\n\n\n$S$ configuration\n(+)-Lactic acid\nFIGURE 5.9 Assigning configuration to (a)(R)-(-)-lactic acid and (b) (S)-(+)-lactic acid.\n\nFurther examples are provided by naturally occurring (-)-glyceraldehyde and (+)-alanine, which both have the $S$ configuration as shown in FIGURE 5.10. Note that the sign of optical rotation, $(+)$ or $(-)$, is not related to the $R, S$ designation. (S)-Glyceraldehyde happens to be levorotatory (-), and (S)-alanine happens to be dextrorotatory $(+)$. There is no simple correlation between $R, S$ configuration and direction or magnitude of optical rotation.\n(a)\n\n\n(S)-Glyceraldehyde\n[(S)-(-)-2,3-Dihydroxypropanal]\n$[\\alpha]_{D}=-8.7$\n(b)"}
{"id": 284, "contents": "RULE 1 - \n(S)-Glyceraldehyde\n[(S)-(-)-2,3-Dihydroxypropanal]\n$[\\alpha]_{D}=-8.7$\n(b)\n\n(S)-Alanine\n[(S)-(+)-2-Aminopropanoic acid]\n$[\\alpha]_{D}=+8.5$\n\nFIGURE 5.10 Assigning configuration to (a) (-)-glyceraldehyde. (b) (+)-alanine. Both happen to have the Sconfiguration, although one is levorotatory and the other is dextrorotatory.\n\nOne additional point needs to be mentioned-the matter of absolute configuration. How do we know that the assignments of $R$ and $S$ configuration are correct in an absolute sense, rather than a relative, sense? Since\nthere is no correlation between the $R, S$ configuration and the direction or magnitude of optical rotation, how do we know that the $R$ configuration belongs to the levorotatory enantiomer of lactic acid? This difficult question was finally solved in 1951, when an X-ray diffraction method was found for determining the absolute spatial arrangement of atoms in a molecule. Based on those results, we can say with certainty that the $R, S$ conventions are correct."}
{"id": 285, "contents": "Assigning Configuration to Chirality Centers - \nOrient each of the following drawings so that the lowest-ranked group is toward the rear, and then assign $R$ or $S$ configuration to each:\n(a)\n\n(b)"}
{"id": 286, "contents": "Strategy - \nIt takes practice to be able to visualize and orient a chirality center in three dimensions. You might start by indicating where the observer must be located $-180^{\\circ}$ opposite the lowest-ranked group. Then imagine yourself in the position of the observer, and redraw what you would see."}
{"id": 287, "contents": "Solution - \nIn (a), you would be located in front of the page toward the top right of the molecule, and you would see group 2 to your left, group 3 to your right, and group 1 below you. This corresponds to an $R$ configuration.\n\n\nIn (b), you would be located behind the page toward the top left of the molecule from your point of view, and you would see group 3 to your left, group 1 to your right, and group 2 below you. This also corresponds to an $R$ configuration."}
{"id": 288, "contents": "Drawing the Three-Dimensional Structure of a Specific Enantiomer - \nDraw a tetrahedral representation of ( $R$ )-2-chlorobutane."}
{"id": 289, "contents": "Strategy - \nBegin by ranking the four substituents bonded to the chirality center: (1) -Cl , (2) $-\\mathrm{CH}_{2} \\mathrm{CH}_{3}$, (3) $-\\mathrm{CH}_{3}$, (4) -H . To draw a tetrahedral representation of the molecule, orient the lowest-ranked group ( -H ) away from you and imagine that the other three groups are coming out of the page toward you. Then, place the remaining three substituents such that the direction of travel $1 \\rightarrow 2 \\rightarrow 3$ is clockwise (right turn), and tilt the molecule toward you to bring the rear hydrogen into view. Using molecular models is a real help in working problems of this sort."}
{"id": 290, "contents": "Solution - \n(R)-2-Chlorobutane\n\nPROBLEM Which member in each of the following sets ranks higher?\n5-7 (a) -H or -Br\n(b) -Cl or -Br\n(c) $-\\mathrm{CH}_{3}$ or $-\\mathrm{CH}_{2} \\mathrm{CH}_{3}$\n(d) $-\\mathrm{NH}_{2}$ or -OH\n(e) $-\\mathrm{CH}_{2} \\mathrm{OH}$ or $-\\mathrm{CH}_{3}$\n(f) $-\\mathrm{CH}_{2} \\mathrm{OH}$ or $-\\mathrm{CH}=\\mathrm{O}$\n\nPROBLEM Rank each of the following sets of substituents:\n5-8 (a) $-\\mathrm{H},-\\mathrm{OH},-\\mathrm{CH}_{2} \\mathrm{CH}_{3},-\\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}$\n(b) $-\\mathrm{CO}_{2} \\mathrm{H},-\\mathrm{CO}_{2} \\mathrm{CH}_{3},-\\mathrm{CH}_{2} \\mathrm{OH},-\\mathrm{OH}$\n(c) $-\\mathrm{CN},-\\mathrm{CH}_{2} \\mathrm{NH}_{2},-\\mathrm{CH}_{2} \\mathrm{NHCH}_{3},-\\mathrm{NH}_{2}$ (d) $-\\mathrm{SH},-\\mathrm{CH}_{2} \\mathrm{SCH}_{3},-\\mathrm{CH}_{3},-\\mathrm{SSCH}_{3}$\n\nPROBLEM Orient each of the following drawings so that the lowest-ranked group is toward the rear, and then\n5-9 assign $R$ or $S$ configuration:\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Assign $R$ or $S$ configuration to the chirality center in each of the following molecules:\n5-10 (a)\n\n(b)\n\n(c)\n\n\nPROBLEM Draw a tetrahedral representation of ( $S$ )-2-pentanol (2-hydroxypentane).\n5-11\nPROBLEM Assign $R$ or $S$ configuration to the chirality center in the following molecular model of the amino 5-12 acid methionine (blue $=\\mathrm{N}$, yellow $=\\mathrm{S}$ ):"}
{"id": 291, "contents": "Solution - 5.6 Diastereomers\nMolecules like lactic acid, alanine, and glyceraldehyde are relatively simple because each has only one chirality center and thus only two stereoisomers. The situation becomes more complex, however, with molecules that have more than one chirality center. As a general rule, a molecule with $n$ chirality centers can have up to\n$2^{n}$ stereoisomers (although it may have fewer, as we'll see below). Take the amino acid threonine (2-amino-3-hydroxybutanoic acid), for example. Since threonine has two chirality centers (C2 and C3), there are four possible stereoisomers, as shown in FIGURE 5.11. Check for yourself that the $R, S$ configurations of all stereoisomers are correct.\nEnantiomers\n\n\n\nEnantiomers\n\nFIGURE 5.11 The four stereoisomers of 2-amino-3-hydroxybutanoic acid.\nThe four stereoisomers of 2-amino-3-hydroxybutanoic acid can be grouped into two pairs of enantiomers. The $2 R, 3 R$ stereoisomer is the mirror image of $2 S, 3 S$, and the $2 R, 3 S$ stereoisomer is the mirror image of $2 S, 3 R$. But what is the relationship between any two molecules that are not mirror images? What, for instance, is the relationship between the $2 R, 3 R$ isomer and the $2 R, 3 S$ isomer? They are stereoisomers, yet they aren't enantiomers. To describe such a relationship, we need a new term-diastereomer.\n\nDiastereomers (dia-stair-e-oh-mers) are stereoisomers that are not mirror images. Since we used the right-hand/left-hand analogy to describe the relationship between two enantiomers, we might extend the analogy by saying that the relationship between diastereomers is like that of hands from different people. Your hand and your friend's hand look similar, but they aren't identical and they aren't mirror images. The same is true of diastereomers: they're similar, but they aren't identical and they aren't mirror images."}
{"id": 292, "contents": "Solution - 5.6 Diastereomers\nNote carefully the difference between enantiomers and diastereomers: enantiomers have opposite configurations at all chirality centers, whereas diastereomers have opposite configurations at some (one or more) chirality centers but the same configuration at others. A full description of the four stereoisomers of threonine is given in TABLE 5.2. Of the four, only the $2 S, 3 R$ isomer, $[\\alpha]_{\\mathrm{D}}=-28.3$, occurs naturally in plants and animals and is an essential nutrient for humans. This result is typical: most biological molecules are chiral, and usually only one stereoisomer is found in nature.\n\n| TABLE 5.2 Relationships among the Four Stereoisomers of <br> Threonine |\n| :--- | :--- |\n| Stereoisomer Enantiomer Diastereomer <br> $2 R, 3 R$ $2 S, 3 S$ $2 R, 3 S$ and $2 S, 3 R$ <br> $2 S, 3 S$ $2 R, 3 R$ $2 R, 3 S$ and $2 S, 3 R$ |\n\nTABLE 5.2 Relationships among the Four Stereoisomers of\nThreonine\n\n| Stereoisomer | Enantiomer | Diastereomer |\n| :--- | :--- | :--- |\n| $2 R, 3 S$ | $2 S, 3 R$ | $2 R, 3 R$ and $2 S, 3 S$ |\n| $2 S, 3 R$ | $2 R, 3 S$ | $2 R, 3 R$ and $2 S, 3 S$ |\n\nIn the special case where two diastereomers differ at only one chirality center but are the same at all others, we say that the compounds are epimers. Cholestanol and coprostanol, for instance, are both found in human feces, and both have nine chirality centers. Eight of the nine are identical, but the one at C5 is different. Thus, cholestanol and coprostanol are epimeric at C5.\n\n\nCholestanol\n\n\nCoprostanol"}
{"id": 293, "contents": "Solution - 5.6 Diastereomers\nCholestanol\n\n\nCoprostanol\n\nEpimers\nPROBLEM One of the following molecules (a)-(d) is D-erythrose 4-phosphate, an intermediate in the Calvin\n5-13 photosynthetic cycle by which plants incorporate $\\mathrm{CO}_{2}$ into carbohydrates. If D-erythrose 4-phosphate has $R$ stereochemistry at both chirality centers, which of the structures is it? Which of the remaining three structures is the enantiomer of D-erythrose 4-phosphate, and which are diastereomers?\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM How many chirality centers does morphine have? How many stereoisomers of morphine are 5-14 possible in principle?"}
{"id": 294, "contents": "Morphine - \nPROBLEM Assign $R$ or $S$ configuration to each chirality center in the following molecular model of the amino\n5-15 acid isoleucine (blue $=\\mathrm{N}$ ):"}
{"id": 295, "contents": "Morphine - 5.7 Meso Compounds\nLet's look at another example (Section 5.4) of a compound with more than one chirality center: the tartaric acid used by Pasteur. The four stereoisomers can be drawn as follows:\n\n\n2R,3R\n\n\n2S,3S\n\n\n\nThe $2 R, 3 R$ and $2 S, 3 S$ structures are nonsuperimposable mirror images and therefore represent a pair of enantiomers. A close look at the $2 R, 3 S$ and $2 S, 3 R$ structures, however, shows that they are superimposable, and thus identical, as can be seen by rotating one structure $180^{\\circ}$.\n\n\nIdentical\nThe $2 R, 3 S$ and $2 S, 3 R$ structures are identical because the molecule has a plane of symmetry and is therefore achiral. The symmetry plane cuts through the C2-C3 bond, making one half of the molecule a mirror image of the other half (FIGURE 5.12). Because of the plane of symmetry, the molecule is achiral, despite the fact that it has two chirality centers. Such compounds, which are achiral, yet contain chirality centers, are called meso compounds (me-zo). Thus, tartaric acid exists in three stereoisomeric forms: two enantiomers and one meso form.\n\n\n\nFIGURE 5.12 A symmetry plane through the C2-C3 bond of meso-tartaric acid makes the molecule achiral.\nSome physical properties of the three stereoisomers are listed in TABLE 5.3. The (+)- and (-)-tartaric acids have identical melting points, solubilities, and densities, but they differ in the sign of their rotation of planepolarized light. The meso isomer, by contrast, is diastereomeric with the (+) and (-) forms. It has no mirrorimage relationship to (+)- and (-)-tartaric acids, is a different compound altogether, and thus has different\nphysical properties.\nTABLE 5.3 Some Properties of the Stereoisomers of Tartaric Acid Some Properties of the Stereoisomers of Tartaric Acid"}
{"id": 296, "contents": "Morphine - 5.7 Meso Compounds\n| Stereoisomer | Melting point $\\left({ }^{\\circ} \\mathrm{C}\\right)$ | $[\\alpha]_{\\mathrm{D}}$ | Density $\\left(\\mathrm{g} / \\mathrm{cm}^{3}\\right)$ | Solubility at $20{ }^{\\circ} \\mathrm{C}$ <br> $(\\mathrm{g} / 100 \\mathrm{~mL} \\mathrm{H}$ |\n| :--- | :--- | ---: | :--- | :--- |\n| $(+)$ | $168-170$ | +12 | 1.7598 | 139.0 |\n| $(-)$ | $168-170$ | -12 | 1.7598 | 139.0 |\n| Meso | $146-148$ | 0 | 1.6660 | 125.0 |"}
{"id": 297, "contents": "Distinguishing Chiral Compounds from Meso Compounds - \nDoes cis-1,2-dimethylcyclobutane have any chirality centers? Is it chiral?"}
{"id": 298, "contents": "Strategy - \nTo see whether a chirality center is present, look for a carbon atom bonded to four different groups. To see whether the molecule is chiral, look for the presence or absence of a symmetry plane. Not all molecules with chirality centers are chiral overall-meso compounds are an exception."}
{"id": 299, "contents": "Solution - \nA look at the structure of cis-1,2-dimethylcyclobutane shows that both methyl-bearing ring carbons (C1 and C2) are chirality centers. Overall, though, the compound is achiral because there is a symmetry plane bisecting the ring between C 1 and C 2 . Thus, the molecule is a meso compound.\n\n\nPROBLEM Which of the following structures represent meso compounds?\n5-16\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM Which of the following have a meso form? (Recall that the -ol suffix refers to an alcohol, ROH.)\n5-17\n(a) 2,3-Butanediol\n(b) 2,3-Pentanediol\n(c) 2,4-Pentanediol\n\nPROBLEM Does the following structure represent a meso compound? If so, indicate the symmetry plane.\n5-18"}
{"id": 300, "contents": "Solution - 5.8 Racemic Mixtures and the Resolution of Enantiomers\nTo end this discussion of stereoisomerism, let's return for a last look at Pasteur's pioneering work, described in Section 5.4. Pasteur took an optically inactive tartaric acid salt and found that he could crystallize from it two optically active forms having what we would now call $2 R, 3 R$ and $2 S, 3 S$ configurations. But what was the optically inactive form he started with? It couldn't have been meso-tartaric acid, because meso-tartaric acid is a different chemical compound and can't interconvert with the two chiral enantiomers without breaking and re-forming chemical bonds.\n\nThe answer is that Pasteur started with a $50: 50$ mixture of the two chiral tartaric acid enantiomers. Such a mixture is called a racemate (rass-uh-mate), or racemic mixture, and is denoted by either the symbol $( \\pm)$ or the prefix $d, l$ to indicate an equal mixture of dextrorotatory and levorotatory forms. Racemates show no optical rotation because the (+) rotation from one enantiomer exactly cancels the ( - ) rotation from the other. Through good luck, Pasteur was able to separate, or resolve, racemic tartaric acid into its (+) and (-) enantiomers. Unfortunately, the fractional crystallization technique he used doesn't work for most racemates, so other methods are needed.\n\nThe most common method for resolving the racemate of a chiral carboxylic acid $\\left(\\mathrm{RCO}_{2} \\mathrm{H}\\right)$ is to carry out an acidbase reaction between the acid and an amine base $\\left(\\mathrm{RNH}_{2}\\right)$ to yield an ammonium salt:"}
{"id": 301, "contents": "Solution - 5.8 Racemic Mixtures and the Resolution of Enantiomers\nTo understand how this method of resolution works, let's see what happens when a racemic mixture of chiral acids, such as (+)- and (-)-lactic acids, reacts with an achiral amine base, such as methylamine, $\\mathrm{CH}_{3} \\mathrm{NH}_{2}$. The situation is analogous to what happens when left and right hands (chiral) pick up a ball (achiral). Both left and right hands pick up the ball equally well, and the products-ball in right hand versus ball in left hand-are mirror images. In the same way, both (+)- and (-)-lactic acid react with methylamine equally well, and the product is a racemic mixture of the two enantiomers methylammonium (+)-lactate and methylammonium (-)-lactate (FIGURE 5.13).\n(R)\n\n(S)\n\nRacemic lactic acid\n(50\\% R, 50\\% S)\n\n\nRacemic ammonium salt\n(50\\% R, 50\\% S)\n\nFIGURE 5.13 Reaction of racemic lactic acid with achiral methylamine gives a racemic mixture of ammonium salts.\nNow let's see what happens when the racemic mixture of (+)- and (-)-lactic acids reacts with a single enantiomer of a chiral amine base, such as ( $R$ )-1-phenylethylamine. The situation is analogous to what happens when left and right hands (chiral) put on a right-handed glove (also chiral). Left and right hands don't put on the righthanded glove in the same way, so the products-right hand in right glove versus left hand in right glove-are not mirror images; they're similar but different."}
{"id": 302, "contents": "Solution - 5.8 Racemic Mixtures and the Resolution of Enantiomers\nIn the same way, (+)- and (-)-lactic acids react with ( $R$ )-1-phenylethylamine to give two different products (FIGURE 5.14). ( $R$ )-Lactic acid reacts with ( $R$ )-1-phenylethylamine to give the $R, R$ salt, and ( $S$ )-lactic acid reacts with the $R$ amine to give the $S, R$ salt. The two salts are diastereomers, not enantiomers. They have different chemical and physical properties, and it may therefore be possible to separate them by crystallization or some other means. Once separated, acidification of the two diastereomeric salts with a strong acid makes it possible to isolate the two pure enantiomers of lactic acid and to recover the chiral amine for reuse.\n\n\nFIGURE 5.14 Reaction of racemic lactic acid with ( $R$ )-1-phenylethylamine yields a mixture of diastereomeric ammonium salts, which have different properties and can be separated."}
{"id": 303, "contents": "Predicting the Chirality of a Reaction Product - \nWe'll see in Section 21.3 that carboxylic acids $\\left(\\mathrm{RCO}_{2} \\mathrm{H}\\right)$ react with alcohols ( $\\mathrm{R}^{\\prime} \\mathrm{OH}$ ) to form esters ( $\\mathrm{RCO}_{2} \\mathrm{R}^{\\prime}$ ).\n\nSuppose that ( $\\pm$ )-lactic acid reacts with $\\mathrm{CH}_{3} \\mathrm{OH}$ to form the ester, methyl lactate. What stereochemistry would you expect the product(s) to have? What is the relationship of the products?"}
{"id": 304, "contents": "Solution - \nReaction of a racemic acid with an achiral alcohol such as methanol yields a racemic mixture of mirror-image (enantiomeric) products.\n\n\nPROBLEM Suppose that acetic acid $\\left(\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right)$ reacts with (S)-2-butanol to form an ester (see Worked Example 5-19 5.6). What stereochemistry would you expect the product(s) to have? What is the relationship of the products?\n\n\nPROBLEM What stereoisomers would result from reaction of ( $\\pm$ )-lactic acid with ( $S$ )-1-phenylethylamine, and 5-20 what is the relationship between them?"}
{"id": 305, "contents": "Solution - 5.9 A Review of Isomerism\nAs noted on several previous occasions, isomers are compounds with the same chemical formula but different structures. We've seen several kinds of isomers in the past few chapters, and it's a good idea at this point to see how they relate to one another (FIGURE 5.15).\n\n\nFIGURE 5.15 A summary of the different kinds of isomers.\nThere are two fundamental types of isomers, both of which we've now encountered: constitutional isomers and stereoisomers.\n\nConstitutional isomers (Section 3.2) are compounds whose atoms are connected differently. Among the kinds of constitutional isomers we've seen are skeletal, functional, and positional isomers.\n\n| Different carbon skeletons | <smiles>CC(C)C</smiles> | and | $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$ |\n| :---: | :---: | :---: | :---: |\n|  | 2-Methylpropane |  | Butane |\n| Different functional groups | $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$ | and | $\\mathrm{CH}_{3} \\mathrm{OCH}_{3}$ |\n|  | Ethyl alcohol |  | Dimethyl ether |\n| Different position of functional groups | <smiles>CC(C)N</smiles> | and | $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}$ |\n|  | Isopropylamine |  | Propylamine |\n\nStereoisomers (Section 4.2) are compounds whose atoms are connected in the same order but with a different spatial arrangement. Among the kinds of stereoisomers we've seen are enantiomers, diastereomers, and cis-trans isomers of cycloalkanes. Actually, cis-trans isomers are just a subclass of diastereomers because they are non-mirror-image stereoisomers:\n\n```\nEnantiomers (nonsuperimposable mirror-image stereoisomers)\n```\n\n```\nDiastereomers\n(nonsuperimposable\nnon-mirror-image\nstereoisomers)\n```\n\nConfigurational diastereomers"}
{"id": 306, "contents": "Solution - 5.9 A Review of Isomerism\n```\nDiastereomers\n(nonsuperimposable\nnon-mirror-image\nstereoisomers)\n```\n\nConfigurational diastereomers\n\nCis-trans diastereomers (substituents on same side or opposite side of double bond or ring)\n\n(R)-Lactic acid\n\n(2R,3R)-2-Amino-3hydroxybutanoic acid\n\ntrans-1,3-Dimethylcyclopentane\n\n(S)-Lactic acid\n\n(2R,3S)-2-Amino-3hydroxybutanoic acid\n\ncis-1,3-Dimethylcyclopentane\n\nPROBLEM What kinds of isomers are the following pairs?\n5-21 (a) (S)-5-Chloro-2-hexene and chlorocyclohexane\n(b) $(2 R, 3 R)$-Dibromopentane and $(2 S, 3 R)$-dibromopentane"}
{"id": 307, "contents": "Solution - 5.10 Chirality at Nitrogen, Phosphorus, and Sulfur\nAs noted previously, the most common cause of chirality in a molecule is the presence of four different substituents bonded to a tetrahedral atom. Although that atom is usually carbon, it doesn't necessarily have to be. Nitrogen, phosphorus, and sulfur atoms are all commonly encountered in organic molecules, and can all be chirality centers. We know, for instance, that trivalent nitrogen is tetrahedral, with its lone pair of electrons acting as the fourth \"substituent\" (Section 1.10). Is trivalent nitrogen chiral? Does a compound such as ethylmethylamine exist as a pair of enantiomers?\n\nThe answer is both yes and no. Yes in principle, but no in practice. It turns out that most trivalent nitrogen compounds undergo a rapid umbrella-like inversion that interconverts enantiomers, so we can't isolate individual enantiomers except in special cases.\n\n\nA similar situation occurs in trivalent phosphorus compounds, called phosphines, but the inversion at phosphorus is substantially slower than inversion at nitrogen, so stable chiral phosphines can be isolated. ( $R$ )and ( $S$ )-methylpropylphenylphosphine, for instance, are configurationally stable for several hours at $100{ }^{\\circ} \\mathrm{C}$. We'll see the importance of phosphine chirality in Section 26.7 in connection with the synthesis of chiral amino acids.\n\n(R)-Methylpropylphenylphosphine (configurationally stable)"}
{"id": 308, "contents": "Solution - 5.10 Chirality at Nitrogen, Phosphorus, and Sulfur\n(R)-Methylpropylphenylphosphine (configurationally stable)\n\nDivalent sulfur compounds are achiral, but trivalent sulfur compounds called sulfonium salts $\\left(\\mathrm{R}_{3} \\mathrm{~S}^{+}\\right)$can be chiral. Like phosphines, sulfonium salts undergo relatively slow inversion, so chiral sulfonium salts are configurationally stable and can be isolated. Perhaps the best known example is the coenzyme $S$-adenosylmethionine, the so-called biological methyl donor, which is involved in many metabolic pathways as a source of $\\mathrm{CH}_{3}$ groups. (The \" $S$ \" in the name $S$-adenosylmethionine stands for sulfur and means that the adenosyl group is attached to the sulfur atom of the amino acid methionine.) The molecule has $S$ stereochemistry at sulfur and is configurationally stable for several days at room temperature. Its $R$ enantiomer is also known but is not biologically active.\n\n(S)-S-Adenosylmethionine\n\nAdenosine"}
{"id": 309, "contents": "Solution - 5.11 Prochirality\nClosely related to the concept of chirality, and particularly important in biological chemistry, is the notion of prochirality. A molecule is said to be prochiral if it can be converted from achiral to chiral in a single chemical step. For instance, an unsymmetrical ketone like 2-butanone is prochiral because it can be converted to the chiral alcohol 2-butanol by the addition of hydrogen, as we'll see in Section 17.4.\n\n\nWhich enantiomer of 2-butanol is produced depends on which face of the planar carbonyl group undergoes reaction. To distinguish between the possibilities, we use the stereochemical descriptors Re and Si. Rank the\nthree groups attached to the trigonal, $s p^{2}$-hybridized carbon, and imagine curved arrows from the highest to second-highest to third-highest ranked substituents. The face on which the arrows curve clockwise is designated the Re face (similar to $R$ ), and the face on which the arrows curve counterclockwise is designated the $\\boldsymbol{S i}$ face (similar to $S$ ). In this example, addition of hydrogen from the Re face gives ( $S$ )-2-butane, and addition from the Si face gives ( $R$ )-2-butane.\n\n\nIn addition to compounds with planar, $s p^{2}$-hybridized atoms, compounds with tetrahedral, $s p^{3}$-hybridized atoms can also be prochiral. An $s p^{3}$-hybridized atom is said to be a prochirality center if, by changing one of its attached groups, it becomes a chirality center. The $-\\mathrm{CH}_{2} \\mathrm{OH}$ carbon atom of ethanol, for instance, is a prochirality center because changing one of its attached -H atoms converts it into a chirality center."}
{"id": 310, "contents": "Solution - 5.11 Prochirality\nEthanol\nTo distinguish between the two identical atoms (or groups of atoms) on a prochirality center, we imagine a change that will raise the ranking of one atom over the other without affecting its rank with respect to other attached groups. On the $-\\mathrm{CH}_{2} \\mathrm{OH}$ carbon of ethanol, for instance, we might imagine replacing one of the ${ }^{1} \\mathrm{H}$ atoms (protium) by ${ }^{2} \\mathrm{H}$ (deuterium). The newly introduced ${ }^{2} \\mathrm{H}$ atom ranks higher than the remaining ${ }^{1} \\mathrm{H}$ atom, but it remains lower than other groups attached to the carbon. Of the two identical atoms in the original compound, the atom whose replacement leads to an $R$ chirality center is said to be pro- $\\boldsymbol{R}$ and the atom whose replacement leads to an $S$ chirality center is pro-S.\n\n\nA large number of biological reactions involve prochiral compounds. One of the steps in the citric acid cycle by which food is metabolized, for instance, is the addition of $\\mathrm{H}_{2} \\mathrm{O}$ to fumarate to give malate. Addition of -OH occurs on the Si face of a fumarate carbon and gives $(S)$-malate as product.\n\n\nAs another example, studies with deuterium-labeled substrates have shown that the reaction of ethanol with the coenzyme nicotinamide adenine dinucleotide $\\left(\\mathrm{NAD}^{+}\\right)$, catalyzed by yeast alcohol dehydrogenase, occurs with exclusive removal of the pro- $R$ hydrogen from ethanol and with addition only to the Re face of $\\mathrm{NAD}^{+}$.\n\n\nEthanol\nNAD ${ }^{+}$\nAcetaldehyde\nNADH\nDetermining the stereochemistry of reactions at prochirality centers is a powerful method for studying detailed mechanisms in biochemical reactions. As just one example, the conversion of citrate to cis-aconitate in the citric acid cycle has been shown to occur with loss of a pro- $R$ hydrogen, implying that the OH and H groups leave from opposite sides of the molecule."}
{"id": 311, "contents": "Solution - 5.11 Prochirality\nPROBLEM Identify the indicated hydrogens in the following molecules as pro- $R$ or pro- $S$ :\n\n5-22 (a)\n\n(S)-Glyceraldehyde\n(b)\n\n\nPhenylalanine\n\nPROBLEM Identify the indicated faces of carbon atoms in the following molecules as Re or Si :\n\n5-23 (a)\n\n\nHydroxyacetone\n(b)\n\n\nCrotyl alcohol\n\nPROBLEM The lactic acid that builds up in tired muscles is formed from pyruvate. If the reaction occurs with\n5-24 addition of hydrogen to the Re face of pyruvate, what is the stereochemistry of the product?\n\n\nPROBLEM The aconitase-catalyzed addition of water to cis-aconitate in the citric acid cycle occurs with the\n5-25 following stereochemistry. Does the addition of the OH group occur on the Re or Si face of the substrate? What about the addition of the H? Do the H and OH groups add from the same side of the double bond or from opposite sides?\n\ncis-Aconitate\n(2R,3S)-Isocitrate"}
{"id": 312, "contents": "Solution - 5.12 Chirality in Nature and Chiral Environments\nAlthough the different enantiomers of a chiral molecule have the same physical properties, they usually have different biological properties. For example, a change in chirality can affect the biological properties of many drugs, such as fluoxetine, a heavily prescribed medication sold under the trade name Prozac. Racemic fluoxetine is an effective antidepressant but has no activity against migraine. The pure $S$ enantiomer, however, works remarkably well in preventing migraine. Other examples of how chirality affects biological properties are given in the Chapter 5 Chemistry Matters at the end of this chapter.\n\n(S)-Fluoxetine (prevents migraine)\n\n\nWhy do different enantiomers have different biological properties? To have a biological effect, a substance typically must fit into an appropriate receptor that has a complementary shape. But because biological receptors are chiral, only one enantiomer of a chiral substrate can fit, just as only a right hand can fit into a right-handed glove. The mirror-image enantiomer will be a misfit, like a left hand in a right-handed glove. A representation of the interaction between a chiral molecule and a chiral biological receptor is shown in FIGURE 5.16: one enantiomer fits the receptor perfectly, but the other does not.\n\n\nFIGURE 5.16 Interaction of a chiral object with a chiral receptor. A left hand interacts with a chiral object, much as a biological receptor interacts with a chiral molecule. (a) One enantiomer fits into the hand perfectly: green thumb, red palm, and gray pinkie finger, with the blue substituent exposed. (b) The other enantiomer, however, can't fit into the hand. When the green thumb and gray pinkie finger interact appropriately, the palm holds a blue substituent rather than a red one, with the red substituent exposed.\n\nThe hand-in-glove fit of a chiral substrate into a chiral receptor is relatively straightforward, but it's less obvious how a prochiral substrate can undergo a selective reaction. Take the reaction of ethanol with $\\mathrm{NAD}^{+}$catalyzed by yeast alcohol dehydrogenase. As we saw at the end of Section 5.11, this reaction occurs with exclusive removal of the pro- $R$ hydrogen from ethanol and with addition only to the Re face of the $\\mathrm{NAD}^{+}$carbon."}
{"id": 313, "contents": "Solution - 5.12 Chirality in Nature and Chiral Environments\nWe can understand this result by imagining that the chiral enzyme receptor again has three binding sites, as in\n\nFIGURE 5.16. When green and gray substituents of a prochiral substrate are held appropriately, however, only one of the two red substituents-say, the pro- $S$ one-is also held while the other, pro- $R$, substituent is exposed for reaction.\n\nWe describe the situation by saying that the receptor provides a chiral environment for the substrate. In the absence of a chiral environment, the two red substituents are chemically identical, but in the presence of a chiral environment, they are chemically distinctive (FIGURE 5.17a). The situation is similar to what happens when you pick up a coffee mug. By itself, the mug has a plane of symmetry and is achiral. When you pick up the mug, however, your hand provides a chiral environment so that one side becomes much more accessible and easier to drink from than the other (FIGURE 5.17b).\n\n\nFIGURE 5.17 (a) When a prochiral molecule is held in a chiral environment, the two seemingly identical substituents are distinguishable. (b) Similarly, when an achiral coffee mug is held in the chiral environment of your hand, it's much easier to drink from one side than the other because the two sides of the mug are now distinguishable."}
{"id": 314, "contents": "Chiral Drugs - \nThe many hundreds of different pharmaceutical agents approved for use by the U.S. Food and Drug Administration come from many sources. Many drugs are isolated directly from plants or bacteria, and others are made by chemical modification of naturally occurring compounds. An estimated $33 \\%$, however, are made entirely in the laboratory and have no relatives in nature.\n\n\nFIGURE 5.18 The $S$ enantiomer of ibuprofen soothes the aches and pains of athletic injuries much more effectively than the $R$ enantiomer. (credit: \"World Athletic Championships 2007 in Osaka\" by Eckhard Pecher/Wikimedia Commons, CC BY 2.5)\n\nThose drugs that come from natural sources, either directly or after laboratory modification, are usually chiral and are generally found only as a single enantiomer rather than as a racemate. Penicillin V, for example, an antibiotic isolated from the Penicillium mold, has the $2 S, 5 R, 6 R$ configuration. Its enantiomer, which does not occur naturally but can be made in the laboratory, has no antibiotic activity.\n\n\nPenicillin V ( $2 S, 5 R, 6 R$ configuration)\nIn contrast to drugs from natural sources, drugs that are made entirely in the laboratory are either achiral or, if chiral, are often produced and sold as racemates. Ibuprofen, for example, has one chirality center and is sold commercially under such trade names as Advil, Nuprin, and Motrin as a $50: 50$ mixture of $R$ and $S$. It turns out, however, that only the $S$ enantiomer is active as an analgesic and anti-inflammatory agent. The $R$ enantiomer of ibuprofen is inactive, although it is slowly converted in the body to the active $S$ form.\n\n(S)-Ibuprofen (an active analgesic agent)"}
{"id": 315, "contents": "Chiral Drugs - \n(S)-Ibuprofen (an active analgesic agent)\n\n\nNot only is it chemically wasteful to synthesize and administer an enantiomer that does not serve the intended purpose, many instances are now known where the presence of the \"wrong\" enantiomer in a racemic mixture either affects the body's ability to utilize the \"right\" enantiomer or has unintended pharmacological effects of its own. The presence of $(R)$-ibuprofen in the racemic mixture, for instance, slows the rate at which the $S$ enantiomer takes effect in the body, from 12 minutes to 38 minutes.\n\nTo get around this problem, pharmaceutical companies attempt to devise methods of enantioselective synthesis, which allow them to prepare only a single enantiomer rather than a racemic mixture. Viable methods have been developed for the preparation of (S)-ibuprofen, which is marketed in Europe. We'll look further into enantioselective synthesis in the Chapter 19 Chemistry Matters."}
{"id": 316, "contents": "Key Terms - \n- absolute configuration\n- achiral\n- Cahn-Ingold-Prelog rules\n- chiral\n- chiral environment\n- chirality center\n- configuration\n- dextrorotatory\n- diastereomer\n- enantiomer\n- epimer\n- levorotatory\n- meso compound\n- optically active\n- prochiral\n- prochirality center\n- pro-R configuration\n- pro-S configuration\n- $R$ configuration\n- racemate\n- Re face\n- resolution\n- S configuration\n- Si face\n- specific rotation, $[\\alpha]_{D}$"}
{"id": 317, "contents": "Summary - \nIn this chapter, we've looked at some of the causes and consequences of molecular handedness-a topic of particular importance in understanding biological chemistry. The subject can be a bit complex but is so important that it's worthwhile spending time to become familiar with it.\n\nAn object or molecule that is not superimposable on its mirror image is said to be chiral, meaning \"handed.\" A chiral molecule is one that does not have a plane of symmetry cutting through it so that one half is a mirror image of the other half. The most common cause of chirality in organic molecules is the presence of a tetrahedral, $s p^{3}$-hybridized carbon atom bonded to four different groups-a so-called chirality center. Chiral compounds can exist as a pair of nonsuperimposable mirror-image stereoisomers called enantiomers. Enantiomers are identical in all physical properties except for the direction in which they rotate plane-polarized light.\n\nThe stereochemical configuration of a chirality center can be specified as either $\\boldsymbol{R}$ (rectus) or $\\boldsymbol{S}$ (sinister) by using the Cahn-Ingold-Prelog rules. First rank the four substituents on the chiral carbon atom, and then orient the molecule so that the lowest-ranked group points directly back. If a curved arrow drawn in the direction of decreasing rank $(1 \\rightarrow 2 \\rightarrow 3)$ for the remaining three groups is clockwise, the chirality center has the $R$ configuration. If the direction is counterclockwise, the chirality center has the $S$ configuration.\n\nSome molecules have more than one chirality center. Enantiomers have opposite configuration at all chirality centers, whereas diastereomers have the same configuration in at least one center but opposite configurations at the others. Epimers are diastereomers that differ in configuration at only one chirality center. A compound with $n$ chirality centers can have a maximum of $2^{n}$ stereoisomers.\n\nA meso compound contains a chirality center but is achiral overall because it has a plane of symmetry. Racemic mixtures, or racemates, are 50:50 mixtures of (+) and (-) enantiomers. Racemates and individual diastereomers differ in their physical properties, such as solubility, melting point, and boiling point."}
{"id": 318, "contents": "Summary - \nA molecule is prochiral if it can be converted from achiral to chiral in a single chemical step. A prochiral $s p^{2}$-hybridized atom has two faces, described as either $\\mathbf{R e}$ or $\\mathbf{S i}$. An $s p^{3}$-hybridized atom is a prochirality center if, by changing one of its attached atoms, a chirality center results. The atom whose replacement leads to an $R$ chirality center is pro- $\\boldsymbol{R}$, and the atom whose replacement leads to an $S$ chirality center is pro- $\\boldsymbol{S}$."}
{"id": 319, "contents": "Visualizing Chemistry - \nPROBLEM Which of the following structures are identical? (Green $=\\mathrm{Cl}$.)\n5-26\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM Assign $R$ or $S$ configurations to the chirality centers in the following molecules (blue $=\\mathrm{N}$ ):\n\n5-27 (a)\n\n\nSerine\n(b)\n\n\n\nAdrenaline\n\nPROBLEM Which, if any, of the following structures represent meso compounds? (Blue $=\\mathrm{N}$, green $=\\mathrm{Cl}$.)\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Assign $R$ or $S$ configuration to each chirality center in pseudoephedrine, an over-the-counter 5-29 decongestant found in cold remedies (blue $=\\mathrm{N}$ ).\n\n\nPROBLEM Orient each of the following drawings so that the lowest-ranked group is toward the rear, and then 5-30 assign $R$ or $S$ configuration:\n(a)\n\n(b)\n\n(c)\n\n\nChirality and Optical Activity\nPROBLEM Which of the following objects are chiral?\n5-31\n(a) A basketball\n(b) A fork\n(c) A wine glass\n(d) A golf club\n(e) A spiral staircase\n(f) A snowflake\n\nPROBLEM Which of the following compounds are chiral? Draw them, and label the chirality centers.\n5-32 (a) 2,4-Dimethylheptane (b) 5-Ethyl-3,3-dimethylheptane (c) cis-1,4-Dichlorocyclohexane\nPROBLEM Draw chiral molecules that meet the following descriptions:\n5-33 (a) A chloroalkane, $\\mathrm{C}_{5} \\mathrm{H}_{11} \\mathrm{Cl}$\n(b) An alcohol, $\\mathrm{C}_{6} \\mathrm{H}_{14} \\mathrm{O}$\n(c) An alkene, $\\mathrm{C}_{6} \\mathrm{H}_{12}$\n(d) An alkane, $\\mathrm{C}_{8} \\mathrm{H}_{18}$"}
{"id": 320, "contents": "Visualizing Chemistry - \nPROBLEM Eight alcohols have the formula $\\mathrm{C}_{5} \\mathrm{H}_{12} \\mathrm{O}$. Draw them. Which are chiral?\n5-34\nPROBLEM Draw compounds that fit the following descriptions:\n5-35 (a) A chiral alcohol with four carbons (b) A chiral carboxylic acid with the formula $\\mathrm{C}_{5} \\mathrm{H}_{10} \\mathrm{O}_{2}$\n(c) A compound with two chirality centers\n(d) A chiral aldehyde with the formula $\\mathrm{C}_{3} \\mathrm{H}_{5} \\mathrm{BrO}$\n\nPROBLEM Erythronolide B is the biological precursor of erythromycin, a broad-spectrum antibiotic. How 5-36 many chirality centers does erythronolide B have? Identify them.\n\n\nErythronolide B"}
{"id": 321, "contents": "Assigning Configuration to Chirality Centers - \nPROBLEM Which of the following pairs of structures represent the same enantiomer, and which represent\n5-37 different enantiomers?\n(a)\n\n\n(b)\n\n\n(c)\n\n\n\n\n\nPROBLEM What is the relationship between the specific rotations of ( $2 R, 3 R$ )-dichloropentane and ( $2 S, 3 S$ )-\n5-38 dichloropentane? Between $(2 R, 3 S)$-dichloropentane and $(2 R, 3 R)$-dichloropentane?\nPROBLEM What is the stereochemical configuration of the enantiomer of ( $2 S, 4 R$ )-2,4-octanediol?\n5-39\nPROBLEM What are the stereochemical configurations of the two diastereomers of ( $2 S, 4 R$ )-2,4-octanediol? (A\n5-40 diol is a compound with two - OH groups.)\nPROBLEM Orient each of the following drawings so that the lowest-ranked group is toward the rear, and then\n5-41 assign $R$ or $S$ configuration:\n(a)\n\n(b)\n\n(c)"}
{"id": 322, "contents": "Assigning Configuration to Chirality Centers - \n(b)\n\n(c)\n\n\nPROBLEM Assign Cahn-Ingold-Prelog rankings to the following sets of substituents:\n5-42 (a)\n(a) $-\\mathrm{CH}=\\mathrm{CH}_{2},-\\mathrm{CH}\\left(\\mathrm{CH}_{3}\\right)_{2},-\\mathrm{C}\\left(\\mathrm{CH}_{3}\\right)_{3},-\\mathrm{CH}_{2} \\mathrm{CH}_{3}$\n(b) $-\\mathrm{C} \\equiv \\mathrm{CH},-\\mathrm{CH}=\\mathrm{CH}_{2},-\\mathrm{C}\\left(\\mathrm{CH}_{3}\\right)_{3}$,\n(c) $-\\mathrm{CO}_{2} \\mathrm{CH}_{3},-\\mathrm{COCH}_{3},-\\mathrm{CH}_{2} \\mathrm{OCH}_{3},-\\mathrm{CH}_{2} \\mathrm{CH}_{3}$\n(d) $-\\mathrm{C} \\equiv \\mathrm{N},-\\mathrm{CH}_{2} \\mathrm{Br},-\\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{Br},-\\mathrm{Br}$\n\n\nPROBLEM Assign $R$ or $S$ configurations to each chirality center in the following molecules:\n\n5-43 (a)\n\n(b)\n\n(c)\n\n\nPROBLEM Assign $R$ or $S$ configuration to each chirality center in the following molecules:\n5-44 (a)\n\n(b)\n\n(c)\n\n\nPROBLEM Assign $R$ or $S$ configuration to each chirality center in the following biological molecules:\n\n5-45 (a)\n\n\nBiotin\n(b)\n\n\nProstaglandin $\\mathrm{E}_{\\mathbf{1}}$\n\nPROBLEM Draw tetrahedral representations of the following molecules:\n5-46 (a) (S)-2-Chlorobutane\n(b) (R)-3-Chloro-1-pentene $\\left[\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}(\\mathrm{Cl}) \\mathrm{CH}_{2} \\mathrm{CH}_{3}\\right]$\n\nPROBLEM Assign $R$ or $S$ configuration to each chirality center in the following molecules:\n5-47 (a)\n\n(b)"}
{"id": 323, "contents": "Assigning Configuration to Chirality Centers - \nPROBLEM Assign $R$ or $S$ configuration to each chirality center in the following molecules:\n5-47 (a)\n\n(b)\n\n\nPROBLEM Assign $R$ or $S$ configurations to the chirality centers in ascorbic acid (vitamin C).\n5-48"}
{"id": 324, "contents": "Ascorbic acid - \nPROBLEM Assign $R$ or $S$ stereochemistry to the chirality centers in the following Newman projections:\n5-49 (a)\n\n\n(b)\n\n\nPROBLEM Xylose is a common sugar found in many types of wood, including maple and cherry. Because it\n5-50 is much less prone to cause tooth decay than sucrose, xylose has been used in candy and chewing gum. Assign $R$ or $S$ configurations to the chirality centers in xylose.\n\n(+)-Xylose"}
{"id": 325, "contents": "Meso Compounds - \nPROBLEM Draw examples of the following:\n5-51 (a) A meso compound with the formula $\\mathrm{C}_{8} \\mathrm{H}_{18}$ (b) A meso compound with the formula $\\mathrm{C}_{9} \\mathrm{H}_{20}$\n(c) A compound with two chirality centers, one $R$ and the other $S$\n\nPROBLEM Draw the meso form of each of the following molecules, and indicate the plane of symmetry in each: 5-52\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Draw the structure of a meso compound that has five carbons and three chirality centers. 5-53\n\nPROBLEM Ribose, an essential part of ribonucleic acid (RNA), has the following structure:\n5-54\n\n\nRibose\n(a) How many chirality centers does ribose have? Identify them.\n(b) How many stereoisomers of ribose are there?\n(c) Draw the structure of the enantiomer of ribose.\n(d) Draw the structure of a diastereomer of ribose.\n\nPROBLEM On reaction with hydrogen gas in the presence of a platinum catalyst, ribose (Problem 5-54) is 5-55 converted into ribitol. Is ribitol optically active or inactive? Explain.\n\n\nRibitol\n\nProchirality\nPROBLEM Identify the indicated hydrogens in the following molecules as pro- $R$ or pro- $S$ :\n\n-56 (a)\n\n\nMalic acid\n(b)\n\n\nMethionine\n(c)\n\n\nCysteine\n\nPROBLEM Identify the indicated faces in the following molecules as Re or Si :\n\n5-57 (a)\n\n\nPyruvate\n(b)\n\n\nCrotonate\n\nPROBLEM One of the steps in fat metabolism is the hydration of crotonate to yield 3-hydroxybutyrate. The\n5-58 reaction occurs by addition of -OH to the Si face at C 3 , followed by protonation at C 2 , also from the Si face. Draw the product of the reaction, showing the stereochemistry of each step.\n\n\nCrotonate\n\n\n3-Hydroxybutyrate"}
{"id": 326, "contents": "Meso Compounds - \nCrotonate\n\n\n3-Hydroxybutyrate\n\nPROBLEM The dehydration of citrate to yield cis-aconitate, a step in the citric acid cycle, involves the pro- $R$ 5-59 \"arm\" of citrate rather than the pro- $S$ arm. Which of the following two products is formed?\n\n\nPROBLEM The first step in the metabolism of glycerol, formed by digestion of fats, is phosphorylation of the\n5-60 pro- $R-\\mathrm{CH}_{2} \\mathrm{OH}$ group by reaction with adenosine triphosphate (ATP) to give the corresponding glycerol phosphate plus adenosine diphosphate (ADP). Show the stereochemistry of the product.\n\n\nGlycerol\nGlycerol phosphate\nPROBLEM One of the steps in fatty-acid biosynthesis is the dehydration of ( $R$ )-3-hydroxybutyryl ACP to give\n5-61 trans-crotonyl ACP. Does the reaction remove the pro- $R$ or the pro- $S$ hydrogen from C2?\n\n(R)-3-Hydroxybutyryl ACP\n\ntrans-Crotonyl ACP"}
{"id": 327, "contents": "General Problems - \nPROBLEM Draw all possible stereoisomers of 1,2-cyclobutanedicarboxylic acid, and indicate the\n5-62 interrelationships. Which, if any, are optically active? Do the same for 1,3-cyclobutanedicarboxylic acid.\n\nPROBLEM Draw tetrahedral representations of the two enantiomers of the amino acid cysteine,\n5-63 $\\mathrm{HSCH}_{2} \\mathrm{CH}\\left(\\mathrm{NH}_{2}\\right) \\mathrm{CO}_{2} \\mathrm{H}$, and identify each as $R$ or $S$.\nPROBLEM The naturally occurring form of the amino acid cysteine (Problem 5-63) has the $R$ configuration\n5-64 at its chirality center. On treatment with a mild oxidizing agent, two cysteines join to give cystine, a disulfide. Assuming that the chirality center is not affected by the reaction, is cystine optically active? Explain.\n\n\nCysteine\nCystine\nPROBLEM Draw tetrahedral representations of the following molecules:\n5-65 (a) The $2 S, 3 R$ enantiomer of 2,3-dibromopentane\n(b) The meso form of 3,5-heptanediol\n\nPROBLEM Assign $R$ or $S$ configurations to the chiral centers in cephalexin, trade-named Keflex, the most\n5-66 widely prescribed antibiotic in the United States."}
{"id": 328, "contents": "Cephalexin - \nPROBLEM Chloramphenicol, a powerful antibiotic isolated in 1947 from the Streptomyces venezuelae\n5-67 bacterium, is active against a broad spectrum of bacterial infections and is particularly valuable against typhoid fever. Assign $R$ or $S$ configurations to the chirality centers in chloramphenicol."}
{"id": 329, "contents": "Chloramphenicol - \nPROBLEM Allenes are compounds with adjacent carbon-carbon double bonds. Many allenes are chiral, even\n5-68 though they don't contain chirality centers. Mycomycin, for example, a naturally occurring antibiotic isolated from the bacterium Nocardia acidophilus, is chiral and has $[\\alpha]_{D}=-130$. Explain why mycomycin is chiral.\n$\\mathrm{HC} \\equiv \\mathrm{C}-\\mathrm{C} \\equiv \\mathrm{C}-\\mathrm{CH}=\\mathrm{C}=\\mathrm{CH}-\\mathrm{CH}=\\mathrm{CH}-\\mathrm{CH}=\\mathrm{CH}-\\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$\nMycomycin\nPROBLEM Long before chiral allenes were known (Problem 5-68), the resolution of 5-69 4-methylcyclohexylideneacetic acid into two enantiomers had been carried out. Why is it chiral? What geometric similarity does it have to allenes?\n\n\n4-Methylcyclohexylideneacetic acid\nPROBLEM ( $S$ )-1-Chloro-2-methylbutane undergoes light-induced reaction with $\\mathrm{Cl}_{2}$ to yield a mixture of 5-70 products, among which are 1,4-dichloro-2-methylbutane and 1,2-dichloro-2-methylbutane.\n(a) Write the reaction, showing the correct stereochemistry of the reactant.\n(b) One of the two products is optically active, but the other is optically inactive. Which is which?"}
{"id": 330, "contents": "Chloramphenicol - \nPROBLEM How many stereoisomers of 2,4-dibromo-3-chloropentane are there? Draw them, and indicate\n5-71 which are optically active.\nPROBLEM Draw both cis- and trans-1,4-dimethylcyclohexane in their more stable chair conformations.\n5-72 (a) How many stereoisomers are there of cis-1,4-dimethylcyclohexane, and how many of trans-1,4-dimethylcyclohexane?\n(b) Are any of the structures chiral?\n(c) What are the stereochemical relationships among the various stereoisomers of 1,4-dimethylcyclohexane?\n\nPROBLEM Draw both cis- and trans-1,3-dimethylcyclohexane in their more stable chair conformations.\n5-73 (a) How many stereoisomers are there of cis-1,3-dimethylcyclohexane, and how many of trans-1,3-dimethylcyclohexane?\n(b) Are any of the structures chiral?\n(c) What are the stereochemical relationships among the various stereoisomers of 1,3-dimethylcyclohexane?\n\nPROBLEM cis-1,2-Dimethylcyclohexane is optically inactive even though it has two chirality centers. Explain. 5-74\n\nPROBLEM We'll see in Chapter 11 that alkyl halides react with hydrosulfide ion ( $\\mathrm{HS}^{-}$) to give a product whose\n5-75 stereochemistry is inverted from that of the reactant.\nDraw the reaction of (S)-2-bromobutane with $\\mathrm{HS}^{-}$ion to yield 2-butanethiol, $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}(\\mathrm{SH}) \\mathrm{CH}_{3}$. Is the stereochemistry of the product $R$ or $S$ ?"}
{"id": 331, "contents": "Chloramphenicol - \nAn alkyl\nbromide\nPROBLEM Ketones react with sodium acetylide (the sodium salt of acetylene, $\\mathrm{Na}^{+-}: \\mathrm{C} \\equiv \\mathrm{CH}$ ) to give alcohols.\n5-76 For example, the reaction of sodium acetylide with 2-butanone yields 3-methyl-1-pentyn-3-ol:\n\n(a) Is the product chiral?\n(b) Assuming that the reaction takes place with equal likelihood from both Re and Si faces of the carbonyl group, is the product optically active? Explain.\n\nPROBLEM Imagine that a reaction similar to that in Problem 5-76 is carried out between sodium acetylide and\n5-77 (R)-2-phenylpropanal to yield 4-phenyl-1-pentyn-3-ol:\n\n( $R$ )-2-Phenylpropanal\n4-Phenyl-1-pentyn-3-ol\n(a) Is the product chiral?\n(b) Draw both major and minor reaction products, assuming that the reaction takes place preferentially from the Re face of the carbonyl group. Is the product mixture optically active? Explain."}
{"id": 332, "contents": "An Overview of Organic Reactions - \nFIGURE 6.1 Many chemical reactions are like pole vaulters going over the bar. They need a big, initial push of activation energy. (credit: modification of work \"UChicago Pole Vault\" by Eric Guo/Flickr, CC BY 2.0)"}
{"id": 333, "contents": "CHAPTER CONTENTS - 6.1 Kinds of Organic Reactions\n6.2 How Organic Reactions Occur: Mechanisms\n6.3 Polar Reactions\n6.4 An Example of a Polar Reaction: Addition of HBr to Ethylene\n6.5 Using Curved Arrows in Polar Reaction Mechanisms\n6.6 Radical Reactions\n6.7 Describing a Reaction: Equilibria, Rates, and Energy Changes\n6.8 Describing a Reaction: Bond Dissociation Energies\n6.9 Describing a Reaction: Energy Diagrams and Transition States\n6.10 Describing a Reaction: Intermediates\n6.11 A Comparison Between Biological Reactions and Laboratory Reactions\n\nWHY THIS CHAPTER? All chemical reactions, whether they take place in the laboratory or in living organisms, follow the same \"rules.\" Reactions in living organisms often look more complex than laboratory reactions because of the size of the biomolecules and the involvement of biological catalysts called enzymes, but the principles governing all chemical reactions are the same.\n\nTo understand both organic and biological chemistry, it's necessary to know not just what occurs but also why and how chemical reactions take place. In this chapter, we'll start with an overview of the fundamental kinds of organic reactions, we'll see why reactions occur, and we'll see how reactions can be described. Once this\nbackground is out of the way, we'll then be ready to begin studying the details of organic chemistry in future chapters.\n\nWhen first approached, organic chemistry might seem overwhelming. It's not so much that any one part is difficult to understand, it's that there are so many parts: tens of millions of compounds, dozens of functional groups, and an apparently endless number of reactions. With study, though, it becomes evident that there are only a few fundamental ideas that underlie all organic reactions. Far from being a collection of isolated facts, organic chemistry is a beautifully logical subject that is unified by a few broad themes. When these themes are understood, learning organic chemistry becomes much easier and memorization is minimized. The aim of this book is to describe the themes and clarify the patterns that unify organic chemistry in future chapters."}
{"id": 334, "contents": "CHAPTER CONTENTS - 6.1 Kinds of Organic Reactions\nOrganic chemical reactions can be organized broadly in two ways-by what kinds of reactions occur and by how those reactions occur. Let's look first at the kinds of reactions that take place. There are four general types of organic reactions: additions, eliminations, substitutions, and rearrangements.\n\n- Addition reactions occur when two reactants add together to form a single product with no atoms \"left over.\" An example that we'll be studying soon is the reaction of an alkene, such as ethylene, with HBr to yield an alkyl bromide.\n\n- Elimination reactions are, in a sense, the opposite of addition reactions. They occur when a single reactant splits into two products, often with the formation of a small molecule such as water or HBr . An example is the acid-catalyzed reaction of an alcohol to yield water and an alkene."}
{"id": 335, "contents": "This one - \nreactant...\n\nEthylene (an alkene)\n\n- Substitution reactions occur when two reactants exchange parts to give two new products. An example is the reaction of an ester such as methyl acetate with water to yield a carboxylic acid plus an alcohol. Similar reactions occur in many biological pathways, including the metabolism of dietary fats.\n\n- Rearrangement reactions occur when a single reactant undergoes a reorganization of bonds and atoms to yield an isomeric product. An example is the conversion of dihydroxyacetone phosphate into its constitutional isomer glyceraldehyde 3-phosphate, a step in the glycolysis pathway by which carbohydrates are metabolized.\n\n\nPROBLEM Classify each of the following reactions as an addition, elimination, substitution, or rearrangement:\n6-1 (a) $\\mathrm{CH}_{3} \\mathrm{Br}+\\mathrm{KOH} \\longrightarrow \\mathrm{CH}_{3} \\mathrm{OH}+\\mathrm{KBr}$\n(b) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{Br} \\longrightarrow \\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}_{2}+\\mathrm{HBr}$\n(c) $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}_{2}+\\mathrm{H}_{2} \\longrightarrow \\mathrm{CH}_{3} \\mathrm{CH}_{3}$"}
{"id": 336, "contents": "This one - 6.2 How Organic Reactions Occur: Mechanisms\nHaving looked at the kinds of reactions that take place, let's now see how they occur. An overall description of how a reaction occurs is called a reaction mechanism. A mechanism describes in detail exactly what takes place at each stage of a chemical transformation-which bonds are broken and in what order, which bonds are formed and in what order, and what the relative rates are for each step. A complete mechanism must also account for all reactants used and all products formed.\n\nAll chemical reactions involve bond-breaking and bond-making. When two molecules come together, react, and yield products, specific bonds in the reactant molecules are broken and specific bonds in the product molecules are formed. Fundamentally, there are two ways in which a covalent two-electron bond can break. A bond can break in an electronically unsymmetrical way so that both bonding electrons remain with one product fragment, leaving the other with a vacant orbital, or a bond can break in an electronically symmetrical way so that one electron remains with each product fragment. The unsymmetrical cleavage is said to be heterolytic, and the symmetrical cleavage is said to be homolytic.\n\nWe'll develop this point in more detail later, but note for now that the movement of two electrons in the unsymmetrical process is indicated using a full-headed curved arrow ( $\\curvearrowright$ ), whereas the movement of one electron in the symmetrical process is indicated using a half-headed, or \"fishhook,\" arrow ( $\\AA$ ).\n\n\nJust as there are two ways in which a bond can break, there are two ways in which a covalent two-electron bond can form. A bond can form in an electronically unsymmetrical way if both bonding electrons are donated to the new bond by one reactant, or in a symmetrical way if one electron is donated by each reactant.\n\n\nProcesses that involve unsymmetrical bond-breaking and bond-making are called polar reactions. Polar reactions involve species that have an even number of electrons and thus have only electron pairs in their orbitals. Polar processes are by far the more common reaction type in both organic and biological chemistry, and a large part of this book is devoted to their description."}
{"id": 337, "contents": "This one - 6.2 How Organic Reactions Occur: Mechanisms\nProcesses that involve symmetrical bond-breaking and bond-making are called radical reactions. A radical, often called a free radical, is a neutral chemical species that contains an odd number of electrons and thus has a single, unpaired electron in one of its orbitals.\n\nIn addition to polar and radical reactions, there is a third, less commonly encountered process called a pericyclic reaction. Rather than explain pericyclic reactions now, though, we'll look at them more carefully in Chapter 30."}
{"id": 338, "contents": "This one - 6.3 Polar Reactions\nPolar reactions occur because of the electrical attraction between positively polarized and negatively polarized centers on functional groups in molecules. To see how these reactions take place, let's first recall the discussion of polar covalent bonds in Section 2.1 and then look more deeply into the effects of bond polarity on organic molecules.\n\nMost organic compounds are electrically neutral; they have no net charge, either positive or negative. We saw in Section 2.1, however, that certain bonds within a molecule, particularly the bonds in functional groups, are polar. Bond polarity is a consequence of an unsymmetrical electron distribution in a bond and is due to the difference in electronegativity of the bonded atoms.\n\nElements such as oxygen, nitrogen, fluorine, and chlorine are more electronegative than carbon, so a carbon atom bonded to one of these atoms has a partial positive charge ( $\\delta+$ ). Metals are less electronegative than carbon, so a carbon atom bonded to a metal has a partial negative charge ( $\\delta-$ ). Electrostatic potential maps of chloromethane and methyllithium illustrate these charge distributions, showing that the carbon atom in chloromethane is electron-poor (blue) while the carbon in methyllithium is electron-rich (red).\n\n\n\nChloromethane\n\n\nMethyllithium\n\n\nThe polarity patterns of some common functional groups are shown in TABLE 6.1. Note that carbon is always positively polarized except when bonded to a metal.\n\nTABLE 6.1 Polarity Patterns in Some Common Functional\nGroups\n\n| Compound type | Functional group structure |\n| :--- | :--- |\n| Alcohol | $\\_{\\mathrm{C}}^{\\mathrm{C+}-\\mathrm{O-}}$ |\n| Alkene | C= |\n\nSymmetrical, nonpolar"}
{"id": 339, "contents": "This one - 6.3 Polar Reactions\nSymmetrical, nonpolar\n\n| Alkyl halide | <smiles>[X]C(C)(C)C</smiles> |\n| :---: | :---: |\n| Amine | $\\begin{aligned} & \\backslash \\delta+\\delta- \\\\ & -\\mathrm{C}-\\mathrm{NH}_{2} \\end{aligned}$ |\n| Ether | <smiles></smiles> |\n| Thiol | <smiles>CC(C)(C)[SnH3]</smiles> |\n| Nitrile | $-{ }^{\\delta+} \\equiv{ }^{\\delta-} \\mathrm{N}$ |\n| Grignard reagent | <smiles></smiles> |\n| Alkyllithium | <smiles>CC(C)(C)[Te]</smiles> |\n\n\n\nThis discussion of bond polarity is oversimplified in that we've considered only bonds that are inherently polar due to differences in electronegativity. Polar bonds can also result from the interaction of functional groups with acids or bases. Take an alcohol such as methanol, for example. In neutral methanol, the carbon atom is somewhat electron-poor because the electronegative oxygen attracts the electrons in the $\\mathrm{C}-\\mathrm{O}$ bond. On protonation of the methanol oxygen by an acid, however, a full positive charge on oxygen attracts the electrons in the C-O bond much more strongly and makes the carbon much more electron-poor. We'll see numerous examples throughout this book of reactions that are catalyzed by acids because of the resultant increase in bond polarity upon protonation."}
{"id": 340, "contents": "This one - 6.3 Polar Reactions\nMethanol-weakly\nelectron-poor carbon\nProtonated methanol-\nstrongly electron-poor carbon\nYet a further consideration is the polarizability (as opposed to polarity) of atoms in a molecule. As the electric field around a given atom changes because of changing interactions with solvent or other polar molecules nearby, the electron distribution around that atom also changes. The measure of this response to an external electrical influence is called the polarizability of the atom. Larger atoms with more loosely held electrons are more polarizable, and smaller atoms with fewer, tightly held electrons are less polarizable. Thus, sulfur is more polarizable than oxygen, and iodine is more polarizable than chlorine. The effect of this higher polarizability of sulfur and iodine is that carbon-sulfur and carbon-iodine bonds, although nonpolar according to electronegativity values (FIGURE 2.3), nevertheless usually react as if they were polar.\n\n\nWhat does functional-group polarity mean with respect to chemical reactivity? Because unlike charges attract, the fundamental characteristic of all polar organic reactions is that electron-rich sites react with electron-poor sites. Bonds are made when an electron-rich atom donates a pair of electrons to an electron-poor atom, and bonds are broken when one atom leaves with both electrons from the former bond.\n\nAs we saw in Section 2.11, the movement of an electron pair during a polar reaction is indicated using a curved, full-headed arrow to show where electrons move when reactant bonds are broken and product bonds are formed during the reaction."}
{"id": 341, "contents": "This one - 6.3 Polar Reactions\nAs we saw in Section 2.11, the movement of an electron pair during a polar reaction is indicated using a curved, full-headed arrow to show where electrons move when reactant bonds are broken and product bonds are formed during the reaction.\n\n\nIn referring to the electron-rich and electron-poor species involved in polar reactions, chemists use the words nucleophile and electrophile. A nucleophile is a substance that is \"nucleus-loving.\" (Remember that a nucleus is positively charged.) A nucleophile has a negatively polarized, electron-rich atom and can form a bond by donating a pair of electrons to a positively polarized, electron-poor atom. Nucleophiles can be either neutral or negatively charged; ammonia, water, hydroxide ion, and chloride ion are examples. An electrophile, by contrast, is \"electron-loving.\" An electrophile has a positively polarized, electron-poor atom and can form a bond by accepting a pair of electrons from a nucleophile. Electrophiles can be either neutral or positively charged. Acids ( $\\mathrm{H}^{+}$donors), alkyl halides, and carbonyl compounds are examples (FIGURE 6.2).\n\n\nFIGURE 6.2 Some nucleophiles and electrophiles. Electrostatic potential maps identify the nucleophilic (negative) and electrophilic (positive) atoms.\n\nNote that neutral compounds can often react either as nucleophiles or as electrophiles, depending on the circumstances. After all, if a compound is neutral yet has an electron-rich nucleophilic site, it must also have a corresponding electron-poor electrophilic site. Water, for instance, acts as an electrophile when it donates $\\mathrm{H}^{+}$ but acts as a nucleophile when it donates a nonbonding pair of electrons. Similarly, a carbonyl compound acts as an electrophile when it reacts at its positively polarized carbon atom, yet acts as a nucleophile when it reacts at its negatively polarized oxygen atom."}
{"id": 342, "contents": "This one - 6.3 Polar Reactions\nIf the definitions of nucleophiles and electrophiles sound similar to those given in Section 2.11 for Lewis acids and Lewis bases, that's because there is indeed a correlation. Lewis bases are electron donors and behave as nucleophiles, whereas Lewis acids are electron acceptors and behave as electrophiles. Thus, much of organic chemistry is explainable in terms of acid-base reactions. The main difference is that the words acid and base are used broadly in all fields of chemistry, while the words nucleophile and electrophile are used primarily in organic chemistry when carbon bonding is involved."}
{"id": 343, "contents": "Identifying Electrophiles and Nucleophiles - \nWhich of the following species is likely to behave as a nucleophile and which as an electrophile?\n(a) $\\mathrm{NO}_{2}{ }^{+}$\n(b) $\\mathrm{CN}^{-}$\n(c) $\\mathrm{CH}_{3} \\mathrm{NH}_{2}$\n(d) $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{~S}^{+}$"}
{"id": 344, "contents": "Strategy - \nA nucleophile has an electron-rich site, either because it is negatively charged or because it has a functional group containing an atom that has a lone pair of electrons. An electrophile has an electron-poor site, either because it is positively charged or because it has a functional group containing an atom that is positively polarized."}
{"id": 345, "contents": "Solution - \n(a) $\\mathrm{NO}_{2}{ }^{+}$(nitronium ion) is likely to be an electrophile because it is positively charged.\n(b) : $\\mathrm{C} \\equiv \\mathrm{N}^{-}$(cyanide ion) is likely to be a nucleophile because it is negatively charged.\n(c) $\\mathrm{CH}_{3} \\mathrm{NH}_{2}$ (methylamine) might be either a nucleophile or an electrophile, depending on the circumstances. The lone pair of electrons on the nitrogen atom makes methylamine a potential nucleophile, while positively polarized $\\mathrm{N}-\\mathrm{H}$ hydrogens make methylamine a potential acid (electrophile).\n(d) $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{~S}^{+}$(trimethylsulfonium ion) is likely to be an electrophile because it is positively charged.\n\nPROBLEM Which of the following species are likely to be nucleophiles and which electrophiles? Which might 6-2 be both?\n(a) $\\mathrm{CH}_{3} \\mathrm{Cl}$ (b) $\\mathrm{CH}_{3} \\mathrm{~S}^{-}$\n(c)\n\n(d)\n\n\nPROBLEM An electrostatic potential map of boron trifluoride is shown. Is $\\mathrm{BF}_{3}$ likely to be a nucleophile or an 6-3 electrophile? Draw a Lewis structure for $\\mathrm{BF}_{3}$, and explain your answer."}
{"id": 346, "contents": "Solution - 6.4 An Example of a Polar Reaction: Addition of HBr to Ethylene\nLet's look at a typical polar process-the addition reaction of an alkene, such as ethylene, with hydrogen bromide. When ethylene is treated with HBr at room temperature, bromoethane is produced. Overall, the reaction can be formulated as\n\n\nThe reaction is an example of a polar reaction type known as an electrophilic addition reaction and can be understood using the general ideas discussed in the previous section. Let's begin by looking at the two reactants.\n\nWhat do we know about ethylene? We know from Section 1.8 that a carbon-carbon double bond results from the orbital overlap of two $s p^{2}$-hybridized carbon atoms. The $\\sigma$ part of the double bond results from $s p^{2}-s p^{2}$ overlap, and the $\\pi$ part results from $p-p$ overlap.\n\nWhat kind of chemical reactivity might we expect from a $\\mathrm{C}=\\mathrm{C}$ bond? We know that alkanes, such as ethane, are relatively inert because all valence electrons are tied up in strong, nonpolar, $\\mathrm{C}-\\mathrm{C}$ and $\\mathrm{C}-\\mathrm{H}$ bonds. Furthermore, the bonding electrons in alkanes are relatively inaccessible to approaching reactants because they are sheltered in $\\sigma$ bonds between nuclei. The electronic situation in alkenes is quite different, however. For one thing, double\nbonds have a greater electron density than single bonds-four electrons in a double bond versus only two in a single bond. In addition, the electrons in the $\\pi$ bond are accessible to approaching reactants because they are located above and below the plane of the double bond rather than being sheltered between the nuclei (FIGURE 6.3). As a result, the double bond is nucleophilic and the chemistry of alkenes is dominated by reactions with electrophiles.\n\n\n> Carbon-carbon $\\sigma$ bond: stronger; less accessible bonding electrons\n\n\nCarbon-carbon $\\pi$ bond: weaker; more accessible electrons\n\nFIGURE 6.3 A comparison of carbon-carbon single and double bonds. A double bond is both more accessible to approaching reactants than a single bond and more electron-rich (more nucleophilic). An electrostatic potential map of ethylene indicates that the double bond is the region of highest negative charge."}
{"id": 347, "contents": "Solution - 6.4 An Example of a Polar Reaction: Addition of HBr to Ethylene\nWhat about the second reactant, HBr ? As a strong acid, HBr is a powerful proton $\\left(\\mathrm{H}^{+}\\right)$donor and electrophile. Thus, the reaction between HBr and ethylene is a typical electrophile-nucleophile combination, characteristic of all polar reactions.\n\nWe'll see more details about alkene electrophilic addition reactions shortly, but for the present we can imagine the reaction as taking place by the pathway shown in FIGURE 6.4. The reaction begins when the alkene nucleophile donates a pair of electrons from its $\\mathrm{C}=\\mathrm{C}$ bond to HBr to form a new $\\mathrm{C}-\\mathrm{H}$ bond plus $\\mathrm{Br}^{-}$, as indicated by the path of the curved arrows in the first step of FIGURE 6.4. One curved arrow begins at the middle of the double bond (the source of the electron pair) and points to the hydrogen atom in HBr (the atom to which a bond will form). This arrow indicates that a new $\\mathrm{C}-\\mathrm{H}$ bond forms using electrons from the former $\\mathrm{C}=\\mathrm{C}$ bond. Simultaneously, a second curved arrow begins in the middle of the $\\mathrm{H}-\\mathrm{Br}$ bond and points to the Br , indicating that the $\\mathrm{H}-\\mathrm{Br}$ bond breaks and the electrons remain with the Br atom, giving $\\mathrm{Br}^{-}$."}
{"id": 348, "contents": "FIGURE 6.4 MECHANISM - \nThe electrophilic addition reaction of ethylene and HBr . The reaction takes place in two steps, both of which involve electrophile-nucleophile interactions.\n(1) A hydrogen atom on the electrophile HBr is attacked by $\\pi$ electrons from the nucleophilic double bond, forming a new $\\mathrm{C}-\\mathrm{H}$ bond. This leaves the other carbon atom with a + charge and a vacant $p$ orbital. Simultaneously, two electrons from the $\\mathrm{H}-\\mathrm{Br}$ bond move onto bromine, giving bromide anion.\n\n2 Bromide ion donates an electron pair to the positively charged carbon atom, forming a $\\mathrm{C}-\\mathrm{Br}$ bond and yielding the neutral addition product.\n\n\nEthylene\n(1)\n\n\n\nBromoethane\n\nWhen one of the alkene carbon atoms bonds to the incoming hydrogen, the other carbon atom, having lost its share of the double-bond electrons, now has only six valence electrons and is left with a positive charge. This positively charged species-a carbon-cation, or carbocation-is itself an electrophile that can accept an electron pair from nucleophilic $\\mathrm{Br}^{-}$anion in a second step, forming a $\\mathrm{C}-\\mathrm{Br}$ bond and yielding the observed addition product. Once again, a curved arrow in FIGURE 6.4 shows the electron-pair movement from $\\mathrm{Br}^{-}$to the positively charged carbon.\n\nThe electrophilic addition of HBr to ethylene is only one example of a polar process; there are many others that we'll study in depth in later chapters. But regardless of the details of individual reactions, all polar reactions take place between an electron-poor site and an electron-rich site and involve the donation of an electron pair from a nucleophile to an electrophile.\n\nPROBLEM What product would you expect from reaction of cyclohexene with HBr ? With HCl ?\n\n\nPROBLEM Reaction of HBr with 2-methylpropene yields 2-bromo-2-methylpropane. What is the structure of 6-5 the carbocation formed during the reaction? Show the mechanism of the reaction."}
{"id": 349, "contents": "FIGURE 6.4 MECHANISM - 6.5 Using Curved Arrows in Polar Reaction Mechanisms\nIt takes practice to use curved arrows properly in reaction mechanisms, but there are a few rules and a few common patterns you should look for that will help you become more proficient:\n\nRULE 1\nElectrons move from a nucleophilic source ( Nu : or $\\mathrm{Nu}^{-}{ }^{-}$) to an electrophilic sink ( $\\mathbf{E}$ or $\\mathbf{E}^{+}$). The nucleophilic source must have an electron pair available, usually either as a lone pair or in a multiple bond. For example:\n\n\nThe electrophilic sink must be able to accept an electron pair, usually because it has either a positively charged atom or a positively polarized atom in a functional group. For example:\n\nElectrons usually flow to one of these electrophiles.\n\n\n\nRULE 2\nThe nucleophile can be either negatively charged or neutral. If the nucleophile is negatively charged, the atom that donates an electron pair becomes neutral. For example:\n\n\nIf the nucleophile is neutral, the atom that donates the electron pair acquires a positive charge. For example:\n\n\nRULE 3\nThe electrophile can be either positively charged or neutral. If the electrophile is positively charged, the atom bearing that charge becomes neutral after accepting an electron pair. For example:\n\n\nIf the electrophile is neutral, the atom that ultimately accepts the electron pair acquires a negative charge. For this to happen, however, the negative charge must be stabilized by being on an electronegative atom such as oxygen, nitrogen, or a halogen. Carbon and hydrogen do not typically stabilize a negative charge. For example:\n\n\nThe result of Rules 2 and 3 together is that charge is conserved during the reaction. A negative charge in one of the reactants gives a negative charge in one of the products, and a positive charge in one of the reactants gives a positive charge in one of the products."}
{"id": 350, "contents": "RULE 4 - \nThe octet rule must be followed. That is, no second-row atom can be left with ten electrons (or four for hydrogen). If an electron pair moves to an atom that already has an octet (or two electrons for hydrogen), another electron pair must simultaneously move from that atom to maintain the octet. When two electrons move from the $\\mathrm{C}=\\mathrm{C}$ bond of ethylene to the hydrogen atom of $\\mathrm{H}_{3} \\mathrm{O}^{+}$, for instance, two electrons must leave that hydrogen. This means that the $\\mathrm{H}-\\mathrm{O}$ bond must break and the electrons must stay with the oxygen, giving neutral water.\n\n\nWorked Example 6.2 gives another example of drawing curved arrows."}
{"id": 351, "contents": "Using Curved Arrows in Reaction Mechanisms - \nAdd curved arrows to the following polar reaction to show the flow of electrons:"}
{"id": 352, "contents": "Strategy - \nLook at the reaction, and identify the bonding changes that have occurred. In this case, a $\\mathrm{C}-\\mathrm{Br}$ bond has broken and a $\\mathrm{C}-\\mathrm{C}$ bond has formed. The formation of the $\\mathrm{C}-\\mathrm{C}$ bond involves donation of an electron pair from the nucleophilic carbon atom of the reactant on the left to the electrophilic carbon atom of $\\mathrm{CH}_{3} \\mathrm{Br}$, so we draw a curved arrow originating from the lone pair on the negatively charged C atom and pointing to the C atom of $\\mathrm{CH}_{3} \\mathrm{Br}$. At the same time that the $\\mathrm{C}-\\mathrm{C}$ bond forms, the $\\mathrm{C}-\\mathrm{Br}$ bond must break so that the octet rule is not violated. We therefore draw a second curved arrow from the $\\mathrm{C}-\\mathrm{Br}$ bond to Br . The bromine is now a stable $\\mathrm{Br}^{-}$ion."}
{"id": 353, "contents": "Solution - \nPROBLEM Add curved arrows to the following polar reactions to indicate the flow of electrons in each: 6-6\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Predict the products of the following polar reaction, a step in the citric acid cycle for food\n6-7 metabolism, by interpreting the flow of electrons indicated by the curved arrows:"}
{"id": 354, "contents": "Solution - 6.6 Radical Reactions\nRadical reactions are much less common than polar reactions but are nevertheless important in some industrial processes and biological pathways. We'll look at them in more detail in Sections $\\mathbf{1 0 . 2}$ and $\\mathbf{1 0 . 3}$ but will briefly see how they occur at this point.\n\nA radical is highly reactive because it contains an atom with an odd number of electrons (usually seven) in its valence shell, rather than a noble-gas octet. The radical can achieve a valence-shell octet in several ways however. For instance, it might abstract an atom and one bonding electron from another reactant, leaving behind a new radical. The net result is a radical substitution reaction.\n\n\nAlternatively, a reactant radical might add to a double bond, taking one electron from the double bond and yielding a new radical. The net result is a radical addition reaction.\n\n\nAn example of an industrially useful radical reaction is the reaction of chlorine with methane to yield chloromethane, which is used to manufacture the solvents dichloromethane $\\left(\\mathrm{CH}_{2} \\mathrm{Cl}_{2}\\right)$ and chloroform $\\left(\\mathrm{CHCl}_{3}\\right)$. The process begins with irradiation of $\\mathrm{Cl}_{2}$ with ultraviolet light to break the relatively weak $\\mathrm{Cl}-\\mathrm{Cl}$ bond of $\\mathrm{Cl}_{2}$ and produce chlorine radicals $(\\cdot \\mathrm{Cl})$.\n\n$$\n: \\ddot{C}: \\ddot{C l}: \\ddot{\\mathrm{C}}: \\quad \\xrightarrow{\\text { Light }} 2: \\ddot{\\mathrm{c}} \\cdot\n$$\n\nChlorine radicals then react with methane by abstracting a hydrogen atom to give HCl and a methyl radical $\\left(\\cdot \\mathrm{CH}_{3}\\right)$ that reacts further with $\\mathrm{Cl}_{2}$ to give chloromethane plus a new chlorine radical that cycles back and repeats the first step. Thus, once the sequence has started, it becomes a self-sustaining cycle of repeating steps (a) and (b), making the overall process a chain reaction."}
{"id": 355, "contents": "Solution - 6.6 Radical Reactions\nAs a biological example of a radical reaction, look at the synthesis of prostaglandins, a large class of molecules found in virtually all body tissues and fluids. A number of pharmaceuticals are based on or derived from prostaglandins, including medicines that induce labor during childbirth, reduce intraocular pressure in glaucoma, control bronchial asthma, and help treat congenital heart defects.\n\nProstaglandin biosynthesis is initiated by the abstraction of a hydrogen atom from arachidonic acid by an iron-oxygen radical, thereby generating a carbon radical in a substitution reaction. Don't be intimidated by the size of the molecules; focus on the changes that occur in each step. (To help you do that, the unchanged part of the molecule is \"ghosted,\" with only the reactive part clearly visible.)\n\n\nFollowing the initial abstraction of a hydrogen atom, the carbon radical then reacts with $\\mathrm{O}_{2}$ to give an oxygen radical, which reacts with a $\\mathrm{C}=\\mathrm{C}$ bond within the same molecule in an addition reaction. Several further transformations ultimately yield prostaglandin $\\mathrm{H}_{2}$.\n\n\nPROBLEM Radical chlorination of alkanes is not generally useful because mixtures of products often result 6-8 when more than one kind of $\\mathrm{C}-\\mathrm{H}$ bond is present in the substrate. Draw and name all monochloro substitution products $\\mathrm{C}_{6} \\mathrm{H}_{13} \\mathrm{Cl}$ you might obtain by reaction of 2-methylpentane with $\\mathrm{Cl}_{2}$.\n\nPROBLEM Using curved fishhook arrows, propose a mechanism for the formation of the cyclopentane ring of 6-9 prostaglandin $\\mathrm{H}_{2}$."}
{"id": 356, "contents": "Solution - 6.7 Describing a Reaction: Equilibria, Rates, and Energy Changes\nEvery chemical reaction can go in either a forward or reverse direction. Reactants can go forward to products, and products can revert to reactants. As you may remember from your general chemistry course, the position of the resulting chemical equilibrium is expressed by an equation in which $K_{\\text {eq }}$, the equilibrium constant, is equal to the product concentrations multiplied together, divided by the reactant concentrations multiplied together, with each concentration raised to the power of its coefficient in the balanced equation. For the generalized reaction\n\n$$\na \\mathrm{~A}+b \\mathrm{~B} \\rightleftharpoons c \\mathrm{C}+d \\mathrm{D}\n$$\n\nwe have\n\n$$\nK_{\\mathrm{eq}}=\\frac{[\\mathrm{C}]^{c}[\\mathrm{D}]^{d}}{[\\mathrm{~A}]^{a}[\\mathrm{~B}]^{b}}\n$$\n\nThe value of the equilibrium constant tells which side of the reaction arrow is energetically favored. If $K_{\\text {eq }}$ is much larger than 1, then the product concentration term [C] ${ }^{c}[\\mathrm{D}]^{d}$ is much larger than the reactant concentration term $[\\mathrm{A}]^{a}[\\mathrm{~B}]^{b}$, and the reaction proceeds as written from left to right. If $K_{\\text {eq }}$ is near 1, appreciable amounts of both reactant and product are present at equilibrium. And if $K_{\\text {eq }}$ is much smaller than 1, the reaction does not take place as written but instead goes in the reverse direction, from right to left.\n\nIn the reaction of ethylene with HBr , for example, we can write the following equilibrium expression and determine experimentally that the equilibrium constant at room temperature is approximately $7.1 \\times 10^{7}$ :"}
{"id": 357, "contents": "Solution - 6.7 Describing a Reaction: Equilibria, Rates, and Energy Changes\nIn the reaction of ethylene with HBr , for example, we can write the following equilibrium expression and determine experimentally that the equilibrium constant at room temperature is approximately $7.1 \\times 10^{7}$ :\n\n$$\n\\begin{aligned}\n& \\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}_{2}+\\mathrm{HBr} \\rightleftarrows \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{Br} \\\\\n& K_{\\text {eq }}=\\frac{\\left[\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{Br}\\right]}{\\left\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}_{2}\\right}=7.1 \\times 10^{7}\n\\end{aligned}\n$$\n\nBecause $K_{\\text {eq }}$ is relatively large, the reaction proceeds as written and more than $99.99999 \\%$ of the ethylene is converted into bromoethane. For practical purposes, an equilibrium constant greater than about $10^{3}$ means that the amount of reactant left over will be barely detectable (less than $0.1 \\%$ ).\n\nWhat determines the magnitude of the equilibrium constant? For a reaction to have a favorable equilibrium constant and proceed as written, the energy of the products must be lower than the energy of the reactants. In other words, energy must be released. This situation is analogous to that of a rock poised precariously in a highenergy position near the top of a hill. When it rolls downhill, the rock releases energy until it reaches a more stable, low-energy position at the bottom."}
{"id": 358, "contents": "Solution - 6.7 Describing a Reaction: Equilibria, Rates, and Energy Changes\nThe energy change that occurs during a chemical reaction is called the Gibbs free-energy change ( $\\boldsymbol{\\Delta} \\boldsymbol{G} \\boldsymbol{G}$ ), which is equal to the free energy of the products minus the free energy of the reactants: $\\Delta G=G_{\\text {products }}-G_{\\text {reactants }}$. For a favorable reaction, $\\Delta G$ has a negative value, meaning that energy is lost by the chemical system and released to the surroundings, usually as heat. Such reactions are said to be exergonic. For an unfavorable reaction, $\\Delta G$ has a positive value, meaning that energy is absorbed by the chemical system from the surroundings. Such reactions are said to be endergonic.\n\nYou might also recall from general chemistry that the standard free-energy change for a reaction is denoted as $\\Delta G^{\\circ}$, where the superscript ${ }^{\\circ}$ means that the reaction is carried out under standard conditions, with pure substances in their most stable form at 1 atm pressure and a specified temperature, usually 298 K . For biological reactions, the standard free-energy change is denoted as $\\Delta G^{\\circ \\prime}$ and refers to a reaction carried out at\n$\\mathrm{pH}=7.0$ with solute concentrations of 1.0 M .\n\n\nBecause the equilibrium constant, $K_{\\text {eq }}$, and the standard free-energy change, $\\Delta G^{\\circ}$, both measure whether a reaction is favorable, they are mathematically related by the equation\n\n$$\n\\Delta G^{\\circ}=-R T \\ln K_{\\mathrm{eq}} \\quad \\text { or } \\quad K_{\\mathrm{eq}}=e^{-\\Delta G^{\\circ} / R T}\n$$\n\nwhere\n\n$$\n\\begin{aligned}\n& R=8.314 \\mathrm{~J} /(\\mathrm{K} \\cdot \\mathrm{~mol})=1.987 \\mathrm{cal} /(\\mathrm{K} \\cdot \\mathrm{~mol}) \\\\\n& T=\\text { Kelvin temperature } \\\\\n& e=2.718\n\\end{aligned}\n$$"}
{"id": 359, "contents": "Solution - 6.7 Describing a Reaction: Equilibria, Rates, and Energy Changes\n$\\ln K_{\\text {eq }}=$ natural logarithm of $K_{\\text {eq }}$\nFor example, the reaction of ethylene with HBr has $K_{\\mathrm{eq}}=7.1 \\times 10^{7}$, so $\\Delta G^{\\circ}=-44.8 \\mathrm{~kJ} / \\mathrm{mol}(-10.7 \\mathrm{kcal} / \\mathrm{mol})$ at 298 K:\n\n$$\n\\begin{aligned}\nK_{\\mathrm{eq}} & =7.1 \\times 10^{7} \\quad \\text { and } \\quad \\ln K_{\\mathrm{eq}}=18.08 \\\\\n\\Delta G^{\\circ} & =-R T \\ln K_{\\mathrm{eq}}=-8.314 \\mathrm{~J} /(\\mathrm{K} \\cdot \\mathrm{~mol})(18.08) \\\\\n& =-44,800 \\mathrm{~J} / \\mathrm{mol}=-44.8 \\mathrm{~kJ} / \\mathrm{mol}\n\\end{aligned}\n$$\n\nThe free-energy change $\\Delta G$ is made up of two terms, an enthalpy term, $\\Delta H$, and a temperature-dependent entropy term, $T \\Delta S$. Of the two terms, the enthalpy term is often larger and more dominant.\n\n$$\n\\Delta G^{\\circ}=\\Delta H^{\\circ}-T \\Delta S^{\\circ}\n$$\n\nFor the reaction of ethylene with HBr at room temperature ( 298 K ), the approximate values are"}
{"id": 360, "contents": "Solution - 6.7 Describing a Reaction: Equilibria, Rates, and Energy Changes\n$$\n\\Delta G^{\\circ}=\\Delta H^{\\circ}-T \\Delta S^{\\circ}\n$$\n\nFor the reaction of ethylene with HBr at room temperature ( 298 K ), the approximate values are\n\n$$\n\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}_{2}+\\mathrm{HBr} \\rightleftarrows \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{Br} \\quad\\left\\{\\begin{array}{l}\n\\Delta G^{\\circ}=-44.8 \\mathrm{~kJ} / \\mathrm{mol} \\\\\n\\Delta H^{\\circ}=-84.1 \\mathrm{~kJ} / \\mathrm{mol} \\\\\n\\Delta S^{\\circ}=-0.132 \\mathrm{~kJ} /(\\mathrm{K} \\cdot \\mathrm{~mol}) \\\\\nK_{\\mathrm{eq}}=7.1 \\times 10^{7}\n\\end{array}\\right.\n$$\n\nThe enthalpy change $(\\boldsymbol{\\Delta H})$, also called the heat of reaction, is a measure of the change in total bonding energy during a reaction. If $\\Delta H$ is negative, as in the reaction of HBr with ethylene, the products have less energy than the reactants. Thus, the products are more stable and have stronger bonds than the reactants, heat is released, and the reaction is said to be exothermic. If $\\Delta H$ is positive, the products are less stable and have weaker bonds than the reactants, heat is absorbed, and the reaction is said to be endothermic. For example, if a reaction breaks reactant bonds with a total strength of $380 \\mathrm{~kJ} / \\mathrm{mol}$ and forms product bonds with a total strength of 400 $\\mathrm{kJ} / \\mathrm{mol}$, then $\\Delta H$ for the reaction is $400 \\mathrm{~kJ} / \\mathrm{mol}-380 \\mathrm{~kJ} / \\mathrm{mol}=-20 \\mathrm{~kJ} / \\mathrm{mol}$ and the reaction is exothermic."}
{"id": 361, "contents": "Solution - 6.7 Describing a Reaction: Equilibria, Rates, and Energy Changes\nThe entropy change ( $\\boldsymbol{\\Delta} \\boldsymbol{S}$ ) is a measure of the change in the amount of molecular randomness, or freedom of motion, that accompanies a reaction. For example, in an elimination reaction of the type\n\n$$\n\\mathbf{A} \\longrightarrow \\mathbf{B}+\\mathbf{C}\n$$\n\nthere is more freedom of movement and molecular randomness in the products than in the reactant because one molecule has split into two. Thus, there is a net increase in entropy during the reaction and $\\Delta S$ has a positive value.\n\nOn the other hand, for an addition reaction of the type\n\n$$\n\\mathbf{A}+\\mathbf{B} \\longrightarrow \\mathbf{C}\n$$\n\nthe opposite is true. Because such reactions restrict the freedom of movement of two molecules by joining them together, the product has less randomness than the reactants and $\\Delta S$ has a negative value. The reaction of ethylene and HBr to yield bromoethane, which has $\\Delta S^{\\circ}=-0.132 \\mathrm{~kJ} /(\\mathrm{K} \\cdot \\mathrm{mol})$, is an example. TABLE 6.2 describes the thermodynamic terms more fully.\n\nTABLE 6.2 Explanation of Thermodynamic Quantities: $\\Delta G^{\\circ}=\\Delta H^{\\circ}-T \\Delta S^{\\circ}$"}
{"id": 362, "contents": "Solution - 6.7 Describing a Reaction: Equilibria, Rates, and Energy Changes\nTABLE 6.2 Explanation of Thermodynamic Quantities: $\\Delta G^{\\circ}=\\Delta H^{\\circ}-T \\Delta S^{\\circ}$\n\n| Term | Name | Explanation |\n| :--- | :--- | :--- |\n| $\\Delta G^{\\circ}$ | Gibbs <br> free- <br> energy <br> change | The energy difference between reactants and products. When $\\Delta G^{\\circ}$ is negative, the reaction <br> is exergonic, has a favorable equilibrium constant, and can occur spontaneously. When $\\Delta G^{\\circ}$ <br> is positive, the reaction is endergonic, has an unfavorable equilibrium constant, and cannot <br> occur spontaneously. |\n| $\\Delta \\boldsymbol{H}^{\\circ}$ | Enthalpy <br> change | The heat of reaction, or difference in strength between the bonds broken in a reaction and <br> the bonds formed. When $\\Delta H^{\\circ}$ is negative, the reaction releases heat and is exothermic. <br> When $\\Delta H^{\\circ}$ is positive, the reaction absorbs heat and is endothermic. |\n| $\\Delta \\boldsymbol{S}^{\\circ}$ | Entropy <br> change | The change in molecular randomness during a reaction. When $\\Delta S^{\\circ}$ is negative, randomness <br> decreases. When $\\Delta S^{\\circ}$ is positive, randomness increases. |\n\nKnowing the value of $K_{\\text {eq }}$ for a reaction is useful, but it's important to realize its limitations. An equilibrium constant tells only the position of the equilibrium, or how much product is theoretically possible. It doesn't tell the rate of reaction, or how fast the equilibrium is established. Some reactions are extremely slow even though they have favorable equilibrium constants. Gasoline is stable at room temperature, for instance, because the rate of its reaction with oxygen is slow at 298 K . Only at higher temperatures, such as contact with a lighted match, does gasoline react rapidly with oxygen and undergo complete conversion to the equilibrium products water and carbon dioxide. Rates (how fast a reaction occurs) and equilibria (how much a reaction occurs) are entirely different."}
{"id": 363, "contents": "Equilibrium $\\rightarrow$ In what direction does the reaction proceed? - \nPROBLEM Which reaction is more energetically favored, one with $\\Delta G^{\\circ}=-44 \\mathrm{~kJ} / \\mathrm{mol}$ or one with $\\Delta G^{\\circ}=+44 \\mathrm{~kJ} /$ 6-10 mol?\n\nPROBLEM Which reaction is more exergonic, one with $K_{\\text {eq }}=1000$ or one with $K_{\\text {eq }}=0.001$ ? 6-11"}
{"id": 364, "contents": "Equilibrium $\\rightarrow$ In what direction does the reaction proceed? - 6.8 Describing a Reaction: Bond Dissociation Energies\nWe've just seen that heat is released (negative $\\Delta H$ ) when a bond is formed because the products are more stable and have stronger bonds than the reactants. Conversely, heat is absorbed (positive $\\Delta H$ ) when a bond is broken because the products are less stable and have weaker bonds than the reactants. The amount of energy needed to break a given bond to produce two radical fragments when the molecule is in the gas phase at $25^{\\circ} \\mathrm{C}$ is a quantity called the bond strength, or bond dissociation energy ( $D$ ).\n\n$$\n\\mathrm{A}: \\mathrm{B} \\xrightarrow[\\text { energy }]{\\text { Bond dissociation }} \\mathrm{A} \\cdot+\\cdot \\mathrm{B}\n$$\n\nEach specific bond has its own characteristic strength, and extensive tables of such data are available. For example, a $\\mathrm{C}-\\mathrm{H}$ bond in methane has a bond dissociation energy $D=439.3 \\mathrm{~kJ} / \\mathrm{mol}(105.0 \\mathrm{kcal} / \\mathrm{mol})$, meaning that $439.3 \\mathrm{~kJ} / \\mathrm{mol}$ must be added to break a $\\mathrm{C}-\\mathrm{H}$ bond of methane to give the two radical fragments $\\cdot \\mathrm{CH}_{3}$ and $\\cdot$ H. Conversely, $439.3 \\mathrm{~kJ} / \\mathrm{mol}$ of energy is released when a methyl radical and a hydrogen atom combine to form methane. TABLE 6.3 lists some other bond strengths.\n\nTABLE 6.3 Some Bond Dissociation Energies, $\\boldsymbol{D}$"}
{"id": 365, "contents": "Equilibrium $\\rightarrow$ In what direction does the reaction proceed? - 6.8 Describing a Reaction: Bond Dissociation Energies\n| Bond | D (kJ/mol) | Bond | D (kJ/mol) | Bond | D (kJ/mol) |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| $\\mathrm{H}-\\mathrm{H}$ | 436 | $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CH}-\\mathrm{H}$ | 410 | $\\mathrm{C}_{2} \\mathrm{H}_{5}-\\mathrm{CH}_{3}$ | 370 |\n| H-F | 570 | $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CH}-\\mathrm{CI}$ | 354 | $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CH}-\\mathrm{CH}_{3}$ | 369 |\n| H-CI | 431 | $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CH}-\\mathrm{Br}$ | 299 | $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{C}-\\mathrm{CH}_{3}$ | 363 |\n| $\\mathrm{H}-\\mathrm{Br}$ | 366 | $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{C}-\\mathrm{H}$ | 400 | $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}-\\mathrm{CH}_{3}$ | 426 |\n| $\\mathrm{H}-\\mathrm{I}$ | 298 | $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{C}-\\mathrm{CI}$ | 352 | $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}_{2}-\\mathrm{CH}_{3}$ | 318 |\n| $\\mathrm{CI}-\\mathrm{CI}$ | 242 | $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{C}-\\mathrm{Br}$ | 293 | $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}_{2}$ | 728 |"}
{"id": 366, "contents": "Equilibrium $\\rightarrow$ In what direction does the reaction proceed? - 6.8 Describing a Reaction: Bond Dissociation Energies\n| $\\mathrm{Br}-\\mathrm{Br}$ | 194 | $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{C}-\\mathrm{I}$ | 227 | <smiles>Cc1ccccc1</smiles> | 427 |\n| I-I | 152 | $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}-\\mathrm{H}$ | 464 | <smiles>CCc1ccccc1</smiles> | 325 |\n| $\\mathrm{CH}_{3}-\\mathrm{H}$ | 439 | $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}-\\mathrm{CI}$ | 396 | <smiles>CC=O</smiles> | 374 |\n| $\\mathrm{CH}_{3}-\\mathrm{CI}$ | 350 | $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}_{2}-\\mathrm{H}$ | 369 | $\\mathrm{HO}-\\mathrm{H}$ | 497 |\n| $\\mathrm{CH}_{3}-\\mathrm{Br}$ | 294 | $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}_{2}-\\mathrm{CI}$ | 298 | $\\mathrm{HO}-\\mathrm{OH}$ | 211 |\n| $\\mathrm{CH}_{3}-\\mathrm{I}$ | 239 | <smiles>c1ccccc1</smiles> | 472 | $\\mathrm{CH}_{3} \\mathrm{O}-\\mathrm{H}$ | 440 |\n| $\\mathrm{CH}_{3}-\\mathrm{OH}$ | 385 | <smiles>Clc1ccccc1</smiles> | 400 | $\\mathrm{CH}_{3} \\mathrm{~S}-\\mathrm{H}$ | 366 |\n| $\\mathrm{CH}_{3}-\\mathrm{NH}_{2}$ | 386 | <smiles>Cc1ccccc1</smiles> | 375 | $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{O}-\\mathrm{H}$ | 441 |"}
{"id": 367, "contents": "Equilibrium $\\rightarrow$ In what direction does the reaction proceed? - 6.8 Describing a Reaction: Bond Dissociation Energies\n| $\\mathrm{C}_{2} \\mathrm{H}_{5}-\\mathrm{H}$ | 421 | <smiles>ClCc1ccccc1</smiles> | 300 | <smiles>CC(C)=O</smiles> | 352 |\n| $\\mathrm{C}_{2} \\mathrm{H}_{5}-\\mathrm{CI}$ | 352 | <smiles>Brc1ccccc1</smiles> | 336 | $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{O}-\\mathrm{CH}_{3}$ | 355 |"}
{"id": 368, "contents": "Equilibrium $\\rightarrow$ In what direction does the reaction proceed? - 6.8 Describing a Reaction: Bond Dissociation Energies\nTABLE 6.3 Some Bond Dissociation Energies, $\\boldsymbol{D}$\n\n| Bond | $\\boldsymbol{D}(\\mathrm{kJ} / \\mathrm{mol})$ | Bond | $\\boldsymbol{D}(\\mathrm{kJ} / \\mathrm{mol})$ | Bond | $\\boldsymbol{D}(\\mathrm{kJ} / \\mathrm{mol})$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n| $\\mathrm{C}_{2} \\mathrm{H}_{5}-\\mathrm{Br}$ | 293 |  | 464 | $\\mathrm{NH}_{2}-\\mathrm{H}$ | 450 |\n| $\\mathrm{C}_{2} \\mathrm{H}_{5}-\\mathrm{I}$ | 233 | $\\mathrm{HC} \\equiv \\mathrm{C}-\\mathrm{H}$ | 558 | $\\mathrm{H}-\\mathrm{CN}$ | 528 |\n| $\\mathrm{C}_{2} \\mathrm{H}_{5}-\\mathrm{OH}$ | 391 | $\\mathrm{CH}_{3}-\\mathrm{CH}_{3}$ | 377 |  |  |\n\nThink again about the connection between bond strengths and chemical reactivity. In an exothermic reaction, more heat is released than is absorbed. But because making bonds in the products releases heat and breaking bonds in the reactants absorbs heat, the bonds in the products must be stronger than the bonds in the reactants. In other words, exothermic reactions are favored by products with strong bonds and by reactants with weak, easily broken bonds."}
{"id": 369, "contents": "Equilibrium $\\rightarrow$ In what direction does the reaction proceed? - 6.8 Describing a Reaction: Bond Dissociation Energies\nSometimes, particularly in biochemistry, reactive substances that undergo highly exothermic reactions, such as ATP (adenosine triphosphate), are referred to as \"energy-rich\" or \"high-energy\" compounds. Such a label doesn't mean that ATP is special or different from other compounds, it only means that ATP has relatively weak bonds that require a relatively small amount of heat to break, thus leading to a larger release of heat when a strong new bond forms in a reaction. When a typical organic phosphate such as glycerol 3-phosphate reacts with water, for instance, only $9 \\mathrm{~kJ} / \\mathrm{mol}$ of heat is released ( $\\Delta H^{\\circ}=-9 \\mathrm{~kJ} / \\mathrm{mol}$ ), but when ATP reacts with water, $30 \\mathrm{~kJ} / \\mathrm{mol}$ of heat is released $\\left(\\Delta H^{\\circ \\prime}=-30 \\mathrm{~kJ} / \\mathrm{mol}\\right)$. The difference between the two reactions is due to the fact that the bond broken in ATP is substantially weaker than the bond broken in glycerol 3-phosphate. We'll see the metabolic importance of this reaction in later chapters.\n$\\Delta H^{0^{\\prime}}=-9 \\mathrm{~kJ} / \\mathrm{mol}$\n\n\nAdenosine triphosphate (ATP)\nAdenosine diphosphate (ADP)"}
{"id": 370, "contents": "Equilibrium $\\rightarrow$ In what direction does the reaction proceed? - 6.9 Describing a Reaction: Energy Diagrams and Transition States\nFor a reaction to take place, reactant molecules must collide and reorganization of atoms and bonds must occur. Let's again look at the addition reaction of HBr and ethylene.\n\n\nAs the reaction proceeds, ethylene and HBr approach each other, the ethylene $\\pi$ bond and the $\\mathrm{H}-\\mathrm{Br}$ bond break, a new $\\mathrm{C}-\\mathrm{H}$ bond forms in step 1 and a new $\\mathrm{C}-\\mathrm{Br}$ bond forms in step 2.\n\nTo depict graphically the energy changes that occur during a reaction, chemists use energy diagrams, such as that in FIGURE 6.5. The vertical axis of the diagram represents the total energy of all reactants, and the horizontal axis, called the reaction coordinate, represents the progress of the reaction from beginning to end. Let's see how the addition of HBr to ethylene can be described in an energy diagram.\n\n\nFIGURE 6.5 An energy diagram for the first step in the reaction of ethylene with HBr. The energy difference between reactants and the transition state, $\\Delta G^{\\ddagger}$, defines the reaction rate. The energy difference between reactants and carbocation product, $\\Delta G^{\\circ}$, defines the position of the equilibrium.\n\nAt the beginning of the reaction, ethylene and HBr have the total amount of energy indicated by the reactant level on the left side of the diagram in FIGURE 6.5. As the two reactants collide and reaction commences, their electron clouds repel each other, causing the energy level to rise. If the collision has occurred with enough force and proper orientation, however, the reactants continue to approach each other despite the rising repulsion until the new $\\mathrm{C}-\\mathrm{H}$ bond starts to form. At some point, a structure of maximum energy is reached, a structure called the transition state."}
{"id": 371, "contents": "Equilibrium $\\rightarrow$ In what direction does the reaction proceed? - 6.9 Describing a Reaction: Energy Diagrams and Transition States\nThe transition state represents the highest-energy structure involved in this step of the reaction. It is unstable and can't be isolated, but we can imagine it to be an activated complex of the two reactants in which both the $\\mathrm{C}=\\mathrm{C} \\pi$ bond and $\\mathrm{H}-\\mathrm{Br}$ bond are partially broken and the new $\\mathrm{C}-\\mathrm{H}$ bond is partially formed (FIGURE 6.6).\n\n\nFIGURE 6.6 A hypothetical transition-state structure for the first step of the reaction of ethylene with HBr . The $\\mathrm{C}=\\mathrm{C} \\pi$ bond and $\\mathrm{H}-\\mathrm{Br}$ bond are just beginning to break, and the $\\mathrm{C}-\\mathrm{H}$ bond is just beginning to form.\n\nThe energy difference between reactants and the transition state is called the activation energy, $\\boldsymbol{\\Delta} \\boldsymbol{G}^{*}$, and determines how rapidly the reaction occurs at a given temperature. (The double-dagger superscript, ${ }^{\\ddagger}$, always\nrefers to the transition state.) A high activation energy results in a slow reaction because few collisions occur with enough energy for the reactants to reach the transition state. A low activation energy results in a rapid reaction because almost all collisions occur with enough energy for the reactants to reach the transition state.\n\nAs an analogy, you might think of reactants that need enough energy to climb the activation barrier to the transition state as hikers who need enough energy to climb to the top of a mountain pass. If the pass is a high one, the hikers need a lot of energy and surmount the barrier with difficulty. If the pass is low, the hikers need less energy and reach the top easily."}
{"id": 372, "contents": "Equilibrium $\\rightarrow$ In what direction does the reaction proceed? - 6.9 Describing a Reaction: Energy Diagrams and Transition States\nAs a rough generalization, many organic reactions have activation energies in the range 40 to $150 \\mathrm{~kJ} / \\mathrm{mol}$ ( $10-35 \\mathrm{kcal} / \\mathrm{mol}$ ). The reaction of ethylene with HBr , for example, has an activation energy of approximately $140 \\mathrm{~kJ} / \\mathrm{mol}(34 \\mathrm{kcal} / \\mathrm{mol})$. Reactions with activation energies less than $80 \\mathrm{~kJ} / \\mathrm{mol}$ take place at or below room temperature, while reactions with higher activation energies normally require a higher temperature to give the reactants enough energy to climb the activation barrier.\n\nOnce the transition state is reached, the reaction can either continue on to give the carbocation product or revert back to reactants. When reversion to reactants occurs, the transition-state structure comes apart and an amount of free energy corresponding to $-\\Delta G^{\\ddagger}$ is released. When the reaction continues on to give the carbocation, the new $\\mathrm{C}-\\mathrm{H}$ bond forms fully and an amount of energy is released corresponding to the difference between the transition state and carbocation product. The net energy change for the step, $\\Delta G^{\\circ}$, is represented in the diagram as the difference in level between reactant and product. Since the carbocation is higher in energy than the starting alkene, the step is endergonic, has a positive value of $\\Delta G^{\\circ}$, and absorbs energy.\n\nNot all energy diagrams are like that shown for the reaction of ethylene and HBr . Each reaction has its own energy profile. Some reactions are fast (small $\\Delta G^{\\ddagger}$ ) and some are slow (large $\\Delta G^{\\dagger}$ ); some have a negative $\\Delta G^{\\circ}$, and some have a positive $\\Delta G^{\\circ}$. FIGURE 6.7 illustrates some different possibilities."}
{"id": 373, "contents": "Equilibrium $\\rightarrow$ In what direction does the reaction proceed? - 6.9 Describing a Reaction: Energy Diagrams and Transition States\nFIGURE 6.7 Some hypothetical energy diagrams: (a) a fast exergonic reaction (low $\\Delta G^{\\ddagger}$, negative $\\Delta G^{\\circ}$ ); (b) a slow exergonic reaction (high $\\Delta G^{\\ddagger}$, negative $\\Delta G^{\\circ}$ ); (c) a fast endergonic reaction (small $\\Delta G^{\\ddagger}$, small positive $\\Delta G^{\\circ}$ ); (d) a slow endergonic reaction (high $\\Delta G^{\\ddagger}$, positive $\\Delta G^{\\circ}$ ).\nPROBLEM Which reaction is faster, one with $\\Delta G^{\\ddagger}=+45 \\mathrm{~kJ} / \\mathrm{mol}$ or one with $\\Delta G^{\\ddagger}=+70 \\mathrm{~kJ} / \\mathrm{mol}$ ?"}
{"id": 374, "contents": "Equilibrium $\\rightarrow$ In what direction does the reaction proceed? - 6.10 Describing a Reaction: Intermediates\nHow can we describe the carbocation formed in the first step of the reaction of ethylene with HBr ? The carbocation is clearly different from the reactants, yet it isn't a transition state and it isn't a final product.\n\n\nWe call the carbocation, which exists only transiently during the course of the multistep reaction, a reaction intermediate. As soon as the intermediate is formed in the first step by reaction of ethylene with $\\mathrm{H}^{+}$, it reacts further with $\\mathrm{Br}^{-}$in a second step to give the final product, bromoethane. This second step has its own activation energy $\\left(\\Delta G^{\\ddagger}\\right)$, its own transition state, and its own energy change ( $\\Delta G^{\\circ}$ ). We can picture the second transition state as an activated complex between the electrophilic carbocation intermediate and the nucleophilic bromide anion, in which $\\mathrm{Br}^{-}$donates a pair of electrons to the positively charged carbon atom as the new $\\mathrm{C}-\\mathrm{Br}$ bond just starts to form.\n\nA complete energy diagram for the overall reaction of ethylene with HBr is shown in FIGURE 6.8. In essence, we draw a diagram for each of the individual steps and then join them so that the carbocation product of step 1 is the reactant for step 2. As indicated in FIGURE 6.8, the reaction intermediate lies at an energy minimum between steps. Because the energy level of the intermediate is higher than the level of either the reactant that formed it or the product it yields, the intermediate can't normally be isolated. It is, however, more stable than its two neighboring transition states."}
{"id": 375, "contents": "Equilibrium $\\rightarrow$ In what direction does the reaction proceed? - 6.10 Describing a Reaction: Intermediates\nFIGURE 6.8 An energy diagram for the reaction of ethylene with $\\mathbf{H B r}$. Two separate steps are involved, each with its own activation energy $\\left(\\Delta G^{\\ddagger}\\right)$ and free-energy change ( $\\Delta G^{\\circ}$ ). The overall $\\Delta G^{\\ddagger}$ for the complete reaction is the energy difference between reactants and the highest transition state (which corresponds to $\\Delta G 1^{\\ddagger}$ in this case), and the overall $\\Delta G^{\\circ}$ for the reaction is the energy difference between reactants and final products.\nEach step in a multistep process can always be considered separately. Each step has its own $\\Delta G^{\\ddagger}$ and its own $\\Delta G^{\\circ}$. The overall activation energy that controls the rate of the reaction, however, is the energy difference between initial reactants and the highest transition state, regardless of which step it occurs in. The overall $\\Delta G^{\\circ}$ of the reaction is the energy difference between reactants and final products.\n\nThe biological reactions that take place in living organisms have the same energy requirements as reactions that take place in the laboratory and can be described in similar ways. They are, however, constrained by the fact that they must have low enough activation energies to occur at moderate temperatures, and they must release\nenergy in relatively small amounts to avoid overheating the organism. These constraints can be met through the use of large, structurally complex, enzyme catalysts that alter the mechanism of a reaction to a pathway that can proceed through a series of small steps rather than one or two large steps. Thus, an energy diagram for a biological reaction might look like that in FIGURE 6.9.\n\n\nFIGURE 6.9 Energy diagrams for a typical, enzyme-catalyzed biological reaction and an uncatalyzed laboratory reaction. The biological reaction involves many steps, each of which has a relatively small activation energy and small energy change. The end result is the same, however."}
{"id": 376, "contents": "Drawing a Reaction Energy Diagram - \nSketch an energy diagram for a one-step reaction that is fast and highly exergonic."}
{"id": 377, "contents": "Strategy - \nA fast reaction has a small $\\Delta G^{\\ddagger}$, and a highly exergonic reaction has a large negative $\\Delta G^{\\circ}$."}
{"id": 378, "contents": "Drawing a Reaction Energy Diagram - \nSketch an energy diagram for a two-step exergonic reaction whose second step has a higher-energy transition state than its first step. Show $\\Delta G^{\\ddagger}$ and $\\Delta G^{\\circ}$ for the overall reaction."}
{"id": 379, "contents": "Strategy - \nA two-step reaction has two transition states and an intermediate between them. The $\\Delta G^{\\ddagger}$ for the overall reaction is the energy change between reactants and the highest-energy transition state-the second one in this case. An exergonic reaction has a negative overall $\\Delta G^{\\circ}$."}
{"id": 380, "contents": "Solution - \nPROBLEM Sketch an energy diagram for a two-step reaction in which both steps are exergonic and in which\n6-13 the second step has a higher-energy transition state than the first. Label the parts of the diagram corresponding to reactant, product, intermediate, overall $\\Delta G^{\\ddagger}$, and overall $\\Delta G^{\\circ}$."}
{"id": 381, "contents": "Solution - 6.11 A Comparison Between Biological Reactions and Laboratory Reactions\nBeginning in the next chapter, we'll be seeing a lot of reactions, some that are important in laboratory chemistry yet don't occur in nature and others that have counterparts in biological pathways. In comparing laboratory reactions with biological reactions, several differences are apparent. For one, laboratory reactions are usually carried out in an organic solvent such as diethyl ether or dichloromethane to dissolve the reactants and bring them into contact, whereas biological reactions occur in the aqueous medium within cells. For another, laboratory reactions often take place over a wide range of temperatures without catalysts, while biological reactions take place at the temperature of the organism and are catalyzed by enzymes.\n\nWe'll look at enzymes in more detail in Section 26.10, but you may already be aware that an enzyme is a large, globular, protein molecule that contains in its structure a protected pocket called its active site. The active site is lined by acidic or basic groups as needed for catalysis and has precisely the right shape to bind and hold a substrate molecule in the orientation necessary for reaction. FIGURE 6.10 shows a molecular model of hexokinase, along with an X-ray crystal structure of the glucose substrate and adenosine diphosphate (ADP) bound in the active site. Hexokinase is an enzyme that catalyzes the initial step of glucose metabolism-the transfer of a phosphate group from ATP to glucose, giving glucose 6-phosphate and ADP. The structures of ATP and ADP were shown at the end of Section 6.8.\n\n\nNote how the hexokinase-catalyzed phosphorylation reaction of glucose is written. It's common when writing biological equations to show only the structures of the primary reactant and product, while abbreviating the structures of various biological \"reagents\" and by-products such as ATP and ADP. A curved arrow intersecting the straight reaction arrow indicates that ATP is also a reactant and ADP also a product.\n\n\nFIGURE 6.10 Models of hexokinase in space-filling and wire-frame formats, showing the cleft that contains the active site where substrate binding and reaction catalysis occur. At the bottom is an X-ray crystal structure of the enzyme active site, showing the positions of both glucose and ADP as well as a lysine amino acid that acts as a base to deprotonate glucose."}
{"id": 382, "contents": "Solution - 6.11 A Comparison Between Biological Reactions and Laboratory Reactions\nYet a third difference between laboratory and biological reactions is that laboratory reactions are often done using relatively small, simple reagents such as $\\mathrm{Br}_{2}, \\mathrm{HCl}, \\mathrm{NaBH}_{4}, \\mathrm{CrO}_{3}$, and so forth, while biological reactions usually involve relatively complex \"reagents\" called coenzymes. In the hexokinase-catalyzed phosphorylation of glucose just shown, ATP is the coenzyme. As another example, compare the $\\mathrm{H}_{2}$ molecule, a laboratory reagent that adds to a carbon-carbon double bond to yield an alkane, with the reduced nicotinamide adenine dinucleotide (NADH) molecule, a coenzyme that effects an analogous addition of hydrogen to a double bond in many biological pathways. Of all the atoms in the coenzyme, only the one hydrogen atom shown in red is transferred to the double-bond substrate.\n\n\nReduced nicotinamide adenine dinucleotide, NADH\n(a coenzyme)\nDon't be intimidated by the size of the ATP or NADH molecule; most of the structure is there to provide an overall shape for binding to the enzyme and to provide appropriate solubility behavior. When looking at biological molecules, focus on the small part of the molecule where the chemical change takes place.\n\nOne final difference between laboratory and biological reactions is in their specificity. A catalyst might be used in the laboratory to catalyze the reaction of thousands of different substances, but an enzyme, because it can only bind a specific substrate molecule having a specific shape, will usually catalyze only a specific reaction. It's this exquisite specificity that makes biological chemistry so remarkable and that makes life possible. TABLE 6.4 summarizes some of the differences between laboratory and biological reactions.\n\nTABLE 6.4 A Comparison of Typical Laboratory and Biological Reactions A Comparison of Typical Laboratory and Biological Reactions"}
{"id": 383, "contents": "Solution - 6.11 A Comparison Between Biological Reactions and Laboratory Reactions\nTABLE 6.4 A Comparison of Typical Laboratory and Biological Reactions A Comparison of Typical Laboratory and Biological Reactions\n\n|  | Laboratory reaction | Biological reaction |\n| :--- | :--- | :--- |\n| Solvent | Organic liquid, such as ether | Aqueous environment in cells |\n| Temperature | Wide range; -80 to $150^{\\circ} \\mathrm{C}$ | Temperature of organism |\n| Catalyst | Either none, or very simple | Large, complex enzymes needed |\n| Reagent size | Usually small and simple | Relatively complex coenzymes |\n| Specificity | Little specificity for substrate | Very high specificity for substrate |"}
{"id": 384, "contents": "Where Do Drugs Come From? - \nIt has been estimated that major pharmaceutical companies in the United States spent some $\\$ 200$ billion on drug research and development in 2020, while government agencies and private foundations spent another $\\$ 28$ billion. What does this money buy? From 1983 to 2022, the money resulted in a total of 1237 new molecular entities (NMEs) - new biologically active chemical substances approved for sale as drugs by the U.S. Food and Drug Administration (FDA).\n\nWhere do the new drugs come from? According to a study carried out several years ago at the U.S. National Cancer Institute, only about $33 \\%$ of new drugs are entirely synthetic and completely unrelated to any naturally occurring substance. The remaining $67 \\%$ take their lead, to a greater or lesser extent, from nature. Vaccines and genetically engineered proteins of biological origin account for $15 \\%$ of NMEs, but most new drugs come from natural products, a catchall term generally taken to mean small molecules found in bacteria, plants, algae, and other living organisms. Unmodified natural products isolated directly from the producing organism account for $24 \\%$ of NMEs, while natural products that have been chemically modified in the laboratory account for the remaining $28 \\%$.\n\nOrigin of New Drugs 1981-2014\n\n\nMany years of work go into screening many thousands of substances to identify a single compound that might ultimately gain approval as an NME. But after that single compound has been identified, the work has just begun because it takes an average of 9 to 10 years for a drug to make it through the approval process. First, the safety of the drug in animals must be demonstrated and an economical method of manufacture must be devised. With these preliminaries out of the way, an Investigational New Drug (IND) application is submitted to the FDA for permission to begin testing in humans.\n\n\nFIGURE 6.11 Introduced in June, 2006, Gardasil is the first vaccine ever approved for the prevention of cancer. Where do new drugs like this come from? (credit: modification of work \"COVIran Barekat vaccine production\" by Sadegh Nikgostar/Wikimedia Commons, CC BY 4.0)"}
{"id": 385, "contents": "Where Do Drugs Come From? - \nHuman testing takes, or should take, 5 to 7 years and is divided into three phases. Phase I clinical trials are\ncarried out on a small group of healthy volunteers to establish safety and look for side effects. Several months to a year are needed, and only about $70 \\%$ of drugs pass at this point. Phase II clinical trials next test the drug for 1 to 2 years in several hundred patients with the target disease or condition, looking both for safety and efficacy, and only about $33 \\%$ of the original group pass. Finally, phase III trials are undertaken on a large sample of patients to document definitively the drug's safety, dosage, and efficacy. If the drug is one of the $25 \\%$ of the original group that make it to the end of phase III, all the data are then gathered into a New Drug Application (NDA) and sent to the FDA for review and approval, which can take another 2 years. Ten years have elapsed and at least $\\$ 500$ million has been spent, with only a $20 \\%$ success rate for the drugs that began testing. Finally, though, the drug will begin to appear in medicine cabinets. The following timeline shows the process."}
{"id": 386, "contents": "Key Terms - \n- activation energy $\\left(\\Delta G^{\\ddagger}\\right)$\n- active site\n- addition reaction\n- bond dissociation energy (D)\n- carbocation\n- electrophile\n- elimination reaction\n- endergonic\n- endothermic\n- enthalpy change ( $\\Delta H$ )\n- entropy change ( $\\Delta S$ )\n- enzyme\n- exergonic\n- exothermic\n- Gibbs free-energy change ( $\\Delta G$ )\n- heat of reaction\n- nucleophile\n- polar reaction\n- polarizability\n- radical\n- radical reaction\n- reaction coordinate\n- reaction intermediate\n- reaction mechanism\n- rearrangement reaction\n- substitution reaction\n- transition state"}
{"id": 387, "contents": "Summary - \nAll chemical reactions, whether in the laboratory or in living organisms, follow the same chemical rules. To understand both organic and biological chemistry, it's necessary to know not just what occurs but also why and how chemical reactions take place. In this chapter, we've taken a brief look at the fundamental kinds of organic reactions, we've seen why reactions occur, and we've seen how reactions can be described.\n\nThere are four common kinds of reactions: addition reactions take place when two reactants add together to give a single product; elimination reactions take place when one reactant splits apart to give two products; substitution reactions take place when two reactants exchange parts to give two new products; and rearrangement reactions take place when one reactant undergoes a reorganization of bonds and atoms to give an isomeric product.\n\nA full description of how a reaction occurs is called its mechanism. There are two general kinds of mechanisms by which most reactions take place: radical mechanisms and polar mechanisms. Polar reactions, the more common type, occur because of an attractive interaction between a nucleophilic (electron-rich) site in one molecule and an electrophilic (electron-poor) site in another molecule. A bond is formed in a polar reaction when the nucleophile donates an electron pair to the electrophile. This transfer of electrons is indicated by a curved arrow showing the direction of electron travel from the nucleophile to the electrophile. Radical reactions involve species that have an odd number of electrons. A bond is formed when each reactant donates one\nelectron."}
{"id": 388, "contents": "Summary - \nThe energy changes that take place during reactions can be described by considering both rates (how fast the reactions occur) and equilibria (how much the reactions occur). The position of a chemical equilibrium is determined by the value of the free-energy change $(\\boldsymbol{\\Delta} \\boldsymbol{G})$ for the reaction, where $\\Delta G=\\Delta H-T \\Delta S$. The enthalpy term $(\\Delta H)$ corresponds to the net change in strength of chemical bonds broken and formed during the reaction; the entropy term $(\\Delta S)$ corresponds to the change in the amount of molecular randomness during the reaction. Reactions that have negative values of $\\Delta G$ release energy, are said to be exergonic, and have favorable equilibria. Reactions that have positive values of $\\Delta G$ absorb energy, are said to be endergonic, and have unfavorable equilibria.\n\nA reaction can be described pictorially using an energy diagram that follows the reaction course from reactants through transition state to product. The transition state is an activated complex occurring at the highestenergy point of a reaction. The amount of energy needed by reactants to reach this high point is the activation energy, $\\Delta G^{*}$. The higher the activation energy, the slower the reaction.\n\nMany reactions take place in more than one step and involve the formation of a reaction intermediate. An intermediate is a species that lies at an energy minimum between steps on the reaction curve and is formed briefly during the course of a reaction."}
{"id": 389, "contents": "Visualizing Chemistry - \nPROBLEM The following alkyl halide can be prepared by addition of HBr to two different alkenes. Draw the 6-14 structures of both (reddish-brown $=\\mathrm{Br}$ ).\n\n\nPROBLEM The following structure represents the carbocation intermediate formed in the addition reaction of 6-15 HBr to two different alkenes. Draw the structures of both.\n\n\nPROBLEM Electrostatic potential maps of (a) formaldehyde $\\left(\\mathrm{CH}_{2} \\mathrm{O}\\right)$ and $(\\mathbf{b})$ methanethiol $\\left(\\mathrm{CH}_{3} \\mathrm{SH}\\right)$ are shown.\n6-16 Is the formaldehyde carbon atom likely to be electrophilic or nucleophilic? What about the methanethiol sulfur atom? Explain.\n(a)\n\n\nFormaldehyde\n(b)\n\n\nMethanethiol\n\nPROBLEM Look at the following energy diagram:\n6-17\n\n(a) Is $\\Delta G^{\\circ}$ for the reaction positive or negative? Label it on the diagram.\n(b) How many steps are involved in the reaction?\n(c) How many transition states are there? Label them on the diagram.\n\nPROBLEM Look at the following energy diagram for an enzyme-catalyzed reaction:\n6-18\n\n(a) How many steps are involved? (b) Which step is most exergonic?\n(c) Which step is slowest?"}
{"id": 390, "contents": "Energy Diagrams and Reaction Mechanisms - \nPROBLEM What is the difference between a transition state and an intermediate?\n6-19\nPROBLEM Draw an energy diagram for a one-step reaction with $K_{\\mathrm{eq}}<1$. Label the parts of the diagram 6-20 corresponding to reactants, products, transition state, $\\Delta G^{\\circ}$, and $\\Delta G^{\\ddagger}$. Is $\\Delta G^{\\circ}$ positive or negative?\n\nPROBLEM Draw an energy diagram for a two-step reaction with $K_{\\mathrm{eq}}>1$. Label the overall $\\Delta G^{\\circ}$, transition 6-21 states, and intermediate. Is $\\Delta G^{\\circ}$ positive or negative?\n\nPROBLEM Draw an energy diagram for a two-step exergonic reaction whose second step is faster than its first 6-22 step.\n\nPROBLEM Draw an energy diagram for a reaction with $K_{\\text {eq }}=1$. What is the value of $\\Delta G^{\\circ}$ in this reaction? 6-23\n\nPROBLEM The addition of water to ethylene to yield ethanol has the following thermodynamic parameters: 6-24\n$\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}_{2}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftarrows \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}\\left\\{\\begin{array}{l}\\Delta H^{\\circ}=-44 \\mathrm{~kJ} / \\mathrm{mol} \\\\ \\Delta S^{\\circ}=-0.12 \\mathrm{~kJ} /(\\mathrm{K} \\cdot \\mathrm{mol}) \\\\ K_{\\text {eq }}=24\\end{array}\\right.$\n(a) Is the reaction exothermic or endothermic?\n(b) Is the reaction favorable (spontaneous) or unfavorable (nonspontaneous) at room temperature ( 298 K )?\n\nPROBLEM When isopropylidenecyclohexane is treated with strong acid at room temperature, isomerization\n6-25 occurs by the mechanism shown below to yield 1-isopropylcyclohexene:"}
{"id": 391, "contents": "Energy Diagrams and Reaction Mechanisms - \nPROBLEM When isopropylidenecyclohexane is treated with strong acid at room temperature, isomerization\n6-25 occurs by the mechanism shown below to yield 1-isopropylcyclohexene:\n\n\nIsopropylidenecyclohexane\n1-Isopropylcyclohexene\nAt equilibrium, the product mixture contains about $30 \\%$ isopropylidenecyclohexane and about 70\\% 1-isopropylcyclohexene.\n(a) What is an approximate value of $K_{\\text {eq }}$ for the reaction?\n(b) Since the reaction occurs slowly at room temperature, what is its approximate $\\Delta G^{\\ddagger}$ ?\n(c) Draw an energy diagram for the reaction.\n\nPROBLEM Add curved arrows to the mechanism shown in Problem 6-25 to indicate the electron movement in 6-26 each step.\n\nPROBLEM Draw the complete mechanism for each of the following polar reactions.\n6-27 (a)\n\n(b)\n\n\n\n(c)\n\n\nPROBLEM Use curved arrows to show the flow of electrons, and draw the carbon radical that is formed when 6-28 the halogen radicals below add to the corresponding alkenes.\n(a)\n\n$\\qquad$ (b)\n\n(c)"}
{"id": 392, "contents": "Polar Reactions - \nPROBLEM Identify the functional groups in the following molecules, and show the polarity of each:\n6-29 (a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{C} \\equiv \\mathrm{N}$\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n\nPROBLEM Identify the following reactions as additions, eliminations, substitutions, or rearrangements:\n6-30 (a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{Br}+\\mathrm{NaCN} \\longrightarrow \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CN}(+\\mathrm{NaBr})$\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM Identify the likely electrophilic and nucleophilic sites in each of the following molecules:\n\n6-31 (a)\n\n\n\nTestosterone\n(b)\n\n\nMethamphetamine\n\nPROBLEM Identify the electrophile and the nucleophile.\n\n6-32 (a)\n\n(b)\n\n(c)\n\n\nPROBLEM Add curved arrows to the following polar reactions to indicate the flow of electrons in each:\n6-33 (a)\n\n(b)\n\n\nPROBLEM Follow the flow of electrons indicated by the curved arrows in each of the following polar reactions,\n6-34 and predict the products that result:\n(a)\n\n(b)\n\n(c)\n\n(d)"}
{"id": 393, "contents": "Radical Reactions - \nPROBLEM Radical chlorination of pentane is a poor way to prepare 1-chloropentane, but radical chlorination\n6-35 of neopentane, $\\left(\\mathrm{CH}_{3}\\right)_{4} \\mathrm{C}$, is a good way to prepare neopentyl chloride, $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CCH}_{2} \\mathrm{Cl}$. Explain.\nPROBLEM Despite the limitations of radical chlorination of alkanes, the reaction is still useful for synthesizing\n6-36 certain halogenated compounds. For which of the following compounds does radical chlorination give a single monochloro product?\n(a) $\\mathrm{CH}_{3} \\mathrm{CH}_{3}$\n(b) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$\n(c)\n\n(d)\n\n(e) $\\mathrm{CH}_{3} \\mathrm{C} \\equiv \\mathrm{CCH}_{3}$\n(f)\n\n\nPROBLEM Draw the different monochlorinated constitutional isomers you would obtain by the radical\n6-37 chlorination of the following compounds.\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Answer question 6-37 taking all stereoisomers into account.\n6-38\nPROBLEM Show the structure of the carbocation that would result when each of the following alkenes reacts 6-39 with an acid, $\\mathrm{H}^{+}$.\n(a)\n\n(b)\n\n(c)"}
{"id": 394, "contents": "General Problems - \nPROBLEM 2-Chloro-2-methylpropane reacts with water in three steps to yield 2-methyl-2-propanol. The first\n6-40 step is slower than the second, which in turn is much slower than the third. The reaction takes place slowly at room temperature, and the equilibrium constant is approximately 1.\n\n\n2-Chloro-2-\n2-Methyl-2-propanol methylpropane\n(a) Give approximate values for $\\Delta G^{\\ddagger}$ and $\\Delta G^{\\circ}$ that are consistent with the above information.\n(b) Draw an energy diagram for the reaction, labeling all points of interest and placing relative energy levels on the diagram consistent with the information given.\n\nPROBLEM Add curved arrows to the mechanism shown in Problem 6-40 to indicate the electron movement in\n6-41 each step.\nPROBLEM The reaction of hydroxide ion with chloromethane to yield methanol and chloride ion is an example\n6-42 of a general reaction type called a nucleophilic substitution reaction:\n$\\mathrm{HO}^{-}+\\mathrm{CH}_{3} \\mathrm{Cl} \\rightleftarrows \\mathrm{CH}_{3} \\mathrm{OH}+\\mathrm{Cl}^{-}$\nThe value of $\\Delta H^{\\circ}$ for the reaction is $-75 \\mathrm{~kJ} / \\mathrm{mol}$, and the value of $\\Delta S^{\\circ}$ is $+54 \\mathrm{~J} /(\\mathrm{K} \\cdot \\mathrm{mol})$. What is the value of $\\Delta G^{\\circ}$ (in $\\mathrm{kJ} / \\mathrm{mol}$ ) at 298 K ? Is the reaction exothermic or endothermic? Is it exergonic or endergonic?\n\nPROBLEM Methoxide ion $\\left(\\mathrm{CH}_{3} \\mathrm{O}^{-}\\right)$reacts with bromoethane in a single step according to the following\n6-43 equation:\n\n\nIdentify the bonds broken and formed, and draw curved arrows to represent the flow of electrons during the reaction."}
{"id": 395, "contents": "General Problems - \nIdentify the bonds broken and formed, and draw curved arrows to represent the flow of electrons during the reaction.\n\nPROBLEM Ammonia reacts with acetyl chloride $\\left(\\mathrm{CH}_{3} \\mathrm{COCl}\\right)$ to give acetamide $\\left(\\mathrm{CH}_{3} \\mathrm{CONH}_{2}\\right)$. Identify the bonds\n6-44 broken and formed in each step of the reaction, and draw curved arrows to represent the flow of electrons in each step.\n\n\nAcetyl chloride\n\n\n\nPROBLEM The naturally occurring molecule $\\alpha$-terpineol is biosynthesized by a route that includes the 6-45 following step:\n\n(a) Propose a likely structure for the isomeric carbocation intermediate.\n(b) Show the mechanism of each step in the biosynthetic pathway, using curved arrows to indicate electron flow.\n\nPROBLEM Predict the product(s) of each of the following biological reactions by interpreting the flow of 6-46 electrons as indicated by the curved arrows:\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Reaction of 2-methylpropene with HBr might, in principle, lead to a mixture of two alkyl bromide 6-47 addition products. Name them, and draw their structures.\n\nPROBLEM Draw the structures of the two carbocation intermediates that might form during the reaction 6-48 of 2-methylpropene with HBr (Problem 6-47). We'll see in the next chapter that the stability of carbocations depends on the number of alkyl substituents attached to the positively charged carbon-the more alkyl substituents there are, the more stable the cation. Which of the two carbocation intermediates you drew is more stable?"}
{"id": 396, "contents": "Alkenes: Structure and Reactivity - \nFIGURE 7.1 Tomatoes are good for you. Their red color is due to lycopene, which has 13 double bonds. (credit: modification of work \"Tomatoes\" by Jeremy Keith/Flickr, CC BY 2.0)"}
{"id": 397, "contents": "CHAPTER CONTENTS - \n7.1 Industrial Preparation and Use of Alkenes\n7.2 Calculating the Degree of Unsaturation\n7.3 Naming Alkenes\n7.4 Cis-Trans Isomerism in Alkenes\n7.5 Alkene Stereochemistry and the $E, Z$ Designation\n7.6 Stability of Alkenes\n7.7 Electrophilic Addition Reactions of Alkenes\n7.8 Orientation of Electrophilic Additions: Markovnikov's Rule\n7.9 Carbocation Structure and Stability\n7.10 The Hammond Postulate\n7.11 Evidence for the Mechanism of Electrophilic Additions: Carbocation Rearrangements\n\nWHY THIS CHAPTER? Carbon-carbon double bonds are present in most organic and biological molecules, so a good understanding of their behavior is needed. In this chapter, we'll look at some consequences of alkene stereoisomerism and then focus on the broadest and most general class of alkene reactions, the electrophilic addition reaction. Carbon-carbon triple bonds, by contrast, occur much less commonly, so we'll not spend much time on their chemistry.\n\nAn alkene, sometimes called an olefin from the German term for oil forming, is a hydrocarbon that contains a carbon-carbon double bond, while an alkyne is a hydrocarbon that contains a carbon-carbon triple bond. Alkenes occur abundantly in nature. Ethylene, for instance, is a plant hormone that induces ripening in fruit, and $\\alpha$-pinene is the major component of turpentine. Lycopene, found in fruits such as watermelon and papaya\nas well as tomatoes, is an antioxidant with numerous health benefits such as sun protection and cardiovascular protection.\n\n\nLycopene, a conjugated polyene"}
{"id": 398, "contents": "CHAPTER CONTENTS - 7.1 Industrial Preparation and Use of Alkenes\nEthylene and propylene, the simplest alkenes, are the two most important organic chemicals produced industrially. Approximately 220 million tons of ethylene and 138 million tons of propylene are produced worldwide each year for use in the synthesis of polyethylene, polypropylene, ethylene glycol, acetic acid, acetaldehyde, and a host of other substances (FIGURE 7.2).\n\n\nEthylene, propylene, and butene are synthesized from ( $\\mathrm{C}_{2}-\\mathrm{C}_{8}$ ) alkanes by a process called steam cracking at temperatures up to $900^{\\circ} \\mathrm{C}$.\n\n$$\n\\begin{gathered}\n\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{n} \\mathrm{CH}_{3} \\quad[n=0-6] \\\\\n\\downarrow \\begin{array}{l}\n850-900{ }^{\\circ} \\mathrm{C}, \\\\\n\\text { steam }\n\\end{array} \\\\\n\\mathrm{H}_{2}+\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}_{2}+\\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{CH}_{2}+\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}=\\mathrm{CH}_{2}\n\\end{gathered}\n$$\n\nThe cracking process is complex, although it undoubtedly involves radical reactions. The high-temperature\nreaction conditions cause spontaneous breaking of $\\mathrm{C}-\\mathrm{C}$ and $\\mathrm{C}-\\mathrm{H}$ bonds, with the resultant formation of smaller fragments. We might imagine, for instance, that a molecule of butane splits into two ethyl radicals, each of which then loses a hydrogen atom to generate two molecules of ethylene."}
{"id": 399, "contents": "CHAPTER CONTENTS - 7.1 Industrial Preparation and Use of Alkenes\nSteam cracking is an example of a reaction whose energetics are dominated by entropy ( $\\Delta S^{\\circ}$ ) rather than by enthalpy $\\left(\\Delta H^{\\circ}\\right)$ in the free-energy equation $\\Delta G^{\\circ}=\\Delta H^{\\circ}-T \\Delta S^{\\circ}$ discussed in Section 6.7. Although the bond dissociation energy $D$ for a carbon-carbon single bond is relatively high (about $370 \\mathrm{~kJ} / \\mathrm{mol}$ ) and cracking is endothermic, the large positive entropy change resulting from the fragmentation of one large molecule into several smaller pieces, together with the high temperature, makes the $T \\Delta S^{\\circ}$ term larger than the $\\Delta H^{\\circ}$ term, thereby favoring the cracking reaction."}
{"id": 400, "contents": "CHAPTER CONTENTS - 7.2 Calculating the Degree of Unsaturation\nBecause of its double bond, an alkene has fewer hydrogens than an alkane with the same number of carbons $-\\mathrm{C}_{n} \\mathrm{H}_{2 n}$ for an alkene versus $\\mathrm{C}_{n} \\mathrm{H}_{2 n+2}$ for an alkane-and is therefore referred to as unsaturated. Ethylene, for example, has the formula $\\mathrm{C}_{2} \\mathrm{H}_{4}$, whereas ethane has the formula $\\mathrm{C}_{2} \\mathrm{H}_{6}$.\n\n\nEthylene: $\\mathrm{C}_{2} \\mathrm{H}_{4}$ (Fewer hydrogens-Unsaturated)\n\n\nEthane: $\\mathrm{C}_{2} \\mathrm{H}_{6}$\n(More hydrogens-Saturated)\n\nIn general, each ring or double bond in a molecule corresponds to a loss of two hydrogens from the alkane formula $\\mathrm{C}_{n} \\mathrm{H}_{2 n+2}$. Knowing this relationship, it's possible to work backward from a molecular formula to calculate a molecule's degree of unsaturation-the number of rings and/or multiple bonds present in the molecule.\n\nLet's assume that we want to find the structure of an unknown hydrocarbon. A molecular weight determination yields a value of 82 amu , which corresponds to a molecular formula of $\\mathrm{C}_{6} \\mathrm{H}_{10}$. Since the saturated $\\mathrm{C}_{6}$ alkane (hexane) has the formula $\\mathrm{C}_{6} \\mathrm{H}_{14}$, the unknown compound has two fewer pairs of hydrogens $\\left(\\mathrm{H}_{14}-\\mathrm{H}_{10}=\\mathrm{H}_{4}=\\right.$ $2 \\mathrm{H}_{2}$ ) so its degree of unsaturation is 2 . The unknown therefore contains either two double bonds, one ring and one double bond, two rings, or one triple bond. There's still a long way to go to establish its structure, but the simple calculation has told us a lot about the molecule.\n\n\n4-Methyl-1,3-pentadiene (two double bonds)\n\n\nCyclohexene (one ring, one double bond)\n\n\nBicyclo[3.1.0]hexane (two rings)"}
{"id": 401, "contents": "CHAPTER CONTENTS - 7.2 Calculating the Degree of Unsaturation\n4-Methyl-1,3-pentadiene (two double bonds)\n\n\nCyclohexene (one ring, one double bond)\n\n\nBicyclo[3.1.0]hexane (two rings)\n\n\n4-Methyl-2-pentyne (one triple bond)\n\n$$\n\\mathrm{C}_{6} \\mathrm{H}_{10}\n$$\n\nSimilar calculations can be carried out for compounds containing elements other than just carbon and hydrogen.\n\n- Organohalogen compounds (C, H, X, where $\\mathbf{X}=\\mathbf{F}, \\mathbf{C l}, \\mathbf{B r}$, or I) A halogen substituent acts as a replacement for hydrogen in an organic molecule, so we can add the number of halogens and hydrogens to arrive at an equivalent hydrocarbon formula from which the degree of unsaturation can be found. For example, the formula $\\mathrm{C}_{4} \\mathrm{H}_{6} \\mathrm{Br}_{2}$ is equivalent to the hydrocarbon formula $\\mathrm{C}_{4} \\mathrm{H}_{8}$ and thus corresponds to one degree of unsaturation.\n\n- Organooxygen compounds ( $\\mathbf{C}, \\mathbf{H}, \\mathbf{0}$ ) Oxygen forms two bonds, so it doesn't affect the formula of an equivalent hydrocarbon and can be ignored when calculating the degree of unsaturation. You can convince yourself of this by seeing what happens when an oxygen atom is inserted into an alkane bond: $\\mathrm{C}-\\mathrm{C}$ becomes $\\mathrm{C}-\\mathrm{O}-\\mathrm{C}$ or $\\mathrm{C}-\\mathrm{H}$ becomes $\\mathrm{C}-\\mathrm{O}-\\mathrm{H}$, and there is no change in the number of hydrogen atoms. For example, the formula $\\mathrm{C}_{5} \\mathrm{H}_{8} \\mathrm{O}$ is equivalent to the hydrocarbon formula $\\mathrm{C}_{5} \\mathrm{H}_{8}$ and thus corresponds to two degrees of unsaturation."}
{"id": 402, "contents": "CHAPTER CONTENTS - 7.2 Calculating the Degree of Unsaturation\n$$\n\\begin{aligned}\n& \\text { O removed from here } \\\\\n& \\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}=\\mathrm{CHCH}_{2} \\mathrm{OH}=\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}=\\mathrm{CHCH}_{2}-\\mathrm{H} \\\\\n& \\mathrm{C}_{5} \\mathrm{H}_{8} \\mathrm{O}=\\text { \" } \\mathrm{C}_{5} \\mathrm{H}_{8} \" \\quad \\begin{array}{l}\n\\text { Two unsaturations: } \\\\\n\\text { two double bonds }\n\\end{array}\n\\end{aligned}\n$$\n\n- Organonitrogen compounds ( $\\mathbf{C}, \\mathbf{H}, \\mathbf{N}$ ) Nitrogen forms three bonds, so an organonitrogen compound has one more hydrogen than a related hydrocarbon. We therefore subtract the number of nitrogens from the number of hydrogens to arrive at the equivalent hydrocarbon formula. Again, you can convince yourself of this by seeing what happens when a nitrogen atom is inserted into an alkane bond: $\\mathrm{C}-\\mathrm{C}$ becomes $\\mathrm{C}-\\mathrm{NH}-\\mathrm{C}$ or $\\mathrm{C}-\\mathrm{H}$ becomes $\\mathrm{C}-\\mathrm{NH}_{2}$, meaning that one additional hydrogen atom has been added. We must therefore subtract this extra hydrogen atom to arrive at the equivalent hydrocarbon formula. For example, the formula $\\mathrm{C}_{5} \\mathrm{H}_{9} \\mathrm{~N}$ is equivalent to $\\mathrm{C}_{5} \\mathrm{H}_{8}$ and thus has two degrees of unsaturation.\n\n\nTo summarize:\n\n- Add the number of halogens to the number of hydrogens.\n- Ignore the number of oxygens.\n- Subtract the number of nitrogens from the number of hydrogens."}
{"id": 403, "contents": "CHAPTER CONTENTS - 7.2 Calculating the Degree of Unsaturation\nTo summarize:\n\n- Add the number of halogens to the number of hydrogens.\n- Ignore the number of oxygens.\n- Subtract the number of nitrogens from the number of hydrogens.\n\nPROBLEM Calculate the degree of unsaturation in each of the following formulas, and then draw as many\n7-1 structures as you can for each:\n(a) $\\mathrm{C}_{4} \\mathrm{H}_{8}$\n(b) $\\mathrm{C}_{4} \\mathrm{H}_{6}$\n(c) $\\mathrm{C}_{3} \\mathrm{H}_{4}$\n\nPROBLEM Calculate the degree of unsaturation in each of the following formulas:\n7-2 (a) $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{~N}$\n(b) $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NO}_{2}$\n(c) $\\mathrm{C}_{8} \\mathrm{H}_{9} \\mathrm{Cl}_{3}$\n(d) $\\mathrm{C}_{9} \\mathrm{H}_{16} \\mathrm{Br}_{2}$\n(e) $\\mathrm{C}_{10} \\mathrm{H}_{12} \\mathrm{~N}_{2} \\mathrm{O}_{3}$\n(f) $\\mathrm{C}_{20} \\mathrm{H}_{32} \\mathrm{ClN}$\n\nPROBLEM Diazepam, marketed as an antianxiety medication under the name Valium, has three rings, eight\n7-3 double bonds, and the formula $\\mathrm{C}_{16} \\mathrm{H}_{?} \\mathrm{ClN}_{2} \\mathrm{O}$. How many hydrogens does diazepam have? (Calculate the answer; don't count hydrogens in the structure.)\n\n\nDiazepam"}
{"id": 404, "contents": "CHAPTER CONTENTS - 7.3 Naming Alkenes\nAlkenes are named using a series of rules similar to those for alkanes (Section 3.4), with the suffix -ene used instead of -ane to identify the functional group. There are three steps to this process."}
{"id": 405, "contents": "STEP 1 - \nName the parent hydrocarbon. Find the longest carbon chain containing the double bond, and name the compound accordingly, using the suffix -ene:\n\n\nNamed as a pentene\n\nas a hexene, since the double bond is not contained in the six-carbon chain\n\nSTEP 2\nNumber the carbon atoms in the chain. Begin at the end nearer the double bond or, if the double bond is equidistant from the two ends, begin at the end nearer the first branch point. This rule ensures that the doublebond carbons receive the lowest possible numbers.\n\n$$\n\\underset{6}{\\mathrm{CH}_{3}} \\mathrm{CH}_{5} \\mathrm{C}_{4} \\mathrm{CH}_{2} \\underset{3}{\\mathrm{C}} \\mathrm{H}=\\underset{2}{\\mathrm{C}} \\mathrm{HC}_{1} \\mathrm{H}_{3}\n$$\n\n\n\nSTEP 3\nWrite the full name. Number the substituents according to their positions in the chain, and list them alphabetically. Indicate the position of the double bond by giving the number of the first alkene carbon and placing that number directly before the parent name. If more than one double bond is present, indicate the position of each and use one of the suffixes -diene, -triene, and so on.\n\n\n2-Hexene\n\n\n2-Ethyl-1-pentene\n\n\n2-Methyl-3-hexene\n\n\n2-Methyl-1,3-butadiene\n\nWe should also note that IUPAC changed their naming recommendations in 1993 to place the locant indicating the position of the double bond immediately before the -ene suffix rather than before the parent name: but-2-ene rather than 2 -butene, for instance. This change has not been widely accepted by the chemical community in the United States, however, so we'll stay with the older but more commonly used names. Be aware, though, that you may occasionally encounter the newer system."}
{"id": 406, "contents": "STEP 1 - \n|  | <smiles>CCC(C)C=CC(C)C(C)C</smiles> | <smiles>C=CC(C=CC)CCC</smiles> |\n| :---: | :---: | :---: |\n| Newer naming system: | 2,5-Dimethylhept-3-ene | 3-Propylhexa-1,4-diene |\n| (Older naming system: | 2,5-Dimethyl-3-heptene | 3-Propyl-1,4-hexadiene) |\n\nCycloalkenes are named similarly, but because there is no chain end to begin from, we number the cycloalkene so that the double bond is between C 1 and C 2 and the first substituent has as low a number as possible. It's not necessary to indicate the position of the double bond in the name because it's always between C1 and C2. As with open-chain alkenes, the newer but not yet widely accepted naming rules place the locant immediately before the suffix in a cyclic alkene.\n\n\n1-Methylcyclohexene\n\n\n1,4-Cyclohexadiene\n\n\n1,5-Dimethylcyclopentene\n(New: Cyclohexa-1,4-diene)\nFor historical reasons, there are a few alkenes whose names are firmly entrenched in common usage but don't conform to the rules. For example, the alkene derived from ethane should be called ethene, but the name ethylene has been used for so long that it is accepted by IUPAC. TABLE 7.1 lists several other common names that are often used and are recognized by IUPAC. Note also that a $=\\mathrm{CH}_{2}$ substituent is called a methylene group, a $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}$ - substituent is called a vinyl group, and a $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}_{2}$ - substituent is called an allyl group."}
{"id": 407, "contents": "STEP 1 - \n$$\n\\mathrm{H}_{2} \\mathrm{C} \\geqslant \\quad \\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}<\\quad \\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}-\\mathrm{CH}_{2} \\geqslant\n$$\n\nA methylene group A vinyl group An allyl group\n\n| Compound | Systematic name | Common name |\n| :---: | :---: | :---: |\n| $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}_{2}$ | Ethene | Ethylene |\n| $\\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{CH}_{2}$ | Propene | Propylene |\n| <smiles>C=C(C)C</smiles> | 2-Methylpropene | Isobutylene |\n| <smiles>C=CC(=C)C</smiles> | 2-Methyl-1,3-butadiene | Isoprene |\n\nPROBLEM Give IUPAC names for the following compounds:\n\n7-4 (a)\n\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM Draw structures corresponding to the following IUPAC names:\n7-5 (a) 2-Methyl-1,5-hexadiene (b) 3-Ethyl-2,2-dimethyl-3-heptene\n(c) 2,3,3-Trimethyl-1,4,6-octatriene (d) 3,4-Diisopropyl-2,5-dimethyl-3-hexene\n\nPROBLEM Name the following cycloalkenes:\n7-6 (a)\n\n(b)\n\n(c)\n\n\nPROBLEM Change the following old names to new, post-1993 names, and draw the structure of each 7-7 compound:\n(a) 2,5,5-Trimethyl-2-hexene\n(b) 2,3-Dimethyl-1,3-cyclohexadiene"}
{"id": 408, "contents": "STEP 1 - 7.4 Cis-Trans Isomerism in Alkenes\nWe saw in the chapter on Structure and Bonding that the carbon-carbon double bond can be described in two ways. In valence bond language (Section 1.8), the carbons are $s p^{2}$-hybridized and have three equivalent hybrid orbitals that lie in a plane at angles of $120^{\\circ}$ to one another. The carbons form a $\\sigma$ bond by a head-on overlap of $s p^{2}$ orbitals and form a $\\pi$ bond by sideways overlap of unhybridized $p$ orbitals oriented perpendicular to the $s p^{2}$ plane, as shown in FIGURE 1.15.\n\nIn molecular orbital language (Section 1.11), interaction between the p orbitals leads to one bonding and one antibonding $\\pi$ molecular orbital. The $\\pi$ bonding MO has no node between nuclei and results from a combination of $p$ orbital lobes with the same algebraic sign. The $\\pi$ antibonding MO has a node between nuclei and results from a combination of lobes with different algebraic signs, as shown in FIGURE 1.19.\n\nAlthough essentially free rotation around single bonds is possible (Section 3.6), the same is not true of double bonds. For rotation to occur around a double bond, the $\\pi$ bond must break and re-form (FIGURE 7.3). Thus, the barrier to double-bond rotation must be at least as great as the strength of the $\\pi$ bond itself, an estimated 350 $\\mathrm{kJ} / \\mathrm{mol}(84 \\mathrm{kcal} / \\mathrm{mol})$. Recall that the barrier to bond rotation in ethane is only $12 \\mathrm{~kJ} / \\mathrm{mol}$."}
{"id": 409, "contents": "STEP 1 - 7.4 Cis-Trans Isomerism in Alkenes\nFIGURE 7.3 The $\\boldsymbol{\\pi}$ bond must break for rotation to take place around a carbon-carbon double bond.\nThe lack of rotation around carbon-carbon double bonds is of more than just theoretical interest; it also has chemical consequences. Imagine the situation for a disubstituted alkene such as 2-butene. (Disubstituted means that two substituents other than hydrogen are bonded to the double-bond carbons.) The two methyl groups in 2-butene can either be on the same side of the double bond or on opposite sides, a situation similar to that in disubstituted cycloalkanes (Section 4.2).\n\nSince bond rotation can't occur, the two 2-butenes can't spontaneously interconvert; they are different, isolable compounds. As with disubstituted cycloalkanes, we call such compounds cis-trans stereoisomers. The compound with substituents on the same side of the double bond is called cis-2-butene, and the isomer with substituents on opposite sides is trans-2-butene (FIGURE 7.4).\n\ncis-2-Butene\n\ntrans-2-Butene\n\nFIGURE 7.4 Cis and trans isomers of 2-butene. The cis isomer has the two methyl groups on the same side of the double bond, and the trans isomer has methyl groups on opposite sides.\n\nCis-trans isomerism is not limited to disubstituted alkenes. It can occur whenever both double-bond carbons are attached to two different groups. If one of the double-bond carbons is attached to two identical groups, however, cis-trans isomerism is not possible (FIGURE 7.5).\n\n\n\nThese two compounds are not identical; they are cis-trans isomers.\n\nFIGURE 7.5 The requirement for cis-trans isomerism in alkenes. Compounds that have one of their carbons bonded to two identical groups can't exist as cis-trans isomers. Only when both carbons are bonded to two different groups is cis-trans isomerism possible."}
{"id": 410, "contents": "STEP 1 - 7.4 Cis-Trans Isomerism in Alkenes\nPROBLEM The sex attractant of the common housefly is an alkene named cis-9-tricosene. Draw its structure.\n7-8 (Tricosane is the straight-chain alkane $\\mathrm{C}_{23} \\mathrm{H}_{48}$.)\nPROBLEM Which of the following compounds can exist as pairs of cis-trans isomers? Draw each pair, and\n7-9 indicate the geometry of each isomer.\n(a) $\\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{CH}_{2}$\n(b) $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{C}=\\mathrm{CHCH}_{3}$\n(c) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}=\\mathrm{CHCH}_{3}$\n(d) $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{C}=\\mathrm{C}\\left(\\mathrm{CH}_{3}\\right) \\mathrm{CH}_{2} \\mathrm{CH}_{3}$\n(e) $\\mathrm{ClCH}=\\mathrm{CHCl}$ (f) $\\mathrm{BrCH}=\\mathrm{CHCl}$\n\nPROBLEM Name the following alkenes, including a cis or trans designation:"}
{"id": 411, "contents": "7-10 (a) - \n(b)"}
{"id": 412, "contents": "7-10 (a) - 7.5 Alkene Stereochemistry and the E,Z Designation\nThe cis-trans naming system used in the previous section works only with disubstituted alkenes-compounds that have two substituents other than hydrogen on the double bond. With trisubstituted and tetrasubstituted double bonds, a more general method is needed for describing double-bond geometry. (Trisubstituted means three substituents other than hydrogen on the double bond; tetrasubstituted means four substituents other than hydrogen.)\n\nThe method used for describing alkene stereochemistry is called the $\\boldsymbol{E}, \\boldsymbol{Z}$ system and employs the same Cahn-Ingold-Prelog sequence rules given in Section 5.5 for specifying the configuration of a chirality center. Let's briefly review the sequence rules and then see how they're used to specify double-bond geometry. For a more thorough review, reread Section 5.5."}
{"id": 413, "contents": "RULE 1 - \nConsidering each of the double-bond carbons separately, look at the two substituents attached and rank them according to the atomic number of the first atom in each ( 8 for oxygen, 6 for carbon, 1 for hydrogen, and so forth). An atom with higher atomic number ranks higher than an atom with lower atomic number.\n\nRULE 2\nIf a decision can't be reached by ranking the first atoms in the two substituents, look at the second, third, or fourth atoms away from the double-bond until the first difference is found."}
{"id": 414, "contents": "RULE 3 - \nMultiple-bonded atoms are equivalent to the same number of single-bonded atoms.\nOnce the two groups attached to each double-bonded carbon have been ranked as either higher or lower, look at the entire molecule. If the higher-ranked groups on each carbon are on the same side of the double bond, the alkene is said to have a $\\boldsymbol{Z}$ configuration, for the German zusammen, meaning \"together.\" If the higher-ranked groups are on opposite sides, the alkene has an $\\boldsymbol{E}$ configuration, for the German entgegen, meaning \"opposite.\" (For a simple way to remember which is which, note that the groups are on \"ze zame zide\" in the $Z$ isomer.)\n\n$E$ double bond\n(Higher-ranked groups\nare on opposite sides.)\n\n\nZ double bond\n(Higher-ranked groups are on the same side.)\n\nAs an example, look at the following two isomers of 2-chloro-2-butene. Because chlorine has a higher atomic number than carbon, a -Cl substituent is ranked higher than a $-\\mathrm{CH}_{3}$ group. Methyl is ranked higher than hydrogen, however, so isomer (a) is designated $E$ because the higher-ranked groups are on opposite sides of the double bond. Isomer (b) has a $Z$ configuration because its higher-ranked groups are on ze zame zide of the double bond.\n\n(a) (E)-2-Chloro-2-butene\n(b) (Z)-2-Chloro-2-butene\n\nFor further practice, work through each of the following examples to convince yourself that the assignments are correct:\n\n(E)-3-Methyl-1,3-pentadiene\n\n(E)-1-Bromo-2-isopropyl-1,3-butadiene\n\n(Z)-2-Hydroxymethyl-2-butenoic acid"}
{"id": 415, "contents": "Assigning E and Z Configurations to Alkenes - \nAssign $E$ or $Z$ configuration to the double bond in the following compound:"}
{"id": 416, "contents": "Strategy - \nLook at the two substituents connected to each double-bonded carbon, and determine their ranking using the Cahn-Ingold-Prelog rules. Then, check whether the two higher-ranked groups are on the same or opposite sides of the double bond."}
{"id": 417, "contents": "Solution - \nThe left-hand carbon has -H and $-\\mathrm{CH}_{3}$ substituents, of which $-\\mathrm{CH}_{3}$ ranks higher by sequence rule 1 . The righthand carbon has $-\\mathrm{CH}\\left(\\mathrm{CH}_{3}\\right)_{2}$ and $-\\mathrm{CH}_{2} \\mathrm{OH}$ substituents, which are equivalent by rule 1 . By rule 2 , however, $-\\mathrm{CH}_{2} \\mathrm{OH}$ ranks higher than $-\\mathrm{CH}\\left(\\mathrm{CH}_{3}\\right)_{2}$ because the substituent $-\\mathrm{CH}_{2} \\mathrm{OH}$ has an oxygen as its highest second atom, but $-\\mathrm{CH}\\left(\\mathrm{CH}_{3}\\right)_{2}$ has a carbon as its highest second atom. The two higher-ranked groups are on the same side of the double bond, so we assign a $Z$ configuration.\n\n\nZ configuration\n\nPROBLEM Which member in each of the following sets ranks higher?\n7-11 (a) -H or $-\\mathrm{CH}_{3}$\n(b) -Cl or $-\\mathrm{CH}_{2} \\mathrm{Cl}$\n(c) $-\\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{Br}$ or $-\\mathrm{CH}=\\mathrm{CH}_{2}$\n(d) $-\\mathrm{NHCH}_{3}$ or $-\\mathrm{OCH}_{3}$\n(e) $-\\mathrm{CH}_{2} \\mathrm{OH}$ or $-\\mathrm{CH}=\\mathrm{O}$ (f) $-\\mathrm{CH}_{2} \\mathrm{OCH}_{3}$ or $-\\mathrm{CH}=\\mathrm{O}$"}
{"id": 418, "contents": "Solution - \nPROBLEM Rank the substituents in each of the following sets according to the sequence rules:\n7-12 (a) $-\\mathrm{CH}_{3},-\\mathrm{OH},-\\mathrm{H},-\\mathrm{Cl}$ (b) $-\\mathrm{CH}_{3},-\\mathrm{CH}_{2} \\mathrm{CH}_{3},-\\mathrm{CH}=\\mathrm{CH}_{2},-\\mathrm{CH}_{2} \\mathrm{OH}$\n(c) $-\\mathrm{CO}_{2} \\mathrm{H},-\\mathrm{CH}_{2} \\mathrm{OH},-\\mathrm{C} \\equiv \\mathrm{N},-\\mathrm{CH}_{2} \\mathrm{NH}_{2}$ (d) $-\\mathrm{CH}_{2} \\mathrm{CH}_{3},-\\mathrm{C} \\equiv \\mathrm{CH},-\\mathrm{C} \\equiv \\mathrm{N},-\\mathrm{CH}_{2} \\mathrm{OCH}_{3}$\n\nPROBLEM Assign $E$ or $Z$ configuration to the following alkenes:\n7-13 (a)\n\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM Assign stereochemistry ( $E$ or $Z$ ) to the double bond in the following compound, and convert the 7-14 drawing into a skeletal structure (red = O):"}
{"id": 419, "contents": "Solution - 7.6 Stability of Alkenes\nAlthough the cis-trans interconversion of alkene isomers does not occur spontaneously, it can often be made to happen by treating the alkene with a strong acid catalyst. If we interconvert cis-2-butene with trans-2-butene and allow them to reach equilibrium, we find that they aren't of equal stability. The trans isomer is more stable than the cis isomer by $2.8 \\mathrm{~kJ} / \\mathrm{mol}(0.66 \\mathrm{kcal} / \\mathrm{mol})$ at room temperature, corresponding to a $76: 24$ ratio.\n\n\nCis alkenes are less stable than their trans isomers because of steric strain between the two larger substituents on the same side of the double bond. This is the same kind of steric interference that we saw previously in the axial conformation of methylcyclohexane (Section 4.7).\n\ncis-2-Butene\n\ntrans-2-Butene\n\nAlthough it's sometimes possible to find relative stabilities of alkene isomers by establishing a cis-trans equilibrium through treatment with strong acid, a more general method is to take advantage of the fact that alkenes undergo a hydrogenation reaction to give the corresponding alkane when treated with $\\mathrm{H}_{2}$ gas in the presence of a catalyst such as palladium or platinum.\n\n\nEnergy diagrams for the hydrogenation reactions of cis- and trans-2-butene are shown in FIGURE 7.6. Because cis-2-butene is less stable than trans-2-butene by $2.8 \\mathrm{~kJ} / \\mathrm{mol}$, the energy diagram shows the cis alkene at a higher energy level. After reaction, however, both curves are at the same energy level (butane). It therefore follows that $\\Delta G^{\\circ}$ for reaction of the cis isomer must be larger than $\\Delta G^{\\circ}$ for reaction of the trans isomer by 2.8 $\\mathrm{kJ} / \\mathrm{mol}$. In other words, more energy is released in the hydrogenation of the cis isomer than the trans isomer because the cis isomer has more energy to begin with."}
{"id": 420, "contents": "Solution - 7.6 Stability of Alkenes\nFIGURE 7.6 Energy diagrams for hydrogenation of cis- and trans-2-butene. The cis isomer is higher in energy than the trans isomer by about $2.8 \\mathrm{~kJ} / \\mathrm{mol}$ and therefore releases more energy when hydrogenated.\n\nIf we were to measure the so-called heats of hydrogenation ( $\\Delta H^{\\circ}{ }_{\\text {hydrog }}$ ) for two double-bond isomers and find their difference, we could determine the relative stabilities of cis and trans isomers without having to measure an equilibrium position. cis-2-Butene, for instance, has $\\Delta H^{\\circ}{ }_{\\text {hydrog }}=-119 \\mathrm{~kJ} / \\mathrm{mol}(-28.3 \\mathrm{kcal} / \\mathrm{mol})$, while trans-2-butene has $\\Delta H^{\\circ}$ hydrog $=-115 \\mathrm{~kJ} / \\mathrm{mol}(-27.4 \\mathrm{kcal} / \\mathrm{mol})-$ a difference of $4 \\mathrm{~kJ} / \\mathrm{mol}$.\n\n\nCis isomer\n$\\Delta H^{\\circ}{ }_{\\text {hydrog }}=-120 \\mathrm{~kJ} / \\mathrm{mol}$\n\n\nTrans isomer\n$\\Delta H^{\\circ}$ hydrog $=-116 \\mathrm{~kJ} / \\mathrm{mol}$\n\nThe $4 \\mathrm{~kJ} / \\mathrm{mol}$ energy difference between the 2-butene isomers calculated from heats of hydrogenation agrees reasonably well with the $2.8 \\mathrm{~kJ} / \\mathrm{mol}$ energy difference calculated from equilibrium data, but the values aren't exactly the same for two reasons. First, there is probably some experimental error, because heats of hydrogenation are difficult to measure accurately. Second, heats of reaction and equilibrium constants don't measure exactly the same thing. Heats of reaction measure enthalpy changes, $\\Delta H^{\\circ}$, whereas equilibrium constants measure free-energy changes, $\\Delta G^{\\circ}$, so we might expect a slight difference between the two.\n\nTABLE 7.2 Heats of Hydrogenation of Some Alkenes"}
{"id": 421, "contents": "Solution - 7.6 Stability of Alkenes\nTABLE 7.2 Heats of Hydrogenation of Some Alkenes\n\n| Substitution | Alkene | $\\Delta H^{\\circ}$ hydrog |  |\n| :--- | :--- | :--- | :--- | :--- |\n|  |  | $(\\mathrm{kJ} / \\mathrm{mol})$ | $(\\mathrm{kcal} / \\mathrm{mol})$ |\n| Ethylene | $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}_{2}$ | -136 | -32.6 |\n| Monosubstituted | $\\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{CH}_{2}$ | -125 | -29.9 |\n| Disubstituted | $\\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{CHCH}_{3}$ (cis) <br> $\\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{CHCH}_{3}$ (trans) <br> $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{C}=\\mathrm{CH}_{2}$ | -119 <br> -115 <br> -118 | -28.3 <br> -28.4 |\n| Trisubstituted | $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{C}=\\mathrm{CHCH}_{3}$ | -112 | -26.7 |\n| Tetrasubstituted | $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{C}=\\mathrm{C}\\left(\\mathrm{CH}_{3}\\right)_{2}$ | -110 | -26.4 |\n\nTABLE 7.2 lists some representative data for the hydrogenation of different alkenes and shows that alkenes become more stable with increasing substitution. That is, alkenes follow the stability order:"}
{"id": 422, "contents": "Solution - 7.6 Stability of Alkenes\nTABLE 7.2 lists some representative data for the hydrogenation of different alkenes and shows that alkenes become more stable with increasing substitution. That is, alkenes follow the stability order:\n\n\nThe stability order of substituted alkenes is due to a combination of two factors. One is a stabilizing interaction between the $\\mathrm{C}=\\mathrm{C} \\pi$ orbital and adjacent $\\mathrm{C}-\\mathrm{H} \\sigma$ bonds on substituents. In valence-bond language, the interaction is called hyperconjugation. In a molecular orbital description, there is a bonding MO that extends over the four-atom $\\mathrm{C}=\\mathrm{C}-\\mathrm{C}-\\mathrm{H}$ grouping, as shown in FIGURE 7.7. The more substituents present on the double bond, the more hyperconjugation occurs and the more stable the alkene.\n\n\nFIGURE 7.7 Hyperconjugation is a stabilizing interaction between the $\\mathbf{C}=\\mathbf{C} \\boldsymbol{\\pi}$ orbital and adjacent $\\mathbf{C}-\\mathbf{H} \\boldsymbol{\\sigma}$ bonds on substituents. The more substituents there are, the greater the stabilization of the alkene.\nA second factor that contributes to alkene stability involves bond strengths. A bond between a $s p^{2}$ carbon and a $s p^{3}$ carbon is somewhat stronger than a bond between two $s p^{3}$ carbons. Thus, in comparing 1-butene and 2-butene, the monosubstituted isomer has one $s p^{3}-s p^{3}$ bond and one $s p^{3}-s p^{2}$ bond, while the disubstituted isomer has two $s p^{3}-s p^{2}$ bonds. More highly substituted alkenes always have a higher ratio of $s p^{3}-s p^{2}$ bonds to $s p^{3}-s p^{3}$ bonds than less highly substituted alkenes and are therefore more stable.\n\n\n2-Butene\n(more stable)\n\n\n1-Butene\n(less stable)\n\nPROBLEM Name the following alkenes, and tell which compound in each pair is more stable:\n7-15 (a)\n\nor"}
{"id": 423, "contents": "Solution - 7.6 Stability of Alkenes\n2-Butene\n(more stable)\n\n\n1-Butene\n(less stable)\n\nPROBLEM Name the following alkenes, and tell which compound in each pair is more stable:\n7-15 (a)\n\nor\n\n(b)\n\nor\n\n(c)\n\nor"}
{"id": 424, "contents": "Solution - 7.7 Electrophilic Addition Reactions of Alkenes\nBefore beginning a detailed discussion of alkene reactions, let's review briefly some conclusions from the previous chapter. We said in Section 6.5 that alkenes behave as nucleophiles (Lewis bases) in polar reactions, donating a pair of electrons from their electron-rich $\\mathrm{C}=\\mathrm{C}$ bond to an electrophile (Lewis acid). For example, reaction of 2-methylpropene with HBr yields 2-bromo-2-methylpropane. A careful study of this and similar reactions by the British chemist Christopher Ingold and others in the 1930s led to the generally accepted mechanism shown in FIGURE 7.8 for an electrophilic addition reaction."}
{"id": 425, "contents": "FIGURE 7.8 MECHANISM - \nMechanism of the electrophilic addition of HBr to 2-methylpropene. The reaction occurs in two steps, protonation and bromide addition, and involves a carbocation intermediate.\n(1) A hydrogen atom on the electrophile HBr is attacked by $\\pi$ electrons from the nucleophilic double bond, forming a new $\\mathrm{C}-\\mathrm{H}$ bond. This leaves the other carbon atom with a + charge and a vacant $p$ orbital. Simultaneously, two electrons from the $\\mathrm{H}-\\mathrm{Br}$ bond move onto bromine, giving bromide anion.\n(2) The bromide ion donates an electron pair to the positively charged carbon atom, forming a $\\mathrm{C}-\\mathrm{Br}$ bond and yielding the neutral addition product.\n\n\nCarbocation intermediate\n\n\n2-Bromo-2-methylpropane\n\nThe reaction begins with an attack on the hydrogen of the electrophile HBr by the electrons of the nucleophilic $\\pi$ bond. Two electrons from the $\\pi$ bond form a new $\\sigma$ bond between the entering hydrogen and an alkene carbon, as shown by the curved arrow at the top of FIGURE 7.8. The resulting carbocation intermediate is itself an electrophile, which can accept an electron pair from nucleophilic $\\mathrm{Br}^{-}$ion to form a $\\mathrm{C}-\\mathrm{Br}$ bond and yield a neutral addition product.\n\nAn energy diagram for the overall electrophilic addition reaction (FIGURE 7.9) has two peaks (transition states) separated by a valley (carbocation intermediate). The energy level of the intermediate is higher than that of the starting alkene, but the reaction as a whole is exergonic (negative $\\Delta G^{\\circ}$ ). The first step, protonation of the alkene to yield the intermediate cation, is relatively slow. But once the cation intermediate is formed, it rapidly reacts to yield the final alkyl bromide product. The relative rates of the two steps are indicated in FIGURE 7.9 by the fact that $\\Delta G_{1}{ }^{\\ddagger}$ is larger than $\\Delta G_{2}{ }^{\\dagger}$."}
{"id": 426, "contents": "FIGURE 7.8 MECHANISM - \nFIGURE 7.9 Energy diagram for the two-step electrophilic addition of $\\mathbf{H B r}$ to 2-methylpropene. The first step is slower than the second step.\n\nElectrophilic addition to alkenes is successful not only with HBr but with $\\mathrm{HCl}, \\mathrm{HI}$, and $\\mathrm{H}_{2} \\mathrm{O}$ as well. Note that HI is usually generated in the reaction mixture by treating potassium iodide with phosphoric acid and that a strong acid catalyst is needed for the addition of water.\n\n\n2-Methylpropene\n2-Chloro-2-methylpropane\n(94\\%)\n\n\n1-Pentene\n(HI)\n2-Iodopentane\n\n\n1-Methylcyclohexene\n1-Methylcyclohexanol"}
{"id": 427, "contents": "WRITING ORGANIC REACTIONS - \nThis is a good place to mention that the equations for organic reactions are sometimes written in different ways to emphasize different points. In describing a laboratory process, for instance, the reaction of 2-methylpropene with HCl might be written in the format $\\mathrm{A}+\\mathrm{B} \\longrightarrow \\mathrm{C}$ to emphasize that both reactants are equally important for the purposes of the discussion. The solvent and notes about other reaction conditions such as temperature are written either above or below the reaction arrow.\n\n\nAlternatively, we might write the same reaction in a format to emphasize that 2 -methylpropene is the reactant whose chemistry is of greater interest. The second reactant, HCl , is placed above the reaction arrow together with notes about solvent and reaction conditions.\n\n\nIn describing a biological process, the reaction is almost always written to show only the structures of the primary reactant and product, while abbreviating the structures of various biological \"reagents\" and byproducts with a curved arrow that intersects the straight reaction arrow. As discussed in Section 6.11, the reaction of glucose with ATP to give glucose 6-phosphate plus ADP would then be written as"}
{"id": 428, "contents": "WRITING ORGANIC REACTIONS - 7.8 Orientation of Electrophilic Additions: Markovnikov's Rule\nLook carefully at the electrophilic addition reactions shown in the previous section. In each case, an unsymmetrically substituted alkene has given a single addition product rather than the mixture that might be expected. For example, 2-methylpropene might react with HCl to give both 2-chloro-2-methylpropane and 1-chloro-2-methylpropane, but it doesn't. It gives only 2-chloro-2-methylpropane as the sole product. Similarly, it's invariably the case in biological alkene addition reactions that only a single product is formed. We say that such reactions are regiospecific (ree-jee-oh-specific) when only one of two possible orientations of an addition occurs.\n\n\nAfter looking at the results of many such reactions, the Russian chemist Vladimir Markovnikov proposed in 1869 what has become known as:\n\nMarkovnikov's rule\nIn the addition of HX to an alkene, the H attaches to the carbon with fewer alkyl substituents and the X\nattaches to the carbon with more alkyl substituents.\n\n\n\n1-Methylcyclohexene\n1-Bromo-1-methylcyclohexane\nWhen both double-bonded carbon atoms have the same degree of substitution, a mixture of addition products results.\n\n\nBecause carbocations are involved as intermediates in these electrophilic addition reactions, Markovnikov's rule can be restated in the following way:\n\nMarkovnikov's rule restated\nIn the addition of HX to an alkene, the more highly substituted carbocation is formed as the intermediate rather than the less highly substituted one.\n\nFor example, addition of $\\mathrm{H}^{+}$to 2-methylpropene yields the intermediate tertiary carbocation rather than the alternative primary carbocation, and addition to 1-methylcyclohexene yields a tertiary cation rather than a secondary one. Why should this be?"}
{"id": 429, "contents": "Predicting the Product of an Electrophilic Addition Reaction - \nWhat product would you expect from reaction of HCl with 1-ethylcyclopentene?"}
{"id": 430, "contents": "Strategy - \nWhen solving a problem that asks you to predict a reaction product, begin by looking at the functional group(s) in the reactants and deciding what kind of reaction is likely to occur. In the present instance, the reactant is an alkene that will probably undergo an electrophilic addition reaction with HCl . Next, recall what you know about electrophilic addition reactions to predict the product. You know that electrophilic addition reactions follow Markovnikov's rule, so $\\mathrm{H}^{+}$will add to the double-bond carbon that has one alkyl group ( C 2 on the ring) and the Cl will add to the double-bond carbon that has two alkyl groups ( C 1 on the ring)."}
{"id": 431, "contents": "Solution - \nThe expected product is 1-chloro-1-ethylcyclopentane."}
{"id": 432, "contents": "Synthesizing a Specific Compound - \nWhat alkene would you start with to prepare the following alkyl halide? There may be more than one possibility."}
{"id": 433, "contents": "Strategy - \nWhen solving a problem that asks how to prepare a given product, always work backward. Look at the product, identify the functional group(s) it contains, and ask yourself, \"How can I prepare that functional group?\" In the present instance, the product is a tertiary alkyl chloride, which can be prepared by reaction of an alkene with HCl . The carbon atom bearing the -Cl atom in the product must be one of the double-bond carbons in the reactant. Draw and evaluate all possibilities."}
{"id": 434, "contents": "Solution - \nThere are three possibilities, all of which could give the desired product according to Markovnikov's rule.\n\nor\nor\n\n\n\nPROBLEM Predict the products of the following reactions:\n7-16 (a)\n\n(d)\n\n(b)\n\n(c)\n\n(Addition of $\\mathrm{H}_{2} \\mathrm{O}$ occurs.)\n\nPROBLEM What alkenes would you start with to prepare the following products?\n\n7-17 (a)\n\n\n(b)\n\n\n(c)\n\n(d)"}
{"id": 435, "contents": "Solution - 7.9 Carbocation Structure and Stability\nTo understand why Markovnikov's rule works, we need to learn more about the structure and stability of carbocations and about the general nature of reactions and transition states. The first point to explore involves structure.\n\nA great deal of experimental evidence has shown that carbocations are planar. The trivalent carbon is $s p^{2}$-hybridized, and the three substituents are oriented toward the corners of an equilateral triangle, as indicated in FIGURE 7.10. Because there are only six valence electrons on carbon and all six are used in the three $\\sigma$ bonds, the $p$ orbital extending above and below the plane is unoccupied.\n\n\nFIGURE 7.10 The structure of a carbocation. The trivalent carbon is $s p^{2}$-hybridized and has a vacant $p$ orbital perpendicular to the plane of the carbon and three attached groups.\n\nThe second point to explore involves carbocation stability. 2-Methylpropene might react with $\\mathrm{H}^{+}$to form a carbocation having three alkyl substituents (a tertiary ion, $3^{\\circ}$ ), or it might react to form a carbocation having one alkyl substituent (a primary ion, $1^{\\circ}$ ). Since the tertiary alkyl chloride, 2 -chloro-2-methylpropane, is the only product observed, formation of the tertiary cation is evidently favored over formation of the primary cation. Thermodynamic measurements show that, indeed, the stability of carbocations increases with increasing substitution so that the stability order is tertiary > secondary > primary > methyl.\n\n\nMethyl\n\n\nPrimary ( $\\mathbf{1}^{\\circ}$ )\n\n\nSecondary ( $2^{\\circ}$ )\n\n\nTertiary ( $\\mathbf{3}^{\\circ}$ )"}
{"id": 436, "contents": "Stability - \nOne way of determining carbocation stabilities is to measure the amount of energy required to form a carbocation by dissociation of the corresponding alkyl halide, $R-X \\rightarrow R^{+}+: X^{-}$. As shown in FIGURE 7.11, tertiary alkyl halides dissociate to give carbocations more easily than secondary or primary ones. Thus, trisubstituted carbocations are more stable than disubstituted ones, which are more stable than monosubstituted ones. The data in FIGURE 7.11 are taken from measurements made in the gas phase, but a similar stability order is found for carbocations in solution. The dissociation enthalpies are much lower in solution because polar solvents can stabilize the ions, but the order of carbocation stability remains the same.\n\n\nFIGURE 7.11 A plot of dissociation enthalpy versus substitution pattern for the gas-phase dissociation of alkyl chlorides to yield carbocations. More highly substituted alkyl halides dissociate more easily than less highly substituted ones.\n\nWhy are more highly substituted carbocations more stable than less highly substituted ones? There are at least two reasons. Part of the answer has to do with inductive effects, and part has to do with hyperconjugation. Inductive effects, discussed in Section 2.1 in connection with polar covalent bonds, result from the shifting of electrons in a $\\sigma$ bond in response to the electronegativity of nearby atoms. In the present instance, electrons from a relatively larger and more polarizable alkyl group can shift toward a neighboring positive charge more easily than the electron from a hydrogen. Thus, the more alkyl groups attached to the positively charged carbon, the more electron density shifts toward the charge and the more inductive stabilization of the cation occurs (FIGURE 7.12).\n\n\n\nMethyl:\nNo alkyl groups donating electrons\n\n\n\nPrimary:\nOne alkyl group donating electrons\n\n\n\nSecondary:\nTwo alkyl groups donating electrons\n\n\n\nTertiary:\nThree alkyl groups donating electrons\n\nFIGURE 7.12 A comparison of inductive stabilization for methyl, primary, secondary, and tertiary carbocations. The more alkyl groups that are bonded to the positively charged carbon, the more electron density shifts toward the charge, making the charged carbon less electron-poor (blue in electrostatic potential maps)."}
{"id": 437, "contents": "Stability - \nHyperconjugation, discussed in Section 7.6 in connection with the stabilities of substituted alkenes, is the stabilizing interaction between a $p$ orbital and properly oriented $\\mathrm{C}-\\mathrm{H} \\sigma$ bonds on neighboring carbons that are roughly parallel to the $p$ orbital. The more alkyl groups there are on the carbocation, the more possibilities there are for hyperconjugation and the more stable the carbocation. FIGURE 7.13 shows the molecular orbital for the ethyl carbocation, $\\mathrm{CH}_{3} \\mathrm{CH}_{2}{ }^{+}$, and indicates the difference between the $\\mathrm{C}-\\mathrm{H}$ bond perpendicular to the cation $p$ orbital and the two $\\mathrm{C}-\\mathrm{H}$ bonds more parallel to the cation $p$ orbital. Only these roughly parallel $\\mathrm{C}-\\mathrm{H}$ bonds are oriented properly to take part in hyperconjugation.\n\n\nFIGURE 7.13 Stabilization of the ethyl carbocation, $\\mathbf{C H}_{\\mathbf{3}} \\mathbf{C H}_{\\mathbf{2}}{ }^{\\boldsymbol{+}}$, through hyperconjugation. Interaction of neighboring $\\mathrm{C}-\\mathrm{H} \\sigma$ bonds with the vacant $p$ orbital stabilizes the cation and lowers its energy. The molecular orbital shows that only the two $\\mathrm{C}-\\mathrm{H}$ bonds more parallel to the cation $p$ orbital are oriented properly. The $\\mathrm{C}-\\mathrm{H}$ bond perpendicular to the cation $p$ orbital can't take part properly.\n\nPROBLEM Show the structures of the carbocation intermediates you would expect in the following reactions:\n\n7-18 (a)\n\n(b)\n\n\nPROBLEM Draw a skeletal structure of the following carbocation. Identify it as primary, secondary, or tertiary, 7-19 and identify the hydrogen atoms that have the proper orientation for hyperconjugation in the conformation shown."}
{"id": 438, "contents": "Stability - 7.10 The Hammond Postulate\nLet's summarize what we've learned of electrophilic addition reactions to this point:\n\n- Electrophilic addition to an unsymmetrically substituted alkene gives the more substituted carbocation intermediate. A more substituted carbocation forms faster than a less substituted one and, once formed, rapidly goes on to give the final product.\n- A more substituted carbocation is more stable than a less substituted one. That is, the stability order of carbocations is tertiary > secondary > primary > methyl.\n\nWhat we have not yet seen is how these two points are related. Why does the stability of the carbocation intermediate affect the rate at which it's formed and thereby determine the structure of the final product? After all, carbocation stability is determined by the free-energy change $\\Delta G^{\\circ}$, but reaction rate is determined by the activation energy $\\Delta G^{\\ddagger}$. The two quantities aren\u2019t directly related.\n\nAlthough there is no simple quantitative relationship between the stability of a carbocation intermediate and the rate of its formation, there is an intuitive relationship. It's generally true when comparing two similar reactions that the more stable intermediate forms faster than the less stable one. The situation is shown graphically in FIGURE 7.14, where the energy profile in part (a) represents the typical situation, as opposed to the profile in part (b). That is, the curves for two similar reactions don't cross one another.\n\n\nFIGURE 7.14 Energy diagrams for two similar competing reactions. In (a), the faster reaction yields the more stable intermediate. In (b), the slower reaction yields the more stable intermediate. The curves shown in (a) represent the typical situation.\n\nCalled the Hammond postulate, the explanation of the relationship between reaction rate and intermediate stability goes like this: Transition states represent energy maxima. They are high-energy activated complexes that occur transiently during the course of a reaction and immediately go on to a more stable species. Although we can't actually observe transition states because they have no finite lifetime, the Hammond postulate says that we can get an idea of a particular transition state's structure by looking at the structure of the nearest stable species. Imagine the two cases shown in FIGURE 7.15, for example. The reaction profile in part (a) shows the energy curve for an endergonic reaction step, and the profile in part (b) shows the curve for an exergonic step."}
{"id": 439, "contents": "Stability - 7.10 The Hammond Postulate\nFIGURE 7.15 Energy diagrams for endergonic and exergonic steps. (a) In an endergonic step, the energy levels of transition state and product are closer. (b) In an exergonic step, the energy levels of transition state and reactant are closer.\n\nIn an endergonic reaction (FIGURE 7.15a), the energy level of the transition state is closer to that of the product than that of the reactant. Since the transition state is closer energetically to the product, we make the natural assumption that it's also closer structurally. In other words, the transition state for an endergonic reaction step structurally resembles the product of that step. Conversely, the transition state for an exergonic reaction (FIGURE 7.15b), is closer energetically, and thus structurally, to the reactant than to the product. We therefore say that the transition state for an exergonic reaction step structurally resembles the reactant for that step."}
{"id": 440, "contents": "Hammond postulate: - \nThe structure of a transition state resembles the structure of the nearest stable species. Transition states for endergonic steps structurally resemble products, and transition states for exergonic steps structurally resemble reactants.\n\nHow does the Hammond postulate apply to electrophilic addition reactions? The formation of a carbocation by protonation of an alkene is an endergonic step. Thus, the transition state for alkene protonation structurally resembles the carbocation intermediate, and any factor that stabilizes the carbocation will also stabilize the nearby transition state. Since increasing alkyl substitution stabilizes carbocations, it also stabilizes the transition states leading to those ions, thus resulting in a faster reaction. In other words, more stable carbocations form faster because their greater stability is reflected in the lower-energy transition state leading to them (FIGURE 7.16).\n\n\nFIGURE 7.16 Energy diagrams for carbocation formation. The more stable tertiary carbocation is formed faster (green curve) because its increased stability lowers the energy of the transition state leading to it.\n\nWe can imagine the transition state for alkene protonation to be a structure in which one of the alkene carbon atoms has almost completely rehybridized from $s p^{2}$ to $s p^{3}$ and the remaining alkene carbon bears much of the positive charge (FIGURE 7.17). This transition state is stabilized by hyperconjugation and inductive effects in the same way as the product carbocation. The more alkyl groups that are present, the greater the extent of stabilization and the faster the transition state forms.\n\n\nFIGURE 7.17 The hypothetical structure of a transition state for alkene protonation. The transition state is closer in both energy and structure to the carbocation than to the alkene. Thus, an increase in carbocation stability (lower $\\Delta G^{\\circ}$ ) also causes an increase in transitionstate stability (lower $\\Delta G^{\\ddagger}$ ), thereby increasing the rate of its formation."}
{"id": 441, "contents": "Hammond postulate: - \nPROBLEM What about the second step in the electrophilic addition of HCl to an alkene-the reaction of\n$\\mathbf{7 - 2 0}$ chloride ion with the carbocation intermediate? Is this step exergonic or endergonic? Does the transition state for this second step resemble the reactant (carbocation) or product (alkyl chloride)? Make a rough drawing of what the transition-state structure might look like."}
{"id": 442, "contents": "Hammond postulate: - 7.11 Evidence for the Mechanism of Electrophilic Additions: Carbocation Rearrangements\nHow do we know that the carbocation mechanism for electrophilic addition reactions of alkenes is correct? The answer is that we don't know it's correct; at least we don't know with complete certainty. Although an incorrect reaction mechanism can be disproved by demonstrating that it doesn't account for observed data, a correct reaction mechanism can never be entirely proven. The best we can do is to show that a proposed mechanism is consistent with all known facts. If enough facts are accounted for, the mechanism is probably correct.\n\nOne of the best pieces of evidence supporting the carbocation mechanism for the electrophilic addition reaction was discovered during the 1930s by F. C. Whitmore of Pennsylvania State University, who found that structural rearrangements often occur during the reaction of HX with an alkene. For example, reaction of HCl with 3 -methyl-1-butene yields a substantial amount of 2 -chloro-2-methylbutane in addition to the \"expected\" product, 2-chloro-3-methylbutane.\n\n\nIf the reaction takes place in a single step, it would be difficult to account for rearrangement, but if the reaction takes place in several steps, rearrangement is more easily explained. Whitmore suggested that it is a carbocation intermediate that undergoes rearrangement. The secondary carbocation intermediate formed by protonation of 3-methyl-1-butene rearranges to a more stable tertiary carbocation by a hydride shift-the shift of a hydrogen atom and its electron pair (a hydride ion, $: \\mathrm{H}^{-}$) between neighboring carbons.\n\n\n2-Chloro-3-methylbutane\n2-Chloro-2-methylbutane\nCarbocation rearrangements can also occur by the shift of an alkyl group with its electron pair. For example, reaction of 3,3-dimethyl-1-butene with HCl leads to an equal mixture of unrearranged 3 -chloro-2,2-dimethylbutane and rearranged 2 -chloro-2,3-dimethylbutane. In this instance, a secondary carbocation rearranges to a more stable tertiary carbocation by the shift of a methyl group."}
{"id": 443, "contents": "Hammond postulate: - 7.11 Evidence for the Mechanism of Electrophilic Additions: Carbocation Rearrangements\n3-Chloro-2,2-dimethylbutane 2-Chloro-2,3-dimethylbutane\nNote the similarities between the two carbocation rearrangements: in both cases, a group (: $\\mathrm{H}^{-}$or : $\\mathrm{CH}_{3}{ }^{-}$) moves to an adjacent positively charged carbon, taking its bonding electron pair with it. Also in both cases, a less stable carbocation rearranges to a more stable ion. Rearrangements of this kind are a common feature of carbocation chemistry and are particularly important in the biological pathways by which steroids and related substances are synthesized. An example is the following hydride shift that occurs during the biosynthesis of cholesterol.\n\n\nAs always, when looking at any complex chemical transformation, whether biochemical or not, focus on the part of the molecule where the change is occurring and don't worry about the rest. The tertiary carbocation just pictured looks complicated, but all the chemistry is taking place in the small part of the molecule inside the red circle.\n\nPROBLEM On treatment with HBr , vinylcyclohexane undergoes addition and rearrangement to yield\n7-21 1-bromo-1-ethylcyclohexane. Using curved arrows, propose a mechanism to account for this result.\n\n\nVinylcyclohexane 1-Bromo-1-ethylcyclohexane"}
{"id": 444, "contents": "Bioprospecting: Hunting for Natural Products - \nMost people know the names of the common classes of biomolecules-proteins, carbohydrates, lipids, and nucleic acids-but there are many more kinds of compounds in living organisms than just those four. All living organisms also contain a vast diversity of substances usually grouped under the heading natural products. The term natural product really refers to any naturally occurring substance but is generally taken to mean a so-called secondary metabolite-a small molecule that is not essential to the growth and development of the producing organism and is not classified by structure.\n\n\nFIGURE 7.18 Rapamycin, an immunosuppressant natural product used during organ transplants, was originally isolated from a soil sample found on Rapa Nui (Easter Island), an island 2200 miles off the coast of Chile known for its giant Moai statues. (credit: modification of work \"Moai facing inland at Ahu Tongariki\" by Ian Sewell/Wikimedia Commons, CC BY 2.5)\n\nIt has been estimated that well over 300,000 secondary metabolites exist, and it's thought that their primary function is to increase the likelihood of an organism's survival by repelling or attracting other organisms. Alkaloids, such as morphine; antibiotics, such as erythromycin and the penicillins; and immunosuppressive agents, such as rapamycin (sirolimus) prescribed for liver transplant recipients, are examples."}
{"id": 445, "contents": "Bioprospecting: Hunting for Natural Products - \nWhere do these natural products come from, and how are they found? Although most chemists and biologists spend their working time in the laboratory, a few spend their days scuba diving on South Pacific islands or trekking through the rainforests of South America and Southeast Asia at work as bioprospectors. Their job is to\nhunt for new and unusual natural products that might be useful as drugs.\nAs noted in the Chapter 6 Chemistry Matters, more than half of all new drug candidates come either directly or indirectly from natural products. Morphine from the opium poppy, prostaglandin $\\mathrm{E}_{1}$ from sheep prostate glands, erythromycin A from a Streptomyces erythreus bacterium cultured from a Philippine soil sample, and benzylpenicillin from the mold Penicillium notatum are examples. The immunosuppressive agent rapamycin, whose structure is shown previously, was first isolated from a Streptomyces hygroscopicus bacterium found in a soil sample from Rapa Nui (Easter Island), located 2200 miles off the coast of Chile.\n\nWith less than $1 \\%$ of living organisms yet investigated, bioprospectors have a lot of work to do. But there is a race going on. Rainforests throughout the world are being destroyed at an alarming rate, causing many species of both plants and animals to become extinct before they can even be examined. The governments in many countries seem aware of the problem, but there is as yet no international treaty on biodiversity that could help preserve vanishing species."}
{"id": 446, "contents": "Key Terms - \n- alkene $\\left(\\mathrm{R}_{2} \\mathrm{C}=\\mathrm{CR}_{2}\\right)$\n- allyl group\n- degree of unsaturation\n- Econfiguration\n- E,Z system\n- electrophilic addition reaction\n- Hammond postulate\n- hydride shift\n- hyperconjugation\n- Markovnikov's rule\n- methylene group\n- regiospecific reaction\n- unsaturated\n- vinyl group\n- Z configuration"}
{"id": 447, "contents": "Summary - \nCarbon-carbon double bonds are present in most organic and biological molecules, so a good understanding of their behavior is needed. In this chapter, we've looked at some consequences of alkene stereoisomerism and at the details of the broadest class of alkene reactions-the electrophilic addition reaction.\n\nAn alkene is a hydrocarbon that contains a carbon-carbon double bond. Because they contain fewer hydrogens than alkanes with the same number of carbons, alkenes are said to be unsaturated.\n\nBecause rotation around the double bond can't occur, substituted alkenes can exist as cis-trans stereoisomers. The configuration of a double bond can be specified by applying the Cahn-Ingold-Prelog sequence rules, which rank the substituents on each double-bond carbon. If the higher-ranking groups on each carbon are on the same side of the double bond, the configuration is $\\boldsymbol{Z}$ (zusammen, \"together\"); if the higher-ranking groups on each carbon are on opposite sides of the double bond, the configuration is $\\boldsymbol{E}$ (entgegen, \"apart\").\n\nAlkene chemistry is dominated by electrophilic addition reactions. When HX reacts with an unsymmetrically substituted alkene, Markovnikov's rule predicts that the $H$ will add to the carbon having fewer alkyl substituents and the X group will add to the carbon having more alkyl substituents. Electrophilic additions to alkenes take place through carbocation intermediates formed by reaction of the nucleophilic alkene $\\pi$ bond with electrophilic $\\mathrm{H}^{+}$. Carbocation stability follows the order"}
{"id": 448, "contents": "Summary - \nTertiary $\\left(3^{\\circ}\\right)>$ Secondary $\\left(2^{\\circ}\\right)>\\operatorname{Primary}\\left(1^{\\circ}\\right)>$ Methyl\n$\\mathrm{R}_{3} \\mathrm{C}^{+}>\\mathrm{R}_{2} \\mathrm{CH}^{+}>\\mathrm{RCH}_{2}^{+}>\\mathrm{CH}_{3}{ }^{+}$\nMarkovnikov's rule can be restated by saying that, in the addition of HX to an alkene, a more stable carbocation intermediate is formed. This result is explained by the Hammond postulate, which says that the transition state of an exergonic reaction step structurally resembles the reactant, whereas the transition state of an endergonic reaction step structurally resembles the product. Since an alkene protonation step is endergonic, the stability of the more highly substituted carbocation is reflected in the stability of the transition state leading to its formation.\n\nEvidence in support of a carbocation mechanism for electrophilic additions comes from the observation that structural rearrangements often take place during reaction. Rearrangements occur by shift of either a hydride ion, : $\\mathrm{H}^{-}$(a hydride shift), or an alkyl anion, $: \\mathrm{R}^{-}$, from a carbon atom to the neighboring positively charged carbon. This results in isomerization of a less stable carbocation to a more stable one."}
{"id": 449, "contents": "Visualizing Chemistry - \nPROBLEM Name the following alkenes, and convert each drawing into a skeletal structure:\n\n7-22 (a)\n\n(b)\n\n\nPROBLEM Assign $E$ or $Z$ stereochemistry to the double bonds in each of the following alkenes, and convert\n7-23 each drawing into a skeletal structure (red $=0$, green $=\\mathrm{Cl}$ ):\n(a)\n\n(b)\n\n\nPROBLEM The following carbocation is an intermediate in the electrophilic addition reaction of HCl with\n7-24 two different alkenes. Identify both, and tell which $\\mathrm{C}-\\mathrm{H}$ bonds in the carbocation are aligned for hyperconjugation with the vacant $p$ orbital on the positively charged carbon.\n\n\nPROBLEM The following alkyl bromide can be made by HBr addition to three different alkenes. Show their 7-25 structures."}
{"id": 450, "contents": "Mechanism Problems - \nPROBLEM Predict the major product and show the complete mechanism for each of the following electrophilic 7-26 addition reactions.\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Each of the following electrophilic addition reactions involves a carbocation rearrangement.\n7-27 Predict the product and draw the complete mechanism of each using curved arrows.\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM When 1,3-butadiene reacts with 1 mol of HBr , two isolable products result. Propose mechanisms 7-28 for both.\n\n\nPROBLEM When methyl vinyl ether reacts with a strong acid, $\\mathrm{H}^{+}$adds to C 2 instead of C 1 or the oxygen atom.\n7-29 Explain."}
{"id": 451, "contents": "Methyl vinyl ether - \nPROBLEM Addition of HCl to 1-isopropylcyclohexene yields a rearranged product. Propose a mechanism,\n7-30 showing the structures of the intermediates and using curved arrows to indicate electron flow in each step.\n\n\nPROBLEM Addition of HCl to 1-isopropenyl-1-methylcyclopentane yields\n7-31 1-chloro-1,2,2-trimethylcyclohexane. Propose a mechanism, showing the structures of the intermediates and using curved arrows to indicate electron flow in each step.\n\n\nPROBLEM Limonene, a fragrant hydrocarbon found in lemons and oranges, is biosynthesized from geranyl\n7-32 diphosphate by the following pathway. Add curved arrows to show the mechanism of each step. Which step involves an alkene electrophilic addition? (The ion $\\mathrm{OP}_{2} \\mathrm{O}_{6}{ }^{4-}$ is the diphosphate ion, and \"Base\" is an unspecified base in the enzyme that catalyzes the reaction.)\n\n\nPROBLEM epi-Aristolochene, a hydrocarbon found in both pepper and tobacco, is biosynthesized by the\n7-33 following pathway. Add curved arrows to show the mechanism of each step. Which steps involve alkene electrophilic addition(s), and which involve carbocation rearrangement(s)? (The abbreviation $\\mathrm{H}-\\mathrm{A}$ stands for an unspecified acid, and \"Base\" is an unspecified base in the enzyme.)"}
{"id": 452, "contents": "Calculating a Degree of Unsaturation - \nPROBLEM Calculate the degree of unsaturation in the following formulas, and draw five possible structures for\n7-34 each:\n(a) $\\mathrm{C}_{10} \\mathrm{H}_{16}$\n(b) $\\mathrm{C}_{8} \\mathrm{H}_{8} \\mathrm{O}$\n(c) $\\mathrm{C}_{7} \\mathrm{H}_{10} \\mathrm{Cl}_{2}$\n(d) $\\mathrm{C}_{10} \\mathrm{H}_{16} \\mathrm{O}_{2}$\n(e) $\\mathrm{C}_{5} \\mathrm{H}_{9} \\mathrm{NO}_{2}$\n(f) $\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{ClNO}$\n\nPROBLEM How many hydrogens does each of the following compounds have?\n7-35 (a) $\\mathrm{C}_{8} \\mathrm{H}_{?} \\mathrm{O}_{2}$, has two rings and one double bond (b) $\\mathrm{C}_{7} \\mathrm{H}_{?} \\mathrm{~N}$, has two double bonds\n(c) $\\mathrm{C}_{9} \\mathrm{H}_{?} \\mathrm{NO}$, has one ring and three double bonds\n\nPROBLEM Loratadine, marketed as an antiallergy medication under the brand name Claritin, has four rings,\n7-36 eight double bonds, and the formula $\\mathrm{C}_{22} \\mathrm{H}_{?} \\mathrm{ClN}_{2} \\mathrm{O}_{2}$. How many hydrogens does loratadine have? (Calculate your answer; don't count hydrogens in the structure.)\n\n\nLoratadine"}
{"id": 453, "contents": "Naming Alkenes - \nPROBLEM Name the following alkenes:\n\n7-37 (a)\n\n(d)\n\n(b)\n\n(e)\n\n(c)\n\n(f)\n\n\nPROBLEM Draw structures corresponding to the following systematic names:\n7-38 (a) (4E)-2,4-Dimethyl-1,4-hexadiene (b) cis-3,3-Dimethyl-4-propyl-1,5-octadiene\n(c) 4-Methyl-1,2-pentadiene (d) (3E,5Z)-2,6-Dimethyl-1,3,5,7-octatetraene\n(e) 3-Butyl-2-heptene (f) trans-2,2,5,5-Tetramethyl-3-hexene\n\nPROBLEM Name the following cycloalkenes:\n\n7-39 (a)\n\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n\nPROBLEM Ocimene is a triene found in the essential oils of many plants. What is its IUPAC name, including\n7-40 stereochemistry?\n\n\nOcimene\n\nPROBLEM $\\alpha$-Farnesene is a constituent of the natural wax found on apples. What is its IUPAC name, including\n7-41 stereochemistry?\n\n$\\alpha$-Farnesene\nPROBLEM Menthene, a hydrocarbon found in mint plants, has the systematic name\n7-42 1-isopropyl-4-methylcyclohexene. Draw its structure.\nPROBLEM Draw and name the six alkene isomers, $\\mathrm{C}_{5} \\mathrm{H}_{10}$, including $E, Z$ isomers.\n7-43\nPROBLEM Draw and name the 17 alkene isomers, $\\mathrm{C}_{6} \\mathrm{H}_{12}$, including $E, Z$ isomers.\n7-44"}
{"id": 454, "contents": "Naming Alkenes - \nAlkene Isomers and Their Stability\nPROBLEM Rank the following sets of substituents according to the Cahn-Ingold-Prelog sequence rules:\n7-45\n(a) $-\\mathrm{CH}_{3},-\\mathrm{Br},-\\mathrm{H},-\\mathrm{I}$\n(b) $-\\mathrm{OH},-\\mathrm{OCH}_{3},-\\mathrm{H},-\\mathrm{CO}_{2} \\mathrm{H}$\n(c) $-\\mathrm{CO}_{2} \\mathrm{H},-\\mathrm{CO}_{2} \\mathrm{CH}_{3},-\\mathrm{CH}_{2} \\mathrm{OH},-\\mathrm{CH}_{3}$\n(d)\n(e) $-\\mathrm{CH}=\\mathrm{CH}_{2},-\\mathrm{CN},-\\mathrm{CH}_{2} \\mathrm{NH}_{2},-\\mathrm{CH}_{2} \\mathrm{Br}$\n(f)\n\n\n\nPROBLEM Assign $E$ or $Z$ configuration to each of the following compounds:\n7-46 (a)\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM Which of the following $E, Z$ designations are correct, and which are incorrect?\n7-47 (a)\n\n(b)\n\n(c)\n\nZ\n(d)\n\n(e)\n\nZ\n(f)\n\nE\n\nPROBLEM Rank the double bonds according to their increasing stability.\n7-48 (a)\n\n\n\n\n(b)\n\n\n\n(c)\n\n\n\n\nPROBLEM trans-2-Butene is more stable than cis-2-butene by only $4 \\mathrm{~kJ} / \\mathrm{mol}$, but 7-49 trans-2,2,5,5-tetramethyl-3-hexene is more stable than its cis isomer by $39 \\mathrm{~kJ} / \\mathrm{mol}$. Explain."}
{"id": 455, "contents": "Naming Alkenes - \nPROBLEM Cyclodecene can exist in both cis and trans forms, but cyclohexene cannot. Explain.\n7-50\nPROBLEM Normally, a trans alkene is more stable than its cis isomer. trans-Cyclooctene, however, is less stable\n7-51 than cis-cyclooctene by $38.5 \\mathrm{~kJ} / \\mathrm{mol}$. Explain.\nPROBLEM trans-Cyclooctene is less stable than cis-cyclooctene by $38.5 \\mathrm{~kJ} / \\mathrm{mol}$, but trans-cyclononene is less\n7-52 stable than cis-cyclononene by only $12.2 \\mathrm{~kJ} / \\mathrm{mol}$. Explain.\nPROBLEM Tamoxifen, a drug used in the treatment of breast cancer, and clomiphene, a drug used in fertility\n7-53 treatment, have similar structures but very different effects. Assign $E$ or $Z$ configuration to the double bonds in both compounds.\n\n\nTamoxifen (anticancer)\n\n\nClomiphene\n(fertility treatment)\n\nCarbocations and Electrophilic Addition Reactions\nPROBLEM Rank the following carbocations according to their increasing stability.\n\n7-54 (a)\n\n\n\n(b)\n$+$\n\n\n(c)\n\n\n\n\nPROBLEM Use the Hammond Postulate to determine which alkene in each pair would be expected to form a 7-55 carbocation faster in an electrophilic addition reaction.\n(a)\n\nvs.\n\n(b)\n\nvs.\n\n(c)\n\n\nPROBLEM The following carbocations can be stabilized by resonance. Draw all the resonance forms that would 7-56 stabilize each carbocation.\n(a)\n\n(b)\n$\\longleftrightarrow$\n(c)\n\n$\\longleftrightarrow$ ?\n\nPROBLEM Predict the major product in each of the following reactions:\n7-57 (a)\n\n(b)\n\n(Addition of $\\mathrm{H}_{2} \\mathrm{O}$ occurs.)\n(c)\n\n(d)\n\n\nPROBLEM Predict the major product from addition of HBr to each of the following alkenes:\n7-58 (a)\n\n(b)\n\n(c)"}
{"id": 456, "contents": "Naming Alkenes - \n(d)\n\n\nPROBLEM Predict the major product from addition of HBr to each of the following alkenes:\n7-58 (a)\n\n(b)\n\n(c)\n\n\nPROBLEM Alkenes can be converted into alcohols by acid-catalyzed addition of water. Assuming that 7-59 Markovnikov's rule is valid, predict the major alcohol product from each of the following alkenes.\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Each of the following carbocations can rearrange to a more stable ion. Propose structures for the\n7-60 likely rearrangement products.\n(a)\n\n(b)\n\n(c)"}
{"id": 457, "contents": "General Problems - \nPROBLEM Allene (1,2-propadiene), $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{C}=\\mathrm{CH}_{2}$, has two adjacent double bonds. What kind of hybridization\n7-61 must the central carbon have? Sketch the bonding $\\pi$ orbitals in allene. What shape do you predict for allene?\n\nPROBLEM The heat of hydrogenation for allene (Problem 7-61) to yield propane is $-295 \\mathrm{~kJ} / \\mathrm{mol}$, and the heat\n7-62 of hydrogenation for a typical monosubstituted alkene, such as propene, is $-125 \\mathrm{~kJ} / \\mathrm{mol}$. Is allene more stable or less stable than you might expect for a diene? Explain.\n\nPROBLEM Retin A, or retinoic acid, is a medication commonly used to reduce wrinkles and treat severe acne.\n7-63 How many different isomers arising from $E, Z$ double-bond isomerizations are possible?\n\n\nRetin A (retinoic acid)\nPROBLEM Fucoserratene and ectocarpene are sex pheromones produced by marine brown algae. What are\n7-64 their systematic names? (Ectocarpene is difficult; make your best guess, and then check your answer in the Student Solutions Manual.)\n\n\nFucoserratene\n\n\nEctocarpene\n\nPROBLEM tert-Butyl esters $\\left[\\mathrm{RCO}_{2} \\mathrm{C}\\left(\\mathrm{CH}_{3}\\right)_{3}\\right]$ are converted into carboxylic acids $\\left(\\mathrm{RCO}_{2} \\mathrm{H}\\right)$ by reaction with\n7-65 trifluoroacetic acid, a reaction useful in protein synthesis (Section 26.7). Assign $E, Z$ designation to the double bonds of both reactant and product in the following scheme, and explain why there is an apparent change in double-bond stereochemistry:\n\n\nPROBLEM Vinylcyclopropane reacts with HBr to yield a rearranged alkyl bromide. Follow the flow of electrons\n7-66 as represented by the curved arrows, show the structure of the carbocation intermediate in brackets, and show the structure of the final product."}
{"id": 458, "contents": "Vinylcyclopropane - \nPROBLEM Calculate the degree of unsaturation in each of the following formulas:\n7-67 (a) Cholesterol, $\\mathrm{C}_{27} \\mathrm{H}_{46} \\mathrm{O}$ (b) DDT, $\\mathrm{C}_{14} \\mathrm{H}_{9} \\mathrm{Cl}_{5}$ (c) Prostaglandin $\\mathrm{E}_{1}, \\mathrm{C}_{20} \\mathrm{H}_{34} \\mathrm{O}_{5}$\n(d) Caffeine, $\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2}$\n(e) Cortisone, $\\mathrm{C}_{21} \\mathrm{H}_{28} \\mathrm{O}_{5}$ (f) Atropine, $\\mathrm{C}_{17} \\mathrm{H}_{23} \\mathrm{NO}_{3}$\n\nPROBLEM The isobutyl cation spontaneously rearranges to the tert-butyl cation by a hydride shift. Is the\n7-68 rearrangement exergonic or endergonic? Draw what you think the transition state for the hydride shift might look like according to the Hammond postulate."}
{"id": 459, "contents": "Vinylcyclopropane - \nPROBLEM Draw an energy diagram for the addition of HBr to 1-pentene. Let one curve on your diagram\n7-69 show the formation of 1-bromopentane product and another curve on the same diagram show the formation of 2-bromopentane product. Label the positions for all reactants, intermediates, and products. Which curve has the higher-energy carbocation intermediate? Which curve has the higher-energy first transition state?\nPROBLEM Sketch the transition-state structures involved in the reaction of HBr with 1-pentene (Problem\n7-70 7-69). Tell whether each structure resembles reactant or product.\nPROBLEM Aromatic compounds such as benzene react with alkyl chlorides in the presence of $\\mathrm{AlCl}_{3}$ catalyst to\n7-71 yield alkylbenzenes. This reaction occurs through a carbocation intermediate, formed by reaction of the alkyl chloride with $\\mathrm{AlCl}_{3}\\left(\\mathrm{R}-\\mathrm{Cl}+\\mathrm{AlCl}_{3} \\longrightarrow \\mathrm{R}^{+}+\\mathrm{AlCl}_{4}{ }^{-}\\right)$. How can you explain the observation that reaction of benzene with 1-chloropropane yields isopropylbenzene as the major product?\n\n\nPROBLEM Reaction of 2,3-dimethyl-1-butene with HBr leads to an alkyl bromide, $\\mathrm{C}_{6} \\mathrm{H}_{13} \\mathrm{Br}$. On treatment of\n7-72 this alkyl bromide with KOH in methanol, elimination of HBr occurs and a hydrocarbon that is isomeric with the starting alkene is formed. What is the structure of this hydrocarbon, and how do you think it is formed from the alkyl bromide?"}
{"id": 460, "contents": "Alkenes: Reactions and Synthesis - \nFIGURE 8.1 The Spectra fiber used to make the bulletproof vests used by police and military is made of ultra-high-molecular-weight polyethylene, a simple alkene polymer. (credit: modification of work \"US Navy 081028-N-3857R-007 Seabees participate in a chemical, biological and radiological warfare drill\" by U.S. Navy photo by Mass Communication Specialist 1st Class Chad Runge/Wikimedia Commons, Public Domain)"}
{"id": 461, "contents": "CHAPTER CONTENTS - \n8.1 Preparing Alkenes: A Preview of Elimination Reactions\n8.2 Halogenation of Alkenes: Addition of $X_{2}$\n8.3 Halohydrins from Alkenes: Addition of HO-X\n8.4 Hydration of Alkenes: Addition of $\\mathrm{H}_{2} \\mathrm{O}$ by Oxymercuration\n8.5 Hydration of Alkenes: Addition of $\\mathrm{H}_{2} \\mathrm{O}$ by Hydroboration\n8.6 Reduction of Alkenes: Hydrogenation\n8.7 Oxidation of Alkenes: Epoxidation and Hydroxylation\n8.8 Oxidation of Alkenes: Cleavage to Carbonyl Compounds\n8.9 Addition of Carbenes to Alkenes: Cyclopropane Synthesis\n8.10 Radical Additions to Alkenes: Chain-Growth Polymers\n8.11 Biological Additions of Radicals to Alkenes\n8.12 Reaction Stereochemistry: Addition of $\\mathrm{H}_{2} \\mathrm{O}$ to an Achiral Alkene\n8.13 Reaction Stereochemistry: Addition of $\\mathrm{H}_{2} \\mathrm{O}$ to a Chiral Alkene\n\nWHY THIS CHAPTER? Much of the background needed to understand organic reactions has now been covered, and it's time to begin a systematic description of the major functional groups. In this chapter on alkenes, and in future chapters on other functional groups, we'll discuss a variety of reactions, but try to focus on the general principles and patterns of reactivity that tie organic chemistry together. There are no shortcuts; you have to know the reactions to understand organic and biological chemistry."}
{"id": 462, "contents": "CHAPTER CONTENTS - \nAlkene addition reactions occur widely, both in the laboratory and in living organisms. Although we've studied\nonly the addition of HX thus far, many closely related reactions also take place. In this chapter, we'll see briefly how alkenes are prepared and we'll discuss further examples of alkene addition reactions. Particularly important are the addition of a halogen $\\left(\\mathrm{X}_{2}\\right)$ to give a 1,2-dihalide, addition of a hypohalous acid (HOX) to give a halohydrin, addition of water to give an alcohol, addition of hydrogen to give an alkane, addition of a single oxygen to give a three-membered cyclic ether called an epoxide, and addition of two hydroxyl groups to give a 1,2-diol.\n\n\nFIGURE 8.2 Some useful alkene reactions."}
{"id": 463, "contents": "CHAPTER CONTENTS - 8.1 Preparing Alkenes: A Preview of Elimination Reactions\nBefore getting to the main subject of this chapter-the reactions of alkenes-let's take a brief look at how alkenes are prepared. The subject is a bit complex, though, so we'll return to it in Chapter 11 for a more detailed study. For the present, it's enough to realize that alkenes are readily available from simple precursors-usually alcohols in biological systems and either alcohols or alkyl halides in the laboratory.\n\nJust as the chemistry of alkenes is dominated by addition reactions, the preparation of alkenes is dominated by elimination reactions. Additions and eliminations are, in many respects, two sides of the same coin. That is, an addition reaction might involve the addition of HBr or $\\mathrm{H}_{2} \\mathrm{O}$ to an alkene to form an alkyl halide or alcohol, whereas an elimination reaction might involve the loss of HBr or $\\mathrm{H}_{2} \\mathrm{O}$ from an alkyl halide or alcohol to form an alkene.\n\n\nThe two most common elimination reactions are dehydrohalogenation-the loss of HX from an alkyl halide-and dehydration-the loss of water from an alcohol. Dehydrohalogenation usually occurs by reaction of an alkyl halide with strong base such as potassium hydroxide. For example, bromocyclohexane yields cyclohexene when treated with KOH in ethanol solution.\n\n\nBromocyclohexane Cyclohexene (81\\%)\nDehydration is often carried out in the laboratory by treatment of an alcohol with a strong acid. For example, when 1-methylcyclohexanol is warmed with aqueous sulfuric acid in tetrahydrofuran (THF) solvent, loss of\nwater occurs and 1-methylcyclohexene is formed.\n\n\n1-Methylcyclohexanol\n1-Methylcyclohexene (91\\%)"}
{"id": 464, "contents": "CHAPTER CONTENTS - 8.1 Preparing Alkenes: A Preview of Elimination Reactions\n1-Methylcyclohexanol\n1-Methylcyclohexene (91\\%)\n\n\nIn biological pathways, dehydrations rarely occur with isolated alcohols. Instead, they normally take place on substrates in which the -OH is positioned two carbons away from a $\\mathrm{C}=\\mathrm{O}$ group. In the biosynthesis of fats, for instance, $\\beta$-hydroxybutyryl ACP is converted by dehydration to trans-crotonyl ACP, where ACP is an abbreviation for acyl carrier protein. We'll see the reason for this requirement in Section 11.10.\n\n$\\boldsymbol{\\beta}$-Hydroxybutyryl ACP\ntrans-Crotonyl ACP\nPROBLEM One problem with elimination reactions is that mixtures of products are often formed. For example,\n8-1 treatment of 2-bromo-2-methylbutane with KOH in ethanol yields a mixture of two alkene products. What are their likely structures?\n\nPROBLEM How many alkene products, including $E, Z$ isomers, might be obtained by dehydration of 8-2 3-methyl-3-hexanol with aqueous sulfuric acid?"}
{"id": 465, "contents": "3-Methyl-3-hexanol - 8.2 Halogenation of Alkenes: Addition of $X_{2}$\nBromine and chlorine add rapidly to alkenes to yield 1,2-dihalides, a process called halogenation. For example, nearly 50 million tons of 1,2-dichloroethane (ethylene dichloride) are synthesized worldwide each year, much of it by addition of $\\mathrm{Cl}_{2}$ to ethylene. The product is used both as a solvent and as starting material for the manufacture of poly(vinyl chloride), PVC, the third most widely synthesized polymer in the world afterpolyethelyne and polypropolyne. Fluorine is too reactive and difficult to control for most laboratory applications, and iodine does not react with most alkenes.\n\n\nEthylene\n1,2-Dichloroethane (ethylene dichloride)\nBased on what we've seen thus far, a possible mechanism for the reaction of bromine with alkenes might involve electrophilic addition of $\\mathrm{Br}^{+}$to the alkene, giving a carbocation intermediate that could undergo further reaction with $\\mathrm{Br}^{-}$to yield the dibromo addition product."}
{"id": 466, "contents": "Possible - \nmechanism?\n\nAlthough this mechanism seems plausible, it's not fully consistent with known facts. In particular, it doesn't explain the stereochemistry of the addition reaction. That is, the mechanism doesn't account for which product stereoisomer is formed.\n\nWhen the halogenation reaction is carried out on a cycloalkene, such as cyclopentene, only the trans stereoisomer of the dihalide addition product is formed, rather than the mixture of cis and trans isomers that might have been expected if a planar carbocation intermediate were involved. We say that the reaction occurs with anti stereochemistry, meaning that the two bromine atoms come from opposite faces of the double bond-one from the top face and one from the bottom face.\n\n\nAn explanation for the observed stereochemistry of addition was suggested in 1937 by George Kimball and Irving Roberts, who proposed that the reaction intermediate is not a carbocation but is instead a bromonium ion, $\\mathbf{R}_{\\mathbf{2}} \\mathbf{B r}^{+}$, formed by electrophilic addition of $\\mathrm{Br}^{+}$to the alkene. (Similarly, a chloronium ion contains a positively charged, divalent chlorine, $\\mathrm{R}_{2} \\mathrm{Cl}^{+}$.) The bromonium ion is formed in a single step by interaction of the alkene with $\\mathrm{Br}_{2}$ and the simultaneous loss of $\\mathrm{Br}^{-}$.\n\n\nHow does the formation of a bromonium ion account for the observed anti stereochemistry of addition to cyclopentene? If a bromonium ion is formed as an intermediate, we can imagine that the large bromine atom might \"shield\" one side of the molecule. Reaction with $\\mathrm{Br}^{-}$ion in the second step could then occur only from the opposite, unshielded side to give the trans product."}
{"id": 467, "contents": "Possible - \nThe bromonium ion postulate, made more than 85 years ago to explain the stereochemistry of halogen addition to alkenes, is a remarkable example of deductive logic in chemistry. Arguing from experimental results, chemists were able to make a hypothesis about the intimate mechanistic details of alkene electrophilic reactions. Subsequently, strong evidence supporting the mechanism came from the work of George Olah at the University of Southern California, who prepared and studied stable solutions of cyclic bromonium ions in liquid $\\mathrm{SO}_{2}$. There's no question that bromonium ions exist.\n\n\nAlkene halogenation reactions occur in nature just as they do in the laboratory but are limited primarily to marine organisms living in halide-rich environments. These biological halogenation reactions are carried out by enzymes called haloperoxidases, which use $\\mathrm{H}_{2} \\mathrm{O}_{2}$ to oxidize $\\mathrm{Br}^{-}$or $\\mathrm{Cl}^{-}$ions to a biological equivalent of $\\mathrm{Br}^{+}$or $\\mathrm{Cl}^{+}$. Electrophilic addition to the double bond of a substrate molecule then yields a bromonium or chloronium ion intermediate just as in the laboratory, and reaction with another halide ion completes the process. Halomon, for example, an antitumor pentahalide isolated from red alga, is thought to arise by a route that involves twofold addition of BrCl through the corresponding bromonium ions.\n\n\n\nHalomon\nPROBLEM What product would you expect to obtain from addition of $\\mathrm{Cl}_{2}$ to 1,2 -dimethylcyclohexene? Show 8-3 the stereochemistry of the product.\n\nPROBLEM Addition of HCl to 1,2-dimethylcyclohexene yields a mixture of two products. Show the 8-4 stereochemistry of each, and explain why a mixture is formed."}
{"id": 468, "contents": "Possible - 8.3 Halohydrins from Alkenes: Addition of HO-X\nAnother example of an electrophilic addition is the reaction an alkene with either $\\mathrm{Br}_{2}$ or $\\mathrm{Cl}_{2}$ in the presence of\nwater to yield a 1,2-halo alcohol, called a halohydrin.\n\n\nAn alkene\n\n$+\\mathrm{HX}$\n\nA halohydrin\n\nWe saw in the previous section that when $\\mathrm{Br}_{2}$ reacts with an alkene, the cyclic bromonium ion intermediate reacts with the only nucleophile present, $\\mathrm{Br}^{-}$ion. If the reaction is carried out in the presence of an additional nucleophile, however, the intermediate bromonium ion can be intercepted by the added nucleophile and diverted to a different product. In the presence of a high concentration of water, for instance, water competes with $\\mathrm{Br}^{-}$ion as a nucleophile and reacts with the bromonium ion intermediate to yield a bromohydrin. The net effect is addition of $\\mathrm{HO}-\\mathrm{Br}$ to the alkene by the pathway shown in FIGURE 8.3.\n\nFIGURE 8.3 MECHANISM\n\nBromohydrin formation by reaction of an alkene with $\\mathrm{Br}_{2}$ in the presence of water.\nWater acts as a nucleophile in step 2 to react with the intermediate bromonium ion.\n(1) Reaction of the alkene with $\\mathrm{Br}_{2}$ yields a bromonium ion intermediate, as previously discussed.\n(2) Water acts as a nucleophile, using a lone pair of electrons to open the bromonium ion ring and form a bond to carbon. Since oxygen donates its electrons in this step, it now has the positive charge.\n\n(1) $\\mathrm{Br}_{2}$\n\n\n2\n\n\nLoss of a proton $\\left(\\mathrm{H}^{+}\\right)$from oxygen then gives $\\mathrm{H}_{3} \\mathrm{O}^{+}$and the neutral bromohydrin addition product."}
{"id": 469, "contents": "(3) - \n3-Bromo-2-butanol\n\nIn practice, few alkenes are soluble in water, and bromohydrin formation is often carried out in a solvent such as aqueous dimethyl sulfoxide, $\\mathrm{CH}_{3} \\mathrm{SOCH}_{3}$ (DMSO), using a reagent called N -bromosuccinimide (NBS) as a source of $\\mathrm{Br}_{2}$. NBS is a stable, easily handled compound that slowly decomposes in water to yield $\\mathrm{Br}_{2}$ at a controlled rate. Bromine itself can also be used in the addition reaction, but it is more dangerous and more difficult to handle than NBS.\n\n\nNotice that the aromatic ring in the above example does not react with $\\mathrm{Br}_{2}$, even though it appears to have three carbon-carbon double bonds. As we'll see in Section 15.2, aromatic rings are a good deal more stable and less reactive than might be expected.\n\nThere are a number of biological examples of halohydrin formation, particularly in marine organisms. As with halogenation (Section 8.2), halohydrin formation is carried out by haloperoxidases. For example:\n\n\nPROBLEM What product would you expect from the reaction of cyclopentene with NBS and water? Show the 8-5 stereochemistry.\n\nPROBLEM When an unsymmetrical alkene such as propene is treated with $N$-bromosuccinimide in aqueous\n8-6 dimethyl sulfoxide, the major product has the bromine atom bonded to the less highly substituted carbon atom. Is this Markovnikov or non-Markovnikov orientation? (Section 7.8) Explain."}
{"id": 470, "contents": "(3) - 8.4 Hydration of Alkenes: Addition of $\\mathrm{H}_{2} \\mathrm{O}$ by Oxymercuration\nWater adds to alkenes to yield alcohols, a process called hydration. The reaction takes place on treatment of the alkene with water and a strong acid catalyst, such as $\\mathrm{H}_{2} \\mathrm{SO}_{4}$, by a mechanism similar to that of HX addition. Thus, as shown in FIGURE 8.4, protonation of an alkene double bond yields a carbocation intermediate, which reacts with water to yield a protonated alcohol product, $\\mathrm{ROH}_{2}{ }^{+}$. Loss of $\\mathrm{H}^{+}$from this protonated alcohol gives the neutral alcohol and regenerates the acid catalyst."}
{"id": 471, "contents": "FIGURE 8.4 MECHANISM - \nMechanism of the acid-catalyzed hydration of an alkene to yield an alcohol. Protonation of the alkene gives a carbocation intermediate, which reacts with water. The initial product is then deprotonated.\n\nA hydrogen atom on the electrophile $\\mathrm{H}_{3} \\mathrm{O}^{+}$ is attacked by $\\pi$ electrons from the nucleophilic double bond, forming a new $\\mathrm{C}-\\mathrm{H}$ bond. This leaves the other carbon atom with $\\mathrm{a}+$ charge and a vacant $p$ orbital. Simultaneously, two electrons from the $\\mathrm{H}-\\mathrm{O}$ bond move onto oxygen, giving neutral water.\n\n\n2-Methylpropene\n(1)\n\n\n2 The nucleophile $\\mathrm{H}_{2} \\mathrm{O}$ donates an electron pair to the positively charged carbon atom, forming a $\\mathrm{C}-\\mathrm{O}$ bond and leaving a positive charge on oxygen in the protonated alcohol addition product.\n\n\nProtonated alcohol\n(3) Water acts as a base to remove $\\mathrm{H}^{+}$, regenerating $\\mathrm{H}_{3} \\mathrm{O}^{+}$and yielding the neutral alcohol addition product.\n\n(3)\n\n\n2-Methyl-2-propanol\n\nMost ethanol throughout the world is now made by fermentation of biological precursors, such as corn and sugar, but acid-catalyzed alkene hydration is particularly suited to large-scale industrial procedures, and approximately 90,000 tons of ethanol is manufactured each year in the United States by hydration of ethylene. The reaction is of little value in the laboratory, however, because it requires high temperatures- $250{ }^{\\circ} \\mathrm{C}$ in the case of ethylene-and strongly acidic conditions."}
{"id": 472, "contents": "Ethylene - \nAcid-catalyzed hydration of double bonds is also uncommon in biological pathways. Instead, biological\nhydrations usually require that the double bond be adjacent to a carbonyl group for reaction to proceed. Fumarate, for instance, is hydrated to give malate as one step in the citric acid cycle of food metabolism. Note that the requirement for an adjacent carbonyl group in the addition of water is the same as in Section 8.1 for the elimination of water. We'll see the reason for this requirement in Section 19.13, but will note for now that the reaction is not an electrophilic addition but instead occurs through a mechanism that involves formation of an anion intermediate followed by protonation by an acid HA.\n\n\nWhen it comes to circumventing problems like those with acid-catalyzed alkene hydrations, laboratory chemists have a great advantage over the cellular \"chemists\" in living organisms. Laboratory chemists are not constrained to carry out their reactions in water solution; they can choose from any of a large number of solvents. Laboratory reactions don't need to be carried out at a fixed temperature; they can take place over a wide range of temperatures. And laboratory reagents aren't limited to containing carbon, oxygen, nitrogen, and a few other elements; they can contain any element in the periodic table.\n\nIn the laboratory, alkenes are often hydrated by the oxymercuration-demercuration procedure, which involves electrophilic addition of $\\mathrm{Hg}^{2+}$ to the alkene on reaction with mercury(II) acetate $\\left\\left(\\mathrm{CH}_{3} \\mathrm{CO}_{2}\\right)_{2} \\mathrm{Hg}\\right.$. The intermediate organomercury compound is then treated with sodium borohydride, $\\mathrm{NaBH}_{4}$, and demercuration occurs to produce an alcohol. For example:\n!["}
{"id": 473, "contents": "1-Methylcyclopentene - \n1-Methylcyclopentanol\n(92\\%)\n\nAlkene oxymercuration is closely analogous to halohydrin formation. The reaction is initiated by electrophilic addition of $\\mathrm{Hg}^{2+}$ (mercuric) ion to the alkene to give an intermediate mercurinium ion, whose structure resembles that of a bromonium ion (FIGURE 8.5). Nucleophilic addition of water as in halohydrin formation, followed by the loss of a proton, then yields a stable organomercury product. The final step, demercuration of the organomercury compound by reaction with sodium borohydride, is complex and involves radicals. Note that the regiochemistry of the reaction corresponds to Markovnikov addition of water; that is, the - OH group attaches to the more highly substituted carbon atom, and the -H attaches to the less highly substituted carbon. The hydrogen that replaces mercury in the demercuration step can attach from either side of the molecule depending on the exact circumstances.\n\n\nFIGURE 8.5 Mechanism of the oxymercuration of an alkene to yield an alcohol. (1) Electrophilic addition of $\\mathrm{Hg}^{2+}$ gives a mercurinium ion, which (2) reacts with water as in halohydrin formation. Loss of a proton gives an organomercury product, and (3) reaction with $\\mathrm{NaBH}_{4}$ removes the mercury. The product of the reaction is a more highly substituted alcohol, corresponding to Markovnikov regiochemistry.\n\nPROBLEM What products would you expect from oxymercuration-demercuration of the following alkenes?\n8-7 (a)\n$\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}=\\mathrm{CH}_{2}$\n(b)\n\n\nPROBLEM From what alkenes might the following alcohols have been prepared?\n8-8 (a)\n\n(b)"}
{"id": 474, "contents": "1-Methylcyclopentene - 8.5 Hydration of Alkenes: Addition of $\\mathrm{H}_{2} \\mathrm{O}$ by Hydroboration\nIn addition to the oxymercuration-demercuration method, which yields the Markovnikov product, a complementary method that yields the non-Markovnikov product is also useful. Discovered in 1959 by H.C. Brown at Purdue University and called hydroboration, the reaction involves addition of a B-H bond of borane, $\\mathrm{BH}_{3}$, to an alkene to yield an organoborane intermediate, $\\mathrm{RBH}_{2}$. Oxidation of the organoborane by reaction with basic hydrogen peroxide, $\\mathrm{H}_{2} \\mathrm{O}_{2}$, then gives an alcohol. For example:\n\n\nBorane is very reactive as a Lewis acid because the boron atom has only six electrons in its valence shell rather than an octet. In tetrahydrofuran solution, $\\mathrm{BH}_{3}$ accepts an electron pair from a solvent molecule in a Lewis acid-base reaction to complete its octet and form a stable $\\mathrm{BH}_{3}-$ THF complex.\n\n\nWhen an alkene reacts with $\\mathrm{BH}_{3}$ in THF solution, rapid addition to the double bond occurs three times and a trialkylborane, $\\mathrm{R}_{3} \\mathrm{~B}$, is formed. For example, 1 molar equivalent of $\\mathrm{BH}_{3}$ adds to 3 molar equivalents of cyclohexene to yield tricyclohexylborane. When tricyclohexylborane is then treated with aqueous hydrogen $\\mathrm{H}_{2} \\mathrm{O}_{2}$ in basic solution, an oxidation takes place. The three $\\mathrm{C}-\\mathrm{B}$ bonds are broken, -OH groups bond to the three carbons, and 3 equivalents of cyclohexanol are produced. The net effect of the two-step hydroboration-oxidation sequence is hydration of the alkene double bond."}
{"id": 475, "contents": "1-Methylcyclopentene - 8.5 Hydration of Alkenes: Addition of $\\mathrm{H}_{2} \\mathrm{O}$ by Hydroboration\nOne of the features that makes the hydroboration reaction so useful is the regiochemistry that results when an unsymmetrical alkene is hydroborated. For example, hydroboration-oxidation of 1-methylcyclopentene yields trans-2-methylcyclopentanol. In this process, boron and hydrogen add to the alkene from the same face of the double bond-that is, with syn stereochemistry, the opposite of anti-with boron attaching to the less highly substituted carbon. During the oxidation step, the boron is replaced by an -OH with the same stereochemistry, resulting in an overall syn non-Markovnikov addition of water. This stereochemical result is particularly useful because it is complementary to the Markovnikov regiochemistry observed for oxymercuration-demercuration.\n\n\nWhy does alkene hydroboration take place with syn, non-Markovnikov regiochemistry to yield the less highly substituted alcohol? Hydroboration differs from many other alkene addition reactions in that it occurs in a single step without a carbocation intermediate (FIGURE 8.6). Because the $\\mathrm{C}-\\mathrm{H}$ and $\\mathrm{C}-\\mathrm{B}$ bonds form at the same time and from the same face of the alkene, syn stereochemistry results. Non-Markovnikov regiochemistry occurs because attachment of boron is favored at the less sterically crowded carbon atom of the alkene.\n\n\nFIGURE 8.6 Mechanism of alkene hydroboration. The reaction occurs in a single step in which the $\\mathrm{C}-\\mathrm{H}$ and $\\mathrm{C}-\\mathrm{B}$ bonds form at the same time and on the same face of the double bond. The lower energy, more rapidly formed transition state is the one with less steric crowding, leading to non-Markovnikov regiochemistry."}
{"id": 476, "contents": "Predicting the Products of a Hydration Reaction - \nWhat products would you obtain from reaction of 2-methyl-2-pentene with:\n(a) $\\mathrm{BH}_{3}$, followed by $\\mathrm{H}_{2} \\mathrm{O}_{2}, \\mathrm{OH}^{-}$\n(b) $\\mathrm{Hg}(\\mathrm{OAc})_{2}$, followed by $\\mathrm{NaBH}_{4}$"}
{"id": 477, "contents": "Strategy - \nWhen predicting the product of a reaction, you have to recall what you know about the kind of reaction being carried out and apply that knowledge to the specific case you're dealing with. In the present instance, recall that the two methods of hydration-hydroboration-oxidation and oxymercuration-demercuration-give complementary products. Hydroboration-oxidation occurs with syn stereochemistry and gives the nonMarkovnikov addition product; oxymercuration-demercuration gives the Markovnikov product."}
{"id": 478, "contents": "Solution - \n(a) 2-Methyl-2-pentene\n\n\n\n2-Methyl-3-pentanol\n\nPROBLEM What alkenes might be used to prepare the following alcohols by hydroboration-oxidation?\n8-10 (a)\n\n\n(b)\n\n(c)\n\n\nPROBLEM The following cycloalkene gives a mixture of two alcohols on hydroboration followed by oxidation.\n8-11 Draw the structures of both, and explain the result."}
{"id": 479, "contents": "Solution - 8.6 Reduction of Alkenes: Hydrogenation\nAlkenes react with $\\mathrm{H}_{2}$ in the presence of a metal catalyst such as palladium or platinum to yield the corresponding saturated alkanes. We describe the result by saying that the double bond has been hydrogenated, or reduced. Note that the word reduction is used somewhat differently in organic chemistry from what you might have learned previously. In general chemistry, a reduction is defined as the gain of one or more electrons by an atom. In organic chemistry, however, a reduction is a reaction that results in a gain of electron density for carbon, caused either by bond formation between carbon and a less electronegative atom-usually hydrogen-or by bond-breaking between carbon and a more electronegative atom-usually oxygen, nitrogen, or a halogen. We'll explore this topic in more detail in Section 10.8.\n\nReduction Increases electron density on carbon by:\n-forming this: $\\mathrm{C}-\\mathrm{H}$\n\n- or breaking one of these: $\\mathrm{C}-\\mathrm{O} \\quad \\mathrm{C}-\\mathrm{N} \\quad \\mathrm{C}-\\mathrm{X}$"}
{"id": 480, "contents": "A reduction: - \nAn alkene\nAn alkane\nPlatinum and palladium are the most common laboratory catalysts for alkene hydrogenations. Palladium is normally used as a very fine powder \"supported\" on an inert material such as charcoal ( $\\mathrm{Pd} / \\mathrm{C}$ ) to maximize surface area. Platinum is normally used as $\\mathrm{PtO}_{2}$, a reagent known as Adams' catalyst after its discoverer, Roger Adams at the University of Illinois.\n\nCatalytic hydrogenation, unlike most other organic reactions, is a heterogeneous process rather than a homogeneous one. That is, the hydrogenation reaction does not occur in a homogeneous solution but instead takes place on the surface of solid catalyst particles. Hydrogenation usually occurs with syn stereochemistry: both hydrogens add to the double bond from the same face.\n\n\nAs shown in FIGURE 8.7, hydrogenation begins with adsorption of $\\mathrm{H}_{2}$ onto the catalyst surface. Complexation between catalyst and alkene then occurs as a vacant orbital on the metal interacts with the filled alkene $\\pi$ orbital\non the alkene. In the final steps, hydrogen is inserted into the double bond and the saturated product diffuses away from the catalyst. The stereochemistry of hydrogenation is syn because both hydrogens add to the double bond from the same catalyst surface.\n\nFIGURE 8.7 MECHANISM\n\nMechanism of alkene hydrogenation. The reaction takes place with syn stereochemistry on the surface of insoluble catalyst particles.\ncatalyst surface and dissociates into hydrogen atoms.\n\n(2) The alkene adsorbs to the catalyst surface, using its $\\pi$ bond to complex to the metal atoms.\n(3) A hydrogen atom is transferred from the metal to one of the alkene carbon atoms, forming a partially reduced intermediate with a $\\mathrm{C}-\\mathrm{H}$ bond and carbon-metal $\\sigma$ bond.\n(4) A second hydrogen is transferred from the metal to the second carbon, giving the alkane product and regenerating the catalyst. Because both hydrogens are transferred to the same face of the alkene, the reduction has syn stereochemistry.\n\n> Alkane plus regenerated catalyst"}
{"id": 481, "contents": "A reduction: - \n> Alkane plus regenerated catalyst\n\nAn interesting feature of catalytic hydrogenation is that the reaction is extremely sensitive to the steric environment around the double bond. As a result, the catalyst usually approaches the more accessible face of an alkene, giving rise to a single product. In $\\alpha$-pinene, for example, one of the methyl groups attached to the fourmembered ring hangs over the top face of the double bond and blocks approach of the hydrogenation catalyst from that side. Reduction therefore occurs exclusively from the bottom face to yield the product shown.\n\n\nAlkenes are much more reactive toward catalytic hydrogenation than most other unsaturated functional groups, and the reaction is therefore quite selective. Other functional groups, such as aldehydes, ketones, esters, and nitriles, often survive alkene hydrogenation conditions unchanged, although reaction with these groups does occur under more vigorous conditions. Note that, particularly in the hydrogenation of methyl 3-phenylpropenoate shown below, the aromatic ring is not reduced by hydrogen and palladium even though it contains apparent double bonds.\n\n\nCyclohex-2-enone\nCyclohexanone\n(ketone not reduced)\n\n\n\nCyclohexylideneacetonitrile\n\n\nCyclohexylacetonitrile\n(nitrile not reduced)\n\nIn addition to its usefulness in the laboratory, catalytic hydrogenation is also important in the food industry, where unsaturated vegetable oils are reduced on a large scale to produce the saturated fats used in margarine and cooking products (FIGURE 8.8). As we'll see in Section 27.1, vegetable oils are triesters of glycerol, $\\mathrm{HOCH}_{2} \\mathrm{CH}(\\mathrm{OH}) \\mathrm{CH}_{2} \\mathrm{OH}$, with three long-chain carboxylic acids called fatty acids. The fatty acids are generally polyunsaturated, and their double bonds have cis stereochemistry. Complete hydrogenation yields the corresponding saturated fatty acids, but incomplete hydrogenation often results in partial cis-trans isomerization of a remaining double bond. When eaten and digested, the free trans fatty acids are released, raising blood cholesterol levels and potentially contributing to coronary problems.\n\n\n\nA polyunsaturated\nfatty acid in\nvegetable oil\n\nA vegetable oil"}
{"id": 482, "contents": "A saturated fatty - \nacid in margarine\n\nA trans fatty acid\n\nFIGURE 8.8 Catalytic hydrogenation of polyunsaturated fats leads primarily to saturated products, along with a small amount of isomerized trans fats.\n\nDouble-bond reductions are very common in biological pathways, although the mechanism is completely different from that of laboratory catalytic hydrogenation over palladium. As with biological hydrations (Section 8.4), biological reductions usually occur in two steps and require that the double bond be adjacent to a carbonyl group. In the first step, the biological reducing agent NADPH (reduced nicotinamide adenine dinucleotide phosphate), adds a hydride ion ( $\\mathrm{H}:^{-}$) to the double bond to give an anion. In the second, the anion is protonated by acid HA, leading to overall addition of $\\mathrm{H}_{2}$. An example is the reduction of trans-crotonyl ACP to yield butyryl ACP, a step involved in the biosynthesis of fatty acids (FIGURE 8.9).\n\n\n\nNADPH\nFIGURE 8.9 Reduction of the carbon-carbon double bond in trans-crotonyl ACP, a step in the biosynthesis of fatty acids. One hydrogen is delivered from NADPH as a hydride ion, $\\mathrm{H}^{-}$; the other hydrogen is delivered by protonation of the anion intermediate with an acid, HA.\n\nPROBLEM What product would you obtain from catalytic hydrogenation of the following alkenes?\n8-12 (a)\n\n(b)\n\n(c)\n\n="}
{"id": 483, "contents": "A saturated fatty - 8.7 Oxidation of Alkenes: Epoxidation and Hydroxylation\nLike the word reduction used in the previous section for the addition of hydrogen to a double bond, the word oxidation has a slightly different meaning in organic chemistry than what you might have previously learned. In general chemistry, an oxidation is defined as the loss of one or more electrons by an atom. In organic chemistry, however, an oxidation is a reaction that results in a loss of electron density for carbon, caused either by bond formation between carbon and a more electronegative atom-usually oxygen, nitrogen, or a halogen-or by bond-breaking between carbon and a less electronegative atom-usually hydrogen. Note that an oxidation often adds oxygen, while a reduction often adds hydrogen.\n\nOxidation $\\quad \\begin{array}{lll}\\text { Decreases electron density on carbon by: } \\\\ & - \\text { forming one of these: } \\mathrm{C}-\\mathrm{O} \\quad \\mathrm{C}-\\mathrm{N} & \\mathrm{C}-\\mathrm{X} \\\\ & - \\text { or breaking this: } \\mathrm{C}-\\mathrm{H} & \\end{array}$\nIn the laboratory, alkenes are oxidized to give epoxides on treatment with a peroxyacid, $\\mathrm{RCO}_{3} \\mathrm{H}$, such as meta-chloroperoxybenzoic acid. An epoxide, also called an oxirane, is a cyclic ether with an oxygen atom in a three-membered ring. For example:\n\n\nPeroxyacids transfer an oxygen atom to the alkene with syn stereochemistry-both C-O bonds form on the same face of the double bond-through a one-step mechanism without intermediates. The oxygen atom farthest from the carbonyl group is the one transferred.\n\n\nAnother method for the synthesis of epoxides involves the use of halohydrins, prepared by electrophilic addition of HO-X to alkenes (Section 8.3). When a halohydrin is treated with base, HX is eliminated and an epoxide is produced."}
{"id": 484, "contents": "A saturated fatty - 8.7 Oxidation of Alkenes: Epoxidation and Hydroxylation\nEpoxides undergo an acid-catalyzed ring-opening reaction with water (a hydrolysis) to give the corresponding 1,2-dialcohol, or diol, also called a glycol. Thus, the net result of the two-step alkene epoxidation/hydrolysis is hydroxylation-the addition of an - OH group to each of the two double-bond carbons. In fact, approximately 204 million tons of ethylene glycol, $\\mathrm{HOCH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}$, most of it used for automobile antifreeze, are produced worldwide each year by the epoxidation of ethylene and subsequent hydrolysis.\n\n\nAcid-catalyzed epoxide opening begins with protonation of the epoxide to increase its reactivity, followed by nucleophilic addition of water. This nucleophilic addition is analogous to the final step of alkene bromination, in which a cyclic bromonium ion is opened by a nucleophile (Section 8.2). That is, a trans-1,2-diol results when an epoxycycloalkane is opened by aqueous acid, just as a trans-1,2-dibromide results when a cycloalkene is brominated. We'll look at epoxide chemistry in more detail in Section 18.6.\n\n\nHydroxylation can also be carried out directly (without going through an intermediate epoxide) by treating an alkene with osmium tetroxide, $\\mathrm{OsO}_{4}$. The reaction occurs with syn stereochemistry and does not involve a carbocation intermediate. Instead, it takes place through an intermediate cyclic osmate, which is formed in a single step by addition of $\\mathrm{OsO}_{4}$ to the alkene. This cyclic osmate is then cleaved using aqueous sodium bisulfite, $\\mathrm{NaHSO}_{3}$.\n\n\n1,2-Dimethylcyclopentene\n\nA cyclic osmate intermediate\ncis-1,2-Dimethyl-1,2-cyclopentanediol (87\\%)"}
{"id": 485, "contents": "A saturated fatty - 8.7 Oxidation of Alkenes: Epoxidation and Hydroxylation\n1,2-Dimethylcyclopentene\n\nA cyclic osmate intermediate\ncis-1,2-Dimethyl-1,2-cyclopentanediol (87\\%)\n\nBecause $\\mathrm{OSO}_{4}$ is both very expensive and very toxic, the reaction is usually carried out using only a small, catalytic amount of $\\mathrm{OsO}_{4}$ in the presence of a stoichiometric amount of a safe and inexpensive co-oxidant such as $N$-methylmorpholine $N$-oxide, abbreviated NMO. The initially formed osmate intermediate reacts rapidly with NMO to yield the product diol plus $N$-methylmorpholine and reoxidized $\\mathrm{OsO}_{4}$, which reacts with more alkene in a catalytic cycle.\n\n\nPROBLEM What product would you expect from reaction of cis-2-butene with meta-chloroperoxybenzoic 8-13 acid? Show the stereochemistry.\n\nPROBLEM Starting with an alkene, how would you prepare each of the following compounds?\n8-14 (a)\n\n(b)\n\n\n\n(c)"}
{"id": 486, "contents": "A saturated fatty - 8.8 Oxidation of Alkenes: Cleavage to Carbonyl Compounds\nIn all the alkene addition reactions we've seen thus far, the carbon-carbon double bond has been converted into a single bond but the carbon skeleton has been unchanged. There are, however, powerful oxidizing reagents that will cleave $\\mathrm{C}=\\mathrm{C}$ bonds and produce two carbonyl-containing fragments.\n\nOzone $\\left(\\mathrm{O}_{3}\\right)$ is perhaps the most useful double-bond cleavage reagent. Prepared by passing a stream of oxygen through a high-voltage electrical discharge, ozone adds rapidly to a $\\mathrm{C}=\\mathrm{C}$ bond at low temperature to give a cyclic intermediate called a molozonide. Once formed, the molozonide spontaneously rearranges to form an ozonide. Although we won't study the mechanism of this rearrangement in detail, it involves the molozonide coming apart into two fragments that then recombine in a different way.\n\n$$\n3 \\mathrm{O}_{2} \\xrightarrow[\\text { discharge }]{\\text { Electric }} 2 \\mathrm{O}_{3}\n$$\n\n\n\nLow-molecular-weight ozonides are explosive and therefore not isolated. Instead, the ozonide is immediately treated with a reducing agent, such as zinc metal in acetic acid, to produce carbonyl compounds. The net result of the ozonolysis/reduction sequence is that the $\\mathrm{C}=\\mathrm{C}$ bond is cleaved and an oxygen atom becomes doubly bonded to each of the original alkene carbons. If an alkene with a tetrasubstituted double bond is ozonized, two ketone fragments result; if an alkene with a trisubstituted double bond is ozonized, one ketone and one aldehyde result; and so on.\n\n\nIsopropylidenecyclohexane\n(tetrasubstituted)"}
{"id": 487, "contents": "Cyclohexanone Acetone - \n84\\%; two ketones\n\n\nSeveral oxidizing reagents other than ozone also cause double-bond cleavage, although such reactions are not often used. For example, potassium permanganate $\\left(\\mathrm{KMnO}_{4}\\right)$ in neutral or acidic solution cleaves alkenes to give carbonyl-containing products. If hydrogens are present on the double bond, carboxylic acids are produced; if two hydrogens are present on one carbon, $\\mathrm{CO}_{2}$ is formed.\n\n\n3,7-Dimethyl-1-octene 2,6-Dimethylheptanoic acid (45\\%)\nIn addition to direct cleavage with ozone or $\\mathrm{KMnO}_{4}$, an alkene can also be cleaved in a two-step process by initial hydroxylation to a 1,2-diol, as discussed in the previous section, followed by treatment of the diol with periodic acid, $\\mathrm{HIO}_{4}$. If the two -OH groups are in an open chain, two carbonyl compounds result. If the two -OH groups are on a ring, a single, open-chain dicarbonyl compound is formed. As indicated in the following examples, the cleavage reaction takes place through a cyclic periodate intermediate."}
{"id": 488, "contents": "Predicting the Reactant in an Ozonolysis Reaction - \nWhat alkene would yield a mixture of cyclopentanone and propanal on treatment with ozone followed by reduction with zinc?"}
{"id": 489, "contents": "Strategy - \nReaction of an alkene with ozone, followed by reduction with zinc, cleaves the $\\mathrm{C}=\\mathrm{C}$ bond and becomes two $\\mathrm{C}=\\mathrm{O}$ bonds. Working backward from the carbonyl-containing products, the alkene precursor can be found by removing the oxygen from each product and joining the two carbon atoms.\n\nSolution\n\n\nPROBLEM What products would you expect from reaction of 1-methylcyclohexene with the following 8-15 reagents?\n(a) Aqueous acidic $\\mathrm{KMnO}_{4}$\n(b) $\\mathrm{O}_{3}$, followed by $\\mathrm{Zn}, \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$\n\nPROBLEM Propose structures for alkenes that yield the following products on reaction with ozone followed by 8-16 treatment with Zn :\n(a) $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{C}=\\mathrm{O}+\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{O}$\n(b) 2 equiv $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}=\\mathrm{O}$"}
{"id": 490, "contents": "Strategy - 8.9 Addition of Carbenes to Alkenes: Cyclopropane Synthesis\nYet another kind of alkene addition is the reaction with a carbene to yield a cyclopropane. A carbene, $\\mathbf{R}_{\\mathbf{2}} \\mathbf{C}$ :, is a neutral molecule containing a divalent carbon with only six electrons in its valence shell. It is therefore highly reactive and generated only as a reaction intermediate, rather than as an isolable molecule. Because they're electron-deficient, carbenes behave as electrophiles and react with nucleophilic $\\mathrm{C}=\\mathrm{C}$ bonds. The reaction occurs in a single step without intermediates."}
{"id": 491, "contents": "An alkene A carbene A cyclopropane - \nOne of the simplest methods for generating a substituted carbene is by treatment of chloroform, $\\mathrm{CHCl}_{3}$, with a strong base such as KOH . As shown in FIGURE 8.10, the loss of a proton from $\\mathrm{CHCl}_{3}$ gives trichloromethanide anion, ${ }^{-}: \\mathrm{CCl}_{2}$, which spontaneously expels a $\\mathrm{Cl}^{-}$ion to yield dichlorocarbene, $: \\mathrm{CCl}_{2}$."}
{"id": 492, "contents": "FIGURE 8.10 MECHANISM - \nMechanism of the formation of dichlorocarbene by reaction of chloroform with strong base. Deprotonation of $\\mathrm{CHCl}_{3}$ gives the trichloromethanide anion, ${ }^{-}: \\mathrm{CCl}_{3}$, which spontaneously expels a $\\mathrm{Cl}^{-}$ion.\n\n\n1. Base abstracts the hydrogen from chloroform, leaving behind the electron pair from the $\\mathrm{C}-\\mathrm{H}$ bond and forming the trichloromethanide anion.\n\nChloroform\n(1)\n\nrichloromethanide\nanion\nSpontaneous loss of chloride ion then yields the neutral dichlorocarbene.\n\n\u0e51]\n\n\nDichlorocarbene\n\nThe carbon atom in dichlorocarbene is $s p^{2}$-hybridized, with a vacant $p$ orbital extending above and below the plane of the three atoms and with an unshared pair of electrons occupying the third $s p^{2}$ lobe. Note that this electronic description of dichlorocarbene is similar to that of a carbocation (Section 7.9) with respect to both the $s p^{2}$ hybridization of carbon and the vacant $p$ orbital. Electrostatic potential maps further illustrate the similarity (FIGURE 8.11).\n\n\nFIGURE 8.11 The structure of dichlorocarbene. Electrostatic potential maps show how the positive region coincides with the empty $p$ orbital in both dichlorocarbene and a carbocation $\\left(\\mathrm{CH}_{3}{ }^{+}\\right)$. The negative region in the dichlorocarbene map coincides with the lone-pair electrons.\n\nIf dichlorocarbene is generated in the presence of an alkene, addition to the double bond occurs and a dichlorocyclopropane is formed. As the reaction of dichlorocarbene with cis-2-pentene demonstrates, the addition is stereospecific, meaning that only a single stereoisomer is formed as product. Starting from a cis alkene, for instance, only cis-disubstituted cyclopropane is produced; starting from a trans alkene, only transdisubstituted cyclopropane is produced."}
{"id": 493, "contents": "FIGURE 8.10 MECHANISM - \nThe best method for preparing nonhalogenated cyclopropanes is by a process called the Simmons-Smith reaction. First investigated at the DuPont company, this reaction does not involve a free carbene. Rather, it utilizes a carbenoid-a metal-complexed reagent with carbene-like reactivity. When diiodomethane is treated with a specially prepared zinc-copper mix, (iodomethyl)zinc iodide, $\\mathrm{ICH}_{2} \\mathrm{ZnI}$, is formed. In the presence of an alkene, $\\mathrm{ICH}_{2} \\mathrm{ZnI}$ transfers a $\\mathrm{CH}_{2}$ group to the double bond to yield cyclopropane. For example, cyclohexene reacts cleanly and with good yield to give the corresponding cyclopropane. Although we won't discuss the mechanistic details, carbene addition to an alkene is one of a general class of reactions called cycloadditions, which we'll study more carefully in Chapter 30.\n\n$$\n\\mathrm{CH}_{2} \\mathrm{I}_{2}+\\mathrm{Zn}(\\mathrm{Cu}) \\longrightarrow \\mathrm{ICH} 2-\\mathrm{ZnI} \\quad\\left[\": \\mathrm{CH}_{2}^{\\prime \\prime \\prime}\\right]\n$$\n\nDiiodomethane\n(Iodomethyl)zinc iodide\n(a carbenoid)\n\n\nCyclohexene\n\nPROBLEM What products would you expect from the following reactions?\n\n8-17 (a)\n\n\n(b)"}
{"id": 494, "contents": "FIGURE 8.10 MECHANISM - 8.10 Radical Additions to Alkenes: Chain-Growth Polymers\nIn our brief introduction to radical reactions in Section 6.6 , we said that radicals can add to $\\mathrm{C}=\\mathrm{C}$ bonds, taking one electron from the double bond and leaving one behind to yield a new radical. Let's now look at the process in more detail, focusing on the industrial synthesis of alkene polymers. A polymer is a large-sometimes very large-molecule, built up by repetitive joining together of many smaller molecules, called monomers.\n\nNature makes wide use of biological polymers. Cellulose, for instance, is a polymer built of repeating glucose monomer units; proteins are polymers built of repeating amino acid monomers; and nucleic acids are polymers built of repeating nucleotide monomers."}
{"id": 495, "contents": "Cellulose-a glucose polymer - \nProtein-an amino acid polymer\n\n\nAn amino acid\nNucleic acid-a nucleotide polymer\n\n\nA nucleic acid\nSynthetic polymers, such as polyethylene, are much simpler chemically than biopolymers, but there is still a great diversity to their structures and properties, depending on the identity of the monomers and on the reaction conditions used for polymerization. The simplest synthetic polymers are those that result when an alkene is treated with a small amount of a suitable catalyst. Ethylene, for example, yields polyethylene, an enormous alkane that may have a molecular weight up to 6 million $u$ and may contain as many as 200,000 monomer units. Worldwide production of polyethylene is approximately 88 million tons per year."}
{"id": 496, "contents": "Polyethylene-a synthetic alkene polymer - \nPolyethylene and other simple alkene polymers are called chain-growth polymers because they are formed in a chain-reaction process in which an initiator adds to a carbon-carbon double bond to yield a reactive intermediate. The intermediate then reacts with a second molecule of monomer to yield a new intermediate, which reacts with a third monomer unit, and so on.\n\nHistorically, ethylene polymerization was carried out at high pressure (1000-3000 atm) and high temperature $\\left(100-250{ }^{\\circ} \\mathrm{C}\\right)$ in the presence of a radical initiator such as benzoyl peroxide. Like many radical reactions, the mechanism of ethylene polymerization occurs in three steps: initiation, propagation, and termination:\n\n- Initiation The polymerization reaction is initiated when a few radicals are generated on heating a small\namount of benzoyl peroxide catalyst to break the weak $\\mathrm{O}-\\mathrm{O}$ bond. The initially formed benzoyloxy radical loses $\\mathrm{CO}_{2}$ and gives a phenyl radical ( $\\mathrm{Ph} \\cdot$ ), which adds to the $\\mathrm{C}=\\mathrm{C}$ bond of ethylene to start the polymerization process. One electron from the ethylene double bond pairs up with the odd electron on the phenyl radical to form a new $\\mathrm{C}-\\mathrm{C}$ bond, and the other electron remains on carbon.\n\n\n- Propagation Polymerization occurs when the carbon radical formed in the initiation step adds to another ethylene molecule to yield another radical. Repetition of the process for hundreds or thousands of times builds the polymer chain.\n\n- Termination The chain process is eventually ended by a reaction that consumes the radical. The combination of two growing chains is one possible chain-terminating reaction.\n\n$$\n2 \\mathrm{R}-\\mathrm{CH}_{2} \\mathrm{CH}_{2} \\cdot \\longrightarrow \\mathrm{R}-\\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2}-\\mathrm{R}\n$$"}
{"id": 497, "contents": "Polyethylene-a synthetic alkene polymer - \nEthylene is not unique in its ability to form a polymer. Many substituted ethylenes, called vinyl monomers, also undergo polymerization to yield polymers with substituent groups regularly spaced on alternating carbon atoms along the chain. Propylene, for example, yields polypropylene, and styrene yields polystyrene.\n\n\nWhen an unsymmetrically substituted vinyl monomer such as propylene or styrene is polymerized, the radical addition steps can take place at either end of the double bond to yield either a primary radical intermediate $\\left(\\mathrm{RCH}_{2} \\cdot\\right)$ or a secondary radical $\\left(\\mathrm{R}_{2} \\mathrm{CH} \\cdot\\right)$. Just as in electrophilic addition reactions, however, we find that only the more highly substituted, secondary radical is formed.\n\n\nTABLE 8.1 shows some commercially important alkene polymers, their uses, and the monomers from which they are made.\n\nTABLE 8.1 Some Alkene Polymers and Their Uses\n\n| Monomer |  | Trade or common name of <br> polymer |  |\n| :--- | :--- | :--- | :--- |\n| Ethylene | $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}_{2}$ | Polyethylene | Uses |\n| Propene (propylene) | $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}_{3}$ | Polypropylene | Moldings, rope, <br> carpets |\n| Chloroethylene (vinyl <br> chloride) | $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCl}$ | Poly(vinyl chloride) | Insulation, films, pipes |\n| Styrene | $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHC}_{6} \\mathrm{H}_{5}$ | Polystyrene | Foam, moldings |\n| Tetrafluoroethylene | $\\mathrm{F}_{2} \\mathrm{C}=\\mathrm{CF}_{2}$ | Teflon | Gaskets, nonstick |\n| coatings |  |  |  |"}
{"id": 498, "contents": "Predicting the Structure of a Polymer - \nShow the structure of poly(vinyl chloride), a polymer made from $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCl}$, by drawing several repeating units."}
{"id": 499, "contents": "Strategy - \nMentally break the carbon-carbon double bond in the monomer unit, and form single bonds by connecting numerous units together."}
{"id": 500, "contents": "Solution - \nThe general structure of poly(vinyl chloride) is\n\n\nPROBLEM Show the monomer units you would use to prepare the following polymers:"}
{"id": 501, "contents": "8-18 (a) - \n(b)\n\n\nPROBLEM One of the chain-termination steps that sometimes occurs to interrupt polymerization is the\n8-19 following reaction between two radicals. Propose a mechanism for the reaction, using fishhook arrows to indicate electron flow.\n\n2\n\n$$\n\\geqslant \\mathrm{CH}_{2} \\dot{\\mathrm{C}} \\mathrm{H}_{2} \\longrightarrow \\quad \\geqslant \\mathrm{CH}_{2} \\mathrm{CH}_{3}+\\frac{>}{<} \\mathrm{CH}=\\mathrm{CH}_{2}\n$$"}
{"id": 502, "contents": "8-18 (a) - 8.11 Biological Additions of Radicals to Alkenes\nThe same high reactivity of radicals that enables the alkene polymerization we saw in the previous section also makes it difficult to carry out controlled radical reactions on complex molecules. As a result, there are severe limitations on the usefulness of radical addition reactions in the laboratory. In contrast to an electrophilic addition, where reaction occurs once and the reactive cation intermediate is rapidly quenched by a nucleophile, the reactive intermediate in a radical reaction is not usually quenched. Instead, it reacts again and again in a largely uncontrollable way."}
{"id": 503, "contents": "Radical addition - \n(Intermediate is not quenched, so reaction does not stop.)\n\n\n\nIn biological reactions, the situation is different from that in the laboratory. Only one substrate molecule at a time is present in the active site of an enzyme, and that molecule is held in a precise position, with other necessary reacting groups nearby. As a result, biological radical reactions are more controlled and more common than laboratory or industrial radical reactions. A particularly impressive example occurs in the biosynthesis of prostaglandins from arachidonic acid, where a sequence of four radical additions take place. Its reaction mechanism was discussed briefly in Section 6.6.\n\nAs shown in FIGURE 8.12, prostaglandin biosynthesis begins with abstraction of a hydrogen atom from C13 of arachidonic acid by an iron-oxy radical to give a carbon radical that reacts with $\\mathrm{O}_{2}$ at C 11 through a resonance form. The oxygen radical that results adds to the $\\mathrm{C} 8-\\mathrm{C} 9$ double bond to give a carbon radical at C8, which adds to the $\\mathrm{C} 12-\\mathrm{C} 13$ double bond and gives a carbon radical at C 13 . A resonance form of this carbon radical adds at C 15 to a second $\\mathrm{O}_{2}$ molecule, completing the prostaglandin skeleton. Reduction of the $\\mathrm{O}-\\mathrm{O}$ bond then gives prostaglandin $\\mathrm{H}_{2}$, called $\\mathrm{PGH}_{2}$. The pathway looks complicated, but the entire process is catalyzed with exquisite control by a single enzyme.\n\n\nArachidonic acid\n\n\n\n\n\nFIGURE 8.12 Pathway for the biosynthesis of prostaglandins from arachidonic acid. Steps 2 and 5 are radical addition reactions to $\\mathrm{O}_{2}$; steps 3 and 4 are radical additions to carbon-carbon double bonds."}
{"id": 504, "contents": "Radical addition - 8.12 Reaction Stereochemistry: Addition of $\\mathrm{H}_{2} \\mathrm{O}$ to an Achiral Alkene\nMost of the biochemical reactions that take place in the body, as well as many organic reactions in the laboratory, yield products with chirality centers. For example, acid-catalyzed addition of $\\mathrm{H}_{2} \\mathrm{O}$ to 1-butene in the laboratory yields 2 -butanol, a chiral alcohol. What is the stereochemistry of this chiral product? If a single enantiomer is formed, is it $R$ or $S$ ? If a mixture of enantiomers is formed, how much of each? In fact, the 2 -butanol produced is a racemic mixture of $R$ and $S$ enantiomers. Let's see why.\n\n\nTo understand why a racemic product results from the reaction of $\\mathrm{H}_{2} \\mathrm{O}$ with 1-butene, think about the reaction\nmechanism. 1-Butene is first protonated to yield an intermediate secondary carbocation. Because the trivalent carbon is $s p^{2}$-hybridized and planar, the cation has a plane of symmetry and is achiral. As a result, it can react with $\\mathrm{H}_{2} \\mathrm{O}$ equally well from either the top or the bottom. Reaction from the top leads to ( $S$ )-2-butanol through transition state 1 (TS 1) in FIGURE 8.13, and reaction from the bottom leads to ( $R$ )-2-butanol through TS 2. But the two transition states are mirror images, so they have identical energies, form at identical rates, and are equally likely to occur.\n\n\nFIGURE 8.13 Reaction of $\\mathbf{H}_{2} \\mathbf{O}$ with the carbocation resulting from protonation of 1-butene. Reaction from the top leads to $S$ product and is the mirror image of reaction from the bottom, which leads to $R$ product. Because they are energetically identical, they are equally likely and lead to a racemic mixture of products. The dotted $\\mathrm{C} \\cdots \\mathrm{O}$ bond in the transition state indicates partial bond formation."}
{"id": 505, "contents": "Radical addition - 8.12 Reaction Stereochemistry: Addition of $\\mathrm{H}_{2} \\mathrm{O}$ to an Achiral Alkene\nAs a general rule, the formation of a new chirality center by achiral reactants always leads to a racemic mixture of enantiomeric products. Put another way, optical activity can't appear from nowhere; an optically active product can only result by starting with an optically active reactant or chiral environment (Section 5.12).\n\nIn contrast to laboratory reactions, enzyme-catalyzed biological reactions often give a single enantiomer of a chiral product, even when the substrate is achiral. One step in the citric acid cycle of food metabolism, for instance, is the aconitase-catalyzed addition of water to (Z)-aconitate (usually called cis-aconitate) to give isocitrate.\n\n\nEven though cis-aconitate is achiral, only the ( $2 R, 3 S$ ) enantiomer of the product is formed. As discussed in Section 5.11 and Section 5.12, cis-aconitate is a prochiral molecule, which is held in a chiral environment by the aconitase enzyme during the reaction. In this environment, the two faces of the double bond are chemically distinct, and addition occurs on only the Re face at C2."}
{"id": 506, "contents": "Radical addition - 8.13 Reaction Stereochemistry: Addition of $\\mathrm{H}_{2} \\mathrm{O}$ to a Chiral Alkene\nThe reaction discussed in the previous section involves an addition to an achiral reactant and forms an optically inactive, racemic mixture of two enantiomeric products. What would happen, though, if we were to carry out the reaction on a single enantiomer of a chiral reactant? For example, what stereochemical result would be obtained from addition of $\\mathrm{H}_{2} \\mathrm{O}$ to a chiral alkene, such as $(R)$-4-methyl-1-hexene? The product of the reaction, 4-methyl-2-hexanol, has two chirality centers and so has four possible stereoisomers.\n\n\nLet's think about the two chirality centers separately. What about the configuration at C4, the methyl-bearing carbon atom? Since C 4 has the $R$ configuration in the starting material and this chirality center is unaffected by the reaction, its configuration is unchanged. Thus, the configuration at C 4 in the product remains $R$ (assuming that the relative rankings of the four attached groups are not changed by the reaction).\n\nWhat about the configuration at C2, the newly formed chirality center? As shown in FIGURE 8.14, the stereochemistry at C 2 is established by reaction of $\\mathrm{H}_{2} \\mathrm{O}$ with a carbocation intermediate in the usual manner. But this carbocation doesn't have a plane of symmetry; it is chiral because of the chirality center at C4. Because the carbocation is chiral and has no plane of symmetry, it doesn't react equally well from the top and bottom faces. One of the two faces is likely, for steric reasons, to be a bit more accessible than the other, leading to a mixture of $R$ and $S$ products in some ratio other than $50: 50$. Thus, two diastereomeric products, ( $2 R, 4 R$ )-4-methyl-2-hexanol and ( $2 S, 4 R$ )-4-methyl-2-hexanol, are formed in unequal amounts, and the mixture is optically active."}
{"id": 507, "contents": "Radical addition - 8.13 Reaction Stereochemistry: Addition of $\\mathrm{H}_{2} \\mathrm{O}$ to a Chiral Alkene\n(2S,4R)-4-Methyl-2-hexanol\n(2R,4R)-4-Methyl-2-hexanol\nFIGURE 8.14 Stereochemistry of the acid-catalyzed addition of $\\mathbf{H}_{\\mathbf{2}} \\mathbf{O}$ to the chiral alkene, ( $R$ )-4-methyl-1-hexene. A mixture of diastereomeric $2 R, 4 R$ and $2 S, 4 R$ products is formed in unequal amounts because reaction of the chiral carbocation intermediate is not equally likely from top and bottom. The product mixture is optically active.\n\nAs a general rule, the formation of a new chirality center by a chiral reactant leads to unequal amounts of diastereomeric products. If the chiral reactant is optically active because only one enantiomer is used rather than a racemic mixture, then the products are also optically active.\n\nPROBLEM What products are formed from acid-catalyzed hydration of racemic ( $\\pm$ )-4-methyl-1-hexene? What 8-20 can you say about the relative amounts of the products? Is the product mixture optically active?\n\nPROBLEM What products are formed from hydration of 4-methylcyclopentene? What can you say about the 8-21 relative amounts of the products?"}
{"id": 508, "contents": "Terpenes: Naturally Occurring Alkenes - \nEver since its discovery in Persia around 1000 A.D., it has been known that steam distillation, the distillation of plant materials together with water, produces a fragrant mixture of liquids called essential oils. The resulting oils have long been used as medicines, spices, and perfumes, and their investigation played a major role in the emergence of organic chemistry as a science during the 19th century.\n\n\nFIGURE 8.15 The wonderful fragrance of leaves from the California bay laurel is due primarily to myrcene, a simple terpene. (credit: \"California Bay Umbellularia californica\" by Don Loaire/Flickr, CC BY 2.0)\n\nChemically, plant essential oils consist largely of mixtures of compounds called terpenoids-small organic molecules with an immense diversity of structure. More than 60,000 different terpenoids are known. Some are open-chain molecules, and others contain rings; some are hydrocarbons, and others contain oxygen. Hydrocarbon terpenoids, in particular, are known as terpenes, and all contain double bonds. For example:\n\n\nMyrcene (oil of bay)\n\n$\\alpha$-Pinene (turpentine)\n\n\nHumulene (oil of hops)\n\n$\\beta$-Santalene (sandalwood oil)\nthe isoprene rule, they can be thought of as arising from head-to-tail joining of 5 -carbon isoprene units ( 2 -methyl-1,3-butadiene). Carbon 1 is the head of the isoprene unit, and carbon 4 is the tail. For example, myrcene contains two isoprene units joined head to tail, forming an 8-carbon chain with two 1-carbon branches. $\\alpha$-Pinene similarly contains two isoprene units assembled into a more complex cyclic structure, and humulene contains three isoprene units. See if you can identify the isoprene units in $\\alpha$-pinene, humulene, and $\\beta$-santalene."}
{"id": 509, "contents": "Terpenes: Naturally Occurring Alkenes - \nTerpenes (and terpenoids) are further classified according to the number of 5 -carbon units they contain. Thus, monoterpenes are 10-carbon substances derived from two isoprene units, sesquiterpenes are 15-carbon molecules derived from three isoprene units, diterpenes are 20-carbon substances derived from four isoprene units, and so on. Monoterpenes and sesquiterpenes are found primarily in plants, but the higher terpenoids occur in both plants and animals, and many have important biological roles. The triterpenoid lanosterol, for instance, is the biological precursor from which all steroid hormones are made.\n\n\nLanosterol\n(a triterpene, $\\mathrm{C}_{30}$ )\nIsoprene itself is not the true biological precursor of terpenoids. Nature instead uses two \"isoprene equivalents\"-isopentenyl diphosphate and dimethylallyl diphosphate-which are themselves made by two different routes depending on the organism. Lanosterol, in particular, is biosynthesized from acetic acid by a complex pathway that has been worked out in great detail. We'll look at the subject more closely in Sections 27.5 and 27.7 .\n\n\nIsopentenyl diphosphate\n\n\nDimethylallyl diphosphate"}
{"id": 510, "contents": "Key Terms - \n- anti stereochemistry\n- bromonium ion\n- carbene, $\\mathrm{R}_{2} \\mathrm{C}$\n- chain-growth polymer\n- epoxide\n- glycol\n- halohydrin\n- hydroboration\n- hydrogenated\n- hydroxylation\n- monomer\n- oxidation\n- oxirane\n- oxymercuration-demercuration\n- ozonide\n- polymer\n- reduction\n- Simmons-Smith reaction\n- stereospecific\n- syn stereochemistry"}
{"id": 511, "contents": "Summary - \nWith the background needed to understand organic reactions now covered, this chapter has begun the systematic description of major functional groups.\nAlkenes are generally prepared by an elimination reaction, such as dehydrohalogenation, the elimination of HX from an alkyl halide, or dehydration, the elimination of water from an alcohol. The converse of this elimination reaction is the addition of various substances to the alkene double bond to give saturated products.\n$\\mathrm{HCl}, \\mathrm{HBr}$, and HI add to alkenes by a two-step electrophilic addition mechanism. Initial reaction of the nucleophilic double bond with $\\mathrm{H}^{+}$gives a carbocation intermediate, which then reacts with halide ion. Bromine and chlorine add to alkenes via three-membered-ring bromonium ion or chloronium ion intermediates to give addition products having anti stereochemistry. If water is present during the halogen addition reaction, a halohydrin is formed.\n\nHydration of an alkene-the addition of water-is carried out by either of two procedures, depending on the product desired. Oxymercuration-demercuration involves electrophilic addition of $\\mathrm{Hg}^{2+}$ to an alkene, followed by trapping of the cation intermediate with water and subsequent treatment with $\\mathrm{NaBH}_{4}$. Hydroboration involves addition of borane $\\left(\\mathrm{BH}_{3}\\right)$ followed by oxidation of the intermediate organoborane with alkaline $\\mathrm{H}_{2} \\mathrm{O}_{2}$. The two hydration methods are complementary: oxymercuration-demercuration gives the product of Markovnikov addition, whereas hydroboration-oxidation gives the product with non-Markovnikov syn stereochemistry."}
{"id": 512, "contents": "Summary - \nAlkenes are reduced by addition of $\\mathrm{H}_{2}$ in the presence of a catalyst such as platinum or palladium to yield alkanes, a process called catalytic hydrogenation. Alkenes are also oxidized by reaction with a peroxyacid to give epoxides, which can be converted into trans-1,2-diols by acid-catalyzed hydrolysis. The corresponding cis-1,2-diols can be made directly from alkenes by hydroxylation with $\\mathrm{OsO}_{4}$. Alkenes can also be cleaved to produce carbonyl compounds by reaction with ozone, followed by reduction with zinc metal. In addition, alkenes react with divalent substances called carbenes, $\\mathbf{R}_{\\mathbf{2}} \\mathbf{C}$ :, to give cyclopropanes. Nonhalogenated cyclopropanes are best prepared by treatment of the alkene with $\\mathrm{CH}_{2} \\mathrm{I}_{2}$ and zinc-copper, a process called the Simmons-Smith reaction.\n\nAlkene polymers-large molecules resulting from repetitive bonding of many hundreds or thousands of small monomer units-are formed by chain-reaction polymerization of simple alkenes. Polyethylene, polypropylene, and polystyrene are examples. As a general rule, radical addition reactions are not common in the laboratory but occur frequently in biological pathways.\n\nMany reactions give chiral products. If the reactants are optically inactive, the products are also optically inactive. If one or both of the reactants is optically active, the products can also be optically active."}
{"id": 513, "contents": "LEARNING REACTIONS - \nWhat's seven times nine? Sixty-three, of course. You didn't have to stop and figure it out; you knew the answer immediately because you long ago learned the multiplication tables. Learning the reactions of organic chemistry requires the same approach: reactions have to be learned for quick recall if they are to be useful.\n\nDifferent people take different approaches to learning reactions. Some people make flashcards; others find studying with friends to be helpful. To help guide your study, most chapters in this book end with a summary of the reactions just presented. In addition, the accompanying Study Guide and Student Solutions Manual has several appendixes that organize organic reactions from other perspectives. Fundamentally, though, there are no shortcuts. Learning organic chemistry does take effort."}
{"id": 514, "contents": "Summary of Reactions - \nNo stereochemistry is implied unless specifically indicated with wedged, solid, and dashed lines.\n\n1. Addition reactions of alkenes\na. Addition of $\\mathrm{HCl}, \\mathrm{HBr}$, and HI (Section 7.7 and Section 7.8)\n\nMarkovnikov regiochemistry occurs, with H adding to the less highly substituted alkene carbon and halogen adding to the more highly substituted carbon.\n\nb. Addition of halogens $\\mathrm{Cl}_{2}$ and $\\mathrm{Br}_{2}$ (Section 8.2)\n\nAnti addition is observed through a halonium ion intermediate.\n\nc. Halohydrin formation (Section 8.3)\n\nMarkovnikov regiochemistry and anti stereochemistry occur.\n\nd. Addition of water by oxymercuration-demercuration (Section 8.4) Markovnikov regiochemistry occurs.\n\n\ne. Addition of water by hydroboration-oxidation (Section 8.5) Non-Markovnikov syn addition occurs.\n\n\nf. Catalytic hydrogenation (Section 8.6)\n\nSyn addition occurs.\n\ng. Epoxidation with a peroxyacid (Section 8.7)\n\nSyn addition occurs.\n\nh. Hydroxylation with $\\mathrm{OsO}_{4}$ (Section 8.7)\n\nSyn addition occurs.\n\ni. Addition of carbenes to yield cyclopropanes (Section 8.9)\n(1) Dichlorocarbene addition\n\n(2) Simmons-Smith reaction\n\n2. Hydroxylation by acid-catalyzed epoxide hydrolysis (Section 8.7)\n\nAnti stereochemistry occurs.\n\n3. Oxidative cleavage of alkenes (Section 8.8)\na. Reaction with ozone followed by zinc in acetic acid\n\nb. Reaction with $\\mathrm{KMnO}_{4}$ in acidic solution\n\n\n4. Cleavage of 1,2 -diols (Section 8.8)"}
{"id": 515, "contents": "Visualizing Chemistry - \nPROBLEM Name the following alkenes, and predict the products of their reaction with (1)\n8-22 meta-chloroperoxybenzoic acid, (2) $\\mathrm{KMnO}_{4}$ in aqueous acid, (3) $\\mathrm{O}_{3}$, followed by Zn in acetic acid:\n(a)\n\n(b)\n\n\nPROBLEM Draw the structures of alkenes that would yield the following alcohols on hydration (red = 0). Tell in\n8-23 each case whether you would use hydroboration-oxidation or oxymercuration-demercuration.\n(a)\n\n(b)\n\n\nPROBLEM The following alkene undergoes hydroboration-oxidation to yield a single product rather than a\n8-24 mixture. Explain the result, and draw the product showing its stereochemistry.\n\n\nPROBLEM From what alkene was the following 1,2-diol made, and what method was used, epoxide hydrolysis $\\mathbf{8 - 2 5}$ or $\\mathrm{OsO}_{4}$ ?"}
{"id": 516, "contents": "Mechanism Problems - \nPROBLEM Predict the products for the following reactions, showing the complete mechanism and appropriate 8-26 stereochemistry:\n(a)\n\n\n(b)\n\n(c)\n\n\nPROBLEM Draw the structures of the organoboranes formed when borane reacts with the following alkenes,\n8-27 including the regiochemistry and stereochemistry as appropriate. Propose a mechanism for each reaction.\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM meta-Chlorobenzoic acid is not the only peroxyacid capable of epoxide formation. For each reaction 8-28 below, predict the products and show the mechanism.\n\n(a)\n\n$+\\mathrm{CF}_{3} \\mathrm{CO}_{3} \\mathrm{H} \\longrightarrow$\n(b)\n\n\nPROBLEM Give the mechanism and products for the following acid-catalyzed epoxide-opening reactions,\n8-29 including appropriate stereochemistry.\n(a)\n\n(b)\n\n(c)\n\n$\\qquad$ ?\n\nPROBLEM Which of the reactions below would result in a product mixture that would rotate plane-polarized 8-30 light?\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Reaction of 2-methylpropene with $\\mathrm{CH}_{3} \\mathrm{OH}$ in the presence of $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ catalyst yields methyl tert-butyl\n8-31 ether, $\\mathrm{CH}_{3} \\mathrm{OC}\\left(\\mathrm{CH}_{3}\\right)_{3}$, by a mechanism analogous to that of acid-catalyzed alkene hydration. Write the mechanism, using curved arrows for each step.\n\nPROBLEM Iodine azide, $\\mathrm{IN}_{3}$, adds to alkenes by an electrophilic mechanism similar to that of bromine. If a\n8-32 monosubstituted alkene such as 1-butene is used, only one product results:"}
{"id": 517, "contents": "Mechanism Problems - \n(a) Add lone-pair electrons to the structure shown for $\\mathrm{IN}_{3}$, and draw a second resonance form for the molecule.\n(b) Calculate formal charges for the atoms in both resonance structures you drew for $\\mathrm{IN}_{3}$ in part (a).\n(c) In light of the result observed when $\\mathrm{IN}_{3}$ adds to 1-butene, what is the polarity of the $\\mathrm{I}-\\mathrm{N}_{3}$ bond? Propose a mechanism for the reaction using curved arrows to show the electron flow in each step.\n\nPROBLEM 10-Bromo- $\\alpha$-chamigrene, a compound isolated from marine algae, is thought to be biosynthesized\n8-33 from $\\gamma$-bisabolene by the following route:\n\n\nDraw the structures of the intermediate bromonium and cyclic carbocation, and propose mechanisms for all three steps.\n\nPROBLEM Isolated from marine algae, prelaureatin is thought to be biosynthesized from laurediol by the\n8-34 following route. Propose a mechanism.\n\n\nPROBLEM Dichlorocarbene can be generated by heating sodium trichloroacetate. Propose a mechanism for\n8-35 the reaction, and use curved arrows to indicate the movement of electrons in each step. What relationship does your mechanism bear to the base-induced elimination of HCl from chloroform?\n\n\nPROBLEM Reaction of cyclohexene with mercury(II) acetate in $\\mathrm{CH}_{3} \\mathrm{OH}$ rather than $\\mathrm{H}_{2} \\mathrm{O}$, followed by treatment 8-36 with $\\mathrm{NaBH}_{4}$, yields cyclohexyl methyl ether rather than cyclohexanol. Suggest a mechanism.\n\n\nCyclohexene\nCyclohexyl methyl ether\n\nPROBLEM Use your general knowledge of alkene chemistry to suggest a mechanism for the following reaction.\n8-37\n\n\nPROBLEM Treatment of 4-penten-1-ol with aqueous $\\mathrm{Br}_{2}$ yields a cyclic bromo ether rather than the expected\n8-38 bromohydrin. Suggest a mechanism, using curved arrows to show electron movement."}
{"id": 518, "contents": "Mechanism Problems - \n4-Penten-1-ol 2-(Bromomethyl)tetrahydrofuran\nPROBLEM Hydroboration of 2-methyl-2-pentene at $25{ }^{\\circ} \\mathrm{C}$, followed by oxidation with alkaline $\\mathrm{H}_{2} \\mathrm{O}_{2}$, yields\n8-39 2-methyl-3-pentanol, but hydroboration at $160{ }^{\\circ} \\mathrm{C}$ followed by oxidation yields 4-methyl-1-pentanol. Suggest a mechanism."}
{"id": 519, "contents": "Reactions of Alkenes - \nPROBLEM Predict the products of the following reactions (the aromatic ring is unreactive in all cases). Indicate\n8-40 regiochemistry when relevant.\n\n\nPROBLEM Suggest structures for alkenes that give the following reaction products. There may be more than 8-41 one answer for some cases.\n(a)\n? $\\xrightarrow{\\mathrm{H}_{2} / \\mathrm{Pd}}$\n\n(b)\n\n(c)\n$? \\xrightarrow{\\mathrm{Br}_{2}}$\n\n(d)\n\n\n(e)\n\n(f)\n\n\n\nPROBLEM Predict the products of the following reactions, showing both regiochemistry and stereochemistry\n8-42 where appropriate:\n(a)\n\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM Which reaction would you expect to be faster, addition of HBr to cyclohexene or to\n8-43 1-methylcyclohexene? Explain.\nPROBLEM What product will result from hydroboration-oxidation of 1-methylcyclopentene with deuterated\n8-44 borane, $\\mathrm{BD}_{3}$ ? Show both the stereochemistry (spatial arrangement) and the regiochemistry (orientation) of the product.\n\nPROBLEM The cis and trans isomers of 2-butene give different cyclopropane products in the Simmons-Smith\n8-45 reaction. Show the structures of both, and explain the difference.\n\n\nPROBLEM Predict the products of the following reactions. Don't worry about the size of the molecule;\n8-46 concentrate on the functional groups.\n\n\nPROBLEM Addition of HCl to 1-methoxycyclohexene yields 1-chloro-1-methoxycyclohexane as a sole product.\n8-47 Use resonance structures of the carbocation intermediate to explain why none of the alternate regioisomer is formed.\n\n$\\xrightarrow{\\mathrm{HCl}}$\n\n\n1-Methoxycyclohexene\n\n\n1-Chloro-1-methoxy-\ncyclohexane\n\nSynthesis Using Alkenes\nPROBLEM How would you carry out the following transformations? What reagents would you use in each case?\n\n8-48 (a)\n\n\n(b)\n\n(c)\n\n(d)"}
{"id": 520, "contents": "Reactions of Alkenes - \nSynthesis Using Alkenes\nPROBLEM How would you carry out the following transformations? What reagents would you use in each case?\n\n8-48 (a)\n\n\n(b)\n\n(c)\n\n(d)\n\n\n\n\nPROBLEM Draw the structure of an alkene that yields only acetone, $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{C}=\\mathrm{O}$, on ozonolysis followed by\n8-49 treatment with Zn .\nPROBLEM Show the structures of alkenes that give the following products on oxidative cleavage with $\\mathrm{KMnO}_{4}$\n8-50 in acidic solution:\n(a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}+\\mathrm{CO}_{2}$\n(b) $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{C}=\\mathrm{O}+\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$\n(c)\n\n(d)\n\n\nPROBLEM In planning the synthesis of one compound from another, it's just as important to know what not to\n8-51 do as to know what to do. The following reactions all have serious drawbacks to them. Explain the potential problems of each.\n(a)\n\n(b)\n\n\n(c)\n\n(d)\n\n\n\n\nPROBLEM Which of the following alcohols could not be made selectively by hydroboration-oxidation of an 8-52 alkene? Explain.\n(a)\n\n(b)\n\n(c)\n\n(d)"}
{"id": 521, "contents": "Polymers - \nPROBLEM Plexiglas, a clear plastic used to make many molded articles, is made by polymerization of methyl\n8-53 methacrylate. Draw a representative segment of Plexiglas.\n\n\nMethyl methacrylate\n\nPROBLEM Poly(vinyl pyrrolidone), prepared from $N$-vinylpyrrolidone, is used both in cosmetics and as a\n8-54 component of a synthetic substitute for blood. Draw a representative segment of the polymer.\n\n$N$-Vinylpyrrolidone\n\nPROBLEM When a single alkene monomer, such as ethylene, is polymerized, the product is a homopolymer.\n8-55 If a mixture of two alkene monomers is polymerized, however, a copolymer often results. The following structure represents a segment of a copolymer called Saran. What two monomers were copolymerized to make Saran?\n\n\nSaran"}
{"id": 522, "contents": "General Problems - \nPROBLEM Compound A has the formula $\\mathrm{C}_{10} \\mathrm{H}_{16}$. On catalytic hydrogenation over palladium, it reacts with\n8-56 only 1 molar equivalent of $\\mathrm{H}_{2}$. Compound $\\mathbf{A}$ also undergoes reaction with ozone, followed by zinc treatment, to yield a symmetrical diketone, $\\mathbf{B}\\left(\\mathrm{C}_{10} \\mathrm{H}_{16} \\mathrm{O}_{2}\\right)$.\n(a) How many rings does $\\mathbf{A}$ have? (b) What are the structures of $\\mathbf{A}$ and $\\mathbf{B}$ ?\n(c) Write the reactions.\n\nPROBLEM An unknown hydrocarbon $\\mathbf{A}$ with the formula $\\mathrm{C}_{6} \\mathrm{H}_{12}$ reacts with 1 molar equivalent of $\\mathrm{H}_{2}$ over a\n8-57 palladium catalyst. Hydrocarbon $\\mathbf{A}$ also reacts with $\\mathrm{OsO}_{4}$ to give diol $\\mathbf{B}$. When oxidized with $\\mathrm{KMnO}_{4}$\nin acidic solution, A gives two fragments. One fragment is propanoic acid, $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$, and the other fragment is ketone $\\mathbf{C}$. What are the structures of $\\mathbf{A}, \\mathbf{B}$, and $\\mathbf{C}$ ? Write all reactions.\n\nPROBLEM Using an oxidative cleavage reaction, explain how you would distinguish between the following two\n8-58 isomeric dienes:"}
{"id": 523, "contents": "General Problems - \nPROBLEM Using an oxidative cleavage reaction, explain how you would distinguish between the following two\n8-58 isomeric dienes:\n\n\nPROBLEM Compound $\\mathbf{A}, \\mathrm{C}_{10} \\mathrm{H}_{18} \\mathrm{O}$, undergoes reaction with dilute $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ at $50{ }^{\\circ} \\mathrm{C}$ to yield a mixture of two\n8-59 alkenes, $\\mathrm{C}_{10} \\mathrm{H}_{16}$. The major alkene product, $\\mathbf{B}$, gives only cyclopentanone after ozone treatment followed by reduction with zinc in acetic acid. Identify $\\mathbf{A}$ and $\\mathbf{B}$, and write the reactions."}
{"id": 524, "contents": "Cyclopentanone - \nPROBLEM Draw the structure of a hydrocarbon that absorbs 2 molar equivalents of $\\mathrm{H}_{2}$ on catalytic 8-60 hydrogenation and gives only butanedial on ozonolysis.\n\n\nButanedial\n\nPROBLEM Simmons-Smith reaction of cyclohexene with diiodomethane gives a single cyclopropane product,\n8-61 but the analogous reaction of cyclohexene with 1,1-diiodoethane gives (in low yield) a mixture of two isomeric methylcyclopropane products. What are the two products, and how do they differ?\n\nPROBLEM The sex attractant of the common housefly is a hydrocarbon with the formula $\\mathrm{C}_{23} \\mathrm{H}_{46}$. On treatment\n8-62 with aqueous acidic $\\mathrm{KMnO}_{4}$, two products are obtained, $\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{12} \\mathrm{CO}_{2} \\mathrm{H}$ and $\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{7} \\mathrm{CO}_{2} \\mathrm{H}$. Propose a structure.\n\nPROBLEM Compound A has the formula $\\mathrm{C}_{8} \\mathrm{H}_{8}$. It reacts rapidly with $\\mathrm{KMnO}_{4}$ to give $\\mathrm{CO}_{2}$ and a carboxylic\n8-63 acid, B $\\left(\\mathrm{C}_{7} \\mathrm{H}_{6} \\mathrm{O}_{2}\\right)$, but reacts with only 1 molar equivalent of $\\mathrm{H}_{2}$ on catalytic hydrogenation over a palladium catalyst. On hydrogenation under conditions that reduce aromatic rings, 4 equivalents of $\\mathrm{H}_{2}$ are taken up and hydrocarbon $\\mathbf{C}\\left(\\mathrm{C}_{8} \\mathrm{H}_{16}\\right)$ is produced. What are the structures of $\\mathbf{A}, \\mathbf{B}$, and $\\mathbf{C}$ ? Write the reactions."}
{"id": 525, "contents": "Cyclopentanone - \nPROBLEM How would you distinguish between the following pairs of compounds using simple chemical tests?\n8-64 Tell what you would do and what you would see.\n(a) Cyclopentene and cyclopentane\n(b) 2-Hexene and benzene\n\nPROBLEM $\\alpha$-Terpinene, $\\mathrm{C}_{10} \\mathrm{H}_{16}$, is a pleasant-smelling hydrocarbon that has been isolated from oil of\n8-65 marjoram. On hydrogenation over a palladium catalyst, $\\alpha$-terpinene reacts with 2 molar equivalents of $\\mathrm{H}_{2}$ to yield a hydrocarbon, $\\mathrm{C}_{10} \\mathrm{H}_{20}$. On ozonolysis, followed by reduction with zinc and acetic acid, $\\alpha$-terpinene yields two products, glyoxal and 6-methyl-2,5-heptanedione.\n\n\nGlyoxal\n\n\n6-Methyl-2,5-heptanedione\n(a) How many degrees of unsaturation does $\\alpha$-terpinene have?\n(b) How many double bonds and how many rings does it have?\n(c) Propose a structure for $\\alpha$-terpinene.\n\nPROBLEM Evidence that cleavage of 1,2-diols by $\\mathrm{HIO}_{4}$ occurs through a five-membered cyclic periodate\n8-66 intermediate is based on the measurement of reaction rates. When diols $\\mathbf{A}$ and $\\mathbf{B}$ were prepared and the rates of their reaction with $\\mathrm{HIO}_{4}$ were measured, it was found that diol A cleaved approximately 1 million times faster than diol B. Make molecular models of $\\mathbf{A}$ and $\\mathbf{B}$ and of potential cyclic\nperiodate intermediates, and then explain the results.\n\nA\n(cis diol)\n\nB\n(trans diol)"}
{"id": 526, "contents": "Cyclopentanone - \nA\n(cis diol)\n\nB\n(trans diol)\n\nPROBLEM Reaction of HBr with 3-methylcyclohexene yields a mixture of four products: cis- and\n8-67 trans-1-bromo-3-methylcyclohexane and cis- and trans-1-bromo-2-methylcyclohexane. The analogous reaction of HBr with 3-bromocyclohexene yields trans-1,2-dibromocyclohexane as the sole product. Draw structures of the possible intermediates, and then explain why only a single product is formed in this reaction.\n\n\nPROBLEM We'll see in the next chapter that alkynes undergo many of the same reactions that alkenes do. What\n8-68 product might you expect from each of the following reactions?\n\n$\\begin{cases}\\text { (a) } \\xrightarrow{\\text { 1 equiv } \\mathrm{Br}_{2}} & \\text { ? } \\\\ \\text { (b) } \\xrightarrow[\\text { equiv } \\mathrm{H}_{2}, \\mathrm{Pd} / \\mathrm{C}]{ } & \\text { ? } \\\\ \\text { (c) } \\xrightarrow{\\text { 1 equiv } \\mathrm{HBr}} & \\text { ? }\\end{cases}$\nPROBLEM Hydroxylation of cis-2-butene with $\\mathrm{OsO}_{4}$ yields a different product than hydroxylation of\n8-69 trans-2-butene. Draw the structure, show the stereochemistry of each product, and explain the difference between them."}
{"id": 527, "contents": "Cyclopentanone - \nPROBLEM Compound $\\mathbf{A}, \\mathrm{C}_{11} \\mathrm{H}_{16} \\mathrm{O}$, was found to be an optically active alcohol. Despite its apparent\n8-70 unsaturation, no hydrogen was absorbed on catalytic reduction over a palladium catalyst. On treatment of $\\mathbf{A}$ with dilute sulfuric acid, dehydration occurred and an optically inactive alkene $\\mathbf{B}, \\mathrm{C}_{11} \\mathrm{H}_{14}$, was the major product. Alkene $\\mathbf{B}$, on ozonolysis, gave two products. One product was identified as propanal, $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CHO}$. Compound $\\mathbf{C}$, the other product, was shown to be a ketone, $\\mathrm{C}_{8} \\mathrm{H}_{8} \\mathrm{O}$. How many degrees of unsaturation does $\\mathbf{A}$ have? Write the reactions, and identify $\\mathbf{A}, \\mathbf{B}$, and C."}
{"id": 528, "contents": "Alkynes: An Introduction to Organic Synthesis - \nFIGURE 9.1 Synthesizing organic compounds is like conducting a musical group. When in tune, chemists can create highly complex organic compounds. (credit: modification of work \"Jazz great visits Navy\" by U.S. Navy, Michael Worner/Wikimedia Commons, Public Domain)"}
{"id": 529, "contents": "CHAPTER CONTENTS - 9.1 Naming Alkynes\n9.2 Preparation of Alkynes: Elimination Reactions of Dihalides\n9.3 Reactions of Alkynes: Addition of HX and $\\mathrm{X}_{2}$\n9.4 Hydration of Alkynes\n9.5 Reduction of Alkynes\n9.6 Oxidative Cleavage of Alkynes\n9.7 Alkyne Acidity: Formation of Acetylide Anions\n9.8 Alkylation of Acetylide Anions\n9.9 An Introduction to Organic Synthesis\n\nWHY THIS CHAPTER? Alkynes are less common than alkenes, both in the laboratory and in living organisms, so we won't cover them in great detail. The real importance of this chapter is that we'll use alkyne chemistry as a vehicle to begin looking at some of the general strategies used in organic synthesis-the construction of complex molecules in the laboratory. Without the ability to design and synthesize new molecules in the laboratory, many of the medicines we take for granted would not exist and few new ones would be made.\n\nAn alkyne is a hydrocarbon that contains a carbon-carbon triple bond. Acetylene, $\\mathrm{H}-\\mathrm{C} \\equiv \\mathrm{C}-\\mathrm{H}$, the simplest alkyne, was once widely used in industry as a starting material for the preparation of acetaldehyde, acetic acid, vinyl chloride, and other high-volume chemicals, but more efficient routes to these substances using ethylene as starting material are now available. Acetylene is still used in the preparation of acrylic polymers, such as\n\nPlexiglas and Lucite, but is probably best known as the gas burned in high-temperature oxy-acetylene welding torches.\n\nIn addition to simple alkynes with one triple bond, research is also being carried out on polyynes-linear carbon chains of alternating single and triple bonds. Polyynes with up to eight triple bonds are thought to be present in interstellar space, and evidence has been presented for the existence of carbyne, an allotrope of carbon consisting of alternating single and triple bonds in long chains of indefinite length. The electronic properties of polyynes are being explored for potential use in nanotechnology applications.\n\n\nA polyyne detected in interstellar space"}
{"id": 530, "contents": "CHAPTER CONTENTS - 9.1 Naming Alkynes\nAlkyne nomenclature follows the general rules for hydrocarbons discussed in Section 3.4 and Section 7.3. The suffix -yne is used, and the position of the triple bond is indicated by giving the number of the first alkyne carbon in the chain. Numbering the main chain begins at the end nearer the triple bond so that the triple bond receives as low a number as possible.\n\n\nBegin numbering at the end nearer the triple bond.\n\n> 6-Methyl-3-octyne\n> (New: 6-Methyloct-3-yne)\n\nCompounds with more than one triple bond are called diynes, triynes, and so forth; compounds containing both double and triple bonds are called enynes (not ynenes). Numbering of an enyne chain starts from the end nearer the first multiple bond, whether double or triple. When there is a choice in numbering, double bonds receive lower numbers than triple bonds. For example:\n\n\n1-Hepten-6-yne\n(New: Hept-1-en-6-yne)\n\n\n4-Methyl-7-nonen-1-yne\n(New: 4-Methylnon-7-en-1-yne)\n\nAs with alkyl and alkenyl substituents derived from alkanes and alkenes, respectively, alkynyl groups are also possible.\n\n\n\n1-Butynyl\n(an alkynyl group)\n(New: But-1-ynyl)\nPROBLEM Name the following alkynes:\n9-1 (a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n$\\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{CHCH}=\\mathrm{CHC} \\equiv \\mathrm{CCH}_{3}$\n\nPROBLEM There are seven isomeric alkynes with the formula $\\mathrm{C}_{6} \\mathrm{H}_{10}$. Draw and name them.\n9-2"}
{"id": 531, "contents": "CHAPTER CONTENTS - 9.2 Preparation of Alkynes: Elimination Reactions of Dihalides\nAlkynes can be prepared by the elimination of HX from alkyl halides in a similar manner as alkenes (Section 8.1). Treatment of a 1,2-dihaloalkane (called a vicinal dihalide) with an excess amount of a strong base such as KOH or $\\mathrm{NaNH}_{2}$ results in a twofold elimination of HX and formation of an alkyne. As with the elimination of HX to form an alkene, we'll defer a full discussion of this topic and the relevant reaction mechanisms to Chapter 11.\n\nThe starting vicinal dihalides are themselves readily available by addition of $\\mathrm{Br}_{2}$ or $\\mathrm{Cl}_{2}$ to alkenes. Thus, the overall halogenation/dehydrohalogenation sequence makes it possible to go from an alkene to an alkyne. For example, diphenylethylene is converted into diphenylacetylene by reaction with $\\mathrm{Br}_{2}$ and subsequent base treatment.\n\n\nThe twofold dehydrohalogenation takes place through a vinylic halide intermediate, which suggests that vinylic halides themselves should give alkynes when treated with strong base. (A vinylic substituent is one that is attached to a double-bond.) This is indeed the case. For example:"}
{"id": 532, "contents": "CHAPTER CONTENTS - 9.3 Reactions of Alkynes: Addition of HX and $\\mathrm{X}_{2}$\nYou might recall from Section 1.9 that a carbon-carbon triple bond results from the interaction of two $s p$-hybridized carbon atoms. The two $s p$ hybrid orbitals of carbon lie at an angle of $180^{\\circ}$ to each other along an axis perpendicular to the axes of the two unhybridized $2 p_{\\mathrm{y}}$ and $2 p_{\\mathrm{z}}$ orbitals. When two $s p$-hybridized carbons approach each other, one $s p-s p \\sigma$ bond and two $p-p \\pi$ bonds are formed. The two remaining $s p$ orbitals form bonds to other atoms at an angle of $180^{\\circ}$ from the carbon-carbon bond. Thus, acetylene is a linear molecule with $\\mathrm{H}-\\mathrm{C} \\equiv \\mathrm{C}$ bond angles of $180^{\\circ}$ (FIGURE 9.2). The length of the $\\mathrm{C} \\equiv \\mathrm{C}$ bond is 120 pm , and its strength is approximately $965 \\mathrm{~kJ} / \\mathrm{mol}(231 \\mathrm{kcal} / \\mathrm{mol})$, making it the shortest and strongest known carbon-carbon bond.\n\n\nFIGURE 9.2 The structure of acetylene, $\\mathbf{H}-\\mathbf{C} \\equiv \\mathbf{C}-\\mathbf{H}$. The $\\mathbf{H}-\\mathbf{C} \\equiv \\mathbf{C}$ bond angles are $\\mathbf{1 8 0}^{\\circ}$, and the $\\mathbf{C} \\equiv \\mathbf{C}$ bond length is $\\mathbf{1 2 0}$ pm. The electrostatic potential map shows that the $\\pi$ bonds create a negative belt around the molecule."}
{"id": 533, "contents": "CHAPTER CONTENTS - 9.3 Reactions of Alkynes: Addition of HX and $\\mathrm{X}_{2}$\nAs a general rule, electrophiles undergo addition reactions with alkynes much as they do with alkenes. Take the reaction of alkynes with HX, for instance. The reaction often can be stopped with the addition of 1 equivalent of HX, but reaction with an excess of HX leads to a dihalide product. For example, reaction of 1-hexyne with 2 equivalents of HBr yields 2,2-dibromohexane. As the following examples indicate, the regiochemistry of addition follows Markovnikov's rule, with halogen adding to the more highly substituted side of the alkyne bond\nand hydrogen adding to the less highly substituted side. Trans stereochemistry of H and X normally, although not always, occurs in the product.\n\nHBr addition\n\n\n1-Hexyne\n2-Bromo-1-hexene\n2,2-Dibromohexane\nHCl addition\n\n\nBromine and chlorine also add to alkynes to give addition products, and trans stereochemistry again results.\n$\\mathrm{Br}_{2}$ addition\n\n\n1-Butyne (E)-1,2-Dibromo-1-butene 1,1,2,2-Tetrabromobutane\nThe mechanism of alkyne addition is similar but not identical to that of alkene addition. When an electrophile such as HBr adds to an alkene, the reaction takes place in two steps and involves an alkyl carbocation intermediate (Section 7.7 and Section 7.8). If HBr were to add by the same mechanism to an alkyne, an analogous vinylic carbocation would be formed as the intermediate.\n\n\nA vinylic carbocation has an $s p$-hybridized carbon and generally forms less readily than an alkyl carbocation (FIGURE 9.3). As a rule, a secondary vinylic carbocation forms about as readily as a primary alkyl carbocation, but a primary vinylic carbocation is so difficult to form that there is no clear evidence it even exists. Thus, many alkyne additions occur through more complex mechanistic pathways.\n\n\n\nA $2^{\\circ}$ vinylic carbocation\n\n\nA $\\mathbf{2}^{\\circ}$ alkyl carbocation"}
{"id": 534, "contents": "CHAPTER CONTENTS - 9.3 Reactions of Alkynes: Addition of HX and $\\mathrm{X}_{2}$\nA $2^{\\circ}$ vinylic carbocation\n\n\nA $\\mathbf{2}^{\\circ}$ alkyl carbocation\n\nFIGURE 9.3 The structure of a secondary vinylic carbocation. The cationic carbon atom is sp-hybridized and has a vacant $p$ orbital perpendicular to the plane of the $\\pi$ bond orbitals. Only one R group is attached to the positively charged carbon rather than two, as in a secondary alkyl carbocation. The electrostatic potential map shows that the most positive regions coincide with lobes of the vacant $p$ orbital and are perpendicular to the most negative regions associated with the $\\pi$ bond.\n\nPROBLEM What products would you expect from the following reactions?"}
{"id": 535, "contents": "CHAPTER CONTENTS - 9.4 Hydration of Alkynes\nLike alkenes (Section 8.4 and Section 8.5), alkynes can be hydrated by either of two methods. Direct addition of water catalyzed by mercury(II) ion yields the Markovnikov product, and indirect addition of water by a hydroboration-oxidation sequence yields the non-Markovnikov product."}
{"id": 536, "contents": "Mercury (II)-Catalyzed Hydration of Alkynes - \nAlkynes don't react directly with aqueous acid but will undergo hydration readily in the presence of mercury(II) sulfate as a Lewis acid catalyst. The reaction occurs with Markovnikov regiochemistry, so the - OH group adds to the more highly substituted carbon and the -H attaches to the less highly substituted one."}
{"id": 537, "contents": "An enol - \n2-Hexanone (78\\%)\nInterestingly, the actual product isolated from alkyne hydration is not a vinylic alcohol, or enol (ene $+o l$ ), but is instead a ketone. Although the enol is an intermediate in the reaction, it immediately rearranges into a ketone by a process called keto-enol tautomerism. The individual keto and enol forms are said to be tautomers, a word used to describe two isomers that undergo spontaneous interconversion accompanied by the change in position of a hydrogen. With few exceptions, the keto-enol tautomeric equilibrium lies on the side of the ketone; enols are almost never isolated. We'll look more closely at this equilibrium in Section 22.1.\n\n\nAs shown in FIGURE 9.4, the mechanism of the mercury(II)-catalyzed alkyne hydration reaction is analogous to the oxymercuration reaction of alkenes (Section 8.4). Electrophilic addition of mercury(II) ion to the alkyne gives a vinylic cation, which reacts with water and loses a proton to yield a mercury-containing enol intermediate. In contrast with alkene oxymercuration, however, no treatment with $\\mathrm{NaBH}_{4}$ is necessary to remove the mercury. The acidic reaction conditions alone are sufficient to effect replacement of mercury by\nhydrogen. Tautomerization then gives the ketone.\nFIGURE 9.4 MECHANISM\nMechanism of the mercury(II)-catalyzed hydration of an alkyne to yield a\nketone. The reaction occurs through initial formation of an intermediate enol, which tautomerizes to the ketone.\n(1) The alkyne uses a pair of electrons to attack the electrophilic mercury(II) ion, yielding a mercury-containing\n\nvinylic carbocation intermediate.\n(2) Nucleophilic attack of water on the carbocation forms a $\\mathrm{C}-\\mathrm{O}$ bond and yields a protonated mercurycontaining enol.\n\n$\\bigcirc$\n\n-\n\n(4) Replacement of $\\mathrm{Hg}^{2+}$ by $\\mathrm{H}^{+}$ occurs to give a neutral enol."}
{"id": 538, "contents": "An enol - \n$\\bigcirc$\n\n-\n\n(4) Replacement of $\\mathrm{Hg}^{2+}$ by $\\mathrm{H}^{+}$ occurs to give a neutral enol.\n\n$$\n\\text { (4) } \\mathrm{H}_{3} \\mathrm{O}^{+}\n$$\n\n\n(5)\n\n\nA mixture of both possible ketones results when an unsymmetrically substituted internal alkyne ( $\\mathrm{RC} \\equiv \\mathrm{CR}^{\\prime}$ ) is hydrated. The reaction is therefore most useful when applied to a terminal alkyne ( $\\mathrm{RC} \\equiv \\mathrm{CH}$ ) because only a methyl ketone is formed."}
{"id": 539, "contents": "A terminal alkyne - \nA methyl ketone\nPROBLEM What products would you obtain by mercury-catalyzed hydration of the following alkynes?\n\n\nPROBLEM What alkynes would you start with to prepare the following ketones?\n\n(b)"}
{"id": 540, "contents": "Hydroboration-Oxidation of Alkynes - \nBorane adds rapidly to an alkyne just as it does to an alkene, and the resulting vinylic borane can be oxidized by $\\mathrm{H}_{2} \\mathrm{O}_{2}$ to give an enol, which tautomerizes to either a ketone or an aldehyde, depending on the alkyne. Hydroboration-oxidation of an internal alkyne such as 3-hexyne is straightforward and gives a ketone, but hydroboration-oxidation of a terminal alkyne is more complex because two molecules of borane often add to the triple bond, complicating the situation. To prevent this double addition, a bulky, sterically encumbered borane such as bis(1,2-dimethylpropyl)borane, known commonly as disiamylborane is used in place of $\\mathrm{BH}_{3}$. When a terminal alkyne such as 1-butene reacts with disiamylborane, addition to the triple bond occurs normally, but a second addition is hindered by the bulk of the dialkylborane. Oxidation with $\\mathrm{H}_{2} \\mathrm{O}_{2}$ then gives an enol, which tautomerizes to the aldehyde.\n\n\nThe hydroboration-oxidation sequence is complementary to the direct, mercury(II)-catalyzed hydration reaction of a terminal alkyne because different products result. Direct hydration with aqueous acid and mercury(II) sulfate leads to a methyl ketone, whereas hydroboration-oxidation of the same terminal alkyne\nleads to an aldehyde.\n\n\nPROBLEM What alkyne would you start with to prepare each of the following compounds by a 9-6 hydroboration-oxidation reaction?\n(a)\n\n(b)\n\n\nPROBLEM How would you prepare the following carbonyl compounds starting from an alkyne (reddish brown\n9-7 = Br)?"}
{"id": 541, "contents": "Hydroboration-Oxidation of Alkynes - 9.5 Reduction of Alkynes\nAlkynes are reduced to alkanes by addition of $\\mathrm{H}_{2}$ over a metal catalyst. The reaction occurs in two steps through an alkene intermediate, and measurements show that the first step in the reaction is more exothermic than the second.\n\n\nComplete reduction to the alkane occurs when palladium on carbon ( $\\mathrm{Pd} / \\mathrm{C}$ ) is used as catalyst, but hydrogenation can be stopped at the alkene stage if the less active Lindlar catalyst is used. The Lindlar catalyst is a finely divided palladium metal that has been precipitated onto a calcium carbonate support and then deactivated by treatment with lead acetate and quinoline, an aromatic amine. The hydrogenation occurs with syn stereochemistry (Section 8.5), giving a cis alkene product.\n\n\n\nThe alkyne hydrogenation reaction has been explored extensively by the Hoffmann-LaRoche pharmaceutical company, where it is used in the commercial synthesis of vitamin A. The cis isomer of vitamin A produced initially on hydrogenation is converted to the trans isomer by heating.\n\n\n7-cis-Retinol\n(7-cis-vitamin A; vitamin A has\na trans double bond at C7)\nAn alternative method for the conversion of an alkyne to an alkene uses sodium or lithium metal as the reducing agent in liquid ammonia as solvent. This method is complementary to the Lindlar reduction because it produces trans rather than cis alkenes. For example, 5-decyne gives trans-5-decene on treatment with lithium in liquid ammonia. The mechanism is explained below."}
{"id": 542, "contents": "5-Decyne - \ntrans-5-Decene (78\\%)\nAlkali metals dissolve in liquid ammonia at $-33^{\\circ} \\mathrm{C}$ to produce a deep blue solution containing the metal cation and ammonia-solvated electrons. When an alkyne is then added to the solution, reduction occurs by the mechanism shown in FIGURE 9.5. An electron first adds to the triple bond to yield an intermediate anion radical-a species that is both an anion (has a negative charge) and a radical (has an odd number of electrons). This anion radical is a strong base, able to remove $\\mathrm{H}^{+}$from ammonia to give a vinylic radical. Addition of a second electron to the vinylic radical gives a vinylic anion, which abstracts a second $\\mathrm{H}^{+}$from ammonia to give trans alkene product.\n\nFIGURE 9.5 MECHANISM\nMechanism of the lithium/ammonia reduction of an alkyne to produce a trans alkene.\n(1) Lithium metal donates an electron to the alkyne to give an anion radical . . .\n2. . . which abstracts a proton from ammonia solvent to yield a vinylic radical.\n\n\nThe vinylic radical accepts another electron from a second lithium atom to produce a vinylic anion . . .\n4. . . which abstracts another proton from ammonia solvent to yield the final trans alkene product.\n\n\n\nA trans alkene\n\nTrans stereochemistry of the alkene product is established during the second reduction step (3) when the less-hindered trans vinylic anion is formed from the vinylic radical. Vinylic radicals undergo rapid cis-trans equilibration, but vinylic anions equilibrate much less rapidly. Thus, the more stable trans vinylic anion is formed rather than the less stable cis anion and is then protonated without equilibration.\n\nPROBLEM Using any alkyne needed, how would you prepare the following alkenes?\n9-8 (a) trans-2-Octene\n(b) cis-3-Heptene\n(c) 3-Methyl-1-pentene"}
{"id": 543, "contents": "5-Decyne - 9.6 Oxidative Cleavage of Alkynes\nAlkynes, like alkenes, can be cleaved by reaction with powerful oxidizing agents such as ozone or $\\mathrm{KMnO}_{4}$, although the reaction is of little value and it is mentioned only for completeness. A triple bond is generally less reactive than a double bond, and yields of cleavage products can be low. The products obtained from cleavage of an internal alkyne are carboxylic acids; from a terminal alkyne, $\\mathrm{CO}_{2}$ is formed as one product."}
{"id": 544, "contents": "A terminal alkyne - 9.7 Alkyne Acidity: Formation of Acetylide Anions\nThe most striking difference between alkenes and alkynes is that terminal alkynes are relatively acidic. When a terminal alkyne is treated with a strong base, such as sodium amide, $\\mathrm{Na}^{+}{ }^{-} \\mathrm{NH}_{2}$, the terminal hydrogen is removed and the corresponding acetylide anion is formed."}
{"id": 545, "contents": "Acetylide anion - \nAccording to the Br\u00f8nsted-Lowry definition (Section 2.7), an acid is a substance that donates $\\mathrm{H}^{+}$. Although we usually think of oxyacids $\\left(\\mathrm{H}_{2} \\mathrm{SO}_{4}, \\mathrm{HNO}_{3}\\right)$ or halogen acids $(\\mathrm{HCl}, \\mathrm{HBr})$ in this context, any compound containing a hydrogen atom can be an acid under the right circumstances. By measuring dissociation constants of different acids and expressing the results as $\\mathrm{p} K_{\\mathrm{a}}$ values, an acidity order can be established. Recall from Section 2.8 that a lower $\\mathrm{p} K_{\\mathrm{a}}$ corresponds to a stronger acid and a higher $\\mathrm{p} K_{\\mathrm{a}}$ corresponds to a weaker one.\n\nWhere do hydrocarbons lie on the acidity scale? As the data in TABLE 9.1 show, both methane ( $\\mathrm{p} K_{\\mathrm{a}} \\approx 60$ ) and ethylene $\\left(\\mathrm{p} K_{\\mathrm{a}}=44\\right)$ are very weak acids and thus do not react with any of the common bases. Acetylene, however, has $\\mathrm{p} K_{\\mathrm{a}}=25$ and can be deprotonated by the conjugate base of any acid whose $\\mathrm{p} K_{\\mathrm{a}}$ is greater than 25 . Amide ion $\\left(\\mathrm{NH}_{2}{ }^{-}\\right.$), for example, the conjugate base of ammonia ( $\\mathrm{p} K_{\\mathrm{a}}=35$ ), is often used to deprotonate terminal alkynes.\n\nTABLE 9.1 Acidity of Simple Hydrocarbons"}
{"id": 546, "contents": "Acetylide anion - \nTABLE 9.1 Acidity of Simple Hydrocarbons\n\n| Family | Example | $K_{\\mathrm{a}}$ |  | $\\mathrm{p} K_{\\mathrm{a}}$ |\n| :--- | :--- | :--- | :--- | :---: |\n| Alkyne | $\\mathrm{HC} \\equiv \\mathrm{CH}$ | $10^{-25}$ | 25 | Stronger <br> acid |\n| Alkene | $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}_{2}$ | $10^{-44}$ | 44 |  |\n| Alkane | $\\mathrm{CH}_{4}$ | $10^{-60}$ | 60 | Weaker <br> acid |\n\nWhy are terminal alkynes more acidic than alkenes or alkanes? In other words, why are acetylide anions more stable than vinylic or alkyl anions? The simplest explanation involves the hybridization of the negatively charged carbon atom. An acetylide anion has an sp-hybridized carbon, so the negative charge resides in an orbital that has $50 \\%$ \" $s$ character.\" A vinylic anion has a $s p^{2}$-hybridized carbon with $33 \\% s$ character, and an alkyl anion $\\left(s p^{3}\\right)$ has only $25 \\% s$ character. Because $s$ orbitals are nearer the positive nucleus and lower in energy than $p$ orbitals, the negative charge is stabilized to a greater extent in an orbital with higher $s$ character (FIGURE 9.6).\n\n\nFIGURE 9.6 A comparison of alkyl, vinylic, and acetylide anions. The acetylide anion, with $s p$ hybridization, has more $s$ character and is more stable. Electrostatic potential maps show that placing the negative charge closer to the carbon nucleus makes carbon appear less negative (red)."}
{"id": 547, "contents": "Acetylide anion - \nPROBLEM The $\\mathrm{p} K_{\\mathrm{a}}$ of acetone, $\\mathrm{CH}_{3} \\mathrm{COCH}_{3}$, is 19.3. Which of the following bases is strong enough to 9-9 deprotonate acetone?\n(a) $\\mathrm{KOH}\\left(\\mathrm{p} K_{\\mathrm{a}}\\right.$ of $\\left.\\mathrm{H}_{2} \\mathrm{O}=15.7\\right)$\n(b) $\\mathrm{Na}^{+-} \\mathrm{C} \\equiv \\mathrm{CH}\\left(\\mathrm{p} K_{\\mathrm{a}}\\right.$ of $\\left.\\mathrm{C}_{2} \\mathrm{H}_{2}=25\\right)$\n(c) $\\mathrm{NaHCO}_{3}\\left(\\mathrm{p} K_{\\mathrm{a}}\\right.$ of $\\left.\\mathrm{H}_{2} \\mathrm{CO}_{3}=6.4\\right)$\n(d) $\\mathrm{NaOCH}_{3}\\left(\\mathrm{p} K_{\\mathrm{a}}\\right.$ of $\\left.\\mathrm{CH}_{3} \\mathrm{OH}=15.6\\right)$"}
{"id": 548, "contents": "Acetylide anion - 9.8 Alkylation of Acetylide Anions\nThe negative charge and unshared electron pair on carbon make an acetylide anion strongly nucleophilic. As a result, an acetylide anion can react with electrophiles, such as alkyl halides, in a process that replaces the halide and yields a new alkyne product.\n\n\nWe won't study the details of this substitution reaction until Chapter 11, but for now you can picture it as happening by the pathway shown in FIGURE 9.7. The nucleophilic acetylide ion uses an electron pair to form a bond to the positively polarized, electrophilic carbon atom of bromomethane. As the new $\\mathrm{C}-\\mathrm{C}$ bond forms, $\\mathrm{Br}^{-}$ departs, taking with it the electron pair from the former $\\mathrm{C}-\\mathrm{Br}$ bond and yielding propyne as product. We call such a reaction an alkylation because a new alkyl group has become attached to the starting alkyne."}
{"id": 549, "contents": "A mechanism for the alkylation reaction of acetylide anion with bromomethane to give propyne. - \n(1) The nucleophilic acetylide anion uses its electron lone pair to form a bond to the positively polarized, electrophilic carbon atom of bromomethane. As the new $C-C$ bond begins to form, the $\\mathrm{C}-\\mathrm{Br}$ bond begins to break in the transition state.\n\n(1)\n\n\nTransition state\n(2) The new $\\mathrm{C}-\\mathrm{C}$ bond is fully formed and the old $\\mathrm{C}-\\mathrm{Br}$ bond is fully broken at the end of the reaction.\n(2)\n\n\nAlkyne alkylation is not limited to acetylene itself. Any terminal alkyne can be converted into its corresponding anion and then allowed to react with an alkyl halide to give an internal alkyne product. Hex-1-yne, for instance, gives dec-5-yne when treated first with $\\mathrm{NaNH}_{2}$ and then with 1-bromobutane.\n\n\nBecause of its generality, acetylide alkylation is a good method for preparing substituted alkynes from simpler precursors. A terminal alkyne can be prepared by alkylation of acetylene itself, and an internal alkyne can be prepared by further alkylation of a terminal alkyne.\n\n\nAcetylene\nA terminal alkyne"}
{"id": 550, "contents": "A terminal alkyne - \nAn internal alkyne\nThe only limit to the alkylation reaction is that it can only use primary alkyl bromides and alkyl iodides because acetylide ions are sufficiently strong bases to cause elimination instead of substitution when they react with secondary and tertiary alkyl halides. For example, reaction of bromocyclohexane with propyne anion yields the elimination product cyclohexene rather than the substitution product 1-propynylcyclohexane.\n\n\nPROBLEM Show the terminal alkyne and alkyl halide from which the following products can be obtained. If 9-10 two routes look feasible, list both.\n(a)\n$\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{C} \\equiv \\mathrm{CCH}_{3}$\n(b)\n$\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CHC} \\equiv \\mathrm{CCH}_{2} \\mathrm{CH}_{3}$\n(c)\n\n\nPROBLEM How would you prepare cis-2-butene starting from propyne, an alkyl halide, and any other reagents\n9-11 needed? This problem can't be worked in a single step. You'll have to carry out more than one reaction."}
{"id": 551, "contents": "A terminal alkyne - 9.9 An Introduction to Organic Synthesis\nAs mentioned in the introduction, one of the purposes of this chapter is to use alkyne chemistry as a vehicle to begin looking at some of the general strategies used in organic synthesis-the construction of complex molecules in the laboratory. There are many reasons for carrying out the laboratory synthesis of an organic compound. In the pharmaceutical industry, new molecules are designed and synthesized in the hope that some might be useful new drugs. In the chemical industry, syntheses are done to devise more economical routes to known compounds. In academic laboratories, the synthesis of extremely complex molecules is sometimes done just for the intellectual challenge involved in mastering so difficult a subject. The successful synthesis route is a highly creative work that is sometimes described by such subjective terms as elegant or beautiful.\n\nIn this book, too, we will often devise syntheses of molecules from simpler precursors, but the purpose here is to learn. The ability to plan a successful multistep synthetic sequence requires a working knowledge of the uses and limitations of many different organic reactions. Furthermore, it requires the practical ability to piece together the steps in a sequence such that each reaction does only what is desired without causing changes elsewhere in the molecule. Planning a synthesis makes you approach a chemical problem in a logical way, draw on your knowledge of chemical reactions, and organize that knowledge into a workable plan-it helps you learn organic chemistry.\n\nThere's no secret to planning an organic synthesis: all it takes is a knowledge of the different reactions and some practice. The only real trick is to work backward in what is often called a retrosynthetic direction. Don't look at a potential starting material and ask yourself what reactions it might undergo. Instead, look at the final product and ask, \"What was the immediate precursor of that product?\" For example, if the final product is an alkyl halide, the immediate precursor might be an alkene, to which you could add HX. If the final product is a cis alkene, the immediate precursor might be an alkyne, which you could hydrogenate using the Lindlar catalyst. Having found an immediate precursor, work backward again, one step at a time, until you get back to the starting material. You have to keep the starting material in mind, of course, so that you can work back to it, but you don't want that starting material to be your main focus.\n\nLet's work several examples of increasing complexity."}
{"id": 552, "contents": "Devising a Synthesis Route - \nHow would you synthesize cis-2-hexene from 1-pentyne and an alkyl halide? More than one step is needed."}
{"id": 553, "contents": "Strategy - \nWhen undertaking any synthesis problem, you should look at the product, identify the functional groups it contains, and then ask yourself how those functional groups can be prepared. Always work retrosynthetically, one step at a time.\n\nThe product in this case is a cis-disubstituted alkene, so the first question is, \"What is an immediate precursor of a cis-disubstituted alkene?\" We know that an alkene can be prepared from an alkyne by reduction and that the right choice of experimental conditions will allow us to prepare either a trans-disubstituted alkene (using lithium in liquid ammonia) or a cis-disubstituted alkene (using catalytic hydrogenation over the Lindlar catalyst). Thus, reduction of 2 -hexyne by catalytic hydrogenation using the Lindlar catalyst should yield cis-2-hexene."}
{"id": 554, "contents": "cis-2-Hexene - \nNext ask, \"What is an immediate precursor of 2-hexyne?\" We've seen that an internal alkyne can be prepared by alkylation of a terminal alkyne anion. In the present instance, we're told to start with 1-pentyne and an alkyl halide. Thus, alkylation of the anion of 1-pentyne with iodomethane should yield 2-hexyne."}
{"id": 555, "contents": "Solution - \ncis-2-Hexene can be synthesized from the given starting materials in three steps.\n\ncis-2-Hexene"}
{"id": 556, "contents": "Devising a Synthesis Route - \nHow would you synthesize 2-bromopentane from acetylene and an alkyl halide? More than one step is needed."}
{"id": 557, "contents": "Strategy - \nIdentify the functional group in the product (an alkyl bromide) and work the problem retrosynthetically. What is an immediate precursor of an alkyl bromide? Perhaps an alkene plus HBr. Of the two possibilities, Markovnikov addition of HBr to 1-pentene looks like a better choice than addition to 2 -pentene because the latter reaction would give a mixture of isomers.\n\n\nWhat is an immediate precursor of an alkene? Perhaps an alkyne, which could be reduced.\n\n$$\n\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{C} \\equiv \\mathrm{CH} \\xrightarrow[\\text { Lindlar catalyst }]{\\mathrm{H}_{2}} \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}=\\mathrm{CH}_{2}\n$$\n\nWhat is an immediate precursor of a terminal alkyne? Perhaps sodium acetylide and an alkyl halide.\n\n$$\n\\mathrm{Na}^{+}: \\overline{\\mathrm{C}} \\equiv \\mathrm{CH}+\\mathrm{BrCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3} \\longrightarrow \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{C} \\equiv \\mathrm{CH}\n$$"}
{"id": 558, "contents": "Solution - \nThe desired product can be synthesized in four steps from acetylene and 1-bromopropane."}
{"id": 559, "contents": "Devising a Synthesis Route - \nHow would you synthesize 5-methyl-1-hexanol (5-methyl-1-hydroxyhexane) from acetylene and an alkyl halide?\n\n\nAcetylene Alkyl 5-Methyl-1-hexanol halide"}
{"id": 560, "contents": "Strategy - \nWhat is an immediate precursor of a primary alcohol? Perhaps a terminal alkene, which could be hydrated with non-Markovnikov regiochemistry by reaction with borane followed by oxidation with $\\mathrm{H}_{2} \\mathrm{O}_{2}$.\n\n\nWhat is an immediate precursor of a terminal alkene? Perhaps a terminal alkyne, which could be reduced.\n\n\nWhat is an immediate precursor of 5-methyl-1-hexyne? Perhaps acetylene and 1-bromo-3-methylbutane."}
{"id": 561, "contents": "Solution - \nThe synthesis can be completed in four steps from acetylene and 1-bromo-3-methylbutane:\n\n\n5-Methyl-1-hexanol\n\nPROBLEM Beginning with 4-octyne as your only source of carbon, and using any inorganic reagents necessary,\n9-12 how would you synthesize the following compounds?\n(a) cis-4-Octene\n(b) Butanal (c) 4-Bromooctane\n(d) 4-Octanol\n(e) 4,5-Dichlorooctane\n(f) Butanoic acid\n\nPROBLEM Beginning with acetylene and any alkyl halide needed, how would you synthesize the following 9-13 compounds?\n(a) Decane\n(b) 2,2-Dimethylhexane\n(c) Hexanal\n(d) 2-Heptanone"}
{"id": 562, "contents": "CRO CHEMISTRY MATTERS - \nThe Art of Organic Synthesis\n\n\nFIGURE 9.8 Vitamin $B_{12}$ has been synthesized from scratch in the laboratory, but the bacteria growing on sludge from municipal sewage plants do a much better job. (credit: \"Aeration and sludge-wasting\" by U.S. Department of Agriculture/Flickr, Public Domain)\n\nIf you think some of the synthesis problems at the end of this chapter are difficult, try devising a synthesis of vitamin $\\mathrm{B}_{12}$ starting only from simple substances you can buy in a chemical catalog. This extraordinary achievement was reported in 1973 as the culmination of a collaborative effort headed by Robert B. Woodward of Harvard University and Albert Eschenmoser of the Swiss Federal Institute of Technology in Z\u00fcrich. More than 100 graduate students and postdoctoral associates contributed to the work, which took more than a decade to complete.\n\n\nVitamin $B_{12}$\nWhy put such extraordinary effort into the laboratory synthesis of a molecule so easily obtained from natural sources? There are many reasons. On a basic human level, a chemist might be motivated primarily by the challenge, much as a climber might be challenged by the ascent of a difficult peak. Beyond the pure challenge, the completion of a difficult synthesis is also valuable in that it establishes new standards and raises the field to a new level. If vitamin $B_{12}$ can be made, then why can't any molecule found in nature be made? Indeed, the decades that have passed since the work of Woodward and Eschenmoser have seen the laboratory synthesis of many enormously complex and valuable substances. Sometimes these substances-for instance, the anticancer compound paclitaxel, trade named Taxol-are not easily available in nature, so laboratory synthesis is the only\nmethod for obtaining larger quantities."}
{"id": 563, "contents": "CRO CHEMISTRY MATTERS - \nBut perhaps the most important reason for undertaking a complex synthesis is that, in so doing, new reactions and new chemistry are discovered. It invariably happens in a complex synthesis that a point is reached at which the planned route fails. At such a time, the only alternatives are either to quit or to devise a way around the difficulty. New reactions and new principles come from such situations, and it is in this way that the science of organic chemistry grows richer. In the synthesis of vitamin $B_{12}$, for example, unexpected findings emerged that led to the understanding of an entire new class of reactions-the pericyclic reactions that are the subject of Chapter 30 in this book. From synthesizing vitamin $\\mathrm{B}_{12}$ to understanding pericyclic reactions-no one could have possibly predicted such a link at the beginning of the synthesis, but that is the way of science."}
{"id": 564, "contents": "Key Terms - \n- acetylide anion - Lindlar Catalyst\n- alkylation - retrosynthetic\n- alkyne\n- tautomer\n- enol"}
{"id": 565, "contents": "Summary - \nAlkynes are less common than alkenes, both in the laboratory and in living organisms, so we haven't covered them in great detail. The real importance of this chapter is that alkyne chemistry is a useful vehicle for looking at the general strategies used in organic synthesis-the construction of complex molecules in the laboratory.\n\nAn alkyne is a hydrocarbon that contains a carbon-carbon triple bond. Alkyne carbon atoms are $s p$-hybridized, and the triple bond consists of one $s p-s p \\sigma$ bond and two $p-p \\pi$ bonds. There are relatively few general methods of alkyne synthesis. Two favorable ones are the alkylation of an acetylide anion with a primary alkyl halide and the twofold elimination of HX from a vicinal dihalide.\n\nThe chemistry of alkynes is dominated by electrophilic addition reactions, similar to those of alkenes. Alkynes react with HBr and HCl to yield vinylic halides and with $\\mathrm{Br}_{2}$ and $\\mathrm{Cl}_{2}$ to yield 1,2-dihalides (vicinal dihalides). Alkynes can be hydrated by reaction with aqueous sulfuric acid in the presence of mercury(II) catalyst. The reaction leads to an intermediate enol that immediately tautomerizes to yield a ketone. Because the addition reaction occurs with Markovnikov regiochemistry, a methyl ketone is produced from a terminal alkyne. Alternatively, hydroboration-oxidation of a terminal alkyne yields an aldehyde.\n\nAlkynes can be reduced to yield alkenes and alkanes. Complete reduction of the triple bond over a palladium hydrogenation catalyst yields an alkane; partial reduction by catalytic hydrogenation over a Lindlar catalyst yields a cis alkene. Reduction of the alkyne with lithium in ammonia yields a trans alkene."}
{"id": 566, "contents": "Summary - \nTerminal alkynes are weakly acidic. The alkyne hydrogen can be removed by a strong base such as $\\mathrm{Na}^{+-} \\mathrm{NH}_{2}$ to yield an acetylide anion. An acetylide anion acts as a nucleophile and can displace a halide ion from a primary alkyl halide in an alkylation reaction. Acetylide anions are more stable than either alkyl anions or vinylic anions because their negative charge is in a hybrid orbital with $50 \\% s$ character, allowing the charge to be closer to the nucleus."}
{"id": 567, "contents": "Summary of Reactions - \nNo stereochemistry is implied unless specifically indicated with wedged, solid, and dashed lines.\n\n1. Preparation of alkynes\na. Dehydrohalogenation of vicinal dihalides (Section 9.2)\n\n\nb. Alkylation of acetylide anions (Section 9.8)\n\n2. Reactions of alkynes\na. Addition of HCl and HBr (Section 9.3)\n\nb. Addition of $\\mathrm{Cl}_{2}$ and $\\mathrm{Br}_{2}$ (Section 9.3)\n\nc. Hydration (Section 9.4)\n(1) Mercuric sulfate catalyzed\n\n\nAn enol\nA methyl ketone\n(2) Hydroboration-oxidation\n\nd. Reduction (Section 9.5)\n(1) Catalytic hydrogenation\n\n\n(2) Lithium in liquid ammonia\n\n\nA trans alkene\ne. Conversion into acetylide anions (Section 9.7)\n\n$$\n\\mathrm{R}-\\mathrm{C} \\equiv \\mathrm{C}-\\mathrm{H} \\xrightarrow[\\mathrm{NH}_{3}]{\\mathrm{NaNH}_{2}} \\mathrm{R}-\\mathrm{C} \\equiv \\mathrm{C}:-\\mathrm{Na}^{+}+\\mathrm{NH}_{3}\n$$"}
{"id": 568, "contents": "Visualizing Chemistry - \nPROBLEM Name the following alkynes, and predict the products of their reaction with (1) $\\mathrm{H}_{2}$ in the presence\n9-14 of a Lindlar catalyst and (2) $\\mathrm{H}_{3} \\mathrm{O}^{+}$in the presence of $\\mathrm{HgSO}_{4}$ :\n(a)\n\n(b)\n\n\nPROBLEM From what alkyne might each of the following substances have been made? (Green = Cl.)\n\n9-15 (a)\n\n(b)\n\n\nPROBLEM How would you prepare the following substances, starting from any compounds having four\n9-16 carbons or fewer?\n(a)\n\n(b)\n\n\nPROBLEM The following cycloalkyne is too unstable to exist. Explain.\n9-17"}
{"id": 569, "contents": "Mechanism Problems - \nPROBLEM Assuming that halogens add to alkynes in the same manner as they add to alkenes, propose a\n9-18 mechanism for and predict the product(s) of the reaction of phenylpropyne with $\\mathrm{Br}_{2}$.\nPROBLEM Assuming that strong acids add to alkynes in the same manner as they add to alkenes, propose a\n9-19 mechanism for each of the following reactions.\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM The mercury-catalyzed hydration of alkynes involves the formation of an organomercury enol\n9-20 intermediate. Draw the electron-pushing mechanism to show how each of the following intermediates is formed.\n(a)\n\n(b)\n\n\n\nPROBLEM The final step in the hydration of an alkyne under acidic conditions is the tautomerization of an enol\n9-21 intermediate to give the corresponding ketone. The mechanism involves a protonation followed by a deprotonation. Show the mechanism for each of the following tautomerizations.\n(a)\n\n(b)\n\n$\\xrightarrow{\\mathrm{H}_{3} \\mathrm{O}^{+}}$\n\n(c)\n\n\nPROBLEM Predict the product(s) and show the complete electron-pushing mechanism for each of the\n9-22 following dissolving metal reductions.\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Identify the mechanisms for the following reactions as polar, radical, or both.\n9-23 (a)\n\n(b)\n\n(c)\n\n\nPROBLEM Predict the product and provide the complete electron-pushing mechanism for the following two-\n9-24 step synthetic processes.\n(a)\n\n(b)\n\n(c)\n\n$$\n\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{C} \\equiv \\mathrm{CH} \\xrightarrow{\\text { 2. } \\mathrm{PhCH}_{2} \\mathrm{Br}}\n$$\n\nPROBLEM Reaction of acetone with $\\mathrm{D}_{3} \\mathrm{O}^{+}$yields hexadeuterioacetone. That is, all the hydrogens in acetone are\n9-25 exchanged for deuterium. Review the mechanism of mercuric-ion-catalyzed alkyne hydration, and then propose a mechanism for this deuterium incorporation."}
{"id": 570, "contents": "Acetone - \nHexadeuterioacetone"}
{"id": 571, "contents": "Naming Alkynes - \nPROBLEM Give IUPAC names for the following compounds:\n9-26 (a)\n\n(b) $\\mathrm{CH}_{3} \\mathrm{C} \\equiv \\mathrm{CCH}_{2} \\mathrm{C} \\equiv \\mathrm{CCH}_{2} \\mathrm{CH}_{3}$\n(c)\n\n(d)\n\n(e)\n$\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}=\\mathrm{CHC} \\equiv \\mathrm{CH}$\n(f)\n\n\nPROBLEM Draw structures corresponding to the following names:\n9-27 (a) 3,3-Dimethyl-4-octyne (b) 3-Ethyl-5-methyl-1,6,8-decatriyne\n(c) 2,2,5,5-Tetramethyl-3-hexyne (d) 3,4-Dimethylcyclodecyne\n(e) 3,5-Heptadien-1-yne\n(f) 3-Chloro-4,4-dimethyl-1-nonen-6-yne (g) 3-sec-Butyl-1-heptyne\n(h) 5-tert-Butyl-2-methyl-3-octyne\n\nPROBLEM The following two hydrocarbons have been isolated from various plants in the sunflower family.\n9-28 Name them according to IUPAC rules.\n(a) $\\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{CHC} \\equiv \\mathrm{CC} \\equiv \\mathrm{CCH}=\\mathrm{CHCH}=\\mathrm{CHCH}=\\mathrm{CH}_{2}$ (all trans)\n(b) $\\mathrm{CH}_{3} \\mathrm{C} \\equiv \\mathrm{CC} \\equiv \\mathrm{CC} \\equiv \\mathrm{CC} \\equiv \\mathrm{CC} \\equiv \\mathrm{CCH}=\\mathrm{CH}_{2}$"}
{"id": 572, "contents": "Reactions of Alkynes - \nPROBLEM Terminal alkynes react with $\\mathrm{Br}_{2}$ and water to yield bromo ketones. For example:\n9-29\n\n\nPropose a mechanism for the reaction. To what reaction of alkenes is the process analogous?\nPROBLEM Predict the products of the following reactions:\n9-30\n\n\nPROBLEM Predict the products from reaction of 1-hexyne with the following reagents:\n9-31 (a) 1 equiv HBr (b) 1 equiv $\\mathrm{Cl}_{2}$ (c) $\\mathrm{H}_{2}$, Lindlar catalyst (d) $\\mathrm{NaNH}_{2}$ in $\\mathrm{NH}_{3}$, then $\\mathrm{CH}_{3} \\mathrm{Br}$\n(e) $\\mathrm{H}_{2} \\mathrm{O}, \\mathrm{H}_{2} \\mathrm{SO}_{4}, \\mathrm{HgSO}_{4}$ (f) 2 equiv HCl\n\nPROBLEM Predict the products from reaction of 5-decyne with the following reagents:\n9-32 (a) $\\mathrm{H}_{2}$, Lindlar catalyst\n(b) Li in $\\mathrm{NH}_{3}$\n(c) 1 equiv $\\mathrm{Br}_{2}$\n(d) $\\mathrm{BH}_{3}$ in THF, then $\\mathrm{H}_{2} \\mathrm{O}_{2}, \\mathrm{OH}^{-}$\n(e) $\\mathrm{H}_{2} \\mathrm{O}, \\mathrm{H}_{2} \\mathrm{SO}_{4}, \\mathrm{HgSO}_{4}$\n(f) Excess $\\mathrm{H}_{2}, \\mathrm{Pd} / \\mathrm{C}$ catalyst\n\nPROBLEM Predict the products from reaction of 2-hexyne with the following reagents:\n9-33 (a) 2 equiv $\\mathrm{Br}_{2}$\n(b) 1 equiv HBr\n(c) Excess HBr\n(d) Li in $\\mathrm{NH}_{3}$\n(e) $\\mathrm{H}_{2} \\mathrm{O}, \\mathrm{H}_{2} \\mathrm{SO}_{4}, \\mathrm{HgSO}_{4}$"}
{"id": 573, "contents": "Reactions of Alkynes - \nPROBLEM Propose structures for hydrocarbons that give the following products on oxidative cleavage by\n9-34 $\\mathrm{KMnO}_{4}$ or $\\mathrm{O}_{3}$ :\n(a) $\\mathrm{CO}_{2}+\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{5} \\mathrm{CO}_{2} \\mathrm{H}$\n(b)\n\n(c) $\\mathrm{HO}_{2} \\mathrm{C}\\left(\\mathrm{CH}_{2}\\right)_{8} \\mathrm{CO}_{2} \\mathrm{H}$\n(d)\n\n(e)\n\n\nPROBLEM Identify the reagents a-c in the following scheme: 9-35\n\n\nOrganic Synthesis\nPROBLEM How would you carry out the following multistep conversions? More than one step may be needed 9-36 in some instances.\n\n\nPROBLEM How would you carry out the following reactions?\n9-37 (a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n\nPROBLEM Each of the following syntheses requires more than one step. How would you carry them out?\n9-38 (a)\n\n(b)\n\n\nPROBLEM How would you carry out the following multistep transformation?\n9-39\n\n\nPROBLEM How would you carry out the following multistep conversions?"}
{"id": 574, "contents": "9-40 - \nPROBLEM Synthesize the following compounds using 1-butyne as the only source of carbon, along with any\n9-41 inorganic reagents you need. More than one step may be needed.\n(a) 1,1,2,2-Tetrachlorobutane (b) 1,1-Dichloro-2-ethylcyclopropane\n\nPROBLEM How would you synthesize the following compounds from acetylene and any alkyl halides with four 9-42 or fewer carbons? More than one step may be needed.\n(a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{C} \\equiv \\mathrm{CH}$\n(b) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{C} \\equiv \\mathrm{CCH}_{2} \\mathrm{CH}_{3}$\n(c)\n\n(d)\n\n(e) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CHO}$\n\nPROBLEM How would you carry out the following reactions to introduce deuterium into organic molecules?\n9-43 (a)\n\n(c)\n$\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{C} \\equiv \\mathrm{CH} \\xrightarrow{?} \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{C} \\equiv \\mathrm{CD}$\n(b)\n\n(d)\n\n\nPROBLEM How would you prepare cyclodecyne starting from acetylene and any required alkyl halide?\n\nPROBLEM The sex attractant given off by the common housefly is an alkene named muscalure. Propose a\n9-45 synthesis of muscalure starting from acetylene and any alkyl halides needed. What is the IUPAC name for muscalure?\n\n\nMuscalure"}
{"id": 575, "contents": "General Problems - \nPROBLEM A hydrocarbon of unknown structure has the formula $\\mathrm{C}_{8} \\mathrm{H}_{10}$. On catalytic hydrogenation over the\n9-46 Lindlar catalyst, 1 equivalent of $\\mathrm{H}_{2}$ is absorbed. On hydrogenation over a palladium catalyst, 3 equivalents of $\\mathrm{H}_{2}$ are absorbed.\n(a) How many degrees of unsaturation are present in the unknown structure?\n(b) How many triple bonds are present? (c) How many double bonds are present?\n(d) How many rings are present? (e) Draw a structure that fits the data.\n\nPROBLEM Compound $\\mathbf{A}\\left(\\mathrm{C}_{9} \\mathrm{H}_{12}\\right)$ absorbed 3 equivalents of $\\mathrm{H}_{2}$ on catalytic reduction over a palladium catalyst\n9-47 to give $\\mathbf{B}\\left(\\mathrm{C}_{9} \\mathrm{H}_{18}\\right)$. On ozonolysis, compound $\\mathbf{A}$ gave, among other things, a ketone that was identified as cyclohexanone. On treatment with $\\mathrm{NaNH}_{2}$ in $\\mathrm{NH}_{3}$, followed by addition of iodomethane, compound $\\mathbf{A}$ gave a new hydrocarbon, $\\mathbf{C}\\left(\\mathrm{C}_{10} \\mathrm{H}_{14}\\right)$. What are the structures of $\\mathbf{A}, \\mathbf{B}$, and $\\mathbf{C}$ ?\n\nPROBLEM Hydrocarbon $A$ has the formula $\\mathrm{C}_{12} \\mathrm{H}_{8}$. It absorbs 8 equivalents of $\\mathrm{H}_{2}$ on catalytic reduction over\n9-48 a palladium catalyst. On ozonolysis, only two products are formed: oxalic acid $\\left(\\mathrm{HO}_{2} \\mathrm{CCO}_{2} \\mathrm{H}\\right)$ and succinic acid $\\left(\\mathrm{HO}_{2} \\mathrm{CCH}_{2} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}\\right)$. Write the reactions, and propose a structure for $\\mathbf{A}$."}
{"id": 576, "contents": "General Problems - \nPROBLEM Occasionally, a chemist might need to invert the stereochemistry of an alkene-that is, to convert\n9-49 a cis alkene to a trans alkene, or vice versa. There is no one-step method for doing an alkene inversion, but the transformation can be carried out by combining several reactions in the proper sequence. How would you carry out the following reactions?\n(a)\n(b) cis-5-Decene $\\xrightarrow{?}$ trans-5-Decene\ntrans-5-Decene $\\xrightarrow{?}$ cis-5-Decene\n\nPROBLEM Organometallic reagents such as sodium acetylide undergo an addition reaction with ketones,\n9-50 giving alcohols:\n\n\nHow might you use this reaction to prepare 2-methyl-1,3-butadiene, the starting material used in the manufacture of synthetic rubber?\n\nPROBLEM The oral contraceptive agent Mestranol is synthesized using a carbonyl addition reaction like that\n9-51 shown in Problem 9-50. Draw the structure of the ketone needed."}
{"id": 577, "contents": "Mestranol - \nPROBLEM 1-Octen-3-ol, a potent mosquito attractant commonly used in mosquito traps, can be prepared in 9-52 two steps from hexanal, $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CHO}$. The first step is an acetylide-addition reaction like that described in Problem 9-50. What is the structure of the product from the first step, and how can it be converted into 1-octen-3-ol?\n\n\n1-Octen-3-ol\n\nPROBLEM Erythrogenic acid, $\\mathrm{C}_{18} \\mathrm{H}_{26} \\mathrm{O}_{2}$, is an acetylenic fatty acid that turns a vivid red on exposure to\n9-53 light. On catalytic hydrogenation over a palladium catalyst, 5 equivalents of $\\mathrm{H}_{2}$ are absorbed, and stearic acid, $\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{16} \\mathrm{CO}_{2} \\mathrm{H}$, is produced. Ozonolysis of erythrogenic acid gives four products: formaldehyde, $\\mathrm{CH}_{2} \\mathrm{O}$; oxalic acid, $\\mathrm{HO}_{2} \\mathrm{CCO}_{2} \\mathrm{H}$; azelaic acid, $\\mathrm{HO}_{2} \\mathrm{C}\\left(\\mathrm{CH}_{2}\\right)_{7} \\mathrm{CO}_{2} \\mathrm{H}$; and the aldehyde acid $\\mathrm{OHC}\\left(\\mathrm{CH}_{2}\\right)_{4} \\mathrm{CO}_{2} \\mathrm{H}$. Draw two possible structures for erythrogenic acid, and suggest a way to tell them apart by carrying out some simple reactions."}
{"id": 578, "contents": "Mestranol - \nPROBLEM Hydrocarbon $\\mathbf{A}$ has the formula $\\mathrm{C}_{9} \\mathrm{H}_{12}$ and absorbs 3 equivalents of $\\mathrm{H}_{2}$ to yield $\\mathbf{B}, \\mathrm{C}_{9} \\mathrm{H}_{18}$, when\n9-54 hydrogenated over a $\\mathrm{Pd} / \\mathrm{C}$ catalyst. On treatment of $\\mathbf{A}$ with aqueous $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ in the presence of mercury(II), two isomeric ketones, $\\mathbf{C}$ and $\\mathbf{D}$, are produced. Oxidation of $\\mathbf{A}$ with $\\mathrm{KMnO}_{4}$ gives a mixture of acetic acid $\\left(\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right)$ and the tricarboxylic acid $\\mathbf{E}$. Propose structures for compounds $\\mathbf{A}-\\mathbf{D}$, and write the reactions.\n\n\nE\nPROBLEM A cumulene is a compound with three adjacent double bonds. Draw an orbital picture of a\n9-55 cumulene. What kind of hybridization do the two central carbon atoms have? What is the geometric relationship of the substituents on one end to the substituents on the other end? What kind of isomerism is possible? Make a model to help see the answer.\n$\\mathrm{R}_{2} \\mathrm{C}=\\mathrm{C}=\\mathrm{C}=\\mathrm{CR}_{2}$"}
{"id": 579, "contents": "A cumulene - \nPROBLEM Which of the following bases could be used to deprotonate 1-butyne?"}
{"id": 580, "contents": "9-56 - \n(a) ${ }_{\\mathrm{KOH}}$\n(b)\n\n(c) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{Li}$\n(d)\n\n\nPROBLEM Arrange the following carbocations in order of increasing stability.\n9-57\n(a)\n\n\n(b)"}
{"id": 581, "contents": "CHAPTER 10 <br> Organohalides - \nFIGURE 10.1 The gases released during volcanic eruptions contain large amounts of organohalides, including chloromethane, chloroform, dichlorodifluoromethane, and many others. (credit: modification of work \"Tavurvur volcano\" by Taro Taylor, Richard Bartz/ Wikimedia Commons, CC BY 2.0)"}
{"id": 582, "contents": "CHAPTER CONTENTS - \n10.1 Names and Structures of Alkyl Halides\n10.2 Preparing Alkyl Halides from Alkanes: Radical Halogenation\n10.3 Preparing Alkyl Halides from Alkenes: Allylic Bromination\n10.4 Stability of the Allyl Radical: Resonance Revisited\n10.5 Preparing Alkyl Halides from Alcohols\n10.6 Reactions of Alkyl Halides: Grignard Reagents\n10.7 Organometallic Coupling Reactions\n10.8 Oxidation and Reduction in Organic Chemistry\n\nWHY THIS CHAPTER? Alkyl halides are encountered less frequently than their oxygen-containing relatives and are not often involved in the biochemical pathways of terrestrial organisms, but some of the kinds of reactions they undergo-nucleophilic substitutions and eliminations-are encountered frequently. Thus, alkyl halide chemistry is a relatively simple model for many mechanistically similar but structurally more complex reactions found in biomolecules. We'll begin this chapter with a look at how to name and prepare alkyl halides, and we'll see several of their reactions. Then, in the next chapter, we'll make a detailed study of the substitution and elimination reactions of alkyl halides-two of the most important and well-studied reaction types in organic chemistry.\n\nNow that we've covered the chemistry of hydrocarbons, it's time to start looking at more complex substances that contain elements in addition to C and H . We'll begin by discussing the chemistry of organohalides,\ncompounds that contain one or more halogen atoms.\nHalogen-substituted organic compounds are widespread in nature, and more than 5000 organohalides have been found in algae and various other marine organisms. Chloromethane, for instance, is released in large amounts by ocean kelp, as well as by forest fires and volcanoes. Halogen-containing compounds also have an array of industrial applications, including their use as solvents, inhaled anesthetics in medicine, refrigerants, and pesticides.\n\n\nTrichloroethylene (a solvent)\n\n\nHalothane (an inhaled anesthetic)\n\n\nTetrafluoropropene (a refrigerant)\n\n\nBromomethane (a fumigant)"}
{"id": 583, "contents": "CHAPTER CONTENTS - \nTrichloroethylene (a solvent)\n\n\nHalothane (an inhaled anesthetic)\n\n\nTetrafluoropropene (a refrigerant)\n\n\nBromomethane (a fumigant)\n\nStill other halo-substituted compounds are used as medicines and food additives. The nonnutritive sweetener sucralose, marketed as Splenda, contains three chlorine atoms, for instance. Sucralose is about 600 times as sweet as sucrose, so only 1 mg is equivalent to an entire teaspoon of table sugar.\n\n\nSucralose\nA large variety of organohalides are known. The halogen might be bonded to an alkynyl group ( $\\mathrm{C} \\equiv \\mathrm{C}-\\mathrm{X}$ ), a vinylic group $(C=C-X)$, an aromatic ring ( $\\mathrm{Ar}-\\mathrm{X}$ ), or an alkyl group. In this chapter, however, we'll be primarily concerned with alkyl halides, compounds with a halogen atom bonded to a saturated, $s p^{3}$-hybridized carbon atom."}
{"id": 584, "contents": "CHAPTER CONTENTS - 10.1 Names and Structures of Alkyl Halides\nAlthough commonly called alkyl halides, halogen-substituted alkanes are named systematically as haloalkanes (Section 3.4), treating the halogen as a substituent on a parent alkane chain. There are three steps:\n\nSTEP 1\nFind the longest chain, and name it as the parent. If a double or triple bond is present, the parent chain must contain it.\n\nSTEP 2\nNumber the carbons of the parent chain beginning at the end nearer the first substituent, whether alkyl or halo. Assign each substituent a number according to its position on the chain.\n\n\n5-Bromo-2,4-dimethylheptane\n\n\n2-Bromo-4,5-dimethylheptane\n\nIf different halogens are present, number each one and list them in alphabetical order when writing the name.\n\n\n1-Bromo-3-chloro-4-methylpentane\nSTEP 3\nIf the parent chain can be properly numbered from either end by step 2 , begin at the end nearer the"}
{"id": 585, "contents": "substituent that has alphabetical precedence. - \n2-Bromo-5-methylhexane\n(Not 5-bromo-2-methylhexane)\n\nIn addition to their systematic names, many simple alkyl halides are also named by identifying first the alkyl group and then the halogen. For example, $\\mathrm{CH}_{3} \\mathrm{I}$ can be called either iodomethane or methyl iodide. Such names are well entrenched in the chemical literature and in daily usage, but they won't be used in this book.\n\n\nHalogens increase in size going down the periodic table, so the lengths of the corresponding carbon-halogen bonds increase accordingly (TABLE 10.1). In addition, $C-X$ bond strengths decrease going down the periodic table. As we've been doing thus far, we'll continue using an X to represent any of the halogens $\\mathrm{F}, \\mathrm{Cl}, \\mathrm{Br}$, or I .\n\n| Halomethane | Bond length (pm) | Bond strength |  | Dipole moment (D) |\n| :---: | :---: | :---: | :---: | :---: |\n|  |  | ( $\\mathrm{kJ} / \\mathrm{mol}$ ) | ( $\\mathrm{kcal} / \\mathrm{mol}$ ) |  |\n| $\\mathrm{CH}_{3} \\mathrm{~F}$ | 139 | 460 | 110 | 1.85 |\n| $\\mathrm{CH}_{3} \\mathrm{Cl}$ | 178 | 350 | 84 | 1.87 |\n| $\\mathrm{CH}_{3} \\mathrm{Br}$ | 193 | 294 | 70 | 1.81 |\n| $\\mathrm{CH}_{3} \\mathrm{I}$ | 214 | 239 | 57 | 1.62 |"}
{"id": 586, "contents": "substituent that has alphabetical precedence. - \nIn our discussion of bond polarity in functional groups in Section 6.3, we noted that halogens are more electronegative than carbon. The $\\mathrm{C}-\\mathrm{X}$ bond is therefore polar, with the carbon atom bearing a slight positive charge ( $\\delta+$ ) and the halogen a slight negative charge ( $\\delta-$ ). This polarity results in a dipole moment for all the halomethanes (TABLE 10.1) and implies that the alkyl halide $C-X$ carbon atom should behave as an electrophile in polar reactions. We'll soon see that this is indeed the case.\n\n\n\nPROBLEM Give IUPAC names for the following alkyl halides:\n10-1 (a)\n$\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{I}$\n(b)\n\n(c)\n\n(d)\n\n\n\nPROBLEM Draw structures corresponding to the following IUPAC names:\n10-2 (a) 2-Chloro-3,3-dimethylhexane (b) 3,3-Dichloro-2-methylhexane\n(c) 3-Bromo-3-ethylpentane (d) 1,1-Dibromo-4-isopropylcyclohexane\n(e) 4-sec-Butyl-2-chlorononane (f) 1,1-Dibromo-4-tert-butylcyclohexane"}
{"id": 587, "contents": "substituent that has alphabetical precedence. - 10.2 Preparing Alkyl Halides from Alkanes: Radical Halogenation\nAs we saw briefly in Section 6.6, simple alkyl halides can sometimes be prepared by the radical reaction of an alkane with $\\mathrm{Cl}_{2}$ or $\\mathrm{Br}_{2}$ in the presence of ultraviolet light. The detailed mechanism is shown in FIGURE 10.2 for chlorination.\nInitiation step\n$\\begin{aligned} & \\text { Propagation steps } \\\\ & \\text { (a repeating cycle) }\\end{aligned}$\n\n\n\nFIGURE 10.2 Mechanism of the radical chlorination of methane. Three kinds of steps are required in radical substitution reactions: initiation, propagation, and termination. The propagation steps are a repeating cycle, with $\\mathrm{Cl} \\cdot$ a reactant in step 1 and a product in step 2 , and with $\\cdot \\mathrm{CH}_{3}$ a product in step 1 and a reactant in step 2 . The symbol hv shown in the initiation step is the standard way of indicating irradiation with light ( $v$ is the lowercase Greek letter nu).\n\nRadical substitution reactions require three kinds of steps: initiation, propagation, and termination. Once an initiation step has started the process by producing radicals, the reaction continues in a self-sustaining cycle. The cycle requires two repeating propagation steps in which a radical, the halogen, and the alkane yield alkyl halide product plus more radical to carry on the chain. The chain is occasionally terminated by the combination of two radicals.\n\nUnfortunately, alkane halogenation is a poor synthetic method for preparing alkyl halides because mixtures of products invariably result. For example, chlorination of methane does not stop cleanly at the monochlorinated stage but continues to give a mixture of dichloro, trichloro, and even tetrachloro products.\n\n\nThe situation is even worse for chlorination of alkanes that have more than one kind of hydrogen. Chlorination of butane, for instance, gives two monochlorinated products in a $30: 70$ ratio in addition to multiply chlorinated products such as dichlorobutane, trichlorobutane, and so on."}
{"id": 588, "contents": "substituent that has alphabetical precedence. - 10.2 Preparing Alkyl Halides from Alkanes: Radical Halogenation\nAs another example, 2-methylpropane yields 2-chloro-2-methylpropane and 1-chloro-2-methylpropane in a $35: 65$ ratio, along with more highly chlorinated products.\n\n\nFrom these and similar reactions, it's possible to calculate a reactivity order toward chlorination for different kinds of hydrogen atoms in a molecule. Take the butane chlorination, for instance. Butane has six equivalent primary hydrogens $\\left(-\\mathrm{CH}_{3}\\right)$ and four equivalent secondary hydrogens $\\left(-\\mathrm{CH}_{2}-\\right)$. The fact that butane yields $30 \\%$ of 1-chlorobutane product means that each one of the six primary hydrogens is responsible for $30 \\% \\div 6=5 \\%$ of the product. Similarly, the fact that $70 \\%$ of 2 -chlorobutane is formed means that each of the four secondary hydrogens is responsible for $70 \\% \\div 4=17.5 \\%$ of the product. Thus, a secondary hydrogen reacts $17.5 \\% \\div 5 \\%=$ 3.5 times as often as a primary hydrogen.\n\nA similar calculation for the chlorination of 2-methylpropane indicates that each of the nine primary hydrogens accounts for $65 \\% \\div 9=7.2 \\%$ of the product, while the single tertiary hydrogen ( $\\mathrm{R}_{3} \\mathrm{CH}$ ) accounts for $35 \\%$ of the product. Thus, a tertiary hydrogen is $35 \\% \\div 7.2 \\%=5$ times as reactive as a primary hydrogen toward chlorination."}
{"id": 589, "contents": "substituent that has alphabetical precedence. - 10.2 Preparing Alkyl Halides from Alkanes: Radical Halogenation\nThe observed reactivity order of alkane hydrogens toward radical chlorination can be explained by looking at the bond dissociation energies given previously in TABLE 6.3. The data show that a tertiary $\\mathrm{C}-\\mathrm{H}$ bond ( $400 \\mathrm{~kJ} /$ mol ; $96 \\mathrm{kcal} / \\mathrm{mol}$ ) is weaker than a secondary $\\mathrm{C}-\\mathrm{H}$ bond ( $410 \\mathrm{~kJ} / \\mathrm{mol} ; 98 \\mathrm{kcal} / \\mathrm{mol}$ ), which is in turn weaker than a primary C-H bond ( $421 \\mathrm{~kJ} / \\mathrm{mol} ; 101 \\mathrm{kcal} / \\mathrm{mol}$ ). Since less energy is needed to break a tertiary C-H bond than to break a primary or secondary $\\mathrm{C}-\\mathrm{H}$ bond, the resultant tertiary radical is more stable than a primary or secondary radical.\n\n\nPROBLEM Draw and name all monochloro products you would expect to obtain from radical chlorination of 10-3 2-methylpentane. Which, if any, are chiral?\n\nPROBLEM Taking the relative reactivities of $1^{\\circ}, 2^{\\circ}$, and $3^{\\circ}$ hydrogen atoms into account, what product(s) would 10-4 you expect to obtain from monochlorination of 2 -methylbutane? What would the approximate percentage of each product be? (Don't forget to take into account the number of each kind of"}
{"id": 590, "contents": "hydrogen.) - 10.3 Preparing Alkyl Halides from Alkenes: Allylic Bromination\nWe've already seen several methods for preparing alkyl halides from alkenes, including the reactions of HX and $\\mathrm{X}_{2}$ with alkenes in electrophilic addition reactions (Section 7.7 and Section 8.2). The hydrogen halides HCl , HBr , and HI react with alkenes by a polar mechanism to give the product of Markovnikov addition. Bromine and chlorine undergo anti addition through halonium ion intermediates to give 1,2-dihalogenated products.\n\n\nAnother laboratory method for preparing alkyl halides from alkenes is by reaction with $N$-bromosuccinimide (abbreviated NBS), in the presence of ultraviolet light, to give products resulting from substitution of hydrogen by bromine at the position next to the double bond-the allylic position. Cyclohexene, for example, gives 3-bromocyclohexene.\n\n\nThis allylic bromination with NBS is analogous to the alkane chlorination reaction discussed in the previous section and occurs by a radical chain-reaction pathway (FIGURE 10.3). As in alkane halogenation, a Br\u2022 radical abstracts an allylic hydrogen atom, forming an allylic radical plus HBr . The HBr then reacts with NBS to form $\\mathrm{Br}_{2}$, which in turn reacts with the allylic radical to yield the brominated product and a $\\mathrm{Br} \\cdot$ radical that cycles back into the first step and carries on the chain.\n\n\nFIGURE 10.3 Mechanism of allylic bromination of an alkene with NBS. The process is a radical chain reaction in which (1) a $\\mathrm{Br} \\cdot \\mathrm{radical}$ abstracts an allylic hydrogen atom of the alkene and gives an allylic radical plus HBr . (2) The HBr then reacts with NBS to form $\\mathrm{Br}_{2}$, which (3) reacts with the allylic radical to yield the bromoalkene product and a $\\mathrm{Br} \\cdot$ radical that continues the chain."}
{"id": 591, "contents": "hydrogen.) - 10.3 Preparing Alkyl Halides from Alkenes: Allylic Bromination\nWhy does bromination with NBS occur exclusively at an allylic position rather than elsewhere in the molecule? The answer, once again, is found by looking at bond dissociation energies to see the relative stabilities of various\nkinds of radicals. Although a typical secondary alkyl C-H bond has a strength of about $410 \\mathrm{~kJ} / \\mathrm{mol}(98 \\mathrm{kcal} / \\mathrm{mol})$ and a typical vinylic $\\mathrm{C}-\\mathrm{H}$ bond has a strength of $465 \\mathrm{~kJ} / \\mathrm{mol}$ ( $111 \\mathrm{kcal} / \\mathrm{mol}$ ), an allylic $\\mathrm{C}-\\mathrm{H}$ bond has a strength of only about $370 \\mathrm{~kJ} / \\mathrm{mol}(88 \\mathrm{kcal} / \\mathrm{mol})$. An allylic radical is therefore more stable than a typical alkyl radical with the same substitution by about $40 \\mathrm{~kJ} / \\mathrm{mol}(9 \\mathrm{kcal} / \\mathrm{mol})$.\n\n\nWe can thus expand the stability ordering to include vinylic and allylic radicals."}
{"id": 592, "contents": "hydrogen.) - 10.4 Stability of the Allyl Radical: Resonance Revisited\nTo see why an allylic radical is so stable, look at the orbital picture in FIGURE 10.4. The radical carbon atom with an unpaired electron can adopt $s p^{2}$ hybridization, placing the unpaired electron in a $p$ orbital and giving a structure that is electronically symmetrical. The $p$ orbital on the central carbon can therefore overlap equally well with a $p$ orbital on either of the two neighboring carbons.\n\n\n\nFIGURE 10.4 An orbital view of the allyl radical. The $p$ orbital on the central carbon can overlap equally well with a $p$ orbital on either neighboring carbon, giving rise to two equivalent resonance structures.\n\nBecause the allyl radical is electronically symmetrical, it has two resonance forms-one with the unpaired electron on the left and the double bond on the right and another with the unpaired electron on the right and the double bond on the left. Neither structure is correct by itself; the true structure of the allyl radical is a resonance hybrid of the two. (You might want to review Section 2.4 to Section 2.6 to brush up on resonance.) As noted in Section 2.5, the greater the number of resonance forms, the greater the stability of a compound, because bonding electrons are attracted to more nuclei. An allyl radical, with two resonance forms, is therefore more stable than a typical alkyl radical, which has only a single structure.\n\nIn molecular orbital terms, the stability of the allyl radical is due to the fact that the unpaired electron is delocalized, or spread out, over an extended $\\pi$-orbital network rather than localized at only one site, as shown\nby the computer-generated MO in FIGURE 10.4. This delocalization is particularly apparent in the so-called spin-density surface in FIGURE 10.5, which shows the calculated location of the unpaired electron. The two terminal carbons share the unpaired electron equally.\n\n\nFIGURE 10.5 The spin density surface of the allyl radical locates the position of the unpaired electron and shows that it is equally shared between the two terminal carbons."}
{"id": 593, "contents": "hydrogen.) - 10.4 Stability of the Allyl Radical: Resonance Revisited\nFIGURE 10.5 The spin density surface of the allyl radical locates the position of the unpaired electron and shows that it is equally shared between the two terminal carbons.\n\nIn addition to its effect on stability, delocalization of the unpaired electron in the allyl radical has other chemical consequences. Because the unpaired electron is delocalized over both ends of the $\\pi$ orbital system, reaction with $\\mathrm{Br}_{2}$ can occur at either end. As a result, allylic bromination of an unsymmetrical alkene often leads to a mixture of products. For example, bromination of 1-octene gives a mixture of 3-bromo-1-octene and 1-bromo-2-octene. The two products are not formed in equal amounts, however, because the intermediate allylic radical is not symmetrical and reaction at the two ends is not equally likely. Reaction at the less hindered, primary end is favored.\n\n\nThe products of allylic bromination reactions are useful for conversion into dienes by dehydrohalogenation with base. Cyclohexene can be converted into 1,3-cyclohexadiene, for example."}
{"id": 594, "contents": "Predicting the Product of an Allylic Bromination Reaction - \nWhat products would you expect from the reaction of 4,4-dimethylcyclohexene with NBS?"}
{"id": 595, "contents": "Strategy - \nDraw the alkene reactant, and identify the allylic positions. In this case, there are two different allylic positions; we'll label them $\\mathbf{A}$ and $\\mathbf{B}$. Now abstract an allylic hydrogen from each position to generate the two corresponding allylic radicals. Each of the two allylic radicals can add a Br atom at either end ( $\\mathbf{A}$ or $\\mathbf{A}^{\\prime} ; \\mathbf{B}$ or $\\mathbf{B}^{\\prime}$ ), to give a mixture of up to four products. Draw and name the products. In the present instance, the \"two\" products from reaction at position $\\mathbf{B}$ are identical, so only three products are formed in this reaction.\n\nSolution\n\n\nPROBLEM Draw three resonance forms for the cyclohexadienyl radical.\n10-5\nCyclohexadienyl radical\n\nPROBLEM The major product of the reaction of methylenecyclohexane with $N$-bromosuccinimide is 10-6 1-(bromomethyl)cyclohexene. Explain.\n\n\nPROBLEM What products would you expect from reaction of the following alkenes with NBS? If more than one 10-7 product is formed, show the structures of all.\n(a)\n\n(b)"}
{"id": 596, "contents": "Strategy - 10.5 Preparing Alkyl Halides from Alcohols\nThe most generally useful method for preparing alkyl halides is to make them from alcohols, which themselves can be obtained from carbonyl compounds as we'll see in Sections $\\mathbf{1 7 . 4}$ and 17.5. Because of the importance of this process, many different methods have been developed to transform alcohols into alkyl halides. The simplest method is to treat the alcohol with $\\mathrm{HCl}, \\mathrm{HBr}$, or HI . For reasons that will be discussed in Section 11.5, this reaction works best with tertiary alcohols, $\\mathrm{R}_{3} \\mathrm{COH}$. Primary and secondary alcohols react much more slowly and at higher temperatures.\n\n\nThe reaction of HX with a tertiary alcohol is so rapid that it's often carried out simply by bubbling pure HCl or HBr gas into a cold ether solution of the alcohol. 1-Methylcyclohexanol, for example, is converted into 1-chloro-1-methylcyclohexane by treatment with HCl ."}
{"id": 597, "contents": "1-Methylcyclohexanol 1-Chloro-1-methylcyclohexane - \n(90\\%)\nPrimary and secondary alcohols are best converted into alkyl halides by treatment with either thionyl chloride $\\left(\\mathrm{SOCl}_{2}\\right)$ or phosphorus tribromide $\\left(\\mathrm{PBr}_{3}\\right)$. These reactions, which normally take place readily under mild conditions, are less acidic and less likely to cause acid-catalyzed rearrangements than the HX method.\n\n\n\nAs the preceding examples indicate, the yields of these $\\mathrm{SOCl}_{2}$ and $\\mathrm{PBr}_{3}$ reactions are generally high and other functional groups such as ethers, carbonyls, and aromatic rings don't usually interfere. We'll look at the mechanisms of these and other related substitution reactions in Section 11.3.\n\nAlkyl fluorides can also be prepared from alcohols. Numerous alternative reagents are used for such reactions, including diethylaminosulfur trifluoride $\\left[\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2}\\right)_{2} \\mathrm{NSF}_{3}\\right]$ and HF in pyridine solvent.\n\n\nCyclohexanol"}
{"id": 598, "contents": "Fluorocyclohexane - \n(99\\%)\nPROBLEM How would you prepare the following alkyl halides from the corresponding alcohols?\n10-8 (a)\n\n(b)\n\n(c)\n\n(d)"}
{"id": 599, "contents": "Fluorocyclohexane - 10.6 Reactions of Alkyl Halides: Grignard Reagents\nAlkyl halides, RX, react with magnesium metal in ether or tetrahydrofuran (THF) solvent to yield alkylmagnesium halides, RMgX . The products, called Grignard reagents (RMgX) after their discoverer, Francois Auguste Victor Grignard, who received the 1912 Nobel Prize in Chemistry, are examples of organometallic compounds because they contain a carbon-metal bond. In addition to alkyl halides, Grignard reagents can also be made from alkenyl (vinylic) and aryl (aromatic) halides. The halogen can be Cl, Br, or I, although chlorides are less reactive than bromides and iodides. Organofluorides rarely react with magnesium.\n\n\nAs you might expect from the discussion of electronegativity and bond polarity in Section 6.3, the carbon-magnesium bond is polarized, making the carbon atom of Grignard reagents both nucleophilic and basic. An electrostatic potential map of methylmagnesium iodide, for instance, indicates the electron-rich (red) character of the carbon bonded to magnesium.\n\n\nA Grignard reagent is formally the magnesium salt, $\\mathrm{R}_{3} \\mathrm{C}^{-+} \\mathrm{MgX}$, of a carbon acid, $\\mathrm{R}_{3} \\mathrm{C}-\\mathrm{H}$, and is thus a carbon anion, or carbanion. But because hydrocarbons are such weak acids, with $\\mathbf{p} \\mathbf{K}_{\\mathbf{a}}$ 's in the range 44 to 60 (Section 9.7), carbon anions are very strong bases. Grignard reagents must therefore be protected from atmospheric moisture to prevent their being protonated and destroyed in acid-base reactions: $\\mathrm{R}-\\mathrm{Mg}-\\mathrm{X}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow \\mathrm{R}-\\mathrm{H}+$ HO-Mg-X."}
{"id": 600, "contents": "Fluorocyclohexane - 10.6 Reactions of Alkyl Halides: Grignard Reagents\nGrignard reagents themselves don't occur in living organisms, but they serve as useful carbon-based nucleophiles in several important laboratory reactions, which we'll look at in detail in Section 17.5. In addition, they act as a simple model for other, more complex carbon-based nucleophiles that are important in biological chemistry. We'll see many examples of these in Chapter 29.\n\nPROBLEM How strong a base would you expect a Grignard reagent to be? Look at Table 9.1 and predict $\\mathbf{1 0 - 9}$ whether the following reactions will occur as written. (The $\\mathrm{p} K_{\\mathrm{a}}$ of $\\mathrm{NH}_{3}$ is 35.)\n(a) $\\mathrm{CH}_{3} \\mathrm{MgBr}+\\mathrm{H}-\\mathrm{C} \\equiv \\mathrm{C}-\\mathrm{H} \\longrightarrow \\mathrm{CH}_{4}+\\mathrm{H}-\\mathrm{C} \\equiv \\mathrm{C}-\\mathrm{MgBr}$\n(b) $\\mathrm{CH}_{3} \\mathrm{MgBr}+\\mathrm{NH}_{3} \\longrightarrow \\mathrm{CH}_{4}+\\mathrm{H}_{2} \\mathrm{~N}-\\mathrm{MgBr}$\n\nPROBLEM How might you replace a halogen substituent by a deuterium atom if you wanted to prepare a 10-10 deuterated compound?"}
{"id": 601, "contents": "Fluorocyclohexane - 10.7 Organometallic Coupling Reactions\nMany other kinds of organometallic compounds can be prepared in a manner similar to that of Grignard reagents. For instance, alkyllithium reagents, RLi, can be prepared by the reaction of an alkyl halide with lithium metal. Alkyllithiums are both nucleophiles and strong bases, and their chemistry is similar in many respects to that of alkylmagnesium halides.\n\n\nOne particularly valuable reaction of alkyllithiums occurs when making lithium diorganocopper compounds, $\\mathrm{R}_{2} \\mathrm{CuLi}$, by reaction with copper(I) iodide in diethyl ether as solvent. Often called Gilman reagents ( $\\mathbf{L i R}_{\\mathbf{2}} \\mathbf{C u}$ ), lithium diorganocopper compounds are useful because they undergo a coupling reaction with organochlorides, bromides, and iodides (but not fluorides). One of the alkyl groups from the lithium diorganocopper reagent replaces the halogen of the organohalide, forming a new carbon-carbon bond and yielding a hydrocarbon product. Lithium dimethylcopper, for instance, reacts with 1-iododecane to give undecane in a $90 \\%$ yield.\n\n\nThis organometallic coupling reaction is useful in organic synthesis because it forms carbon-carbon bonds, thereby allowing the preparation of larger molecules from smaller ones. As the following examples indicate, the coupling reaction can be carried out on aryl and vinylic halides as well as on alkyl halides.\n\ntrans-1-Iodo-1-nonene\ntrans-5-Tridecene (71\\%)\n\n\nIodobenzene\nToluene (91\\%)\nAn organocopper coupling reaction is carried out commercially to synthesize muscalure, (9Z)-tricosene, the sex attractant secreted by the common housefly. Minute amounts of muscalure greatly increase the lure of insecticide-treated fly bait and provide an effective and species-specific means of insect control.\n\n\nThe mechanism of the coupling reaction involves initial formation of a triorganocopper intermediate, followed\nby coupling and loss of a mono-organocopper, RCu. The coupling is not a typical polar nucleophilic substitution reaction of the sort considered in the next chapter."}
{"id": 602, "contents": "Fluorocyclohexane - 10.7 Organometallic Coupling Reactions\nIn addition to the coupling reaction of diorganocopper reagents with organohalides, related processes also occur with other organometallic reagents, particularly organopalladium compounds. One of the most commonly used procedures is the coupling reaction of an aromatic or vinyl substituted boronic acid $\\left[\\mathrm{R}-\\mathrm{B}(\\mathrm{OH})_{2}\\right]$ with an aromatic or vinyl substituted organohalide in the presence of a base and a palladium catalyst. This reaction is less general than the diorganocopper reaction because it doesn't work with alkyl substrates, but it is preferred when possible because it uses only a catalytic amount of metal rather than a full equivalent and because palladium compounds are less toxic than copper compounds. For example:\n\n\nCalled the Suzuki-Miyaura reaction, this process is particularly useful for preparing so-called biaryl compounds, which have two linked aromatic rings. A large number of commonly used drugs fit this description, so the Suzuki-Miyaura reaction is much-used in the pharmaceutical industry. As an example, valsartan, marketed as Diovan, is widely prescribed to treat high blood pressure, heart failure, and diabetic kidney disease. Its synthesis begins with a Suzuki-Miyaura coupling of ortho-chlorobenzonitrile with para-methylbenzeneboronic acid.\n\n\nShown in a simplified form in FIGURE 10.6, the mechanism of the Suzuki-Miyaura reaction involves initial reaction of the aromatic halide with the palladium catalyst to form an organopalladium intermediate, followed by reaction of that intermediate with the aromatic boronic acid. The resultant diorganopalladium complex then decomposes to the coupled biaryl product plus regenerated catalyst.\n\n\nFIGURE 10.6 Mechanism of the Suzuki-Miyaura coupling reaction of an aromatic boronic acid with an aromatic halide to give a biaryl. The reaction takes place by (1) reaction of the aromatic halide, ArX, with the catalyst to form an organopalladium intermediate, followed by (2) reaction with the aromatic boronic acid. (3) Subsequent decomposition of the diarylpalladium intermediate gives the biaryl product.\n\nPROBLEM How would you carry out the following transformations using an organocopper coupling reaction? 10-11 More than one step is required in each case.\n(a)\n\n(b)"}
{"id": 603, "contents": "Fluorocyclohexane - 10.8 Oxidation and Reduction in Organic Chemistry\nWe've pointed out on several occasions that some of the reactions discussed in this and earlier chapters are either oxidations or reductions. As noted in Section 8.7, an organic oxidation results in a loss of electron density by carbon, caused either by bond formation between carbon and a more electronegative atom (usually $\\mathrm{O}, \\mathrm{N}$, or a halogen) or by bond-breaking between carbon and a less electronegative atom (usually H). Conversely, an organic reduction results in a gain of electron density by carbon, caused either by bond formation between carbon and a less electronegative atom or by bond-breaking between carbon and a more electronegative atom (Section 8.6).\n\nOxidation $\\quad$| Decreases electron density on carbon by: |  |  |\n| :--- | :--- | :--- |\n|  | - forming one of these: $\\mathrm{C}-\\mathrm{O} \\quad \\mathrm{C}-\\mathrm{N}$ | $\\mathrm{C}-\\mathrm{X}$ |\n|  | $\\quad$ or breaking this: $\\mathrm{C}-\\mathrm{H}$ |  |\n\nBased on these definitions, the chlorination reaction of methane to yield chloromethane is an oxidation because a $\\mathrm{C}-\\mathrm{H}$ bond is broken and a $\\mathrm{C}-\\mathrm{Cl}$ bond is formed. The conversion of an alkyl chloride to an alkane via a Grignard reagent followed by protonation is a reduction, however, because a $\\mathrm{C}-\\mathrm{Cl}$ bond is broken and a $\\mathrm{C}-\\mathrm{H}$ bond is formed.\n\n\n\nReduction: $\\mathrm{C}-\\mathrm{Cl}$ bond broken and $\\mathrm{C}-\\mathrm{H}$ bond formed\n\nMethane\n\nAs other examples, the reaction of an alkene with $\\mathrm{Br}_{2}$ to yield a 1,2-dibromide is an oxidation because two $\\mathrm{C}-\\mathrm{Br}$ bonds are formed, but the reaction of an alkene with HBr to yield an alkyl bromide is neither an oxidation nor a reduction because both a $\\mathrm{C}-\\mathrm{H}$ and a $\\mathrm{C}-\\mathrm{Br}$ bond are formed."}
{"id": 604, "contents": "Fluorocyclohexane - 10.8 Oxidation and Reduction in Organic Chemistry\nOxidation: Two new bonds formed between carbon and a more electronegative element\n\n\n> Neither oxidation nor reduction:\n> One new $\\mathrm{C}-\\mathrm{H}$ bond and one new $\\mathrm{C}-\\mathrm{Br}$ bond formed\n\nEthylene"}
{"id": 605, "contents": "Bromoethane - \nA list of compounds of increasing oxidation level is shown in FIGURE 10.7. Alkanes are at the lowest oxidation level because they have the maximum possible number of $\\mathrm{C}-\\mathrm{H}$ bonds per carbon, and $\\mathrm{CO}_{2}$ is at the highest level because it has the maximum possible number of $\\mathrm{C}-\\mathrm{O}$ bonds per carbon. Any reaction that converts a compound from a lower level to a higher level is an oxidation, any reaction that converts a compound from a higher level to a lower level is a reduction, and any reaction that doesn't change the level is neither an oxidation nor a reduction.\n\n| $\\mathrm{CH}_{3} \\mathrm{CH}_{3}$ | $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}_{2}$ | $\\mathrm{HC} \\equiv \\mathrm{CH}$ |  |  |\n| :---: | :---: | :---: | :---: | :---: |\n|  | $\\mathrm{CH}_{3} \\mathrm{OH}$ | $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{O}$ | $\\mathrm{HCO}_{2} \\mathrm{H}$ | $\\mathrm{CO}_{2}$ |\n|  | $\\mathrm{CH}_{3} \\mathrm{Cl}$ | $\\mathrm{CH}_{2} \\mathrm{Cl}_{2}$ | $\\mathrm{CHCl}_{3}$ | $\\mathrm{CCl}_{4}$ |\n|  | $\\mathrm{CH}_{3} \\mathrm{NH}_{2}$ | $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{NH}$ | $\\mathrm{HC} \\equiv \\mathrm{N}$ |  |\n| Low oxidation level |  |  |  | High oxidation level |\n\nWorked Example 10.2 shows how to compare the oxidation levels of different compounds with the same number of carbon atoms."}
{"id": 606, "contents": "Comparing Oxidation Levels - \nRank the following compounds in order of increasing oxidation level:\n\n$$\n\\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{CH}_{2}\n$$\n\n\n\n\n$$\n\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{3}\n$$"}
{"id": 607, "contents": "Strategy - \nCompounds that have the same number of carbon atoms can be compared by adding the number of $\\mathrm{C}-\\mathrm{O}, \\mathrm{C}-\\mathrm{N}$, and $\\mathrm{C}-\\mathrm{X}$ bonds in each and then subtracting the number of $\\mathrm{C}-\\mathrm{H}$ bonds. The larger the resultant value, the higher the oxidation level."}
{"id": 608, "contents": "Solution - \nThe first compound (propene) has six $\\mathrm{C}-\\mathrm{H}$ bonds, giving an oxidation level of -6 ; the second (2-propanol) has one $\\mathrm{C}-\\mathrm{O}$ bond and seven $\\mathrm{C}-\\mathrm{H}$ bonds, giving an oxidation level of -6 ; the third (acetone) has two $\\mathrm{C}-\\mathrm{O}$ bonds and six $\\mathrm{C}-\\mathrm{H}$ bonds, giving an oxidation level of -4 ; and the fourth (propane) has eight $\\mathrm{C}-\\mathrm{H}$ bonds, giving an oxidation level of -8 . Thus, the order of increasing oxidation level is\n\n\nPROBLEM Rank both sets of compounds in order of increasing oxidation level:\n10-12 (a)\n\n\n\n\n(b) $\\mathrm{CH}_{3} \\mathrm{CN}$\n$\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{NH}_{2}$\n$\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}$\nPROBLEM Tell whether each of the following reactions is an oxidation, a reduction, or neither.\n\n10-13 (a)\n\n(b)"}
{"id": 609, "contents": "Naturally Occurring Organohalides - \nJust forty years ago in 1980, only about 30 naturally occurring organohalides were known. It was simply assumed that chloroform, halogenated phenols, chlorinated aromatic compounds called PCBs, and other such substances found in the environment were industrial pollutants. Now, less than half a century later, the situation is quite different. More than 5000 organohalides have been found to occur naturally, and tens of thousands more surely exist. From a simple compound like chloromethane to an extremely complex one like the antibiotic vancomycin, a remarkably diverse range of organohalides exists in plants, bacteria, and animals. Many even have valuable physiological activity. The pentahalogenated alkene halomon, for instance, has been isolated from the red alga Portieria hornemannii and found to have anticancer activity against several human tumor cell lines.\n\n\nSome naturally occurring organohalides are produced in massive quantities. Forest fires, volcanic eruptions, and marine kelp release up to 5 million tons of $\\mathrm{CH}_{3} \\mathrm{Cl}$ per year, for example, while annual industrial emissions total about 26,000 tons. Termites are thought to release as much as $10^{8} \\mathrm{~kg}$ of chloroform per year. A detailed examination of the Okinawan acorn worm Ptychodera flava found that the 64 million worms living in a $1 \\mathrm{~km}^{2}$ study area excreted nearly 8000 pounds per year of bromophenols and bromoindoles, compounds previously thought to be non-natural pollutants."}
{"id": 610, "contents": "Naturally Occurring Organohalides - \nWhy do organisms produce organohalides, many of which are undoubtedly toxic? The answer seems to be that many organisms use organohalogen compounds for self-defense, either as feeding deterrents, irritants to predators, or natural pesticides. Marine sponges, coral, and sea hares, for example, release foul-tasting organohalides that deter fish, starfish, and other predators. Even humans appear to produce halogenated compounds as part of their defense against infection. The human immune system contains a peroxidase enzyme capable of carrying out halogenation reactions on fungi and bacteria, thereby killing the pathogen. And most remarkable of all, even free chlorine $-\\mathrm{Cl}_{2}$-has been found to be present in humans.\n\n\nFIGURE 10.8 Marine corals secrete organohalogen compounds that act as a feeding deterrent to fish. (credit: \"Coral reef\" by Qui Nguyen, United Nations Environment Programme/Flickr, Public Domain)\n\nMuch remains to be learned-only a few hundred of the more than 500,000 known species of marine organisms have been examined-but it's clear that organohalides are an integral part of the world around us."}
{"id": 611, "contents": "Key Terms - \n```\n- alkyl halide\n- Gilman reagent (LiR2Cu)\n- allylic\n- Grignard reagent (RMgX)\n- carbanion\n- organohalide\n- delocalized\n- organometallic\n```"}
{"id": 612, "contents": "Summary - \nAlkyl halides are not often found in terrestrial organisms, but the kinds of reactions they undergo are among the most important and well-studied reaction types in organic chemistry. In this chapter, we saw how to name and prepare alkyl halides, and we'll soon make a detailed study of their substitution and elimination reactions.\n\nSimple alkyl halides can be prepared by radical halogenation of alkanes, but mixtures of products usually result. The reactivity order of alkanes toward halogenation is identical to the stability order of radicals: $\\mathrm{R}_{3} \\mathrm{C} \\cdot>\\mathrm{R}_{2} \\mathrm{CH} \\cdot$ $>\\mathrm{RCH}_{2}$. Alkyl halides can also be prepared from alkenes by reaction with $N$-bromosuccinimide (NBS) to give the product of allylic bromination. The NBS bromination of alkenes takes place through an intermediate allylic radical, which is stabilized by resonance.\n\nAlcohols react with HX to form alkyl halides, but the reaction works well only for tertiary alcohols, $\\mathrm{R}_{3} \\mathrm{COH}$. Primary and secondary alkyl halides are normally prepared from alcohols using either $\\mathrm{SOCl}_{2}$, $\\mathrm{PBr}_{3}$, or HF in pyridine. Alkyl halides react with magnesium in ether solution to form organomagnesium halides, called Grignard reagents (RMgX), which are both nucleophilic and strongly basic.\n\nAlkyl halides also react with lithium metal to form organolithium reagents, RLi. In the presence of CuI, these form diorganocoppers, or Gilman reagents ( $\\mathbf{L i R}_{\\mathbf{2}} \\mathbf{C u}$ ). Gilman reagents react with organohalides to yield coupled hydrocarbon products."}
{"id": 613, "contents": "Summary of Reactions - \nNo stereochemistry is implied unless specifically indicated with wedged, solid, and dashed lines.\n\n1. Preparation of alkyl halides\na. From alkenes by allylic bromination (Section 10.3)\n\nb. From alcohols (Section 10.5)\n(1) Reaction with HX\n\n\nReactivity order: $3^{\\circ}>2^{\\circ}>1^{\\circ}$\n(2) Reaction of $1^{\\circ}$ and $2^{\\circ}$ alcohols with $\\mathrm{SOCl}_{2}$\n\n(3) Reaction of $1^{\\circ}$ and $2^{\\circ}$ alcohols with $\\mathrm{PBr}_{3}$\n\n(4) Reaction of $1^{\\circ}$ and $2^{\\circ}$ alcohols with HF-pyridine\n\n2. Reactions of alkyl halides\na. Formation of Grignard (organomagnesium) reagents (Section 10.6)\n\n$$\n\\mathrm{R}-\\mathrm{X} \\underset{\\text { Ether }}{\\mathrm{Mg}} \\mathrm{R}-\\mathrm{Mg}-\\mathrm{X}\n$$\n\nb. Formation of Gilman (diorganocopper) reagents (Section 10.7)\n\n$$\n\\begin{gathered}\n\\mathrm{R}-\\mathrm{X} \\\\\n2 \\mathrm{R}-\\mathrm{Li}+\\mathrm{CuI} \\xrightarrow{\\substack{\\text { Pentane } \\\\\n\\text { In ether }}} \\mathrm{R}-\\mathrm{Li}+\\mathrm{LiX} \\\\\n{[\\mathrm{R}-\\mathrm{Cu}-\\mathrm{R}]^{-} \\mathrm{Li}^{+}+\\mathrm{LiI}}\n\\end{gathered}\n$$\n\nc. Organometallic coupling (Section 10.7)\n(1) Diorganocopper reaction\n\n$$\n\\mathrm{R}_{2} \\mathrm{CuLi}+\\mathrm{R}^{\\prime}-\\mathrm{X} \\xrightarrow{\\text { In ether }} \\mathrm{R}-\\mathrm{R}^{\\prime}+\\mathrm{RCu}+\\mathrm{LiX}\n$$\n\n(2) Palladium-catalyzed Suzuki-Miyaura reaction"}
{"id": 614, "contents": "Visualizing Chemistry - \nPROBLEM Give IUPAC names for the following alkyl halides (green $=\\mathrm{Cl}$ ):"}
{"id": 615, "contents": "10-14 (a) - \n(b)\n\n\nPROBLEM Show the product(s) of reaction of the following alkenes with NBS:"}
{"id": 616, "contents": "10-15 (a) - \n(b)\n\n\nPROBLEM The following alkyl bromide can be prepared by reaction of the alcohol ( $S$ )-2-pentanol with $\\mathrm{PBr}_{3}$.\n10-16 Name the compound, assign $(R)$ or $(S)$ stereochemistry, and tell whether the reaction of the alcohol results in the same stereochemistry or a change in stereochemistry (reddish brown $=\\mathrm{Br}$ )."}
{"id": 617, "contents": "Mechanism Problems - \nPROBLEM In light of the fact that tertiary alkyl halides undergo spontaneous dissociation to yield a\n10-17 carbocation plus halide ion (see Problem 10-41), propose a mechanism for the following reaction."}
{"id": 618, "contents": "Naming Alkyl Halides - \nPROBLEM Name the following alkyl halides:\n\n10-18 (a)\n\n(b)\n\n(c)\n\n\n(d)\n\n(e) $\\mathrm{ClCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{C} \\equiv \\mathrm{CCH}_{2} \\mathrm{Br}$\n\nPROBLEM Draw structures corresponding to the following IUPAC names:\n10-19 (a) 2,3-Dichloro-4-methylhexane (b) 4-Bromo-4-ethyl-2-methylhexane\n(c) 3-Iodo-2,2,4,4-tetramethylpentane (d) cis-1-Bromo-2-ethylcyclopentane\n\nPROBLEM Draw and name all the monochlorination products you might obtain from radical chlorination of\n10-20 the following compounds. Which of the products are chiral? Are any of the products optically active?\n(a) 2-methylbutane\n(b) methylcyclopropane\n(c) 2,2-dimethylpentane"}
{"id": 619, "contents": "Synthesizing Alkyl Halides - \nPROBLEM How would you prepare the following compounds, starting with cyclopentene and any other\n10-21 reagents needed?\n(a) Chlorocyclopentane\n(b) Methylcyclopentane\n(c) 3-Bromocyclopentene\n(d) Cyclopentanol\n(e) Cyclopentylcyclopentane\n(f) 1,3-Cyclopentadiene\n\nPROBLEM Predict the product(s) of the following reactions:\n\n10-22 (a)\n\n(b)\n\n(c)\n\n(d)\n\n\n(f)\n\n(g)\n\n\nPROBLEM A chemist requires a large amount of 1-bromo-2-pentene as starting material for a synthesis and\n10-23 decides to carry out an NBS allylic bromination reaction. What is wrong with the following synthesis plan? What side products would form in addition to the desired product?\n\n\nPROBLEM What product(s) would you expect from the reaction of 1-methylcyclohexene with NBS? Would you\n10-24 use this reaction as part of a synthesis?\n\n\nPROBLEM What product(s) would you expect from the reaction of 1,4-hexadiene with NBS? What is the\n10-25 structure of the most stable radical intermediate?\nPROBLEM What product would you expect from the reaction of 1-phenyl-2-butene with NBS? Explain.\n10-26\n\n\n1-Phenyl-2-butene\n\nOxidation and Reduction\nPROBLEM Rank the compounds in each of the following series in order of increasing oxidation level:\n10-27 (a)\n\n(b)\n\n\nPROBLEM Which of the following compounds have the same oxidation level, and which have different levels?\n10-28\n\n1\n\n2\n\n3\n\n4\n\n5\n\nPROBLEM Tell whether each of the following reactions is an oxidation, a reduction, or neither:\n10-29 (a)\n\n\n(b)\n\n(c)"}
{"id": 620, "contents": "General Problems - \nPROBLEM Arrange the following radicals from most stable to least stable.\n\n10-30 (a)\n\n\n\n(b)\n\n(c)\n\n\nPROBLEM Alkylbenzenes such as toluene (methylbenzene) react with NBS to give products in which bromine 10-31\nsubstitution has occurred at the position next to the aromatic ring (the benzylic position). Explain, based on the bond dissociation energies in Table 6.3.\n\n\nPROBLEM Draw resonance structures for the benzyl radical, $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CH}_{2} \\cdot$, the intermediate produced in the NBS\n10-32 bromination reaction of toluene (Problem 10-31).\nPROBLEM Draw resonance structures for the following species:\n10-33 (a)\n\n(b)\n\n(c)\n\n\nPROBLEM (S)-3-Methylhexane undergoes radical bromination to yield optically inactive 10-34 3-bromo-3-methylhexane as the major product. Is the product chiral? What conclusions can you draw about the radical intermediate?\n\nPROBLEM Assume that you have carried out a radical chlorination reaction on ( $R$ )-2-chloropentane and have\n10-35 isolated (in low yield) 2,4-dichloropentane. How many stereoisomers of the product are formed, and in what ratio? Are any of the isomers optically active? (See Problem 10-34.)\n\nPROBLEM How would you carry out the following syntheses?\n10-36\n\n\nPROBLEM The syntheses shown here are unlikely to occur as written. What is wrong with each?\n10-37 (a)\n\n(b)\n\n\n\n\n\nPROBLEM Why do you suppose it's not possible to prepare a Grignard reagent from a bromo alcohol such\n$\\mathbf{1 0 - 3 8}$ as 4-bromo-1-pentanol? Give another example of a molecule that is unlikely to form a Grignard reagent."}
{"id": 621, "contents": "General Problems - \nPROBLEM Addition of HBr to a double bond with an ether (-OR) substituent occurs regiospecifically to give\n10-39 a product in which the -Br and -OR are bonded to the same carbon. Draw the two possible carbocation intermediates in this electrophilic addition reaction, and explain using resonance why the observed product is formed.\n\n\nHBr\n\n\n\nPROBLEM Identify the reagents $\\mathbf{a}-\\mathbf{c}$ in the following scheme:\n10-40\n\n\nPROBLEM Tertiary alkyl halides, $\\mathrm{R}_{3} \\mathrm{CX}$, undergo spontaneous dissociation to yield a carbocation, $\\mathrm{R}_{3} \\mathrm{C}^{+}$, plus\n10-41 halide ion. Which do you think reacts faster, $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CBr}$ or $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHC}\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{Br}$ ? Explain.\nPROBLEM Carboxylic acids $\\left(\\mathrm{RCO}_{2} \\mathrm{H} ; \\mathrm{p} K_{\\mathrm{a}} \\approx 5\\right)$ are approximately $10^{11}$ times more acidic than alcohols (ROH;\n10-42 $\\mathrm{p} K_{\\mathrm{a}} \\approx 16$ ). In other words, a carboxylate ion $\\left(\\mathrm{RCO}_{2}^{-}\\right)$is more stable than an alkoxide ion ( $\\mathrm{RO}^{-}$). Explain, using resonance.\n\nPROBLEM How might you use a Suzuki-Miyaura reaction to prepare the biaryl compounds below? In each 10-43 case, show the two potential reaction partners.\n(a)\n\n(b)\n\n(c)"}
{"id": 622, "contents": "General Problems - \nPROBLEM How might you use a Suzuki-Miyaura reaction to prepare the biaryl compounds below? In each 10-43 case, show the two potential reaction partners.\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM The relative rate of radical bromination is $1: 82: 1640$ for $1^{\\circ}: 2^{\\circ}: 3^{\\circ}$ hydrogens, respectively. Draw\n10-44 all of the monobrominated products that you might obtain from the radical bromination of the compounds below. Calculate the relative percentage of each.\n(a) methylcyclobutane\n(b) 3,3-dimethylpentane\n(c) 3-methylpentane\n\nPROBLEM Choose the alcohol from each pair below that would react faster with HX to form the corresponding\n10-45 alkyl halide.\n(a)\nor\n\n(b)\nor\n\n(c)\n\n\nPROBLEM Predict the product and provide the entire catalytic cycle for the following Suzuki-Miyaura 10-46 reactions."}
{"id": 623, "contents": "CHAPTER 11 - \nReactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations\n\n\nFIGURE 11.1 Competition occurs throughout nature. In chemistry, competition often occurs between alternative reaction pathways, such as in the substitution and elimination reactions of alkyl halides. (credit: modification of work \"Bull moose fight\" by Grand Teton, National Parks Service/Flickr, Public Domain)"}
{"id": 624, "contents": "CHAPTER CONTENTS - \n11.1 The Discovery of Nucleophilic Substitution Reactions\n11.2 The $\\mathbf{S}_{\\mathrm{N}} 2$ Reaction\n11.3 Characteristics of the $\\mathbf{S}_{\\mathrm{N}} \\mathbf{2}$ Reaction\n11.4 The $\\mathrm{S}_{\\mathrm{N}} 1$ Reaction\n11.5 Characteristics of the $\\mathrm{S}_{\\mathrm{N}} 1$ Reaction\n11.6 Biological Substitution Reactions\n11.7 Elimination Reactions: Zaitsev's Rule\n11.8 The E2 Reaction and the Deuterium Isotope Effect\n11.9 The E2 Reaction and Cyclohexane Conformation\n11.10 The E1 and E1cB Reactions\n11.11 Biological Elimination Reactions\n11.12 A Summary of Reactivity: $\\mathrm{S}_{\\mathrm{N}} 1, \\mathrm{~S}_{\\mathrm{N}} 2, \\mathrm{E} 1, \\mathrm{E} 1 \\mathrm{CB}$, and E 2\n\nWHY THIS CHAPTER? Nucleophilic substitution and base-induced elimination are two of the most widely occurring and versatile reactions in organic chemistry, both in the laboratory and in biological pathways. We'll look at them closely in this chapter to see how they occur, what their characteristics are, and how they can be used. We'll begin with substitution reactions.\n\nWe saw in the preceding chapter that the carbon-halogen bond in an alkyl halide is polar and that the carbon atom is electron-poor. Thus, alkyl halides are electrophiles, and much of their chemistry involves polar reactions with nucleophiles and bases. Alkyl halides do one of two things when they react with a nucleophile/ base such as hydroxide ion: they either undergo substitution of the X group by the nucleophile, or they undergo\nelimination of HX to yield an alkene.\n\n\nElimination"}
{"id": 625, "contents": "CHAPTER CONTENTS - 11.1 The Discovery of Nucleophilic Substitution Reactions\nDiscovery of the nucleophilic substitution reaction of alkyl halides dates back to work carried out by the German chemist Paul Walden in 1896. Walden found that the pure enantiomeric (+)- and ( - )-malic acids could be interconverted through a series of simple substitution reactions. When Walden treated (-)-malic acid with $\\mathrm{PCl}_{5}$, he isolated (+)-chlorosuccinic acid. This, on treatment with wet $\\mathrm{Ag}_{2} \\mathrm{O}$, gave (+)-malic acid. Similarly, reaction of $(+)$-malic acid with $\\mathrm{PCl}_{5}$ gave (-)-chlorosuccinic acid, which was converted into (-)-malic acid when treated with wet $\\mathrm{Ag}_{2} \\mathrm{O}$. The full cycle of reactions is shown in FIGURE 11.2.\n\n(-)-Malic acid\n$[\\alpha]_{D}=-2.3$\n$\\uparrow \\mathrm{Ag}_{2} \\mathrm{O}, \\mathrm{H}_{2} \\mathrm{O}$\n\n(-)-Chlorosuccinic acid\n(+)-Chlorosuccinic acid\n\n\n(+)-Malic acid\n$[\\alpha]_{D}=+2.3$\n\nFIGURE 11.2 Walden's cycle of reactions interconverting (+)- and (-)-malic acids.\nAt the time, the results were astonishing. The eminent chemist Emil Fischer called Walden's discovery \"the most remarkable observation made in the field of optical activity since the fundamental observations of Pasteur.\" Because (-)-malic acid was converted into (+)-malic acid, some reactions in the cycle must have occurred with a change, or inversion, of configuration at the chirality center. But which ones, and how? (Remember from Section 5.5 that the direction of light rotation and the configuration of a chirality center aren't directly related. You can't tell by looking at the sign of rotation whether a change in configuration has occurred during a reaction.)"}
{"id": 626, "contents": "CHAPTER CONTENTS - 11.1 The Discovery of Nucleophilic Substitution Reactions\nToday, we refer to the transformations taking place in Walden's cycle as nucleophilic substitution reactions because each step involves the substitution of one nucleophile (chloride ion, $\\mathrm{Cl}^{-}$, or hydroxide ion, $\\mathrm{HO}^{-}$) by another. Nucleophilic substitution reactions are one of the most common and versatile reaction types in organic chemistry.\n\n$$\n\\mathrm{R}-\\mathrm{X}+\\mathrm{Nu}:^{-} \\longrightarrow \\mathrm{R}-\\mathrm{Nu}+\\mathrm{X}:^{-}\n$$\n\nFollowing the work of Walden, further investigations were undertaken during the 1920s and 1930s to clarify the mechanism of nucleophilic substitution reactions and to find out how inversions of configuration occur. Among the first series studied was one that interconverted the two enantiomers of 1-phenyl-2-propanol (FIGURE 11.3). Although this particular series of reactions involves nucleophilic substitution of an alkyl para-toluenesulfonate (called a tosylate) rather than an alkyl halide, exactly the same type of reaction is involved as that studied by Walden. For all practical purposes, the entire tosylate group acts as if it were simply a halogen substituent. (In fact, when you see a tosylate substituent in a molecule, do a mental substitution and tell yourself that you're\ndealing with an alkyl halide.)\n\n\nFIGURE 11.3 A Walden cycle interconverting (+) and (-) enantiomers of 1-phenyl-2-propanol. Chirality centers are marked by asterisks, and the bonds broken in each reaction are indicated by red wavy lines. The inversion of chirality occurs in step 2, where acetate ion substitutes for tosylate ion."}
{"id": 627, "contents": "CHAPTER CONTENTS - 11.1 The Discovery of Nucleophilic Substitution Reactions\nIn the three-step reaction sequence shown in FIGURE 11.3, (+)-1-phenyl-2-propanol is interconverted with its (-) enantiomer, so at least one of the three steps must involve an inversion of configuration at the chirality center. Step 1, formation of a tosylate, occurs by breaking the $\\mathrm{O}-\\mathrm{H}$ bond of the alcohol rather than the $\\mathrm{C}-\\mathrm{O}$ bond to the chiral carbon, so the configuration around the carbon is unchanged. Similarly, step 3, hydroxide-ion cleavage of the acetate, takes place without breaking the $\\mathrm{C}-\\mathrm{O}$ bond at the chirality center. Thus, the inversion of stereochemical configuration must take place in step 2, the nucleophilic substitution of tosylate ion by acetate ion.\n\n\nFrom this and nearly a dozen other series of similar reactions, researchers concluded that the nucleophilic substitution reaction of a primary or secondary alkyl halide or tosylate always proceeds with inversion of configuration. (Tertiary alkyl halides and tosylates, as we'll see shortly, give different stereochemical results and react by a different mechanism than the primary and secondary ones.)"}
{"id": 628, "contents": "Predicting the Stereochemistry of a Nucleophilic Substitution Reaction - \nWhat product would you expect from a nucleophilic substitution reaction of ( $R$ )-1-bromo-1-phenylethane with cyanide ion, ${ }^{-} \\mathrm{C} \\equiv \\mathrm{N}$, as nucleophile? Show the stereochemistry of both reactant and product, assuming that inversion of configuration occurs."}
{"id": 629, "contents": "Strategy - \nDraw the $R$ enantiomer of the reactant, and then change the configuration of the chirality center while replacing the ${ }^{-} \\mathrm{Br}$ with $\\mathrm{a}^{-} \\mathrm{CN}$.\n\nSolution\n\n(R)-1-Bromo-1-phenylethane\n(S)-2-PhenyIpropanenitrile\n\nPROBLEM What product would you expect from a nucleophilic substitution reaction of ( $S$ )-2-bromohexane\n11-1 with acetate ion, $\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}$? Assume that inversion of configuration occurs, and show the stereochemistry of both the reactant and product."}
{"id": 630, "contents": "Strategy - 11.2 The $\\mathrm{S}_{\\mathrm{N}} 2$ Reaction\nIn almost all chemical reactions, there is a direct relationship between the rate at which the reaction occurs and the concentrations of the reactants. When we measure this relationship, we measure the kinetics of the reaction. For example, let's look at the kinetics of a simple nucleophilic substitution-the reaction of $\\mathrm{CH}_{3} \\mathrm{Br}$ with $\\mathrm{OH}^{-}$to yield $\\mathrm{CH}_{3} \\mathrm{OH}$ plus $\\mathrm{Br}^{-}$.\n\n\nWith a given temperature, solvent, and concentration of reactants, the substitution occurs at a certain rate. If we double the concentration of $\\mathrm{OH}^{-}$, the frequency of encounters between reaction partners doubles and we find that the reaction rate also doubles. Similarly, if we double the concentration of $\\mathrm{CH}_{3} \\mathrm{Br}$, the reaction rate again doubles. We call such a reaction, in which the rate is linearly dependent on the concentrations of two species, a second-order reaction. Mathematically, we can express this second-order dependence of the nucleophilic substitution reaction by setting up a rate equation. As either [ RX ] or [ ${ }^{-} \\mathrm{OH}$ ] changes, the rate of the reaction changes proportionately.\n\n$$\n\\begin{aligned}\n\\text { Reaction rate } & =\\text { Rate of disappearance of reactant } \\\\\n& =k \\times[\\mathrm{RX}] \\times\\left[{ }^{-} \\mathrm{OH}\\right]\n\\end{aligned}\n$$\n\nwhere\n$[\\mathrm{RX}]=\\mathrm{CH}_{3} \\mathrm{Br}$ concentration in molarity\n$[-\\mathrm{OH}]=-\\mathrm{OH}$ concentration in molarity\n$k=$ a constant value (the rate constant)\nA mechanism that accounts for both the inversion of configuration and the second-order kinetics that are observed with nucleophilic substitution reactions was suggested in 1937 by the British chemists E. D. Hughes\nand Christopher Ingold, who formulated what they called the $\\mathbf{S}_{\\mathbf{N}} \\mathbf{2}$ reaction-short for substitution, nucleophilic, bimolecular. (Bimolecular means that two molecules, nucleophile and alkyl halide, take part in the step whose kinetics are measured.)"}
{"id": 631, "contents": "Strategy - 11.2 The $\\mathrm{S}_{\\mathrm{N}} 2$ Reaction\nThe essential feature of the $\\mathrm{S}_{\\mathrm{N}} 2$ mechanism is that it takes place in a single step, without intermediates, when the incoming nucleophile reacts with the alkyl halide or tosylate (the substrate) from a direction opposite the group that is displaced (the leaving group). As the nucleophile comes in on one side of the substrate and bonds to the carbon, the halide or tosylate departs from the other side, thereby inverting the stereochemical configuration. The process is shown in FIGURE 11.4 for the reaction of ( $S$ ) -2-bromobutane with $\\mathrm{HO}^{-}$to give ( $R$ )-2-butanol.\n\nFIGURE 11.4 MECHANISM\nThe mechanism of the $\\mathbf{S}_{\\mathbf{N}} \\mathbf{2}$ reaction. The reaction takes place in a single step when the incoming nucleophile approaches from a direction $180^{\\circ}$ away from the leaving halide ion, thereby inverting the stereochemistry at carbon.\n(1) The nucleophile ${ }^{-} \\mathrm{OH}$ uses its lone-pair electrons to attack the alkyl halide carbon $180^{\\circ}$ away from the departing halogen. This leads to a transition state with a partially formed $\\mathrm{C}-\\mathrm{OH}$ bond and a partially broken $\\mathrm{C}-\\mathrm{Br}$ bond.\n(2) The stereochemistry at carbon is inverted as the $\\mathrm{C}-\\mathrm{OH}$ bond forms fully and the bromide ion departs with the electron pair from the former $\\mathrm{C}-\\mathrm{Br}$ bond.\n\n(S)-2-Bromobutane\n-\n\n\nTransition state\n\n(R)-2-Butanol"}
{"id": 632, "contents": "Strategy - 11.2 The $\\mathrm{S}_{\\mathrm{N}} 2$ Reaction\n(S)-2-Bromobutane\n-\n\n\nTransition state\n\n(R)-2-Butanol\n\nAs shown in FIGURE 11.4, the $\\mathrm{S}_{\\mathrm{N}} 2$ reaction occurs when an electron pair on the nucleophile Nu : ${ }^{-}$forces out the group $\\mathrm{X}:^{-}$, which takes with it the electron pair from the former $\\mathrm{C}-\\mathrm{X}$ bond. This occurs through a transition state in which the new $\\mathrm{Nu}-\\mathrm{C}$ bond is partially formed at the same time that the old $\\mathrm{C}-\\mathrm{X}$ bond is partially broken and in which the negative charge is shared by both the incoming nucleophile and the outgoing halide ion. The transition state for this inversion has the remaining three bonds to carbon in a planar arrangement (FIGURE 11.5).\n\n\nFIGURE 11.5 The transition state of an $\\mathbf{S}_{\\mathbf{N}} 2$ reaction has a planar arrangement of the carbon atom and the remaining three groups. Electrostatic potential maps show that negative charge is delocalized in the transition state.\n\nThe mechanism proposed by Hughes and Ingold is fully consistent with experimental results, explaining both stereochemical and kinetic data. Thus, the requirement for a backside approach of the entering nucleophile ( $180^{\\circ}$ away from the departing X group) causes the stereochemistry of the substrate to invert, much like an umbrella turning inside-out in the wind. The Hughes-Ingold mechanism also explains why second-order kinetics are observed: the $\\mathrm{S}_{\\mathrm{N}} 2$ reaction occurs in a single step that involves both alkyl halide and nucleophile. Two molecules are involved in the step whose rate is measured.\n\nPROBLEM What product would you expect to obtain from $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of $\\mathrm{OH}^{-}$with $(R)$-2-bromobutane? Show\n11-2 the stereochemistry of both the reactant and product.\nPROBLEM Assign configuration to the following substance, and draw the structure of the product that would 11-3 result from nucleophilic substitution reaction with $\\mathrm{HS}^{-}$(reddish brown $=\\mathrm{Br}$ ):"}
{"id": 633, "contents": "Strategy - 11.3 Characteristics of the $\\mathrm{S}_{\\mathrm{N}} 2$ Reaction\nNow that we know how $\\mathrm{S}_{\\mathrm{N}} 2$ reactions occur, we need to see how they can be used and what variables affect them. Some $\\mathrm{S}_{\\mathrm{N}} 2$ reactions are fast, and some are slow; some take place in high yield and others in low yield. Understanding the factors involved can be of tremendous value. Let's begin by recalling a few things about reaction rates in general.\n\nThe rate of a chemical reaction is determined by the activation energy $\\Delta G^{\\ddagger}$, the energy difference between reactant ground state and transition state. A change in reaction conditions can affect $\\Delta G^{\\ddagger}$ either by changing the reactant energy level or by changing the transition-state energy level. Lowering the reactant energy or raising the transition-state energy increases $\\Delta G^{\\ddagger}$ and decreases the reaction rate; raising the reactant energy or\ndecreasing the transition-state energy decreases $\\Delta G^{\\ddagger}$ and increases the reaction rate (FIGURE 11.6). We'll see examples of all these effects as we look at $\\mathrm{S}_{\\mathrm{N}} 2$ reaction variables.\n\n\nFIGURE 11.6 The effects of changes in reactant and transition-state energy levels on reaction rate. (a) A higher reactant energy level (red curve) corresponds to a faster reaction (smaller $\\Delta G^{\\ddagger}$ ). (b) A higher transition-state energy level (red curve) corresponds to a slower reaction (larger $\\Delta G^{\\ddagger}$ )."}
{"id": 634, "contents": "Steric Effects in the $\\mathrm{S}_{\\mathrm{N}} 2$ Reaction - \nThe first $\\mathrm{S}_{\\mathrm{N}} 2$ reaction variable to look at is the structure of the substrate. Because the $\\mathrm{S}_{\\mathrm{N}} 2$ transition state involves partial bond formation between the incoming nucleophile and the alkyl halide carbon atom, it seems reasonable that a hindered, bulky substrate should prevent easy approach of the nucleophile, making bond formation difficult. In other words, the transition state for reaction of a sterically hindered substrate, whose carbon atom is \"shielded\" from the approach of the incoming nucleophile, is higher in energy and forms more slowly than the corresponding transition state for a less hindered substrate (FIGURE 11.7).\n(a)\n\n\n(c)\n\n(b)\n\n\n\n(d)\n\n\n\nFIGURE 11.7 Steric hindrance to the $\\mathbf{S}_{\\mathbf{N}} \\mathbf{2}$ reaction. As the models indicate, the carbon atom in (a) bromomethane is readily accessible, resulting in a fast $\\mathrm{S}_{\\mathrm{N}} 2$ reaction. The carbon atoms in (b) bromoethane (primary), (c) 2-bromopropane (secondary), and (d) 2-bromo-2-methylpropane (tertiary) are successively more hindered, resulting in successively slower $\\mathrm{S}_{\\mathrm{N}} 2$ reactions."}
{"id": 635, "contents": "Steric Effects in the $\\mathrm{S}_{\\mathrm{N}} 2$ Reaction - \nAs FIGURE 11.7 shows, the difficulty of nucleophile approach increases as the three substituents bonded to the halo-substituted carbon atom increase in size. Methyl halides are by far the most reactive substrates in $\\mathrm{S}_{\\mathrm{N}} 2$ reactions, followed by primary alkyl halides such as ethyl and propyl. Alkyl branching at the reacting center, as in isopropyl halides $\\left(2^{\\circ}\\right)$, slows the reaction greatly, and further branching, as in tert-butyl halides $\\left(3^{\\circ}\\right)$, effectively halts the reaction. Even branching one carbon away from the reacting center, as in 2,2-dimethylpropyl (neopentyl) halides, greatly hinders nucleophilic displacement. As a result, $\\mathrm{S}_{\\mathrm{N}} 2$ reactions occur only at relatively unhindered sites and are normally useful only with methyl halides, primary halides, and a few simple secondary halides. Relative reactivities for some different substrates are as follows:\n\n\nVinylic halides $\\left(\\mathrm{R}_{2} \\mathrm{C}=\\mathrm{CRX}\\right)$ and aryl halides are not shown on this reactivity list because they are unreactive toward $\\mathrm{S}_{\\mathrm{N}} 2$ displacement. This lack of reactivity is due to steric factors: the incoming nucleophile would have to approach in the plane of the carbon-carbon double bond and burrow through part of the molecule to carry out a backside displacement.\n\n\nVinylic halide"}
{"id": 636, "contents": "The Nucleophile - \nAnother variable that has a major effect on the $S_{N} 2$ reaction is the nature of the nucleophile. Any species, either neutral or negatively charged, can act as a nucleophile as long as it has an unshared pair of electrons; that is, as long as it is a Lewis base. If the nucleophile is negatively charged, the product is neutral; if the nucleophile is neutral, the product is positively charged.\n\n\nA wide array of substances can be prepared using nucleophilic substitution reactions. In fact, we've already seen examples in previous chapters. For instance, the reaction of an acetylide anion with an alkyl halide, discussed in Section 9.8, is an $\\mathrm{S}_{\\mathrm{N}} 2$ reaction in which the acetylide nucleophile displaces a halide leaving group.\n\n$$\n\\mathrm{R}-\\mathrm{C} \\equiv \\mathrm{C}:^{-}+\\mathrm{CH}_{3} \\mathrm{Br} \\xrightarrow[\\text { reaction }]{\\mathrm{S}_{\\mathrm{N}^{2}}} \\mathrm{R}-\\mathrm{C} \\equiv \\mathrm{C}-\\mathrm{CH}_{3}+\\mathrm{Br}^{-}\n$$"}
{"id": 637, "contents": "An acetylide anion - \nTABLE 11.1 lists some nucleophiles in the order of their reactivity, shows the products of their reactions with bromomethane, and gives the relative rates of their reactions. There are large differences in the rates at which various nucleophiles react.\n\nTABLE 11.1 Some $S_{N} 2$ Reactions with Bromomethane"}
{"id": 638, "contents": "An acetylide anion - \nTABLE 11.1 Some $S_{N} 2$ Reactions with Bromomethane\n\n| $\\mathrm{Nu}:^{-}+\\mathrm{CH}_{3} \\mathrm{Br} \\rightarrow \\mathrm{CH}_{3} \\mathrm{Nu}+\\mathrm{Br}^{-}$ |  |  |  |  |\n| :---: | :---: | :---: | :---: | :---: |\n| Nucleophile |  | Product |  | Relative rate of reaction |\n| Formula | Name | Formula | Name |  |\n| $\\mathrm{H}_{2} \\mathrm{O}$ | Water | $\\mathrm{CH}_{3} \\mathrm{OH}_{2}{ }^{+}$ | Methylhydronium ion | 1 |\n| $\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}$ | Acetate | $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{CH}_{3}$ | Methyl acetate | 500 |\n| $\\mathrm{NH}_{3}$ | Ammonia | $\\mathrm{CH}_{3} \\mathrm{NH}_{3}{ }^{+}$ | Methylammonium ion | 700 |\n| $\\mathrm{Cl}^{-}$ | Chloride | $\\mathrm{CH}_{3} \\mathrm{Cl}$ | Chloromethane | 1,000 |\n| $\\mathrm{HO}^{-}$ | Hydroxide | $\\mathrm{CH}_{3} \\mathrm{OH}$ | Methanol | 10,000 |\n| $\\mathrm{CH}_{3} \\mathrm{O}^{-}$ | Methoxide | $\\mathrm{CH}_{3} \\mathrm{OCH}_{3}$ | Dimethyl ether | 25,000 |\n| $\\mathrm{I}^{-}$ | Iodide | $\\mathrm{CH}_{3} \\mathrm{I}$ | Iodomethane | 100,000 |\n| ${ }^{-} \\mathrm{CN}$ | Cyanide | $\\mathrm{CH}_{3} \\mathrm{CN}$ | Acetonitrile | 125,000 |\n| HS ${ }^{-}$ | Hydrosulfide | $\\mathrm{CH}_{3} \\mathrm{SH}$ | Methanethiol | 125,000 |"}
{"id": 639, "contents": "An acetylide anion - \nWhat are the reasons for the reactivity differences observed in TABLE 11.1? Why do some reactants appear to be much more \"nucleophilic\" than others? The answers to these questions aren't straightforward. Part of the problem is that the term nucleophilicity is imprecise. The term is usually taken to be a measure of the affinity of a nucleophile for a carbon atom in the $\\mathrm{S}_{\\mathrm{N}} 2$ reaction, but the reactivity of a given nucleophile can change from one reaction to the next. The exact nucleophilicity of a species in a given reaction depends on the substrate, the solvent, and even the reactant concentrations. Detailed explanations for the observed nucleophilicities aren't always simple, but some trends can be detected from the data of TABLE 11.1."}
{"id": 640, "contents": "An acetylide anion - \n- Nucleophilicity roughly parallels basicity when comparing nucleophiles that have the same reacting atom. Thus, $\\mathrm{OH}^{-}$is both more basic and more nucleophilic than acetate ion, $\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}$, which in turn is more basic and more nucleophilic than $\\mathrm{H}_{2} \\mathrm{O}$. Since \"nucleophilicity\" is usually taken as the affinity of a Lewis base for a carbon atom in the $\\mathrm{S}_{\\mathrm{N}} 2$ reaction and \"basicity\" is the affinity of a base for a proton, it's easy to see why there might be a correlation between the two kinds of behavior.\n- Nucleophilicity usually increases going down a column of the periodic table. Thus, $\\mathrm{HS}^{-}$is more nucleophilic than $\\mathrm{HO}^{-}$, and the halide reactivity order is $\\mathrm{I}^{-}>\\mathrm{Br}^{-}>\\mathrm{Cl}^{-}$. Going down the periodic table, elements have their valence electrons in successively larger shells where they are successively farther from the nucleus, less tightly held, and consequently more reactive. This matter is complex, though, and the nucleophilicity order can change depending on the solvent.\n- Negatively charged nucleophiles are usually more reactive than neutral ones. As a result, $\\mathrm{S}_{\\mathrm{N}} 2$ reactions are often carried out under basic conditions rather than neutral or acidic conditions.\n\nPROBLEM What product would you expect from $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of 1-bromobutane with each of the following?\n11-4 (a) NaI\n(b) KOH\n(c) $\\mathrm{H}-\\mathrm{C} \\equiv \\mathrm{C}-\\mathrm{Li}$\n(d) $\\mathrm{NH}_{3}$"}
{"id": 641, "contents": "An acetylide anion - \nPROBLEM Which substance in each of the following pairs is more reactive as a nucleophile? Explain.\n11-5 (a) $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{~N}^{-}$or $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NH}$\n(b) $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{~B}$ or $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{~N}$\n(c) $\\mathrm{H}_{2} \\mathrm{O}$ or $\\mathrm{H}_{2} \\mathrm{~S}$"}
{"id": 642, "contents": "The Leaving Group - \nStill another variable that can affect the $S_{N} 2$ reaction is the nature of the group displaced by the incoming\nnucleophile, the leaving group. Because the leaving group is expelled with a negative charge in most $\\mathrm{S}_{\\mathrm{N}} 2$ reactions, the best leaving groups are those that best stabilize the negative charge in the transition state. The greater the extent of charge stabilization by the leaving group, the lower the energy of the transition state and the more rapid the reaction. But as we saw in Section 2.8 , the groups that best stabilize a negative charge are also the weakest bases. Thus, weak bases such as $\\mathrm{Cl}^{-}, \\mathrm{Br}^{-}$, and tosylate ion make good leaving groups, while strong bases such as $\\mathrm{OH}^{-}$and $\\mathrm{NH}_{2}{ }^{-}$make poor leaving groups.\n\n\nIt's just as important to know which are poor leaving groups as to know which are good, and the preceding data clearly indicate that $\\mathrm{F}^{-}, \\mathrm{HO}^{-}, \\mathrm{RO}^{-}$, and $\\mathrm{H}_{2} \\mathrm{~N}^{-}$are not displaced by nucleophiles. In other words, alkyl fluorides, alcohols, ethers, and amines do not typically undergo $\\mathrm{S}_{\\mathrm{N}} 2$ reactions. To carry out an $\\mathrm{S}_{\\mathrm{N}} 2$ reaction with an alcohol, it's necessary to convert the ${ }^{-}$OH into a better leaving group. This, in fact, is just what happens when a primary or secondary alcohol is converted into either an alkyl chloride by reaction with $\\mathrm{SOCl}_{2}$ or an alkyl bromide by reaction with $\\mathrm{PBr}_{3}$ (Section 10.5).\n\n\nAlternatively, an alcohol can be made more reactive toward nucleophilic substitution by treating it with para-toluenesulfonyl chloride to form a tosylate. As noted previously, tosylates are even more reactive than halides in nucleophilic substitutions. Note that tosylate formation does not change the configuration of the oxygen-bearing carbon because the $\\mathrm{C}-\\mathrm{O}$ bond is not broken."}
{"id": 643, "contents": "The Leaving Group - \nA $1^{\\circ}$ or $2^{\\circ}$ alcohol\n\n\nA tosylate\n\nThe one general exception to the rule that ethers don't typically undergo $\\mathrm{S}_{\\mathrm{N}} 2$ reactions pertains to epoxides, the three-membered cyclic ethers that we saw in Section 8.7. Because of the angle strain in their three-membered ring, epoxides are much more reactive than other ethers. They react with aqueous acid to give 1,2-diols, as we saw in Section 8.7, and they react readily with many other nucleophiles as well. Propene oxide, for instance, reacts with HCl to give 1-chloro-2-propanol by an $\\mathrm{S}_{\\mathrm{N}} 2$ backside attack on the less hindered primary carbon atom. We'll look at the process in more detail in Section 18.5.\n\n\nPROBLEM Rank the following compounds in order of their expected reactivity toward $\\mathrm{S}_{\\mathrm{N}} 2$ reaction:\n11-6\n$\\mathrm{CH}_{3} \\mathrm{Br}, \\mathrm{CH}_{3} \\mathrm{OTos}$, $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CCl}$, $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CHCl}$"}
{"id": 644, "contents": "The Solvent - \nThe rates of $\\mathrm{S}_{\\mathrm{N}} 2$ reactions are strongly affected by the solvent. Protic solvents-those that contain an -OH or -NH group-are generally the worst for $\\mathrm{S}_{\\mathrm{N}} 2$ reactions, while polar aprotic solvents, which are polar but don't have an - OH or -NH group, are the best.\n\nProtic solvents, such as methanol and ethanol, slow down $\\mathrm{S}_{\\mathrm{N}} 2$ reactions by solvation of the reactant nucleophile. The solvent molecules hydrogen-bond to the nucleophile and form a cage around it, thereby lowering its energy and reactivity.\n\n\nenhanced ground-state stability)"}
{"id": 645, "contents": "The Solvent - \nenhanced ground-state stability)\n\nIn contrast with protic solvents-which decrease the rates of $S_{N} 2$ reactions by lowering the ground-state energy of the nucleophile-polar aprotic solvents increase the rates of $\\mathrm{S}_{\\mathrm{N}} 2$ reactions by raising the ground-state energy of the nucleophile. Acetonitrile $\\left(\\mathrm{CH}_{3} \\mathrm{CN}\\right)$, dimethylformamide [ $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NCHO}$, abbreviated DMF], and dimethyl sulfoxide $\\left[\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{SO}\\right.$, abbreviated DMSO] are particularly useful. A solvent known as hexamethylphosphoramide $\\left\\{\\left[\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{~N}\\right]_{3} \\mathrm{PO}\\right.$, abbreviated HMPA $\\}$ can also be useful but it should only be handled with great care and not be allowed to touch the eyes or skin. These solvents can dissolve many salts because of their high polarity, but they tend to solvate metal cations rather than nucleophilic anions. As a result, the bare, unsolvated anions have a greater nucleophilicity and $\\mathrm{S}_{\\mathrm{N}} 2$ reactions take place at correspondingly increased rates. For instance, a rate increase of 200,000 has been observed on changing from methanol to HMPA for the reaction of azide ion with 1-bromobutane.\n\n\nPROBLEM Organic solvents like benzene, ether, and chloroform are neither protic nor strongly polar. What 11-7 effect would you expect these solvents to have on the reactivity of a nucleophile in $\\mathrm{S}_{\\mathrm{N}} 2$ reactions?"}
{"id": 646, "contents": "A Summary of $\\mathrm{S}_{\\mathrm{N}} 2$ Reaction Characteristics - \nThe effects on $\\mathrm{S}_{\\mathrm{N}} 2$ reactions of the four variables-substrate structure, nucleophile, leaving group, and solvent-are summarized in the following statements and in the energy diagrams of FIGURE 11.8:\n\nSubstrate $\\quad$ Steric hindrance raises the energy of the $S_{N} 2$ transition state, increasing $\\Delta G^{\\ddagger}$ and decreasing the reaction rate (FIGURE 11.8a). As a result, $\\mathrm{S}_{\\mathrm{N}} 2$ reactions are best for methyl and primary substrates. Secondary substrates react slowly, and tertiary substrates do not react by an $\\mathrm{S}_{\\mathrm{N}} 2$ mechanism.\nNucleophile Basic, negatively charged nucleophiles are less stable and have a higher ground-state energy than neutral ones, decreasing $\\Delta G^{\\ddagger}$ and increasing the $\\mathrm{S}_{\\mathrm{N}} 2$ reaction rate (FIGURE 11.8b).\n\nLeaving Good leaving groups (more stable anions) lower the energy of the transition state,\ngroup\nSolvent decreasing $\\Delta G^{\\ddagger}$ and increasing the $S_{N} 2$ reaction rate (FIGURE 11.8c). Protic solvents solvate the nucleophile, thereby lowering its ground-state energy, increasing $\\Delta G^{\\ddagger}$, and decreasing the $S_{N} 2$ reaction rate. Polar aprotic solvents surround the accompanying cation but not the nucleophilic anion, thereby raising the ground-state energy of the nucleophile, decreasing $\\Delta G^{\\ddagger}$, and increasing the reaction rate (FIGURE 11.8d).\n\n\nFIGURE 11.8 Energy diagrams showing the effects of (a) substrate, (b) nucleophile, (c) leaving group, and (d) solvent on $\\mathbf{S}_{\\mathbf{N}} \\mathbf{2}$ reaction rates. Substrate and leaving group effects are felt primarily in the transition state. Nucleophile and solvent effects are felt primarily in the reactant ground state."}
{"id": 647, "contents": "A Summary of $\\mathrm{S}_{\\mathrm{N}} 2$ Reaction Characteristics - 11.4 The $\\mathrm{S}_{\\mathrm{N}} 1$ Reaction\nMost nucleophilic substitutions take place by the $S_{N} 2$ pathway just discussed. The reaction is favored when carried out with an unhindered substrate and a negatively charged nucleophile in a polar aprotic solvent, but is disfavored when carried out with a hindered substrate and a neutral nucleophile in a protic solvent. You might therefore expect the reaction of a tertiary substrate (hindered) with water (neutral, protic) to be among the slowest of substitution reactions. Remarkably, however, the opposite is true. The reaction of the tertiary halide 2-bromo-2-methylpropane $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CBr}$ with $\\mathrm{H}_{2} \\mathrm{O}$ to give the alcohol 2-methyl-2-propanol is more than 1 million times faster than the corresponding reaction of $\\mathrm{CH}_{3} \\mathrm{Br}$ to give methanol.\n\n\nWhat's going on here? A nucleophilic substitution reaction is occurring-a hydroxyl group is replacing a halogen-yet the reactivity order seems backward. These reactions can't be taking place by the $S_{N} 2$ mechanism we've been discussing, so we must therefore conclude that they are occurring by an alternative substitution\nmechanism. This alternative mechanism is called the $\\mathbf{S}_{\\mathbf{N}} \\mathbf{1}$ reaction, for substitution, nucleophilic, unimolecular.\n\nIn contrast to the $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of $\\mathrm{CH}_{3} \\mathrm{Br}$ with $\\mathrm{OH}^{-}$, the $\\mathrm{S}_{\\mathrm{N}} 1$ reaction of $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CBr}$ with $\\mathrm{H}_{2} \\mathrm{O}$ has a rate that depends only on the alkyl halide concentration and is independent of the $\\mathrm{H}_{2} \\mathrm{O}$ concentration. In other words, the process is a first-order reaction; the concentration of the nucleophile does not appear in the rate equation."}
{"id": 648, "contents": "A Summary of $\\mathrm{S}_{\\mathrm{N}} 2$ Reaction Characteristics - 11.4 The $\\mathrm{S}_{\\mathrm{N}} 1$ Reaction\n$$\n\\begin{aligned}\n\\text { Reaction rate } & =\\text { Rate of disappearance of alkyl halide } \\\\\n& =k \\times[\\mathrm{RX}]\n\\end{aligned}\n$$\n\nTo explain this result, we need to know more about kinetics measurements. Many organic reactions occur in several steps, one of which usually has a higher-energy transition state than the others and is therefore slower. We call this step with the highest transition-state energy the rate-limiting step, or rate-determining step. No reaction can proceed faster than its rate-limiting step, which acts as a kind of traffic jam, or bottleneck. In the $\\mathrm{S}_{\\mathrm{N}} 1$ reaction of $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CBr}$ with $\\mathrm{H}_{2} \\mathrm{O}$, the fact that the nucleophile concentration does not appear in the firstorder rate equation means that it is not involved in the rate-limiting step and must therefore be involved in some other, non-rate-limiting step. The mechanism shown in FIGURE 11.9 accounts for these observations."}
{"id": 649, "contents": "FIGURE 11.9 MECHANISM - \nThe mechanism of the $\\mathrm{S}_{\\mathrm{N}} 1$ reaction of 2-bromo-2-methylpropane with $\\mathrm{H}_{2} \\mathrm{O}$ involves three steps. Step 1 -the spontaneous, unimolecular dissociation of the alkyl bromide to yield a carbocation-is rate-limiting.\n(1) Spontaneous dissociation of the alkyl bromide occurs in a slow, rate-limiting step to generate a carbocation intermediate plus bromide ion.\n2) The carbocation intermediate reacts with water as a nucleophile in a fast step to yield protonated alcohol as product.\n(3) Loss of a proton from the protonated alcohol intermediate then gives the neutral alcohol product.\n\n\n\nUnlike what occurs in an $\\mathrm{S}_{\\mathrm{N}} 2$ reaction, where the leaving group is displaced while the incoming nucleophile approaches, an $\\mathrm{S}_{\\mathrm{N}} 1$ reaction takes place by loss of the leaving group before the nucleophile approaches. 2-Bromo-2-methylpropane spontaneously dissociates to the tert-butyl carbocation $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{C}^{+}$, plus $\\mathrm{Br}^{-}$in a slow, rate-limiting step, and the intermediate carbocation is then immediately trapped by the nucleophile water in\na faster second step. Thus, water is not a reactant in the step whose rate is measured. The energy diagram is shown in FIGURE 11.10.\n\n\nFIGURE 11.10 An energy diagram for an $\\mathbf{S}_{\\mathbf{N}} \\mathbf{1}$ reaction. The rate-limiting step is the spontaneous dissociation of the alkyl halide to give a carbocation intermediate. Reaction of the carbocation with a nucleophile then occurs in a second, faster step."}
{"id": 650, "contents": "FIGURE 11.9 MECHANISM - \nBecause an $\\mathrm{S}_{\\mathrm{N}} 1$ reaction occurs through a carbocation intermediate, its stereochemical outcome is different from that of an $S_{N} 2$ reaction. Carbocations, as we've seen, are planar, $s p^{2}$-hybridized, and achiral. Thus, if we carry out an $\\mathrm{S}_{\\mathrm{N}} 1$ reaction on one enantiomer of a chiral reactant and go through an achiral carbocation intermediate, the product loses its optical activity (Section 8.12). That is, the symmetrical intermediate carbocation can react with a nucleophile equally well from either side, leading to a racemic, 50 : 50 mixture of enantiomers (FIGURE 11.11).\n\n\nFIGURE 11.11 Stereochemistry of the $\\mathbf{S}_{\\mathbf{N}} \\mathbf{1}$ reaction. Because the reaction goes through an achiral intermediate, an enantiomerically pure reactant gives an optically inactive racemic product.\n\nThe conclusion that $\\mathrm{S}_{\\mathrm{N}} 1$ reactions on enantiomerically pure substrates should give racemic products is nearly, but not exactly, what is found. In fact, few $\\mathrm{S}_{\\mathrm{N}} 1$ displacements occur with complete racemization. Most give a minor ( $0-20 \\%$ ) excess of inversion. The reaction of ( $R$ )-6-chloro-2,6-dimethyloctane with $\\mathrm{H}_{2} \\mathrm{O}$, for example, leads to an alcohol product that is approximately $80 \\%$ racemized and $20 \\%$ inverted $(80 \\% R, S+20 \\% S$ is equivalent to $40 \\% R+60 \\% S$ ).\n\n(R)-6-Chloro-2,6-dimethyloctane\n\n60\\% S (inversion)\n\n40\\% R (retention)"}
{"id": 651, "contents": "FIGURE 11.9 MECHANISM - \n(R)-6-Chloro-2,6-dimethyloctane\n\n60\\% S (inversion)\n\n40\\% R (retention)\n\nThis lack of complete racemization in $\\mathrm{S}_{\\mathrm{N}} 1$ reactions is due to the fact that ion pairs are involved. According to this explanation, first proposed by Saul Winstein at UCLA, dissociation of the substrate occurs to give a structure in which the two ions are still loosely associated and in which the carbocation is effectively shielded from reaction on one side by the departing anion. If a certain amount of substitution occurs before the two ions fully diffuse apart, then a net inversion of configuration will be observed (FIGURE 11.12).\n\n\nInversion\nRacemization\nFIGURE 11.12 Ion pairs in an $\\mathbf{S}_{\\mathbf{N}} \\mathbf{1}$ reaction. The leaving group shields one side of the carbocation intermediate from reaction with the nucleophile, thereby leading to some inversion of configuration rather than complete racemization.\n\nPROBLEM What product(s) would you expect from reaction of (S)-3-chloro-3-methyloctane with acetic acid?\n11-8 Show the stereochemistry of both reactant and product.\nPROBLEM Among the many examples of $\\mathrm{S}_{\\mathrm{N}} 1$ reactions that occur with incomplete racemization, the optically 11-9 pure tosylate of 2,2-dimethyl-1-phenyl-1-propanol $\\left([\\alpha]_{D}=-30.3\\right)$ gives the corresponding acetate $\\left([\\alpha]_{D}=+5.3\\right)$ when heated in acetic acid. If complete inversion had occurred, the optically pure acetate would have had $[\\alpha]_{D}=+53.6$. What percentage racemization and what percentage inversion occurred in this reaction?\n\n\nPROBLEM Assign configuration to the following substrate, and show the stereochemistry and identity of the\n11-10 product you would obtain by $S_{N} 1$ reaction with water (reddish brown $=B r$ ):"}
{"id": 652, "contents": "FIGURE 11.9 MECHANISM - 11.5 Characteristics of the $\\mathrm{S}_{\\mathrm{N}} 1$ Reaction\nJust as the $S_{N} 2$ reaction is strongly influenced by the structure of the substrate, the leaving group, the nucleophile, and the solvent, the $S_{N} 1$ reaction is similarly influenced. Factors that lower $\\Delta G^{\\ddagger}$, either by lowering the energy level of the transition state or by raising the energy level of the ground state, favor faster $\\mathrm{S}_{\\mathrm{N}} 1$ reactions. Conversely, factors that raise $\\Delta G^{\\ddagger}$, either by raising the energy level of the transition state or by lowering the energy level of the reactant, slow down the $\\mathrm{S}_{\\mathrm{N}} 1$ reaction."}
{"id": 653, "contents": "The Substrate - \nAccording to the Hammond postulate (Section 7.10), any factor that stabilizes a high-energy intermediate also stabilizes the transition state leading to that intermediate. Because the rate-limiting step in an $\\mathrm{S}_{\\mathrm{N}} 1$ reaction is the spontaneous, unimolecular dissociation of the substrate to yield a carbocation, the reaction is favored whenever a stabilized carbocation intermediate is formed. The more stable the carbocation intermediate, the faster the $\\mathrm{S}_{\\mathrm{N}} 1$ reaction.\n\nWe saw in Section 7.9 that the stability order of alkyl carbocations is $3^{\\circ}>2^{\\circ}>1^{\\circ}>$ methyl. To this list we should also add the resonance-stabilized allyl and benzyl cations. Just as allylic radicals are unusually stable because the unpaired electron can be delocalized over an extended $\\pi$ orbital system (Section 10.4), so allylic and benzylic carbocations are unusually stable. (The word benzylic means \"next to an aromatic ring.\") As FIGURE 11.13 indicates, an allylic cation has two resonance forms. In one form, the double bond is on the \"left\"; in the other form it's on the \"right.\" A benzylic cation has five resonance forms, all of which contribute to the overall resonance hybrid.\n\n\nFIGURE 11.13 Resonance forms of allylic and benzylic carbocations. The positive charge is delocalized over the $\\pi$ system in both. Electron-poor atoms are indicated by blue arrows.\n\nBecause of resonance stabilization, a primary allylic or benzylic carbocation is about as stable as a secondary\nalkyl carbocation, and a secondary allylic or benzylic carbocation is about as stable as a tertiary alkyl carbocation. This stability order of carbocations is the same as the order of $\\mathrm{S}_{\\mathrm{N}} 1$ reactivity for alkyl halides and tosylates."}
{"id": 654, "contents": "Carbocation stability - \nWe should also note parenthetically that primary allylic and benzylic substrates are particularly reactive in $\\mathrm{S}_{\\mathrm{N}} 2$ reactions as well as in $\\mathrm{S}_{\\mathrm{N}} 1$ reactions. Allylic and benzylic $\\mathrm{C}-\\mathrm{X}$ bonds are about $50 \\mathrm{~kJ} / \\mathrm{mol}$ ( $12 \\mathrm{kcal} / \\mathrm{mol}$ ) weaker than the corresponding saturated bonds and are therefore more easily broken.\n\n\nPROBLEM Rank the following substances in order of their expected $\\mathrm{S}_{\\mathrm{N}} 1$ reactivity: 11-11\n\n\nPROBLEM 3-Bromo-1-butene and 1-bromo-2-butene undergo $\\mathrm{S}_{\\mathrm{N}} 1$ reaction at nearly the same rate, even 11-12 though one is a secondary halide and the other is primary. Explain."}
{"id": 655, "contents": "The Leaving Group - \nWe said during the discussion of $S_{N} 2$ reactivity that the best leaving groups are those that are most stable; that is, those that are the conjugate bases of strong acids. An identical reactivity order is found for the $\\mathrm{S}_{\\mathrm{N}} 1$ reaction because the leaving group is directly involved in the rate-limiting step. Thus, the $\\mathrm{S}_{\\mathrm{N}} 1$ reactivity order is\n\n$$\n\\mathrm{HO}^{-}<\\mathrm{Cl}^{-}<\\mathrm{Br}^{-}<\\mathrm{I}^{-} \\approx \\mathrm{TosO}^{-}<\\mathrm{H}_{2} \\mathrm{O}\n$$"}
{"id": 656, "contents": "Leaving group reactivity - \nNote that in the $S_{N} 1$ reaction, which is often carried out under acidic conditions, neutral water is sometimes the leaving group. This occurs, for example, when an alkyl halide is prepared from a tertiary alcohol by reaction with HBr or HCl (Section 10.5). As shown in FIGURE 11.14, the alcohol is first protonated and then spontaneously loses $\\mathrm{H}_{2} \\mathrm{O}$ to generate a carbocation, which reacts with halide ion to give the alkyl halide. Knowing that an $\\mathrm{S}_{\\mathrm{N}} 1$ reaction is involved in the conversion of alcohols to alkyl halides explains why the reaction works well only for tertiary alcohols. Tertiary alcohols react fastest because they give the most stable carbocation intermediates."}
{"id": 657, "contents": "FIGURE 11.14 MECHANISM - \nThe mechanism of the $\\mathbf{S}_{\\mathbf{N}} 1$ reaction of a tertiary alcohol with HBr to yield an alkyl halide. Neutral water is the leaving group (step 2).\n(1) The -OH group is first protonated by HBr .\n(2) Spontaneous dissociation of the protonated alcohol occurs in a slow, rate-limiting step to yield a carbocation intermediate plus water.\n\nCarbocation\n(3) The carbocation intermediate reacts with bromide ion in a fast step to yield the neutral substitution product.\n\n(1)\n\n\u00a9 ||\n\n(3)"}
{"id": 658, "contents": "The Nucleophile - \nThe nature of the nucleophile plays a major role in the $S_{N} 2$ reaction but does not affect an $S_{N} 1$ reaction. Because the $\\mathrm{S}_{\\mathrm{N}} 1$ reaction occurs through a rate-limiting step in which the added nucleophile has no part, the nucleophile can't affect the reaction rate. The reaction of 2-methyl-2-propanol with HX, for instance, occurs at the same rate regardless of whether X is $\\mathrm{Cl}, \\mathrm{Br}$, or I. Furthermore, neutral nucleophiles are just as effective as negatively charged ones, so $\\mathrm{S}_{\\mathrm{N}} 1$ reactions frequently occur under neutral or acidic conditions.\n\n\n2-Methyl-2-propanol\n(Same rate for $\\mathrm{X}=\\mathrm{Cl}, \\mathrm{Br}, \\mathrm{I}$ )"}
{"id": 659, "contents": "The Solvent - \nWhat about the solvent? Do solvents have the same effect in $\\mathrm{S}_{\\mathrm{N}} 1$ reactions that they have in $\\mathrm{S}_{\\mathrm{N}} 2$ reactions? The answer is both yes and no. Yes, solvents have a large effect on $\\mathrm{S}_{\\mathrm{N}} 1$ reactions, but no, the reasons for the effects on $S_{N} 1$ and $S_{N} 2$ reactions are not the same. Solvent effects in the $S_{N} 2$ reaction are due largely to stabilization or destabilization of the nucleophile reactant, while solvent effects in the $\\mathrm{S}_{\\mathrm{N}} 1$ reaction are due largely to stabilization or destabilization of the transition state.\n\nThe Hammond postulate says that any factor stabilizing the intermediate carbocation should increase the rate of an $S_{N} 1$ reaction. Solvation of the carbocation-the interaction of the ion with solvent molecules-has such an\neffect. Solvent molecules orient around the carbocation so that the electron-rich ends of the solvent dipoles face the positive charge (FIGURE 11.15), thereby lowering the energy of the ion and favoring its formation.\n\n\nFIGURE 11.15 Solvation of a carbocation by water. The electron-rich oxygen atoms of solvent molecules orient around the positively charged carbocation and thereby stabilize it.\n\nThe properties of a solvent that contribute to its ability to stabilize ions by solvation are related to the solvent's polarity. $\\mathrm{S}_{\\mathrm{N}} 1$ reactions take place much more rapidly in strongly polar solvents, such as water and methanol, than in less polar solvents, such as ether and chloroform. In the reaction of 2-chloro-2-methylpropane, for example, a rate increase of 100,000 is observed upon going from ethanol (less polar) to water (more polar). The rate increases when going from a hydrocarbon solvent to water are so large they can't be measured accurately."}
{"id": 660, "contents": "The Solvent - \nIt should be emphasized again that both the $\\mathrm{S}_{\\mathrm{N}} 1$ and the $\\mathrm{S}_{\\mathrm{N}} 2$ reaction show solvent effects, but that they do so for different reasons. $\\mathrm{S}_{\\mathrm{N}} 2$ reactions are disfavored in protic solvents because the ground-state energy of the nucleophile is lowered by solvation. $\\mathrm{S}_{\\mathrm{N}} 1$ reactions are favored in protic solvents because the transition-state energy leading to carbocation intermediate is lowered by solvation."}
{"id": 661, "contents": "A Summary of $\\mathrm{S}_{\\mathrm{N}} 1$ Reaction Characteristics - \nThe effects on $\\mathrm{S}_{\\mathrm{N}} 1$ reactions of the four variables-substrate, leaving group, nucleophile, and solvent-are summarized in the following statements:\n\n| Substrate | The best substrates yield the most stable carbocations. As a result, $\\mathrm{S}_{\\mathrm{N}} 1$ reactions are best <br> for tertiary, allylic, and benzylic halides. |\n| :--- | :--- |\n| Leaving | Good leaving groups increase the reaction rate by lowering the energy level of the <br> group <br> transition state for carbocation formation. |\n| Nucleophile | The nucleophile must be nonbasic to prevent a competitive elimination of HX (Section <br> 11.7), but otherwise does not affect the reaction rate. Neutral nucleophiles work well. |\n| Solvent | Polar solvents stabilize the carbocation intermediate by solvation, thereby increasing the <br> reaction rate. |"}
{"id": 662, "contents": "Predicting the Mechanism of a Nucleophilic Substitution Reaction - \nPredict whether each of the following substitution reactions is likely to be $\\mathrm{S}_{\\mathrm{N}} 1$ or $\\mathrm{S}_{\\mathrm{N}} 2$ :"}
{"id": 663, "contents": "Strategy - \nLook at the substrate, leaving group, nucleophile, and solvent. Then decide from the summaries at the ends of Section 11.3 and Section 11.5 whether an $S_{N} 1$ or an $S_{N} 2$ reaction is favored. $S_{N} 1$ reactions are favored by tertiary, allylic, or benzylic substrates, by good leaving groups, by nonbasic nucleophiles, and by protic solvents. $\\mathrm{S}_{\\mathrm{N}} 2$ reactions are favored by primary substrates, by good leaving groups, by good nucleophiles, and by polar aprotic solvents."}
{"id": 664, "contents": "Solution - \n(a) This is likely to be an $\\mathrm{S}_{\\mathrm{N}} 1$ reaction because the substrate is secondary and benzylic, the nucleophile is weakly basic, and the solvent is protic.\n(b) This is likely to be an $\\mathrm{S}_{\\mathrm{N}} 2$ reaction because the substrate is primary, the nucleophile is a good one, and the solvent is polar aprotic.\n\nPROBLEM Predict whether each of the following substitution reactions is likely to be $S_{N} 1$ or $S_{N} 2$ :"}
{"id": 665, "contents": "11-13 (a) - \n(b)"}
{"id": 666, "contents": "11-13 (a) - 11.6 Biological Substitution Reactions\nBoth $\\mathrm{S}_{\\mathrm{N}} 1$ and $\\mathrm{S}_{\\mathrm{N}} 2$ reactions are common in biological chemistry, particularly in the pathways for biosynthesis of the many thousands of plant-derived substances called terpenoids, which we saw briefly in the Chapter 8 Chemistry Matters and will discuss in Section 27.5. Unlike what typically happens in the laboratory, however, the substrate in a biological substitution reaction is usually an organodiphosphate rather than an alkyl halide. Thus, the leaving group is the diphosphate ion, abbreviated $\\mathrm{PP}_{\\mathrm{i}}$, rather than a halide ion. In fact, it's useful to think of the diphosphate group as the \"biological equivalent\" of a halide. The dissociation of an organodiphosphate in a biological reaction is typically assisted by complexation to a divalent metal cation such as $\\mathrm{Mg}^{2+}$ to help neutralize charge and make the diphosphate a better leaving group."}
{"id": 667, "contents": "An organodiphosphate - \nDiphosphate ion\nAs an example, two $S_{N} 1$ reactions occur during the biosynthesis of geraniol, a fragrant alcohol found in roses and used in perfumery. Geraniol biosynthesis begins with dissociation of dimethylallyl diphosphate to give an allylic carbocation, which reacts with isopentenyl diphosphate (FIGURE 11.16). From the viewpoint of isopentenyl diphosphate, the reaction is an electrophilic alkene addition, but from the viewpoint of dimethylallyl diphosphate, the process is an $S_{N} 1$ reaction in which the carbocation intermediate reacts with a double bond as the nucleophile.\n\n\nFIGURE 11.16 Biosynthesis of geraniol from dimethylallyl diphosphate. Two $\\mathrm{S}_{\\mathrm{N}} 1$ reactions occur, both with diphosphate ion as the leaving group.\n\nFollowing this initial $\\mathrm{S}_{\\mathrm{N}} 1$ reaction, loss of the pro- $R$ hydrogen gives geranyl diphosphate, itself an allylic diphosphate that dissociates a second time. Reaction of the geranyl carbocation with water in a second $\\mathrm{S}_{\\mathrm{N}} 1$ reaction, followed by loss of a proton, then yields geraniol."}
{"id": 668, "contents": "An organodiphosphate - \nAs another example, $\\mathrm{S}_{\\mathrm{N}} 2$ reactions are involved in almost all biological methylations, which transfer a $-\\mathrm{CH}_{3}$ group from an electrophilic donor to a nucleophile. The donor is $S$-adenosylmethionine (abbreviated SAM), which contains a positively charged sulfur (a sulfonium ion, Section 5.12), and the leaving group is the neutral $S$-adenosylhomocysteine molecule. In the biosynthesis of epinephrine (adrenaline) from norepinephrine, for\ninstance, the nucleophilic nitrogen atom of norepinephrine attacks the electrophilic methyl carbon atom of $S$-adenosylmethionine in an $\\mathrm{S}_{\\mathrm{N}} 2$ reaction, displacing $S$-adenosylhomocysteine (FIGURE 11.17). In effect, $S$-adenosylmethionine is simply a biological equivalent of $\\mathrm{CH}_{3} \\mathrm{Cl}$.\n\n\nFIGURE 11.17 The biosynthesis of epinephrine from norepinephrine occurs by an $\\mathbf{S}_{\\mathbf{N}} \\mathbf{2}$ reaction with $\\boldsymbol{S}$-adenosylmethionine.\nPROBLEM Review the mechanism of geraniol biosynthesis shown in Figure 11.16, and propose a mechanism\n11-14 for the biosynthesis of limonene from linalyl diphosphate.\n\n\nLinalyl diphosphate\nLimonene"}
{"id": 669, "contents": "An organodiphosphate - 11.7 Elimination Reactions: Zaitsev's Rule\nWe said at the beginning of this chapter that two kinds of reactions can take place when a nucleophile/Lewis base reacts with an alkyl halide. The nucleophile can either substitute for the halide by reaction at carbon or can cause elimination of HX by reaction at a neighboring hydrogen:\n\n\nElimination reactions are more complex than substitution reactions for several reasons. One is the problem of regiochemistry. What product results by loss of HX from an unsymmetrical halide? In fact, elimination reactions almost always give mixtures of alkene products, and the best we can usually do is to predict which will be the major product.\n\nAccording to Zaitsev's rule, formulated in 1875 by the Russian chemist Alexander Zaitsev, base-induced elimination reactions generally (although not always) give the more stable alkene product-that is, the alkene with more alkyl substituents on the double-bond carbons. In the following two cases, for example, the more highly substituted alkene product predominates."}
{"id": 670, "contents": "ZAITSEV'S RULE - \nIn the elimination of HX from an alkyl halide, the more highly substituted alkene product predominates."}
{"id": 671, "contents": "2-Bromo-2-methylbutane - \n2-Methyl-2-butene\n(70\\%)\n\n2-Methyl-1-butene\n(30\\%)\n\nAnother factor that complicates a study of elimination reactions is that they can take place by different mechanisms, just as substitutions can. We'll consider three of the most common mechanisms-the E1, E2, and E1cB reactions-which differ in the timing of $\\mathrm{C}-\\mathrm{H}$ and $\\mathrm{C}-\\mathrm{X}$ bond-breaking.\n\nIn the E1 reaction, the $\\mathrm{C}-\\mathrm{X}$ bond breaks first to give a carbocation intermediate, which undergoes subsequent base abstraction of $\\mathrm{H}^{+}$to yield the alkene. In the E 2 reaction, base-induced $\\mathrm{C}-\\mathrm{H}$ bond cleavage is simultaneous with $\\mathrm{C}-\\mathrm{X}$ bond cleavage, giving the alkene in a single step. In the E1cB reaction (cB for \"conjugate base\"), base abstraction of the proton occurs first, giving a carbanion ( $\\mathrm{R}:^{-}$) intermediate. This anion, the conjugate base of the reactant \"acid,\" then undergoes loss of $\\mathrm{X}^{-}$in a subsequent step to give the alkene. All three mechanisms occur frequently in the laboratory, but the E1cB mechanism predominates in biological pathways.\n\nE1 Reaction: $C-X$ bond breaks first to give a carbocation intermediate, followed by base removal of a proton to yield the alkene.\n\n\nE2 Reaction: $\\mathrm{C}-\\mathrm{H}$ and $\\mathrm{C}-\\mathrm{X}$ bonds break simultaneously, giving the alkene in a single step without intermediates.\n\n\nE1cB Reaction: $\\mathrm{C}-\\mathrm{H}$ bond breaks first, giving a carbanion intermediate that loses $\\mathrm{X}^{-}$to form the alkene."}
{"id": 672, "contents": "Predicting the Product of an Elimination Reaction - \nWhat product would you expect from reaction of 1-chloro-1-methylcyclohexane with KOH in ethanol?"}
{"id": 673, "contents": "Strategy - \nTreatment of an alkyl halide with a strong base such as KOH yields an alkene. To find the products in a specific case, locate the hydrogen atoms on each carbon next to the leaving group, and then generate the potential alkene products by removing HX in as many ways as possible. The major product will be the one that has the most highly substituted double bond-in this case, 1-methylcyclohexene."}
{"id": 674, "contents": "Solution - \nPROBLEM Ignoring double-bond stereochemistry, what products would you expect from elimination reactions\n11-15 of the following alkyl halides? Which product will be the major product in each case?\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM What alkyl halides might the following alkenes have been made from?\n11-16 (a)\n\n(b)"}
{"id": 675, "contents": "Solution - 11.8 The E2 Reaction and the Deuterium Isotope Effect\nThe E2 reaction (for elimination, bimolecular) occurs when an alkyl halide is treated with a strong base, such as hydroxide ion or alkoxide ion ( $\\mathrm{RO}^{-}$). It is the most commonly occurring pathway for elimination and can be formulated as shown in FIGURE 11.18."}
{"id": 676, "contents": "FIGURE 11.18 MECHANISM - \nMechanism of the E2 reaction of an alkyl halide. The reaction takes place in a single step through a transition state in which the double bond begins to form at the same time the H and X groups are leaving.\n(1) Base (B:) attacks a neighboring hydrogen and begins to remove the H at the same time as the alkene double bond starts to form and the X group starts to leave.\n(2) Neutral alkene is produced when the $\\mathrm{C}-\\mathrm{H}$ bond is fully broken and the X group has departed with the $\\mathrm{C}-\\mathrm{X}$ bond electron pair.\n\n(1)\n\n\nTransition state\n\n\n\nLike the $\\mathrm{S}_{\\mathrm{N}} 2$ reaction, the E2 reaction takes place in one step without intermediates. As the base begins to abstract $\\mathrm{H}^{+}$from a carbon next to the leaving group, the $\\mathrm{C}-\\mathrm{H}$ bond begins to break, a $\\mathrm{C}=\\mathrm{C}$ bond begins to form, and the leaving group begins to depart, taking with it the electron pair from the $\\mathrm{C}-\\mathrm{X}$ bond. Among the pieces of evidence supporting this mechanism is the fact that E2 reactions show second-order kinetics and follow the rate law: rate $=k \\times[\\mathrm{RX}] \\times[\\mathrm{Base}]$. That is, both the base and alkyl halide take part in the rate-limiting step."}
{"id": 677, "contents": "FIGURE 11.18 MECHANISM - \nA second piece of evidence in support of the E2 mechanism is provided by a phenomenon known as the deuterium isotope effect. For reasons that we won't go into, a carbon-hydrogen bond is weaker by about $5 \\mathrm{~kJ} / \\mathrm{mol}(1.2 \\mathrm{kcal} / \\mathrm{mol})$ than the corresponding carbon-deuterium bond. Thus, a $\\mathrm{C}-\\mathrm{H}$ bond is more easily broken than an equivalent $\\mathrm{C}-\\mathrm{D}$ bond, and the rate of $\\mathrm{C}-\\mathrm{H}$ bond cleavage is faster. For instance, the baseinduced elimination of HBr from 1-bromo-2-phenylethane proceeds 7.11 times faster than the corresponding elimination of DBr from 1-bromo-2, 2-dideuterio-2-phenylethane. This result tells us that the $\\mathrm{C}-\\mathrm{H}$ (or $\\mathrm{C}-\\mathrm{D}$ ) bond is broken in the rate-limiting step, consistent with our picture of the E 2 reaction as a one-step process. If it were otherwise, we wouldn't observe a rate difference."}
{"id": 678, "contents": "(H)-Faster reaction <br> (D)-Slower reaction - \nYet a third piece of mechanistic evidence involves the stereochemistry of E2 eliminations. As shown by a large number of experiments, E2 reactions occur with periplanar geometry, meaning that all four reacting atoms-the hydrogen, the two carbons, and the leaving group-lie in the same plane. Two such geometries are possible: syn periplanar geometry, in which the H and the X are on the same side of the molecule, and anti periplanar geometry, in which the H and the X are on opposite sides of the molecule. Of the two, anti periplanar geometry is energetically preferred because it allows the substituents on the two carbons to adopt a staggered relationship, whereas syn geometry requires that the substituents be eclipsed.\n\n\nAnti periplanar geometry (staggered, lower energy)\n\n\n\n\nSyn periplanar geometry\n(eclipsed, higher energy)\nWhat's so special about periplanar geometry? Because the $s p^{3} \\sigma$ orbitals in the reactant $\\mathrm{C}-\\mathrm{H}$ and $\\mathrm{C}-\\mathrm{X}$ bonds must overlap and become $p \\pi$ orbitals in the alkene product, there must also be some overlap in the transition state. This can occur most easily if all the orbitals are in the same plane to begin with-that is, if they're periplanar (FIGURE 11.19).\n\n\nFIGURE 11.19 The transition state for the E2 reaction of an alkyl halide with base. Overlap of the developing $p$ orbitals in the transition state requires periplanar geometry of the reactant.\n\nYou can think of E2 elimination reactions with periplanar geometry as being similar to $\\mathrm{S}_{\\mathrm{N}} 2$ reactions with $180^{\\circ}$ geometry. In an $\\mathrm{S}_{\\mathrm{N}} 2$ reaction, an electron pair from the incoming nucleophile pushes out the leaving group on the opposite side of the molecule. In an E2 reaction, an electron pair from a neighboring $\\mathrm{C}-\\mathrm{H}$ bond also pushes out the leaving group on the opposite side of the molecule.\n\n$\\mathrm{S}_{\\mathrm{N}} 2$ reaction (backside attack)"}
{"id": 679, "contents": "(H)-Faster reaction <br> (D)-Slower reaction - \n$\\mathrm{S}_{\\mathrm{N}} 2$ reaction (backside attack)\n\n\nE2 reaction (anti periplanar)\n\nAnti periplanar geometry for E2 eliminations has specific stereochemical consequences that provide strong evidence for the proposed mechanism. To take just one example, meso-1,2-dibromo-1,2-diphenylethane undergoes E2 elimination on treatment with base to give only the $E$ alkene. None of the isomeric $Z$ alkene is formed because the transition state leading to the $Z$ alkene would have to have syn periplanar geometry and\nwould thus be higher in energy."}
{"id": 680, "contents": "WORKED EXAMPLE 11.4 - \nPredicting the Double-Bond Stereochemistry of the Product in an E2 Reaction\nWhat stereochemistry do you expect for the alkene obtained by E2 elimination of (1S,2S)-1,2-dibromo-1,2-diphenylethane?"}
{"id": 681, "contents": "Strategy - \nDraw ( $1 S, 2 S$ )-1,2-dibromo-1,2-diphenylethane so that you can see its stereochemistry and so that the -H and -Br groups to be eliminated are anti periplanar. Then carry out the elimination while keeping all substituents in approximately the same positions, and see what alkene results."}
{"id": 682, "contents": "Solution - \nAnti periplanar elimination of HBr gives (Z)-1-bromo-1,2-diphenylethylene.\n\n\n\nPROBLEM What stereochemistry do you expect for the alkene obtained by E2 elimination of 11-17 ( $1 R, 2 R$ )-1,2-dibromo-1,2-diphenylethane? Draw a Newman projection of the reacting conformation.\n\nPROBLEM What stereochemistry do you expect for the trisubstituted alkene obtained by E2 elimination of the\n11-18 following alkyl halide on treatment with KOH ? (reddish brown = Br.)"}
{"id": 683, "contents": "Solution - 11.9 The E2 Reaction and Cyclohexane Conformation\nAnti periplanar geometry for E2 reactions is particularly important in cyclohexane rings, where chair geometry forces a rigid relationship between the substituents on neighboring carbon atoms (Section 4.8). The anti periplanar requirement for E2 reactions overrides Zaitsev's rule and can be met in cyclohexanes only if the hydrogen and the leaving group are trans diaxial (FIGURE 11.20). If either the leaving group or the hydrogen is\nequatorial, E2 elimination can't occur."}
{"id": 684, "contents": "Equatorial chlorine: H and Cl are not anti periplanar - \nFIGURE 11.20 The geometric requirement for an E2 reaction in a substituted cyclohexane. The leaving group and the hydrogen must both be axial for anti periplanar elimination to occur.\n\nThe elimination of HCl from the isomeric menthyl and neomenthyl chlorides shown in FIGURE 11.21 gives a good illustration of this trans-diaxial requirement. Neomenthyl chloride undergoes elimination of HCl on reaction with ethoxide ion 200 times faster than menthyl chloride. Furthermore, neomenthyl chloride yields 3 -menthene as the major alkene product, whereas menthyl chloride yields 2-menthene.\n\n\nNeomenthyl chloride\n(b)\n\n\nMenthyl chloride\n\n$\\downarrow \\mid$ Ring-flip\n\n\n\nFIGURE 11.21 Dehydrochlorination of menthyl and neomenthyl chlorides. (a) Neomenthyl chloride loses HCl directly from its more stable conformation, but (b) menthyl chloride must first ring-flip to a higher energy conformation before HCl loss can occur. The abbreviation \"Et\" represents an ethyl group.\n\nThe difference in reactivity between the isomeric menthyl chlorides is due to the difference in their conformations. Neomenthyl chloride has the conformation shown in FIGURE 11.21a, with the methyl and isopropyl groups equatorial and the chlorine axial-a perfect geometry for E2 elimination. Loss of the hydrogen atom at C 4 occurs easily to yield the more substituted alkene product, 3-menthene, as predicted by Zaitsev's rule.\n\nMenthyl chloride, by contrast, has a conformation in which all three substituents are equatorial (FIGURE 11.21b). To achieve the necessary geometry for elimination, menthyl chloride must first ring-flip to a higherenergy chair conformation, in which all three substituents are axial. E2 elimination then occurs with loss of\nthe only trans-diaxial hydrogen available, leading to the non-Zaitsev product 2 -menthene. The net effect of the simple change in chlorine stereochemistry is a 200 -fold change in reaction rate and a complete change of product. The chemistry of the molecule is controlled by its conformation."}
{"id": 685, "contents": "Equatorial chlorine: H and Cl are not anti periplanar - \nPROBLEM Which isomer would you expect to undergo E2 elimination faster, 11-19 trans-1-bromo-4-tert-butylcyclohexane or cis-1-bromo-4-tert-butylcyclohexane? Draw each molecule in its more stable chair conformation, and explain your answer."}
{"id": 686, "contents": "Equatorial chlorine: H and Cl are not anti periplanar - 11.10 The E1 and E1cB Reactions\nThe E1 Reaction\nJust as the E2 reaction is analogous to the $\\mathrm{S}_{\\mathrm{N}} 2$ reaction, the $\\mathrm{S}_{\\mathrm{N}} 1$ reaction has a close analog called the E1 reaction (for elimination, unimolecular). The E1 reaction can be formulated as shown in FIGURE 11.22, with the elimination of HCl from 2-chloro-2-methylpropane.\n\nFIGURE 11.22 MECHANISM\nMechanism of the E1 reaction. Two steps are involved, the first of which is ratelimiting, and a carbocation intermediate is present.\n\n\n1 Spontaneous dissociation of the tertiary alkyl chloride yields an intermediate carbocation in a slow, rate-limiting step.\n(1) $\\| \\begin{aligned} & \\text { Rate- } \\\\ & \\text { limiting }\\end{aligned}$\n\nCarbocation\n\n2 Loss of a neighboring $\\mathrm{H}^{+}$in a fast step yields the neutral alkene product. The electron pair from the $\\mathrm{C}-\\mathrm{H}$ bond goes to form the alkene $\\pi$ bond.\n\n(2) $\\downarrow$ Fast\n\n\nE1 eliminations begin with the same unimolecular dissociation to give a carbocation that we saw in the $\\mathrm{S}_{\\mathrm{N}} 1$ reaction, but the dissociation is followed by loss of $\\mathrm{H}^{+}$from the adjacent carbon rather than by substitution. In fact, the E1 and $\\mathrm{S}_{\\mathrm{N}} 1$ reactions normally occur together whenever an alkyl halide is treated in a protic solvent with a nonbasic nucleophile. Thus, the best E1 substrates are also the best $\\mathrm{S}_{\\mathrm{N}} 1$ substrates, and mixtures of substitution and elimination products are usually obtained. For example, when 2-chloro-2-methylpropane is warmed to $65^{\\circ} \\mathrm{C}$ in $80 \\%$ aqueous ethanol, a 64 : 36 mixture of 2-methyl-2-propanol ( $\\mathrm{S}_{\\mathrm{N}} 1$ ) and 2-methylpropene (E1) results."}
{"id": 687, "contents": "Equatorial chlorine: H and Cl are not anti periplanar - 11.10 The E1 and E1cB Reactions\n2-Chloro-2-methylpropane\n\n\n2-Methyl-2-propanol (64\\%)\n$+$\n\n\n2-Methylpropene (36\\%)\n\nMuch evidence has been obtained in support of the E1 mechanism. For example, E1 reactions show first-order kinetics, consistent with a rate-limiting, unimolecular dissociation process. Furthermore, E1 reactions show no\ndeuterium isotope effect because rupture of the $\\mathrm{C}-\\mathrm{H}($ or $\\mathrm{C}-\\mathrm{D}$ ) bond occurs after the rate-limiting step rather than during it. Thus, we can't measure a rate difference between a deuterated and nondeuterated substrate.\n\nA final piece of evidence involves the stereochemistry of elimination. Unlike the E2 reaction, where anti periplanar geometry is required, there is no geometric requirement on the E1 reaction because the halide and the hydrogen are lost in separate steps. We might therefore expect to obtain the more stable (Zaitsev's rule) product from E1 reaction, which is just what we find. To return to a familiar example, menthyl chloride loses HCl under E1 conditions in a polar solvent to give a mixture of alkenes in which the Zaitsev product, 3-menthene, predominates (FIGURE 11.23).\n\n\nFIGURE 11.23 Elimination reactions of menthyl chloride. E2 conditions (1, strong base in 100\\% ethanol) lead to 2-menthene through an anti periplanar elimination, whereas E1 conditions ( 2 , dilute base in $80 \\%$ aqueous ethanol) lead to a mixture of 2 -menthene and 3-menthene."}
{"id": 688, "contents": "The E1cB Reaction - \nIn contrast to the E1 reaction, which involves a carbocation intermediate, the E1cB reaction takes place through a carbanion intermediate. Base-induced abstraction of a proton in a slow, rate-limiting step gives an anion, which expels a leaving group on the adjacent carbon. The reaction is particularly common in substrates that have a poor leaving group, such as -OH , two carbons removed from a carbonyl group, as in $\\mathrm{HOC}-\\mathrm{CH}-\\mathrm{C}=\\mathrm{O}$. The poor leaving group disfavors the alternative E1 and E2 possibilities, and the carbonyl group makes the adjacent hydrogen unusually acidic by resonance stabilization of the anion intermediate. We'll look at this acidifying effect of a carbonyl group in Section 22.5."}
{"id": 689, "contents": "The E1cB Reaction - 11.11 Biological Elimination Reactions\nAll three elimination reactions-E2, E1, and E1cB-occur in biological pathways, but the E1cB mechanism is particularly common. The substrate is usually an alcohol rather than an alkyl halide, and the H atom removed is usually adjacent to a carbonyl group, just as in laboratory reactions. Thus, 3-hydroxy carbonyl compounds are frequently converted to unsaturated carbonyl compounds by elimination reactions. A typical example occurs during the biosynthesis of fats and oils when a 3-hydroxybutyryl thioester is dehydrated to the corresponding unsaturated (crotonyl) thioester. The base in this reaction is a histidine amino acid in the enzyme, and the loss of the -OH group is assisted by simultaneous protonation."}
{"id": 690, "contents": "3-Hydroxybutyryl - \nthioester"}
{"id": 691, "contents": "3-Hydroxybutyryl - 11.12 A Summary of Reactivity: $\\mathrm{S}_{\\mathrm{N}} 1, \\mathrm{~S}_{\\mathrm{N}} 2$, E1, E1cB, and E2\n$\\mathrm{S}_{\\mathrm{N}} 1, \\mathrm{~S}_{\\mathrm{N}} 2$, E1, E1cB, E2-how can you keep it all straight and predict what will happen in any given case? Will substitution or elimination occur? Will the reaction be bimolecular or unimolecular? There are no rigid answers to these questions, but it's possible to recognize some trends and make some generalizations.\n\n- Primary alkyl halides $\\mathrm{S}_{\\mathrm{N}} 2$ substitution occurs if a good nucleophile is used, E2 elimination occurs if a strong, sterically hindered base is used, and E1cB elimination occurs if the leaving group is two carbons away from a carbonyl group.\n- Secondary alkyl halides $\\mathrm{S}_{\\mathrm{N}} 2$ substitution occurs if a weakly basic nucleophile is used in a polar aprotic solvent, E2 elimination predominates if a strong base is used, and E1cB elimination takes place if the leaving group is two carbons away from a carbonyl group. Secondary allylic and benzylic alkyl halides can also undergo $\\mathrm{S}_{\\mathrm{N}} 1$ and E1 reactions if a weakly basic nucleophile is used in a protic solvent.\n- Tertiary alkyl halides E2 elimination occurs when a base is used, but $\\mathrm{S}_{\\mathrm{N}} 1$ substitution and E1 elimination occur together under neutral conditions, such as in pure ethanol or water. E1cB elimination takes place if the leaving group is two carbons away from a carbonyl group."}
{"id": 692, "contents": "Predicting the Product and Mechanism of Reactions - \nTell whether each of the following reactions is likely to be $\\mathrm{S}_{\\mathrm{N}} 1, \\mathrm{~S}_{\\mathrm{N}} 2$, E1, E1cB, or E2, and predict the product of each:\n\n(b)"}
{"id": 693, "contents": "Strategy - \nLook carefully in each reaction at the structure of the substrate, the leaving group, the nucleophile, and the solvent. Then decide from the preceding summary which kind of reaction is likely to be favored."}
{"id": 694, "contents": "Solution - \n(a) A secondary, nonallylic substrate can undergo an $\\mathrm{S}_{\\mathrm{N}} 2$ reaction with a good nucleophile in a polar aprotic solvent but will undergo an E2 reaction on treatment with a strong base in a protic solvent. In this case, E2 reaction is likely to predominate.\n\n(b) A secondary benzylic substrate can undergo an $\\mathrm{S}_{\\mathrm{N}} 2$ reaction on treatment with a nonbasic nucleophile in a polar aprotic solvent and will undergo an E2 reaction on treatment with a base. Under protic conditions, such as aqueous formic acid $\\left(\\mathrm{HCO}_{2} \\mathrm{H}\\right)$, an $\\mathrm{S}_{\\mathrm{N}} 1$ reaction is likely, along with some E1 reaction.\n\n\nPROBLEM Tell whether each of the following reactions is likely to be $\\mathrm{S}_{\\mathrm{N}} 1, \\mathrm{~S}_{\\mathrm{N}} 2$, E1, E1cB, or E2:\n11-20 (a)\n\n(b)\n\n(c)\n\n\n(d)"}
{"id": 695, "contents": "Green Chemistry - \nOrganic chemistry in the 20th century changed the world, giving us new medicines, food preservatives, insecticides, adhesives, textiles, dyes, building materials, composites, and all manner of polymers. But these advances did not come without a cost: Almost every chemical process produces waste that must be dealt with, including reaction solvents and toxic by-products that might evaporate into the air or be leached into groundwater if not disposed of properly. Even apparently harmless by-products must be safely buried or otherwise sequestered. As always, there's no such thing as a free lunch. With the good also comes the bad.\n\nIt may never be possible to make organic chemistry completely benign, but awareness of the environmental problems caused by many chemical processes has grown dramatically in recent years, giving rise to a movement called green chemistry. Green chemistry is the design and implementation of chemical products and processes that reduce waste and attempt to eliminate the generation of hazardous substances. There are 12 principles of green chemistry:\n\nPrevent waste - Waste should be prevented rather than treated or cleaned up after it has been created.\nMaximize atom economy - Synthetic methods should maximize the incorporation of all materials used in a process into the final product so that waste is minimized.\nUse less hazardous processes - Synthetic methods should use reactants and generate wastes with minimal toxicity to health and the environment.\nDesign safer chemicals - Chemical products should be designed to have minimal toxicity.\nUse safer solvents - Minimal use should be made of solvents, separation agents, and other auxiliary substances in a reaction.\nDesign for energy efficiency - Energy requirements for chemical processes should be minimized, with reactions carried out at room temperature if possible.\nUse renewable feedstocks - Raw materials should come from renewable sources when feasible.\nMinimize derivatives - Syntheses should be designed with minimal use of protecting groups to avoid extra steps and reduce waste.\nUse catalysis - Reactions should be catalytic rather than stoichiometric.\nDesign for degradation - Products should be designed to be biodegradable at the end of their useful lifetimes.\nMonitor pollution in real time - Processes should be monitored in real time for the formation of hazardous substances.\nPrevent accidents - Chemical substances and processes should minimize the potential for fires, explosions, or other accidents."}
{"id": 696, "contents": "Green Chemistry - \nThe foregoing 12 principles may not all be met in most real-world applications, but they provide a worthy goal and they can make chemists think more carefully about the environmental implications of their work. Real success stories have occurred, and more are in progress. Approximately 7 million pounds per year of ibuprofen ( 6 billion tablets!) are now made by a \"green\" process that produces approximately $99 \\%$ less waste than the process it replaces. Only three steps are needed, the anhydrous HF solvent used in the first step is recovered and reused, and the second and third steps are catalytic."}
{"id": 697, "contents": "Key Terms - \n- anti periplanar\n- benzylic\n- bimolecular\n- deuterium isotope effect\n- E1 reaction\n- E1cB reaction\n- E2 reaction\n- first-order reaction\n- ion pair\n- kinetics\n- leaving group\n- periplanar\n- rate-determining step\n- rate-limiting step\n- second-order reaction\n- $\\mathrm{S}_{\\mathrm{N}} 1$ reaction\n- $\\mathrm{S}_{\\mathrm{N}} 2$ reaction\n- solvation\n- syn periplanar\n- tosylate\n- unimolecular\n- Zaitsev\u2019s rule\n- nucleophilic substitution reaction"}
{"id": 698, "contents": "Summary - \nThe reaction of an alkyl halide or tosylate with a nucleophile/base results either in substitution or in elimination. The resultant nucleophilic substitution and base-induced elimination reactions are two of the most widely occurring and versatile reaction types in organic chemistry, both in the laboratory and in biological pathways.\n\nNucleophilic substitutions are of two types: $\\mathbf{S}_{\\mathbf{N}} \\mathbf{2}$ reactions and $\\mathbf{S}_{\\mathbf{N}} \\mathbf{1}$ reactions. In the $\\mathrm{S}_{\\mathrm{N}} 2$ reaction, the entering nucleophile approaches the halide from a direction $180^{\\circ}$ away from the leaving group, resulting in an umbrellalike inversion of configuration at the carbon atom. The reaction is kinetically second-order and is strongly inhibited by increasing steric bulk of the reactants. Thus, $\\mathrm{S}_{\\mathrm{N}} 2$ reactions are favored for primary and secondary substrates.\n\nIn the $\\mathrm{S}_{\\mathrm{N}} 1$ reaction, the substrate spontaneously dissociates to a carbocation in a slow rate-limiting step, followed by a rapid reaction with the nucleophile. As a result, $\\mathrm{S}_{\\mathrm{N}} 1$ reactions are kinetically first-order and take place with substantial racemization of configuration at the carbon atom. They are most favored for tertiary substrates. Both $\\mathrm{S}_{\\mathrm{N}} 1$ and $\\mathrm{S}_{\\mathrm{N}} 2$ reactions occur in biological pathways, although the leaving group is typically a diphosphate ion rather than a halide."}
{"id": 699, "contents": "Summary - \nEliminations of alkyl halides to yield alkenes occur by three mechanisms: E2 reactions, E1 reactions, and E1cB reactions, which differ in the timing of $\\mathrm{C}-\\mathrm{H}$ and $\\mathrm{C}-\\mathrm{X}$ bond-breaking. In the E 2 reaction, $\\mathrm{C}-\\mathrm{H}$ and $\\mathrm{C}-\\mathrm{X}$ bondbreaking occur simultaneously when a base abstracts $\\mathrm{H}^{+}$from one carbon while the leaving group departs from the neighboring carbon. The reaction takes place preferentially through an anti periplanar transition state in which the four reacting atoms-hydrogen, two carbons, and leaving group-are in the same plane. The reaction shows second-order kinetics and a deuterium isotope effect, and occurs when a secondary or tertiary substrate is treated with a strong base. These elimination reactions usually give a mixture of alkene products in which the more highly substituted alkene predominates (Zaitsev's rule).\n\nIn the E1 reaction, $\\mathrm{C}-\\mathrm{X}$ bond-breaking occurs first. The substrate dissociates to yield a carbocation in the\nslow rate-limiting step before losing $\\mathrm{H}^{+}$from an adjacent carbon in a second step. The reaction shows firstorder kinetics and no deuterium isotope effect and occurs when a tertiary substrate reacts in polar, nonbasic solution.\n\nIn the E1cB reaction, $\\mathrm{C}-\\mathrm{H}$ bond-breaking occurs first. A base abstracts a proton to give a carbanion, followed by loss of the leaving group from the adjacent carbon in a second step. The reaction is favored when the leaving group is two carbons removed from a carbonyl, which stabilizes the intermediate anion by resonance. Biological elimination reactions typically occur by this E1cB mechanism.\n\nIn general, substrates react in the following way:"}
{"id": 700, "contents": "Summary - \nIn general, substrates react in the following way:\n\n$$\n\\begin{array}{lll}\n\\begin{array}{l}\n\\mathrm{RCH}_{2} \\mathrm{X} \\\\\n\\text { (primary) }\n\\end{array} & \\longrightarrow & {\\text { Mostly } \\mathrm{S}_{\\mathrm{N}} 2 \\text { substitution }} \\\\\n\\begin{array}{l}\n\\mathrm{R}_{2} \\mathrm{CHX} \\\\\n\\text { (secondary) }\n\\end{array} & \\longrightarrow & \\begin{array}{l}\n\\mathrm{S}_{\\mathrm{N}} 2 \\text { substitution with nonbasic nucleophiles } \\\\\n\\mathrm{E} 2 \\text { elimination with strong bases }\n\\end{array} \\\\\n\\begin{array}{ll}\n\\mathrm{R}_{3} \\mathrm{CX} \\\\\n\\text { (tertiary) }\n\\end{array} & \\longrightarrow & \\begin{array}{l}\n\\text { Mostly E2 elimination } \\\\\n\\text { (S } \\mathrm{S}_{\\mathrm{N}} 1 \\text { substitution and E1 elimination in nonbasic solvents) }\n\\end{array}\n\\end{array}\n$$"}
{"id": 701, "contents": "Summary of Reactions - \n1. Nucleophilic substitutions\na. $\\mathrm{S}_{\\mathrm{N}} 1$ reaction of $3^{\\circ}$, allylic, and benzylic halides (Section 11.4 and Section 11.5)\n\nb. $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of $1^{\\circ}$ and simple $2^{\\circ}$ halides (Section 11.2 and Section 11.3)\n\n2. Eliminations\na. E1 reaction (Section 11.10)\n\nb. E1cB reaction (Section 11.10)\n\nc. E2 reaction (Section 11.8)"}
{"id": 702, "contents": "Visualizing Chemistry - \nPROBLEM Write the product you would expect from reaction of each of the following alkyl halides with (1) $\\mathrm{Na}^{+}$\n11-21 ${ }^{-} \\mathrm{SCH}_{3}$ and (2) $\\mathrm{Na}^{+-} \\mathrm{OH}($ green $=\\mathrm{Cl})$ :\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM From what alkyl bromide was the following alkyl acetate made by $\\mathrm{S}_{\\mathrm{N}} 2$ reaction? Write the reaction,\n11-22 showing all stereochemistry.\n\n\nPROBLEM Assign $R$ or $S$ configuration to the following molecule, write the product you would expect from $\\mathrm{S}_{\\mathrm{N}} 2$\n11-23 reaction with NaCN , and assign $R$ or $S$ configuration to the product (green $=\\mathrm{Cl}$ ):\n\n\nPROBLEM Draw the structure and assign $Z$ or $E$ stereochemistry to the product you expect from E2 reaction of\n11-24 the following molecule with $\\mathrm{NaOH}($ green $=\\mathrm{Cl})$ :"}
{"id": 703, "contents": "Mechanism Problems - \nPROBLEM Predict the product(s) and show the mechanism for each of the following reactions. What do the\n11-25 mechanisms have in common? Why?\n(a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{OTos}+\\mathrm{NaBr}$ DMF\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM Show the mechanism for each of the following reactions. What do the mechanisms have in 11-26 common? Why?\n(a)\n\n\n(b)\n\n(c)\n\n\nPROBLEM Predict the product(s) for each of the following elimination reactions. In each case show the\n11-27 mechanism. What do the mechanisms have in common? Why?\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Predict the product(s) for each of the following elimination reactions. In each case show the\n11-28 mechanism. What do the mechanisms have in common? Why?\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Predict the product(s) for each of the following elimination reactions. In each case show the\n11-29 mechanism. What do the mechanisms have in common? Why?\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Predict the product of each of the following reactions, and indicate if the mechanism is likely to be 11-30 $S_{N} 1, S_{N} 2$, E1, E2, or E1cB.\n(a)\n\n(c)\n\n$+$\n\n(b)\n\n\n(d)\n\n\nPROBLEM We saw in Section 8.7 that bromohydrins are converted into epoxides when treated with base.\n11-31 Propose a mechanism, using curved arrows to show the electron flow.\n\n$\\xrightarrow[\\text { Ethanol }]{\\mathrm{NaOH}}$\n\n\nPROBLEM The following tertiary alkyl bromide does not undergo a nucleophilic substitution reaction by either\n11-32 $\\mathrm{S}_{\\mathrm{N}} 1$ or $\\mathrm{S}_{\\mathrm{N}} 2$ mechanisms. Explain.\n\n\nPROBLEM Metabolism of $S$-adenosylhomocysteine (Section 11.6) involves the following sequence. Propose a\n11-33 mechanism for the second step."}
{"id": 704, "contents": "Mechanism Problems - \nPROBLEM Metabolism of $S$-adenosylhomocysteine (Section 11.6) involves the following sequence. Propose a\n11-33 mechanism for the second step.\n\n\nPROBLEM Reaction of iodoethane with $\\mathrm{CN}^{-}$yields a small amount of isonitrile, $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{~N} \\equiv \\mathrm{C}$, along with the 11-34 nitrile $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{C} \\equiv \\mathrm{N}$ as the major product. Write electron-dot structures for both products, assign formal charges as necessary, and propose mechanisms to account for their formation.\n\nPROBLEM One step in the urea cycle for ridding the body of ammonia is the conversion of argininosuccinate\n11-35 to the amino acid arginine plus fumarate. Propose a mechanism for the reaction, and show the structure of arginine.\n\n\nPROBLEM Methyl esters $\\left(\\mathrm{RCO}_{2} \\mathrm{CH}_{3}\\right)$ undergo a cleavage reaction to yield carboxylate ions plus iodomethane\n11-36 on heating with LiI in dimethylformamide:\n\n\nThe following evidence has been obtained: (1) The reaction occurs much faster in DMF than in ethanol. (2) The corresponding ethyl ester $\\left(\\mathrm{RCO}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}\\right)$ cleaves approximately 10 times more slowly than the methyl ester. Propose a mechanism for the reaction. What other kinds of experimental evidence could you gather to support your hypothesis?\n\nPROBLEM $\\mathrm{S}_{\\mathrm{N}} 2$ reactions take place with inversion of configuration, and $\\mathrm{S}_{\\mathrm{N}} 1$ reactions take place with\n11-37 racemization. The following substitution reaction, however, occurs with complete retention of configuration. Propose a mechanism. (Hint: two inversions = retention.)\n\n\nPROBLEM Propose a mechanism for the following reaction, an important step in the laboratory synthesis of 11-38 proteins:"}
{"id": 705, "contents": "Nucleophilic Substitution Reactions - \nPROBLEM Draw all isomers of $\\mathrm{C}_{4} \\mathrm{H}_{9} \\mathrm{Br}$, name them, and arrange them in order of decreasing reactivity in the 11-39 $\\mathrm{S}_{\\mathrm{N}} 2$ reaction.\n\nPROBLEM The following Walden cycle has been carried out: Explain the results, and indicate where inversion 11-40 occurs.\n\n\nPROBLEM Which compound in each of the following pairs will react faster in an $\\mathrm{S}_{\\mathrm{N}} 2$ reaction with $\\mathrm{OH}^{-}$?\n11-41 (a) $\\mathrm{CH}_{3} \\mathrm{Br}$ or $\\mathrm{CH}_{3} \\mathrm{I}$ (b) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{I}$ in ethanol or in dimethyl sulfoxide (c) $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CCl}$ or $\\mathrm{CH}_{3} \\mathrm{Cl}$\n(d) $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHBr}$ or $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}_{2} \\mathrm{Br}$\n\nPROBLEM Which reactant in each of the following pairs is more nucleophilic? Explain.\n11-42\n(a) ${ }^{-} \\mathrm{NH}_{2}$ or $\\mathrm{NH}_{3}$\n(b) $\\mathrm{H}_{2} \\mathrm{O}$ or $\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}$\n(c) $\\mathrm{BF}_{3}$ or $\\mathrm{F}^{-}$\n(d) $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{P}$ or $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{~N}$\n(e) $\\mathrm{I}^{-}$or $\\mathrm{Cl}^{-}$\n(f) ${ }^{-} \\mathrm{C} \\equiv \\mathrm{N}$ or ${ }^{-} \\mathrm{OCH}_{3}$"}
{"id": 706, "contents": "Nucleophilic Substitution Reactions - \nPROBLEM What effect would you expect the following changes to have on the rate of the $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of\n11-43 1-iodo-2-methylbutane with cyanide ion?\n(a) The $\\mathrm{CN}^{-}$concentration is halved, and the 1-iodo-2-methylbutane concentration is doubled.\n(b) Both the $\\mathrm{CN}^{-}$and the 1-iodo-2-methylbutane concentrations are tripled.\n\nPROBLEM What effect would you expect the following changes to have on the rate of the reaction of ethanol\n11-44 with 2 -iodo-2-methylbutane?\n(a) The concentration of the halide is tripled.\n(b) The concentration of the ethanol is halved by adding diethyl ether as an inert solvent.\n\nPROBLEM How might you prepare each of the following using a nucleophilic substitution reaction at some\n11-45 step?\n(a)\n\n(b)\n\n(c) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CN}$\n(d) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}$"}
{"id": 707, "contents": "Nucleophilic Substitution Reactions - \nPROBLEM Which reaction in each of the following pairs would you expect to be faster?\n11-46 (a) The $\\mathrm{S}_{\\mathrm{N}} 2$ displacement by $\\mathrm{I}^{-}$on $\\mathrm{CH}_{3} \\mathrm{Cl}$ or on $\\mathrm{CH}_{3} \\mathrm{OTos}$\n(b) The $\\mathrm{S}_{\\mathrm{N}} 2$ displacement by $\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}$on bromoethane or on bromocyclohexane\n(c) The $\\mathrm{S}_{\\mathrm{N}} 2$ displacement on 2-bromopropane by $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{O}^{-}$or by $\\mathrm{CN}^{-}$\n(d) The $\\mathrm{S}_{\\mathrm{N}} 2$ displacement by $\\mathrm{HC} \\equiv \\mathrm{C}^{-}$on bromomethane in benzene or in acetonitrile\n\nPROBLEM Predict the product and give the stereochemistry resulting from reaction of each of the following\n11-47 nucleophiles with $(R)$-2-bromooctane:\n(a) ${ }^{-} \\mathrm{CN}$\n(b) $\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}$\n(c) $\\mathrm{CH}_{3} \\mathrm{~S}^{-}$\n\nPROBLEM ( $R$ )-2-Bromooctane undergoes racemization to give ( $\\pm$ )-2-bromooctane when treated with NaBr in\n11-48 dimethyl sulfoxide. Explain."}
{"id": 708, "contents": "Elimination Reactions - \nPROBLEM Propose structures for compounds that fit the following descriptions:\n11-49 (a) An alkyl halide that gives a mixture of three alkenes on E2 reaction\n(b) An organohalide that will not undergo nucleophilic substitution\n(c) An alkyl halide that gives the non-Zaitsev product on E2 reaction\n(d) An alcohol that reacts rapidly with HCl at $0^{\\circ} \\mathrm{C}$\n\nPROBLEM What products would you expect from the reaction of 1-bromopropane with each of the following?\n11-50 (a) $\\mathrm{NaNH}_{2}$\n(b) $\\mathrm{KOC}\\left(\\mathrm{CH}_{3}\\right)_{3}$\n(c) NaI\n(d) NaCN\n(e) $\\mathrm{NaC} \\equiv \\mathrm{CH}$\n(f) Mg , then $\\mathrm{H}_{2} \\mathrm{O}$\n\nPROBLEM 1-Chloro-1,2-diphenylethane can undergo E2 elimination to give either cis- or\n11-51 trans-1,2-diphenylethylene (stilbene). Draw Newman projections of the reactive conformations leading to both possible products, and suggest a reason why the trans alkene is the major product.\n\n\n1-Chloro-1,2-diphenylethane\n\ntrans-1,2-Diphenylethylene\n\nPROBLEM Predict the major alkene product of the following E1 reaction:\n11-52\n\n\nPROBLEM There are eight diastereomers of 1,2,3,4,5,6-hexachlorocyclohexane. Draw each in its more stable\n11-53 chair conformation. One isomer loses HCl in an E2 reaction nearly 1000 times more slowly than the others. Which isomer reacts so slowly, and why?"}
{"id": 709, "contents": "General Problems - \nPROBLEM The following reactions are unlikely to occur as written. Tell what is wrong with each, and predict\n11-54 the actual product.\n(a)\n\n\n(b)\n\n(c)\n\n\nPROBLEM Arrange the following carbocations in order of increasing stability.\n11-55 (a)\n\n\n$+$\n\n(b)\n\n\n\n\n(c)\n\n\n\n\nPROBLEM Order each of the following sets of compounds with respect to $\\mathrm{S}_{\\mathrm{N}} 1$ reactivity:\n11-56 (a)\n\n\n\n(b) $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CCl}$\n$\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CBr}$\n$\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{COH}$\n(c)\n\n\n\n\nPROBLEM Order each of the following sets of compounds with respect to $\\mathrm{S}_{\\mathrm{N}} 2$ reactivity:\n11-57 (a)\n\n$\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{Cl}$\n\n(b)\n\n\n\n(c) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{OCH}_{3}$\n$\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{OTos}$\n$\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{Br}$\nPROBLEM Predict the major product(s) of each of the following reactions. Identify those reactions where you 11-58 would expect the product mixture to rotate plane-polarized light.\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Reaction of the following $S$ tosylate with cyanide ion yields a nitrile product that also has $S$\n11-59 stereochemistry. Explain.\n\n( $S$ stereochemistry)\nPROBLEM Ethers can often be prepared by $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of alkoxide ions, $\\mathrm{RO}^{-}$, with alkyl halides. Suppose you\n11-60 wanted to prepare cyclohexyl methyl ether. Which of the following two possible routes would you choose? Explain."}
{"id": 710, "contents": "General Problems - \nPROBLEM How can you explain the fact that trans-1-bromo-2-methylcyclohexane yields the non-Zaitsev\n11-61 elimination product 3-methylcyclohexene on treatment with base?\n\ntrans-1-Bromo-2-methylcyclohexane\n3-Methylcyclohexene\nPROBLEM Predict the product(s) of the following reaction, indicating stereochemistry where necessary:\n11-62\n\n\nPROBLEM Alkynes can be made by dehydrohalogenation of vinylic halides in a reaction that is essentially\n11-63 an E2 process. In studying the stereochemistry of this elimination, it was found that (Z)-2-chloro-2-butenedioic acid reacts 50 times as fast as the corresponding $E$ isomer. What conclusion can you draw about the stereochemistry of eliminations in vinylic halides? How does this result compare with eliminations of alkyl halides?\n\n\nPROBLEM Based on your answer to Problem 11-63, predict the product(s) and show the mechanism for each\n11-64 of the following reactions.\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM (S)-2-Butanol slowly racemizes on standing in dilute sulfuric acid. Explain.\n11-65\n\n\nPROBLEM Reaction of HBr with ( $R$ )-3-methyl-3-hexanol leads to racemic 3-bromo-3-methylhexane. Explain.\n11-66\n\n\n3-Methyl-3-hexanol\n\nPROBLEM Treatment of 1-bromo-2-deuterio-2-phenylethane with strong base leads to a mixture of\n11-67 deuterated and nondeuterated phenylethylenes in an approximately 7:1 ratio. Explain.\n\n\nPROBLEM Propose a structure for an alkyl halide that gives only ( $E$ )-3-methyl-2-phenyl-2-pentene on E2 11-68 elimination. Make sure you indicate the stereochemistry."}
{"id": 711, "contents": "General Problems - \nPROBLEM Although anti periplanar geometry is preferred for E2 reactions, it isn't absolutely necessary. The\n11-69 following deuterated bromo compound reacts with strong base to yield an undeuterated alkene. A syn elimination has occurred. Make a molecular model of the reactant, and explain the result.\n\n\nPROBLEM The reaction of 1-chlorooctane with $\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}$to give octyl acetate is greatly accelerated by adding 11-70 a small quantity of iodide ion. Explain.\n\nPROBLEM Compound $\\mathbf{X}$ is optically inactive and has the formula $\\mathrm{C}_{16} \\mathrm{H}_{16} \\mathrm{Br}_{2}$. On treatment with strong base,\n11-71 $\\mathbf{X}$ gives hydrocarbon $\\mathbf{Y}, \\mathrm{C}_{16} \\mathrm{H}_{14}$. Compound $\\mathbf{Y}$ absorbs 2 equivalents of hydrogen when reduced over a palladium catalyst and reacts with ozone to give two fragments. One fragment, $\\mathbf{Z}$, is an aldehyde with formula $\\mathrm{C}_{7} \\mathrm{H}_{6} \\mathrm{O}$. The other fragment is glyoxal, $(\\mathrm{CHO})_{2}$. Write the reactions involved, and suggest structures for $\\mathbf{X}, \\mathbf{Y}$, and $\\mathbf{Z}$. What is the stereochemistry of $\\mathbf{X}$ ?\n\nPROBLEM When a primary alcohol is treated with $p$-toluenesulfonyl chloride at room temperature in the\n11-72 presence of an organic base such as pyridine, a tosylate is formed. When the same reaction is carried out at higher temperature, an alkyl chloride is often formed. Explain."}
{"id": 712, "contents": "General Problems - \nPROBLEM The amino acid methionine is formed by a methylation reaction of homocysteine with\n11-73 $N$-methyltetrahydrofolate. The stereochemistry of the reaction has been probed by carrying out the transformation using a donor with a \"chiral methyl group,\" which contains protium (H), deuterium (D), and tritium (T) isotopes of hydrogen. Does the methylation reaction occur with inversion or retention of configuration?\n\n$N$-Methyltetrahydrofolate\nTetrahydrofolate\nPROBLEM Amines are converted into alkenes by a two-step process called the Hofmann elimination. $\\mathrm{S}_{\\mathrm{N}} 2$\n11-74 reaction of the amine with an excess of $\\mathrm{CH}_{3} \\mathrm{I}$ in the first step yields an intermediate that undergoes E2 reaction when treated with silver oxide as base. Pentylamine, for example, yields 1-pentene. Propose a structure for the intermediate, and explain why it readily undergoes elimination.\n$\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2} \\xrightarrow[\\text { 2. } \\mathrm{Ag}_{2} \\mathrm{O}, \\mathrm{H}_{2} \\mathrm{O}]{\\text { 1. Excess } \\mathrm{CH}_{3} \\mathrm{I}} \\quad \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}=\\mathrm{CH}_{2}$\nPROBLEM The antipsychotic drug flupentixol is prepared by the following scheme:\n11-75"}
{"id": 713, "contents": "General Problems - \n(a) What alkyl chloride $\\mathbf{B}$ reacts with amine $\\mathbf{A}$ to form $\\mathbf{C}$ ?\n(b) Compound $\\mathbf{C}$ is treated with $\\mathrm{SOCl}_{2}$, and the product is allowed to react with magnesium metal to give a Grignard reagent $\\mathbf{D}$. What is the structure of $\\mathbf{D}$ ?\n(c) We'll see in Section 19.7 that Grignard reagents add to ketones, such as E, to give tertiary alcohols, such as $\\mathbf{F}$. Because of the newly formed chirality center, compound $\\mathbf{F}$ exists as a pair of enantiomers. Draw both, and assign $R, S$ configurations.\n(d) Two stereoisomers of flupentixol are subsequently formed from $\\mathbf{F}$, but only one is shown. Draw the other isomer, and identify the type of stereoisomerism."}
{"id": 714, "contents": "CHAPTER 12 - \nStructure Determination: Mass Spectrometry and Infrared Spectroscopy\n\n\nFIGURE 12.1 More than a thousand different chemical compounds have been isolated from coffee. Their structures were determined using various spectroscopic techniques. (credit: modification of work \"Coffee\" by Rafael Salda\u00f1a/Flickr, CC BY 2.0)"}
{"id": 715, "contents": "CHAPTER CONTENTS - \n12.1 Mass Spectrometry of Small Molecules: Magnetic-Sector Instruments\n12.2 Interpreting Mass Spectra\n12.3 Mass Spectrometry of Some Common Functional Groups\n12.4 Mass Spectrometry in Biological Chemistry: Time-of-Flight (TOF) Instruments\n12.5 Spectroscopy and the Electromagnetic Spectrum\n12.6 Infrared Spectroscopy\n12.7 Interpreting Infrared Spectra\n12.8 Infrared Spectra of Some Common Functional Groups\n\nWHY THIS CHAPTER? Finding the structures of new molecules, whether small ones synthesized in the laboratory or large proteins and nucleic acids found in living organisms, is central to progress in chemistry and biochemistry. We can only scratch the surface of structure determination in this book, but after reading this and the following two chapters, you should have a good idea of the range of structural techniques available and of how and when each is used.\n\nEvery time a reaction is run, the products must be identified, and every time a new compound is found in nature, its structure must be determined. Determining the structure of an organic compound was a difficult and time-consuming process until the mid-20th century, but powerful techniques and specialized instruments are now routinely used to simplify the problem. In this and the next two chapters, we'll look at four such techniques-mass spectrometry (MS), infrared (IR) spectroscopy, ultraviolet spectroscopy (UV), and nuclear magnetic resonance spectroscopy (NMR)-and we'll see the kind of information that can be obtained from each.\n\n| Mass spectrometry | What is the size and formula? |\n| :--- | :--- |\n| Infrared spectroscopy | What functional groups are present? |\n| Ultraviolet spectroscopy | Is a conjugated $\\pi$ electron system present? |\n| Nuclear magnetic resonance spectroscopy | What is the carbon-hydrogen framework? |"}
{"id": 716, "contents": "CHAPTER CONTENTS - 12.1 Mass Spectrometry of Small Molecules: Magnetic-Sector Instruments\nAt its simplest, mass spectrometry (MS) is a technique for measuring the mass, and therefore the molecular weight (MW), of a molecule. In addition, it's often possible to gain structural information about a molecule by measuring the masses of the fragments produced when molecules are broken apart.\n\nMore than 20 different kinds of commercial mass spectrometers are available depending on the intended application, but all have three basic parts: an ionization source in which sample molecules are given an electrical charge, a mass analyzer in which ions are separated by their mass-to-charge ratio, and a detector in which the separated ions are observed and counted.\n\n\nAmong the most common mass spectrometers used for routine purposes in the laboratory is the electronimpact, magnetic-sector instrument shown schematically in FIGURE 12.2. A small amount of sample is vaporized into the ionization source, where it is bombarded by a stream of high-energy electrons. The energy of the electron beam can be varied but is commonly around 70 electron volts (eV), or $6700 \\mathrm{~kJ} / \\mathrm{mol}$. When a highenergy electron strikes an organic molecule, it dislodges a valence electron from the molecule, producing a cation radical-cation because the molecule has lost an electron and now has a positive charge; radical because the molecule now has an odd number of electrons.\n\n\nElectron bombardment transfers so much energy that most of the cation radicals fragment after formation. They break apart into smaller pieces, some of which retain the positive charge and some of which are neutral. The fragments then flow through a curved pipe in a strong magnetic field, which deflects them into different paths according to their mass-to-charge ratio $(\\mathrm{m} / \\mathrm{z})$. Neutral fragments are not deflected by the magnetic field and are lost on the walls of the pipe, but positively charged fragments are sorted by the mass spectrometer onto a detector, which records them as peaks at the various $m / z$ ratios. Since the number of charges $z$ on each ion is usually 1 , the value of $m / z$ for each ion is simply its mass $m$. Masses up to approximately 2500 atomic mass units (amu) can be analyzed by this type of instrument."}
{"id": 717, "contents": "CHAPTER CONTENTS - 12.1 Mass Spectrometry of Small Molecules: Magnetic-Sector Instruments\nFIGURE 12.2 Representation of an electron-ionization, magnetic-sector mass spectrometer. Molecules are ionized by collision with highenergy electrons, causing some of the molecules to fragment. Passage of the charged fragments through a magnetic field then sorts them according to their mass.\n\nAnother common type of mass spectrometer uses what is called a quadrupole mass analyzer, which has a set of four solid rods is arranged parallel to the direction of the ion beam, with an oscillating electrostatic field is generated in the space between the rods. For a given field, only one $m / z$ value will make it through the quadrupole region. The others will crash into the rods or the walls of the instrument and never reach the detector FIGURE 12.3.\n\n\nFIGURE 12.3 Representation of a quadrupole mass analyzer. Only ions of a certain $m / z$ will reach the detector; other ions will collide with the rods.\n\nThe mass spectrum of a compound is typically presented as a bar graph, with masses ( $m / z$ values) on the $x$ axis and intensity, or relative abundance of ions of a given $\\mathrm{m} / \\mathrm{z}$ striking the detector, on the $y$ axis. The tallest peak, assigned an intensity of $100 \\%$, is called the base peak, and the peak that corresponds to the unfragmented cation radical is called the parent peak, or the molecular ion ( $M^{+}$, or simply $M$ ). FIGURE 12.4 shows the mass spectrum of propane.\n\n\nFIGURE 12.4 Mass spectrum of propane ( $\\mathrm{C}_{3} \\mathrm{H}_{8}$; $\\mathbf{M W}=44$ ).\nMass spectral fragmentation patterns are usually complex, and the molecular ion is often not the base peak. The mass spectrum of propane in FIGURE 12.4, for instance, shows a molecular ion at $m / z=44$ that is only about $30 \\%$ as high as the base peak at $m / z=29$. In addition, many other fragment ions are present."}
{"id": 718, "contents": "CHAPTER CONTENTS - 12.2 Interpreting Mass Spectra\nWhat kinds of information can we get from a mass spectrum? The most obvious information is the molecular weight of the sample, which in itself can be invaluable. If we were given samples of hexane (MW = 86), 1-hexene ( $\\mathrm{MW}=84$ ), and 1-hexyne ( $\\mathrm{MW}=82$ ), for example, mass spectrometry would easily distinguish them.\n\nSome instruments, called double-focusing mass spectrometers, have two magnetic sectors in their mass analyzers, giving these spectrometers have such high resolution that they provide mass measurements accurate to 5 ppm , or about 0.0005 amu , making it possible to distinguish between two formulas with the same nominal mass. For example, both $\\mathrm{C}_{5} \\mathrm{H}_{12}$ and $\\mathrm{C}_{4} \\mathrm{H}_{8} \\mathrm{O}$ have MW $=72$, but they differ slightly beyond the decimal point: $\\mathrm{C}_{5} \\mathrm{H}_{12}$ has an exact mass of 72.0939 amu , whereas $\\mathrm{C}_{4} \\mathrm{H}_{8} \\mathrm{O}$ has an exact mass of 72.0575 amu . A highresolution instrument can easily distinguish between them. Note, however, that exact mass measurements refer to molecules with specific isotopic compositions. Thus, the sum of the exact atomic masses of the specific isotopes in a molecule is measured -1.00783 amu for ${ }^{1} \\mathrm{H}, 12.00000 \\mathrm{amu}$ for ${ }^{12} \\mathrm{C}, 14.00307 \\mathrm{amu}$ for ${ }^{14} \\mathrm{~N}, 15.994$ 91 amu for ${ }^{16} \\mathrm{O}$, and so on-rather than the sum of the average atomic masses of elements, as found on a periodic table."}
{"id": 719, "contents": "CHAPTER CONTENTS - 12.2 Interpreting Mass Spectra\nUnfortunately, not every compound shows a molecular ion in its electron-impact mass spectrum. Although $\\mathrm{M}^{+}$ is usually easy to identify if it's abundant, some compounds, such as 2,2-dimethylpropane, fragment so easily that no molecular ion is observed (FIGURE 12.5). In such cases, alternative \"soft\" ionization methods that don't use electron bombardment can prevent or minimize fragmentation (see Section 12.4).\n\n\nFIGURE 12.5 Mass spectrum of 2,2-dimethylpropane ( $\\mathbf{C}_{\\mathbf{5}} \\mathbf{H}_{\\mathbf{1 2}}$; MW = 72). No molecular ion is observed when electron-impact ionization is used. What do you think is the formula and structure of the $M^{+}$peak at $m / z=57$ ?\n\nKnowing the molecular weight makes it possible to narrow considerably the choices of molecular formula. For example, if the mass spectrum of an unknown compound shows a molecular ion at $m / z=110$, the molecular formula is likely to be $\\mathrm{C}_{8} \\mathrm{H}_{14}, \\mathrm{C}_{7} \\mathrm{H}_{10} \\mathrm{O}, \\mathrm{C}_{6} \\mathrm{H}_{6} \\mathrm{O}_{2}$, or $\\mathrm{C}_{6} \\mathrm{H}_{10} \\mathrm{~N}_{2}$. There are always a number of molecular formulas possible for all but the lowest molecular weights, and a computer can easily generate a list of the choices."}
{"id": 720, "contents": "CHAPTER CONTENTS - 12.2 Interpreting Mass Spectra\nA further point about mass spectrometry, noticeable in the spectra of both propane (FIGURE 12.4) and 2,2-dimethylpropane (FIGURE 12.5), is that the peak for the molecular ion is not at the highest $\\mathrm{m} / \\mathrm{z}$ value. There is also a small peak at $\\mathrm{M}+1$ due to the presence of different isotopes in the molecules. Although ${ }^{12} \\mathrm{C}$ is the most abundant carbon isotope, a small amount ( $1.10 \\%$ natural abundance) of ${ }^{13} \\mathrm{C}$ is also present. Thus, a certain percentage of the molecules analyzed in the mass spectrometer are likely to contain a ${ }^{13} \\mathrm{C}$ atom, giving rise to the observed $\\mathrm{M}+1$ peak. In addition, a small amount of ${ }^{2} \\mathrm{H}$ (deuterium; $0.015 \\%$ natural abundance) is present, making a further contribution to the $\\mathrm{M}+1$ peak.\n\nMass spectrometry would be useful even if molecular weight and formula were the only information that could be obtained, but in fact it provides much more. For one thing, the mass spectrum of a compound serves as a kind of \"molecular fingerprint.\" Every organic compound fragments in a unique way depending on its structure, and the likelihood of two compounds having identical mass spectra is small. Thus, it's sometimes possible to identify an unknown by computer-based matching of its mass spectrum to one of the more than 785,061 searchable spectra recorded in a database called the Registry of Mass Spectral Data.\n\nIt's also possible to derive structural information about a molecule by interpreting its fragmentation pattern. Fragmentation occurs when the high-energy cation radical flies apart by spontaneous cleavage of a chemical bond. One of the two fragments retains the positive charge and is a carbocation, while the other fragment is a neutral radical."}
{"id": 721, "contents": "CHAPTER CONTENTS - 12.2 Interpreting Mass Spectra\nNot surprisingly, the positive charge often remains with the fragment that is best able to stabilize it. In other words, a relatively stable carbocation is often formed during fragmentation. For example, 2,2-dimethylpropane tends to fragment in such a way that the positive charge remains with the tert-butyl group. 2,2-Dimethylpropane therefore has a base peak at $m / z=57$, corresponding to $\\mathrm{C}_{4} \\mathrm{H}_{9}{ }^{+}$(FIGURE 12.5).\n\n$m / z=57$\nBecause mass-spectral fragmentation patterns are usually complex, it's often difficult to assign structures to fragment ions. Most hydrocarbons fragment in many ways, as demonstrated by the mass spectrum of hexane in FIGURE 12.6. The hexane spectrum shows a moderately abundant molecular ion at $m / z=86$ and fragment ions at $m / z=71,57,43$, and 29 . Since all the carbon-carbon bonds of hexane are electronically similar, all break to a similar extent, giving rise to the observed mixture of ions."}
{"id": 722, "contents": "CHAPTER CONTENTS - 12.2 Interpreting Mass Spectra\nFIGURE 12.6 Mass spectrum of hexane $\\left(\\mathbf{C}_{\\mathbf{6}} \\mathbf{H}_{\\mathbf{1 4}} \\mathbf{;} \\mathbf{M W}=\\mathbf{8 6}\\right)$. The base peak is at $m / z=57$, and numerous other ions are present.\nFIGURE 12.7 shows how the hexane fragments might arise. The loss of a methyl radical $\\left(\\mathrm{CH}_{3}, \\mathrm{M}=15\\right)$ from the hexane cation radical $\\left(\\mathrm{M}^{+}=86\\right)$ gives rise to a fragment of mass $86-15=71$; the loss of an ethyl radical $\\left(\\mathrm{C}_{2} \\mathrm{H}_{5}\\right.$, $M=29$ ) accounts for a fragment of mass $86-29=57$; the loss of a propyl radical ( $\\mathrm{C}_{3} \\mathrm{H}_{7}, \\mathrm{M}=43$ ) accounts for a fragment of mass $86-43=43$; and the loss of a butyl radical accounts for a fragment of mass 29 . With practice, it's sometimes possible to analyze the fragmentation pattern of an unknown compound and work backward to a structure that is compatible with the data.\n\n\nFIGURE 12.7 Fragmentation of hexane in a mass spectrometer.\nWe'll see in the next section and in later chapters that specific functional groups, such as alcohols, ketones, aldehydes, and amines, show specific kinds of mass spectral fragmentations that can be interpreted to provide structural information."}
{"id": 723, "contents": "Using Mass Spectra to Identify Compounds - \nAssume that you have two unlabeled samples, one of methylcyclohexane and the other of ethylcyclopentane. How could you use mass spectrometry to tell them apart? The mass spectra of both are shown in FIGURE 12.8.\n\n\nFIGURE 12.8 Mass spectra of unlabeled samples A and B for Worked Example 12.1."}
{"id": 724, "contents": "Strategy - \nLook at the possible structures and decide on how they differ. Then think about how any of these differences in structure might give rise to differences in mass spectra. Methyl cyclohexane, for instance, has a $-\\mathrm{CH}_{3}$ group,\nand ethylcyclopentane has a $-\\mathrm{CH}_{2} \\mathrm{CH}_{3}$ group, which should affect the fragmentation patterns."}
{"id": 725, "contents": "Solution - \nBoth mass spectra show molecular ions at $\\mathrm{M}^{+}=98$, corresponding to $\\mathrm{C}_{7} \\mathrm{H}_{14}$, but they differ in their fragmentation patterns. Sample $A$ has its base peak at $m / z=69$, corresponding to the loss of a $\\mathrm{CH}_{2} \\mathrm{CH}_{3}$ group (29 mass units), but $\\mathbf{B}$ has a rather small peak at $m / z=69$. Sample $\\mathbf{B}$ shows a base peak at $m / z=83$, corresponding to the loss of a $\\mathrm{CH}_{3}$ group ( 15 mass units), but sample $\\mathbf{A}$ has only a small peak at $\\mathrm{m} / \\mathrm{z}=83$. We can therefore be reasonably certain that $\\mathbf{A}$ is ethylcyclopentane and $\\mathbf{B}$ is methylcyclohexane.\n\nPROBLEM The sex hormone testosterone contains only C, H, and O and has a mass of 288.2089 amu , as 12-1 determined by high-resolution mass spectrometry. What is the likely molecular formula of testosterone?\n\nPROBLEM Two mass spectra are shown in Figure 12.9. One spectrum is that of 2-methyl-2-pentene; the other 12-2 is of 2-hexene. Which is which? Explain.\n(a)\n\n\n(b)\n\nFIGURE 12.9 Mass spectra for Problem 12-2."}
{"id": 726, "contents": "Solution - 12.3 Mass Spectrometry of Some Common Functional Groups\nAs each functional group is discussed in future chapters, mass-spectral fragmentations characteristic of that group will be described. As a preview, though, we'll point out some distinguishing features of several common functional groups."}
{"id": 727, "contents": "Alcohols - \nAlcohols undergo fragmentation in a mass spectrometer by two pathways: alpha cleavage and dehydration. In the $\\alpha$-cleavage pathway, a $\\mathrm{C}-\\mathrm{C}$ bond nearest the hydroxyl group is broken, yielding a neutral radical plus a resonance-stabilized, oxygen-containing cation. This type of fragmentation is seen in the spectrum of 2-pentanol in FIGURE 12.10.\n\n\n\nIn the dehydration pathway, water is eliminated, yielding an alkene radical cation with a mass 18 amu less than $\\mathrm{M}^{+}$. For simplicity, we have drawn the dehydration below as an E2-type process. Often the hydrogen that is lost is not beta to the hydroxyl. Only a small peak from dehydration is observed in the spectrum of 2-pentanol (FIGURE 12.10)."}
{"id": 728, "contents": "Amines - \nThe nitrogen rule of mass spectrometry says that a compound with an odd number of nitrogen atoms has an odd-numbered molecular weight. The logic behind the rule comes from the fact that nitrogen is trivalent, thus requiring an odd number of hydrogen atoms. The presence of nitrogen in a molecule is often detected simply by observing its mass spectrum. An odd-numbered molecular ion usually means that the unknown compound has one or three nitrogen atoms, and an even-numbered molecular ion usually means that a compound has either zero or two nitrogen atoms.\n\nAliphatic amines undergo a characteristic $\\alpha$ cleavage in a mass spectrometer, similar to that observed for alcohols. A C-C bond nearest the nitrogen atom is broken, yielding an alkyl radical and a resonance-stabilized, nitrogen-containing cation.\n\n\nThe mass spectrum of triethylamine has a base peak at $m / z=86$, which arises from an alpha cleavage resulting in the loss of a methyl group (FIGURE 12.11).\n\n\nFIGURE 12.11 Mass spectrum of triethylamine."}
{"id": 729, "contents": "Halides - \nThe fact that some elements have two common isotopes gives their mass spectra a distinctive appearance. Chlorine, for example, exists as two isotopes, ${ }^{35} \\mathrm{Cl}$ and ${ }^{37} \\mathrm{Cl}$, in roughly a $3: 1$ ratio. In a sample of chloroethane, three out of four molecules contain a ${ }^{35} \\mathrm{Cl}$ atom and one out of four has a ${ }^{37} \\mathrm{Cl}$ atom. In the mass spectrum of chloroethane (FIGURE 12.12 we see the molecular ion (M) at $m / z=64$ for ions that contain a ${ }^{35} \\mathrm{Cl}$ and another peak at $m / z=66$, called the $M+2$ peak, for ions containing a ${ }^{37} C l$. The ratio of the relative abundance of $M: M+$ 2 is about $3: 1$, a reflection of the isotopic abundances of chlorine.\n\n\nFIGURE 12.12 Mass spectrum of chloroethane.\nIn the case of bromine, the isotopic distribution is $50.7 \\%{ }^{79} \\mathrm{Br}$ and $49.3 \\%{ }^{81} \\mathrm{Br}$. In the mass spectrum of 1-bromohexane (FIGURE 12.13) the molecular ion appears at $\\mathrm{m} / \\mathrm{z}=164$ for ${ }^{79} \\mathrm{Br}$-containing ions and the $\\mathrm{M}+2$ peak is at $m / z=166$ for ${ }^{81} \\mathrm{Br}$-containing ions. The ions at $\\mathrm{m} / \\mathrm{z}=135$ and 137 are informative as well. The two nearly equally large peaks tell us that the ions at those $m / z$ values still contain the bromine atom. The peak at $m / z=85$, on the other hand, does not contain bromine because there is not a large peak at $m / z=87$.\n\n\nFIGURE 12.13 Mass spectrum of 1-bromohexane."}
{"id": 730, "contents": "Carbonyl Compounds - \nKetones and aldehydes that have a hydrogen on a carbon three atoms away from the carbonyl group undergo a characteristic mass-spectral cleavage called the McLafferty rearrangement. The hydrogen atom is transferred to the carbonyl oxygen, a $\\mathrm{C}-\\mathrm{C}$ bond between the alpha and beta carbons is broken, and a neutral alkene fragment is produced. The charge remains with the oxygen-containing fragment.\n\n\nIn addition, ketones and aldehydes frequently undergo $\\alpha$ cleavage of the bond between the carbonyl carbon and the neighboring carbon to yield a neutral radical and a resonance-stabilized acyl cation. Because the carbon neighboring the carbonyl carbon is called the alpha carbon, the reaction is called an alpha cleavage.\n\n(To be more general about neighboring positions in carbonyl compounds, Greek letters are used in alphabetical order: alpha, beta, gamma, delta, and so on.)\n\n\nThe mass spectrum of butyrophenone illustrates both alpha cleavage and the McLafferty rearrangement (FIGURE 12.14). Alpha cleavage of the propyl substituent results in the loss of $\\mathrm{C}_{3} \\mathrm{H}_{7}=43$ mass units from the parent ion at $m / z=148$ to give the fragment ion at $m / z=105$. A McLafferty rearrangement of butyrophenone results in the loss of ethylene, $\\mathrm{C}_{2} \\mathrm{H}_{4}=28$ mass units, from the parent leaving the ion at $\\mathrm{m} / \\mathrm{z}=120$.\n\n\nFIGURE 12.14 Mass spectrum of butyrophenone."}
{"id": 731, "contents": "Identifying Fragmentation Patterns in a Mass Spectrum - \nThe mass spectrum of 2-methyl-3-pentanol is shown in FIGURE 12.15. What fragments can you identify?\n\n\nFIGURE 12.15 Mass spectrum of 2-methyl-3-pentanol, for Worked Example 12.2."}
{"id": 732, "contents": "Strategy - \nCalculate the mass of the molecular ion, and identify the functional groups in the molecule. Then write the fragmentation processes you might expect, and compare the masses of the resultant fragments with the peaks present in the spectrum."}
{"id": 733, "contents": "Solution - \n2-Methyl-3-pentanol, an open-chain alcohol, has $\\mathrm{M}^{+}=102$ and might be expected to fragment by $\\alpha$ cleavage and by dehydration. These processes would lead to fragment ions of $m / z=84,73$, and 59 . Of the three expected fragments, dehydration is not observed (no $m / z=84$ peak), but both $\\alpha$ cleavages take place ( $m / z=73,59$ ).\n\n\nPROBLEM What are the masses of the charged fragments produced in the following cleavage pathways?\n12-3 (a) Alpha cleavage of 2-pentanone $\\left(\\mathrm{CH}_{3} \\mathrm{COCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}\\right)$\n(b) Dehydration of cyclohexanol (hydroxycyclohexane)\n(c) McLafferty rearrangement of 4-methyl-2-pentanone $\\left[\\mathrm{CH}_{3} \\mathrm{COCH}_{2} \\mathrm{CH}\\left(\\mathrm{CH}_{3}\\right)_{2}\\right]$\n(d) Alpha cleavage of triethylamine $\\left[\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2}\\right)_{3} \\mathrm{~N}\\right]$\n\nPROBLEM List the masses of the parent ion and of several fragments you might expect to find in the mass\n12-4 spectrum of the following molecule:"}
{"id": 734, "contents": "Solution - 12.4 Mass Spectrometry in Biological Chemistry: Time-of-Flight (TOF) Instruments\nMS analyses of sensitive biological samples rarely use magnetic sector ionization. Instead, they typically use either electrospray ionization ( $E S I$ ) or matrix-assisted laser desorption ionization (MALDI), typically linked to a time-of-flight (TOF) mass analyzer. Both ESI and MALDI are soft ionization methods that produce charged molecules with little fragmentation, even with sensitive biological samples of very high molecular weight.\n\nIn an ESI source, as a sample solution exits the tube, it is subjected to a high voltage that causes the droplets to become charged. The sample molecules gain one or more protons from charged solvent molecules in the droplet. The volatile solvent quickly evaporates, giving variably protonated sample molecules $\\left(M+H_{n}{ }^{n+}\\right)$. In a MALDI source, the sample is adsorbed onto a suitable matrix compound, such as 2,5-dihydroxybenzoic acid, which is ionized by a short burst of laser light. The matrix compound then transfers the energy to the sample and protonates it, forming $\\mathrm{M}+\\mathrm{H}_{n}{ }^{n+}$ ions.\n\nFollowing ion formation, the variably protonated sample molecules are electrically focused into a small packet with a narrow spatial distribution, and the packet is given a sudden kick of energy by an accelerator electrode. As each molecule in the packet is given the same energy, $E=m v^{2} / 2$, it begins moving with a velocity that depends on the square root of its mass, $v=\\sqrt{2 E / m}$. Lighter molecules move faster, and heavier molecules move slower. The analyzer itself-the drift tube-is simply an electrically grounded metal tube inside which the different charged molecules become separated as they move at different velocities and take different amounts of time to complete their flight."}
{"id": 735, "contents": "Solution - 12.4 Mass Spectrometry in Biological Chemistry: Time-of-Flight (TOF) Instruments\nThe Time of Flight technique is considerably more sensitive than the magnetic sector alternative, and protein samples of up to 100 kilodaltons ( $100,000 \\mathrm{amu}$ ) can be separated with a mass accuracy of 3 ppm. FIGURE 12.16 shows a MALDI-TOF spectrum of chicken egg-white lysozyme, MW = 14,306.7578 daltons. Biochemists generally use the unit dalton, abbreviated Da, instead of amu, although the two are equivalent ( 1 dalton = 1 amu ).\n\n\nFIGURE 12.16 MALDI-TOF mass spectrum of chicken egg-white lysozyme. The peak at $14,306.7578$ daltons (amu) is due to the monoprotonated protein, $\\mathrm{M}+\\mathrm{H}^{+}$, and the peak at $28,614.2188$ daltons is due to an impurity formed by dimerization of the protein. Other peaks at lower $m / z$ values are various protonated species, $M+H_{n}{ }^{n+}$."}
{"id": 736, "contents": "Solution - 12.5 Spectroscopy and the Electromagnetic Spectrum\nInfrared, ultraviolet, and nuclear magnetic resonance spectroscopies differ from mass spectrometry in that they are nondestructive and involve the interaction of molecules with electromagnetic energy rather than with an ionizing source. Before beginning a study of these techniques, however, let's briefly review the nature of radiant energy and the electromagnetic spectrum.\n\nVisible light, X rays, microwaves, radio waves, and so forth are all different kinds of electromagnetic radiation. Collectively, they make up the electromagnetic spectrum, shown in FIGURE 12.17. The electromagnetic spectrum is arbitrarily divided into regions, with the familiar visible region accounting for only a small portion, from $3.8 \\times 10^{-7} \\mathrm{~m}$ to $7.8 \\times 10^{-7} \\mathrm{~m}$ in wavelength. The visible region is flanked by the infrared and ultraviolet regions.\n\n\nFIGURE 12.17 The electromagnetic spectrum covers a continuous range of wavelengths and frequencies, from radio waves at the lowfrequency end to gamma ( $\\gamma$ ) rays at the high-frequency end. The familiar visible region accounts for only a small portion near the middle of the spectrum.\n\nElectromagnetic radiation is often said to have dual behavior. In some respects, it has the properties of a particle, called a photon, yet in other respects it behaves as an energy wave. Like all waves, electromagnetic radiation is characterized by a wavelength, a frequency, and an amplitude (FIGURE 12.18). The wavelength, $\\boldsymbol{\\lambda}$ (Greek lambda), is the distance from one wave maximum to the next. The frequency, $\\boldsymbol{\\nu}$ (Greek nu), is the\nnumber of waves that pass by a fixed point per unit time, usually given in reciprocal seconds $\\left(\\mathrm{s}^{-1}\\right)$, or hertz, $\\mathbf{H z}$ ( $1 \\mathrm{~Hz}=1 \\mathrm{~s}^{-1}$ ). The amplitude is the height of a wave, measured from midpoint to peak. The intensity of radiant energy, whether a feeble glow or a blinding glare, is proportional to the square of the wave's amplitude.\n\n(c)\n\n\nInfrared radiation\n\n$$\n\\left(\\nu=3.75 \\times 10^{14} \\mathrm{~s}^{-1}\\right)\n$$"}
{"id": 737, "contents": "Solution - 12.5 Spectroscopy and the Electromagnetic Spectrum\n(c)\n\n\nInfrared radiation\n\n$$\n\\left(\\nu=3.75 \\times 10^{14} \\mathrm{~s}^{-1}\\right)\n$$\n\nFIGURE 12.18 Electromagnetic waves are characterized by a wavelength, a frequency, and an amplitude. (a) Wavelength ( $\\lambda$ ) is the distance between two successive wave maxima. Amplitude is the height of the wave measured from the center. (b)-(c) What we perceive as different kinds of electromagnetic radiation are simply waves with different wavelengths and frequencies.\nMultiplying the wavelength of a wave in meters ( m ) by its frequency in reciprocal seconds ( $\\mathrm{s}^{-1}$ ) gives the speed of the wave in meters per second ( $\\mathrm{m} / \\mathrm{s}$ ). The rate of travel of all electromagnetic radiation in a vacuum is a constant value, commonly called the \"speed of light\" and abbreviated $c$. Its numerical value is defined as exactly $2.99792458 \\times 10^{8} \\mathrm{~m} / \\mathrm{s}$, usually rounded off to $3.00 \\times 10^{8} \\mathrm{~m} / \\mathrm{s}$.\n\n$$\n\\begin{aligned}\n& \\text { Wavelength } \\times \\text { Frequency }=\\text { Speed } \\\\\n& \\qquad \\begin{array}{c}\n\\lambda(\\mathrm{m}) \\times v\\left(\\mathrm{~s}^{-1}\\right)=c(\\mathrm{~m} / \\mathrm{s}) \\\\\n\\lambda=\\frac{c}{v} \\text { or } v=\\frac{c}{\\lambda}\n\\end{array}\n\\end{aligned}\n$$\n\nJust as matter comes only in discrete units called atoms, electromagnetic energy is transmitted only in discrete amounts called quanta. The amount of energy $\\varepsilon$ corresponding to 1 quantum of energy ( 1 photon) of a given frequency $\\nu$ is expressed by the Planck equation\n\n$$\n\\varepsilon=h v=\\frac{h c}{\\lambda}\n$$"}
{"id": 738, "contents": "Solution - 12.5 Spectroscopy and the Electromagnetic Spectrum\n$$\n\\varepsilon=h v=\\frac{h c}{\\lambda}\n$$\n\nwhere $h=$ Planck's constant $\\left(6.62 \\times 10^{-34} \\mathrm{~J} \\cdot \\mathrm{~s}=1.58 \\times 10^{-34} \\mathrm{cal} \\cdot \\mathrm{s}\\right)$.\nThe Planck equation says that the energy of a given photon varies directly with its frequency $\\nu$ but inversely with its wavelength $\\lambda$. High frequencies and short wavelengths correspond to high-energy radiation such as gamma rays; low frequencies and long wavelengths correspond to low-energy radiation such as radio waves. Multiplying $\\varepsilon$ by Avogadro's number $N_{\\mathrm{A}}$ gives the same equation in more familiar units, where $E$ represents the energy of Avogadro's number (one \"mole\") of photons of wavelength $\\lambda$ :\n\n$$\nE=\\frac{N_{A} h c}{\\lambda}=\\frac{1.20 \\times 10^{-4} \\mathrm{~kJ} / \\mathrm{mol}}{\\lambda(\\mathrm{~m})} \\text { or } \\frac{2.86 \\times 10^{-5} \\mathrm{kcal} / \\mathrm{mol}}{\\lambda(\\mathrm{~m})}\n$$\n\nWhen an organic compound is exposed to a beam of electromagnetic radiation, it absorbs energy of some wavelengths but passes, or transmits, energy of other wavelengths. If we irradiate the sample with energy of many different wavelengths and determine which are absorbed and which are transmitted, we can measure the absorption spectrum of the compound.\n\nAn example of an absorption spectrum-that of ethanol exposed to infrared radiation-is shown in FIGURE\n12.19. The horizontal axis records the wavelength, and the vertical axis records the intensity of the various energy absorptions in percent transmittance. The baseline corresponding to $0 \\%$ absorption (or $100 \\%$ transmittance) runs along the top of the chart, so a downward spike means that energy absorption has occurred at that wavelength."}
{"id": 739, "contents": "Solution - 12.5 Spectroscopy and the Electromagnetic Spectrum\nFIGURE 12.19 An infrared absorption spectrum for ethanol, $\\mathbf{C H}_{\\mathbf{3}} \\mathbf{C H}_{\\mathbf{2}} \\mathbf{O H}$. A transmittance of $100 \\%$ means that all the energy is passing through the sample, whereas a lower transmittance means that some energy is being absorbed. Thus, each downward spike corresponds to an energy absorption.\n\nThe energy a molecule gains when it absorbs radiation must be distributed over the molecule in some way. With infrared radiation, the absorbed energy causes bonds to stretch and bend more vigorously. With ultraviolet radiation, the energy causes an electron to jump from a lower-energy orbital to a higher-energy one. Different radiation frequencies affect molecules in different ways, but each provides structural information when the results are interpreted.\n\nThere are many kinds of spectroscopies, which differ according to the region of the electromagnetic spectrum used. We'll look at three: infrared spectroscopy, ultraviolet spectroscopy, and nuclear magnetic resonance spectroscopy. Let's begin by seeing what happens when an organic sample absorbs infrared energy."}
{"id": 740, "contents": "Correlating Energy and Frequency of Radiation - \nWhich is higher in energy, FM radio waves with a frequency of $1.015 \\times 10^{8} \\mathrm{~Hz}(101.5 \\mathrm{MHz})$ or visible green light with a frequency of $5 \\times 10^{14} \\mathrm{~Hz}$ ?"}
{"id": 741, "contents": "Strategy - \nRemember the equations $\\varepsilon=h \\nu$ and $\\varepsilon=h c / \\lambda$, which say that energy increases as frequency increases and as wavelength decreases."}
{"id": 742, "contents": "Solution - \nSince visible light has a higher frequency than radio waves, it is higher in energy.\n\nPROBLEM Which has higher energy, infrared radiation with $\\lambda=1.0 \\times 10^{-6} \\mathrm{~m}$ or an X ray with $\\lambda=3.0 \\times 10^{-9} \\mathrm{~m}$ ? 12-5 Radiation with $\\nu=4.0 \\times 10^{9} \\mathrm{~Hz}$ or with $\\lambda=9.0 \\times 10^{-6} \\mathrm{~m}$ ?\n\nPROBLEM It's useful to develop a feeling for the amounts of energy that correspond to different parts of 12-6 the electromagnetic spectrum. Calculate the energies in $\\mathrm{kJ} / \\mathrm{mol}$ of each of the following kinds of radiation:\n(a) A gamma ray with $\\lambda=5.0 \\times 10^{-11} \\mathrm{~m}$ (b) An X ray with $\\lambda=3.0 \\times 10^{-9} \\mathrm{~m}$\n(c) Ultraviolet light with $\\nu=6.0 \\times 10^{15} \\mathrm{~Hz}$ (d) Visible light with $\\nu=7.0 \\times 10^{14} \\mathrm{~Hz}$\n(e) Infrared radiation with $\\lambda=2.0 \\times 10^{-5} \\mathrm{~m}$ (f) Microwave radiation with $\\nu=1.0 \\times 10^{11} \\mathrm{~Hz}$"}
{"id": 743, "contents": "Solution - 12.6 Infrared Spectroscopy\nIn infrared (IR) spectroscopy, the IR region of the electromagnetic spectrum covers the range from just above the visible $\\left(7.8 \\times 10^{-7} \\mathrm{~m}\\right)$ to approximately $10^{-4} \\mathrm{~m}$, but only the midportion from $2.5 \\times 10^{-6} \\mathrm{~m}$ to $2.5 \\times 10^{-5} \\mathrm{~m}$\nis used by organic chemists (FIGURE 12.20. Wavelengths within the IR region are usually given in micrometers $\\left(1 \\mu \\mathrm{~m}=10^{-6} \\mathrm{~m}\\right)$, and frequencies are given in wavenumbers rather than in hertz. The wavenumber $\\widetilde{v}$ is the reciprocal of wavelength in centimeters and is therefore expressed in units of $\\mathrm{cm}^{-1}$.\n\n$$\n\\text { Wavenumber: } \\widetilde{v}\\left(\\mathrm{~cm}^{-1}\\right)=\\frac{1}{\\lambda(\\mathrm{~cm})}\n$$\n\nThus, the useful IR region is from 4000 to $400 \\mathrm{~cm}^{-1}$, corresponding to energies of $48.0 \\mathrm{~kJ} / \\mathrm{mol}$ to $4.80 \\mathrm{~kJ} / \\mathrm{mol}$ (11.5-1.15 kcal/mol).\n\n\nFIGURE 12.20 The infrared and adjacent regions of the electromagnetic spectrum.\nWhy does an organic molecule absorb some wavelengths of IR radiation but not others? All molecules have a certain amount of energy and are in constant motion. Their bonds stretch and contract, atoms wag back and forth, and other molecular vibrations occur. Some of the kinds of allowed vibrations are shown below:"}
{"id": 744, "contents": "Solution - 12.6 Infrared Spectroscopy\nThe amount of energy a molecule contains is not continuously variable but is quantized. That is, a molecule can stretch or bend only at specific frequencies corresponding to specific energy levels. Take bond stretching, for example. Although we usually speak of bond lengths as if they were fixed, the numbers given are really averages. In fact, a typical $\\mathrm{C}-\\mathrm{H}$ bond with an average bond length of 110 pm is actually vibrating at a specific frequency, alternately stretching and contracting as if there were a spring connecting the two atoms.\n\nWhen a molecule is irradiated with electromagnetic radiation, energy is absorbed if the frequency of the radiation matches the frequency of the vibration. The result of this energy absorption is an increased amplitude for the vibration; in other words, the \"spring\" connecting the two atoms stretches and compresses a bit further. Since each frequency absorbed by a molecule corresponds to a specific molecular motion, we can find what kinds of motions a molecule has by measuring its IR spectrum. By interpreting these motions, we can find out what kinds of bonds (functional groups) are present in the molecule.\n\n$$\n\\text { IR spectrum } \\rightarrow \\text { What molecular motions? } \\rightarrow \\text { What functional groups? }\n$$"}
{"id": 745, "contents": "Solution - 12.7 Interpreting Infrared Spectra\nThe complete interpretation of an IR spectrum is difficult because most organic molecules have dozens of different bond stretching and bending motions, and thus have dozens of absorptions. On the one hand, this complexity is a problem because it generally limits the laboratory use of IR spectroscopy to pure samples of fairly small molecules-little can be learned from IR spectroscopy about large, complex biomolecules. On the other hand, this complexity is useful because an IR spectrum acts as a unique fingerprint of a compound. In fact,\nthe complex region of the IR spectrum, from $1500 \\mathrm{~cm}^{-1}$ to around $400 \\mathrm{~cm}^{-1}$, is called the fingerprint region. If two samples have identical IR spectra, they are almost certainly identical compounds.\n\nFortunately, we don't need to interpret an IR spectrum fully to get useful structural information. Most functional groups have characteristic IR absorption bands that don't change much from one compound to another. The $\\mathrm{C}=\\mathrm{O}$ absorption of a ketone is almost always in the range 1680 to $1750 \\mathrm{~cm}^{-1}$; the $\\mathrm{O}-\\mathrm{H}$ absorption of an alcohol is almost always in the range 3400 to $3650 \\mathrm{~cm}^{-1}$; the $\\mathrm{C}=\\mathrm{C}$ absorption of an alkene is almost always in the range 1640 to $1680 \\mathrm{~cm}^{-1}$; and so forth. By learning where characteristic functional-group absorptions occur, it's possible to get structural information from IR spectra. TABLE 12.1 lists the characteristic IR bands of some common functional groups.\n\nTABLE 12.1 Characteristic IR Absorptions of Some Functional Groups"}
{"id": 746, "contents": "Solution - 12.7 Interpreting Infrared Spectra\nTABLE 12.1 Characteristic IR Absorptions of Some Functional Groups\n\n| Functional Group |  | Absorption (cm ${ }^{-1}$ )$2850-2960$ | Intensity <br> Medium |\n| :---: | :---: | :---: | :---: |\n| Alkane | $\\mathrm{C}-\\mathrm{H}$ |  |  |\n| Alkene | $=\\mathrm{C}-\\mathrm{H}$ | 3020-3100 | Medium |\n|  | $C=C$ | 1640-1680 | Medium |\n| Alkyne | $\\equiv \\mathrm{C}-\\mathrm{H}$ | 3300 | Strong |\n|  | $\\mathrm{C} \\equiv \\mathrm{C}$ | 2100-2260 | Medium |\n| Alkyl halide | $\\mathrm{C}-\\mathrm{Cl}$ | 600-800 | Strong |\n|  | $\\mathrm{C}-\\mathrm{Br}$ | 500-600 | Strong |\n| Alcohol | $\\mathrm{O}-\\mathrm{H}$ | 3400-3650 | Strong, broad |\n|  | $\\mathrm{C}-\\mathrm{O}$ | 1050-1150 | Strong |\n| Arene | C-H | 3030 | Weak |\n| Aromatic ring |  | 1660-2000 | Weak |\n|  |  | 1450-1600 | Medium |\n| Amine | $\\mathrm{N}-\\mathrm{H}$ | 3300-3500 | Medium |\n|  | $\\mathrm{C}-\\mathrm{N}$ | 1030-1230 | Medium |\n| Carbonyl compound | $\\mathrm{C}=\\mathrm{O}$ | 1670-1780 | Strong |\n|  | Aldehyde | 1730 | Strong |\n|  | Ketone | 1715 | Strong |\n|  | Ester | 1735 | Strong |\n|  | Amide | 1690 | Strong |\n|  | Carboxylic acid | 1710 | Strong |\n\nTABLE 12.1 Characteristic IR Absorptions of Some Functional Groups"}
{"id": 747, "contents": "Solution - 12.7 Interpreting Infrared Spectra\nTABLE 12.1 Characteristic IR Absorptions of Some Functional Groups\n\n| Functional Group | Absorption (cm ${ }^{\\mathbf{1}}$ ) |  | Intensity |\n| :--- | :--- | :---: | :--- |\n| Carboxylic acid | $\\mathrm{O}-\\mathrm{H}$ | $2500-3100$ | Strong, broad |\n| Nitrile | $\\mathrm{C} \\equiv \\mathrm{N}$ | $2210-2260$ | Medium |\n| Nitro | $\\mathrm{NO}_{2}$ | 1540 | Strong |\n\nLook at the IR spectra of hexane, 1-hexene, and 1-hexyne in FIGURE 12.21 to see an example of how IR spectroscopy can be used. Although all three IR spectra contain many peaks, there are characteristic absorptions of $\\mathrm{C}=\\mathrm{C}$ and $\\mathrm{C} \\equiv \\mathrm{C}$ functional groups that allow the three compounds to be distinguished. Thus, 1-hexene shows a characteristic $\\mathrm{C}=\\mathrm{C}$ absorption at $1660 \\mathrm{~cm}^{-1}$ and a vinylic $=\\mathrm{C}-\\mathrm{H}$ absorption at $3100 \\mathrm{~cm}^{-1}$, whereas 1-hexyne has a $\\mathrm{C} \\equiv \\mathrm{C}$ absorption at $2100 \\mathrm{~cm}^{-1}$ and a terminal alkyne $\\equiv \\mathrm{C}-\\mathrm{H}$ absorption at $3300 \\mathrm{~cm}^{-1}$.\n(a)\n\n(b)\n\n(c)\n\n\nFIGURE 12.21 IR spectra of (a) hexane, (b) 1-hexene, and (c) 1-hexyne. Spectra like these are easily obtained from sub-milligram amounts of material in a few minutes using commercially available instruments.\n\nIt helps in remembering the position of specific IR absorptions to divide the IR region from $4000 \\mathrm{~cm}^{-1}$ to 400 $\\mathrm{cm}^{-1}$ into four parts, as shown in FIGURE 12.22."}
{"id": 748, "contents": "Solution - 12.7 Interpreting Infrared Spectra\nFIGURE 12.22 The four regions of the infrared spectrum: single bonds to hydrogen, triple bonds, double bonds, and fingerprint.\n\n- The region from 4000 to $2500 \\mathrm{~cm}^{-1}$ corresponds to absorptions caused by $\\mathrm{N}-\\mathrm{H}, \\mathrm{C}-\\mathrm{H}$, and $\\mathrm{O}-\\mathrm{H}$ singlebond stretching motions. $\\mathrm{N}-\\mathrm{H}$ and $\\mathrm{O}-\\mathrm{H}$ bonds absorb in the 3300 to $3600 \\mathrm{~cm}^{-1}$ range; $\\mathrm{C}-\\mathrm{H}$ bond stretching occurs near $3000 \\mathrm{~cm}^{-1}$.\n- The region from 2500 to $2000 \\mathrm{~cm}^{-1}$ is where triple-bond stretching occurs. Both $\\mathrm{C} \\equiv \\mathrm{N}$ and $\\mathrm{C} \\equiv \\mathrm{C}$ bonds absorb here.\n- The region from 2000 to $1500 \\mathrm{~cm}^{-1}$ is where double bonds $(\\mathrm{C}=\\mathrm{O}, \\mathrm{C}=\\mathrm{N}$, and $\\mathrm{C}=\\mathrm{C})$ absorb. Carbonyl groups generally absorb in the range 1680 to $1750 \\mathrm{~cm}^{-1}$, and alkene stretching normally occurs in the narrow range of 1640 to $1680 \\mathrm{~cm}^{-1}$.\n- The region below $1500 \\mathrm{~cm}^{-1}$ is the fingerprint portion of the IR spectrum. A large number of absorptions due to a variety of $\\mathrm{C}-\\mathrm{C}, \\mathrm{C}-\\mathrm{O}, \\mathrm{C}-\\mathrm{N}$, and $\\mathrm{C}-\\mathrm{X}$ single-bond vibrations occur here."}
{"id": 749, "contents": "Solution - 12.7 Interpreting Infrared Spectra\nWhy do different functional groups absorb where they do? As noted previously, a good analogy is that of two weights (atoms) connected by a spring (a bond). Short, strong bonds vibrate at a higher energy and higher frequency than do long, weak bonds, just as a short, strong spring vibrates faster than a long, weak spring. Thus, triple bonds absorb at a higher frequency than double bonds, which in turn absorb at a higher frequency than single bonds. In addition, $\\mathrm{C}-\\mathrm{H}, \\mathrm{O}-\\mathrm{H}$, and $\\mathrm{N}-\\mathrm{H}$ bonds vibrate at a higher frequency than bonds between heavier $\\mathrm{C}, \\mathrm{O}$, and N atoms."}
{"id": 750, "contents": "Distinguishing Isomeric Compounds by IR Spectroscopy - \nAcetone $\\left(\\mathrm{CH}_{3} \\mathrm{COCH}_{3}\\right)$ and 2-propen-1-ol $\\left(\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}_{2} \\mathrm{OH}\\right)$ are isomers. How could you distinguish them by IR spectroscopy?"}
{"id": 751, "contents": "Strategy - \nIdentify the functional groups in each molecule, and refer to TABLE 12.1."}
{"id": 752, "contents": "Solution - \nAcetone has a strong $\\mathrm{C}=\\mathrm{O}$ absorption at $1715 \\mathrm{~cm}^{-1}$, while 2-propen-1-ol has an -OH absorption at $3500 \\mathrm{~cm}^{-1}$ and a $\\mathrm{C}=\\mathrm{C}$ absorption at $1660 \\mathrm{~cm}^{-1}$.\n\nPROBLEM What functional groups might the following molecules contain?\n12-7 (a) A compound with a strong absorption at $1710 \\mathrm{~cm}^{-1}$\n(b) A compound with a strong absorption at $1540 \\mathrm{~cm}^{-1}$\n(c) A compound with strong absorptions at $1720 \\mathrm{~cm}^{-1}$ and 2500 to $3100 \\mathrm{~cm}^{-1}$\n\nPROBLEM How might you use IR spectroscopy to distinguish between the following pairs of isomers?\n12-8 (a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$ and $\\mathrm{CH}_{3} \\mathrm{OCH}_{3}$ (b) Cyclohexane and 1-hexene\n(c) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ and $\\mathrm{HOCH}_{2} \\mathrm{CH}_{2} \\mathrm{CHO}$"}
{"id": 753, "contents": "Solution - 12.8 Infrared Spectra of Some Common Functional Groups\nAs each functional group is discussed in future chapters, the spectroscopic properties of that group will be described. For the present, we'll point out some distinguishing features of the hydrocarbon functional groups already studied and briefly preview some other common functional groups. We should also point out, however, that in addition to interpreting absorptions that are present in an IR spectrum, it's also possible to get structural information by noticing which absorptions are not present. If the spectrum of a compound has no absorptions at 3300 and $2150 \\mathrm{~cm}^{-1}$, the compound is not a terminal alkyne; if the spectrum has no absorption near 3400 $\\mathrm{cm}^{-1}$, the compound is not an alcohol; and so on."}
{"id": 754, "contents": "Alkanes - \nThe IR spectrum of an alkane is fairly uninformative because no functional groups are present and all absorptions are due to $\\mathrm{C}-\\mathrm{H}$ and $\\mathrm{C}-\\mathrm{C}$ bonds. Alkane $\\mathrm{C}-\\mathrm{H}$ bonds show a strong absorption from 2850 to 2960 $\\mathrm{cm}^{-1}$, and saturated $\\mathrm{C}-\\mathrm{C}$ bonds show a number of bands in the 800 to $1300 \\mathrm{~cm}^{-1}$ range. Since most organic compounds contain saturated alkane-like portions, most organic compounds have these characteristic IR absorptions. The $\\mathrm{C}-\\mathrm{H}$ and $\\mathrm{C}-\\mathrm{C}$ bands are clearly visible in the three spectra shown previously in FIGURE 12.21."}
{"id": 755, "contents": "Alkenes - \nAlkenes show several characteristic stretching absorptions. Vinylic $=\\mathrm{C}-\\mathrm{H}$ bonds absorb from 3020 to 3100 $\\mathrm{cm}^{-1}$, and alkene $\\mathrm{C}=\\mathrm{C}$ bonds usually absorb near $1650 \\mathrm{~cm}^{-1}$, although in some cases their peaks can be rather small and difficult to see clearly when the alkene is symmetric, or nearly so. Both absorptions are visible in the 1-hexene spectrum in FIGURE 12.21b.\n\nAlkenes have characteristic $=\\mathrm{C}-\\mathrm{H}$ out-of-plane bending absorptions in the 700 to $1000 \\mathrm{~cm}^{-1}$ range, thereby allowing the substitution pattern on a double bond to be determined (FIGURE 12.23). For example, monosubstituted alkenes such as 1-hexene show strong characteristic bands at 910 and $990 \\mathrm{~cm}^{-1}$, and 1,1-disubstituted alkenes $\\left(\\mathrm{R}_{2} \\mathrm{C}=\\mathrm{CH}_{2}\\right)$ have an intense band at $890 \\mathrm{~cm}^{-1}$.\n\n| Alkenes | $=\\mathrm{C}-\\mathrm{H}$ | $3020-3100 \\mathrm{~cm}^{-1}$ |\n| :--- | :--- | :--- |\n|  |  |  |\n|  |  | $1640-1680 \\mathrm{~cm}^{-1}$ |\n| $\\mathrm{RCH}=\\mathrm{CH}_{2}$ | 910 and $990 \\mathrm{~cm}^{-1}$ |  |\n| $\\mathrm{R}_{2} \\mathrm{C}=\\mathrm{CH}_{2}$ | $890 \\mathrm{~cm}^{-1}$ |  |\n\n\n\nFIGURE 12.23 C-H out-of-plane bending vibrations for substituted alkenes."}
{"id": 756, "contents": "Alkynes - \nAlkynes show a $\\mathrm{C} \\equiv \\mathrm{C}$ stretching absorption at 2100 to $2260 \\mathrm{~cm}^{-1}$, an absorption that is much more intense for terminal alkynes than for internal alkynes. Terminal alkynes such as 1-hexyne also have a characteristic $\\equiv \\mathrm{C}-\\mathrm{H}$ stretching absorption at $3300 \\mathrm{~cm}^{-1}$ (FIGURE 12.21c). This band is diagnostic for terminal alkynes because it is fairly intense and quite sharp.\n\n$$\n\\begin{array}{lll}\n\\text { Alkynes } & -\\mathrm{C} \\equiv \\mathrm{C}- & 2100-2260 \\mathrm{~cm}^{-1} \\\\\n& \\equiv \\mathrm{C}-\\mathrm{H} & 3300 \\mathrm{~cm}^{-1}\n\\end{array}\n$$"}
{"id": 757, "contents": "Aromatic Compounds - \nAromatic compounds, such as benzene, have a weak C-H stretching absorption at $3030 \\mathrm{~cm}^{-1}$, just to the left of a typical saturated C-H band. In addition, they have a series of weak absorptions in the 1660 to $2000 \\mathrm{~cm}^{-1}$ range and a series of medium-intensity absorptions in the 1450 to $1600 \\mathrm{~cm}^{-1}$ region. These latter absorptions are due to complex molecular motions of the entire ring. The $\\mathrm{C}-\\mathrm{H}$ out-of-plane bending region for benzene derivatives, between 650 to $1000 \\mathrm{~cm}^{-1}$, gives valuable information about the ring's substitution pattern, as it does for the substitution pattern of alkenes (FIGURE 12.24)."}
{"id": 758, "contents": "Aromatic compounds - \nFIGURE 12.24 C-H out-of-plane bending vibrations for substituted benzenes.\nThe IR spectrum of phenylacetylene, shown in Figure 12.29 at the end of this section, gives an example, clearly showing the following absorbances: $\\equiv \\mathrm{C}-\\mathrm{H}$ stretch at $3300 \\mathrm{~cm}^{-1}, \\mathrm{C}-\\mathrm{H}$ stretches from the benzene ring at 3000 to $3100 \\mathrm{~cm}^{-1}, \\mathrm{C}=\\mathrm{C}$ stretches of the benzene ring between 1450 and $1600 \\mathrm{~cm}^{-1}$, and out-of-plane bending of the ring's $\\mathrm{C}-\\mathrm{H}$ groups, indicating monosubstitution at $750 \\mathrm{~cm}^{-1}$."}
{"id": 759, "contents": "Alcohols - \nThe $\\mathrm{O}-\\mathrm{H}$ functional group of alcohols is easy to spot. Alcohols have a characteristic band in the range 3400 to $3650 \\mathrm{~cm}^{-1}$ that is usually broad and intense. Hydrogen bonding between $\\mathrm{O}-\\mathrm{H}$ groups is responsible for making the absorbance so broad. If an $\\mathrm{O}-\\mathrm{H}$ stretch is present, it's hard to miss this band or to confuse it with anything else.\n\n$$\n\\text { Alcohols } \\quad-\\mathrm{O}-\\mathrm{H} \\quad 3400-3650 \\mathrm{~cm}^{-1} \\text { (broad, intense) }\n$$\n\nCyclohexanol (FIGURE 12.25) is a good example."}
{"id": 760, "contents": "Amines - \nThe $\\mathrm{N}-\\mathrm{H}$ functional group of amines is also easy to spot in the IR, with a characteristic absorption in the 3300 to $3500 \\mathrm{~cm}^{-1}$ range. Although alcohols absorb in the same range, an $\\mathrm{N}-\\mathrm{H}$ absorption band is much sharper and less intense than an $\\mathrm{O}-\\mathrm{H}$ band.\n\n\nPrimary amines ( $\\mathrm{R}-\\mathrm{NH}_{2}$ ) have two absorbances-one for the symmetric stretching mode and one for the asymmetric mode (FIGURE 12.26). Secondary amines ( $\\mathrm{R}_{2} \\mathrm{~N}-\\mathrm{H}$ ) only have one $\\mathrm{N}-\\mathrm{H}$ stretching absorbance in this\nregion.\n\n\nFIGURE 12.26 IR spectrum of cyclohexylamine."}
{"id": 761, "contents": "Carbonyl Compounds - \nCarbonyl functional groups are the easiest to identify of all IR absorptions because of their sharp, intense peak in the range 1670 to $1780 \\mathrm{~cm}^{-1}$. Most important, the exact position of absorption within this range can often be used to identify the exact kind of carbonyl functional group-aldehyde, ketone, ester, and so forth."}
{"id": 762, "contents": "ALDEHYDES - \nSaturated aldehydes absorb at $1730 \\mathrm{~cm}^{-1}$; aldehydes next to either a double bond or an aromatic ring absorb at $1705 \\mathrm{~cm}^{-1}$."}
{"id": 763, "contents": "Aldehydes - \n$1730 \\mathrm{~cm}^{-1}$\n$1705 \\mathrm{~cm}^{-1}$\n$1705 \\mathrm{~cm}^{-1}$\nThe $\\mathrm{C}-\\mathrm{H}$ group attached to the carbonyl is responsible for the characteristic IR absorbance for aldehydes at 2750 and $2850 \\mathrm{~cm}^{-1}$ (FIGURE 12.27). Although these are not very intense, the absorbance at $2750 \\mathrm{~cm}^{-1}$ is helpful when trying to distinguish between an aldehyde and a ketone.\n\n\nFIGURE 12.27 The IR spectrum of benzaldehyde."}
{"id": 764, "contents": "KETONES - \nSaturated open-chain ketones and six-membered cyclic ketones absorb at $1715 \\mathrm{~cm}^{-1}$. Ring strain stiffens the $\\mathrm{C}=\\mathrm{O}$ bond, making five-membered cyclic ketones absorb at $1750 \\mathrm{~cm}^{-1}$ and four-membered cyclic ketones absorb at $1780 \\mathrm{~cm}^{-1}$, about 20 to $30 \\mathrm{~cm}^{-1}$ lower than the corresponding saturated ketone."}
{"id": 765, "contents": "ESTERS - \nSaturated esters have a $C=O$ absorbance at $1735 \\mathrm{~cm}^{-1}$ and two strong absorbances in the 1300 to $1000 \\mathrm{~cm}^{-1}$ range from the $\\mathrm{C}-\\mathrm{O}$ portion of the functional group. Like other carbonyl functional groups, esters next to either an aromatic ring or a double bond absorb at $1715 \\mathrm{~cm}^{-1}$, about 20 to $30 \\mathrm{~cm}^{-1}$ lower than a saturated ester."}
{"id": 766, "contents": "Esters - \n$1735 \\mathrm{~cm}^{-1}$\n$1715 \\mathrm{~cm}^{-1}$\n\n$1715 \\mathrm{~cm}^{-1}$"}
{"id": 767, "contents": "Predicting IR Absorptions of Compounds - \nWhere might the following compounds have IR absorptions?"}
{"id": 768, "contents": "Strategy - \nIdentify the functional groups in each molecule, and then check TABLE 12.1 to see where those groups absorb.\nSolution\n(a) Absorptions: 3400 to $3650 \\mathrm{~cm}^{-1}(\\mathrm{O}-\\mathrm{H}), 3020$ to $3100 \\mathrm{~cm}^{-1}(=\\mathrm{C}-\\mathrm{H}), 1640$ to $1680 \\mathrm{~cm}^{-1}(\\mathrm{C}=\\mathrm{C})$. This molecule has an alcohol $\\mathrm{O}-\\mathrm{H}$ group and an alkene double bond.\n(b) Absorptions: $3300 \\mathrm{~cm}^{-1}(\\equiv \\mathrm{C}-\\mathrm{H}), 2100$ to $2260 \\mathrm{~cm}^{-1}(\\mathrm{C} \\equiv \\mathrm{C}), 1735 \\mathrm{~cm}^{-1}(\\mathrm{C}=\\mathrm{O})$. This molecule has a terminal alkyne triple bond and a saturated ester carbonyl group."}
{"id": 769, "contents": "Identifying Functional Groups from an IR Spectrum - \nThe IR spectrum of an unknown compound is shown in FIGURE 12.28. What functional groups does the compound contain?\n\n\nFIGURE 12.28 IR spectrum for Worked Example 12.6."}
{"id": 770, "contents": "Strategy - \nAll IR spectra have many absorptions, but those useful for identifying specific functional groups are usually found in the region from $1500 \\mathrm{~cm}^{-1}$ to $3300 \\mathrm{~cm}^{-1}$. Pay particular attention to the carbonyl region ( 1670 to 1780\n$\\mathrm{cm}^{-1}$ ), the aromatic region ( 1660 to $2000 \\mathrm{~cm}^{-1}$ ), the triple-bond region ( 2000 to $2500 \\mathrm{~cm}^{-1}$ ), and the $\\mathrm{C}-\\mathrm{H}$ region (2500 to $3500 \\mathrm{~cm}^{-1}$ )."}
{"id": 771, "contents": "Solution - \nThe spectrum shows an intense absorption at $1725 \\mathrm{~cm}^{-1}$ due to a carbonyl group (perhaps an aldehyde, -CHO ), a series of weak absorptions from 1800 to $2000 \\mathrm{~cm}^{-1}$ characteristic of aromatic compounds, and a $\\mathrm{C}-\\mathrm{H}$ absorption near $3030 \\mathrm{~cm}^{-1}$, also characteristic of aromatic compounds. In fact, the compound is phenylacetaldehyde.\n\n\nPhenylacetaldehyde\n\nPROBLEM The IR spectrum of phenylacetylene is shown in Figure 12.29. What absorption bands can you 12-9 identify?\n\n\nFIGURE 12.29 The IR spectrum of phenylacetylene, Problem 12-9.\n\nPROBLEM Where might the following compounds have IR absorptions?\n12-10 (a)\n\n(b)\n\n(c)\n\n\nPROBLEM Where might the following compound have IR absorptions?"}
{"id": 772, "contents": "X-Ray Crystallography - \nThe various spectroscopic techniques described in this and the next two chapters are enormously important in chemistry and have been fine-tuned to such a degree that the structure of almost any molecule can be found. Nevertheless, wouldn't it be nice if you could simply look at a molecule and \"see\" its structure with your eyes?\n\nDetermining the three-dimensional shape of an object around you is easy-you just look at it, let your eyes focus the light rays reflected from the object, and let your brain assemble the data into a recognizable image. If the object is small, you use a microscope and let the microscope lens focus the visible light. Unfortunately, there is a limit to what you can see, even with the best optical microscope. Called the diffraction limit, you can't see anything smaller than the wavelength of light you are using for the observation. Visible light has wavelengths of several hundred nanometers, but atoms in molecules have dimensions on the order of 0.1 nm . Thus, to see a molecule-whether a small one in the laboratory or a large, complex enzyme with a molecular weight in the tens of thousands-you need wavelengths in the 0.1 nm range, which corresponds to X rays.\n\nLet's say that we want to determine the structure and shape of an enzyme or other biological molecule. The technique used is called X-ray crystallography. First, the molecule is crystallized (which often turns out to be the most difficult and time-consuming part of the entire process) and a small crystal of 0.4 to 0.5 mm on its longest axis is glued to the end of a glass fiber. The fiber and attached crystal are then mounted in an instrument called an X-ray diffractometer, which consists of a radiation source, a sample positioning and orienting device that can rotate the crystal in any direction, a detector, and a controlling computer.\n\nOnce mounted in the diffractometer, the crystal is irradiated with X rays, usually so-called $\\mathrm{Cu} K_{\\alpha}$ radiation with a wavelength of 0.154 nm . When the X rays strike the enzyme crystal, they interact with electrons in the molecule and are scattered into a diffraction pattern which, when detected and visualized, appears as a series of intense spots against a null background."}
{"id": 773, "contents": "X-Ray Crystallography - \nFIGURE 12.30 The structure of human muscle fructose-1,6-bisphosphate aldolase, as determined by X-ray crystallography. (credit: modification of work Protein Data Bank, 1ALD. PDB ID: 1ALD, Gamblin, S.J. Davies, G.J. Grimes, J.M. Jackson, R.M. Littlechild, J.A. Watson, H.C. (1991) J. Mol. Biol. 219: 573-576, CC BY 1.0.)\n\nManipulation of the diffraction pattern to extract three-dimensional molecular data is a complex process, but the final result is an electron-density map of the molecule. Because electrons are largely localized around atoms, any two centers of electron density located within bonding distance of each other are assumed to represent bonded atoms, leading to a recognizable chemical structure. So important is this structural information for biochemistry that an online database of approximately 145,000 biological substances has been created. Operated by Rutgers University and funded by the U.S. National Science Foundation, the Protein Data Bank ( PDB ) is a worldwide repository for processing and distributing three-dimensional structural data for biological macromolecules. We'll see how to access the PDB in the Chapter 26 Chemistry Matters."}
{"id": 774, "contents": "Key Terms - \n- absorption spectrum\n- amplitude\n- base peak\n- cation radical\n- electromagnetic spectrum\n- frequency, $v$\n- hertz, Hz,\n- infrared (IR) spectroscopy\n- MALDI\n- mass spectrometry (MS)\n- McLafferty Rearrangement\n- parent peak\n- photon\n- quadrupole mass analyzer\n- Time of Flight (TOF)\n- wavelength, $\\lambda$\n- wavenumber, $\\widetilde{v}$"}
{"id": 775, "contents": "Summary - \nFinding the structure of a new molecule, whether a small one synthesized in the laboratory or a large protein found in living organisms, is central to the progression of chemistry and biochemistry. The structure of an organic molecule is usually determined using spectroscopic methods, including mass spectrometry and infrared spectroscopy. Mass spectrometry (MS) tells the molecular weight and formula of a molecule; infrared (IR) spectroscopy identifies the functional groups present in the molecule.\n\nIn small-molecule mass spectrometry, molecules are first ionized by collision with a high-energy electron beam. The ions then fragment into smaller pieces, which are magnetically sorted according to their mass-to-charge ratio $(\\mathrm{m} / \\mathrm{z})$. The ionized sample molecule is called the molecular ion, $M^{+}$, and measurement of its mass gives the molecular weight of the sample. Structural clues about unknown samples can be obtained by interpreting the fragmentation pattern of the molecular ion. Mass-spectral fragmentations are usually complex, however, and interpretation is often difficult. In biological mass spectrometry, molecules are protonated using either electrospray ionization (ESI) or matrix-assisted laser desorption ionization (MALDI), and the protonated molecules are separated by time-of-flight (TOF) mass analysis.\n\nInfrared spectroscopy involves the interaction of a molecule with electromagnetic radiation. When an organic molecule is irradiated with infrared energy, certain frequencies are absorbed by the molecule. The frequencies absorbed correspond to the amounts of energy needed to increase the amplitude of specific molecular vibrations such as bond stretching and bending. Since every functional group has a characteristic combination of bonds, every functional group has a characteristic set of infrared absorptions. For example, the terminal alkyne $\\equiv \\mathrm{C}-\\mathrm{H}$ bond absorbs IR radiation of $3300 \\mathrm{~cm}^{-1}$, and the alkene $\\mathrm{C}=\\mathrm{C}$ bond absorbs in the range 1640 to $1680 \\mathrm{~cm}^{-1}$. By observing which frequencies of infrared radiation are absorbed by a molecule and which are not, it's possible to determine the functional groups a molecule contains."}
{"id": 776, "contents": "Visualizing Chemistry - \nPROBLEM Where in the IR spectrum would you expect each of the following molecules to absorb?\n\n\nPROBLEM Show the structures of the fragments you would expect in the mass spectra of the following 12-13 molecules:\n(a)\n\n(b)\n\n\nMass Spectrometry\nPROBLEM Propose structures for compounds that fit the following mass-spectral data:\n12-14 (a) A hydrocarbon with $\\mathrm{M}^{+}=132$ (b) A hydrocarbon with $\\mathrm{M}^{+}=166$\n(c) A hydrocarbon with $\\mathrm{M}^{+}=84$\n\nPROBLEM Write molecular formulas for compounds that show the following molecular ions in their high-\n12-15 resolution mass spectra, assuming that $\\mathrm{C}, \\mathrm{H}, \\mathrm{N}$, and O might be present. The exact atomic masses are: $1.00783\\left({ }^{1} \\mathrm{H}\\right), 12.00000\\left({ }^{12} \\mathrm{C}\\right), 14.00307\\left({ }^{14} \\mathrm{~N}\\right), 15.99491\\left({ }^{16} \\mathrm{O}\\right)$.\n(a) $\\mathrm{M}^{+}=98.0844$\n(b) $\\mathrm{M}^{+}=123.0320$\n\nPROBLEM Camphor, a saturated monoketone from the Asian camphor tree, is used among other things as a\n12-16 moth repellent and as a constituent of embalming fluid. If camphor has $\\mathrm{M}^{+}=152.1201$ by highresolution mass spectrometry, what is its molecular formula? How many rings does camphor have?\n\nPROBLEM The nitrogen rule of mass spectrometry says that a compound containing an odd number of\n12-17 nitrogens has an odd-numbered molecular ion. Conversely, a compound containing an even number of nitrogens has an even-numbered $\\mathrm{M}^{+}$peak. Explain."}
{"id": 777, "contents": "Visualizing Chemistry - \nPROBLEM In light of the nitrogen rule mentioned in Problem 12-17, what is the molecular formula of pyridine,\n12-18 $\\mathrm{M}^{+}=79$ ?\nPROBLEM Nicotine is a diamino compound isolated from dried tobacco leaves. Nicotine has two rings and\n12-19 $\\mathrm{M}^{+}=162.1157$ by high-resolution mass spectrometry. Give a molecular formula for nicotine, and calculate the number of double bonds.\n\nPROBLEM The hormone cortisone contains $\\mathrm{C}, \\mathrm{H}$, and O , and shows a molecular ion at $\\mathrm{M}^{+}=360.1937$ by\n$\\mathbf{1 2 - 2 0}$ high-resolution mass spectrometry. What is the molecular formula of cortisone? (The degree of unsaturation for cortisone is 8 .)\n\nPROBLEM Halogenated compounds are particularly easy to identify by their mass spectra because both\n12-21 chlorine and bromine occur naturally as mixtures of two abundant isotopes. Recall that chlorine occurs as ${ }^{35} \\mathrm{Cl}$ ( $75.8 \\%$ ) and ${ }^{37} \\mathrm{Cl}(24.2 \\%)$; and bromine occurs as ${ }^{79} \\mathrm{Br}(50.7 \\%)$ and ${ }^{81} \\mathrm{Br}(49.3 \\%)$. At what masses do the molecular ions occur for the following formulas? What are the relative percentages of each molecular ion?\n(a) Bromomethane, $\\mathrm{CH}_{3} \\mathrm{Br}$\n(b) 1-Chlorohexane, $\\mathrm{C}_{6} \\mathrm{H}_{13} \\mathrm{Cl}$\n\nPROBLEM By knowing the natural abundances of minor isotopes, it's possible to calculate the relative heights\n12-22 of $M^{+}$and $M+1$ peaks. If ${ }^{13} \\mathrm{C}$ has a natural abundance of $1.10 \\%$, what are the relative heights of the $M^{+}$and $M+1$ peaks in the mass spectrum of benzene, $C_{6} H_{6}$ ?"}
{"id": 778, "contents": "Visualizing Chemistry - \nPROBLEM Propose structures for compounds that fit the following data:\n12-23 (a) A ketone with $\\mathrm{M}^{+}=86$ and fragments at $m / z=71$ and $m / z=43$\n(b) An alcohol with $\\mathrm{M}^{+}=88$ and fragments at $m / z=73, m / z=70$, and $m / z=59$\n\nPROBLEM 2-Methylpentane $\\left(\\mathrm{C}_{6} \\mathrm{H}_{14}\\right)$ has the mass spectrum shown. Which peak represents $\\mathrm{M}^{+}$? Which is the\n12-24 base peak? Propose structures for fragment ions of $m / z=71,57,43$, and 29 . Why does the base peak have the mass it does?\n\n\nPROBLEM Assume that you are in a laboratory carrying out the catalytic hydrogenation of cyclohexene to 12-25 cyclohexane. How could you use a mass spectrometer to determine when the reaction is finished?\n\nPROBLEM What fragments might you expect in the mass spectra of the following compounds?\n12-26 (a)\n\n(b)\n\n(c)"}
{"id": 779, "contents": "Infrared Spectroscopy - \nPROBLEM How might you use IR spectroscopy to distinguish among the three isomers 1-butyne, 12-27 1,3-butadiene, and 2-butyne?\n\nPROBLEM Would you expect two enantiomers such as ( $R$ )-2-bromobutane and ( $S$ )-2-bromobutane to have\n12-28 identical or different IR spectra? Explain.\nPROBLEM Would you expect two diastereomers such as meso-2,3-dibromobutane and ( $2 R, 3 R$ )-12-29 dibromobutane to have identical or different IR spectra? Explain.\n\nPROBLEM Propose structures for compounds that meet the following descriptions:\n12-30 (a) $\\mathrm{C}_{5} \\mathrm{H}_{8}$, with IR absorptions at 3300 and $2150 \\mathrm{~cm}^{-1}$\n(b) $\\mathrm{C}_{4} \\mathrm{H}_{8} \\mathrm{O}$, with a strong IR absorption at $3400 \\mathrm{~cm}^{-1}$\n(c) $\\mathrm{C}_{4} \\mathrm{H}_{8} \\mathrm{O}$, with a strong IR absorption at $1715 \\mathrm{~cm}^{-1}$\n(d) $\\mathrm{C}_{8} \\mathrm{H}_{10}$, with IR absorptions at 1600 and $1500 \\mathrm{~cm}^{-1}$\n\nPROBLEM How could you use infrared spectroscopy to distinguish between the following pairs of isomers?\n12-31 (a) $\\mathrm{HC} \\equiv \\mathrm{CCH}_{2} \\mathrm{NH}_{2}$ and $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{C} \\equiv \\mathrm{N}$ (b) $\\mathrm{CH}_{3} \\mathrm{COCH}_{3}$ and $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CHO}$\nPROBLEM Two infrared spectra are shown. One is the spectrum of cyclohexane, and the other is the spectrum\n12-32 of cyclohexene. Identify them, and explain your answer.\n(a)\n\n(b)"}
{"id": 780, "contents": "Infrared Spectroscopy - \n(b)\n\n\nPROBLEM At what approximate positions might the following compounds show IR absorptions?\n\n12-33 (a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n\nPROBLEM How would you use infrared spectroscopy to distinguish between the following pairs of 12-34 constitutional isomers?\n(a)\n\nand $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{C} \\equiv \\mathrm{CH}$\n(b)\n\nand\n\n(c)\nand\n$\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CHO}$\nPROBLEM At what approximate positions might the following compounds show IR absorptions?\n12-35 (a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n\nPROBLEM Assume that you are carrying out the dehydration of 1-methylcyclohexanol to yield\n12-36 1-methylcyclohexene. How could you use infrared spectroscopy to determine when the reaction is complete?\n\nPROBLEM Assume that you are carrying out the base-induced dehydrobromination of 12-37 3-bromo-3-methylpentane (Section 11.7) to yield an alkene. How could you use IR spectroscopy to tell which of three possible elimination products is formed, if one includes $E / Z$ isomers?"}
{"id": 781, "contents": "General Problems - \nPROBLEM Which is stronger, the $\\mathrm{C}=\\mathrm{O}$ bond in an ester ( $1735 \\mathrm{~cm}^{-1}$ ) or the $\\mathrm{C}=\\mathrm{O}$ bond in a saturated ketone 12-38 ( $1715 \\mathrm{~cm}^{-1}$ )? Explain.\n\nPROBLEM Carvone is an unsaturated ketone responsible for the odor of spearmint. If carvone has $\\mathrm{M}^{+}=150$ in 12-39 its mass spectrum and contains three double bonds and one ring, what is its molecular formula?\nPROBLEM Carvone (Problem 12-39) has an intense infrared absorption at $1690 \\mathrm{~cm}^{-1}$. What kind of ketone 12-40 does carvone contain?\n\nPROBLEM The mass spectrum (a) and the infrared spectrum (b) of an unknown hydrocarbon are shown.\n12-41 Propose as many structures as you can.\n(a)\n\n(b)\n\n\nPROBLEM The mass spectrum (a) and the infrared spectrum (b) of another unknown hydrocarbon are shown. 12-42 Propose as many structures as you can.\n(a)\n\n(b)\n\n\nPROBLEM Propose structures for compounds that meet the following descriptions:\n12-43 (a) An optically active compound $\\mathrm{C}_{5} \\mathrm{H}_{10} \\mathrm{O}$ with an IR absorption at $1730 \\mathrm{~cm}^{-1}$\n(b) A non-optically active compound $\\mathrm{C}_{5} \\mathrm{H}_{9} \\mathrm{~N}$ with an IR absorption at $2215 \\mathrm{~cm}^{-1}$\n\nPROBLEM 4-Methyl-2-pentanone and 3-methylpentanal are isomers. Explain how you could tell them apart,\n12-44 both by mass spectrometry and by infrared spectroscopy.\n\n\n4-Methyl-2-pentanone\n\n\n3-Methylpentanal"}
{"id": 782, "contents": "General Problems - \n4-Methyl-2-pentanone\n\n\n3-Methylpentanal\n\nPROBLEM Grignard reagents (alkylmagnesium halides) undergo a general and very useful reaction with\n12-45 ketones. Methylmagnesium bromide, for example, reacts with cyclohexanone to yield a product with the formula $\\mathrm{C}_{7} \\mathrm{H}_{14} \\mathrm{O}$. What is the structure of this product if it has an IR absorption at 3400 $\\mathrm{cm}^{-1}$ ?\n\n\nCyclohexanone\nPROBLEM Ketones undergo a reduction when treated with sodium borohydride, $\\mathrm{NaBH}_{4}$. What is the structure 12-46 of the compound produced by reaction of 2-butanone with $\\mathrm{NaBH}_{4}$ if it has an IR absorption at 3400 $\\mathrm{cm}^{-1}$ and $\\mathrm{M}^{+}=74$ in the mass spectrum?\n\n\n2-Butanone\nPROBLEM Nitriles, $\\mathrm{R}-\\mathrm{C} \\equiv \\mathrm{N}$, undergo a hydrolysis reaction when heated with aqueous acid. What is the\n12-47 structure of the compound produced by hydrolysis of propanenitrile, $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{C} \\equiv \\mathrm{N}$, if it has IR absorptions from $2500-3100 \\mathrm{~cm}^{-1}$ and at $1710 \\mathrm{~cm}^{-1}$, and has $\\mathrm{M}^{+}=74$ ?\n\nPROBLEM The infrared spectrum of the compound with the following mass spectrum lacks any significant\n12-48 absorption above $3000 \\mathrm{~cm}^{-1}$. There is a prominent peak near $1740 \\mathrm{~cm}^{-1}$ and another strong peak near $1200 \\mathrm{~cm}^{-1}$. Propose a structure."}
{"id": 783, "contents": "General Problems - \nPROBLEM The infrared spectrum of the compound with the following mass spectrum has a medium-intensity 12-49 peak at about $1650 \\mathrm{~cm}^{-1}$. There is also a C-H out-of-plane bending peak near $880 \\mathrm{~cm}^{-1}$. Propose a structure.\n\n\nPROBLEM The infrared spectrum of the compound with the following mass spectrum has strong absorbances\n$\\mathbf{1 2 - 5 0}$ at 1584,1478 , and $1446 \\mathrm{~cm}^{-1}$. Propose a structure."}
{"id": 784, "contents": "CHAPTER 13 - \nStructure Determination: Nuclear Magnetic Resonance Spectroscopy\n\n\nFIGURE 13.1 NMR spectroscopy is an invaluable aid in carrying out the design and synthesis of new drugs. (credit: modification of work by Unknown/Pxhere, CCO 1.0)"}
{"id": 785, "contents": "CHAPTER CONTENTS - \n13.1 Nuclear Magnetic Resonance Spectroscopy\n13.2 The Nature of NMR Absorptions\n13.3 Chemical Shifts\n13.4 Chemical Shifts in ${ }^{1}$ H NMR Spectroscopy\n13.5 Integration of ${ }^{1} \\mathrm{H}$ NMR Absorptions: Proton Counting\n13.6 Spin-Spin Splitting in ${ }^{1}$ H NMR Spectra\n$13.7^{1} \\mathrm{H}$ NMR Spectroscopy and Proton Equivalence\n13.8 More Complex Spin-Spin Splitting Patterns\n13.9 Uses of ${ }^{1}$ H NMR Spectroscopy\n$13.10{ }^{13} \\mathrm{C}$ NMR Spectroscopy: Signal Averaging and FT-NMR\n13.11 Characteristics of ${ }^{13} \\mathrm{C}$ NMR Spectroscopy\n13.12 DEPT ${ }^{13}$ C NMR Spectroscopy\n13.13 Uses of ${ }^{13} \\mathrm{C}$ NMR Spectroscopy\n\nWHY THIS CHAPTER? Nuclear magnetic resonance (NMR) spectroscopy has far-reaching applications in many scientific fields, particularly in chemical structure determination. Although we'll just give an overview of the subject in this chapter, focusing on NMR applications with small molecules, more advanced NMR techniques are also used in biological chemistry to study protein structure and folding.\n\nNuclear magnetic resonance (NMR) spectroscopy is the most valuable spectroscopic technique available to organic chemists. It's the method of structure determination that organic chemists usually turn to first.\n\nWe saw in the chapter on Structure Determination: Mass Spectrometry and Infrared Spectroscopy that mass spectrometry gives a molecule's formula and infrared spectroscopy identifies a molecule's functional groups. Nuclear magnetic resonance spectroscopy complements these other techniques by mapping a molecule's carbon-hydrogen framework. Taken together, MS, IR, and NMR make it possible to determine the structures of even very complex molecules.\n\n| Mass spectrometry | Molecular size and formula |\n| :--- | :--- |\n| Infrared spectroscopy | Functional groups present |\n| NMR spectroscopy | Map of carbon-hydrogen framework |"}
{"id": 786, "contents": "CHAPTER CONTENTS - 13.1 Nuclear Magnetic Resonance Spectroscopy\nMany kinds of nuclei behave as if they were spinning about an axis, somewhat as the earth spins daily. Because they're positively charged, these spinning nuclei act like tiny magnets and can interact with an external magnetic field, denoted $\\boldsymbol{B}_{0}$. Not all nuclei act this way, but fortunately for organic chemists, both the proton $\\left({ }^{1} \\mathrm{H}\\right)$ and the ${ }^{13} \\mathrm{C}$ nucleus do have spins. The more common ${ }^{12} \\mathrm{C}$ isotope, however, does not have nuclear spin. (In speaking about NMR, the words proton and hydrogen are often used interchangeably, since a hydrogen nucleus is just a proton.) Let's see what the consequences of nuclear spin are and how we can use the results.\n\nIn the absence of an external magnetic field, the spins of magnetic nuclei are oriented randomly. When a sample containing these nuclei is placed between the poles of a strong magnet, however, the nuclei adopt specific orientations, much as a compass needle orients in the earth's magnetic field. A spinning ${ }^{1} \\mathrm{H}$ or ${ }^{13} \\mathrm{C}$ nucleus can orient so that its own tiny magnetic field is aligned either with (parallel to) or against (antiparallel to) the external field. The two orientations don't have the same energy, however, and aren't equally likely. The parallel orientation is slightly lower in energy by an amount that depends on the strength of the external field, making this spin state very slightly favored over the antiparallel orientation (FIGURE 13.2).\n(a)\n\n(b)\n\n\nFIGURE 13.2 (a) Nuclear spins are oriented randomly in the absence of an external magnetic field but (b) have a specific orientation in the presence of an external field, $\\boldsymbol{B}_{\\mathbf{0}}$. Some of the spins (red) are aligned parallel to the external field while others (blue) are antiparallel. The parallel spin state is slightly lower in energy and therefore favored."}
{"id": 787, "contents": "CHAPTER CONTENTS - 13.1 Nuclear Magnetic Resonance Spectroscopy\nIf the oriented nuclei are irradiated with electromagnetic radiation of the proper frequency, energy absorption occurs and the lower-energy spin state \"flips\" to the higher-energy state. When this spin-flip occurs, the magnetic nuclei are said to be in resonance with the applied radiation-hence the name nuclear magnetic resonance.\n\nThe exact frequency necessary for resonance depends both on the strength of the external magnetic field, the identity of the nucleus, and the electronic environment of the nucleus. If a very strong magnetic field is applied, the energy difference between the two spin states is larger and higher-frequency (higher-energy) radiation is required for a spin-flip. If a weaker magnetic field is applied, less energy is required to effect the transition between nuclear spin states (FIGURE 13.3).\n\n\nFIGURE 13.3 The energy difference $\\Delta E$ between nuclear spin states depends on the strength of the applied magnetic field. Absorption of energy with frequency $v$ converts a nucleus from a lower spin state to a higher spin state. (a) Spin states have equal energies in the absence of an applied magnetic field but (b) have unequal energies in the presence of a magnetic field. At $v=200 \\mathrm{MHz}, \\Delta E=8.0 \\times 10^{-5} \\mathrm{~kJ} / \\mathrm{mol}(1.9$ $\\left.\\times 10^{-5} \\mathrm{kcal} / \\mathrm{mol}\\right)$. (c) The energy difference between spin states is greater at larger applied fields. At $v=500 \\mathrm{MHz}, \\Delta E=2.0 \\times 10^{-4} \\mathrm{~kJ} / \\mathrm{mol}$."}
{"id": 788, "contents": "CHAPTER CONTENTS - 13.1 Nuclear Magnetic Resonance Spectroscopy\nIn practice, superconducting magnets that produce enormously powerful fields up to 23.5 tesla (T) are sometimes used, but field strengths in the range of 4.7 to 7.0 T are more common. At a magnetic field strength of 4.7 T , so-called radiofrequency (rf) energy in the 200 MHz range ( $1 \\mathrm{MHz}=10^{6} \\mathrm{~Hz}$ ) brings a ${ }^{1} \\mathrm{H}$ nucleus into resonance, and rf energy of 50 MHz brings a ${ }^{13} \\mathrm{C}$ nucleus into resonance. At the highest field strength currently available in commercial instruments ( 23.5 T ), 1000 MHz energy is required for ${ }^{1} \\mathrm{H}$ spectroscopy. These energies needed for NMR are much smaller than those required for IR spectroscopy; 200 MHz rf energy corresponds to only $8.0 \\times 10^{-5} \\mathrm{~kJ} / \\mathrm{mol}$ versus the 4.8 to $48 \\mathrm{~kJ} / \\mathrm{mol}$ needed for IR spectroscopy.\n${ }^{1} \\mathrm{H}$ and ${ }^{13} \\mathrm{C}$ nuclei are not unique in their ability to exhibit the NMR phenomenon. All nuclei with an odd number of protons ( ${ }^{1} \\mathrm{H},{ }^{2} \\mathrm{H},{ }^{14} \\mathrm{~N},{ }^{19} \\mathrm{~F},{ }^{31} \\mathrm{P}$, for example) and all nuclei with an odd number of neutrons $\\left({ }^{13} \\mathrm{C}\\right.$, for example) show magnetic properties. Only nuclei with even numbers of both protons and neutrons $\\left({ }^{12} \\mathrm{C},{ }^{16} \\mathrm{O},{ }^{32} \\mathrm{~S}\\right)$ do not give rise to magnetic phenomena (TABLE 13.1)."}
{"id": 789, "contents": "CHAPTER CONTENTS - 13.1 Nuclear Magnetic Resonance Spectroscopy\n| TABLE 13.1 The NMR Behavior of Some Common |  |\n| :---: | :---: |\n| Nuclei |  |\n| Magnetic nuclei | Nonmagnetic nuclei |\n| ${ }^{1} \\mathrm{H}$ | ${ }^{12} \\mathrm{C}$ |\n| ${ }^{13} \\mathrm{C}$ | ${ }^{16} \\mathrm{O}$ |\n| ${ }^{2} \\mathrm{H}$ | ${ }^{32} \\mathrm{~S}$ |\n| ${ }^{14} \\mathrm{~N}$ |  |\n| ${ }^{19} \\mathrm{~F}$ |  |\n| ${ }^{31} \\mathrm{P}$ |  |\n\nPROBLEM The amount of energy required to spin-flip a nucleus depends both on the strength of the external 13-1 magnetic field and on the nucleus. At a field strength of 4.7 T , rf energy of 200 MHz is required to bring a ${ }^{1} \\mathrm{H}$ nucleus into resonance, but energy of only 187 MHz will bring a ${ }^{19} \\mathrm{~F}$ nucleus into resonance. Calculate the amount of energy required to spin-flip a ${ }^{19} \\mathrm{~F}$ nucleus. Is this amount greater or less than that required to spin-flip a ${ }^{1} \\mathrm{H}$ nucleus?\n\nPROBLEM Calculate the amount of energy required to spin-flip a proton in a spectrometer operating at 300 13-2 MHz. Does increasing the spectrometer frequency from 200 to 300 MHz increase or decrease the amount of energy necessary for resonance?"}
{"id": 790, "contents": "CHAPTER CONTENTS - 13.2 The Nature of NMR Absorptions\nFrom the description thus far, you might expect all ${ }^{1} \\mathrm{H}$ nuclei in a molecule to absorb energy at the same frequency and all ${ }^{13} \\mathrm{C}$ nuclei to absorb at the same frequency. If so, we would observe only a single NMR absorption signal in the ${ }^{1} \\mathrm{H}$ or ${ }^{13} \\mathrm{C}$ spectrum of a molecule, a situation that would be of little use. In fact, the absorption frequency is not the same for all ${ }^{1} \\mathrm{H}$ or all ${ }^{13} \\mathrm{C}$ nuclei.\n\nAll nuclei in molecules are surrounded by electrons. When an external magnetic field is applied to a molecule, the electrons moving around nuclei set up tiny local magnetic fields of their own. These local magnetic fields act in opposition to the applied field so that the effective field actually felt by the nucleus is a bit weaker than the applied field.\n\n$$\nB_{\\text {effective }}=B_{\\text {applied }}=B_{\\text {local }}\n$$\n\nIn describing this effect of local fields, we say that nuclei experience shielding from the full effect of the applied field by the surrounding electrons. Because each chemically distinct nucleus in a molecule is in a slightly different electronic environment, each nucleus is shielded to a slightly different extent and the effective magnetic field felt by each is slightly different. These tiny differences in the effective magnetic fields experienced by different nuclei can be detected, and we thus see a distinct NMR signal for each chemically distinct ${ }^{13} \\mathrm{C}$ or ${ }^{1} \\mathrm{H}$ nucleus in a molecule. As a result, an NMR spectrum effectively maps the carbon-hydrogen framework of an organic molecule. With practice, it's possible to read this map and derive structural information."}
{"id": 791, "contents": "CHAPTER CONTENTS - 13.2 The Nature of NMR Absorptions\nFIGURE 13.4 shows both the ${ }^{1} \\mathrm{H}$ and the ${ }^{13} \\mathrm{C}$ NMR spectra of methyl acetate, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{CH}_{3}$. The horizontal axis shows the effective field strength felt by the nuclei, and the vertical axis indicates the intensity of absorption of rf energy. Each peak in the NMR spectrum corresponds to a chemically distinct ${ }^{1} \\mathrm{H}$ or ${ }^{13} \\mathrm{C}$ nucleus in the molecule. Note that NMR spectra are formatted with the zero absorption line at the bottom, whereas IR spectra are formatted with the zero absorption line at the top; Section 12.5 . Note also that ${ }^{1} \\mathrm{H}$ and ${ }^{13} \\mathrm{C}$ spectra can't be observed simultaneously on the same spectrometer because different amounts of energy are required to spinflip the different kinds of nuclei. The two spectra must be recorded separately."}
{"id": 792, "contents": "CHAPTER CONTENTS - 13.2 The Nature of NMR Absorptions\nFIGURE 13.4 (a) The ${ }^{\\mathbf{1}} \\mathrm{H}$ NMR spectrum and (b) the proton-decoupled ${ }^{13} \\mathrm{C}$ NMR spectrum of methyl acetate, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{CH}_{3}$. The small peak\nlabeled \"TMS\" at the far right of each spectrum is a calibration peak, as explained in the next section.\nThe ${ }^{13}$ C NMR spectrum of methyl acetate in FIGURE 13.4b shows three peaks, one for each of the three chemically distinct carbon atoms in the molecule. The ${ }^{1} \\mathrm{H}$ NMR spectrum in FIGURE 13.4a shows only two peaks, however, even though methyl acetate has six hydrogens. One peak is due to the $\\mathrm{CH}_{3} \\mathrm{C}=\\mathrm{O}$ hydrogens, and the other to the $-\\mathrm{OCH}_{3}$ hydrogens. Because the three hydrogens in each methyl group have the same electronic environment, they are shielded to the same extent and are said to be equivalent. Chemically equivalent nuclei always show the same absorption. The two methyl groups themselves, however, are not equivalent, so the two sets of hydrogens absorb at different positions.\n\nThe operation of a basic NMR spectrometer is illustrated in FIGURE 13.5. An organic sample is dissolved in a suitable solvent (usually deuteriochloroform, $\\mathrm{CDCl}_{3}$, which has no hydrogens) and placed in a thin glass tube between the poles of a magnet. The strong magnetic field causes the ${ }^{1} \\mathrm{H}$ and ${ }^{13} \\mathrm{C}$ nuclei in the molecule to align in one of the two possible orientations, and the sample is irradiated with rf energy. If the frequency of the rf irradiation is held constant and the strength of the applied magnetic field is varied, each nucleus comes into resonance at a slightly different field strength. A sensitive detector monitors the absorption of rf energy, and its electronic signal is then amplified and displayed as a peak."}
{"id": 793, "contents": "CHAPTER CONTENTS - 13.2 The Nature of NMR Absorptions\nFIGURE 13.5 Schematic operation of a basic NMR spectrometer. A thin glass tube containing the sample solution is placed between the poles of a strong magnet and irradiated with rf energy.\n\nNMR spectroscopy differs from IR spectroscopy (Section 12.6-Section 12.8) in that the timescales of the two techniques are quite different. The absorption of infrared energy by a molecule giving rise to a change in vibrational amplitude is an essentially instantaneous process (about $10^{-13} \\mathrm{~s}$ ), but the NMR process is much slower (about $10^{-3} \\mathrm{~s}$ ). This difference in timescales between IR and NMR spectroscopy is analogous to the difference between cameras operating at very fast and very slow shutter speeds. The fast camera (IR) takes an instantaneous picture and freezes the action. If two rapidly interconverting species are present, IR spectroscopy records the spectrum of both. The slow camera (NMR), however, takes a blurred, time-averaged picture. If two species interconverting faster than $10^{3}$ times per second are present in a sample, NMR records only a single, averaged spectrum, rather than separate spectra of the two discrete species.\n\nBecause of this blurring effect, NMR spectroscopy can be used to measure the rates and activation energies of very fast chemical processes. In cyclohexane, for example, a ring-flip (Section 4.6) occurs so rapidly at room temperature that axial and equatorial hydrogens can't be distinguished by NMR; only a single, averaged ${ }^{1} \\mathrm{H}$ NMR absorption is seen for cyclohexane at $25^{\\circ} \\mathrm{C}$. At $-90^{\\circ} \\mathrm{C}$, however, the ring-flip is slowed down enough that two absorption peaks are visible, one for the six axial hydrogens and one for the six equatorial hydrogens. Knowing the temperature and the rate at which signal blurring begins to occur, it's possible to calculate that the activation energy for the cyclohexane ring-flip is $45 \\mathrm{~kJ} / \\mathrm{mol}(10.8 \\mathrm{kcal} / \\mathrm{mol})$."}
{"id": 794, "contents": "CHAPTER CONTENTS - 13.2 The Nature of NMR Absorptions\nPROBLEM 2-Chloropropene shows signals for three kinds of protons in its ${ }^{1}$ H NMR spectrum. Explain. 13-3"}
{"id": 795, "contents": "CHAPTER CONTENTS - 13.3 Chemical Shifts\nNMR spectra are displayed on charts that show the applied field strength increasing from left to right (FIGURE 13.6). Thus, the left part of the chart is the low-field, or downfield, side, and the right part is the high-field, or upfield, side. Nuclei that absorb on the downfield side of the chart require a lower field strength for resonance, implying that they have less shielding. Nuclei that absorb on the upfield side require a higher field strength for resonance, implying that they have more shielding.\n\n\nFIGURE 13.6 The NMR chart. The downfield, deshielded side is on the left, and the upfield, shielded side is on the right. The tetramethylsilane (TMS) absorption is used as reference point.\n\nTo define the position of an absorption, the NMR chart is calibrated and a reference point is used. In practice, a small amount of tetramethylsilane $\\left[\\mathrm{TMS} ;\\left(\\mathrm{CH}_{3}\\right)_{4} \\mathrm{Si}\\right]$ is added to the sample so that a reference absorption peak is produced when the spectrum is run. TMS is used as reference for both ${ }^{1} \\mathrm{H}$ and ${ }^{13} \\mathrm{C}$ measurements because in both cases it produces a single peak that occurs upfield of other absorptions normally found in organic compounds. The ${ }^{1} \\mathrm{H}$ and ${ }^{13} \\mathrm{C}$ spectra of methyl acetate in FIGURE 13.4 have the TMS reference peak indicated."}
{"id": 796, "contents": "CHAPTER CONTENTS - 13.3 Chemical Shifts\nThe position on the chart at which a nucleus absorbs is called its chemical shift. The chemical shift of TMS is set as the zero point, and other absorptions normally occur downfield, to the left on the chart. NMR charts are calibrated using an arbitrary scale called the delta ( $\\boldsymbol{\\delta}$ ) scale, where $1 \\delta$ equals 1 part-per-million ( 1 ppm ) of the spectrometer operating frequency. For example, if we were measuring the ${ }^{1} \\mathrm{H}$ NMR spectrum of a sample using an instrument operating at $200 \\mathrm{MHz}, 1 \\delta$ would be 1 part per million of $200,000,000 \\mathrm{~Hz}$, or 200 Hz . If we were measuring the spectrum using a 500 MHz instrument, $1 \\delta=500 \\mathrm{~Hz}$. The following equation can be used for any absorption:\n\n$$\n\\delta=\\frac{\\text { Observed chemical shift (number of Hz away from TMS) }}{\\text { Spectrometer frequency in } \\mathrm{MHz}}\n$$\n\nAlthough this method of calibrating NMR charts may seem complex, there's a good reason for it. As we saw earlier, the rf frequency required to bring a given nucleus into resonance depends on the spectrometer's magnetic field strength. But because there are many different kinds of spectrometers with many different magnetic field strengths available, chemical shifts given in frequency units ( Hz ) vary from one instrument to another. Thus, a resonance that occurs at 120 Hz downfield from TMS on one spectrometer might occur at 600 Hz downfield from TMS on another spectrometer with a more powerful magnet.\n\nBy using a system of measurement in which NMR absorptions are expressed in relative terms (parts per million relative to spectrometer frequency) rather than absolute terms (Hz), it's possible to compare spectra obtained\non different instruments. The chemical shift of an NMR absorption in $\\delta$ units is constant, regardless of the operating frequency of the spectrometer. A ${ }^{1} \\mathrm{H}$ nucleus that absorbs at $2.0 \\delta$ on a 200 MHz instrument also absorbs at $2.0 \\delta$ on a 500 MHz instrument."}
{"id": 797, "contents": "CHAPTER CONTENTS - 13.3 Chemical Shifts\nThe range in which most NMR absorptions occur is quite narrow. Almost all ${ }^{1} \\mathrm{H}$ NMR absorptions occur from 0 to $10 \\delta$ downfield from the proton absorption of TMS, and almost all ${ }^{13} \\mathrm{C}$ absorptions occur from 1 to $220 \\delta$ downfield from the carbon absorption of TMS. Thus, there is a likelihood that accidental overlap of nonequivalent signals will occur. The advantage of using an instrument with higher field strength (say, 500 $\\mathrm{MHz})$ rather than lower field strength $(200 \\mathrm{MHz})$ is that different NMR absorptions are more widely separated at the higher field strength. The chances that two signals will accidentally overlap are therefore lessened, and interpretation of spectra becomes easier. For example, two signals that are only 20 Hz apart at $200 \\mathrm{MHz}(0.1$ ppm) are 50 Hz apart at 500 MHz (still 0.1 ppm ).\nPROBLEM The following ${ }^{1} \\mathrm{H}$ NMR peaks were recorded on a spectrometer operating at 200 MHz . Convert each 13-4 into $\\delta$ units.\n(a) $\\mathrm{CHCl}_{3} ; 1454 \\mathrm{~Hz}$\n(b) $\\mathrm{CH}_{3} \\mathrm{Cl} ; 610 \\mathrm{~Hz}$\n(c) $\\mathrm{CH}_{3} \\mathrm{OH} ; 693 \\mathrm{~Hz}$\n(d) $\\mathrm{CH}_{2} \\mathrm{Cl}_{2} ; 1060 \\mathrm{~Hz}$"}
{"id": 798, "contents": "CHAPTER CONTENTS - 13.3 Chemical Shifts\nPROBLEM When the ${ }^{1} \\mathrm{H}$ NMR spectrum of acetone, $\\mathrm{CH}_{3} \\mathrm{COCH}_{3}$, is recorded on an instrument operating at 200 13-5 MHz, a single sharp resonance at $2.1 \\delta$ is seen.\n(a) How many hertz downfield from TMS does the acetone resonance correspond to?\n(b) If the ${ }^{1} \\mathrm{H}$ NMR spectrum of acetone is recorded at 500 MHz , what would the position of the absorption be in $\\delta$ units?\n(c) How many hertz downfield from TMS does this 500 MHz resonance correspond to?"}
{"id": 799, "contents": "CHAPTER CONTENTS - 13.4 Chemical Shifts in ${ }^{1} \\mathrm{H}$ NMR Spectroscopy\nAs mentioned previously, differences in chemical shifts are caused by the small local magnetic field of electrons surrounding different nuclei. Nuclei that are more strongly shielded by electrons require a higher applied field to bring them into resonance so they absorb on the right side of the NMR chart. Nuclei that are less strongly shielded need a lower applied field for resonance so they absorb on the left of the NMR chart.\n\nMost ${ }^{1} \\mathrm{H}$ chemical shifts fall within the range 0 to $10 \\delta$, which can be divided into the five regions shown in TABLE 13.2. By remembering the positions of these regions, it's often possible to tell at a glance what kinds of protons a molecule contains.\n\n\nTABLE 13.3 shows the correlation of ${ }^{1} \\mathrm{H}$ chemical shift with electronic environment in more detail. In general, protons bonded to saturated, $s p^{3}$-hybridized carbons absorb at higher fields, whereas protons bonded to $s p^{2}$-hybridized carbons absorb at lower fields. Protons on carbons that are bonded to electronegative atoms, such as N, O, or halogen, also absorb at lower fields."}
{"id": 800, "contents": "CHAPTER CONTENTS - 13.4 Chemical Shifts in ${ }^{1} \\mathrm{H}$ NMR Spectroscopy\n| Type of hydrogen |  | Chemical shift ( $\\mathbf{8}$ ) |\n| :---: | :---: | :---: |\n| Reference | $\\mathrm{Si}\\left(\\mathrm{CH}_{3}\\right)_{4}$ | 0 |\n| Alkyl (primary) | $-\\mathrm{CH}_{3}$ | 0.7-1.3 |\n| Alkyl (secondary) | $-\\mathrm{CH}_{2}-$ | 1.2-1.6 |\n| Alkyl (tertiary) | <smiles>CC(C)C</smiles> | 1.4-1.8 |\n| Allylic | <smiles>C=CC(C)C</smiles> | 1.6-2.2 |\n| Methyl ketone | <smiles>CC(C)=O</smiles> | 2.0-2.4 |\n| Aromatic methyl | $\\mathrm{Ar}-\\mathrm{CH}_{3}$ | 2.4-2.7 |\n| Alkynyl | $-\\mathrm{C} \\equiv \\mathrm{C}-\\mathrm{H}$ | 2.5-3.0 |\n| Alkyl halide | <smiles>CCC(C)C</smiles> | 2.5-4.0 |\n| Alcohol | <smiles>CC(C)(C)O</smiles> | 2.5-5.0 |\n| Alcohol, ether | <smiles>COC(C)C</smiles> | 3.3-4.5 |\n| Vinylic | <smiles>CC=C(C)C</smiles> | 4.5-6.5 |\n| Aryl | Ar-H | 6.5-8.0 |\n| Aldehyde | <smiles>CC=O</smiles> | $9.7-10.0$ |\n| Carboxylic acid | <smiles>CC(=O)O</smiles> | 11.0-12.0 |"}
{"id": 801, "contents": "Predicting Chemical Shifts in ${ }^{1}$ H NMR Spectra - \nMethyl 2,2-dimethylpropanoate $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CCO}_{2} \\mathrm{CH}_{3}$ has two peaks in its ${ }^{1} \\mathrm{H}$ NMR spectrum. What are their approximate chemical shifts?"}
{"id": 802, "contents": "Strategy - \nIdentify the types of hydrogens in the molecule, and note whether each is alkyl, vinylic, or next to an electronegative atom. Then predict where each absorbs, using TABLE 13.3 if necessary."}
{"id": 803, "contents": "Solution - \nThe $-\\mathrm{OCH}_{3}$ protons absorb around 3.5 to $4.0 \\delta$ because they are on carbon bonded to oxygen. The $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{C}-$ protons absorb near $1.0 \\delta$ because they are typical alkane-like protons.\n\nPROBLEM Each of the following compounds has a single ${ }^{1} \\mathrm{H}$ NMR peak. Approximately where would you 13-6 expect each compound to absorb?\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n\n\n\nPROBLEM Identify the different types of protons in the following molecule, and tell where you would expect\n13-7 each to absorb:"}
{"id": 804, "contents": "Solution - 13.5 Integration of ${ }^{1} \\mathrm{H}$ NMR Absorptions: Proton Counting\nLook at the ${ }^{1} \\mathrm{H}$ NMR spectrum of methyl 2,2-dimethylpropanoate in FIGURE 13.7. There are two peaks, corresponding to the two kinds of protons, but the peaks aren't the same size. The peak at $1.2 \\delta$, due to the $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{C}$ - protons, is larger than the peak at $3.7 \\mathrm{\\delta}$, due to the $-\\mathrm{OCH}_{3}$ protons.\n\n\nFIGURE 13.7 The ${ }^{\\mathbf{1}} \\mathbf{H}$ NMR spectrum of methyl 2,2-dimethylpropanoate. Integrating the peaks in a stair-step manner shows that they have a $1: 3$ ratio, corresponding to the ratio of the numbers of protons $(3: 9)$ responsible for each peak. Modern instruments give a direct digital readout of relative peak areas.\n\nThe area under each peak is proportional to the number of protons causing that peak. By electronically measuring, or integrating, the area under each peak, it's possible to measure the relative numbers of the different kinds of protons in a molecule.\n\nModern NMR instruments provide a digital readout of relative peak areas, but an older, more visual method displays the integrated peak areas as a stair-step line, with the height of each step proportional to the area under the peak, and therefore proportional to the relative number of protons causing the peak. For example, the two steps for the peaks in methyl 2,2-dimethylpropanoate have a $1: 3$ (or $3: 9$ ) height ratio when integrated-exactly what we expect because the three $-\\mathrm{OCH}_{3}$ protons are equivalent and the nine $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{C}$ - protons are equivalent."}
{"id": 805, "contents": "Solution - 13.5 Integration of ${ }^{1} \\mathrm{H}$ NMR Absorptions: Proton Counting\nPROBLEM How many peaks would you expect in the ${ }^{1} \\mathrm{H}$ NMR spectrum of 1,4-dimethylbenzene (para-xylene,\n13-8 or $p$-xylene)? What ratio of peak areas would you expect on integration of the spectrum? Refer to Table 13.3 for approximate chemical shifts, and sketch what the spectrum would look like. (Remember from Section 2.4 that aromatic rings have two resonance forms.)\n\n$p$-Xylene"}
{"id": 806, "contents": "Solution - 13.6 Spin-Spin Splitting in ${ }^{1}$ H NMR Spectra\nIn the ${ }^{1} H$ NMR spectra we've seen thus far, each different kind of proton in a molecule has given rise to a single peak. It often happens, though, that the absorption of a proton splits into multiple peaks, called a multiplet. For example, in the ${ }^{1} \\mathrm{H}$ NMR spectrum of bromoethane shown in FIGURE 13.8, the $-\\mathrm{CH}_{2} \\mathrm{Br}$ protons appear as four peaks (a quartet) centered at $3.42 \\delta$ and the $-\\mathrm{CH}_{3}$ protons appear as three peaks (a triplet) centered at $1.68 \\delta$.\n\n\nFIGURE 13.8 The ${ }^{\\mathbf{1}} \\mathbf{H}$ NMR spectrum of bromoethane, $\\mathbf{C H}_{\\mathbf{3}} \\mathbf{C H}_{\\mathbf{2}} \\mathbf{B r}$. The $-\\mathrm{CH}_{2} \\mathrm{Br}$ protons appear as a quartet at $3.42 \\delta$, and the $-\\mathrm{CH}_{3}$ protons appear as a triplet at $1.68 \\delta$.\n\nCalled spin-spin splitting, multiple absorptions of a nucleus are caused by the interaction, or coupling, of the spins of nearby nuclei. In other words, the tiny magnetic field produced by one nucleus affects the magnetic field felt by a neighboring nucleus. Look at the $-\\mathrm{CH}_{3}$ protons in bromoethane, for example. The three equivalent $-\\mathrm{CH}_{3}$ protons are neighbored by two other magnetic nuclei-the two protons on the adjacent $-\\mathrm{CH}_{2} \\mathrm{Br}$ group. Each of the neighboring $-\\mathrm{CH}_{2} \\mathrm{Br}$ protons has its own nuclear spin, which can align either with or against the applied field, producing a tiny effect that is felt by the $-\\mathrm{CH}_{3}$ protons."}
{"id": 807, "contents": "Solution - 13.6 Spin-Spin Splitting in ${ }^{1}$ H NMR Spectra\nThere are three ways in which the spins of the two $-\\mathrm{CH}_{2} \\mathrm{Br}$ protons can align, as shown in FIGURE 13.9. If both proton spins align with the applied field, the total effective field felt by the neighboring $-\\mathrm{CH}_{3}$ protons is slightly larger than it would be otherwise. Consequently, the applied field necessary to cause resonance is slightly reduced. Alternatively, if one of the $-\\mathrm{CH}_{2} \\mathrm{Br}$ proton spins aligns with the field and one aligns against the field, there is no effect on the neighboring $-\\mathrm{CH}_{3}$ protons. (This arrangement can occur in two ways, depending on which of the two proton spins aligns which way.) Finally, if both $-\\mathrm{CH}_{2} \\mathrm{Br}$ proton spins align against the applied field, the effective field felt by the $-\\mathrm{CH}_{3}$ protons is slightly smaller than it would be otherwise, and the applied field needed for resonance is slightly increased.\n\n\nFIGURE 13.9 The origin of spin-spin splitting in bromoethane. The nuclear spins of neighboring protons, indicated by horizontal arrows, align either with or against the applied field, causing the splitting of absorptions into multiplets.\n\nAny given molecule has only one of the three possible alignments of $-\\mathrm{CH}_{2} \\mathrm{Br}$ spins, but in a large collection of molecules, all three spin states are represented in a 1:2:1 statistical ratio. We therefore find that the neighboring $-\\mathrm{CH}_{3}$ protons come into resonance at three slightly different values of the applied field, and we see a 1:2:1 triplet in the NMR spectrum. One resonance is a little above where it would be without coupling, one is at the same place it would be without coupling, and the third resonance is a little below where it would be without coupling."}
{"id": 808, "contents": "Solution - 13.6 Spin-Spin Splitting in ${ }^{1}$ H NMR Spectra\nIn the same way that the $-\\mathrm{CH}_{3}$ absorption of bromoethane is split into a triplet, the $-\\mathrm{CH}_{2} \\mathrm{Br}$ absorption is split into a quartet. The three spins of the neighboring $-\\mathrm{CH}_{3}$ protons can align in four possible combinations: all three with the applied field, two with and one against (three ways), one with and two against (three ways), or all three against. Thus, four peaks are produced for the $-\\mathrm{CH}_{2} \\mathrm{Br}$ protons in a 1:3:3:1 ratio.\n\nAs a general rule, called the $\\boldsymbol{n}+1$ rule, protons that have $n$ equivalent neighboring protons show $n+1$ peaks in their NMR spectrum. For example, the spectrum of 2-bromopropane in FIGURE 13.10 shows a doublet at 1.71 $\\delta$ and a seven-line multiplet, or septet, at $4.28 \\delta$. The septet is caused by splitting of the $-\\mathrm{CHBr}-$ proton signal by six equivalent neighboring protons on the two methyl groups ( $n=6$ leads to $6+1=7$ peaks). The doublet is due to signal splitting of the six equivalent methyl protons by the single $-\\mathrm{CHBr}-$ proton ( $n=1$ leads to 2 peaks). Integration confirms the expected $6: 1$ ratio.\n\n\nFIGURE 13.10 The ${ }^{\\mathbf{1}} \\mathbf{H}$ NMR spectrum of 2-bromopropane. The $-\\mathrm{CH}_{3}$ proton signal at $1.71 \\delta$ is split into a doublet, and the $-\\mathrm{CHBr}-\\mathrm{proton}$ signal at $4.28 \\delta$ is split into a septet. Note that the distance between peaks is the same in both multiplets. Note also that the two outer peaks of the septet are small enough to be nearly missed."}
{"id": 809, "contents": "Solution - 13.6 Spin-Spin Splitting in ${ }^{1}$ H NMR Spectra\nThe distance between peaks in a multiplet is called the coupling constant and is denoted J. Coupling constants are measured in hertz and generally fall in the range 0 to 18 Hz . The exact value of the coupling constant between two neighboring protons depends on the geometry of the molecule, but a typical value for an openchain alkane is $J=6$ to 8 Hz . The same coupling constant is shared by both groups of hydrogens whose spins are coupled and is independent of spectrometer field strength. In bromoethane, for instance, the $-\\mathrm{CH}_{2} \\mathrm{Br}$ protons\nare coupled to the $-\\mathrm{CH}_{3}$ protons and appear as a quartet with $J=7 \\mathrm{~Hz}$. The $-\\mathrm{CH}_{3}$ protons appear as a triplet with the same $J=7 \\mathrm{~Hz}$ coupling constant.\n\nBecause coupling is a reciprocal interaction between two adjacent groups of protons, it's sometimes possible to tell which multiplets in a complex NMR spectrum are related to each other. If two multiplets have the same coupling constant, they are probably related, and the protons causing those multiplets are therefore adjacent in the molecule.\n\nThe most commonly observed coupling patterns and the relative intensities of lines in their multiplets are listed in TABLE 13.4. Note that it's not possible for a given proton to have five equivalent neighboring protons. (Why not?) A six-line multiplet, or sextet, is therefore found only when a proton has five nonequivalent neighboring protons that coincidentally happen to be coupled with an identical coupling constant $J$.\n\nTABLE 13.4 Some Common Spin Multiplicities"}
{"id": 810, "contents": "Solution - 13.6 Spin-Spin Splitting in ${ }^{1}$ H NMR Spectra\nTABLE 13.4 Some Common Spin Multiplicities\n\n| Number of equivalent adjacent protons | Multiplet | Ratio of intensities |\n| :---: | :--- | :---: |\n| 0 | Singlet | 1 |\n| 1 | Doublet | $1: 1$ |\n| 2 | Triplet | $1: 2: 1$ |\n| 3 | Quartet | $1: 3: 3: 1$ |\n| 4 | Quintet | $1: 4: 6: 4: 1$ |\n| 6 | Septet | $1: 6: 15: 20: 15: 6: 1$ |\n\nSpin-spin splitting in ${ }^{1} \\mathrm{H}$ NMR can be summarized by three rules."}
{"id": 811, "contents": "RULE 1 - \nChemically equivalent protons don't show spin-spin splitting. Equivalent protons may be on the same carbon or on different carbons, but their signals don't split.\n\n\nThree C-H protons are chemically equivalent; no splitting occurs.\n\n\nFour C-H protons are chemically equivalent; no splitting occurs.\n\nRULE 2\nThe signal of a proton with $n$ equivalent neighboring protons is split into a multiplet of $n+1$ peaks with coupling constant $\\boldsymbol{J}$. Protons that are farther than two carbon atoms apart don't usually couple, although they sometimes show weak coupling when they are separated by a $\\pi$ bond.\n\n\nSplitting observed\n\n\nSplitting not usually observed\n\nRULE 3\nTwo groups of protons coupled to each other have the same coupling constant, $J$.\nThe spectrum of para-methoxypropiophenone in FIGURE 13.11 further illustrates these three rules. The downfield absorptions at 6.91 and $7.93 \\delta$ are due to the four aromatic-ring protons. There are two kinds of aromatic protons, each of which gives a signal that is split into a doublet by its neighbor. The $-\\mathrm{OCH}_{3}$ signal is\nunsplit and appears as a sharp singlet at $3.84 \\delta$. The $-\\mathrm{CH}_{2}$ - protons next to the carbonyl group appear at $2.93 \\delta$ in the region expected for protons on carbon next to an unsaturated center, and their signal is split into a quartet by coupling with the protons of the neighboring methyl group. The methyl protons appear as a triplet at $1.20 \\delta$ in the usual upfield region.\n\n\nFIGURE 13.11 The ${ }^{\\mathbf{1}} \\mathbf{H}$ NMR spectrum of para-methoxypropiophenone."}
{"id": 812, "contents": "Assigning a Chemical Structure from a ${ }^{1} \\mathrm{H}$ NMR Spectrum - \nPropose a structure for a compound, $\\mathrm{C}_{5} \\mathrm{H}_{12} \\mathrm{O}$, that fits the following ${ }^{1} \\mathrm{H}$ NMR data: $0.92 \\delta(3 \\mathrm{H}$, triplet, $J=7 \\mathrm{~Hz}$ ), $1.20 \\delta(6 \\mathrm{H}$, singlet), $1.50 \\delta(2 \\mathrm{H}$, quartet, $J=7 \\mathrm{~Hz}$, $1.64 \\delta(1 \\mathrm{H}$, broad singlet)."}
{"id": 813, "contents": "Strategy - \nIt's best to begin solving structural problems by calculating a molecule's degree of unsaturation (we'll see this again in Worked Example 13.4). In the present instance, a formula of $\\mathrm{C}_{5} \\mathrm{H}_{12} \\mathrm{O}$ corresponds to a saturated, openchain molecule, either an alcohol or an ether.\n\nTo interpret the NMR information, let's look at each absorption individually. The three-proton absorption at $0.92 \\delta$ is due to a methyl group in an alkane-like environment, and the triplet-splitting pattern implies that the $\\mathrm{CH}_{3}$ is next to a $\\mathrm{CH}_{2}$. Thus, our molecule contains an ethyl group, $\\mathrm{CH}_{3} \\mathrm{CH}_{2}-$. The six-proton singlet at $1.20 \\delta$ is due to two equivalent alkane-like methyl groups attached to a carbon with no hydrogens, $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{C}$, and the twoproton quartet at $1.50 \\delta$ is due to the $\\mathrm{CH}_{2}$ of the ethyl group. All 5 carbons and 11 of the 12 hydrogens in the molecule are now accounted for. The remaining hydrogen, which appears as a broad one-proton singlet at 1.64 $\\delta$, is probably due to an OH group, since there is no other way to account for it. Putting the pieces together gives the structure: 2-methyl-2-butanol."}
{"id": 814, "contents": "Solution - \n2-Methyl-2-butanol\n\nPROBLEM Predict the splitting patterns you would expect for each proton in the following molecules:\n13-9 (a) $\\mathrm{CHBr}_{2} \\mathrm{CH}_{3}$\n(b)\n(f)\n\n$\\mathrm{CH}_{3} \\mathrm{OCH}_{2} \\mathrm{CH}_{2} \\mathrm{Br}$\n(c) $\\mathrm{ClCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{Cl}$\n(d)\n\n(e)\n\n\nPROBLEM Draw structures for compounds that meet the following descriptions:\n13-10\n(a) $\\mathrm{C}_{2} \\mathrm{H}_{6} \\mathrm{O}$; one singlet\n(b) $\\mathrm{C}_{3} \\mathrm{H}_{7} \\mathrm{Cl}$; one doublet and one septet\n(c) $\\mathrm{C}_{4} \\mathrm{H}_{8} \\mathrm{Cl}_{2} \\mathrm{O}$; two triplets\n(d) $\\mathrm{C}_{4} \\mathrm{H}_{8} \\mathrm{O}_{2}$; one singlet, one triplet, and one quartet\n\nPROBLEM The integrated ${ }^{1} \\mathrm{H}$ NMR spectrum of a compound of formula $\\mathrm{C}_{4} \\mathrm{H}_{10} \\mathrm{O}$ is shown in Figure 13.12.\n13-11 Propose a structure.\n\n\nFIGURE 13.12 An integrated ${ }^{1} \\mathrm{H}$ NMR spectrum for Problem 11."}
{"id": 815, "contents": "13.7 ${ }^{1} \\mathrm{H}$ NMR Spectroscopy and Proton Equivalence - \nBecause each electronically distinct hydrogen in a molecule has its own unique absorption, one use of ${ }^{1} \\mathrm{H}$ NMR is to find out how many kinds of electronically nonequivalent hydrogens are present. In the ${ }^{1} \\mathrm{H}$ NMR spectrum of methyl acetate shown previously in FIGURE 13.4a, for instance, there are two signals, corresponding to the two kinds of nonequivalent protons present, $\\mathrm{CH}_{3} \\mathrm{C}=\\mathrm{O}$ protons and $-\\mathrm{OCH}_{3}$ protons.\n\nFor relatively small molecules, a quick look at the structure is often enough to decide how many kinds of protons are present and thus how many NMR absorptions might appear. If in doubt, though, the equivalence or nonequivalence of two protons can be determined by comparing the structures that would be formed if each hydrogen were replaced by an X group. There are four possibilities.\n\n- One possibility is that the protons are chemically unrelated and thus nonequivalent. If so, the products formed on substitution of H by X would be different constitutional isomers. In butane, for instance, the $-\\mathrm{CH}_{3}$ protons are different from the $-\\mathrm{CH}_{2}$ - protons. They therefore give different products on substitution by X than the $-\\mathrm{CH}_{2}$ protons and would likely show different NMR absorptions.\n\n\n\nThe $-\\mathrm{CH}_{2}$ - and $-\\mathrm{CH}_{3}$ hydrogens are unrelated and have different NMR absorptions.\n\n\nThe two substitution products are constitutional isomers.\n\n- A second possibility is that the protons are chemically identical and thus electronically equivalent. If so, the same product would be formed regardless of which H is substituted by X . In butane, for instance, the six $-\\mathrm{CH}_{3}$ hydrogens on C 1 and C 4 are identical, would give the identical structure on substitution by X , and would show an identical NMR absorption. Such protons are said to be homotopic."}
{"id": 816, "contents": "13.7 ${ }^{1} \\mathrm{H}$ NMR Spectroscopy and Proton Equivalence - \n- The third possibility is a bit more subtle. Although they might at first seem homotopic, the two $-\\mathrm{CH}_{2}-$ hydrogens on C 2 in butane (and the two $-\\mathrm{CH}_{2}$ - hydrogens on C 3 ) are in fact not identical. Substitution by X of a hydrogen at C2 (or C3) would form a new chirality center, so different enantiomers (Section 5.1) would result, depending on whether the pro- $R$ or pro- $S$ hydrogen had been substituted (Section 5.11). Such hydrogens, whose substitution by X would lead to different enantiomers, are said to be enantiotopic. Enantiotopic hydrogens, even though not identical, are nevertheless electronically equivalent and thus have the same NMR absorption.\n\n\nor\n\n\nThe two hydrogens on C2 (and\nThe two possible substitution\nproducts are enantiomers.\non C3) are enantiotopic and have the same NMR absorption.\n\n- The fourth possibility arises in chiral molecules, such as ( $R$ )-2-butanol. The two $-\\mathrm{CH}_{2}$ - hydrogens at C 3 are neither homotopic nor enantiotopic. Because substitution of a hydrogen at C3 would form a second chirality center, different diastereomers (Section 5.6) would result, depending on whether the pro- $R$ or pro- $S$ hydrogen had been substituted. Such hydrogens, whose substitution by X leads to different diastereomers, are said to be diastereotopic. Diastereotopic hydrogens are neither chemically nor electronically equivalent. They are completely different and would likely show different NMR absorptions.\n\n\nThe two hydrogens on C3 are diastereotopic and have different NMR absorptions.\n\nThe two possible substitution products are diastereomers.\n\nPROBLEM Identify the indicated sets of protons as unrelated, homotopic, enantiotopic, or diastereotopic:\n13-12 (a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)"}
{"id": 817, "contents": "13.7 ${ }^{1} \\mathrm{H}$ NMR Spectroscopy and Proton Equivalence - \nPROBLEM Identify the indicated sets of protons as unrelated, homotopic, enantiotopic, or diastereotopic:\n13-12 (a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n\nPROBLEM How many kinds of electronically nonequivalent protons are present in each of the following\n13-13 compounds, and thus how many NMR absorptions might you expect in each?\n(a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{Br}$\n(b) $\\mathrm{CH}_{3} \\mathrm{OCH}_{2} \\mathrm{CH}\\left(\\mathrm{CH}_{3}\\right)_{2}$\n(c) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{NO}_{2}$\n(d) Methylbenzene\n(e) 2-Methyl-1-butene (f) cis-3-Hexene\n\nPROBLEM How many absorptions would you expect ( $S$ )-malate, an intermediate in carbohydrate metabolism, 13-14 to have in its ${ }^{1} \\mathrm{H}$ NMR spectrum? Explain."}
{"id": 818, "contents": "(S)-Malate - 13.8 More Complex Spin-Spin Splitting Patterns\nIn the ${ }^{1} \\mathrm{H}$ NMR spectra we've seen so far, the chemical shifts of different protons have been distinct and the spin-spin splitting patterns have been straightforward. It often happens, however, that different kinds of hydrogens in a molecule have accidentally overlapping signals. The spectrum of toluene (methylbenzene) in FIGURE 13.13, for example, shows that the five aromatic ring protons give a complex, overlapping pattern, even though they aren't all equivalent.\n\n\nFIGURE 13.13 The ${ }^{1}$ H NMR spectrum of toluene, showing the accidental overlap of the five nonequivalent aromatic ring protons.\nYet another complication in ${ }^{1} \\mathrm{H}$ NMR spectroscopy arises when a signal is split by two or more nonequivalent kinds of protons, as is the case with trans-cinnamaldehyde, isolated from oil of cinnamon (FIGURE 13.14). Although the $n+1$ rule predicts splitting caused by equivalent protons, splittings caused by nonequivalent protons are more complex.\n\n\nFIGURE 13.14 The ${ }^{\\mathbf{1}} \\mathbf{H}$ NMR spectrum of trans-cinnamaldehyde. The signal of the proton at C 2 (blue) is split into four peaks-a doublet of doublets-by the two nonequivalent neighboring protons.\n\nTo understand the ${ }^{1} \\mathrm{H}$ NMR spectrum of trans-cinnamaldehyde, we have to isolate the different parts and look at the signal of each proton individually."}
{"id": 819, "contents": "(S)-Malate - 13.8 More Complex Spin-Spin Splitting Patterns\nTo understand the ${ }^{1} \\mathrm{H}$ NMR spectrum of trans-cinnamaldehyde, we have to isolate the different parts and look at the signal of each proton individually.\n\n- The five aromatic proton signals (black in FIGURE 13.14) overlap into a complex pattern with a large peak at $7.42 \\delta$ and a broad absorption at $7.57 \\delta$.\n- The aldehyde proton signal at C1 (red) appears in the normal downfield position at $9.69 \\delta$ and is split into a doublet with $J=6 \\mathrm{~Hz}$ by the adjacent proton at C 2 .\n- The vinylic proton at C3 (green) is next to the aromatic ring and is therefore shifted downfield from the normal vinylic region. This C3 proton signal appears as a doublet centered at $7.49 \\delta$. Because it has one neighbor proton at C 2 , its signal is split into a doublet, with $J=12 \\mathrm{~Hz}$.\n- The C2 vinylic proton signal (blue) appears at $6.73 \\delta$ and shows an interesting, four-line absorption pattern. It is coupled to the two nonequivalent protons at C 1 and C 3 with two different coupling constants: $J_{1-2}=6$ Hz and $J_{2-3}=12 \\mathrm{~Hz}$.\n\nA good way to understand the effect of multiple coupling, such as that occurring for the C2 proton of trans-cinnamaldehyde, is to draw a tree diagram, like that in FIGURE 13.15. The diagram shows the individual effect of each coupling constant on the overall pattern. Coupling with the C3 proton splits the signal of the C2 proton in trans-cinnamaldehyde into a doublet with $J=12 \\mathrm{~Hz}$. Further coupling with the aldehyde proton then splits each peak of the doublet into new doublets with $J=6 \\mathrm{~Hz}$, and we therefore observe a four-line spectrum for the C 2 proton.\n\n\n$6.73 \\delta$"}
{"id": 820, "contents": "(S)-Malate - 13.8 More Complex Spin-Spin Splitting Patterns\n$6.73 \\delta$\n\nFIGURE 13.15 A tree diagram for the C2 proton of trans-cinnamaldehyde shows how it is coupled to the C1 and C3 protons with different coupling constants.\n\nOne further trait evident in the cinnamaldehyde spectrum is that the four peaks of the C 2 proton signal are not all the same size. The two left-hand peaks are somewhat larger than the two right-hand peaks. Such a size difference occurs whenever coupled nuclei have similar chemical shifts-in this case, $7.49 \\delta$ for the C3 proton and $6.73 \\delta$ for the C 2 proton. The peaks nearer the signal of the coupled partner are always larger, and the peaks farther from the signal of the coupled partner are always smaller. Thus, the left-hand peaks of the C2 proton multiplet at $6.73 \\delta$ are closer to the C3 proton absorption at $7.49 \\delta$ and are larger than the right-hand peaks. At the same time, the right-hand peak of the C3 proton doublet at $7.49 \\delta$ is larger than the left-hand peak because it is closer to the C 2 proton multiplet at $6.73 \\delta$. This skewing effect on multiplets can often be useful because it tells where to look in the spectrum to find the coupled partner: look in the direction of the larger peaks.\n\nPROBLEM 3-Bromo-1-phenyl-1-propene shows a complex NMR spectrum in which the vinylic proton at C2 is\n13-15 coupled with both the C 1 vinylic proton $(J=16 \\mathrm{~Hz}$ ) and the C 3 methylene protons ( $J=8 \\mathrm{~Hz}$ ). Draw a tree diagram for the C 2 proton signal, and account for the fact that a five-line multiplet is observed."}
{"id": 821, "contents": "3-Bromo-1-phenyl-1-propene - 13.9 Uses of ${ }^{1} \\mathrm{H}$ NMR Spectroscopy\nNMR is used to help identify the product of nearly every reaction run in the laboratory. For example, we said in Section 8.5 that hydroboration-oxidation of alkenes occurs with non-Markovnikov regiochemistry to yield the less highly substituted alcohol. With the help of NMR, we can now prove this statement.\nDoes hydroboration-oxidation of methylenecyclohexane yield cyclohexylmethanol or 1-methylcyclohexanol?\n\n\nMethylenecyclohexane\nCyclohexylmethanol 1-Methylcyclohexanol\nThe ${ }^{1} \\mathrm{H}$ NMR spectrum of the reaction product is shown in FIGURE 13.16a. The spectrum shows a two-proton peak at $3.40 \\delta$, indicating that the product has a $-\\mathrm{CH}_{2}$ - group bonded to an electronegative oxygen atom $\\left(-\\mathrm{CH}_{2} \\mathrm{OH}\\right)$. Furthermore, the spectrum shows no large three-proton singlet absorption near $1 \\delta$, where we would expect the signal of a quaternary $-\\mathrm{CH}_{3}$ group to appear. (FIGURE 13.16b) gives the spectrum of 1-methylcyclohexanol, the alternative product.) Thus, it's clear that cyclohexylmethanol is the reaction product.\n\n\nFIGURE 13.16 (a) The ${ }^{\\mathbf{1}} \\mathrm{H}$ NMR spectrum of cyclohexylmethanol, the product from hydroboration-oxidation of methylenecyclohexane, and (b) the ${ }^{1} H$ NMR spectrum of 1-methylcyclohexanol, the possible alternative reaction product.\n\nPROBLEM How could you use ${ }^{1} \\mathrm{H}$ NMR to determine the regiochemistry of electrophilic addition to alkenes?\n13-16 For example, does addition of HCl to 1-methylcyclohexene yield 1-chloro-1-methylcyclohexane or 1-chloro-2-methylcyclohexane?"}
{"id": 822, "contents": "$13.10{ }^{13} \\mathrm{C}$ NMR Spectroscopy: Signal Averaging and FT-NMR - \nIn some ways, it's surprising that carbon NMR is even possible. After all, ${ }^{12} \\mathrm{C}$, the most abundant carbon isotope, has no nuclear spin and can't be seen by NMR. Carbon-13 is the only naturally occurring carbon isotope with a nuclear spin, but its natural abundance is only $1.1 \\%$. Thus, only about 1 of every 100 carbon atoms in an organic sample can be observed by NMR. The problem of low abundance has been overcome, however, by the use of signal averaging and Fourier-transform NMR (FT-NMR). Signal averaging increases instrument sensitivity, and FT-NMR increases instrument speed.\n\nThe low natural abundance of ${ }^{13} \\mathrm{C}$ means that any individual NMR spectrum is extremely \"noisy.\" That is, the signals are so weak that they are cluttered with random background electronic noise, as shown in FIGURE 13.17a. If, however, hundreds or thousands of individual runs are added together by a computer and then averaged, a greatly improved spectrum results (FIGURE 13.17b). Background noise, because of its random nature, increases very slowly as the runs are added, while the nonzero signals stand out clearly. Unfortunately,\nthe value of signal averaging is limited when using the method of NMR spectrometer operation described in Section 13.2, because it takes about 5 to 10 minutes to obtain a single spectrum. Thus, a faster way to obtain spectra is needed if signal averaging is to be used.\n\n\nFIGURE 13.17 Carbon-13 NMR spectra of 1-pentanol, $\\mathbf{C H}_{\\mathbf{3}} \\mathbf{C H}_{\\mathbf{2}} \\mathbf{C H}_{\\mathbf{2}} \\mathbf{C H}_{\\mathbf{2}} \\mathbf{C H}_{\\mathbf{2}} \\mathbf{O H}$. Spectrum (a) is a single run, showing the large amount of background noise. Spectrum (b) is an average of 200 runs."}
{"id": 823, "contents": "$13.10{ }^{13} \\mathrm{C}$ NMR Spectroscopy: Signal Averaging and FT-NMR - \nIn the method of NMR spectrometer operation described in Section 13.2, the rf frequency is held constant while the strength of the magnetic field is varied so that all signals in the spectrum are recorded sequentially. In the FT-NMR technique used by modern spectrometers, however, all the signals are recorded simultaneously. A sample is placed in a magnetic field of constant strength and is irradiated with a short pulse of rf energy that covers the entire range of useful frequencies. All ${ }^{1} \\mathrm{H}$ or ${ }^{13} \\mathrm{C}$ nuclei in the sample resonate at once, giving a complex, composite signal that is mathematically manipulated using so-called Fourier transforms and then displayed in the usual way. Because all resonance signals are collected at once, it takes only a few seconds rather than a few minutes to record an entire spectrum.\n\nCombining the speed of FT-NMR with the sensitivity enhancement of signal averaging is what gives modern NMR spectrometers their power. Literally thousands of spectra can be taken and averaged in a few hours, resulting in sensitivity so high that a ${ }^{13} \\mathrm{C}$ NMR spectrum can be obtained from less than 0.1 mg of sample and a ${ }^{1} \\mathrm{H}$ spectrum can be recorded from only a few micrograms.\n\nOne further question needs to be answered before moving forward with our discussion of ${ }^{13} \\mathrm{C}$ NMR. Why is spin-spin splitting seen only for ${ }^{1} \\mathrm{H}$ NMR? Why is there no splitting of carbon signals into multiplets in ${ }^{13} \\mathrm{C}$ NMR? After all, you might expect that the spin of a given ${ }^{13} \\mathrm{C}$ nucleus would couple with the spin of an adjacent magnetic nucleus, either ${ }^{13} \\mathrm{C}$ or ${ }^{1} \\mathrm{H}$."}
{"id": 824, "contents": "$13.10{ }^{13} \\mathrm{C}$ NMR Spectroscopy: Signal Averaging and FT-NMR - \nNo coupling of a ${ }^{13} \\mathrm{C}$ nucleus with nearby carbons is seen because their low natural abundance makes it unlikely that two ${ }^{13} \\mathrm{C}$ nuclei will be adjacent. No coupling of a ${ }^{13} \\mathrm{C}$ nucleus with nearby hydrogens is seen because ${ }^{13} \\mathrm{C}$ spectra are normally recorded using broadband decoupling. At the same time that the sample is irradiated with a pulse of rf energy to cover the carbon resonance frequencies, it is also irradiated by a second band of rf energy covering all the hydrogen resonance frequencies. This second irradiation makes the hydrogens spin-flip so rapidly that their local magnetic fields average to zero and no coupling with carbon spins occurs."}
{"id": 825, "contents": "$13.10{ }^{13} \\mathrm{C}$ NMR Spectroscopy: Signal Averaging and FT-NMR - 13.11 Characteristics of ${ }^{13}$ C NMR Spectroscopy\nAt its simplest, ${ }^{13} \\mathrm{C}$ NMR makes it possible to count the number of different carbon atoms in a molecule. Look at the ${ }^{13} \\mathrm{C}$ NMR spectra of methyl acetate and 1-pentanol shown previously in FIGURE 13.4b and FIGURE 13.17b. In each case, a single sharp resonance line is observed for each different carbon atom.\nMost ${ }^{13} \\mathrm{C}$ resonances are between 0 and 220 ppm downfield from the TMS reference line, with the exact chemical shift of each ${ }^{13} \\mathrm{C}$ resonance dependent on that carbon's electronic environment within the molecule. FIGURE 13.18 shows the correlation of chemical shift with environment.\n\n\nThe factors that determine chemical shifts are complex, but it's possible to make some generalizations from the data in FIGURE 13.18. One trend is that a carbon atom's chemical shift is affected by the electronegativity of nearby atoms. Carbons bonded to oxygen, nitrogen, or halogen absorb downfield (to the left) of typical alkane carbons. Because electronegative atoms attract electrons, they pull electrons away from neighboring carbon atoms, causing those carbons to be deshielded and to come into resonance at a lower field.\n\nAnother trend is that $s p^{3}$-hybridized carbons generally absorb from 0 to $90 \\delta$, while $s p^{2}$ carbons absorb from 110 to 220 . Carbonyl carbons ( $\\mathrm{C}=0$ ) are particularly distinct in ${ }^{13} \\mathrm{C}$ NMR and are always found at the lowfield end of the spectrum, from 160 to $220 \\delta$. FIGURE 13.19 shows the ${ }^{13} \\mathrm{C}$ NMR spectra of 2 -butanone and para-bromoacetophenone and indicates the peak assignments. Note that the $\\mathrm{C}=\\mathrm{O}$ carbons are at the left edge of the spectrum in each case."}
{"id": 826, "contents": "$13.10{ }^{13} \\mathrm{C}$ NMR Spectroscopy: Signal Averaging and FT-NMR - 13.11 Characteristics of ${ }^{13}$ C NMR Spectroscopy\nFIGURE 13.19 Carbon-13 NMR spectra of (a) 2-butanone and (b) para-bromoacetophenone.\nThe ${ }^{13} \\mathrm{C}$ NMR spectrum of para-bromoacetophenone is interesting in several ways. Note particularly that only six carbon absorptions are observed, even though the molecule contains eight carbons. para-bromoacetophenone has a symmetry plane that makes ring carbons 4 and $4^{\\prime}$, and ring carbons 5 and $5^{\\prime}$ equivalent. (Remember from Section 2.4 that aromatic rings have two resonance forms.) Thus, the six ring carbons show only four absorptions in the range 128 to $137 \\delta$.\n\n\npara-Bromoacetophenone\nA second interesting point about both spectra in FIGURE 13.19 is that the peaks aren't uniform in size. Some peaks are larger than others even though they are one-carbon resonances (except for the two 2-carbon peaks of para-bromoacetophenone). This difference in peak size is a general feature of broadband-decoupled ${ }^{13} \\mathrm{C}$ NMR spectra, and explains why we can't integrate ${ }^{13} \\mathrm{C}$ NMR spectra in the same way we integrate the resonances in a ${ }^{1} \\mathrm{H}$ NMR spectrum. The local environment of each carbon atom determines not only its chemical shift but also the time it takes for the nuclei to return to their equilibrium state after receiving a pulse of rf radiation and flipping their spins. Quaternary carbons, regardless of their hybridization state or substituents, typically give smaller resonances than primary, secondary, or tertiary carbons."}
{"id": 827, "contents": "Predicting Chemical Shifts in ${ }^{13} \\mathrm{C}$ NMR Spectra - \nAt what approximate positions would you expect ethyl acrylate, $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCO}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$, to show ${ }^{13} \\mathrm{C}$ NMR absorptions?"}
{"id": 828, "contents": "Strategy - \nIdentify the distinct carbons in the molecule, and note whether each is alkyl, vinylic, aromatic, or in a carbonyl group. Then predict where each absorbs, using FIGURE 13.18 as necessary."}
{"id": 829, "contents": "Solution - \nEthyl acrylate has five chemically distinct carbons: two different $\\mathrm{C}=\\mathrm{C}$, one $\\mathrm{C}=\\mathrm{O}$, one $\\mathrm{O}-\\mathrm{C}$, and one alkyl C . From FIGURE 13.18, the likely absorptions are\n\n\nThe actual absorptions are at $14.1,60.5,128.5,130.3$, and $166.0 \\delta$.\n\nPROBLEM Predict the number of carbon resonance lines you would expect in the ${ }^{13}$ C NMR spectra of the 13-17 following compounds:\n(a) Methylcyclopentane (b) 1-Methylcyclohexene (c) 1,2-Dimethylbenzene\n(d) 2-Methyl-2-butene\n(e)\n\n\nPROBLEM Propose structures for compounds that fit the following descriptions:\n13-18 (a) A hydrocarbon with seven lines in its ${ }^{13} \\mathrm{C}$ NMR spectrum\n(b) A six-carbon compound with only five lines in its ${ }^{13} \\mathrm{C}$ NMR spectrum\n(c) A four-carbon compound with three lines in its ${ }^{13} \\mathrm{C}$ NMR spectrum\n\nPROBLEM Classify the resonances in the ${ }^{13} \\mathrm{C}$ NMR spectrum of methyl propanoate, $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{CH}_{3}$ (Figure 13-19 13.20).\n\n\nFIGURE $13.20{ }^{13} \\mathrm{C}$ NMR spectrum of methyl propanoate, Problem 19."}
{"id": 830, "contents": "Solution - 13.12 DEPT ${ }^{13} \\mathrm{C}$ NMR Spectroscopy\nNumerous techniques developed in recent years have made it possible to obtain enormous amounts of information from ${ }^{13} \\mathrm{C}$ NMR spectra. Among these techniques is one called DEPT-NMR, for distortionless enhancement by polarization transfer, which makes it possible to distinguish between signals due to $\\mathrm{CH}_{3}, \\mathrm{CH}_{2}$, CH , and quaternary carbons. That is, the number of hydrogens attached to each carbon in a molecule can be determined.\n\nA DEPT experiment is usually done in three stages, as shown in FIGURE 13.21 for 6-methyl-5-hepten-2-ol. The first stage is to run an ordinary spectrum (called a broadband-decoupled spectrum) to locate the chemical shifts\nof all carbons. Next, a second spectrum called a DEPT-90 is run, using special conditions under which only signals due to CH carbons appear. Signals due to $\\mathrm{CH}_{3}, \\mathrm{CH}_{2}$, and quaternary carbons are absent. Finally, a third spectrum called a DEPT-135 is run, using conditions under which $\\mathrm{CH}_{3}$ and CH resonances appear as positive signals, $\\mathrm{CH}_{2}$ resonances appear as negative signals-that is, as peaks below the baseline-and quaternary carbons are again absent.\n\n\nFIGURE 13.21 DEPT-NMR spectra for 6-methyl-5-hepten-2-ol. Part (a) is an ordinary broadband-decoupled spectrum, which shows signals for all eight carbons. Part (b) is a DEPT-90 spectrum, which shows only signals for the two CH carbons. Part (c) is a DEPT-135 spectrum, which shows positive signals for the two CH and three $\\mathrm{CH}_{3}$ carbons and negative signals for the two $\\mathrm{CH}_{2}$ carbons."}
{"id": 831, "contents": "Solution - 13.12 DEPT ${ }^{13} \\mathrm{C}$ NMR Spectroscopy\nPutting together the information from all three spectra makes it possible to tell the number of hydrogens attached to each carbon. The CH carbons are identified in the DEPT- 90 spectrum, the $\\mathrm{CH}_{2}$ carbons are identified as negative peaks in the DEPT- 135 spectrum, the $\\mathrm{CH}_{3}$ carbons are identified by subtracting the CH peaks from the positive peaks in the DEPT-135 spectrum, and quaternary carbons are identified by subtracting all peaks in the DEPT-135 spectrum from the peaks in the broadband-decoupled spectrum.\n\nc Subtract DEPT-135 from broadband-decoupled spectrum\nCH DEPT-90\n$\\mathrm{CH}_{2}$ Negative DEPT-135\n$\\mathrm{CH}_{3}$ Subtract DEPT-90 from positive DEPT-135"}
{"id": 832, "contents": "Assigning a Chemical Structure from a ${ }^{13}$ C NMR Spectrum - \nPropose a structure for an alcohol, $\\mathrm{C}_{4} \\mathrm{H}_{10} \\mathrm{O}$, that has the following ${ }^{13} \\mathrm{C}$ NMR spectral data:\nBroadband decoupled ${ }^{13}$ C NMR: 19.0, 31.7, 69.5 \u05e1;\nDEPT-90: 31.7 \u05e1;\nDEPT-135: positive peak at $19.0 \\delta$, negative peak at $69.5 \\delta$."}
{"id": 833, "contents": "Strategy - \nAs noted in Section 7.2, it usually helps with compounds of known formula but unknown structure to calculate the compound's degree of unsaturation. In the present instance, a formula of $\\mathrm{C}_{4} \\mathrm{H}_{10} \\mathrm{O}$ corresponds to a saturated, open-chain molecule.\n\nTo gain information from the ${ }^{13} \\mathrm{C}$ data, let's begin by noting that the unknown alcohol has four carbon atoms, yet has only three NMR absorptions, which implies that two of the carbons must be equivalent. Looking at chemical shifts, two of the absorptions are in the typical alkane region (19.0 and $31.7 \\delta$ ), while one is in the region of a carbon bonded to an electronegative atom ( $69.5 \\delta$ )-oxygen in this instance. The DEPT-90 spectrum tells us that the alkyl carbon at $31.7 \\delta$ is tertiary (CH); the DEPT- 135 spectrum tells us that the alkyl carbon at $19.0 \\delta$ is a methyl $\\left(\\mathrm{CH}_{3}\\right)$ and that the carbon bonded to oxygen ( $69.5 \\delta$ ) is secondary $\\left(\\mathrm{CH}_{2}\\right)$. The two equivalent carbons are probably both methyls bonded to the same tertiary carbon, $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CH}-$. We can now put the pieces together to propose a structure: 2-methyl-1-propanol."}
{"id": 834, "contents": "Solution - \nPROBLEM Assign a chemical shift to each carbon in 6-methyl-5-hepten-2-ol (Figure 13.21)."}
{"id": 835, "contents": "13-20 - \nPROBLEM Estimate the chemical shift of each carbon in the following molecule. Predict which carbons will 13-21 appear in the DEPT-90 spectrum, which will give positive peaks in the DEPT-135 spectrum, and which will give negative peaks in the DEPT-135 spectrum.\n\n\nPROBLEM Propose a structure for an aromatic hydrocarbon, $\\mathrm{C}_{11} \\mathrm{H}_{16}$, that has the following ${ }^{13} \\mathrm{C}$ NMR spectral 13-22 data:\n\nBroadband decoupled: 29.5, 31.8, 50.2, 125.5, 127.5, 130.3, $139.8 \\delta$\nDEPT-90: 125.5, 127.5, $130.3 \\delta$\nDEPT-135: positive peaks at $29.5,125.5,127.5,130.3 \\delta$; negative peak at $50.2 \\delta$"}
{"id": 836, "contents": "13-20 - 13.13 Uses of ${ }^{13} \\mathrm{C}$ NMR Spectroscopy\nThe information derived from ${ }^{13} \\mathrm{C}$ NMR spectroscopy is extraordinarily useful for structure determination. Not only can we count the number of nonequivalent carbon atoms in a molecule, we can also get information about the electronic environment of each carbon and find how many protons are attached to each. As a result, we can address many structural questions that go unanswered by IR spectroscopy or mass spectrometry.\n\nHere's an example: how do we know that the E2 reaction of an alkyl halide follows Zaitsev's rule (Section 11.7)? Does treatment of 1-chloro-1-methylcyclohexane with a strong base give predominantly the trisubstituted alkene 1-methylcyclohexene or the disubstituted alkene methylenecyclohexane?\n\n\n1-Methylcyclohexene will have five $s p^{3}$-carbon resonances in the 20 to $50 \\delta$ range and two $s p^{2}$-carbon resonances in the 100 to $150 \\delta$ range. Methylenecyclohexane, however, because of its symmetry, will have only three $s p^{3}$-carbon resonance peaks and two $s p^{2}$-carbon peaks. The spectrum of the actual reaction product, shown in FIGURE 13.22, clearly identifies 1-methylcyclohexene as the product of this E2 reaction.\n1-chloro-1-methylcyclohexane with base.\n\nPROBLEM We saw in Section 9.3 that addition of HBr to a terminal alkyne leads to the Markovnikov addition\n13-23 product, with the Br bonding to the more highly substituted carbon. How could you use ${ }^{13} \\mathrm{C}$ NMR to identify the product of the addition of 1 equivalent of HBr to 1-hexyne?"}
{"id": 837, "contents": "Magnetic Resonance Imaging (MRI) - \nAs practiced by organic chemists, NMR spectroscopy is a powerful method of structure determination. A small amount of sample, typically a few milligrams or less, is dissolved in a small amount of solvent, the solution is placed in a thin glass tube, and the tube is placed into the narrow ( $1-2 \\mathrm{~cm}$ ) gap between the poles of a strong magnet. Imagine, though, that a much larger NMR instrument were available. Instead of a few milligrams, the sample size could be tens of kilograms; instead of a narrow gap between magnet poles, the gap could be large enough for a whole person to climb into so that an NMR spectrum of body parts could be obtained. That large instrument is exactly what's used for magnetic resonance imaging (MRI), a diagnostic technique of enormous value to the medical community.\n\nLike NMR spectroscopy, MRI takes advantage of the magnetic properties of certain nuclei, typically hydrogen, and of the signals emitted when those nuclei are stimulated by radiofrequency energy. Unlike what happens in NMR spectroscopy, though, MRI instruments use data manipulation techniques to look at the threedimensional location of magnetic nuclei in the body rather than at the chemical nature of the nuclei. As noted, most MRI instruments currently look at hydrogen, present in abundance wherever there is water or fat in the body.\n\nThe signals detected by MRI vary with the density of hydrogen atoms and with the nature of their surroundings, allowing identification of different types of tissue and even allowing the visualization of motion. For example, the volume of blood leaving the heart in a single stroke can be measured, and heart motion can be observed. Soft tissues that don't show up well on X-ray images can be seen clearly, allowing diagnosis of brain tumors, strokes, and other conditions. This technique is also valuable in diagnosing damage to knees or other joints and is a noninvasive alternative to surgical explorations.\n\nSeveral types of atoms in addition to hydrogen can be detected by MRI, and the applications of images based on\n${ }^{31} \\mathrm{P}$ atoms are being explored. This approach holds great promise for studies of metabolism."}
{"id": 838, "contents": "Magnetic Resonance Imaging (MRI) - \nSeveral types of atoms in addition to hydrogen can be detected by MRI, and the applications of images based on\n${ }^{31} \\mathrm{P}$ atoms are being explored. This approach holds great promise for studies of metabolism.\n\n\nFIGURE 13.23 If you're a runner, you really don't want this to happen to you. The MRI of this left knee shows bone formation in the tissue in a patient with osteochondroma. (credit: \"Osteochondroma MRI\" by Michael R Carmont, Sian Davies, Daniel Gey van Pittius and Robin Rees/ Wikimedia Commons, CC BY 2.0)"}
{"id": 839, "contents": "Key Terms - \n```\n\u2022 chemical shift \u2022 homotopic\n- coupling\n- integration\n- coupling constant (J)\n- multiplet\n- delta (\\delta) scale\n- n+1 rule\n- DEPT-NMR\n- nuclear magnetic resonance (NMR) spectroscopy\n- diastereotopic\n- shielding\n- downfield\n- spin-spin splitting\n- enantiotopic\n- upfield\n- FT-NMR\n```"}
{"id": 840, "contents": "Summary - \nNuclear magnetic resonance spectroscopy, or NMR, is the most valuable of the numerous spectroscopic techniques used for structure determination. Although we focused in this chapter on NMR applications with small molecules, more advanced NMR techniques are also used in biological chemistry to study protein structure and folding.\n\nWhen magnetic nuclei, such as ${ }^{1} \\mathrm{H}$ and ${ }^{13} \\mathrm{C}$, are placed in a strong magnetic field, their spins orient either with or against the field. On irradiation with radiofrequency (rf) waves, energy is absorbed and the nuclei \"spin-flip\" from the lower energy state to the higher energy state. This absorption of rf energy is detected, amplified, and displayed as an NMR spectrum.\n\nEach electronically distinct ${ }^{1} \\mathrm{H}$ or ${ }^{13} \\mathrm{C}$ nucleus in a molecule comes into resonance at a slightly different value of the applied field, thereby producing a unique absorption signal. The exact position of each peak is called the chemical shift. Chemical shifts are caused by electrons setting up tiny local magnetic fields that shield a nearby nucleus from the applied field.\n\nThe NMR chart is calibrated in delta units ( $\\boldsymbol{\\delta}$ ), where $1 \\delta=1 \\mathrm{ppm}$ of spectrometer frequency. Tetramethylsilane (TMS) is used as a reference point because it shows both ${ }^{1} \\mathrm{H}$ and ${ }^{13} \\mathrm{C}$ absorptions at unusually high values of applied magnetic field. The TMS absorption occurs on the right-hand (upfield) side of the chart and is arbitrarily assigned a value of $0 \\delta$.\n${ }^{13} \\mathrm{C}$ spectra are run on Fourier-transform NMR (FT-NMR) spectrometers using broadband decoupling of proton spins so that each chemically distinct carbon shows a single unsplit resonance line. As with ${ }^{1} \\mathrm{H}$ NMR, the chemical shift of each ${ }^{13} \\mathrm{C}$ signal provides information about a carbon's chemical environment in the sample. In addition, the number of protons attached to each carbon can be determined using the DEPT-NMR technique."}
{"id": 841, "contents": "Summary - \nIn ${ }^{1} H$ NMR spectra, the area under each absorption peak can be electronically integrated to determine the relative number of hydrogens responsible for each peak. In addition, neighboring nuclear spins can couple, causing the spin-spin splitting of NMR peaks into multiplets. The NMR signal of a hydrogen neighbored by $n$ equivalent adjacent hydrogens splits into $n+1$ peaks (the $\\boldsymbol{n}+\\mathbf{1}$ rule) with coupling constant $\\boldsymbol{J}$."}
{"id": 842, "contents": "Visualizing Chemistry - \nPROBLEM Into how many peaks would you expect the ${ }^{1} \\mathrm{H}$ NMR signals of the indicated protons to be split? 13-24 $($ Green $=\\mathrm{Cl}$.)\n(a)\n\n(b)\n\n\nPROBLEM How many absorptions would you expect the following compound to have in its ${ }^{1} \\mathrm{H}$ and ${ }^{13} \\mathrm{C}$ NMR 13-25 spectra?\n\n\nPROBLEM Sketch what you might expect the ${ }^{1} \\mathrm{H}$ and ${ }^{13} \\mathrm{C}$ NMR spectra of the following compound to look like 13-26 (green $=\\mathrm{Cl}$ ):\n\n\nPROBLEM How many electronically nonequivalent kinds of protons and how many kinds of carbons are 13-27 present in the following compound? Don't forget that cyclohexane rings can ring-flip.\n\n\nPROBLEM Identify the indicated protons in the following molecules as unrelated, homotopic, enantiotopic, or 13-28 diastereotopic.\n(a)\n\n(b)\n\nCysteine"}
{"id": 843, "contents": "Chemical Shifts and NMR Spectroscopy - \nPROBLEM The following ${ }^{1} \\mathrm{H}$ NMR absorptions were obtained on a spectrometer operating at 200 MHz and are 13-29 given in hertz downfield from the TMS standard. Convert the absorptions to $\\delta$ units.\n(a) 436 Hz\n(b) 956 Hz\n(c) 1504 Hz\n\nPROBLEM The following ${ }^{1} \\mathrm{H}$ NMR absorptions were obtained on a spectrometer operating at 300 MHz . Convert 13-30 the chemical shifts from $\\delta$ units to hertz downfield from TMS.\n(a) $2.1 \\mathrm{\\delta}$\n(b) $3.45 \\delta$\n(c) $6.30 \\delta$\n(d) $7.70 \\delta$\n\nPROBLEM When measured on a spectrometer operating at 200 MHz , chloroform $\\left(\\mathrm{CHCl}_{3}\\right)$ shows a single sharp 13-31 absorption at $7.3 \\delta$.\n(a) How many parts per million downfield from TMS does chloroform absorb?\n(b) How many hertz downfield from TMS would chloroform absorb if the measurement were carried out on a spectrometer operating at 360 MHz ?\n(c) What would be the position of the chloroform absorption in $\\delta$ units when measured on a 360 MHz spectrometer?\nPROBLEM Why do you suppose accidental overlap of signals is much more common in ${ }^{1} \\mathrm{H}$ NMR than in ${ }^{13} \\mathrm{C}$ 13-32 NMR?\n\nPROBLEM Is a nucleus that absorbs at $6.50 \\delta$ more shielded or less shielded than a nucleus that absorbs at 13-33 3.20 $\\delta$ ? Does the nucleus that absorbs at $6.50 \\delta$ require a stronger applied field or a weaker applied field to come into resonance than the one that absorbs at $3.20 \\delta$ ?"}
{"id": 844, "contents": "${ }^{\\mathbf{1}} \\mathrm{H}$ NMR Spectroscopy - \nPROBLEM How many types of nonequivalent protons are present in each of the following molecules?\n13-34 (a)\n\n(b) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{OCH}_{3}$\n(c)\n\nNaphthalene\n(d)\n\nStyrene\n(e)"}
{"id": 845, "contents": "Ethyl acrylate - \nPROBLEM The following compounds all show a single line in their ${ }^{1} \\mathrm{H}$ NMR spectra. List them in order of 13-35 expected increasing chemical shift:\n$\\mathrm{CH}_{4}, \\mathrm{CH}_{2} \\mathrm{Cl}_{2}$, cyclohexane, $\\mathrm{CH}_{3} \\mathrm{COCH}_{3}, \\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}_{2}$, benzene\nPROBLEM How many signals would you expect each of the following molecules to have in its ${ }^{1} \\mathrm{H}$ and ${ }^{13} \\mathrm{C}$ 13-36 spectra?\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n\nPROBLEM Propose structures for compounds with the following formulas that show only one peak in their ${ }^{1} \\mathrm{H}$\n13-37 NMR spectra:\n(a) $\\mathrm{C}_{5} \\mathrm{H}_{12}$\n(b) $\\mathrm{C}_{5} \\mathrm{H}_{10}$\n(c) $\\mathrm{C}_{4} \\mathrm{H}_{8} \\mathrm{O}_{2}$\n\nPROBLEM Predict the splitting pattern for each kind of hydrogen in the following molecules:\n13-38\n(a) $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CH}$ (b) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{CH}_{3}$\n(c) trans-2-Butene\n\nPROBLEM Predict the splitting pattern for each kind of hydrogen in isopropyl propanoate,\n13-39 $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{CH}\\left(\\mathrm{CH}_{3}\\right)_{2}$.\nPROBLEM Identify the indicated sets of protons as unrelated, homotopic, enantiotopic, or diastereotopic:\n\n13-40 (a)\n\n(b)\n\n(c)\n\n\nPROBLEM Identify the indicated sets of protons as unrelated, homotopic, enantiotopic, or diastereotopic:"}
{"id": 846, "contents": "13-41 (a) - \n(b)\n\n(c)\n\n\nPROBLEM The acid-catalyzed dehydration of 1-methylcyclohexanol yields a mixture of two alkenes. How\n13-42 could you use ${ }^{1}$ H NMR to help you decide which was which?\n\n\nPROBLEM How could you use ${ }^{1}$ H NMR to distinguish between the following pairs of isomers?\n13-43 (a) $\\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{CHCH}_{2} \\mathrm{CH}_{3}$ and\n\n(b) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OCH}_{2} \\mathrm{CH}_{3}$\nand $\\mathrm{CH}_{3} \\mathrm{OCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$\n(c)\nand\n\n(d)\nand\n\n\nPROBLEM Propose structures for compounds that fit the following ${ }^{1} \\mathrm{H}$ NMR data:\n13-44\n(a) $\\mathrm{C}_{5} \\mathrm{H}_{10} \\mathrm{O}$\n(b) $\\mathrm{C}_{3} \\mathrm{H}_{5} \\mathrm{Br}$\n$0.95 \\delta(6 \\mathrm{H}$, doublet, $J=7 \\mathrm{~Hz})$\n2.32 \u0920 (3 H, singlet)\n$2.10 \\delta(3 \\mathrm{H}$, singlet)\n2.32 ( 3 H , singlet)\n$2.43 \\delta(1 \\mathrm{H}$, multiplet)\n2.32 (3 H, singlet)\n\nPROBLEM Propose structures for the two compounds whose ${ }^{1} \\mathrm{H}$ NMR spectra are shown.\n13-45\n(a) $\\mathrm{C}_{4} \\mathrm{H}_{9} \\mathrm{Br}$\n\n(b) $\\mathrm{C}_{4} \\mathrm{H}_{8} \\mathrm{Cl}_{2}$"}
{"id": 847, "contents": "${ }^{13}$ C NMR Spectroscopy - \nPROBLEM How many ${ }^{13} \\mathrm{C}$ NMR absorptions would you expect for cis-1,3-dimethylcyclohexane? For 13-46 trans-1,3-dimethylcyclohexane? Explain.\n\nPROBLEM How many absorptions would you expect to observe in the ${ }^{13} \\mathrm{C}$ NMR spectra of the following 13-47 compounds?\n(a) 1,1-Dimethylcyclohexane\n(b) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OCH}_{3}$\n(c) tert-Butylcyclohexane\n(d) 3-Methyl-1-pentyne\n(e) cis-1,2-Dimethylcyclohexane\n(f) Cyclohexanone\n\nPROBLEM Suppose you ran a DEPT-135 spectrum for each substance in Problem 13.47. Which carbon atoms 13-48 in each molecule would show positive peaks, and which would show negative peaks?\n\nPROBLEM How could you use ${ }^{1} \\mathrm{H}$ and ${ }^{13} \\mathrm{C}$ NMR to help distinguish the following isomeric compounds of the 13-49 formula $\\mathrm{C}_{4} \\mathrm{H}_{8}$ ?\n\n\nPROBLEM How could you use ${ }^{1} \\mathrm{H}$ NMR, ${ }^{13} \\mathrm{C}$ NMR, and IR spectroscopy to help you distinguish between the 13-50 following structures?\n\n\n3-Methyl-2-cyclohexenone"}
{"id": 848, "contents": "3-Cyclopentenyl methyl ketone - \nPROBLEM Assign as many resonances as you can to specific carbon atoms in the ${ }^{13}$ C NMR spectrum of ethyl 13-51\nbenzoate."}
{"id": 849, "contents": "General Problems - \nPROBLEM Assume that you have a compound with the formula $\\mathrm{C}_{3} \\mathrm{H}_{6} \\mathrm{O}$.\n13-52 (a) How many double bonds and/or rings does your compound contain?\n(b) Propose as many structures as you can that fit the molecular formula.\n(c) If your compound shows an infrared absorption peak at $1715 \\mathrm{~cm}^{-1}$, what functional group does it have?\n(d) If your compound shows a single ${ }^{1} \\mathrm{H}$ NMR absorption peak at $2.1 \\delta$, what is its structure?\n\nPROBLEM The compound whose ${ }^{1} \\mathrm{H}$ NMR spectrum is shown has the molecular formula $\\mathrm{C}_{3} \\mathrm{H}_{6} \\mathrm{Br}_{2}$. Propose a\n13-53 structure.\n\n\nPROBLEM The compound whose ${ }^{1} \\mathrm{H}$ NMR spectrum is shown has the molecular formula $\\mathrm{C}_{4} \\mathrm{H}_{7} \\mathrm{O}_{2} \\mathrm{Cl}$ and has an\n13-54 infrared absorption peak at $1740 \\mathrm{~cm}^{-1}$. Propose a structure."}
{"id": 850, "contents": "General Problems - \nPROBLEM Propose structures for compounds that fit the following ${ }^{1} \\mathrm{H}$ NMR data:\n13-55\n(a) $\\mathrm{C}_{4} \\mathrm{H}_{6} \\mathrm{Cl}_{2}$\n$2.18 \\delta(3 \\mathrm{H}$, singlet)\n$4.16 \\delta(2 \\mathrm{H}$, doublet, $J=7 \\mathrm{~Hz}$ )\n$5.71 \\delta(1 \\mathrm{H}$, triplet, $J=7 \\mathrm{~Hz}$ )\n(d) $\\mathrm{C}_{9} \\mathrm{H}_{11} \\mathrm{Br}$\n$2.15 \\delta(2 \\mathrm{H}$, quintet, $J=7 \\mathrm{~Hz})$\n$2.75 \\delta(2 \\mathrm{H}$, triplet, $J=7 \\mathrm{~Hz})$\n$3.38 \\delta(2 \\mathrm{H}$, triplet, $J=7 \\mathrm{~Hz})$\n7.22 ( 5 H , singlet)\n(b) $\\mathrm{C}_{10} \\mathrm{H}_{14}$\n$1.30 \\delta(9 \\mathrm{H}$, singlet)\n$7.30 \\delta(5 \\mathrm{H}$, singlet)\n(c) $\\mathrm{C}_{4} \\mathrm{H}_{7} \\mathrm{BrO}$\n$2.11 \\delta(3 \\mathrm{H}$, singlet)\n$3.52 \\delta(2 \\mathrm{H}$, triplet, $J=6 \\mathrm{~Hz}$ )\n$4.40 \\delta(2 \\mathrm{H}$, triplet, $J=6 \\mathrm{~Hz})$\n\nPROBLEM Long-range coupling between protons more than two carbon atoms apart is sometimes observed 13-56 when $\\pi$ bonds intervene. An example is found in 1-methoxy-1-buten-3-yne. Not only does the acetylenic proton, $\\mathrm{H}_{a}$, couple with the vinylic proton $\\mathrm{H}_{b}$, it also couples with the vinylic proton $\\mathrm{H}_{\\mathcal{C}}$, four carbon atoms away. The data are:"}
{"id": 851, "contents": "General Problems - \n$$\n\\begin{array}{lll}\n\\mathrm{H}_{a}(3.08 \\delta) & \\mathrm{H}_{b}(4.52 \\delta) & \\mathrm{H}_{c}(6.35 \\delta) \\\\\nJ_{a-b}=3 \\mathrm{~Hz} & J_{a-c}=1 \\mathrm{~Hz} & J_{b-c}=7 \\mathrm{~Hz}\n\\end{array}\n$$"}
{"id": 852, "contents": "1-Methoxy-1-buten-3-yne - \nConstruct tree diagrams that account for the observed splitting patterns of $H_{a}, H_{b}$, and $H_{c}$.\nPROBLEM The ${ }^{1} \\mathrm{H}$ and ${ }^{13} \\mathrm{C}$ NMR spectra of compound $\\mathbf{A}, \\mathrm{C}_{8} \\mathrm{H}_{9} \\mathrm{Br}$, are shown. Propose a structure for $\\mathbf{A}$, and 13-57 assign peaks in the spectra to your structure.\n\n\nPROBLEM Propose structures for the three compounds whose ${ }^{1} \\mathrm{H}$ NMR spectra are shown.\n13-58\n(a) $\\mathrm{C}_{5} \\mathrm{H}_{10} \\mathrm{O}$\n\n(b) $\\mathrm{C}_{7} \\mathrm{H}_{7} \\mathrm{Br}$\n\n(c) $\\mathrm{C}_{8} \\mathrm{H}_{9} \\mathrm{Br}$\n\n\nPROBLEM The mass spectrum and ${ }^{13} \\mathrm{C}$ NMR spectrum of a hydrocarbon are shown. Propose a structure for 13-59 this hydrocarbon, and explain the spectral data.\n\n\nPROBLEM Compound A, a hydrocarbon with $\\mathrm{M}^{+}=96$ in its mass spectrum, has the following ${ }^{13} \\mathrm{C}$ spectral\n13-60 data. On reaction with $\\mathrm{BH}_{3}$, followed by treatment with basic $\\mathrm{H}_{2} \\mathrm{O}_{2}$, $\\mathbf{A}$ is converted into $\\mathbf{B}$, whose ${ }^{13} \\mathbf{C}$ spectral data are also given. Propose structures for A and B."}
{"id": 853, "contents": "Compound A - \nBroadband-decoupled ${ }^{13}$ C NMR: 26.8, 28.7, 35.7, 106.9, $149.7 \\delta$\nDEPT-90: no peaks\nDEPT-135: no positive peaks; negative peaks at 26.8, $28.7,35.7,106.9 \\delta$"}
{"id": 854, "contents": "Compound B - \nBroadband-decoupled ${ }^{13}$ C NMR: 26.1, 26.9, 29.9, 40.5, $68.2 \\delta$\nDEPT-90: 40.5 \u05e1\nDEPT-135: positive peak at $40.5 \\delta$; negative peaks at $26.1,26.9,29.9,68.2 \\delta$\n\nPROBLEM Propose a structure for compound $\\mathbf{C}$, which has $\\mathrm{M}^{+}=86$ in its mass spectrum, an IR absorption at\n13-61 $3400 \\mathrm{~cm}^{-1}$, and the following ${ }^{13} \\mathrm{C}$ NMR spectral data:"}
{"id": 855, "contents": "Compound C - \nBroadband-decoupled ${ }^{13}$ C NMR: 30.2, 31.9, 61.8, 114.7, $138.4 \\delta$\nDEPT-90: $138.4 \\delta$\nDEPT-135: positive peak at $138.4 \\delta$; negative peaks at $30.2,31.9,61.8,114.7 \\delta$\nPROBLEM Compound $\\mathbf{D}$ is isomeric with compound $\\mathbf{C}$ (Problem 13.61) and has the following ${ }^{13}$ C NMR spectral 13-62 data. Propose a structure."}
{"id": 856, "contents": "Compound D - \nBroadband-decoupled ${ }^{13}$ C NMR: 9.7, 29.9, 74.4, 114.4, $141.4 \\delta$\nDEPT-90: 74.4, $141.4 \\delta$\nDEPT-135: positive peaks at $9.7,74.4,141.4 \\delta$; negative peaks at $29.9,114.4 \\delta$\n\nPROBLEM Propose a structure for compound $\\mathbf{E}, \\mathrm{C}_{7} \\mathrm{H}_{12} \\mathrm{O}_{2}$, which has the following ${ }^{13} \\mathrm{C}$ NMR spectral data:\n13-63"}
{"id": 857, "contents": "Compound E - \nBroadband-decoupled ${ }^{13}$ C NMR: 19.1, 28.0, 70.5, 129.0, 129.8, $165.8 \\delta$\nDEPT-90: 28.0, $129.8 \\delta$\nDEPT-135: positive peaks at 19.1, 28.0, $129.8 \\delta$; negative peaks at $70.5,129.0 \\delta$\n\nPROBLEM Compound $\\mathbf{F}$, a hydrocarbon with $\\mathrm{M}^{+}=96$ in its mass spectrum, undergoes reaction with HBr to 13-64 yield compound $\\mathbf{G}$. Propose structures for $\\mathbf{F}$ and $\\mathbf{G}$, whose ${ }^{13} \\mathrm{C}$ NMR spectral data are given below."}
{"id": 858, "contents": "Compound $\\mathbf{F}$ - \nBroadband-decoupled ${ }^{13}$ C NMR: 27.6, 29.3, 32.2, $132.4 \\delta$\nDEPT-90: 132.4 \u05e1\nDEPT-135: positive peak at $132.4 \\delta$; negative peaks at $27.6,29.3,32.2 \\delta$"}
{"id": 859, "contents": "Compound G - \nBroadband-decoupled ${ }^{13}$ C NMR: 25.1, 27.7, 39.9, $56.0 \\delta$\nDEPT-90: 56.0 \u05e1\nDEPT-135: positive peak at $56.0 \\delta$; negative peaks at $25.1,27.7,39.9 \\delta$\nPROBLEM 3-Methyl-2-butanol has five signals in its ${ }^{13}$ C NMR spectrum at 17.90, 18.15, 20.00, 35.05, and\n13-65 72.75 $\\delta$. Why are the two methyl groups attached to C3 nonequivalent? Making a molecular model should be helpful."}
{"id": 860, "contents": "3-Methyl-2-butanol - \nPROBLEM A ${ }^{13} \\mathrm{C}$ NMR spectrum of commercially available 2,4-pentanediol, shows five peaks at 23.3, 23.9,\n13-66 46.5, 64.8, and 68.1 $\\delta$. Explain."}
{"id": 861, "contents": "2,4-Pentanediol - \nPROBLEM Carboxylic acids $\\left(\\mathrm{RCO}_{2} \\mathrm{H}\\right)$ react with alcohols $\\left(\\mathrm{R}^{\\prime} \\mathrm{OH}\\right)$ in the presence of an acid catalyst. The reaction\n13-67 product of propanoic acid with methanol has the following spectroscopic properties. Propose a structure.\n\n\nPropanoic acid\nMS: $\\mathrm{M}^{+}=88$\nIR: $1735 \\mathrm{~cm}^{-1}$\n${ }^{1} \\mathrm{H}$ NMR: $1.11 \\delta(3 \\mathrm{H}$, triplet, $J=7 \\mathrm{~Hz}) ; 2.32 \\delta(2 \\mathrm{H}$, quartet, $J=7 \\mathrm{~Hz})$;\n$3.65 \\delta$ (3 H, singlet)\n${ }^{13}$ C NMR: 9.3, 27.6, 51.4, $174.6 \\delta$\nPROBLEM Nitriles $(R C \\equiv N)$ react with Grignard reagents ( $\\mathrm{R}^{\\prime} \\mathrm{MgBr}$ ). The reaction product from\n13-68 2-methylpropanenitrile with methylmagnesium bromide has the following spectroscopic properties. Propose a structure. MS: M ${ }^{+}=86$ IR: $1715 \\mathrm{~cm}^{-1}{ }^{1} \\mathrm{H}$ NMR: $1.05 \\delta(6 \\mathrm{H}$, doublet, J = 7 Hz ); 2.12 ( 3 H , singlet); $2.67 \\delta\\left(1 \\mathrm{H}\\right.$, septet, $\\mathrm{J}=7 \\mathrm{~Hz}$ ) ${ }^{13} \\mathrm{C}$ NMR: 18.2, 27.2, 41.6, $211.2 \\delta$"}
{"id": 862, "contents": "2,4-Pentanediol - \nPROBLEM The proton NMR spectrum is shown for a compound with the formula $\\mathrm{C}_{5} \\mathrm{H}_{9} \\mathrm{NO}_{4}$. The infrared 13-69 spectrum displays strong bands at 1750 and $1562 \\mathrm{~cm}^{-1}$ and a medium-intensity band at $1320 \\mathrm{~cm}^{-1}$. The normal carbon-13 and the DEPT experimental results are tabulated. Draw the structure of this compound.\n\n| Normal Carbon | DEPT-135 | DEPT-90 |\n| :---: | :---: | :---: |\n| 14 ppm | Positive | No peak |\n| 16 | Positive | No peak |\n| 63 | Negative | No peak |\n| 83 | Positive | Positive |\n| 165 | No peak | No peak |\n\n\n\nPROBLEM The proton NMR spectrum of a compound with the formula $\\mathrm{C}_{5} \\mathrm{H}_{10} \\mathrm{O}$ is shown. The normal carbon-13\n13-70 and the DEPT experimental results are tabulated. The infrared spectrum shows a broad peak at about $3340 \\mathrm{~cm}^{-1}$ and a medium-sized peak at about $1651 \\mathrm{~cm}^{-1}$. Draw the structure of this compound.\n\n| Normal Carbon | DEPT-135 | DEPT-90 |\n| :---: | :---: | :---: |\n| 22.2 ppm | Positive | No peak |\n| 40.9 | Negative | No peak |\n| 60.2 | Negative | No peak |\n| 112.5 | Negative | No peak |\n| 142.3 | No peak | No peak |\n\n\n\nPROBLEM The proton NMR spectrum of a compound with the formula $\\mathrm{C}_{7} \\mathrm{H}_{12} \\mathrm{O}_{2}$ is shown. The infrared\n13-71 spectrum displays a strong band at $1738 \\mathrm{~cm}^{-1}$ and a weak band at $1689 \\mathrm{~cm}^{-1}$. The normal carbon-13 and the DEPT experimental results are tabulated. Draw the structure of this compound."}
{"id": 863, "contents": "2,4-Pentanediol - \n| Normal Carbon | DEPT-135 | DEPT-90 |\n| :---: | :---: | :---: |\n| 18 ppm | Positive | No peak |\n| 21 | Positive | No peak |\n| 26 | Positive | No peak |\n| 61 | Negative | No peak |\n| 119 | Positive | Positive |\n| 139 | No peak | No peak |\n| 171 | No peak | No peak |"}
{"id": 864, "contents": "CHAPTER 14 - \nConjugated Compounds and Ultraviolet Spectroscopy\n\n\nFIGURE 14.1 Saffron, derived from stigmas of the saffron crocus, is the world's most expensive spice. Its color is caused by the presence of alternating single and double bonds. (credit: modification of work \"Welcome to My World! ~ Huron River Watershed\" by j van cise photos/ Flickr, CC BY 2.0)"}
{"id": 865, "contents": "CHAPTER CONTENTS - \n14.1 Stability of Conjugated Dienes: Molecular Orbital Theory\n14.2 Electrophilic Additions to Conjugated Dienes: Allylic Carbocations\n14.3 Kinetic versus Thermodynamic Control of Reactions\n14.4 The Diels-Alder Cycloaddition Reaction\n14.5 Characteristics of the Diels-Alder Reaction\n14.6 Diene Polymers: Natural and Synthetic Rubbers\n14.7 Ultraviolet Spectroscopy\n14.8 Interpreting Ultraviolet Spectra: The Effect of Conjugation\n14.9 Conjugation, Color, and the Chemistry of Vision\n\nWHY THIS CHAPTER? Conjugated compounds of many different sorts are common in nature. Many of the pigments responsible for the brilliant colors of fruits, flowers, and even animals have numerous alternating single and double bonds. $\\beta$-Carotene, for instance, the orange pigment responsible for the color of carrots and an important source of vitamin A , is a conjugated polyene with 11 double bonds. Conjugated enones (alkene + ketone) are common structural features of many biologically important molecules such as progesterone, the hormone that prepares the uterus for implantation of a fertilized ovum. Cyclic conjugated molecules such as benzene are a major field of study in themselves. In this chapter, we'll look at some of the distinctive properties\nof conjugated molecules and at the reasons for those properties.\n\n\nMost of the unsaturated compounds we looked at in the chapters on Alkenes: Structure and Reactivity and Alkenes: Reactions and Synthesis had only one double bond, but many compounds have numerous sites of unsaturation. If the different unsaturations are well separated in a molecule, they react independently, but if they're close together, they may interact. In particular, compounds that have alternating single and double bonds-so-called conjugated compounds-have some distinctive characteristics. The conjugated diene 1,3-butadiene, for instance, has some properties quite different from those of the nonconjugated 1,4-pentadiene.\n\n\n1,3-Butadiene (conjugated; alternating double and single bonds)\n\n\n1,4-Pentadiene (nonconjugated; nonalternating double and single bonds)"}
{"id": 866, "contents": "CHAPTER CONTENTS - 14.1 Stability of Conjugated Dienes: Molecular Orbital Theory\nConjugated dienes can be prepared by some of the methods previously discussed for preparing alkenes (Section 11.7-Section 11.11). The base-induced elimination of HX from an allylic halide is one such reaction."}
{"id": 867, "contents": "Cyclohexene 3-Bromocyclohexene 1,3-Cyclohexadiene (76\\%) - \nSimple conjugated dienes used in polymer synthesis include 1,3-butadiene, chloroprene (2-chloro-1,3-butadiene), and isoprene (2-methyl-1,3-butadiene). Isoprene has been prepared industrially by several methods, including the acid-catalyzed double dehydration of 3-methyl-1,3-butanediol.\n\n\n3-Methyl-1,3-butanediol\n\n\nIsoprene\n2-Methyl-1,3-butadiene\n\nOne of the properties that distinguishes conjugated from nonconjugated dienes is that the central single bond is shorter than might be expected. The C2-C3 single bond in 1,3-butadiene, for instance, has a length of 147 pm , some 6 pm shorter than the C2-C3 bond in butane ( 153 pm ).\n\n\n1,3-Butadiene\n\n\nButane\n\nAnother distinctive property of conjugated dienes is their unusual stability, as evidenced by their heats of hydrogenation (TABLE 14.1). We saw in Section 7.6 that monosubstituted alkenes, such as 1-butene, have $\\Delta H^{\\circ}{ }_{\\text {hydrog }}$ near $-126 \\mathrm{~kJ} / \\mathrm{mol}(-30.1 \\mathrm{kcal} / \\mathrm{mol})$, whereas disubstituted alkenes, such as 2-methylpropene, have $\\Delta H^{\\circ}{ }_{\\text {hydrog }}$ near $-119 \\mathrm{~kJ} / \\mathrm{mol}(-28.4 \\mathrm{kcal} / \\mathrm{mol})$, which is approximately $7 \\mathrm{~kJ} / \\mathrm{mol}$ less negative. We concluded from these data that more highly substituted alkenes are more stable than less substituted ones. That is, more highly substituted alkenes release less heat on hydrogenation because they contain less energy to start with. A similar conclusion can be drawn for conjugated dienes.\n\nTABLE 14.1 Heats of Hydrogenation for Some Alkenes and Dienes"}
{"id": 868, "contents": "Cyclohexene 3-Bromocyclohexene 1,3-Cyclohexadiene (76\\%) - \nTABLE 14.1 Heats of Hydrogenation for Some Alkenes and Dienes\n\n| Alkene or diene | Product | $\\Delta H^{\\circ} \\mathrm{hydrog}$ |  |\n| :---: | :---: | :---: | :---: |\n|  |  | (kJ/mol) | ( $\\mathrm{kcal} / \\mathrm{mol}$ ) |\n| $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}=\\mathrm{CH}_{2}$ | $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$ | -126 | -30.1 |\n| <smiles>C=C(C)C</smiles> | <smiles>CC(C)C</smiles> | -119 | -28.4 |\n| $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}_{2} \\mathrm{CH}=\\mathrm{CH}_{2}$ | $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$ | -253 | -60.5 |\n| $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}-\\mathrm{CH}=\\mathrm{CH}_{2}$ | $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$ | -236 | -56.4 |\n| <smiles>C=CC(=C)C</smiles> | <smiles>CCC(C)C</smiles> | -229 | -54.7 |"}
{"id": 869, "contents": "Cyclohexene 3-Bromocyclohexene 1,3-Cyclohexadiene (76\\%) - \nBecause a monosubstituted alkene has a $\\Delta H^{\\circ}{ }_{\\text {hydrog }}$ of approximately $-126 \\mathrm{~kJ} / \\mathrm{mol}$, we might expect that a compound with two monosubstituted double bonds would have a $\\Delta H^{\\circ}{ }_{\\text {hydrog }}$ approximately twice that value, or $-252 \\mathrm{~kJ} / \\mathrm{mol}$. Nonconjugated dienes, such as 1,4 -pentadiene ( $\\Delta H^{\\circ}{ }_{\\text {hydrog }}=-253 \\mathrm{~kJ} / \\mathrm{mol}$ ), meet this expectation, but the conjugated diene 1,3-butadiene ( $\\Delta H^{\\circ}{ }_{\\text {hydrog }}=-236 \\mathrm{~kJ} / \\mathrm{mol}$ ) does not. 1,3-Butadiene is approximately $16 \\mathrm{~kJ} / \\mathrm{mol}(3.8 \\mathrm{kcal} / \\mathrm{mol})$ more stable than expected.\n\n|  | $\\Delta \\boldsymbol{H}^{\\circ}$ hydrog $(\\mathbf{k J} / \\mathbf{m o l})$ |  |\n| :---: | :--- | :--- |\n| $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}_{2} \\mathrm{CH}=\\mathrm{CH}_{2}$ | $-126+(-126)=-252$ | Expected <br> 1,4-Pentadiene |\n|  | -253 | Observed <br> Difference |\n| $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}=\\mathrm{CH}_{2}$ | $-126+(-126)=-252$ | Expected |\n| 1,3-Butadiene | -236 | Observed |\n|  | -16 | Difference |"}
{"id": 870, "contents": "Cyclohexene 3-Bromocyclohexene 1,3-Cyclohexadiene (76\\%) - \nWhat accounts for the stability of conjugated dienes? According to valence bond theory (Section 1.5 and Section 1.8), their stability is due to orbital hybridization. Typical $\\mathrm{C}-\\mathrm{C}$ single bonds, like those in alkanes, result from $\\sigma$ overlap of $s p^{3}$ orbitals on both carbons, but in a conjugated diene, the central $\\mathrm{C}-\\mathrm{C}$ single bond results from $\\sigma$ overlap of $s p^{2}$ orbitals on both carbons. Because $s p^{2}$ orbitals have more $s$ character ( $33 \\% s$ ) than $s p^{3}$ orbitals $(25 \\% s)$, the electrons in $s p^{2}$ orbitals are closer to the nucleus and the bonds they form are somewhat shorter\nand stronger. Thus, the \"extra\" stability of a conjugated diene results in part from the greater amount of $s$ character in the orbitals forming the $\\mathrm{C}-\\mathrm{C}$ single bond.\n\n\n\nBond formed by overlap of $\\boldsymbol{s p} \\boldsymbol{p}^{2}$ orbitals\nAccording to molecular orbital theory (Section 1.11), the stability of conjugated dienes arises because of an interaction between the $\\pi$ orbitals of the two double bonds. To review briefly, when two $p$ atomic orbitals combine to form a $\\pi$ bond, two $\\pi$ molecular orbitals (MOs) result. One is lower in energy than the starting $p$ orbitals and is therefore bonding; the other is higher in energy, has a node between nuclei, and is antibonding. The two $\\pi$ electrons occupy the low-energy, bonding orbital, resulting in formation of a stable bond between atoms (FIGURE 14.2).\n\n\nFIGURE 14.2 Two $\\boldsymbol{p}$ orbitals combine to form two $\\boldsymbol{\\pi}$ molecular orbitals. Both electrons occupy the low-energy, bonding orbital, leading to a net lowering of energy and formation of a stable bond. The asterisk on $\\psi_{2} *$ indicates an antibonding orbital."}
{"id": 871, "contents": "Cyclohexene 3-Bromocyclohexene 1,3-Cyclohexadiene (76\\%) - \nNow let's combine four adjacent $p$ atomic orbitals, as occurs in a conjugated diene. In so doing, we generate a set of four $\\pi$ molecular orbitals, two of which are bonding and two of which are antibonding (FIGURE 14.3). The four $\\pi$ electrons occupy the two bonding orbitals, leaving the antibonding orbitals vacant.\n\n\nFIGURE 14.3 Four $\\boldsymbol{\\pi}$ molecular orbitals in 1,3-butadiene. Note that the number of nodes between nuclei increases as the energy level of the orbital increases.\n\nThe lowest-energy $\\pi$ molecular orbital (denoted $\\psi_{1}$, Greek psi) has no nodes between the nuclei and is therefore bonding. The $\\pi$ MO of next-lowest energy, $\\psi_{2}$, has one node between nuclei and is also bonding. Above $\\psi_{1}$ and $\\psi_{2}$ in energy are the two antibonding $\\pi \\operatorname{MOs}, \\psi_{3} *$ and $\\psi_{4}{ }^{*}$. (The asterisks indicate antibonding orbitals.) Note that the number of nodes between nuclei increases as the energy level of the orbital increases. The $\\psi_{3} *$ orbital has two nodes between nuclei, and $\\psi_{4}{ }^{*}$, the highest-energy MO, has three nodes between nuclei.\n\nComparing the $\\pi$ molecular orbitals of 1,3-butadiene (two conjugated double bonds) with those of 1,4-pentadiene (two isolated double bonds) shows why the conjugated diene is more stable. In a conjugated diene, the lowest-energy $\\pi \\mathrm{MO}\\left(\\psi_{1}\\right)$ has a favorable bonding interaction between C 2 and C 3 that is absent in a nonconjugated diene. As a result, there is a certain amount of double-bond character to the $\\mathrm{C} 2-\\mathrm{C} 3$ single bond, making that bond both stronger and shorter than a typical single bond. Electrostatic potential maps show clearly the additional electron density in the central single bond (FIGURE 14.4).\n\n\n\n1,3-Butadiene (conjugated)"}
{"id": 872, "contents": "Cyclohexene 3-Bromocyclohexene 1,3-Cyclohexadiene (76\\%) - \n1,3-Butadiene (conjugated)\n\n\n\n1,4-Pentadiene\n(nonconjugated)\n\nFIGURE 14.4 Electrostatic potential maps of 1,3-butadiene (conjugated) and 1,4-pentadiene (nonconjugated). Additional electron density is present in the central $\\mathrm{C}-\\mathrm{C}$ bond of 1,3-butadiene, corresponding to partial double-bond character.\n\nIn describing 1,3-butadiene, we say that the $\\pi$ electrons are spread out, or delocalized, over the entire $\\pi$ framework, rather than localized between two specific nuclei. Delocalization allows the bonding electrons to be closer to more nuclei, thus leading to lower energy and greater stability.\n\nPROBLEM Allene, $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{C}=\\mathrm{CH}_{2}$, has a heat of hydrogenation of $-298 \\mathrm{~kJ} / \\mathrm{mol}(-71.3 \\mathrm{kcal} / \\mathrm{mol})$. Rank a 14-1 conjugated diene, a nonconjugated diene, and an allene in order of stability."}
{"id": 873, "contents": "Cyclohexene 3-Bromocyclohexene 1,3-Cyclohexadiene (76\\%) - 14.2 Electrophilic Additions to Conjugated Dienes: Allylic Carbocations\nOne of the most striking differences between conjugated dienes and typical alkenes is their behavior in electrophilic addition reactions. To review briefly, the addition of an electrophile to a carbon-carbon double bond is a general reaction of alkenes (Section 7.7). Markovnikov regiochemistry is observed because the more stable carbocation is formed as an intermediate. Thus, addition of HCl to 2-methylpropene yields 2-chloro-2-methylpropane rather than 1-chloro-2-methylpropane, and addition of 2 equivalents of HCl to the nonconjugated diene 1,4-pentadiene yields 2,4-dichloropentane.\n\n\n\nConjugated dienes also undergo electrophilic addition reactions readily, but mixtures of products are invariably obtained. Addition of HBr to 1,3 -butadiene, for instance, yields a mixture of two products (not counting cis-trans isomers). 3-Bromo-1-butene is the typical Markovnikov product of $\\mathbf{1 , 2}$-addition to a double bond, but 1-bromo-2-butene seems unusual. The double bond in this product has moved to a position between carbons 2 and 3 , and HBr has added to carbons 1 and 4, a result described as $\\mathbf{1 , 4}$-addition.\n\n\nMany other electrophiles besides HBr add to conjugated dienes, and mixtures of products are usually formed. For example, $\\mathrm{Br}_{2}$ adds to 1,3-butadiene to give a mixture of 3,4-dibromo-1-butene and 1,4-dibromo-2-butene."}
{"id": 874, "contents": "Cyclohexene 3-Bromocyclohexene 1,3-Cyclohexadiene (76\\%) - 14.2 Electrophilic Additions to Conjugated Dienes: Allylic Carbocations\nHow can we account for the formation of 1,4-addition products? The answer is that allylic carbocations are involved as intermediates (recall that the word allylic means \"next to a double bond\"). When 1,3-butadiene reacts with an electrophile such as $\\mathrm{H}^{+}$, two carbocation intermediates are possible-a primary nonallylic carbocation and a secondary allylic cation. Because an allylic cation is stabilized by resonance between two forms (Section 11.5), it is more stable and forms faster than a nonallylic carbocation.\n\n\nWhen the allylic cation reacts with $\\mathrm{Br}^{-}$to complete the electrophilic addition, the reaction can occur either at C 1 or at C3 because both carbons share the positive charge (FIGURE 14.5). Thus, a mixture of 1,2-and 1,4-addition products results. You might recall that a similar product mixture was seen for NBS bromination of alkenes in Section 10.3, a reaction that proceeds through an allylic radical.\n\n\n\nFIGURE 14.5 An electrostatic potential map of the allylic carbocation produced by protonation of 1,3-butadiene shows that the positive charge is shared by carbons 1 and 3. Reaction of $\\mathrm{Br}^{-}$with the more positive carbon (C3) predominantly yields the 1,2-addition product."}
{"id": 875, "contents": "Predicting the Product of an Electrophilic Addition Reaction of a Conjugated Diene - \nGive the structures of the likely products from reaction of 1 equivalent of HCl with 2 -methyl-1,3-cyclohexadiene. Show both 1,2 and 1,4 adducts."}
{"id": 876, "contents": "Strategy - \nElectrophilic addition of HCl to a conjugated diene involves the formation of allylic carbocation intermediate. Thus, the first step is to protonate the two ends of the diene and draw the resonance forms of the two allylic carbocations that result. Then, allow each resonance form to react with $\\mathrm{Cl}^{-}$, generating a maximum of four possible products.\n\nIn the present instance, protonation of the $\\mathrm{C} 1-\\mathrm{C} 2$ double bond gives a carbocation that can react further to give the 1,2 adduct 3 -chloro-3-methylcyclohexene and the 1,4 adduct 3-chloro-1-methylcyclohexene. Protonation of the C3-C4 double bond gives a symmetrical carbocation, whose two resonance forms are equivalent. Thus, the 1,2 adduct and the 1,4 adduct have the same structure: 6 -chloro-1-methylcyclohexene. Of the two possible modes of protonation, the first is more likely because it yields a more stable, tertiary allylic cation rather than a less-stable, secondary allylic cation."}
{"id": 877, "contents": "Solution - \nPROBLEM Give the structures of both 1,2 and 1,4 adducts resulting from reaction of 1 equivalent of HCl with 14-2 1,3-pentadiene.\n\nPROBLEM Look at the possible carbocation intermediates produced during addition of HCl to 1,3-pentadiene 14-3 (Problem 14-2), and predict which 1,2 adduct predominates. Which 1,4 adduct predominates?\n\nPROBLEM Give the structures of both 1,2 and 1,4 adducts resulting from reaction of 1 equivalent of HBr with\n14-4 the following compound:"}
{"id": 878, "contents": "Solution - 14.3 Kinetic versus Thermodynamic Control of Reactions\nElectrophilic addition to a conjugated diene at or below room temperature normally leads to a mixture of products in which the 1,2 adduct predominates over the 1,4 adduct. When the same reaction is carried out at higher temperatures, however, the product ratio often changes and the 1,4 adduct predominates. For example, addition of HBr to 1,3 -butadiene at $0^{\\circ} \\mathrm{C}$ yields a $71: 29$ mixture of 1,2 and 1,4 adducts, but the same reaction carried out at $40^{\\circ} \\mathrm{C}$ yields a $15: 85$ mixture. Furthermore, when the product mixture formed at $0{ }^{\\circ} \\mathrm{C}$ is heated to $40^{\\circ} \\mathrm{C}$ in the presence of HBr , the ratio of adducts slowly changes from $71: 29$ to $15: 85$. Why?\n\n\nTo understand the effect of temperature on product distribution, let's briefly review what we said in Section 6.7 about rates and equilibria. Imagine a reaction that can give either or both of two products, B and C.\n\n\nLet's assume that $\\mathbf{B}$ forms faster than $\\mathbf{C}$ (in other words, $\\Delta G^{\\ddagger} \\mathrm{B}^{<\\Delta G^{\\ddagger}} \\mathrm{C}$ ) but that $\\mathbf{C}$ is more stable than $\\mathbf{B}$ (in other words, $\\Delta G^{\\circ}{ }_{\\mathrm{C}}>\\Delta G^{\\circ}{ }_{\\mathrm{B}}$ ). An energy diagram for the two processes might look like that shown in FIGURE 14.6.\n\n\nFIGURE 14.6 An energy diagram for two competing reactions in which the less stable product $B$ forms faster than the more stable product C ."}
{"id": 879, "contents": "Solution - 14.3 Kinetic versus Thermodynamic Control of Reactions\nFIGURE 14.6 An energy diagram for two competing reactions in which the less stable product $B$ forms faster than the more stable product C .\n\nLet's first carry out the reaction at a lower temperature so that both processes are irreversible and no equilibrium is reached. Because $\\mathbf{B}$ forms faster than $\\mathbf{C}, \\mathbf{B}$ is the major product. It doesn't matter that $\\mathbf{C}$ is more stable than B, because the two are not in equilibrium. The product of an irreversible reaction depends only on relative rates, not on stability. Such reactions are said to be under kinetic control.\n\n\nNow let's carry out the same reaction at some higher temperature so that both processes are readily reversible and an equilibrium is reached. Because $\\mathbf{C}$ is more stable than $\\mathbf{B}, \\mathbf{C}$ is the major product obtained. It doesn't matter that $\\mathbf{C}$ forms more slowly than $\\mathbf{B}$, because the two are in equilibrium. The product of a readily reversible reaction depends only on stability, not on relative rates. Such reactions are said to be under equilibrium control, or thermodynamic control.\n\n\nWe can now explain the effect of temperature on the electrophilic addition reactions of conjugated dienes. At low temperature $\\left(0^{\\circ} \\mathrm{C}\\right)$, HBr adds to 1,3-butadiene under kinetic control to give a $71: 29$ mixture of products, with the more rapidly formed 1,2 adduct predominating. Because these low-temperature conditions don't allow the reaction to reach equilibrium, the product that forms faster predominates. At higher temperature $\\left(40^{\\circ} \\mathrm{C}\\right)$, however, the reaction occurs under thermodynamic control to give a $15: 85$ mixture of products, with the more stable 1,4 adduct predominating. The higher temperature allows the addition process to become reversible, so an equilibrium mixture of products results. FIGURE 14.7 shows this situation in an energy diagram."}
{"id": 880, "contents": "Solution - 14.3 Kinetic versus Thermodynamic Control of Reactions\nFIGURE 14.7 Energy diagram for the electrophilic addition of $\\mathbf{H B r}$ to 1,3-butadiene. The 1,2 adduct is the kinetic product because it forms faster, but the 1,4 adduct is the thermodynamic product because it is more stable.\n\nThe electrophilic addition of HBr to 1,3-butadiene is a good example of how a change in experimental conditions can change the product of a reaction. The concept of thermodynamic control versus kinetic control is a useful one that we can sometimes take advantage of in the laboratory.\n\nPROBLEM The 1,2 adduct and the 1,4 adduct formed by reaction of HBr with 1,3-butadiene are in equilibrium 14-5 at $40^{\\circ} \\mathrm{C}$. Propose a mechanism by which the interconversion of products takes place.\n\nPROBLEM Why do you suppose 1,4 adducts of 1,3-butadiene are generally more stable than 1,2 adducts?\n14-6"}
{"id": 881, "contents": "Solution - 14.4 The Diels-Alder Cycloaddition Reaction\nPerhaps the most striking difference between conjugated and nonconjugated dienes is that conjugated dienes undergo an addition reaction with alkenes to yield substituted cyclohexene products. For example, 1,3-butadiene and 3-buten-2-one give 3-cyclohexenyl methyl ketone.\n\n\nThis process, named the Diels-Alder cycloaddition reaction after its discoverers, is extremely useful in the laboratory because it forms two carbon-carbon bonds in a single step and is one of the few general methods available for making cyclic molecules. (As the name implies, a cycloaddition reaction is one in which two reactants add together to give a cyclic product.) The 1950 Nobel Prize in Chemistry was awarded to Otto Diels and Kurt Alder in recognition of the importance of their discovery.\n\nThe mechanism of Diels-Alder cycloaddition is different from that of other reactions we've studied because it is neither polar nor radical. Rather, the Diels-Alder reaction is a pericyclic process. Pericyclic reactions, which we'll discuss in more detail in Chapter 30, take place in a single step by a cyclic redistribution of bonding electrons. The two reactants simply join together through a cyclic transition state in which the two new $\\mathrm{C}-\\mathrm{C}$ bonds form at the same time.\n\nWe can picture a Diels-Alder addition as occurring by head-on ( $\\sigma$ ) overlap of the two alkene $p$ orbitals with the two $p$ orbitals on carbons 1 and 4 of the diene (FIGURE 14.8). This is, of course, a cyclic orientation of the reactants.\n\n\n\nFIGURE 14.8 Mechanism of the Diels-Alder cycloaddition reaction. The reaction occurs in a single step through a cyclic transition state in which the two new $\\mathrm{C}-\\mathrm{C}$ bonds form simultaneously."}
{"id": 882, "contents": "Solution - 14.4 The Diels-Alder Cycloaddition Reaction\nIn the Diels-Alder transition state, the two alkene carbons and carbons 1 and 4 of the diene rehybridize from $s p^{2}$ to $s p^{3}$ to form two new single bonds, while carbons 2 and 3 of the diene remain $s p^{2}$-hybridized to form the new double bond in the cyclohexene product. We'll study this mechanism in more detail in Section $\\mathbf{3 0 . 5}$ but will concentrate for the present on learning about the characteristics and uses of the Diels-Alder reaction."}
{"id": 883, "contents": "The Dienophile - \nThe Diels-Alder cycloaddition reaction occurs most rapidly if the alkene component, called the dienophile (\"diene lover\"), has an electron-withdrawing substituent group. Thus, ethylene itself reacts sluggishly, but propenal, ethyl propenoate, maleic anhydride, benzoquinone, propenenitrile, and similar compounds are highly reactive. Note also that alkynes, such as methyl propynoate, can act as Diels-Alder dienophiles.\n\n\nIn all cases, the double or triple bond of the dienophile is adjacent to the positively polarized carbon of an electron-withdrawing substituent. As a result, the double-bond carbons in these substances are substantially less electron-rich than the carbons in ethylene, as indicated by the electrostatic potential maps in FIGURE 14.9.\n\n\nFIGURE 14.9 Electrostatic potential maps of ethylene, propenal, and propenenitrile show that electron-withdrawing groups make the double-bond carbons less electron-rich.\n\nOne of the most useful features of the Diels-Alder reaction is that it is stereospecific, meaning that a single product stereoisomer is formed. Furthermore, the stereochemistry of the dienophile is retained. If we carry out cycloaddition with methyl cis-2-butenoate, only the cis-substituted cyclohexene product is formed. With methyl trans-2-butenoate, only the trans-substituted cyclohexene product is formed.\n\n\n1,3-Butadiene Methyl(Z)-2-butenoate\nCis product\n\n\nAnother stereochemical feature of the Diels-Alder reaction is that the diene and dienophile partners orient so that the endo product, rather than the alternative exo product, is formed. The words endo and exo are used to indicate relative stereochemistry when referring to bicyclic structures like substituted norbornanes (Section 4.9). A substituent on one bridge is said to be endo if it is syn (cis) to the larger of the other two bridges and is said to be exo if it is anti (trans) to the larger of the other two."}
{"id": 884, "contents": "The Dienophile - \nEndo products result from Diels-Alder reactions because the amount of orbital overlap between diene and dienophile is greater when the reactants lie directly on top of one another, so that the electron-withdrawing substituent on the dienophile is underneath the diene double bonds. In the reaction of 1,3-cyclopentadiene with maleic anhydride, for instance, the following result is obtained:\n\n\nMaleic anhydride"}
{"id": 885, "contents": "Predicting the Product of a Diels-Alder Reaction - \nPredict the product of the following Diels-Alder reaction:"}
{"id": 886, "contents": "Strategy - \nDraw the diene so that the ends of its two double bonds are near the dienophile double bond. Then form two single bonds between the partners, convert the three double bonds into single bonds, and convert the former single bond of the diene into a double bond. Because the dienophile double bond is cis to begin with, the two attached hydrogens must remain cis in the product."}
{"id": 887, "contents": "Solution - \nPROBLEM Predict the product of the following Diels-Alder reaction:\n14-7"}
{"id": 888, "contents": "The Diene - \nJust as the dienophile component has certain constraints that affect its reactivity, so too does the conjugated diene component. The diene must adopt what is called an $\\boldsymbol{s}$-cis conformation, meaning \"cis-like\" about the single bond, to undergo a Diels-Alder reaction. Only in the $s$-cis conformation are carbons 1 and 4 of the diene close enough to react through a cyclic transition state."}
{"id": 889, "contents": "$s$-Cis conformation - \n$\\boldsymbol{s}$-Trans conformation\nIn the alternative $s$-trans conformation, the ends of the diene partner are too far apart to overlap with the dienophile $p$ orbitals.\n\n\nSuccessful reaction\n\n\nNo reaction (ends too far apart)\n\nTwo examples of dienes that can't adopt an $s$-cis conformation, and thus don't undergo Diels-Alder reactions, are shown in FIGURE 14.10. In the bicyclic diene, the double bonds are rigidly fixed in an $s$-trans arrangement by geometric constraints of the rings. In $(2 Z, 4 Z)-2,4$-hexadiene, steric strain between the two methyl groups prevents the molecule from adopting $s$-cis geometry.\n\n\nA bicyclic diene (rigid $s$-trans diene)\n\n(2Z,4Z)-2,4-Hexadiene (s-trans, more stable)\n\nFIGURE 14.10 Two dienes that can't achieve an $\\boldsymbol{s}$-cis conformation and thus can't undergo Diels-Alder reactions.\nIn contrast to these unreactive dienes that can't achieve an $s$-cis conformation, other dienes are fixed only in the correct $s$-cis geometry and are therefore highly reactive in Diels-Alder cycloaddition. 1,3-Cyclopentadiene, for example, is so reactive that it reacts with itself. At room temperature, 1,3-cyclopentadiene dimerizes. One molecule acts as diene and a second molecule acts as dienophile in a self-Diels-Alder reaction.\n\n\nBiological Diels-Alder reactions are also known but are uncommon. One example occurs in the biosynthesis of the cholesterol-lowering drug lovastatin (trade name Mevacor) isolated from the bacterium Aspergillus terreus. The key step is the intramolecular Diels-Alder reaction of a triene, in which the diene and dienophile components are within the same molecule.\n\n\nPROBLEM Which of the following alkenes would you expect to be good Diels-Alder dienophiles?\n14-8\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)"}
{"id": 890, "contents": "$s$-Cis conformation - \nPROBLEM Which of the following alkenes would you expect to be good Diels-Alder dienophiles?\n14-8\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n\nPROBLEM Which of the following dienes have an $s$-cis conformation, and which have an $s$-trans conformation?\n14-9 Of the $s$-trans dienes, which can readily rotate to $s$-cis?\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Predict the product of the following Diels-Alder reaction:\n14-10"}
{"id": 891, "contents": "$s$-Cis conformation - 14.6 Diene Polymers: Natural and Synthetic Rubbers\nConjugated dienes can be polymerized just as simple alkenes can (Section 8.10). Diene polymers are structurally more complex than simple alkene polymers, however, because double bonds occur every four carbon atoms along the chain, leading to the possibility of cis-trans isomers. The initiator (In) for the reaction can be either a radical, as occurs in ethylene polymerization, or an acid. Note that the polymerization is a 1,4 addition of the growing chain to a conjugated diene monomer.\n\ntrans-Polybutadiene\nRubber is a naturally occurring diene polymer of isoprene (2-methyl-1,3-butadiene) and is produced by more than 400 different plants. The major source is the so-called rubber tree, Hevea brasiliensis, from which the crude material, called latex, is harvested as it drips from a slice made through the bark. The double bonds of rubber have $Z$ stereochemistry, but gutta-percha, the $E$ isomer of rubber, also occurs naturally. Harder and more brittle than rubber, gutta-percha has a variety of applications, including use in dental endodontics and as the covering on some golf balls.\n\n\nA number of different synthetic rubbers are produced commercially by diene polymerization. Both cis- and trans-polyisoprene can be made, and the synthetic rubber thus produced is similar to the natural material. Chloroprene ( 2 -chloro-1,3-butadiene) is polymerized to yield neoprene, an excellent, although expensive, synthetic rubber with good weather resistance. Neoprene is used in the production of industrial hoses and gloves, among other things.\n\n\nBoth natural and synthetic rubbers are too soft and tacky to be useful until they are hardened by heating with elemental sulfur, a process called vulcanization. Vulcanization cross-links the rubber chains by forming carbon-sulfur bonds between them, thereby hardening and stiffening the polymer. The exact degree of hardening can be varied, yielding material soft enough for automobile tires or hard enough for bowling balls (ebonite)."}
{"id": 892, "contents": "$s$-Cis conformation - 14.6 Diene Polymers: Natural and Synthetic Rubbers\nThe unusual ability of rubber to stretch and then contract to its original shape is due to the irregular structure of the polymer chains caused by the double bonds. These double bonds introduce bends and kinks into the polymer chains, thereby preventing neighboring chains from nestling together. When stretched, the randomly coiled chains straighten out and orient along the direction of the pull but are kept from sliding over one another by the cross-links. When the tension is released, the polymer reverts to its original random state.\n\nPROBLEM Draw a segment of the polymer that might be prepared from 2-phenyl-1,3-butadiene. 14-11\n\nPROBLEM Show the mechanism of the acid-catalyzed polymerization of 1,3-butadiene. 14-12"}
{"id": 893, "contents": "$s$-Cis conformation - 14.7 Ultraviolet Spectroscopy\nAs we've seen, mass spectrometry, infrared spectroscopy, and nuclear magnetic resonance spectroscopy are techniques of structure determination applicable to all organic molecules. In addition to these three generally useful methods, there is a fourth-ultraviolet (UV) spectroscopy-that is applicable only to conjugated compounds. UV is less commonly used than the other three spectroscopic techniques because of the specialized information it gives, so we'll only discuss it briefly.\n\n| Mass spectrometry | Molecular size and formula |\n| :--- | :--- |\n| IR spectroscopy | Functional groups present |\n| NMR spectroscopy | Carbon-hydrogen framework |\n| UV spectroscopy | Conjugated $\\boldsymbol{\\pi}$ electron systems |\n\nThe ultraviolet region of the electromagnetic spectrum extends from the short-wavelength end of the visible region $\\left(4 \\times 10^{-7} \\mathrm{~m}\\right)$ to the long-wavelength end of the X-ray region $\\left(10^{-8} \\mathrm{~m}\\right)$, but the narrow range from $2 \\times$ $10^{-7} \\mathrm{~m}$ to $4 \\times 10^{-7} \\mathrm{~m}$ is the part of greatest interest to organic chemists. Absorptions in this region are usually measured in nanometers ( nm ), where $1 \\mathrm{~nm}=10^{-9} \\mathrm{~m}$. Thus, the ultraviolet range of interest is from 200 to 400 nm (FIGURE 14.11).\n\n\nFIGURE 14.11 The ultraviolet (UV) and neighboring regions of the electromagnetic spectrum.\nWe saw in Section 12.5 that when an organic molecule is irradiated with electromagnetic energy, the radiation either passes through the sample or is absorbed, depending on its energy. With IR irradiation, the energy absorbed corresponds to the amount needed to increase molecular vibrations. With UV radiation, the energy absorbed corresponds to the amount needed to promote an electron from a lower-energy orbital to a higherenergy one in a conjugated molecule. The conjugated diene 1,3-butadiene, for instance, has four $\\pi$ molecular orbitals, as shown previously in FIGURE 14.3. The two lower-energy, bonding MOs are occupied in the ground state, and the two higher-energy, antibonding MOs are unoccupied."}
{"id": 894, "contents": "$s$-Cis conformation - 14.7 Ultraviolet Spectroscopy\nOn irradiation with ultraviolet light (hv), 1,3-butadiene absorbs energy and a $\\pi$ electron is promoted from the highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO). Because the electron is promoted from a bonding $\\pi$ molecular orbital to an antibonding $\\pi^{*}$ molecular orbital, we call this a $\\pi \\rightarrow \\pi^{*}$ excitation (read as \"pi to pi star\"). The energy gap between the HOMO and the LUMO of 1,3-butadiene is such that UV light of 217 nm wavelength is required to effect the $\\pi \\rightarrow \\pi^{*}$ electronic transition (FIGURE 14.12).\n\n\nFIGURE 14.12 Ultraviolet excitation of 1,3-butadiene results in the promotion of an electron from $\\Psi_{2}$, the highest occupied molecular orbital (HOMO), to $\\Psi_{3}{ }^{*}$, the lowest unoccupied molecular orbital (LUMO).\n\nAn ultraviolet spectrum is recorded by irradiating a sample with UV light of continuously changing wavelength. When the wavelength corresponds to the energy level required to excite an electron to a higher level, energy is absorbed. This absorption is detected and displayed on a chart that plots wavelength versus absorbance (A), defined as\n\n$$\nA=\\log \\frac{I_{0}}{I}\n$$\n\nwhere $I_{0}$ is the intensity of the incident light and $I$ is the intensity of the light transmitted through the sample.\nNote that UV spectra differ from IR spectra in how they are presented. For historical reasons, IR spectra are usually displayed so that the baseline corresponding to zero absorption runs across the top of the chart and a valley indicates an absorption, whereas UV spectra are displayed with the baseline at the bottom of the chart so that a peak indicates an absorption (FIGURE 14.13).\n\n\nThe amount of UV light absorbed is expressed as the sample's molar absorptivity ( $\\boldsymbol{\\epsilon}$ ), defined by the equation\n\n$$\n\\varepsilon=\\frac{A}{c \\times l}\n$$"}
{"id": 895, "contents": "$s$-Cis conformation - 14.7 Ultraviolet Spectroscopy\nThe amount of UV light absorbed is expressed as the sample's molar absorptivity ( $\\boldsymbol{\\epsilon}$ ), defined by the equation\n\n$$\n\\varepsilon=\\frac{A}{c \\times l}\n$$\n\nwhere\n$A=$ Absorbance\n$c=$ Concentration in $\\mathrm{mol} / \\mathrm{L}$\n$l=$ Sample pathlength in cm\nMolar absorptivity is a physical constant, characteristic of the particular substance being observed and thus characteristic of the particular $\\pi$ electron system in the molecule. Typical values for conjugated dienes are in\nthe range $\\varepsilon=10,000$ to 25,000 . The units for molar absorptivity, $\\mathrm{L} /(\\mathrm{mol} \\cdot \\mathrm{cm})$, are usually dropped.\nA particularly important use of this equation comes from rearranging it to the form $c=A /(\\varepsilon \\cdot I)$, which lets us measure the concentration of a sample in solution when $A, \\varepsilon$, and $l$ are known. As an example, $\\beta$-carotene, the pigment responsible for the orange color of carrots, has $\\varepsilon=138,000 \\mathrm{~L} /(\\mathrm{mol} \\cdot \\mathrm{cm})$. If a sample of $\\beta$-carotene is placed in a cell with a pathlength of 1.0 cm and the UV absorbance reads 0.37 , then the concentration of $\\beta$-carotene in the sample is\n\n$$\n\\begin{aligned}\nc & =\\frac{A}{\\varepsilon l}=\\frac{0.37}{\\left(1.38 \\times 10^{5} \\frac{\\mathrm{~L}}{\\mathrm{~mol} \\cdot \\mathrm{~cm}}\\right)(1.00 \\mathrm{~cm})} \\\\\n& =2.7 \\times 10^{-6} \\mathrm{~mol} / \\mathrm{L}\n\\end{aligned}\n$$"}
{"id": 896, "contents": "$s$-Cis conformation - 14.7 Ultraviolet Spectroscopy\nUnlike IR and NMR spectra, which show many absorptions for a given molecule, UV spectra are usually quite simple-often only a single peak. The peak is usually broad, and we identify its position by noting the wavelength at the top of the peak- $\\lambda_{\\max }$, read as \"lambda max.\"\n\nPROBLEM Calculate the energy range of electromagnetic radiation in the UV region of the spectrum from 200 14-13 to 400 nm (see Section 12.5). How does this value compare with the values calculated previously for IR and NMR spectroscopy?\n\nPROBLEM If pure vitamin A has $\\lambda_{\\max }=325(\\varepsilon=50,100)$, what is the vitamin A concentration in a sample whose 14-14 absorbance at 325 nm is $A=0.735$ in a cell with a pathlength of 1.00 cm ?"}
{"id": 897, "contents": "$s$-Cis conformation - 14.8 Interpreting Ultraviolet Spectra: The Effect of Conjugation\nThe wavelength necessary to effect the $\\pi \\rightarrow \\pi^{*}$ transition in a conjugated molecule depends on the energy gap between HOMO and LUMO, which in turn depends on the nature of the conjugated system. Thus, by measuring the UV spectrum of an unknown, we can derive structural information about the nature of any conjugated $\\pi$ electron system present in a molecule.\n\nOne of the most important factors affecting the wavelength of UV absorption by a molecule is the extent of conjugation. Molecular orbital calculations show that the energy difference between HOMO and LUMO decreases as the extent of conjugation increases. Thus, 1,3-butadiene absorbs at $\\lambda_{\\max }=217 \\mathrm{~nm}$, 1,3,5-hexatriene absorbs at $\\lambda_{\\max }=258 \\mathrm{~nm}$, and 1,3,5,7-octatetraene absorbs at $\\lambda_{\\max }=290 \\mathrm{~nm}$. (Remember: longer wavelength means lower energy.)\n\nOther kinds of conjugated systems, such as conjugated enones and aromatic rings, also have characteristic UV absorptions that are useful in structure determination. The UV absorption maxima of some representative conjugated molecules are given in TABLE 14.2.\n\n| Name | Structure | $\\lambda_{\\text {max }}(\\mathrm{nm})$ |\n| :---: | :---: | :---: |\n| 2-Methyl-1,3-butadiene | <smiles>C=CC(=C)C</smiles> | 220 |\n| 1,3-Cyclohexadiene | <smiles>C1=CCCC=C1</smiles> | 256 |\n| 1,3,5-Hexatriene | $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}-\\mathrm{CH}=\\mathrm{CH}-\\mathrm{CH}=\\mathrm{CH}_{2}$ | 258 |\n| 1,3,5,7-Octatetraene | $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}-\\mathrm{CH}=\\mathrm{CH}-\\mathrm{CH}=\\mathrm{CH}-\\mathrm{CH}=\\mathrm{CH}_{2}$ | 290 |"}
{"id": 898, "contents": "$s$-Cis conformation - 14.8 Interpreting Ultraviolet Spectra: The Effect of Conjugation\nTABLE 14.2 Ultraviolet Absorptions of Some Conjugated Molecules\n\n| Name | Structure | $\\lambda_{\\text {max }}(\\mathrm{nm})$ |\n| :---: | :---: | :---: |\n| 3-Buten-2-one | <smiles>C=CC(C)=O</smiles> | 219 |\n| Benzene | $8$ | 203 |\n\nPROBLEM Which of the following compounds would you expect to show ultraviolet absorptions in the 200 to\n14-15 400 nm range?\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n\nAspirin"}
{"id": 899, "contents": "Indole - 14.9 Conjugation, Color, and the Chemistry of Vision\nWhy are some organic compounds colored while others are not? $\\beta$-Carotene, the pigment in carrots, is yelloworange, for instance, while cholesterol is colorless. The answer involves both the chemical structures of colored molecules and the way we perceive light.\n\nThe visible region of the electromagnetic spectrum is adjacent to the ultraviolet region, extending from approximately 400 to 800 nm . Colored compounds have such elaborate systems of conjugation that their \"UV\" absorptions extend into the visible region. $\\beta$-Carotene, for instance, has 11 double bonds in conjugation, and its absorption occurs at $\\lambda_{\\max }=455 \\mathrm{~nm}$ (FIGURE 14.14).\n\n\nFIGURE 14.14 Ultraviolet spectrum of $\\boldsymbol{\\beta}$-carotene, a conjugated molecule with 11 double bonds. The absorption occurs in the visible region.\n\"White\" light from the sun or from a lamp consists of all wavelengths in the visible region. When white light strikes $\\beta$-carotene, the wavelengths from 400 to 500 nm (blue) are absorbed while all other wavelengths are transmitted and can reach our eyes. We therefore see the white light with the blue removed, which we perceive as a yellow-orange color for $\\beta$-carotene.\n\nConjugation is crucial not only for the colors we see in organic molecules but also for the light-sensitive molecules on which our visual system is based. The key substance for vision is dietary $\\beta$-carotene, which is converted to vitamin A by enzymes in the liver, oxidized to an aldehyde called 11-trans-retinal, and then isomerized by a change in geometry of the C11-C12 double bond to produce 11-cis-retinal."}
{"id": 900, "contents": "Indole - 14.9 Conjugation, Color, and the Chemistry of Vision\nThere are two main types of light-sensitive receptor cells in the retina of the human eye: rod cells and cone cells. The 3 million or so rod cells are primarily responsible for seeing dim light and shades of gray, whereas the 100 million cone cells are responsible for seeing bright light and colors. In the rod cells of the eye, 11-cis-retinal is converted into rhodopsin, a light-sensitive substance formed from the protein opsin and 11-cis-retinal. When light strikes the rod cells, isomerization of the C11-C12 double bond occurs and trans-rhodopsin, called metarhodopsin II, is produced. In the absence of light, this cis-trans isomerization takes approximately 1100 years, but in the presence of light, it occurs within $2 \\times 10^{-13}$ seconds! Isomerization of rhodopsin is accompanied by a change in molecular geometry, which in turn causes a nerve impulse to be sent through the optic nerve to the brain, where it is perceived as vision.\n\n\nRhodopsin\nMetarhodopsin II\nMetarhodopsin II is then recycled back into rhodopsin by a multistep sequence involving cleavage to all-trans-retinal and cis-trans isomerization back to 11-cis-retinal."}
{"id": 901, "contents": "Photolithography - \nFifty years ago, someone interested in owning a computer would have paid approximately $\\$ 150,000$ for 16 megabytes of random-access memory that would have occupied a volume the size of a small desk. Today, anyone can buy 60 times as much computer memory for $\\$ 10$ and fit the small chip into their shirt pocket. The difference between then and now is due to improvements in photolithography, the process by which integratedcircuit chips are made.\n\nPhotolithography begins by coating a layer of $\\mathrm{SiO}_{2}$ onto a silicon wafer and further coating with a thin (0.5-1.0 $\\mu \\mathrm{m})$ film of a light-sensitive organic polymer called a resist. A mask is then used to cover those parts of the chip that will become a circuit, and the wafer is irradiated with UV light. The nonmasked sections of the polymer undergo a chemical change when irradiated, which makes them more soluble than the masked, unirradiated sections. On washing the irradiated chip with solvent, solubilized polymer is selectively removed from the irradiated areas, exposing the $\\mathrm{SiO}_{2}$ underneath. This $\\mathrm{SiO}_{2}$ is then chemically etched away by reaction with hydrofluoric acid, leaving behind a pattern of polymer-coated $\\mathrm{SiO}_{2}$. Further washing removes the remaining polymer, leaving a positive image of the mask in the form of exposed ridges of $\\mathrm{SiO}_{2}$ (FIGURE 14.16). Additional cycles of coating, masking, and etching then produce the completed chips.\n\n\nFIGURE 14.15 Manufacturing the ultrathin circuitry on this computer chip depends on the organic chemical reactions of special polymers. (credit: \"Integrated circuit on microchip\" by Jon Sullivan/Wikimedia Commons, Public Domain)"}
{"id": 902, "contents": "Photolithography - \nFIGURE 14.15 Manufacturing the ultrathin circuitry on this computer chip depends on the organic chemical reactions of special polymers. (credit: \"Integrated circuit on microchip\" by Jon Sullivan/Wikimedia Commons, Public Domain)\n\n\nThe polymer resist currently used in chip manufacturing is based on the two-component diazoquinone-novolac system. Novolac resin is a soft, relatively low-molecular-weight polymer made from methylphenol and formaldehyde, while the diazoquinone is a bicyclic (two-ring) molecule containing a diazo\ngroup $=\\mathrm{N}=\\mathrm{N}$ ) adjacent to a ketone carbonyl ( $\\mathrm{C}=\\mathrm{O}$ ). The diazoquinone-novolac mix is relatively insoluble when fresh, but on exposure to ultraviolet light and water vapor, the diazoquinone component undergoes reaction to yield $\\mathrm{N}_{2}$ and a carboxylic acid, which can be washed away with dilute base. Novolac-diazoquinone technology is capable of producing features as small as $0.5 \\mu \\mathrm{~m}\\left(5 \\times 10^{-7} \\mathrm{~m}\\right)$, but further improvements in miniaturization are still being developed.\n\n\nDiazonaphthoquinone\n\n\nNovolac resin"}
{"id": 903, "contents": "Key Terms - \n- 1,2-addition\n- 1,4-addition\n- conjugated\n- Diels-Alder cycloaddition reaction\n- dienophile\n- endo\n- exo\n- highest occupied molecular orbital (HOMO)\n- kinetic control\n- lowest unoccupied molecular orbital (LUMO)\n- molar absorptivity ( $\\varepsilon$ )\n- $s$-cis conformation\n- stereospecific\n- thermodynamic control\n- ultraviolet (UV) spectroscopy"}
{"id": 904, "contents": "Summary - \nThe unsaturated compounds we've looked at previously have had only one double bond, but many compounds have numerous sites of unsaturation, which gives them some distinctive properties. Many such compounds are common in nature, including pigments and hormones.\n\nA conjugated diene or other compound is one that contains alternating double and single bonds. One characteristic of conjugated dienes is that they are more stable than their nonconjugated counterparts. This stability can be explained by a molecular orbital description in which four $p$ atomic orbitals combine to form four $\\pi$ molecular orbitals. Only the two bonding orbitals are occupied; the two antibonding orbitals are unoccupied. A $\\pi$ bonding interaction in the lowest-energy MO introduces some partial double-bond character between carbons 2 and 3, thereby strengthening the $\\mathrm{C} 2-\\mathrm{C} 3$ bond and stabilizing the molecule.\n\nConjugated dienes undergo several reactions not observed for nonconjugated dienes. One is the 1,4-addition of electrophiles. When a conjugated diene is treated with an electrophile such as $\\mathrm{HCl}, \\mathbf{1 , 2 -}$ and $\\mathbf{1 , 4}$-addition products are formed. Both result from the same resonance-stabilized allylic carbocation intermediate and are produced in varying amounts depending on the reaction conditions. The 1,2 adduct is usually formed faster and is said to be the product of kinetic control. The 1,4 adduct is usually more stable and is said to be the product of thermodynamic control.\n\nAnother reaction unique to conjugated dienes is Diels-Alder cycloaddition. Conjugated dienes react with electron-poor alkenes (dienophiles) in a single step through a cyclic transition state to yield a cyclohexene product. The reaction is stereospecific, meaning that only a single product stereoisomer is formed, and can occur only if the diene is able to adopt an $s$-cis conformation."}
{"id": 905, "contents": "Summary - \nUltraviolet (UV) spectroscopy is a method of structure determination applicable specifically to conjugated $\\pi$-electron systems. When a conjugated molecule is irradiated with ultraviolet light, energy absorption occurs and a $\\pi$ electron is promoted from the highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO). For 1,3-butadiene, radiation of $\\lambda_{\\max }=217 \\mathrm{~nm}$ is required. The greater the extent of conjugation, the less the energy needed and the longer the wavelength of radiation required."}
{"id": 906, "contents": "Summary of Reactions - \n1. Electrophilic addition reactions (Section 14.2 and Section 14.3)\n\n\n\n2. Diels-Alder cycloaddition reaction (Section 14.4 and Section 14.5)"}
{"id": 907, "contents": "Visualizing Chemistry - \nPROBLEM Show the structures of all possible adducts of the following diene with 1 equivalent of HCl :"}
{"id": 908, "contents": "14-16 - \nPROBLEM Show the product of the Diels-Alder reaction of the following diene with 3-buten-2-one,\n14-17 $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCOCH}_{3}$. Make sure you show the full stereochemistry of the reaction product.\n\n\nPROBLEM The following diene does not undergo Diels-Alder reactions. Explain.\n14-18\n\n\nPROBLEM The following model is that of an allylic carbocation intermediate formed by protonation of a\n14-19 conjugated diene with HBr . Show the structure of the diene and the structures of the final reaction products."}
{"id": 909, "contents": "Mechanism Problems - \nPROBLEM Predict the major product(s) from the addition of 1 equivalent of HX and show the mechanism for 14-20 each of the following reactions.\n(a)\n\n(b)\n\n$\\xrightarrow[40^{\\circ} \\mathrm{C}]{\\mathrm{HBr}}$\n(c)\n\n\nPROBLEM We've seen that the Diels-Alder cycloaddition reaction is a one-step, pericyclic process that occurs\n14-21 through a cyclic transition state. Propose a mechanism for the following reaction:\n\n\nPROBLEM In light of your answer to Problem 14-21 propose mechanisms for the following reactions.\n14-22 (a)\n\n(b)\n\n\nPROBLEM Luminol, which is used by forensic scientists to find blood, fluoresces as a result of Diels-Alder-like\n14-23 process. The dianion of luminol reacts with $\\mathrm{O}_{2}$ to form an unstable peroxide intermediate that then loses nitrogen to form a dicarboxylate and emit light. The process is similar to that in Problems 14-21 and 14-22. Propose a mechanism for this process.\n\n\nLuminol dianion\nPROBLEM A useful diene in the synthesis of many naturally occurring substances is known as Danishefsky's\n14-24 diene. It's useful because after the Diels-Alder reaction it can be converted into a product that can't be accessed by a typical Diels-Alder reaction. Show the Diels-Alder adduct and propose a mechanism that accounts for the final products."}
{"id": 910, "contents": "Conjugated Dienes - \nPROBLEM Name the following compounds:\n14-25 (a)\n\n(b)\n$\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}=\\mathrm{CHCH}=\\mathrm{CHCH}_{3}$\n(c) $\\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{C}=\\mathrm{CHCH}=\\mathrm{CHCH}_{3}$\n(d)\n\n\nPROBLEM Draw and name the six possible diene isomers of formula $\\mathrm{C}_{5} \\mathrm{H}_{8}$. Which of the six are conjugated\n14-26 dienes?\nPROBLEM What product(s) would you expect to obtain from reaction of 1,3-cyclohexadiene with each of the\n14-27 following?\n(a) $1 \\mathrm{~mol} \\mathrm{Br}_{2}$ in $\\mathrm{CH}_{2} \\mathrm{Cl}_{2}$\n(b) $\\mathrm{O}_{3}$ followed by Zn\n(c) 1 mol HCl in ether (d) 1 mol DCl in ether\n(e) 3-Buten-2-one $\\left(\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCOCH}_{3}\\right)$ (f) Excess $\\mathrm{OsO}_{4}$, followed by $\\mathrm{NaHSO}_{3}$\n\nPROBLEM Electrophilic addition of $\\mathrm{Br}_{2}$ to isoprene (2-methyl-1,3-butadiene) yields the following product 14-28 mixture:\n\n\nOf the 1,2 -addition products, explain why 3,4-dibromo-3-methyl-1-butene (21\\%) predominates over 3,4-dibromo-2-methyl-1-butene (3\\%).\n\nPROBLEM Propose a structure for a conjugated diene that gives the same product from both 1,2 and\n14-29 1,4-addition of HBr .\nPROBLEM Draw the possible products resulting from addition of 1 equivalent of HCl to\n14-30 1-phenyl-1,3-butadiene. Which would you expect to predominate, and why?"}
{"id": 911, "contents": "Conjugated Dienes - \nDiels-Alder Reactions\nPROBLEM Predict the products of the following Diels-Alder reactions:\n14-31 (a)\n\n(b)\n?\n\n\nPROBLEM 2,3-Di-tert-butyl-1,3-butadiene does not undergo Diels-Alder reactions. Explain.\n14-32"}
{"id": 912, "contents": "2,3-Di-tert-butyl-1,3-butadiene - \nPROBLEM Show the structure, including stereochemistry, of the product from the following Diels-Alder 14-33 reaction:\n\n\nPROBLEM How can you account for the fact that cis-1,3-pentadiene is much less reactive than 14-34 trans-1,3-pentadiene in the Diels-Alder reaction?\n\nPROBLEM Would you expect a conjugated diyne such as 1,3-butadiyne to undergo Diels-Alder reaction with a\n14-35 dienophile? Explain.\nPROBLEM Reaction of isoprene (2-methyl-1,3-butadiene) with ethyl propenoate gives a mixture of two 14-36 Diels-Alder adducts. Show the structure of both, and explain why a mixture is formed.\n\n\nPROBLEM Rank the following dienophiles in order of their expected reactivity in the Diels-Alder reaction.\n14-37\n\n\n\n\n\nPROBLEM 1,3-Cyclopentadiene is very reactive in Diels-Alder cycloaddition reactions, but\n14-38 1,3-cyclohexadiene is less reactive and 1,3-cycloheptadiene is nearly inert. Explain. (Molecular models are helpful.)\n\nPROBLEM 1,3-Pentadiene is much more reactive in Diels-Alder reactions than 2,4-pentadienal. Why might 14-39 this be?"}
{"id": 913, "contents": "1,3-Pentadiene - \n2,4-Pentadienal\n\nPROBLEM How could you use Diels-Alder reactions to prepare the following products? Show the starting diene\n14-40 and dienophile in each case.\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM Show the product of the following reaction.\n14-41"}
{"id": 914, "contents": "Diene Polymers - \nPROBLEM Tires whose sidewalls are made of natural rubber tend to crack and weather rapidly in areas around\n14-42 cities where high levels of ozone and other industrial pollutants are found. Explain.\nPROBLEM 1,3-Cyclopentadiene polymerizes slowly at room temperature to yield a polymer that has no double\n14-43 bonds except on the ends. On heating, the polymer breaks down to regenerate 1,3-cyclopentadiene. Propose a structure for the product."}
{"id": 915, "contents": "UV Spectroscopy - \nPROBLEM Arrange the molecules in each of the following sets according to where you would expect to find\n14-44 their wavelength of maximum absorption in UV spectroscopy, from shortest to longest wavelength.\n(a)\n\n\n\n(b)\n\n(c)\n\n\n\n\nPROBLEM Which of the following compounds would you expect to have a $\\pi \\rightarrow \\pi^{*}$ UV absorption in the 200 to 14-45 400 nm range?\n(a)\n\n(b)\n\n(c) $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{C}=\\mathrm{C}=\\mathrm{O}$\nA ketene\n\nPyridine\nPROBLEM Would you expect allene, $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{C}=\\mathrm{CH}_{2}$, to show a UV absorption in the 200 to 400 nm range?\n14-46 Explain.\nPROBLEM The following ultraviolet absorption maxima have been measured:\n14-47\n\n| 1,3-Butadiene | 217 nm |\n| :--- | :--- |\n| 2-Methyl-1,3-butadiene | 220 nm |\n| 1,3-Pentadiene | 223 nm |\n| 2,3-Dimethyl-1,3-butadiene | 226 nm |\n| 2,4-Hexadiene | 227 nm |\n| 2,4-Dimethyl-1,3-pentadiene | 232 nm |\n| 2,5-Dimethyl-2,4-hexadiene | 240 nm |\n\nWhat conclusion can you draw about the effect of alkyl substitution on UV absorption maxima? Approximately what effect does each added alkyl group have?"}
{"id": 916, "contents": "UV Spectroscopy - \nWhat conclusion can you draw about the effect of alkyl substitution on UV absorption maxima? Approximately what effect does each added alkyl group have?\n\nPROBLEM 1,3,5-Hexatriene has $\\lambda_{\\max }=258 \\mathrm{~nm}$. In light of your answer to Problem 14-47, approximately\n14-48 where would you expect 2,3-dimethyl-1,3,5-hexatriene to absorb?\nPROBLEM $\\beta$-Ocimene is a pleasant-smelling hydrocarbon found in the leaves of certain herbs. It has the\n14-49 molecular formula $\\mathrm{C}_{10} \\mathrm{H}_{16}$ and a UV absorption maximum at 232 nm . On hydrogenation with a palladium catalyst, 2,6-dimethyloctane is obtained. Ozonolysis of $\\beta$-ocimene, followed by treatment with zinc and acetic acid, produces the following four fragments:\n\n\nAcetone\n\n\nFormaldehyde\n\n\nPyruvaldehyde\n\n(a) How many double bonds does $\\beta$-ocimene have?\n(b) Is $\\beta$-ocimene conjugated or nonconjugated? (c) Propose a structure for $\\beta$-ocimene.\n(d) Write the reactions, showing starting material and products."}
{"id": 917, "contents": "General Problems - \nPROBLEM Draw the resonance forms that result when the following dienes are protonated. If the resonance 14-50 forms differ in energy, identify the most stable one.\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Answer the following questions for 1,3,5-cycloheptatriene.\n14-51 (a) How many $p$ atomic orbitals are in the conjugated system?\n(b) How many molecular orbitals describe the conjugated system?\n(c) How many molecular orbitals are bonding molecular orbitals?\n(d) How many molecular orbitals are anti-bonding molecular orbitals?\n(e) Which molecular orbitals are filled with electrons?\n(f) If this molecule were to absorb a photon of UV light an electron would move between which two molecular orbitals (be specific)?\n\nPROBLEM Treatment of 3,4-dibromohexane with strong base leads to loss of 2 equivalents of HBr and\n14-52 formation of a product with formula $\\mathrm{C}_{6} \\mathrm{H}_{10}$. Three products are possible. Name each of the three, and tell how you would use ${ }^{1} \\mathrm{H}$ and ${ }^{13} \\mathrm{C}$ NMR spectroscopy to help identify them. How would you use UV spectroscopy?\n\nPROBLEM Addition of HCl to 1-methoxycyclohexene yields 1-chloro-1-methoxycyclohexane as a sole product.\n14-53 Use resonance structures to explain why none of the other regioisomer is formed.\n\n\nPROBLEM Aldrin, a chlorinated insecticide now banned from use in most countries since 1990, can be made\n14-54 by Diels-Alder reaction of hexachloro-1,3-cyclopentadiene with norbornadiene. What is the structure of aldrin?"}
{"id": 918, "contents": "Norbornadiene - \nPROBLEM Norbornadiene (Problem 14-54) can be prepared by reaction of chloroethylene with\n14-55 1,3-cyclopentadiene, followed by treatment of the product with sodium ethoxide. Write the overall scheme, and identify the two kinds of reactions.\n\nPROBLEM The triene shown here reacts with 2 equivalents of maleic anhydride to yield a product with the\n14-56 formula $\\mathrm{C}_{17} \\mathrm{H}_{16} \\mathrm{O}_{6}$. Predict a structure for the product.\n\n\nPROBLEM Myrcene, $\\mathrm{C}_{10} \\mathrm{H}_{16}$, is found in oil of bay leaves and is isomeric with $\\beta$-ocimene (Problem 14-49).\n14-57 It has an ultraviolet absorption at 226 nm and can be hydrogenated to yield 2,6-dimethyloctane. On ozonolysis followed by zinc/acetic acid treatment, myrcene yields formaldehyde, acetone, and 2-oxopentanedial:\n\n\n2-Oxopentanedial\nPropose a structure for myrcene, and write the reactions, showing starting material and products.\n\nPROBLEM Hydrocarbon $\\mathbf{A}, \\mathrm{C}_{10} \\mathrm{H}_{14}$, has a UV absorption at $\\lambda_{\\max }=236 \\mathrm{~nm}$ and gives hydrocarbon $\\mathbf{B}, \\mathrm{C}_{10} \\mathrm{H}_{18}$, on\n14-58 hydrogenation. Ozonolysis of $\\mathbf{A}$, followed by zinc/acetic acid treatment, yields the following diketo dialdehyde:\n\n\nAn illustration shows the structure of diketo dialdehyde. It shows two carbonyl groups single bonded to each other. Each carbonyl group is bonded to a chain of three methylene groups and then to an aldehyde group.\n(a) Propose two possible structures for $\\mathbf{A}$.\n(b) Hydrocarbon $\\mathbf{A}$ reacts with maleic anhydride to yield a Diels-Alder adduct. Which of your structures for A is correct?\n(c) Write the reactions, showing the starting material and products."}
{"id": 919, "contents": "Norbornadiene - \nPROBLEM Adiponitrile, a starting material used in the manufacture of nylon, can be prepared in three steps\n14-59 from 1,3-butadiene. How would you carry out this synthesis?\n$\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}=\\mathrm{CH}_{2} \\quad \\xrightarrow{3 \\text { steps }} \\mathrm{N} \\equiv \\mathrm{CCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{C} \\equiv \\mathrm{N}$\nAdiponitrile\nPROBLEM Ergosterol, a precursor of vitamin D, has $\\lambda_{\\max }=282 \\mathrm{~nm}$ and molar absorptivity $\\epsilon=11,900$. What is\n14-60 the concentration of ergosterol in a solution whose absorbance $A=0.065$ with a sample pathlength $l=1.00 \\mathrm{~cm}$ ?\n\n\nPROBLEM Dimethyl butynedioate undergoes a Diels-Alder reaction with ( $2 E, 4 E$ )-2,4-hexadiene. Show the\n14-61 structure and stereochemistry of the product.\n\n\nDimethyl butynedioate\nPROBLEM Dimethyl butynedioate also undergoes a Diels-Alder reaction with (2E,4Z)-2,4-hexadiene, but the\n14-62 stereochemistry of the product is different from that of the ( $2 E, 4 E$ ) isomer (Problem 14-61). Explain.\n\nPROBLEM How would you carry out the following synthesis (more than one step is required)? What\n14-63 stereochemical relationship between the $-\\mathrm{CO}_{2} \\mathrm{CH}_{3}$ group attached to the cyclohexane ring and the -CHO groups would your synthesis produce?\n\n\nPROBLEM The double bond of an enamine (alkene + amine) is much more nucleophilic than a typical alkene\n14-64 double bond. Assuming that the nitrogen atom in an enamine is $s p^{2}$-hybridized, draw an orbital picture of an enamine, and explain why the double bond is electron-rich."}
{"id": 920, "contents": "An enamine - \nPROBLEM Benzene has an ultraviolet absorption at $\\lambda_{\\max }=204 \\mathrm{~nm}$, and para-toluidine has $\\lambda_{\\max }=235 \\mathrm{~nm}$.\n14-65 How do you account for this difference?\n\n\nBenzene\n( $\\lambda_{\\text {max }}=204 \\mathrm{~nm}$ )\n\n$p$-Toluidine\n( $\\lambda_{\\text {max }}=\\mathbf{2 3 5} \\mathbf{n m}$ )"}
{"id": 921, "contents": "CHAPTER 15 - \nBenzene and Aromaticity\n\n\nFIGURE 15.1 A fennel plant is an aromatic herb used in cooking. A phenyl group-pronounced exactly the same way-is the characteristic structural unit of \"aromatic\" organic compounds. (credit: modificatio of work \"Fennel\" by Ruth Hartnup/Flickr, CC BY 2.0)"}
{"id": 922, "contents": "CHAPTER CONTENTS - 15.1 Naming Aromatic Compounds\n15.2 Structure and Stability of Benzene\n15.3 Aromaticity and the H\u00fcckel $4 n+2$ Rule\n15.4 Aromatic Ions\n15.5 Aromatic Heterocycles: Pyridine and Pyrrole\n15.6 Polycyclic Aromatic Compounds\n15.7 Spectroscopy of Aromatic Compounds\n\nWHY THIS CHAPTER? The reactivity of substituted aromatic compounds, more than that of any other class of substances, is intimately tied to their exact structure. As a result, aromatic compounds provide an extraordinarily sensitive probe for studying the relationship between structure and reactivity. We'll examine this relationship here and in the following chapter, and we'll find that the lessons learned are applicable to all other organic compounds, including important substances such as the nucleic acids that control our genetic makeup.\n\nIn the early days of organic chemistry, the word aromatic was used to describe such fragrant substances as benzaldehyde (from cherries, peaches, and almonds), toluene (from Tolu balsam), and benzene (from coal distillate). It was soon realized, however, that substances grouped as aromatic differed from most other organic compounds in their chemical behavior.\n\n\nBenzene\n\n\nBenzaldehyde\n\n\nToluene\n\nToday, the association of aromaticity with fragrance has long been lost, and we now use the word aromatic to refer to the class of compounds that contain six-membered rings with three alternating double bonds. Many naturally occurring compounds are aromatic in part, including steroids such as estrone and well-known pharmaceuticals such as the cholesterol-lowering drug atorvastatin, marketed as Lipitor. Benzene itself causes a depressed white blood cell count (leukopenia) on prolonged exposure and should not be used as a laboratory solvent.\n\n\nEstrone\n\n\nAtorvastatin (Lipitor)"}
{"id": 923, "contents": "CHAPTER CONTENTS - 15.1 Naming Aromatic Compounds\nSimple aromatic hydrocarbons come from two main sources: coal and petroleum. Coal is an enormously complex mixture consisting primarily of large arrays of benzene-like rings joined together. Thermal breakdown of coal occurs when heated to $1000^{\\circ} \\mathrm{C}$ in the absence of air, and a mixture of volatile products called coal tar boils off. Fractional distillation of coal tar yields benzene, toluene, xylene (dimethylbenzene), naphthalene, and a host of other aromatic compounds (FIGURE 15.2).\n\n\nBenzene (bp $80^{\\circ} \\mathrm{C}$ )\n\n\nToluene (bp $111{ }^{\\circ} \\mathrm{C}$ )\n\n\nXylene (bp: ortho, $144{ }^{\\circ} \\mathrm{C}$; meta, $139{ }^{\\circ} \\mathrm{C}$; para, $138^{\\circ} \\mathrm{C}$ )\n\n\nIndene (bp $182{ }^{\\circ} \\mathrm{C}$ )\n\n\nPhenanthrene (mp $101{ }^{\\circ} \\mathrm{C}$ )\nFIGURE 15.2 Some aromatic hydrocarbons found in coal tar.\n\nUnlike coal, petroleum contains few aromatic compounds and consists largely of alkanes (see Chapter 3 Chemistry Matters). During petroleum refining, however, aromatic molecules are formed when alkanes are passed over a catalyst at about $500^{\\circ} \\mathrm{C}$ under high pressure.\n\nAromatic substances, more than any other class of organic compounds, have acquired a large number of nonsystematic names. IUPAC rules discourage the use of most such names but do allow some of the more widely used ones to be retained (TABLE 15.1). Thus, methylbenzene is known commonly as toluene; hydroxybenzene as phenol; aminobenzene as aniline; and so on."}
{"id": 924, "contents": "CHAPTER CONTENTS - 15.1 Naming Aromatic Compounds\n| TABLE 15.1 Common Aromatic Compounds | Names of Some |\n| :---: | :---: |\n| Structure | Name |\n| <smiles>Cc1ccccc1</smiles> | Toluene (bp $111^{\\circ} \\mathrm{C}$ ) |\n| <smiles>Oc1ccccc1</smiles> | Phenol $\\left(\\mathrm{mp} 43^{\\circ} \\mathrm{C}\\right)$ |\n| <smiles>Nc1ccccc1</smiles> | Aniline $\\left(\\mathrm{bp} 184^{\\circ} \\mathrm{C}\\right)$ |\n| <smiles>CC(=O)c1ccccc1</smiles> | Acetophenone (mp $21^{\\circ} \\mathrm{C}$ ) |\n| <smiles>O=Cc1ccccc1</smiles> | Benzaldehyde (bp $178{ }^{\\circ} \\mathrm{C}$ ) |\n| <smiles>O=C(O)c1ccccc1</smiles> | Benzoic acid (mp $122^{\\circ} \\mathrm{C}$ ) |\n| <smiles>Cc1ccccc1C</smiles> | ortho-Xylene (bp $144{ }^{\\circ} \\mathrm{C}$ ) |\n| <smiles>C=Cc1ccccc1</smiles> | Styrene <br> (bp $145^{\\circ} \\mathrm{C}$ ) |\n\nMonosubstituted benzenes are named systematically in the same manner as other hydrocarbons, with -benzene as the parent name. Thus, $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{Br}$ is bromobenzene, $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NO}_{2}$ is nitrobenzene, and $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$ is propylbenzene.\n\n\nBromobenzene\n\n\nNitrobenzene\n\n\nPropylbenzene"}
{"id": 925, "contents": "CHAPTER CONTENTS - 15.1 Naming Aromatic Compounds\nBromobenzene\n\n\nNitrobenzene\n\n\nPropylbenzene\n\nAlkyl-substituted benzenes are sometimes referred to as arenes and are named in different ways depending on the size of the alkyl group. If the alkyl substituent is smaller than the ring (six or fewer carbons), the arene is referred to as an alkyl-substituted benzene. If the alkyl substituent is larger than the ring (seven or more carbons), the compound is referred to as a phenyl-substituted alkane. The name phenyl, pronounced fen-nil and sometimes abbreviated as Ph or $\\Phi$ (Greek phi), is used for the $-\\mathrm{C}_{6} \\mathrm{H}_{5}$ unit when the benzene ring\nis considered a substituent. The word is derived from the Greek pheno (\"I bear light\"), commemorating the discovery of benzene by Michael Faraday in 1825 from the oily residue left by the illuminating gas used in London street lamps. In addition, the name benzyl is used for the $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CH}_{2}$ - group.\n\n\nDisubstituted benzenes are named using the prefixes ortho (o), meta (m), or para (p). An ortho-disubstituted benzene has its two substituents in a 1,2 relationship on the ring, a meta-disubstituted benzene has its two substituents in a 1,3 relationship, and a para-disubstituted benzene has its substituents in a 1,4 relationship.\n\northo-Dichlorobenzene 1,2 disubstituted\n\nmeta-Dimethylbenzene (meta-xylene) 1,3 disubstituted\n\npara-Chlorobenzaldehyde\n1,4 disubstituted\n\nThe ortho, meta, para system of nomenclature is also useful when discussing reactions. For example, we might describe the reaction of bromine with toluene by saying, \"Reaction occurs at the para position,\" meaning at the position para to the methyl group already present on the ring."}
{"id": 926, "contents": "CHAPTER CONTENTS - 15.1 Naming Aromatic Compounds\nToluene $\\quad \\boldsymbol{p}$-Bromotoluene\nAs with cycloalkanes (Section 4.1), benzenes with more than two substituents are named by choosing a point of attachment as carbon 1 and numbering the substituents on the ring so that the second substituent has as low a number as possible. If ambiguity still exists, number so that the third or fourth substituent has as low a number as possible, until a point of difference is found. The substituents are listed alphabetically when writing the name.\n\n\nNote in the second and third examples shown that -phenol and -toluene are used as the parent names rather than -benzene. Any of the monosubstituted aromatic compounds shown in TABLE 15.1 can serve as a parent name, with the principal substituent ( -OH in phenol or $-\\mathrm{CH}_{3}$ in toluene) attached to C 1 on the ring.\n\nPROBLEM Tell whether the following compounds are ortho-, meta-, or para-disubstituted:\n15-1 (a)\n\n(b)\n\n(c)\n\n\nPROBLEM Give IUPAC names for the following compounds:\n15-2 (a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n\nPROBLEM Draw structures corresponding to the following IUPAC names:\n15-3 (a) $p$-Bromochlorobenzene (b) $p$-Bromotoluene (c) $m$-Chloroaniline\n(d) 1-Chloro-3,5-dimethylbenzene"}
{"id": 927, "contents": "CHAPTER CONTENTS - 15.2 Structure and Stability of Benzene\nBenzene $\\left(\\mathrm{C}_{6} \\mathrm{H}_{6}\\right)$ has six fewer hydrogens than the six-carbon cycloalkane cyclohexane $\\left(\\mathrm{C}_{6} \\mathrm{H}_{12}\\right)$ and is clearly unsaturated, usually being represented as a six-membered ring with alternating double and single bonds. Yet it has been known since the mid-1800s that benzene is much less reactive than typical alkenes and fails to undergo typical alkene addition reactions. Cyclohexene, for instance, reacts rapidly with $\\mathrm{Br}_{2}$ and gives the addition product 1,2-dibromocyclohexane, but benzene reacts slowly with $\\mathrm{Br}_{2}$ and gives the substitution product $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{Br}$ rather than the addition product.\n\n\nWe can get a quantitative idea of benzene's stability by measuring heats of hydrogenation (Section 7.6). Cyclohexene, an isolated alkene, has $\\Delta H_{\\text {hydrog }}^{\\circ}=-118 \\mathrm{~kJ} / \\mathrm{mol}(-28.2 \\mathrm{kcal} / \\mathrm{mol})$, and 1,3-cyclohexadiene, a conjugated diene, has $\\Delta H^{\\circ}{ }_{\\text {hydrog }}=-230 \\mathrm{~kJ} / \\mathrm{mol}(-55.0 \\mathrm{kcal} / \\mathrm{mol})$. As noted in Section 14.1, this value for 1,3-cyclohexadiene is a bit less than twice that for cyclohexene because conjugated dienes are more stable than isolated dienes."}
{"id": 928, "contents": "CHAPTER CONTENTS - 15.2 Structure and Stability of Benzene\nCarrying the process one step further, we might expect $\\Delta H^{\\circ}{ }_{\\text {hydrog }}$ for \"cyclohexatriene\" (benzene) to be a bit less than $3 \\times 118=354 \\mathrm{~kJ} / \\mathrm{mol}$, or three times the cyclohexene value. The actual value, however, is $-206 \\mathrm{~kJ} /$ mol , some $148 \\mathrm{~kJ} / \\mathrm{mol}(35.4 \\mathrm{kcal} / \\mathrm{mol})$ less than expected. Because of this difference in actual and expected energy released during hydrogenation, benzene must have $148 \\mathrm{~kJ} / \\mathrm{mol}$ less energy to begin with. In other words, benzene is more stable than expected by $148 \\mathrm{~kJ} / \\mathrm{mol}$ (FIGURE 15.3).\n\n\nFIGURE 15.3 A comparison of the heats of hydrogenation for cyclohexene, 1,3-cyclohexadiene, and benzene. Benzene is $148 \\mathrm{~kJ} / \\mathrm{mol}$ $(35.4 \\mathrm{kcal} / \\mathrm{mol})$ more stable than might be expected for \"cyclohexatriene.\"\n\nFurther evidence for the unusual nature of benzene is that all its carbon-carbon bonds have the same length -139 pm , which is intermediate between typical single ( 154 pm ) and double ( 134 pm ) bonds. In addition, an electrostatic potential map shows that the electron density in all six $\\mathrm{C}-\\mathrm{C}$ bonds is identical. Thus, benzene is a planar molecule with the shape of a regular hexagon. All $\\mathrm{C}-\\mathrm{C}-\\mathrm{C}$ bond angles are $120^{\\circ}$, all six carbon atoms are $s p^{2}$-hybridized, and each carbon has a $p$ orbital perpendicular to the plane of the six-membered ring."}
{"id": 929, "contents": "CHAPTER CONTENTS - 15.2 Structure and Stability of Benzene\nBecause all six carbon atoms and all six $p$ orbitals in benzene are equivalent, it's impossible to define three localized $\\pi$ bonds in which a given $p$ orbital overlaps only one neighboring $p$ orbital. Rather, each $p$ orbital overlaps equally well with both neighboring $p$ orbitals, leading to a picture of benzene in which all six $\\pi$ electrons are free to move about the entire ring (FIGURE 15.4b). In resonance terms (Section 2.4 and Section 2.5), benzene is a hybrid of two equivalent forms. Neither form is correct by itself; the true structure of benzene is somewhere in between the two resonance forms but is impossible to draw with our usual conventions. Because of this resonance, benzene is more stable and less reactive than a typical alkene.\n\n\n(b)\n\n\nFIGURE 15.4 (a) An electrostatic potential map of benzene and (b) an orbital picture. Each of the six carbon atoms has a $p$ orbital that can overlap equally well with neighboring $p$ orbitals on both sides. As a result, all $\\mathrm{C}-\\mathrm{C}$ bonds are equivalent and benzene must be represented as a hybrid of two resonance forms.\n\nChemists sometimes represent the two benzene resonance forms by using a circle to indicate the equivalence of the carbon-carbon bonds, but this representation has to be used carefully because it doesn't indicate the number of $\\pi$ electrons in the ring. (How many electrons does a circle represent?) In this book, benzene and\nother aromatic compounds will be represented by single line-bond structures. We'll be able to keep count of $\\pi$ electrons this way but must be aware of the limitations of the drawings.\n\n\nHaving just seen a resonance description of benzene, let's now look at the alternative molecular orbital description. We can construct $\\pi$ molecular orbitals for benzene just as we did for 1,3-butadiene in Section 14.1. If six $p$ atomic orbitals combine in a cyclic manner, six benzene molecular orbitals result, as shown in FIGURE 15.5. The three low-energy molecular orbitals, denoted $\\psi_{1}, \\psi_{2}$, and $\\psi_{3}$, are bonding combinations, and the three high-energy orbitals are antibonding."}
{"id": 930, "contents": "CHAPTER CONTENTS - 15.2 Structure and Stability of Benzene\nSix benzene molecular orbitals\nFIGURE 15.5 The six benzene $\\boldsymbol{\\pi}$ molecular orbitals. The bonding orbitals $\\psi_{2}$ and $\\psi_{3}$ have the same energy and are said to be degenerate, as are the antibonding orbitals $\\psi_{4}{ }^{*}$ and $\\psi_{5}{ }^{*}$. The orbitals $\\psi_{3}$ and $\\psi_{4}{ }^{*}$ have no $\\pi$ electron density on two carbons because of a node passing through these atoms.\n\nNote that the two bonding orbitals $\\psi_{2}$ and $\\psi_{3}$ have the same energy, as do the two antibonding orbitals $\\psi_{4}$ * and $\\psi_{5}{ }^{*}$. Such orbitals with the same energy are said to be degenerate. Note also that the two orbitals $\\psi_{3}$ and $\\psi_{4} *$ have nodes passing through ring carbon atoms, thereby leaving no $\\pi$ electron density on these carbons. The six $p$ electrons of benzene occupy the three bonding molecular orbitals and are delocalized over the entire conjugated system, leading to the observed $150 \\mathrm{~kJ} / \\mathrm{mol}$ stabilization of benzene.\n\nPROBLEM Pyridine is a flat, hexagonal molecule with bond angles of $120^{\\circ}$. It undergoes substitution rather\n15-4 than addition and generally behaves like benzene. Draw a picture of the $\\pi$ orbitals of pyridine to explain its properties. Check your answer by looking ahead to Section 15.5.\n\n\nPyridine"}
{"id": 931, "contents": "CHAPTER CONTENTS - 15.3 Aromaticity and the H\u00fcckel $4 n+2$ Rule\nLet's list what we've said thus far about benzene and, by extension, about other benzene-like aromatic molecules.\n\n- Benzene is cyclic and conjugated.\n- Benzene is unusually stable, having a heat of hydrogenation $150 \\mathrm{~kJ} / \\mathrm{mol}$ less negative than we might expect for a conjugated cyclic triene.\n- Benzene is planar and has the shape of a regular hexagon. All bond angles are $120^{\\circ}$, all carbon atoms are $s p^{2}$-hybridized, and all carbon-carbon bond lengths are 139 pm .\n- Benzene undergoes substitution reactions that retain the cyclic conjugation rather than electrophilic addition reactions that would destroy it.\n- Benzene can be described as a resonance hybrid whose structure is intermediate between two line-bond structures.\n\nThis list might seem to be a good description of benzene and other aromatic molecules, but it isn't enough. Something else, called the H\u00fcckel $4 \\boldsymbol{n}+2$ rule, is needed to complete a description of aromaticity. According to a theory devised in 1931 by the German physicist Erich H\u00fcckel, a molecule is aromatic only if it has a planar, monocyclic system of conjugation and contains a total of $4 n+2 \\pi$ electrons, where $n$ is an integer ( $n=0,1$, 2,3 ,). In other words, only molecules with $2,6,10,14,18, \\pi$ electrons can be aromatic. Molecules with $4 n \\pi$ electrons ( $4,8,12,16$,) can't be aromatic, even though they may be cyclic, planar, and apparently conjugated. In fact, planar, conjugated molecules with $4 n \\pi$ electrons are said to be antiaromatic because delocalization of their $\\pi$ electrons would lead to their destabilization.\n\nLet's look at several examples to see how the H\u00fcckel $4 n+2$ rule works.\n\n- Cyclobutadiene has four $\\pi$ electrons and is antiaromatic. As indicated by the electrostatic potential map, the $\\pi$ electrons are localized in two double bonds rather than delocalized around the ring."}
{"id": 932, "contents": "CHAPTER CONTENTS - 15.3 Aromaticity and the H\u00fcckel $4 n+2$ Rule\n- Cyclobutadiene has four $\\pi$ electrons and is antiaromatic. As indicated by the electrostatic potential map, the $\\pi$ electrons are localized in two double bonds rather than delocalized around the ring.\n\n\nCyclobutadiene is highly reactive and shows none of the properties associated with aromaticity. In fact, it was not even prepared until 1965, when Rowland Pettit at the University of Texas was able to make it at low temperature. Even at $-78^{\\circ} \\mathrm{C}$, however, cyclobutadiene is so reactive that it dimerizes by a Diels-Alder reaction. One molecule behaves as a diene and the other as a dienophile.\n\n\n- Benzene has six $\\pi$ electrons $(4 n+2=6$ when $n=1)$ and is aromatic."}
{"id": 933, "contents": "Benzene - \nThree double bonds; six $\\pi$ electrons\n\n- Cyclooctatetraene has eight $\\pi$ electrons, but when it was first prepared in 1911 by the German chemist Richard Willst\u00e4tter, it was found not to be particularly stable. In fact, its $\\pi$ electrons are localized into four double bonds rather than delocalized around the ring, and the molecule is tub-shaped rather than planar. It has no cyclic conjugation because neighboring $p$ orbitals don't have the necessary parallel alignment for overlap, and it resembles an open-chain polyene in its reactivity.\n\n\n\nX-ray studies show that the $\\mathrm{C}-\\mathrm{C}$ single bonds in cyclooctatetraene are 147 pm long and the double bonds are 134 pm long. In addition, the ${ }^{1} \\mathrm{H}$ NMR spectrum shows a single sharp resonance line at $5.78 \\mathrm{\\delta}$, a value characteristic of an alkene rather than an aromatic molecule.\n\nWhat's so special about $4 n+2 \\pi$ electrons? Why do $2,6,10,14 \\pi$ electrons lead to aromatic stability, while other numbers of electrons do not? The answer comes from molecular orbital theory. When the energy levels of molecular orbitals for cyclic conjugated molecules are calculated, it turns out that there is always a single lowest-lying MO, above which the MOs come in degenerate pairs. Thus, when electrons fill the various molecular orbitals, it takes two electrons, or one pair, to fill the lowest-lying orbital and four electrons, or two pairs, to fill each of $n$ succeeding energy levels-a total of $4 n+2$. Any other number would leave an energy level partially filled.\n\nThe six $\\pi$ molecular orbitals of benzene were shown previously in FIGURE 15.5, and their relative energies are shown again in FIGURE 15.6. The lowest-energy MO, $\\psi_{1}$, occurs singly and contains two electrons. The next two lowest-energy orbitals, $\\psi_{2}$ and $\\psi_{3}$, are degenerate, and it therefore takes four electrons to fill both. The result is a stable six- $\\pi$-electron aromatic molecule with filled bonding orbitals."}
{"id": 934, "contents": "Benzene - \nFIGURE 15.6 Energy levels of the six benzene $\\boldsymbol{\\pi}$ molecular orbitals. There is a single, lowest-energy orbital, above which the orbitals come in degenerate pairs.\n\nPROBLEM To be aromatic, a molecule must have $4 n+2 \\pi$ electrons and must have a planar, monocyclic system 15-5 of conjugation. Cyclodecapentaene fulfills one of these criteria but not the other and has resisted all attempts at synthesis. Explain."}
{"id": 935, "contents": "Benzene - 15.4 Aromatic lons\nAccording to the H\u00fcckel criteria for aromaticity, a molecule must be cyclic, conjugated (nearly planar with a $p$ orbital on each atom), and have $4 n+2 \\pi$ electrons. Nothing in this definition says that the number of $\\pi$ electrons must be the same as the number of atoms in the ring or that the substance must be neutral. In fact, the numbers can differ and the substance can be an ion. Thus, both the cyclopentadienyl anion and the cycloheptatrienyl cation are aromatic even though both are ions and neither contains a six-membered ring.\n\n\nCyclopentadienyl anion\n\n\nCycloheptatrienyl cation\n\nSix $\\pi$ electrons; aromatic ions\nTo understand why the cyclopentadienyl anion and the cycloheptatrienyl cation are aromatic, imagine starting from the related neutral hydrocarbons, 1,3-cyclopentadiene and 1,3,5-cycloheptatriene, and removing one hydrogen from the saturated $\\mathrm{CH}_{2}$ carbon in each. If that carbon then rehybridizes from $s p^{3}$ to $s p^{2}$, the resultant products would be fully conjugated, with a $p$ orbital on every carbon. There are three ways in which the hydrogen might be removed.\n\n- The hydrogen can be removed with both electrons ( $\\mathrm{H}:^{-}$) from the $\\mathrm{C}-\\mathrm{H}$ bond, leaving a carbocation as product.\n- The hydrogen can be removed with one electron (H\u2022) from the $\\mathrm{C}-\\mathrm{H}$ bond, leaving a carbon radical as product.\n- The hydrogen can be removed with no electrons $\\left(\\mathrm{H}^{+}\\right)$from the $\\mathrm{C}-\\mathrm{H}$ bond, leaving a carbanion as product."}
{"id": 936, "contents": "Benzene - 15.4 Aromatic lons\nAll the potential products formed by removing a hydrogen from 1,3-cyclopentadiene and from 1,3,5-cycloheptatriene can be drawn with numerous resonance structures, but H\u00fcckel's rule predicts that only the six- $\\pi$-electron cyclopentadienyl anion and cycloheptatrienyl cation can be aromatic. The other products are predicted by the $4 n+2$ rule to be unstable and antiaromatic (FIGURE 15.7).\n\n\n1,3-Cyclopentadiene\n\n\nCyclopentadienyl cation\n(four $\\pi$ electrons)\nor\n\n\nCyclopentadienyl radical\n(five $\\pi$ electrons)\nor\n\n\nCyclopentadienyl anion (six $\\pi$ electrons)\n\n\n1,3,5-Cycloheptatriene\n\n\n\nCycloheptatrienyl cation\n(six $\\pi$ electrons)\nor\n\n\nCycloheptatrienyl radical (seven $\\pi$ electrons)\nor\n\n\nCycloheptatrienyl anion\n(eight $\\pi$ electrons)\n\nFIGURE 15.7 The aromatic six- $\\pi$-electron cyclopentadienyl anion and the six- $\\pi$-electron cycloheptatrienyl cation. The anion can be formed by removing a hydrogen ion $\\left(\\mathrm{H}^{+}\\right)$from the $\\mathrm{CH}_{2}$ group of 1,3-cyclopentadiene. The cation can be generated by removing a hydride ion $\\left(\\mathrm{H}:^{-}\\right)$from the $\\mathrm{CH}_{2}$ group of 1,3,5-cycloheptatriene."}
{"id": 937, "contents": "Benzene - 15.4 Aromatic lons\nIn practice, both the four- $\\pi$-electron cyclopentadienyl cation and the five- $\\pi$-electron cyclopentadienyl radical are highly reactive and difficult to prepare. Neither shows any sign of the stability expected for an aromatic system. The six- $\\pi$-electron cyclopentadienyl anion, by contrast, is easily prepared and remarkably stable (FIGURE 15.8a). In fact, the anion is so stable and easily formed that 1,3 -cyclopentadiene is one of the most acidic hydrocarbons known, with $\\mathrm{p} K_{\\mathrm{a}}=16$, a value comparable to that of water!\n\nIn the same way, the seven- $\\pi$-electron cycloheptatrienyl radical and eight- $\\pi$-electron anion are reactive and difficult to prepare, while the six- $\\pi$-electron cycloheptatrienyl cation is extraordinarily stable (FIGURE 15.8b).\n\nIn fact, the cycloheptatrienyl cation was first prepared more than a century ago by reaction of 1,3,5-cycloheptatriene with $\\mathrm{Br}_{2}$, although its structure was not recognized at the time.\n(a)\n\n$+\\mathrm{H}_{2} \\mathrm{O}$\n\n1,3-Cyclo-\nCyclopentadienyl anion\n\n\nAromatic cyclopentadienyl anion with six $\\pi$ electrons\n(b)\n\n\n\nCycloheptatrienyl cation six $\\pi$ electrons\n\nFIGURE 15.8 (a) The aromatic cyclopentadienyl anion, showing cyclic conjugation and six $\\boldsymbol{\\pi}$ electrons in five $\\boldsymbol{p}$ orbitals, and (b) the aromatic cycloheptatrienyl cation, showing cyclic conjugation and six $\\boldsymbol{\\pi}$ electrons in seven $\\boldsymbol{p}$ orbitals. Electrostatic potential maps indicate that both ions are symmetrical, with the charge equally shared among all atoms in each ring."}
{"id": 938, "contents": "Benzene - 15.4 Aromatic lons\nPROBLEM Draw the five resonance structures of the cyclopentadienyl anion. Are all carbon-carbon bonds\n15-6 equivalent? How many absorption lines would you expect to see in the ${ }^{1} \\mathrm{H}$ NMR and ${ }^{13} \\mathrm{C}$ NMR spectra of the anion?\n\nPROBLEM Cyclooctatetraene readily reacts with potassium metal to form the stable cyclooctatetraene dianion,\n15-7 $\\mathrm{C}_{8} \\mathrm{H}_{8}{ }^{2-}$. Why do you suppose this reaction occurs so easily? What geometry do you expect for the cyclooctatetraene dianion?\n\n\nPROBLEM The relative energy levels of the five $\\pi$ molecular orbitals of the cyclopentadienyl system are similar\n15-8 to those in benzene. That is, there is a single lowest-energy MO, above which the orbitals come in degenerate pairs. Draw a diagram like that in Figure 15.6, and tell which of the five orbitals are occupied in the cation, radical, and anion."}
{"id": 939, "contents": "Benzene - 15.5 Aromatic Heterocycles: Pyridine and Pyrrole\nLook back once again at the definition of aromaticity in Section 15.3: a cyclic, conjugated molecule containing $4 n+2 \\pi$ electrons. Nothing in this definition says that the atoms in the ring must be carbon. In fact, heterocyclic compounds can also be aromatic. A heterocycle is a cyclic compound that contains atoms of more than one element in its ring, usually carbon along with nitrogen, oxygen, or sulfur. Pyridine and pyrimidine, for example, are six-membered heterocycles with carbon and nitrogen in their rings (FIGURE 15.9).\n\n\nPyridine\n\n\nPyrimidine\n\n\n\nFIGURE 15.9 Pyridine and pyrimidine are nitrogen-containing aromatic heterocycles with $\\pi$ electron arrangements like that of benzene. Both have a lone pair of electrons on nitrogen in an $s p^{2}$ orbital in the plane of the ring.\nPyridine is much like benzene in its $\\pi$ electron structure. Each of the five $s p^{2}$-hybridized carbons has a $p$ orbital perpendicular to the plane of the ring, and each $p$ orbital contains one $\\pi$ electron. The nitrogen atom is also $s p^{2}$-hybridized and has one electron in a $p$ orbital, bringing the total to six $\\pi$ electrons. The nitrogen lone-pair electrons (red in an electrostatic potential map) are in a $s p^{2}$ orbital in the plane of the ring and are not part of the aromatic $\\pi$ system. Pyrimidine, also shown in FIGURE 15.9), is a benzene analog that has two nitrogen atoms in a six-membered, unsaturated ring. Both nitrogens are $s p^{2}$-hybridized, and each contributes one electron to the aromatic $\\pi$ system."}
{"id": 940, "contents": "Benzene - 15.5 Aromatic Heterocycles: Pyridine and Pyrrole\nPyrrole (spelled with two r's and one $I$ ) and imidazole are five-membered heterocycles, yet both have six $\\pi$ electrons and are aromatic. In pyrrole, each of the four $s p^{2}$-hybridized carbons contributes one $\\pi$ electron and the $s p^{2}$-hybridized nitrogen atom contributes the two from its lone pair, which occupies a $p$ orbital (FIGURE 15.10). Imidazole, also shown in (FIGURE 15.10, is an analog of pyrrole that has two nitrogen atoms in a fivemembered, unsaturated ring. Both nitrogens are $s p^{2}$-hybridized, but one is in a double bond and contributes only one electron to the aromatic $\\pi$ system whereas the other is not in a double bond and contributes two from its lone pair.\n\n\nPyrrole\n\n\nImidazole\n\n(Six $\\pi$ electrons)\n\n\nFIGURE 15.10 Pyrrole and imidazole are five-membered, nitrogen-containing heterocycles but have six- $\\pi$-electron arrangements like that of the cyclopentadienyl anion. Both have a lone pair of electrons on nitrogen in a $p$ orbital perpendicular to the ring.\n\nNote that nitrogen atoms have different roles depending on the structure of the molecule. The nitrogen atoms in pyridine and pyrimidine are both in double bonds and contribute only one $\\pi$ electron to the aromatic sextet, just as a carbon atom in benzene does. The nitrogen atom in pyrrole, however, is not in a double bond and contributes two $\\pi$ electrons (its lone pair) to the aromatic sextet. In imidazole, both kinds of nitrogen are present in the same molecule-a double-bonded \"pyridine-like\" nitrogen that contributes one $\\pi$ electron and a \"pyrrolelike\" nitrogen that contributes two.\n\nPyrimidine and imidazole rings are particularly important in biological chemistry. Pyrimidine, for instance, is the parent ring system in cytosine, thymine, and uracil, three of the five heterocyclic amine bases found in nucleic acids. An aromatic imidazole ring is present in histidine, one of the 20 amino acids found in proteins."}
{"id": 941, "contents": "Benzene - 15.5 Aromatic Heterocycles: Pyridine and Pyrrole\nCytosine (in DNA and RNA)\n\n\nThymine (in DNA)\n\n\nUracil (in RNA)\n\n\nHistidine (an amino acid)"}
{"id": 942, "contents": "Accounting for the Aromaticity of a Heterocycle - \nThiophene, a sulfur-containing heterocycle, undergoes typical aromatic substitution reactions rather than addition reactions. Why is thiophene aromatic?\n\n\nThiophene"}
{"id": 943, "contents": "Strategy - \nRecall the requirements for aromaticity-a planar, cyclic, conjugated molecule with $4 n+2 \\pi$ electrons-and see how these requirements apply to thiophene."}
{"id": 944, "contents": "Solution - \nThiophene is the sulfur analog of pyrrole. The sulfur atom is $s p^{2}$-hybridized and has a lone pair of electrons in a $p$ orbital perpendicular to the plane of the ring. Sulfur also has a second lone pair of electrons in the ring plane.\n\n\nThiophene\n\nPROBLEM Draw an orbital picture of furan to show how the molecule is aromatic.\n15-9\n\n\nFuran\n\nPROBLEM Thiamin, or vitamin $B_{1}$, contains a positively charged five-membered nitrogen-sulfur heterocycle\n15-10 called a thiazolium ring. Explain why the thiazolium ring is aromatic.\n\n\nThiamin\n\nThiazolium ring"}
{"id": 945, "contents": "Solution - 15.6 Polycyclic Aromatic Compounds\nThe H\u00fcckel rule is strictly applicable only to monocyclic compounds, but the general concept of aromaticity can be extended to include polycyclic aromatic compounds. Naphthalene, with two benzene-like rings fused together; anthracene, with three rings; benzo[a]pyrene, with five rings; and coronene, with six rings, are all well-known aromatic hydrocarbons. Benzo[a]pyrene is particularly interesting because it is one of the cancercausing substances found in tobacco smoke.\n\nNaphthalene\n\nAnthracene\n\nBenzo[a]pyrene\n\nCoronene\n\nAll polycyclic aromatic hydrocarbons can be represented by a number of different resonance forms. Naphthalene, for instance, has three.\n\n\nNaphthalene\nNaphthalene and other polycyclic aromatic hydrocarbons show many of the chemical properties associated with aromaticity. Thus, measurement of its heat of hydrogenation shows an aromatic stabilization energy of approximately $250 \\mathrm{~kJ} / \\mathrm{mol}$ ( $60 \\mathrm{kcal} / \\mathrm{mol}$ ). Furthermore, naphthalene reacts slowly with electrophiles such as $\\mathrm{Br}_{2}$ to give substitution products rather than double-bond addition products.\n\n\nThe aromaticity of naphthalene is explained by the orbital picture in FIGURE 15.11. Naphthalene has a cyclic, conjugated $\\pi$ electron system, with $p$ orbital overlap both along the ten-carbon periphery of the molecule and across the central bond. Because ten $\\pi$ electrons is a H\u00fcckel number, there is $\\pi$ electron delocalization and consequent aromaticity in naphthalene.\n\nNaphthalene\n\n\nFIGURE 15.11 An orbital picture and electrostatic potential map of naphthalene, showing that the ten $\\boldsymbol{\\pi}$ electrons are fully delocalized"}
{"id": 946, "contents": "throughout both rings. - \nJust as there are heterocyclic analogs of benzene, there are also many heterocyclic analogs of naphthalene. Among the most common are quinoline, isoquinoline, indole, and purine. Quinoline, isoquinoline, and purine all contain pyridine-like nitrogens that are part of a double bond and contribute one electron to the aromatic $\\pi$ system. Indole and purine both contain pyrrole-like nitrogens that contribute two $\\pi$ electrons.\n\n\nQuinoline\n\n\nIsoquinoline\n\n\nIndole\n\n\nPurine\n\nAmong the many biological molecules that contain polycyclic aromatic rings, the amino acid tryptophan contains an indole ring and the antimalarial drug quinine contains a quinoline ring. Adenine and guanine, two of the five heterocyclic amine bases found in nucleic acids, have rings based on purine.\n\n\nTryptophan (an amino acid)\n\n\nAdenine (in DNA and RNA)\n\n\nGuanine (in DNA and RNA)\n\nQuinine (an antimalarial agent)\n\nPROBLEM Azulene, a beautiful blue hydrocarbon, is an isomer of naphthalene. Is azulene aromatic? Draw a\n15-11 second resonance form of azulene in addition to that shown.\n\n\nAzulene\n\nPROBLEM How many electrons does each of the four nitrogen atoms in purine contribute to the aromatic $\\pi$\n15-12 system?\n\n\nPurine"}
{"id": 947, "contents": "Infrared Spectroscopy - \nAs we saw in the brief introduction to infrared spectroscopy (Section 12.8), aromatic rings show a characteristic $\\mathrm{C}-\\mathrm{H}$ stretching absorption at $3030 \\mathrm{~cm}^{-1}$ and a series of peaks in the 1450 to $1600 \\mathrm{~cm}^{-1}$ range of the infrared spectrum. The aromatic $\\mathrm{C}-\\mathrm{H}$ band at $3030 \\mathrm{~cm}^{-1}$ generally has low intensity and occurs just to the left of a typical saturated $\\mathrm{C}-\\mathrm{H}$ band.\n\nAs many as four absorptions are sometimes observed in the 1450 to $1600 \\mathrm{~cm}^{-1}$ region because of the complex molecular motions of the ring itself. Two bands, one at $1500 \\mathrm{~cm}^{-1}$ and one at $1600 \\mathrm{~cm}^{-1}$, are usually the most intense. In addition, aromatic compounds show weak absorptions in the 1660 to $2000 \\mathrm{~cm}^{-1}$ region and strong absorptions in the 690 to $900 \\mathrm{~cm}^{-1}$ range due to $\\mathrm{C}-\\mathrm{H}$ out-of-plane bending. The exact position of both sets of absorptions is diagnostic of the substitution pattern of the aromatic ring (FIGURE 12.24 in Section 12.8)."}
{"id": 948, "contents": "Infrared Spectroscopy - \n| Monosubstituted: | $690-710 \\mathrm{~cm}^{-1}$ | $\\mathbf{1 , 2 , 4 - T r i s u b s t i t u t e d :}$ | $780-830 \\mathrm{~cm}^{-1}$ |\n| :--- | :--- | :--- | :--- |\n|  | $730-770 \\mathrm{~cm}^{-1}$ |  | $870-900 \\mathrm{~cm}^{-1}$ |\n| o-Disubstituted: | $735-770 \\mathrm{~cm}^{-1}$ | $\\mathbf{1 , 2 , 3 - T r i s u b s t i t u t e d :}$ | $670-720 \\mathrm{~cm}^{-1}$ |\n| m-Disubstituted: | $690-710 \\mathrm{~cm}^{-1}$ |  | $750-790 \\mathrm{~cm}^{-1}$ |\n|  | $810-850 \\mathrm{~cm}^{-1}$ | $\\mathbf{1 , 3 , 5 - T r i s u b s t i t u t e d :}$ | $660-700 \\mathrm{~cm}^{-1}$ |\n| p-Disubstituted: | $810-840 \\mathrm{~cm}^{-1}$ |  | $830-900 \\mathrm{~cm}^{-1}$ |\n\nThe IR spectrum of toluene in FIGURE 15.12 shows these characteristic absorptions."}
{"id": 949, "contents": "Ultraviolet Spectroscopy - \nAromatic rings are detectable by ultraviolet spectroscopy because they contain a conjugated $\\pi$ electron system. In general, aromatic compounds show a series of bands, with a fairly intense absorption near 205 nm and a less intense absorption in the 255 to 275 nm range. The presence of these bands in the ultraviolet spectrum of a molecule is a sure indication of an aromatic ring. FIGURE 15.13 shows the ultraviolet spectrum of benzene.\n\n\nFIGURE 15.13 Ultraviolet spectrum of benzene. There are primary bands at 184 and 202 nm and secondary (fine-structure) bands at 255 nm ."}
{"id": 950, "contents": "Nuclear Magnetic Resonance Spectroscopy - \nAromatic hydrogens are strongly deshielded by the ring and absorb between 6.5 and $8.0 \\delta$. The spins of nonequivalent aromatic protons on substituted rings often couple with each other, giving rise to spin-spin splitting patterns that can identify the substitution of the ring.\n\nMuch of the difference in chemical shift between aromatic protons (6.5-8.0 $\\delta$ ) and vinylic protons (4.5-6.5 $\\delta$ ) is due to a property of aromatic rings called ring-current. When an aromatic ring is oriented perpendicular to a strong magnetic field, the electrons circulate around the ring, producing a small local magnetic field. This induced field opposes the applied field in the middle of the ring but reinforces the applied field outside the ring (FIGURE 15.14). Aromatic protons therefore experience an effective magnetic field greater than the applied field and come into resonance at a lower applied field.\n\n\nFIGURE 15.14 The origin of aromatic ring-current. Aromatic protons are deshielded by the induced magnetic field caused by delocalized $\\pi$ electrons circulating around the aromatic ring.\nNote that the aromatic ring-current produces different effects inside and outside the ring. If a ring were large enough to have both \"inside\" and \"outside\" protons, the protons on the outside would be deshielded and absorb\nat a field lower than normal but those on the inside would be shielded and absorb at a field higher than normal. This prediction has been strikingly confirmed by studies on [18]annulene, an $18-\\pi$-electron cyclic conjugated polyene that contains a H\u00fcckel number of electrons $(4 n+2=18$ when $n=4)$. The six inside protons of [18]annulene are strongly shielded by the aromatic ring-current and absorb at $-3.0 \\delta$ (that is, 3.0 ppm upfield from TMS, off the normal chart), while the 12 outside protons are strongly deshielded and absorb in the typical aromatic region at 9.3 ppm downfield from TMS.\n\n\n[18]Annulene\n\n$$\n\\text { Inside } \\mathrm{H}:-3.0 \\delta\n$$"}
{"id": 951, "contents": "Nuclear Magnetic Resonance Spectroscopy - \n[18]Annulene\n\n$$\n\\text { Inside } \\mathrm{H}:-3.0 \\delta\n$$\n\nOutside H: $9.3 \\delta$\nThe presence of a ring-current is characteristic of all H\u00fcckel aromatic molecules and is a good test of aromaticity. For example, benzene, a six- $\\pi$-electron aromatic molecule, absorbs at $7.37 \\delta$ because of its ringcurrent, but cyclooctatetraene, an eight- $\\pi$-electron nonaromatic molecule, absorbs at $5.78 \\delta$.\n\nHydrogens on carbon next to aromatic rings-benzylic hydrogens-also show distinctive absorptions in the NMR spectrum. Benzylic protons normally absorb downfield from other alkane protons in the region from 2.3 to 3.0 $\\delta$.\n\n\nThe ${ }^{1} \\mathrm{H}$ NMR spectrum of $p$-bromotoluene, shown in FIGURE 15.15, displays many of the features just discussed. The aromatic protons appear as two doublets at 7.04 and $7.37 \\mathrm{\\delta}$, and the benzylic methyl protons absorb as a sharp singlet at $2.26 \\delta$. Integration of the spectrum shows the expected $2: 2: 3$ ratio of peak areas.\n\n\nFIGURE 15.15 The ${ }^{\\mathbf{1}} \\mathrm{H}$ NMR spectrum of $\\boldsymbol{p}$-bromotoluene.\nThe carbon atoms in an aromatic ring typically absorb in the range 110 to $140 \\delta$ in the ${ }^{13} \\mathrm{C}$ NMR spectrum, as\nindicated by the examples in FIGURE 15.16. These resonances are easily distinguished from those of alkane carbons but occur in the same range as alkene carbons. Thus, the presence of ${ }^{13} \\mathrm{C}$ absorptions at 110 to $140 \\delta$ does not by itself establish the presence of an aromatic ring. Supporting evidence from infrared, ultraviolet, or ${ }^{1}$ H NMR spectra is needed."}
{"id": 952, "contents": "Nuclear Magnetic Resonance Spectroscopy - \nBenzene\nToluene\nChlorobenzene\nNaphthalene\nFIGURE 15.16 Some ${ }^{13}$ C NMR absorptions of aromatic compounds ( $\\delta$ units)."}
{"id": 953, "contents": "Aspirin, NSAIDs, and COX-2 Inhibitors - \nWhatever the cause-whether tennis elbow, a sprained ankle, or a wrenched knee-pain and inflammation seem to go together. They are, however, different in their origin, and powerful drugs are available for treating each separately. Codeine, for example, is a powerful analgesic, or pain reliever, used in the management of debilitating pain, while cortisone and related steroids are potent anti-inflammatory agents, used for treating arthritis and other crippling inflammations. For minor pains and inflammation, both problems are often treated together by using a common over-the-counter medication called an NSAID, or nonsteroidal antiinflammatory drug.\n\nOne of the most common NSAIDs is aspirin, or acetylsalicylic acid, whose use goes back to the late 1800s. It had been known from before the time of Hippocrates in 400 BC that fevers could be lowered by chewing the bark of willow trees. The active agent in willow bark was found in 1827 to be an aromatic compound called salicin, which could be converted by reaction with water into salicyl alcohol and then oxidized to give salicylic acid. Salicylic acid turned out to be even more effective than salicin for reducing fevers and to have analgesic and anti-inflammatory action as well. Unfortunately, it also turned out to be too corrosive to the walls of the stomach for everyday use. Conversion of the phenol-OH group into an acetate ester, however, yielded acetylsalicylic acid, which proved just as potent as salicylic acid but less corrosive to the stomach.\n\n\nFIGURE 15.17 Many athletes rely on NSAIDs to help with pain and soreness. (credit: \"Dillon Gee delivers a pitch\" by Arturo Pardavila III/ Wikimedia Commons, CC BY 2.0)"}
{"id": 954, "contents": "Aspirin, NSAIDs, and COX-2 Inhibitors - \nAcetylsalicylic acid\n(aspirin)\nAlthough extraordinary in its effect, aspirin is also more dangerous than commonly believed. 15 g of aspirin would be fatal to a child of 9-10 years, and 10 g would be lethal for a 6-year-old, and aspirin can cause stomach bleeding and allergic reactions in long-term users. Even more serious is a condition called Reye's syndrome, a potentially fatal reaction to aspirin sometimes seen in children recovering from the flu. As a result of these problems, numerous other NSAIDs have been developed in the last several decades, most notably ibuprofen and naproxen.\n\nLike aspirin, both ibuprofen and naproxen are relatively simple aromatic compounds containing a side-chain carboxylic acid group. Ibuprofen, sold under the brand names Advil, Nuprin, Motrin, and others, has roughly the same potency as aspirin but is less prone to causing an upset stomach. Naproxen, sold under the names Aleve and Naprosyn, also has about the same potency as aspirin but remains active in the body six times longer.\n\n\nIbuprofen\n(Advil, Nuprin, Motrin)\n\n\nNaproxen\n(Aleve, Naprosyn)\n\nAspirin and other NSAIDs function by blocking the cyclooxygenase (COX) enzymes that carry out the body's synthesis of prostaglandins (Section 8.11 and Section 27.4). There are two forms of the enzyme: COX-1, which carries out the normal physiological production of prostaglandins, and COX-2, which mediates the body's response to arthritis and other inflammatory conditions. Unfortunately, both COX-1 and COX-2 enzymes are blocked by aspirin, ibuprofen, and other NSAIDs, thereby shutting down not only the response to inflammation but also various protective functions, including the control mechanism for production of acid in the stomach."}
{"id": 955, "contents": "Aspirin, NSAIDs, and COX-2 Inhibitors - \nMedicinal chemists have therefore devised a number of drugs that act as selective inhibitors of only the COX-2 enzyme. Inflammation is thereby controlled without blocking protective functions. Originally heralded as a breakthrough in arthritis treatment, the first generation of COX-2 inhibitors, including rofecoxib, (marketed as Vioxx), celecoxib, (marketed as Celebrex), and Valdecoxib (marketed as Bextra), turned out to cause potentially serious heart problems, particularly in elderly or compromised patients. As a result, all three Cox-2 inhibitors were removed from the U.S. and other markets. Further studies, however, concluded that in proper doses the Cox-2 inhibitors were little different from other NSAIDs, and Vioxx returned to the market in 2017 for topical treatment of osteoarthritis.\n\n\nRofecoxib\n(Vioxx)"}
{"id": 956, "contents": "Key Terms - \n- antiaromatic\n- H\u00fcckel $4 n+2$ rule\n- arene\n- meta (m)\n- aromatic\n- ortho (o)\n- benzyl group\n- para (p)\n- degenerate orbitals\n- phenyl group\n- heterocycle\n- ring-current"}
{"id": 957, "contents": "Summary - \nAromatic rings are a common part of many biological structures and are particularly important in nucleic acid chemistry and in the chemistry of several amino acids. In this chapter, we've seen how and why aromatic compounds are different from such apparently related compounds as cycloalkenes.\n\nThe word aromatic is used for historical reasons to refer to the class of compounds related structurally to benzene. Aromatic compounds are systematically named according to IUPAC rules, but many common names\nare also used. Disubstituted benzenes are referred to as ortho ( 1,2 disubstituted), meta ( 1,3 disubstituted), or para (1,4 disubstituted) derivatives. The $\\mathrm{C}_{6} \\mathrm{H}_{5}$ - unit itself is referred to as a phenyl group, and the $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CH}_{2}-$ unit is a benzyl group.\n\nBenzene is described by valence-bond theory as a resonance hybrid of two equivalent structures and is described by molecular orbital theory as a planar, cyclic, conjugated molecule with six $\\pi$ electrons. According to the H\u00fcckel rule, a molecule must have $4 \\boldsymbol{n}+2 \\boldsymbol{\\pi}$ electrons, where $n=0,1,2,3$, and so on, to be aromatic. Planar, cyclic, conjugated molecules with other numbers of $\\pi$ electrons are antiaromatic.\n\nOther substances besides benzene-like compounds are also aromatic. The cyclopentadienyl anion and the cycloheptatrienyl cation, for instance, are aromatic ions. Pyridine and pyrimidine are six-membered, nitrogencontaining, aromatic heterocycles. Pyrrole and imidazole are five-membered, nitrogen-containing heterocycles. Naphthalene, quinoline, indole, and many others are polycyclic aromatic compounds.\n\nAromatic compounds have the following characteristics:"}
{"id": 958, "contents": "Summary - \nAromatic compounds have the following characteristics:\n\n- Aromatic compounds are cyclic, planar, and conjugated.\n- Aromatic compounds are unusually stable. Benzene, for instance, has a heat of hydrogenation $150 \\mathrm{~kJ} / \\mathrm{mol}$ less than we might expect for a cyclic triene.\n- Aromatic compounds react with electrophiles to give substitution products, in which cyclic conjugation is retained, rather than addition products, in which conjugation is destroyed.\n- Aromatic compounds have $4 n+2 \\pi$ electrons, which are delocalized over the ring."}
{"id": 959, "contents": "Visualizing Chemistry - \nPROBLEM Give IUPAC names for the following substances (red $=0$, blue $=\\mathrm{N}$ ):"}
{"id": 960, "contents": "15-13 (a) - \n(b)\n\n\nPROBLEM All-cis cyclodecapentaene is a stable molecule that shows a single absorption in its ${ }^{1} \\mathrm{H}$ NMR 15-14 spectrum at $5.67 \\delta$. Tell whether it is aromatic, and explain its NMR spectrum.\n\n\nPROBLEM 1,6-Methanonaphthalene has an interesting ${ }^{1} \\mathrm{H}$ NMR spectrum in which the eight hydrogens\n15-15 around the perimeter absorb at 6.9 to $7.3 \\delta$, while the two $\\mathrm{CH}_{2}$ protons absorb at $-0.5 \\delta$. Tell whether it is aromatic, and explain its NMR spectrum."}
{"id": 961, "contents": "1,6-Methanonaphthalene - \nPROBLEM The following molecular model is that of a carbocation. Draw two resonance structures for the\n15-16 carbocation, indicating the positions of the double bonds.\n\n\nPROBLEM Azulene, an isomer of naphthalene, has a remarkably large dipole moment for a hydrocarbon ( $\\mu=$\n15-17 1.0 D). Explain, using resonance structures."}
{"id": 962, "contents": "Naming Aromatic Compounds - \nPROBLEM Give IUPAC names for the following compounds:"}
{"id": 963, "contents": "15-18 (a) - \n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n\nPROBLEM Draw structures corresponding to the following names:\n15-19 (a) 3-Methyl-1,2-benzenediamine (b) 1,3,5-Benzenetriol (c) 3-Methyl-2-phenylhexane\n(d) $o$-Aminobenzoic acid (e) $m$-Bromophenol (f) 2,4,6-Trinitrophenol (picric acid)\n\nPROBLEM Draw and name all possible isomers of the following:\n15-20 (a) Dinitrobenzene (b) Bromodimethylbenzene\n(c) Trinitrophenol\n\nPROBLEM Draw and name all possible aromatic compounds with the formula $\\mathrm{C}_{7} \\mathrm{H}_{7} \\mathrm{Cl}$.\n15-21\nPROBLEM Draw and name all possible aromatic compounds with the formula $\\mathrm{C}_{8} \\mathrm{H}_{9} \\mathrm{Br}$. (There are 14.) 15-22"}
{"id": 964, "contents": "Structure of Aromatic Compounds - \nPROBLEM Propose structures for aromatic hydrocarbons that meet the following descriptions:\n15-23 (a) $\\mathrm{C}_{9} \\mathrm{H}_{12}$; gives only one $\\mathrm{C}_{9} \\mathrm{H}_{11} \\mathrm{Br}$ product on substitution of a hydrogen on the aromatic ring with bromine\n(b) $\\mathrm{C}_{10} \\mathrm{H}_{14}$; gives only one $\\mathrm{C}_{10} \\mathrm{H}_{13} \\mathrm{Cl}$ product on substitution of a hydrogen on the aromatic ring with chlorine\n(c) $\\mathrm{C}_{8} \\mathrm{H}_{10}$; gives three $\\mathrm{C}_{8} \\mathrm{H}_{9} \\mathrm{Br}$ products on substitution of a hydrogen on the aromatic ring with bromine\n(d) $\\mathrm{C}_{10} \\mathrm{H}_{14}$; gives two $\\mathrm{C}_{10} \\mathrm{H}_{13} \\mathrm{Cl}$ products on substitution of a hydrogen on the aromatic ring with chlorine\n\nPROBLEM Look at the three resonance structures of naphthalene shown in Section 15.6, and account for the\n15-24 fact that not all carbon-carbon bonds have the same length. The C1-C2 bond is 136 pm long, whereas the $\\mathrm{C} 2-\\mathrm{C} 3$ bond is 139 pm long.\n\nPROBLEM Anthracene has four resonance structures, one of which is shown. Draw the other three.\n15-25\n\n\nAnthracene\n\nPROBLEM Phenanthrene has five resonance structures, one of which is shown. Draw the other four.\n15-26\n\n\nPhenanthrene\n\nPROBLEM Look at the five resonance structures for phenanthrene (Problem 26), and predict which of its\n15-27 carbon-carbon bonds is shortest.\nPROBLEM In 1932, A. A. Levine and A. G. Cole studied the ozonolysis of $o$-xylene and isolated three products:\n15-28 glyoxal, 2,3-butanedione, and pyruvaldehyde:"}
{"id": 965, "contents": "Structure of Aromatic Compounds - \nIn what ratio would you expect the three products to be formed if $o$-xylene is a resonance hybrid of two structures? The actual ratio found was 3 parts glyoxal, 1 part 2,3-butanedione, and 2 parts pyruvaldehyde. What conclusions can you draw about the structure of $o$-xylene?"}
{"id": 966, "contents": "Aromaticity and H\u00fcckel's Rule - \nPROBLEM 3-Chlorocyclopropene, on treatment with $\\mathrm{AgBF}_{4}$, gives a precipitate of AgCl and a stable solution\n15-29 of a product that shows a single ${ }^{1} \\mathrm{H}$ NMR absorption at $11.04 \\delta$. What is a likely structure for the product, and what is its relation to H\u00fcckel's rule?"}
{"id": 967, "contents": "3-Chlorocyclopropene - \nPROBLEM Draw an energy diagram for the three molecular orbitals of the cyclopropenyl system $\\left(\\mathrm{C}_{3} \\mathrm{H}_{3}\\right)$. How\n15-30 are these three molecular orbitals occupied in the cyclopropenyl anion, cation, and radical? Which of the three substances is aromatic according to H\u00fcckel's rule?\n\nPROBLEM Cyclopropanone is highly reactive because of its large amount of angle strain.\n15-31 Methylcyclopropenone, although even more strained than cyclopropanone, is nevertheless quite stable and can even be distilled. Explain, taking the polarity of the carbonyl group into account.\n\n\n\nCyclopropanone"}
{"id": 968, "contents": "Methylcyclopropenone - \nPROBLEM Cycloheptatrienone is stable, but cyclopentadienone is so reactive that it can't be isolated. Explain,\n15-32 taking the polarity of the carbonyl group into account.\n\n\nCycloheptatrienone\n\n\nCyclopentadienone\n\nPROBLEM Which would you expect to be most stable, cyclononatetraenyl radical, cation, or anion? 15-33 (Cyclononatetraene has a ring of nine carbons and four double bonds.)\n\nPROBLEM How might you convert 1,3,5,7-cyclononatetraene to an aromatic substance?\n15-34\nPROBLEM Calicene, like azulene (Problem 17), has an unusually large dipole moment for a hydrocarbon.\n15-35 Explain, using resonance structures."}
{"id": 969, "contents": "Calicene - \nPROBLEM Pentalene is a most elusive molecule that has been isolated only at liquid-nitrogen temperature.\n15-36 The pentalene dianion, however, is well known and quite stable. Explain.\n\n\nPentalene\n\n\nPentalene dianion\n\nPROBLEM Indole is an aromatic heterocycle that has a benzene ring fused to a pyrrole ring. Draw an orbital 15-37 picture of indole.\n(a) How many $\\pi$ electrons does indole have?\n(b) What is the electronic relationship of indole to naphthalene?\n\n\nIndole\n\nPROBLEM Ribavirin, an antiviral agent used against hepatitis C and viral pneumonia, contains a 1,2,4-triazole 15-38 ring. Why is the ring aromatic?"}
{"id": 970, "contents": "Spectroscopy - \nPROBLEM Compound $\\mathbf{A}, \\mathrm{C}_{8} \\mathrm{H}_{10}$, yields three substitution products, $\\mathrm{C}_{8} \\mathrm{H}_{9} \\mathrm{Br}$, on reaction with $\\mathrm{Br}_{2}$. Propose two\n15-39 possible structures for $\\mathbf{A}$. The ${ }^{1} \\mathrm{H}$ NMR spectrum of $\\mathbf{A}$ shows a complex four-proton multiplet at 7.0 $\\delta$ and a six-proton singlet at $2.30 \\delta$. What is the structure of $\\mathbf{A}$ ?\n\nPROBLEM What is the structure of a hydrocarbon that has $\\mathrm{M}^{+}=120$ in its mass spectrum and has the following 15-40 ${ }^{1}$ H NMR spectrum?\n$7.25 \\delta(5 \\mathrm{H}$, broad singlet); $2.90 \\delta(1 \\mathrm{H}$, septet, $J=7 \\mathrm{~Hz}) ; 1.22 \\delta(6 \\mathrm{H}$, doublet, $J=7 \\mathrm{~Hz})$"}
{"id": 971, "contents": "Spectroscopy - \nPROBLEM Propose structures for compounds that fit the following descriptions:\n15-41 (a) $\\mathrm{C}_{10} \\mathrm{H}_{14}$\n${ }^{1} \\mathrm{H}$ NMR: $7.18 \\delta(4 \\mathrm{H}$, broad singlet); $2.70 \\delta(4 \\mathrm{H}$, quartet, $J=7 \\mathrm{~Hz}) ; 1.20 \\delta(6 \\mathrm{H}$, triplet, $J=7 \\mathrm{~Hz})$ IR absorption at $745 \\mathrm{~cm}^{-1}$\n(b) $\\mathrm{C}_{10} \\mathrm{H}_{14}$\n${ }^{1}$ H NMR: $7.0 \\delta(4 \\mathrm{H}$, broad singlet); $2.85 \\delta(1 \\mathrm{H}$, septet, $J=8 \\mathrm{~Hz}) ; 2.28 \\delta(3 \\mathrm{H}$, singlet $) ; 1.20 \\delta(6$ H , doublet, $J=8 \\mathrm{~Hz}$ )\n\nIR absorption at $825 \\mathrm{~cm}^{-1}$"}
{"id": 972, "contents": "General Problems - \nPROBLEM On reaction with acid, 4-pyrone is protonated on the carbonyl-group oxygen to give a stable cationic 15-42 product. Using resonance structures and the H\u00fcckel $4 n+2$ rule, explain why the protonated product is so stable.\n\n\n4-Pyrone\nPROBLEM Bextra, a COX-2 inhibitor once used in the treatment of arthritis, contains an isoxazole ring. Why is $\\mathbf{1 5 - 4 3}$ the ring aromatic?"}
{"id": 973, "contents": "Bextra - \nPROBLEM $N$-Phenylsydnone, so-named because it was first studied at the University of Sydney, Australia, 15-44 behaves like a typical aromatic molecule. Explain, using the H\u00fcckel $4 n+2$ rule.\n\n$N$-Phenylsydnone\nPROBLEM Show the relative energy levels of the seven $\\pi$ molecular orbitals of the cycloheptatrienyl system.\n15-45 Tell which of the seven orbitals are filled in the cation, radical, and anion, and account for the aromaticity of the cycloheptatrienyl cation.\n\nPROBLEM 1-Phenyl-2-butene has an ultraviolet absorption at $\\lambda_{\\max }=208 \\mathrm{~nm}(\\varepsilon=8000)$. On treatment with a\n15-46 small amount of strong acid, isomerization occurs and a new substance with $\\lambda_{\\max }=250 \\mathrm{~nm}(\\varepsilon=$ 15,800 ) is formed. Propose a structure for this isomer, and suggest a mechanism for its formation.\n\nPROBLEM Propose structures for aromatic compounds that have the following ${ }^{1} \\mathrm{H}$ NMR spectra:\n15-47 (a) $\\mathrm{C}_{8} \\mathrm{H}_{9} \\mathrm{Br}$\nIR absorption at $820 \\mathrm{~cm}^{-1}$\n\n(b) $\\mathrm{C}_{9} \\mathrm{H}_{12}$\n\nIR absorption at $750 \\mathrm{~cm}^{-1}$\n\n(c) $\\mathrm{C}_{11} \\mathrm{H}_{16}$\n\nIR absorption at $820 \\mathrm{~cm}^{-1}$\n\n\nPROBLEM Propose a structure for a molecule $\\mathrm{C}_{14} \\mathrm{H}_{12}$ that has the following ${ }^{1} \\mathrm{H}$ NMR spectrum and has IR 15-48 absorptions at 700, 740 , and $890 \\mathrm{~cm}^{-1}$ :"}
{"id": 974, "contents": "Bextra - \nPROBLEM The proton NMR spectrum for a compound with formula $\\mathrm{C}_{10} \\mathrm{H}_{12} \\mathrm{O}_{2}$ is shown. The infrared spectrum\n15-49 has a strong band at $1711 \\mathrm{~cm}^{-1}$. The normal carbon-13 NMR spectral results are tabulated along with the DEPT-135 and DEPT-90 information. Draw the structure of this compound.\n\n\nPROBLEM The proton NMR spectrum of a compound with formula $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NCl}_{2}$ is shown. The normal carbon-13 15-50 and DEPT experimental results are tabulated. The infrared spectrum shows peaks at 3432 and 3313 $\\mathrm{cm}^{-1}$ and a series of medium-sized peaks between 1618 and $1466 \\mathrm{~cm}^{-1}$. Draw the structure of this compound.\n\n| Normal Carbon | DEPT-135 | DEPT-90 |\n| :--- | :--- | :--- |\n| 118.0 ppm | Positive | Positive |\n| 119.5 | No peak | No peak |\n| 128.0 | Positive | Positive |\n| 140.0 | No peak | No peak |\n\n\n\nPROBLEM Aromatic substitution reactions occur by addition of an electrophile such as $\\mathrm{Br}^{+}$to an aromatic\n15-51 ring to yield an allylic carbocation intermediate, followed by loss of $\\mathrm{H}^{+}$. Show the structure of the intermediate formed by reaction of benzene with $\\mathrm{Br}^{+}$.\n\nPROBLEM The substitution reaction of toluene with $\\mathrm{Br}_{2}$ can, in principle, lead to the formation of three\n15-52 isomeric bromotoluene products. In practice, however, only $o$ - and $p$-bromotoluene are formed in substantial amounts. The meta isomer is not formed. Draw the structures of the three possible carbocation intermediates (Problem 51), and explain why ortho and para products predominate over meta products.\n\nPROBLEM Look at the following aromatic anions and their linear counterparts, and draw all of the resonance\n15-53 forms for each. What patterns emerge?\n(a)\n\n\n(b)"}
{"id": 975, "contents": "Bextra - \nPROBLEM Look at the following aromatic anions and their linear counterparts, and draw all of the resonance\n15-53 forms for each. What patterns emerge?\n(a)\n\n\n(b)\n\n\n\nPROBLEM After the following reaction, the chemical shift of $\\mathrm{H}_{\\mathrm{a}}$ moves downfield from 6.98 ppm to 7.30 ppm .\n15-54 Explain.\n\n\nPROBLEM The following compound is the product initially formed in a Claisen rearrangement, which we'll see\n15-55 in Section 18.4. This product is not isolated, but tautomerizes to its enol form. Give the structure of the enol and provide an explanation as to why the enol tautomer is favored.\n\n\nPROBLEM Compounds called azo dyes are the major source of artificial color in textiles and food. Part of the\n15-56 reason for their intense coloring is the conjugation from an electron-donating group through the diazo bridge ( $-\\mathrm{N}=\\mathrm{N}-$ ) to an electron-withdrawing group on the other side. For the following azo dyes, draw a resonance form that shows how the electron-donating group is related to the electronwithdrawing group on the other side of the diazo bridge. Used curved arrows to show how the electrons are reorganized.\n(a)\n\nMethyl Orange\n(b)\nC.I. Acid Red 74"}
{"id": 976, "contents": "CHAPTER 16 - \nChemistry of Benzene: Electrophilic Aromatic Substitution\n\n\nFIGURE 16.1 In the 19th and early-20th centuries, benzene was used as an aftershave lotion because of its pleasant smell and as a solvent to decaffeinate coffee beans. Neither is a good idea. (credit: modification of work \"Ye Old Way (https://loc.getarchive.net/media/ ye-old-way-geo-h-walker-and-co-lith-boston-mass-1)\" by Geo. H. Walker \\& Co./Library of Congress)"}
{"id": 977, "contents": "CHAPTER CONTENTS - 16.1 Electrophilic Aromatic Substitution Reactions: Bromination\n16.2 Other Aromatic Substitutions\n16.3 Alkylation and Acylation of Aromatic Rings: The Friedel-Crafts Reaction\n16.4 Substituent Effects in Electrophilic Substitutions\n16.5 Trisubstituted Benzenes: Additivity of Effects\n16.6 Nucleophilic Aromatic Substitution\n16.7 Benzyne\n16.8 Oxidation of Aromatic Compounds\n16.9 Reduction of Aromatic Compounds\n16.10 Synthesis of Polysubstituted Benzenes\n\nWHY THIS CHAPTER? This chapter continues the coverage of aromatic molecules begun in the preceding chapter, but we'll shift focus to concentrate on reactions, looking at the relationship between aromatic structure and reactivity. This relationship is critical in understanding how many biological molecules and pharmaceutical agents are synthesized and why they behave as they do.\n\nIn the preceding chapter, we looked at aromaticity-the stability associated with benzene and related compounds that contain a cyclic conjugated system of $4 n+2 \\pi$ electrons. In this chapter, we'll look at some of the unique reactions that aromatic molecules undergo.\n\nThe most common reaction of aromatic compounds is electrophilic aromatic substitution, in which an electrophile ( $E^{+}$) reacts with an aromatic ring and substitutes for one of the hydrogens. The reaction is characteristic of all aromatic rings, not just benzene and substituted benzenes. In fact, the ability of a compound to undergo electrophilic substitution is a good test of aromaticity.\n\n\nMany different substituents can be introduced onto an aromatic ring through electrophilic substitution. To list some possibilities, an aromatic ring can be substituted by a halogen $(-\\mathrm{Cl},-\\mathrm{Br},-\\mathrm{I})$, a nitro group $\\left(-\\mathrm{NO}_{2}\\right)$, a sulfonic acid group $\\left(-\\mathrm{SO}_{3} \\mathrm{H}\\right)$, a hydroxyl group $(-\\mathrm{OH})$, an alkyl group ( -R ), or an acyl group ( -COR ). Starting from only a few simple materials, it's possible to prepare many thousands of substituted aromatic compounds.\n\n\nHalogenation\n\n\nNitration\n\n\n\n\nAcylation\n\n\nAlkylation\n\n\nSulfonation\n\nHydroxylation"}
{"id": 978, "contents": "CHAPTER CONTENTS - 16.1 Electrophilic Aromatic Substitution Reactions: Bromination\nBefore seeing how electrophilic aromatic substitutions occur, let's briefly recall what we said in the chapter on Alkenes: Structure and Reactivity about electrophilic alkene additions. When a reagent such as HCl adds to an alkene, the electrophilic hydrogen ion approaches the $\\pi$ electrons of the double bond and forms a bond to one carbon, leaving a positive charge at the other carbon. This carbocation intermediate then reacts with the nucleophilic $\\mathrm{Cl}^{-}$ion to yield the addition product.\n\n\nAn electrophilic aromatic substitution reaction begins in a similar way, but there are a number of differences. One difference is that aromatic rings are less reactive toward electrophiles than alkenes. For example, $\\mathrm{Br}_{2}$ in $\\mathrm{CH}_{2} \\mathrm{Cl}_{2}$ solution reacts instantly with most alkenes but does not react with benzene at room temperature. For bromination of benzene to take place, a catalyst such as $\\mathrm{FeBr}_{3}$ is needed. The catalyst makes the $\\mathrm{Br}_{2}$ molecule more electrophilic by polarizing it to give a $\\mathrm{FeBr}_{4}{ }^{-} \\mathrm{Br}^{+}$species that reacts as if it were $\\mathrm{Br}^{+}$. The polarized $\\mathrm{Br}_{2}$ molecule then reacts with the nucleophilic benzene ring to yield a nonaromatic carbocation intermediate that is doubly allylic (Section 11.5) and has three resonance forms.\n\n\nAlthough more stable than a typical alkyl carbocation because of resonance, the intermediate in electrophilic aromatic substitution is nevertheless much less stable than the starting benzene ring itself, with its $150 \\mathrm{~kJ} / \\mathrm{mol}$ ( $36 \\mathrm{kcal} / \\mathrm{mol}$ ) of aromatic stability. Thus, the reaction of an electrophile with a benzene ring is endergonic, has a substantial activation energy, and is rather slow. FIGURE 16.2 shows an energy diagram comparing the reaction of an electrophile with an alkene and with benzene. The benzene reaction is slower (higher $\\Delta G^{+}$) because the starting material is more stable."}
{"id": 979, "contents": "CHAPTER CONTENTS - 16.1 Electrophilic Aromatic Substitution Reactions: Bromination\nReaction progress\nFIGURE 16.2 A comparison of the reactions of an electrophile ( $E^{+}$) with an alkene and with benzene: $\\Delta G^{\\ddagger}$ alkene $<\\Delta G^{\\ddagger}$ benzene. The benzene reaction is slower than the alkene reaction because of the stability of the aromatic ring.\nAnother difference between alkene addition and aromatic substitution occurs after the carbocation intermediate has formed. Instead of adding $\\mathrm{Br}^{-}$to give an addition product, the carbocation intermediate loses $\\mathrm{H}^{+}$from the bromine-bearing carbon to give a substitution product. Note that this loss of $\\mathrm{H}^{+}$is similar to what occurs in the second step of an E1 reaction (Section 11.10). The net effect of reaction of $\\mathrm{Br}_{2}$ with benzene is the substitution of $\\mathrm{H}^{+}$by $\\mathrm{Br}^{+}$by the overall mechanism shown in FIGURE 16.3."}
{"id": 980, "contents": "FIGURE 16.3 MECHANISM - \nThe mechanism for the electrophilic bromination of benzene. The reaction occurs in two steps and involves a resonance-stabilized carbocation intermediate.\n\n1) An electron pair from the benzene ring attacks the positively polarized bromine, forming a new $\\mathrm{C}-\\mathrm{Br}$ bond and leaving a nonaromatic carbocation intermediate.\n(2) A base removes $\\mathrm{H}^{+}$from the carbocation intermediate, and the neutral substitution product forms as two electrons from the $\\mathrm{C}-\\mathrm{H}$ bond move to re-form the aromatic ring.\n\n\n(1) Slow\n\n(2) Fast\n\n\nWhy does the reaction of $\\mathrm{Br}_{2}$ with benzene take a different course than its reaction with an alkene? The answer is straightforward. If addition occurred, the $150 \\mathrm{~kJ} / \\mathrm{mol}$ stabilization energy of the aromatic ring would be lost and the overall reaction would be endergonic. When substitution occurs, though, the stability of the aromatic ring is retained and the reaction is exergonic. An energy diagram for the overall process is shown in FIGURE 16.4 .\n\n\nFIGURE 16.4 An energy diagram for the electrophilic bromination of benzene. Because the stability of the aromatic ring is retained, the overall process is exergonic.\n\nPROBLEM Monobromination of toluene gives a mixture of three bromotoluene products. Draw and name 16-1 them."}
{"id": 981, "contents": "FIGURE 16.3 MECHANISM - 16.2 Other Aromatic Substitutions\nThere are many other kinds of electrophilic aromatic substitutions besides bromination, and all occur by the same general mechanism. Let's look at some of these other reactions briefly."}
{"id": 982, "contents": "Aromatic Halogenation - \nChlorine and iodine can be introduced into aromatic rings by electrophilic substitution reactions just as bromine can, but fluorine is too reactive and only poor yields of monofluoroaromatic products are obtained by direct fluorination. Instead, other sources of \" $\\mathrm{F}^{+}$\" are used, in which a fluorine atom is bonded to a positively charged nitrogen. One of the most common such reagents goes by the acronym F-TEDA-BF 4 in the presence of trifluoromethanesulfonic acid (TfOH). (You don't need to know the full name of F-TEDA, which is sold under the name Selectfluor.)\n\n\nMore than $20 \\%$ of all pharmaceutical agents sold contain fluorine, including $30 \\%$ of the top 100 drugs sold. Sitagliptin (Januvia), used to treat type 2 diabetes, fluoxetine (Prozac), an antidepressant, and atorvastatin (Lipitor), a statin used to lower cholesterol, are examples.\n\n\nSitagliptin\n(Januvia)\n\n\nFluoxetine\n(Prozac)\n\nAromatic rings react with $\\mathrm{Cl}_{2}$ in the presence of $\\mathrm{FeCl}_{3}$ catalyst to yield chlorobenzenes, just as they react with $\\mathrm{Br}_{2}$ and $\\mathrm{FeBr}_{3}$. This kind of reaction is used in the synthesis of numerous pharmaceutical agents, including the antiallergy medication loratadine, marketed as Claritin.\n\n\nIodine itself is unreactive toward aromatic rings, so an oxidizing agent such as hydrogen peroxide or a copper salt such as $\\mathrm{CuCl}_{2}$ must be added to the reaction. These substances accelerate the iodination reaction by oxidizing $\\mathrm{I}_{2}$ to a more powerful electrophilic species that reacts as if it were $\\mathrm{I}^{+}$. The aromatic ring then reacts with $\\mathrm{I}^{+}$in the typical way, yielding a substitution product.\n\n\nBenzene\n\nIodobenzene (65\\%)"}
{"id": 983, "contents": "Aromatic Halogenation - \nBenzene\n\nIodobenzene (65\\%)\n\nElectrophilic aromatic halogenations also occur in the biosynthesis of many naturally occurring molecules, particularly those produced by marine organisms. In humans, the best-known example occurs in the thyroid gland during the biosynthesis of thyroxine, a hormone involved in regulating growth and metabolism. The amino acid tyrosine is first iodinated by thyroid peroxidase, and two of the iodinated tyrosine molecules then couple. The electrophilic iodinating agent is an $\\mathrm{I}^{+}$species, perhaps hypoiodous acid (HIO), that is formed from iodide ion by oxidation with $\\mathrm{H}_{2} \\mathrm{O}_{2}$.\n\n\n\nThyroxine\n(a thyroid hormone)"}
{"id": 984, "contents": "Aromatic Nitration - \nAromatic rings are nitrated by reaction with a mixture of concentrated nitric and sulfuric acids. The electrophile is the nitronium ion, $\\mathrm{NO}_{2}{ }^{+}$, which is formed from $\\mathrm{HNO}_{3}$ by protonation and loss of water. The nitronium ion reacts with benzene to yield a carbocation intermediate, and loss of $\\mathrm{H}^{+}$from this intermediate gives the neutral substitution product, nitrobenzene (FIGURE 16.5).\n\n\n\nNitrobenzene\n\nFIGURE 16.5 The mechanism for electrophilic nitration of an aromatic ring. An electrostatic potential map of the reactive electrophile $\\mathrm{NO}_{2}{ }^{+}$shows that the nitrogen atom is most positive.\n\nElectrophilic nitration of an aromatic ring does not occur in nature but is particularly important in the laboratory because the nitro-substituted product can be reduced by reagents such as iron, tin, or $\\mathrm{SnCl}_{2}$ to yield the corresponding arylamine, $\\mathrm{ArNH}_{2}$. Attachment of an amino group $\\left(-\\mathrm{NH}_{2}\\right)$ to an aromatic ring by the two-step nitration/reduction sequence is a key part of the industrial synthesis of many dyes and pharmaceutical agents. We'll discuss this reduction and other reactions of aromatic nitrogen compounds in Chapter 24."}
{"id": 985, "contents": "Aromatic Sulfonation - \nAromatic rings can be sulfonated by reaction with so-called fuming sulfuric acid, a mixture of $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ and $\\mathrm{SO}_{3}$. The reactive electrophile is either $\\mathrm{HSO}_{3}{ }^{+}$or neutral $\\mathrm{SO}_{3}$, depending on reaction conditions, and substitution occurs by the same two-step mechanism seen previously for bromination and nitration (FIGURE 16.6). Note, however, that the sulfonation reaction is readily reversible. It can occur either forward or backward, depending on the reaction conditions. Sulfonation is favored in strong acid, but desulfonation is favored in hot, dilute aqueous acid.\n\n\nFIGURE 16.6 The mechanism for electrophilic sulfonation of an aromatic ring. An electrostatic potential map of the reactive electrophile $\\mathrm{HOSO}_{2}{ }^{+}$shows that sulfur and hydrogen are the most positive atoms.\n\nAromatic sulfonation does not occur naturally but is widely used in the preparation of dyes and pharmaceutical agents. For example, the sulfa drugs, such as sulfanilamide, were among the first clinically useful antibiotics. Although largely replaced today by more effective agents, sulfa drugs are still used in the treatment of meningitis and urinary tract infections. These drugs are prepared commercially by a process that involves aromatic sulfonation as its key step.\n\n\nSulfanilamide (an antibiotic)"}
{"id": 986, "contents": "Aromatic Hydroxylation - \nDirect hydroxylation of an aromatic ring to yield a hydroxybenzene (a phenol) is difficult and rarely done in the laboratory but occurs much more frequently in biological pathways. An example is the hydroxylation of $p$-hydroxyphenylacetate to give 3,4 -dihydroxyphenylacetate. The reaction is catalyzed by $p$-hydroxyphenylacetate-3-hydroxylase and requires molecular oxygen plus the coenzyme reduced flavin adenine dinucleotide, abbreviated $\\mathrm{FADH}_{2}$.\n\n\nBy analogy with other electrophilic aromatic substitutions, you might expect that an electrophilic oxygen species acting as an \" $\\mathrm{OH}^{+}$equivalent\" is needed for the hydroxylation reaction. That is just what happens, with the electrophilic oxygen arising by protonation of FAD hydroperoxide, RO-OH (FIGURE 16.7); that is, $\\mathrm{RO}-\\mathrm{OH}+\\mathrm{H}^{+} \\rightarrow \\mathrm{ROH}+\\mathrm{OH}^{+}$. The FAD hydroperoxide itself is formed by reaction of $\\mathrm{FADH}_{2}$ with $\\mathrm{O}_{2}$."}
{"id": 987, "contents": "FIGURE 16.7 MECHANISM - \nMechanism for the electrophilic hydroxylation of $p$-hydroxyphenylacetate, by reaction with FAD hydroperoxide. The hydroxylating species is an \" $\\mathrm{OH}^{+}$equivalent\" that arises by protonation of FAD hydroperoxide, $\\mathrm{RO}-\\mathrm{OH}+\\mathrm{H}^{+} \\rightarrow \\mathrm{ROH}+\\mathrm{OH}^{+}$. The FAD hydroperoxide itself is formed by reaction of $\\mathrm{FADH}_{2}$ with $\\mathrm{O}_{2}$.\n(1) Reduced flavin adenine dinucleotide reacts with molecular oxygen to give a hydroperoxide intermediate.\n\nFAD hydroperoxide\n\nProtonation of a hydroperoxide oxygen by an acid HA makes the neighboring oxygen electrophilic and allows the aromatic ring to react, giving a carbocation intermediate.\n\n\nLoss of $\\mathrm{H}^{+}$from the carbocation gives the hydroxy-substituted aromatic product.\n(3)\n\n\n3,4-Dihydroxyphenylacetate\n\nPROBLEM Propose a mechanism for the electrophilic fluorination of benzene with SelectFluor.\n16-2\nPROBLEM How many products might be formed on chlorination of $o$-xylene ( $o$-dimethylbenzene), $m$-xylene,\n16-3 and $p$-xylene?\nPROBLEM When benzene is treated with $\\mathrm{D}_{2} \\mathrm{SO}_{4}$, deuterium slowly replaces all six hydrogens in the aromatic 16-4 ring. Explain."}
{"id": 988, "contents": "FIGURE 16.7 MECHANISM - 16.3 Alkylation and Acylation of Aromatic Rings: The Friedel-Crafts Reaction\nAmong the most useful electrophilic aromatic substitution reactions in the laboratory is alkylation-the introduction of an alkyl group onto the benzene ring. Called the Friedel-Crafts reaction after its founders in 1877, Charles Friedel and James Crafts, the reaction is carried out by treating an aromatic compound with an alkyl chloride, RCl , in the presence of $\\mathrm{AlCl}_{3}$ to generate a carbocation electrophile, $\\mathrm{R}^{+}$. Aluminum chloride catalyzes the reaction by helping the alkyl halide to generate a carbocation in much the same way that $\\mathrm{FeBr}_{3}$ catalyzes aromatic brominations by polarizing $\\mathrm{Br}_{2}$ (Section 16.1). Loss of $\\mathrm{H}^{+}$then completes the reaction (FIGURE 16.8).\n\nFIGURE 16.8 MECHANISM\nMechanism for the Friedel-Crafts alkylation reaction of benzene with 2-chloropropane to yield isopropylbenzene (cumene). The electrophile is a carbocation, generated by $\\mathbf{A l C l}_{\\mathbf{3}}$-assisted dissociation of an alkyl halide.\n\n(1) An electron pair from the aromatic ring attacks the carbocation, forming a $\\mathrm{C}-\\mathrm{C}$ bond and yielding a new carbocation intermediate.\n\n(1)\n\n\nLoss of a proton then gives the neutral alkylated substitution product.\n\n\n\nDespite its utility, the Friedel-Crafts alkylation has several limitations. For one thing, only alkyl halides can be used. Aromatic (aryl) halides and vinylic halides don't react because aryl and vinylic carbocations are too high in energy to form under Friedel-Crafts conditions.\n\nAn aryl halide\n\n\n\nA vinylic halide"}
{"id": 989, "contents": "FIGURE 16.7 MECHANISM - 16.3 Alkylation and Acylation of Aromatic Rings: The Friedel-Crafts Reaction\nAn aryl halide\n\n\n\nA vinylic halide\n\nNot reactive\nAnother limitation is that Friedel-Crafts reactions don't succeed on aromatic rings that are substituted either by a strongly electron-withdrawing group such as carbonyl $(\\mathrm{C}=0)$ or by a basic amino group that can be protonated. We'll see in the next section that the presence of a substituent group already on a ring can have a dramatic effect on that ring's reactivity to further electrophilic substitution. Rings that contain any of the substituents listed in FIGURE 16.9 do not undergo Friedel-Crafts alkylation.\n\n\nFIGURE 16.9 Limitations on the aromatic substrate in Friedel-Crafts reactions. No reaction occurs if the substrate has either an electronwithdrawing substituent or a basic amino group.\n\nA third limitation to the Friedel-Crafts alkylation is that it's often difficult to stop the reaction after a single substitution. Once the first alkyl group is on the ring, a second substitution reaction is facilitated for reasons we'll discuss in the next section. Thus, we often observe polyalkylation. Reaction of benzene with 1 mol equivalent of 2 -chloro-2-methylpropane, for example, yields $p$-di-tert-butylbenzene as the major product, along with small amounts of tert-butylbenzene and unreacted benzene. A high yield of mono-alkylation product is obtained only when a large excess of benzene is used.\n\n\nA final limitation to the Friedel-Crafts reaction is that a skeletal rearrangement of the alkyl carbocation electrophile sometimes occurs during reaction, particularly when a primary alkyl halide is used. Treatment of benzene with 1 -chlorobutane at $0^{\\circ} \\mathrm{C}$, for instance, gives an approximately $2: 1$ ratio of rearranged (sec-butyl) to unrearranged (butyl) products."}
{"id": 990, "contents": "FIGURE 16.7 MECHANISM - 16.3 Alkylation and Acylation of Aromatic Rings: The Friedel-Crafts Reaction\nThe carbocation rearrangements that accompany Friedel-Crafts reactions are like those that accompany electrophilic additions to alkenes (Section 7.11) and occur either by hydride shift or alkyl shift. For example, the relatively unstable primary butyl carbocation produced by reaction of 1-chlorobutane with $\\mathrm{AlCl}_{3}$ rearranges to the more stable secondary butyl carbocation by the shift of a hydrogen atom and its electron pair (a hydride ion, $\\mathrm{H}:^{-}$) from C 2 to C 1 . Similarly, alkylation of benzene with 1-chloro-2,2-dimethylpropane yields (1,1-dimethylpropyl)benzene. The initially formed primary carbocation rearranges to a tertiary carbocation by shift of a methyl group and its electron pair from C 2 to C 1 .\n\n\nBenzene\n(1,1-Dimethylpropyl)benzene\n\n\nJust as an aromatic ring is alkylated by reaction with an alkyl chloride, it is acylated by reaction with a carboxylic acid chloride, RCOCl , in the presence of $\\mathrm{AlCl}_{3}$. That is, an acyl group ( -COR ; pronounced a-sil) is substituted onto the aromatic ring. For example, reaction of benzene with acetyl chloride yields the ketone acetophenone.\n\n\nThe mechanism of Friedel-Crafts acylation is similar to that of Friedel-Crafts alkylation, and the same limitations on the aromatic substrate noted previously in FIGURE 16.9 for alkylation also apply to acylation. The reactive electrophile is a resonance-stabilized acyl cation, generated by reaction between the acyl chloride and $\\mathrm{AlCl}_{3}$ (FIGURE 16.10). As the resonance structures in the figure indicate, an acyl cation is stabilized by interaction of the vacant orbital on carbon with lone-pair electrons on the neighboring oxygen. Because of this stabilization, no carbocation rearrangement occurs during acylation.\n\n\nFIGURE 16.10 Mechanism of the Friedel-Crafts acylation reaction. The electrophile is a resonance-stabilized acyl cation, whose electrostatic potential map indicates that carbon is the most positive atom."}
{"id": 991, "contents": "FIGURE 16.7 MECHANISM - 16.3 Alkylation and Acylation of Aromatic Rings: The Friedel-Crafts Reaction\nFIGURE 16.10 Mechanism of the Friedel-Crafts acylation reaction. The electrophile is a resonance-stabilized acyl cation, whose electrostatic potential map indicates that carbon is the most positive atom.\n\nUnlike the multiple substitutions that often occur in Friedel-Crafts alkylations, acylations never occur more than once on a ring because the product acylbenzene is less reactive than the nonacylated starting material. We'll account for this reactivity difference in the next section.\n\nAromatic alkylations occur in numerous biological pathways, although there is of course no $\\mathrm{AlCl}_{3}$ present in living systems to catalyze the reaction. Instead, the carbocation electrophile is typically formed by dissociation of an organodiphosphate, as we saw in Section 11.6. The dissociation is usually assisted by complexation to a divalent metal cation such as $\\mathrm{Mg}^{2+}$, just as dissociation of an alkyl chloride is assisted by $\\mathrm{AlCl}_{3}$.\n\n\n\nAn example of a biological Friedel-Crafts reaction occurs during the biosynthesis of phylloquinone, or vitamin $\\mathrm{K}_{1}$, the human blood-clotting factor. Phylloquinone is formed by reaction of 1,4-dihydroxynaphthoic acid with phytyl diphosphate. Phytyl diphosphate first dissociates to a resonance-stabilized allylic carbocation, which then substitutes onto the aromatic ring in the typical way. Several further transformations lead to phylloquinone (FIGURE 16.11).\n\n\nFIGURE 16.11 Biosynthesis of phylloquinone (vitamin $\\mathbf{K}_{\\mathbf{1}}$ ) from 1,4-dihydroxynaphthoic acid. The key step that joins the 20-carbon phytyl side chain to the aromatic ring is a Friedel-Crafts-like electrophilic substitution reaction with a diphosphate ion as the leaving group."}
{"id": 992, "contents": "Predicting the Product of a Carbocation Rearrangement - \nThe Friedel-Crafts reaction of benzene with 2 -chloro-3-methylbutane in the presence of $\\mathrm{AlCl}_{3}$ occurs with a carbocation rearrangement. What is the structure of the product?"}
{"id": 993, "contents": "Strategy - \nA Friedel-Crafts reaction involves initial formation of a carbocation, which can rearrange by either a hydride shift or an alkyl shift to give a more stable carbocation. Draw the initial carbocation, assess its stability, and see if the shift of a hydride ion or an alkyl group from a neighboring carbon will result in increased stability. In the present instance, the initial carbocation is a secondary one that can rearrange to a more stable tertiary one by a hydride shift.\n\n\nSecondary\ncarbocation\nTertiary carbocation\n\nUse this more stable tertiary carbocation to complete the Friedel-Crafts reaction."}
{"id": 994, "contents": "Solution - \nPROBLEM Which of the following alkyl halides would you expect to undergo Friedel-Crafts reaction with 16-5 rearrangement and which without? Explain.\n(a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{Cl}$\n(b) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}(\\mathrm{Cl}) \\mathrm{CH}_{3}$\n(c) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{Cl}$\n(d) $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CCH}_{2} \\mathrm{Cl}$\n(e) Chlorocyclohexane\n\nPROBLEM What is the major monosubstitution product from the Friedel-Crafts reaction of benzene with 16-6 1-chloro-2-methylpropane in the presence of $\\mathrm{AlCl}_{3}$ ?\n\nPROBLEM Identify the carboxylic acid chloride that might be used in a Friedel-Crafts acylation reaction to 16-7 prepare each of the following acylbenzenes:\n(a)\n\n(b)"}
{"id": 995, "contents": "Solution - 16.4 Substituent Effects in Electrophilic Substitutions\nOnly one product can form when an electrophilic substitution occurs on benzene, but what would happen if we were to carry out a reaction on an aromatic ring that already has a substituent? The initial presence of a substituent on the ring has two effects.\n\n- Substituents affect the reactivity of the aromatic ring. Some substituents activate the ring, making it more reactive than benzene, and some deactivate the ring, making it less reactive than benzene. In aromatic nitration, for instance, an -OH substituent makes the ring 1000 times more reactive than benzene, while an $-\\mathrm{NO}_{2}$ substituent makes the ring more than 10 million times less reactive.\n\n\nRelative rate of nitration\n$6 \\times 10^{-8}$\n\n0.033\n\n\n1\n\n\n1000"}
{"id": 996, "contents": "Reactivity - \n- Substituents affect the orientation of the reaction. The three possible disubstituted products-ortho, meta, and para-are usually not formed in equal amounts. Instead, the nature of the substituent initially present on the benzene ring determines the position of the second substitution. An -OH group directs substitution toward the ortho and para positions, for instance, while a carbonyl group such as -CHO directs substitution primarily toward the meta position. TABLE 16.1 lists experimental results for the nitration of some substituted benzenes.\n\n| TABLE 16.1 Orientation Substituted Benzenes |  | of Nitration |  |\n| :---: | :---: | :---: | :---: |\n| <smiles>[Y]c1ccccc1</smiles> | $\\xrightarrow[\\mathrm{O}_{4}, 25^{\\circ} \\mathrm{C}]{ }$ | <smiles>[Y]c1ccccc1</smiles> | $-\\mathrm{NO}_{2}$ |\n|  | Product (\\%) |  |  |\n|  | Ortho | Meta | Para |\n| Meta-directing deactivators |  |  |  |\n| $-\\stackrel{+}{\\mathrm{N}}\\left(\\mathrm{CH}_{3}\\right)_{3}$ | 2 | 87 | 11 |\n| $-\\mathrm{NO}_{2}$ | 7 | 91 | 2 |\n| $-\\mathrm{CO}_{2} \\mathrm{H}$ | 22 | 76 | 2 |\n| -CN | 17 | 81 | 2 |\n| $-\\mathrm{CO}_{2} \\mathrm{CH}_{3}$ | 28 | 66 | 6 |\n| $-\\mathrm{COCH}_{3}$ | 26 | 72 | 2 |\n| - CHO | 19 | 72 | 9 |\n| Ortho- and para-directing deactivators |  |  |  |\n| -F | 13 | 1 | 86 |\n| -Cl | 35 | 1 | 64 |\n| -Br | 43 | 1 | 56 |\n| -I | 45 | 1 | 54 |"}
{"id": 997, "contents": "Reactivity - \nSubstituents can be classified into three groups, as shown in FIGURE 16.12: ortho- and para-directing activators, ortho- and para-directing deactivators, and meta-directing deactivators. There are no metadirecting activators. Notice how the directing effect of a group correlates with its reactivity. All meta-directing groups are strongly deactivating, and most ortho- and para-directing groups are activating. The halogens are unique in being ortho- and para-directing but weakly deactivating.\n\n\nFIGURE 16.12 Classification of substituent effects in electrophilic aromatic substitution. All activating groups are ortho- and paradirecting, and all deactivating groups other than halogen are meta-directing. Halogens are unique in being deactivating but ortho- and paradirecting."}
{"id": 998, "contents": "Predicting the Product of an Electrophilic Aromatic Substitution Reaction - \nPredict the major product of the sulfonation of toluene."}
{"id": 999, "contents": "Strategy - \nIdentify the substituent present on the ring, and decide whether it is ortho- and para-directing or metadirecting. According to FIGURE 16.12, an alkyl substituent is ortho- and para-directing, so sulfonation of toluene will primarily give a mixture of $o$-toluenesulfonic acid and $p$-toluenesulfonic acid."}
{"id": 1000, "contents": "Solution - \nPROBLEM Rank the compounds in each of the following groups in order of their reactivity to electrophilic 16-8 substitution:\n(a) Nitrobenzene, phenol, toluene, benzene (b) Phenol, benzene, chlorobenzene, benzoic acid\n(c) Benzene, bromobenzene, benzaldehyde, aniline\n\nPROBLEM Predict the major products of the following reactions:\n16-9 (a) Nitration of bromobenzene (b) Bromination of nitrobenzene (c) Chlorination of phenol\n(d) Bromination of aniline"}
{"id": 1001, "contents": "Activating and Deactivating Effects - \nWhat makes a group either activating or deactivating? The common characteristic of all activating groups is that they donate electrons to the ring, thereby making the ring more electron-rich, stabilizing the carbocation intermediate, and lowering the activation energy for its formation. Conversely, the common characteristic of all deactivating groups is that they withdraw electrons from the ring, thereby making the ring more electron-poor, destabilizing the carbocation intermediate, and raising the activation energy for its formation.\nReactivity\n\n| is donates electrons; |\n| :--- |\n| isbocation intermediate |\n| is more stable, and ring |\n\nCompare the electrostatic potential maps of benzaldehyde (deactivated), chlorobenzene (weakly deactivated), and phenol (activated) with that of benzene. As shown in FIGURE 16.13, the ring is more positive (yellow-green) when an electron-withdrawing group such as -CHO or -Cl is present and more negative (red) when an electrondonating group such as -OH is present.\n\n\nBenzaldehyde\n\n\n\nChlorobenzene\n\n\n\nBenzene\n\n\n\nPhenol\n\nFIGURE 16.13 Electrostatic potential maps of benzene and several substituted benzenes show that an electron-withdrawing group $(-\\mathrm{CHO}$ or -Cl$)$ makes the ring more electron-poor, while an electron-donating group $(-\\mathrm{OH})$ makes the ring more electron-rich."}
{"id": 1002, "contents": "Activating and Deactivating Effects - \nThe withdrawal or donation of electrons by a substituent group is controlled by an interplay of inductive effects and resonance effects. As we saw in Section 2.1, an inductive effect is the withdrawal or donation of electrons through a $\\sigma$ bond due to electronegativity. Halogens, hydroxyl groups, carbonyl groups, cyano groups, and nitro groups inductively withdraw electrons through the $\\sigma$ bond linking the substituent to a benzene ring. This effect is most pronounced in halobenzenes and phenols, in which the electronegative atom is directly attached to the ring, but is also significant in carbonyl compounds, nitriles, and nitro compounds, in which the electronegative atom is farther removed. Alkyl groups, on the other hand, inductively donate electrons. This is the same hyperconjugative donating effect that causes alkyl substituents to stabilize alkenes (Section 7.6) and carbocations (Section 7.9).\n\n\nInductive electron withdrawal\n\n\nInductive electron\ndonation\nA resonance effect is the withdrawal or donation of electrons through a $\\pi$ bond due to the overlap of a $p$ orbital on the substituent with a $p$ orbital on the aromatic ring. Carbonyl, cyano, and nitro substituents, for example, withdraw electrons from the aromatic ring by resonance. The $\\pi$ electrons flow from the ring to the substituent, leaving a positive charge in the ring. Note that substituents with an electron-withdrawing resonance effect have the general structure $-\\mathrm{Y}=\\mathrm{Z}$, where the Z atom is more electronegative than Y .\n\nConversely, halogen, hydroxyl, alkoxyl (-OR), and amino substituents donate electrons to the aromatic ring by resonance. Lone-pair electrons flow from the substituents to the ring, placing a negative charge on the ring. Substituents with an electron-donating resonance effect have the general structure - $\\ddot{Y}$, where the Y atom has a lone pair of electrons available for donation to the ring."}
{"id": 1003, "contents": "Activating and Deactivating Effects - \nOne further point: inductive effects and resonance effects don't necessarily act in the same direction. Halogen, hydroxyl, alkoxyl, and amino substituents, for instance, have electron-withdrawing inductive effects because of the electronegativity of the $-\\mathrm{X},-\\mathrm{O}$, or -N atom bonded to the aromatic ring but have electron-donating resonance effects because of the lone-pair electrons on those $-\\mathrm{X},-\\mathrm{O}$, or -N atoms. When the two effects act in opposite directions, the stronger effect dominates. Thus, hydroxyl, alkoxyl, and amino substituents are activators because their stronger electron-donating resonance effect outweighs their weaker electronwithdrawing inductive effect. Halogens, however, are deactivators because their stronger electron-withdrawing inductive effect outweighs their weaker electron-donating resonance effect.\n\nPROBLEM Use Figure 16.12 to explain why Friedel-Crafts alkylations often give polysubstitution but\n16-10 Friedel-Crafts acylations do not.\n\n\n(Sole product)\nPROBLEM An electrostatic potential map of (trifluoromethyl)benzene, $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CF}_{3}$, is shown. Would you expect\n16-11 (trifluoromethyl)benzene to be more reactive or less reactive than toluene toward electrophilic substitution? Explain.\n\n(Trifluoromethyl)benzene\n\n\nToluene"}
{"id": 1004, "contents": "Ortho- and Para-Directing Activators: Alkyl Groups - \nInductive and resonance effects account not only for reactivity but also for the orientation of electrophilic aromatic substitutions. Take alkyl groups, for instance, which have an electron-donating inductive effect and are ortho and para directors. The results of toluene nitration are shown in FIGURE 16.14.\n\n\nFIGURE 16.14 Carbocation intermediates in the nitration of toluene. Ortho and para intermediates are more stable than the meta intermediate because the positive charge is on a tertiary carbon rather than a secondary carbon.\n\nNitration of toluene might occur either ortho, meta, or para to the methyl group, giving the three carbocation intermediates shown in in FIGURE 16.14. Although all three intermediates are resonance-stabilized, the ortho and para intermediates are more stabilized than the meta intermediate. For both the ortho and para reactions, but not for the meta reaction, a resonance form places the positive charge directly on the methyl-substituted carbon, where it is in a tertiary position and can be stabilized by the electron-donating inductive effect of the methyl group. The ortho and para intermediates are thus lower in energy than the meta intermediate and form faster."}
{"id": 1005, "contents": "Ortho- and Para-Directing Activators: OH and $\\mathrm{NH}_{2}$ - \nHydroxyl, alkoxyl, and amino groups are also ortho-para activators, but for a different reason than for alkyl groups. As described earlier in this section, hydroxyl, alkoxyl, and amino groups have a strong, electrondonating resonance effect that outweighs a weaker electron-withdrawing inductive effect. When phenol is nitrated, for instance, reaction can occur either ortho, meta, or para to the -OH group, giving the carbocation intermediates shown in FIGURE 16.15. The ortho and para intermediates are more stable than the meta intermediate because they have more resonance forms, including one particularly favorable form that allows the positive charge to be stabilized by electron donation from the substituent oxygen atom. The intermediate from the meta reaction has no such stabilization.\n\n\nFIGURE 16.15 Carbocation intermediates in the nitration of phenol. The ortho and para intermediates are more stable than the meta intermediate because they have more resonance forms, including one particularly favorable form that involves electron donation from the oxygen atom.\n\nPROBLEM Acetanilide is less reactive than aniline toward electrophilic substitution. Explain.\n16-12"}
{"id": 1006, "contents": "Ortho- and Para-Directing Deactivators: Halogens - \nHalogens are deactivating because their stronger electron-withdrawing inductive effect outweighs their weaker electron-donating resonance effect. Although weak, that electron-donating resonance effect is nevertheless felt only at the ortho and para positions and not at the meta position (FIGURE 16.16). Thus, a halogen substituent can stabilize the positive charge of the carbocation intermediates from ortho and para reaction in the same way that hydroxyl and amino substituents can. The meta intermediate, however, has no such stabilization and is therefore formed more slowly.\n\n\nFIGURE 16.16 Carbocation intermediates in the nitration of chlorobenzene. The ortho and para intermediates are more stable than the meta intermediate because of electron donation of the halogen lone-pair electrons.\n\nNote again that halogens, hydroxyl, alkoxyl, and amino groups all withdraw electrons inductively but donate electrons by resonance. Halogens have a stronger electron-withdrawing inductive effect but a weaker electrondonating resonance effect and are thus deactivators. Hydroxyl, alkoxyl, and amino groups have a weaker electron-withdrawing inductive effect but a stronger electron-donating resonance effect and are thus activators. All are ortho and para directors, however, because of the lone pair of electrons on the atom bonded to the aromatic ring."}
{"id": 1007, "contents": "Meta-Directing Deactivators - \nThe influence of meta-directing substituents can be explained using the same kinds of arguments used for ortho and para directors. Look at the nitration of benzaldehyde, for instance (FIGURE 16.17). Of the three possible carbocation intermediates, the meta intermediate has three favorable resonance forms, whereas the ortho and para intermediates have only two. In both ortho and para intermediates, the third resonance form is unfavorable because it places the positive charge directly on the carbon that bears the aldehyde group, where it is disfavored by a repulsive interaction with the positively polarized carbon atom of the $\\mathrm{C}=\\mathrm{O}$ group. Hence, the meta intermediate is more favored and is formed faster than the ortho and para intermediates.\n\n\nFIGURE 16.17 Carbocation intermediates in the nitration of benzaldehyde. The ortho and para intermediates are less stable than the meta intermediate. The meta intermediate is more favorable than ortho and para intermediates because it has three favorable resonance forms rather than two.\n\nIn general, any substituent that has a positively polarized atom $\\left(\\delta^{+}\\right)$directly attached to the ring will make one of the resonance forms of the ortho and para intermediates unfavorable and will thus act as a meta director.\n\nPROBLEM Draw resonance structures for the intermediates from the reaction of an electrophile at the ortho,\n16-13 meta, and para positions of nitrobenzene. Which intermediates are most stable?"}
{"id": 1008, "contents": "A Summary of Substituent Effects in Electrophilic Aromatic Substitution - \nA summary of the activating and directing effects of substituents in electrophilic aromatic substitution is shown in TABLE 16.2.\n\nTABLE 16.2 Substituent Effects in Electrophilic Aromatic Substitution\n\n| Substituent | Reactivity |  | Orienting effect |  |\n| :--- | :--- | :--- | :--- | :--- |\n| -CH 3 | Inductive effect | Resonance effect |  |  |\n| $-\\mathrm{OH},-\\mathrm{NH}_{2}$ | Activating | Ortho, para | Weak withdrawing | Strong donating |\n| $-\\mathrm{F},-\\mathrm{Cl}$ | Deactivating | Ortho, para | Strong withdrawing | Weak donating |\n| $-\\mathrm{Br},-\\mathrm{I}$ |  |  |  |  |\n| $-\\mathrm{NO}_{2},-\\mathrm{CN}$, | Deactivating | Meta | Strong withdrawing | Strong withdrawing |\n| $-\\mathrm{CHO},-\\mathrm{CO}_{2} \\mathrm{R}$ |  |  |  |  |\n| $-\\mathrm{COR},-\\mathrm{CO}_{2} \\mathrm{H}$ |  |  |  |  |"}
{"id": 1009, "contents": "A Summary of Substituent Effects in Electrophilic Aromatic Substitution - 16.5 Trisubstituted Benzenes: Additivity of Effects\nElectrophilic substitution of a disubstituted benzene ring is governed by the same resonance and inductive effects that influence monosubstituted rings. The only difference is that it's necessary to consider the additive effects of two different groups. In practice, this isn't as difficult as it sounds; three rules are usually sufficient.\n\nRULE 1\nIf the directing effects of the two groups reinforce each other, the situation is straightforward. In $p$-nitrotoluene, for example, both the methyl and the nitro group direct further substitution to the same position (ortho to the methyl = meta to the nitro). A single product is thus formed on electrophilic substitution.\n\np-Nitrotoluene\n2-Bromo-4-nitrotoluene\nRULE 2\nIf the directing effects of the two groups oppose each other, the more powerful activating group has the dominant influence, but mixtures of products are often formed. For example, bromination of $p$-methylphenol yields primarily 2-bromo-4-methylphenol because - OH is a more powerful activator than $-\\mathrm{CH}_{3}$.\n\np-Methylphenol\n2-Bromo-4-methylphenol\nRULE 3\nFurther substitution rarely occurs between the two groups in a meta-disubstituted compound because this site is too hindered. Aromatic rings with three adjacent substituents must therefore be prepared by some other route, such as by substitution of an ortho-disubstituted compound.\n\n\nBut:"}
{"id": 1010, "contents": "Predicting the Product of Substitution on a Disubstituted Benzene - \nWhat product would you expect from bromination of $p$-methylbenzoic acid?"}
{"id": 1011, "contents": "Strategy - \nIdentify the two substituents present on the ring, decide the directing effect of each and, if necessary, decide which substituent is the stronger activator. In the present case, the carboxyl group $\\left(-\\mathrm{CO}_{2} \\mathrm{H}\\right)$ is a meta director and the methyl group is an ortho and para director. Both groups direct bromination to the position next to the methyl group, yielding 3-bromo-4-methylbenzoic acid."}
{"id": 1012, "contents": "Solution - \np-Methylbenzoic acid\n3-Bromo-4-methylbenzoic acid\n\nPROBLEM At what position would you expect electrophilic substitution to occur in each of the following 16-14 substances:\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Show the major product(s) from reaction of the following substances with (1) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{Cl}, \\mathrm{AlCl}_{3}$ and 16-15 (2) $\\mathrm{HNO}_{3}, \\mathrm{H}_{2} \\mathrm{SO}_{4}$ :\n(a)\n\n(b)"}
{"id": 1013, "contents": "Solution - 16.6 Nucleophilic Aromatic Substitution\nAlthough aromatic substitution reactions usually occur by an electrophilic mechanism, aryl halides that have electron-withdrawing substituents can also undergo a nucleophilic substitution reaction. For example, $2,4,6$-trinitrochlorobenzene reacts with aqueous NaOH at room temperature to give 2,4,6-trinitrophenol. Here, the nucleophile $\\mathrm{OH}^{-}$substitutes for $\\mathrm{Cl}^{-}$.\n\n\n\n2,4,6-Trinitrochlorobenzene\n\n2,4,6-Trinitrophenol (100\\%)\n\nNucleophilic aromatic substitution is much less common than electrophilic substitution but nevertheless does have certain uses. One such use is the reaction of proteins with 2,4-dinitrofluorobenzene, known as Sanger's reagent, to attach a \"label\" to the terminal $\\mathrm{NH}_{2}$ group of the amino acid at one end of the protein chain.\n\n\n2,4-Dinitrofluorobenzene\nAlthough the reaction appears superficially similar to the $S_{N} 1$ and $S_{N} 2$ nucleophilic substitutions of alkyl halides discussed in the chapter on Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations, it must be different because aryl halides are inert to both $\\mathrm{S}_{\\mathrm{N}} 1$ and $\\mathrm{S}_{\\mathrm{N}} 2$ conditions. $\\mathrm{S}_{\\mathrm{N}} 1$ reactions don't occur with aryl halides because dissociation of the halide is energetically unfavorable, due to the instability of the potential aryl cation product. $\\mathrm{S}_{\\mathrm{N}} 2$ reactions don't occur with aryl halides because the halo-substituted carbon of the aromatic ring is sterically shielded from a backside approach. For a nucleophile to react with an aryl halide, it would have to approach directly through the aromatic ring and invert the stereochemistry of the aromatic ring carbon-a geometric impossibility.\n\n\nDissociation reaction does not occur because the aryl cation is unstable; therefore, no $\\mathrm{S}_{\\mathrm{N}} 1$ reaction."}
{"id": 1014, "contents": "Solution - 16.6 Nucleophilic Aromatic Substitution\nDissociation reaction does not occur because the aryl cation is unstable; therefore, no $\\mathrm{S}_{\\mathrm{N}} 1$ reaction.\n\n\nBackside displacement is sterically blocked; therefore, no $\\mathrm{S}_{\\mathrm{N}} 2$ reaction.\nInstead, nucleophilic substitutions on an aromatic ring proceed by the mechanism shown in FIGURE 16.18. The nucleophile first adds to the electron-deficient aryl halide, forming a resonance-stabilized, negatively charged intermediate called a Meisenheimer complex after its discoverer. Halide ion is then eliminated."}
{"id": 1015, "contents": "FIGURE 16.18 MECHANISM - \nMechanism of nucleophilic aromatic substitution. The reaction occurs in two steps and involves a resonance-stabilized carbanion intermediate.\n(1) Nucleophilic addition of hydroxide ion to the electron-poor aromatic ring takes place, yielding a stabilized carbanion intermediate.\n\n(1)\n\n\n2 The carbanion intermediate undergoes elimination of chloride ion in a second step to give the substitution product.\n(2)\n\n\nNucleophilic aromatic substitution occurs only if the aromatic ring has an electron-withdrawing substituent in a position ortho or para to the leaving group to stabilize the anion intermediate through resonance (FIGURE 16.19). A meta substituent offers no such resonance stabilization. Thus, $p$-chloronitrobenzene and $o$-chloronitrobenzene react with hydroxide ion at $130{ }^{\\circ} \\mathrm{C}$ to yield substitution products, but $m$-chloronitrobenzene is inert to $\\mathrm{OH}^{-}$."}
{"id": 1016, "contents": "Ortho - \nPara\n\n\nMeta\n\n\nFIGURE 16.19 Nucleophilic aromatic substitution on nitrochlorobenzenes. Only in the ortho and para intermediates is the negative charge stabilized by a resonance interaction with the nitro group, so only the ortho and para isomers undergo reaction.\n\nNote the differences between electrophilic and nucleophilic aromatic substitutions. Electrophilic substitutions are favored by electron-donating substituents, which stabilize a carbocation intermediate, while nucleophilic substitutions are favored by electron-withdrawing substituents, which stabilize a carbanion intermediate. Thus, the electron-withdrawing groups that deactivate rings for electrophilic substitution (nitro, carbonyl, cyano, and so forth) activate them for nucleophilic substitution. What's more, these functional groups are meta directors in electrophilic substitution but are ortho-para directors in nucleophilic substitution. And finally, electrophilic substitutions replace hydrogen on the ring, while nucleophilic substitutions replace a leaving group, usually halide ion.\n\nPROBLEM The herbicide oxyfluorfen can be prepared by reaction between a phenol and an aryl fluoride.\n16-16 Propose a mechanism.\n\n\nOxyfluorfen"}
{"id": 1017, "contents": "Ortho - 16.7 Benzyne\nHalobenzenes without electron-withdrawing substituents don't react with nucleophiles under most conditions. At high temperature and pressure, however, even chlorobenzene can be forced to react. Chemists at the Dow Chemical Company discovered in 1928 that phenol could be prepared on an industrial scale by treatment of chlorobenzene with dilute aqueous NaOH at $340^{\\circ} \\mathrm{C}$ under 170 atm pressure.\n\n\nA similar substitution reaction occurs with other strong bases. Treatment of bromobenzene with potassium amide $\\left(\\mathrm{KNH}_{2}\\right)$ in liquid $\\mathrm{NH}_{3}$ solvent, for instance, gives aniline. Curiously, though, when using bromobenzene labeled with radioactive ${ }^{14} \\mathrm{C}$ at the C 1 position, the substitution product has equal amounts of the label at both C 1 and C2, implying the presence of a symmetrical reaction intermediate in which C 1 and C 2 are equivalent.\n\n\nFurther mechanistic evidence comes from trapping experiments. When bromobenzene is treated with $\\mathrm{KNH}_{2}$ in the presence of a conjugated diene, such as furan, a Diels-Alder reaction (Section 14.4) occurs, implying that the symmetrical intermediate is a benzyne, formed by elimination of HBr from bromobenzene. Benzyne is too reactive to be isolated as a pure compound but, in the presence of water, addition occurs to give phenol. In the presence of a diene, Diels-Alder cycloaddition takes place.\n\n\nThe electronic structure of benzyne, shown in FIGURE 16.20, is that of a highly distorted alkyne. Although a typical alkyne triple bond uses $s p$-hybridized carbon atoms, the benzyne triple bond uses $s p^{2}$-hybridized carbons. Furthermore, a typical alkyne triple bond has two mutually perpendicular $\\pi$ bonds formed by $p-p$ overlap, but the benzyne triple bond has one $\\pi$ bond formed by $p-p$ overlap and one $\\pi$ bond formed by $s p^{2}-s p^{2}$ overlap. The latter $\\pi$ bond is in the plane of the ring and is very weak.\n\n\nSide view"}
{"id": 1018, "contents": "Ortho - 16.7 Benzyne\nSide view\n\n\nFIGURE 16.20 An orbital picture and electrostatic potential map of benzyne. The benzyne carbons are $s p^{2}$-hybridized, and the \"third\" bond results from weak overlap of two adjacent $s p^{2}$ orbitals.\n\nPROBLEM Treatment of $p$-bromotoluene with NaOH at $300^{\\circ} \\mathrm{C}$ yields a mixture of two products, but treatment 16-17 of $m$-bromotoluene with NaOH yields a mixture of three products. Explain."}
{"id": 1019, "contents": "Oxidation of Alkyl Side Chains - \nDespite its unsaturation, the benzene ring is inert to strong oxidizing agents such as $\\mathrm{KMnO}_{4}$, which will cleave alkene carbon-carbon bonds (Section 8.8). It turns out, however, that the presence of the aromatic ring has a dramatic effect on the reactivity of alkyl side chains. These side chains react rapidly with oxidizing agents and are converted into carboxyl groups, $-\\mathrm{CO}_{2} \\mathrm{H}$.The net effect is conversion of an alkylbenzene into a benzoic acid, $\\mathrm{Ar}-\\mathrm{R} \\rightarrow \\mathrm{Ar}-\\mathrm{CO}_{2} \\mathrm{H}$. Butylbenzene is oxidized by aqueous $\\mathrm{KMnO}_{4}$ to give benzoic acid, for instance.\n\n\nA similar oxidation is employed industrially for the preparation of the terephthalic acid used in the production of polyester fibers. Worldwide, approximately 118 million tons per year of terephthalic acid is produced by oxidation of $p$-xylene, using air as the oxidant and $\\mathrm{Co}(\\mathrm{III})$ salts as catalyst.\n\n$p$-Xylene $\\quad$ Terephthalic acid\nThe mechanism of side-chain oxidation is complex and involves reaction of $\\mathrm{C}-\\mathrm{H}$ bonds at the position next to the aromatic ring to form intermediate benzylic radicals. tert-Butylbenzene has no benzylic hydrogens, however, and is therefore inert.\n\ntert-Butylbenzene\nAnalogous side-chain oxidations occur in various biosynthetic pathways. The neurotransmitter norepinephrine, for instance, is biosynthesized from dopamine by a benzylic hydroxylation reaction. The process is catalyzed by the copper-containing enzyme dopamine $\\beta$-monooxygenase and occurs by a radical mechanism. A copper-oxygen species in the enzyme first abstracts the pro-R benzylic hydrogen to give a radical, and a hydroxyl is then transferred from copper to carbon.\n\n\nPROBLEM What aromatic products would you obtain from the $\\mathrm{KMnO}_{4}$ oxidation of the following substances? 16-18\n(a)\n\n(b)"}
{"id": 1020, "contents": "Bromination of Alkylbenzene Side Chains - \nSide-chain bromination at the benzylic position occurs when an alkylbenzene is treated with $N$-bromosuccinimide (NBS). For example, propylbenzene gives (1-bromopropyl)benzene in 97\\% yield on reaction with NBS in the presence of benzoyl peroxide, $\\left(\\mathrm{PhCO}_{2}\\right)_{2}$, as a radical initiator. Bromination occurs exclusively in the benzylic position next to the aromatic ring and does not give a mixture of products.\n\n\nThe mechanism of benzylic bromination is similar to that discussed in Section 10.3 for allylic bromination of alkenes. Abstraction of a benzylic hydrogen atom first generates an intermediate benzylic radical, which then reacts with $\\mathrm{Br}_{2}$ in step 2 to yield product and a Br \u2022 radical, which cycles back into the reaction to carry on the chain shown below as a summary. The $\\mathrm{Br}_{2}$ needed for reaction with the benzylic radical is produced in step 3 by a concurrent reaction of HBr with NBS.\n\n\nReaction occurs exclusively at the benzylic position because the benzylic radical intermediate is stabilized by resonance. FIGURE 16.21 shows how the benzyl radical is stabilized by overlap of its $p$ orbital with the ringed $\\pi$ electron system.\n\n\nFIGURE 16.21 A resonance-stabilized benzylic radical. The spin-density surface shows that the unpaired electron is shared by the ortho\nand para carbons of the ring.\nPROBLEM Refer to Table 6.3 for a quantitative idea of the stability of a benzyl radical. How much more stable\n16-19 (in $\\mathrm{kJ} / \\mathrm{mol}$ ) is the benzyl radical than a primary alkyl radical? How does a benzyl radical compare in stability to an allyl radical?\n\nPROBLEM Styrene, the simplest alkenylbenzene, is prepared commercially for use in plastics manufacture\n16-20 by catalytic dehydrogenation of ethylbenzene. How might you prepare styrene from benzene using reactions you've studied?\n\n\nStyrene"}
{"id": 1021, "contents": "Catalytic Hydrogenation of Aromatic Rings - \nJust as aromatic rings are generally inert to oxidation, they're also inert to catalytic hydrogenation under conditions that reduce typical alkene double bonds. As a result, it's possible to reduce an alkene double bond selectively in the presence of an aromatic ring. For example, 4-phenyl-3-buten-2-one is reduced to 4-phenyl-2-butanone using a palladium catalyst at room temperature and atmospheric pressure. Neither the benzene ring nor the ketone carbonyl group is affected.\n\n\nTo hydrogenate an aromatic ring, it's necessary either to use a platinum catalyst with hydrogen gas at a pressure of several hundred atmospheres or to use a more effective catalyst such as rhodium on carbon. Under these conditions, aromatic rings are converted into cyclohexanes. For example, $o$-xylene yields 1,2-dimethylcyclohexane, and 4-tert-butylphenol gives 4-tert-butylcyclohexanol."}
{"id": 1022, "contents": "Reduction of Aryl Alkyl Ketones - \nIn the same way that an aromatic ring activates a neighboring (benzylic) $\\mathrm{C}-\\mathrm{H}$ toward oxidation, it also activates a benzylic carbonyl group toward reduction. Thus, an aryl alkyl ketone prepared by Friedel-Crafts acylation of an aromatic ring can be converted into an alkylbenzene by catalytic hydrogenation over a palladium catalyst.\n\nPropiophenone, for instance, is reduced to propylbenzene by catalytic hydrogenation. Because the net effect of Friedel-Crafts acylation followed by reduction is the preparation of a primary alkylbenzene, this two-step sequence of reactions makes it possible to circumvent the carbocation rearrangement problems associated with direct Friedel-Crafts alkylation using a primary alkyl halide (Section 16.3).\n\n\nThe conversion of a carbonyl group into a methylene group $\\left(\\mathrm{C}=\\mathrm{O} \\rightarrow \\mathrm{CH}_{2}\\right)$ by catalytic hydrogenation is limited to aryl alkyl ketones; dialkyl ketones are not reduced under these conditions. Furthermore, the catalytic reduction of aryl alkyl ketones is not compatible with the presence of a nitro substituent on the aromatic ring because a nitro group is reduced to an amino group under reaction conditions. We'll see a more general method for reducing ketone carbonyl groups to yield alkanes in Section 19.9.\n\n\nPROBLEM How would you prepare diphenylmethane, $(\\mathrm{Ph})_{2} \\mathrm{CH}_{2}$, from benzene and an acid chloride? 16-21"}
{"id": 1023, "contents": "Reduction of Aryl Alkyl Ketones - 16.10 Synthesis of Polysubstituted Benzenes\nAs discussed in the Introduction to Organic Synthesis in Section 9.9, one of the surest ways to learn organic chemistry is to work synthesis problems. The ability to plan a successful multistep synthesis of a complex molecule requires a working knowledge of the uses and limitations of a great many organic reactions. Not only must you know which reactions to use, you must also know when to use them because the order in which reactions are carried out is often critical to the success of the overall scheme.\n\nThe ability to plan a sequence of reactions in the right order is particularly important in the synthesis of substituted aromatic rings, where the introduction of a new substituent is strongly affected by the directing effects of other substituents. Planning syntheses of substituted aromatic compounds is therefore a good way to gain confidence in using the many reactions discussed in the past few chapters.\n\nAs we said in Section 9.9, it's usually best to work a synthesis problem backward, or retrosynthetically. Look at the target molecule and ask yourself, \"What is an immediate precursor of this compound?\" Choose a likely answer and continue working backward, one step at a time, until you arrive at a simple starting material. Let's try some examples."}
{"id": 1024, "contents": "Synthesizing a Polysubstituted Benzene - \nHow would you synthesize 4-bromo-2-nitrotoluene from benzene?"}
{"id": 1025, "contents": "Strategy - \nDraw the target molecule, identify the substituents, and recall how each group can be introduced separately. Then plan retrosynthetically.\n\n\n4-Bromo-2-nitrotoluene\n\nThe three substituents on the ring are a bromine, a methyl group, and a nitro group. A bromine can be introduced by bromination with $\\mathrm{Br}_{2} / \\mathrm{FeBr}_{3}$, a methyl group can be introduced by Friedel-Crafts alkylation with $\\mathrm{CH}_{3} \\mathrm{Cl} / \\mathrm{AlCl}_{3}$, and a nitro group can be introduced by nitration with $\\mathrm{HNO}_{3} / \\mathrm{H}_{2} \\mathrm{SO}_{4}$."}
{"id": 1026, "contents": "Solution - \nAsk yourself, \"What is an immediate precursor of the target?\" The final step will involve introduction of one of three groups-bromine, methyl, or nitro-so we have to consider three possibilities. Of the three, the bromination of o-nitrotoluene could be used because the activating methyl group would dominate the deactivating nitro group and direct bromination to the correct position. Unfortunately, a mixture of product isomers would be formed. A Friedel-Crafts reaction can't be used as the final step because this reaction doesn't work on a nitro-substituted (strongly deactivated) benzene. The best precursor of the desired product is probably $p$-bromotoluene, which can be nitrated ortho to the activating methyl group to give a single product.\n\n$o$-Nitrotoluene\nThis ring will give a mixture of isomers on bromination.\n\n\nm-Bromonitrobenzene\nThis deactivated ring will not undergo a Friedel-Crafts reaction."}
{"id": 1027, "contents": "p-Bromotoluene - \nThis ring will give only the desired isomer on nitration."}
{"id": 1028, "contents": "4-Bromo-2-nitrotoluene - \nNext ask, \"What is an immediate precursor of $p$-bromotoluene?\" Perhaps toluene is an immediate precursor because the methyl group would direct bromination to the ortho and para positions. Alternatively, bromobenzene might be an immediate precursor because we could carry out a Friedel-Crafts methylation and obtain a mixture of ortho and para products. Both answers are satisfactory, although both would also lead unavoidably to a mixture of products that would have to be separated.\n\n\"What is an immediate precursor of toluene?\" Benzene, which could be methylated in a Friedel-Crafts reaction. Alternatively, \"What is an immediate precursor of bromobenzene?\" Benzene, which could be brominated.\n\nThe retrosynthetic analysis has provided two valid routes from benzene to 4-bromo-2-nitrotoluene."}
{"id": 1029, "contents": "Synthesizing a Polysubstituted Benzene - \nSynthesize 4-chloro-2-propylbenzenesulfonic acid from benzene."}
{"id": 1030, "contents": "Strategy - \nDraw the target molecule, identify its substituents, and recall how each of the three can be introduced. Then plan retrosynthetically."}
{"id": 1031, "contents": "4-Chloro-2-propylbenzenesulfonic acid - \nThe three substituents on the ring are a chlorine, a propyl group, and a sulfonic acid group. A chlorine can be introduced by chlorination with $\\mathrm{Cl}_{2} / \\mathrm{FeCl}_{3}$, a propyl group can be introduced by Friedel-Crafts acylation with $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COCl} / \\mathrm{AlCl}_{3}$ followed by reduction with $\\mathrm{H}_{2} / \\mathrm{Pd}$, and a sulfonic acid group can be introduced by sulfonation with $\\mathrm{SO}_{3} / \\mathrm{H}_{2} \\mathrm{SO}_{4}$."}
{"id": 1032, "contents": "Solution - \n\"What is an immediate precursor of the target?\" The final step will involve introduction of one of three groups-chlorine, propyl, or sulfonic acid-so we have to consider three possibilities. Of the three, the chlorination of o-propylbenzenesulfonic acid can't be used because the reaction would occur at the wrong position. Similarly, a Friedel-Crafts reaction can't be used as the final step because this reaction doesn't work on sulfonic-acid-substituted (strongly deactivated) benzenes. Thus, the immediate precursor of the desired product is probably $m$-chloropropylbenzene, which can be sulfonated to give a mixture of product isomers that must then be separated.\n\no-Propylbenzenesulfonic acid\nThis ring will give the wrong isomer on chlorination.\n\np-Chlorobenzenesulfonic acid\nThis deactivated ring will not undergo a Friedel-Crafts reaction.\n\n\nm-Chloropropylbenzene\nThis ring will give the desired product on sulfonation."}
{"id": 1033, "contents": "4-Chloro-2-propylbenzenesulfonic acid - \n\"What is an immediate precursor of $m$-chloropropylbenzene?\" Because the two substituents have a meta relationship, the first substituent placed on the ring must be a meta director so that the second substitution will take place at the proper position. Furthermore, because primary alkyl groups such as propyl can't be introduced directly by Friedel-Crafts alkylation, the precursor of $m$-chloropropylbenzene is probably $m$-chloropropiophenone, which could be catalytically reduced.\n\n\"What is an immediate precursor of $m$-chloropropiophenone?\" Propiophenone, which could be chlorinated in the meta position.\n\n\"What is an immediate precursor of propiophenone?\" Benzene, which could undergo Friedel-Crafts acylation with propanoyl chloride and $\\mathrm{AlCl}_{3}$.\n\n\nThe final synthesis is a four-step route from benzene:\n\n\nPlanning an organic synthesis has been compared with playing chess. There are no tricks; all that's required is a knowledge of the allowable moves (the organic reactions) and the discipline to plan ahead, carefully evaluating the consequences of each move. Practicing may not be easy, but it's a great way to learn organic chemistry.\n\nPROBLEM How might you synthesize the following substances from benzene?\n16-22 (a) $m$-Chloronitrobenzene (b) $m$-Chloroethylbenzene (c) 4-Chloro-1-nitro-2-propylbenzene (d) 3-Bromo-2-methylbenzenesulfonic acid\n\nPROBLEM In planning a synthesis, it's as important to know what not to do as to know what to do. As written,\n16-23 the following reaction schemes have flaws in them. What is wrong with each?\n\n\n(b)"}
{"id": 1034, "contents": "Combinatorial Chemistry - \nTraditionally, organic compounds have been synthesized one at a time. This works well for preparing large amounts of a few substances, but it doesn't work so well for preparing small amounts of a great many substances. This latter goal is particularly important in the pharmaceutical industry, where vast numbers of structurally similar compounds must be synthesized and screened to find an optimum drug candidate.\n\nTo speed the process of drug discovery, combinatorial chemistry has been developed to prepare what are called combinatorial libraries, in which anywhere from a few dozen to several hundred thousand substances are prepared simultaneously. Among the early successes of combinatorial chemistry is the development of a benzodiazepine library, a class of aromatic compounds commonly used as antianxiety agents.\n\n\nBenzodiazepine library\n( $\\mathrm{R}_{\\mathbf{1}}-\\mathrm{R}_{\\mathbf{4}}$ are various organic substituents)\n\nTwo main approaches to combinatorial chemistry are used-parallel synthesis and split synthesis. In parallel synthesis, each compound is prepared independently. Typically, a reactant is first linked to the surface of polymer beads, which are then placed into small wells on a 96 -well glass plate. Programmable robotic instruments add different sequences of building blocks to the different wells, thereby making 96 different products. When the reaction sequences are complete, the polymer beads are washed and their products are released.\n\nIn split synthesis, the initial reactant is again linked to the surface of polymer beads, which are then divided into several groups. A different building block is added to each group of beads, the different groups are combined, and the reassembled mix is again split to form new groups. Another building block is added to each group, the groups are again combined and redivided, and the process continues. If, for example, the beads are divided into four groups at each step, the number of compounds increases in the progression $4 \\rightarrow 16 \\rightarrow 64 \\rightarrow 256$. After 10 steps, more than 1 million compounds have been prepared (FIGURE 16.23)."}
{"id": 1035, "contents": "Combinatorial Chemistry - \nOf course, with so many different final products mixed together, the problem is to identify them. What structure is linked to what bead? Several approaches to this problem have been developed, all of which involve the attachment of encoding labels to each polymer bead to keep track of the chemistry each has undergone. Encoding labels used thus far have included proteins, nucleic acids, halogenated aromatic compounds, and even computer chips.\n\n\nFIGURE 16.22 Organic chemistry by robot means no spilled flasks!\n\n\nFIGURE 16.23 The results of split combinatorial synthesis. Assuming that 4 different building blocks are used at each step, 64 compounds result after 3 steps, and more than one million compounds result after 10 steps."}
{"id": 1036, "contents": "Key Terms - \n- acyl group \u2022 electrophilic aromatic substitution\n- acylation\n- Friedel-Crafts reaction\n- alkylation\n- inductive effect\n- benzyne\n- nucleophilic aromatic substitution\n- combinatorial chemistry\n- resonance effect"}
{"id": 1037, "contents": "Summary - \nWe've continued the coverage of aromatic molecules in this chapter, shifting focus to concentrate on reactions. In particular, we've looked at the relationship between aromatic structure and reactivity, a relationship critical to understanding how numerous biological molecules and pharmaceutical agents are synthesized and why they behave as they do.\n\nAn electrophilic aromatic substitution reaction takes place in two steps-initial reaction of an electrophile, $\\mathbf{E}^{+}$, with the aromatic ring, followed by loss of $\\mathrm{H}^{+}$from the resonance-stabilized carbocation intermediate to regenerate the aromatic ring.\n\n\nMany variations of the reaction can be carried out, including halogenation, nitration, and sulfonation. Friedel-Crafts alkylation and acylation reactions, which involve reaction of an aromatic ring with carbocation electrophiles, are particularly useful. They are limited, however, by the fact that the aromatic ring must be at least as reactive as a halobenzene. In addition, polyalkylation and carbocation rearrangements often occur in Friedel-Crafts alkylation.\n\nSubstituents on the benzene ring affect both the reactivity of the ring toward further substitution and the orientation of that substitution. Groups can be classified as ortho- and para-directing activators, ortho- and para-directing deactivators, or meta-directing deactivators. Substituents influence aromatic rings by a combination of resonance and inductive effects. Resonance effects are transmitted through $\\pi$ bonds; inductive effects are transmitted through $\\sigma$ bonds.\n\nHalobenzenes undergo nucleophilic aromatic substitution through either of two mechanisms. If the halobenzene has a strongly electron-withdrawing substituent in the ortho or para position, substitution occurs by addition of a nucleophile to the ring, followed by elimination of halide from the intermediate anion. If the halobenzene is not activated by an electron-withdrawing substituent, substitution can occur by elimination of HX to give a benzyne, followed by addition of a nucleophile."}
{"id": 1038, "contents": "Summary - \nThe benzylic position of an alkylbenzene can be brominated by reaction with $N$-bromosuccinimide, and the entire side chain can be degraded to a carboxyl group by oxidation with aqueous $\\mathrm{KMnO}_{4}$. Aromatic rings can also be reduced to cyclohexanes by hydrogenation over a platinum or rhodium catalyst, and aryl alkyl ketones are reduced to alkylbenzenes by hydrogenation over a platinum catalyst."}
{"id": 1039, "contents": "Summary of Reactions - \n1. Electrophilic aromatic substitution\na. Fluorination (Section 16.2)\n\nb. Bromination (Section 16.1)\n\nc. Chlorination (Section 16.2)\n\nd. Iodination (Section 16.2)\n\ne. Nitration (Section 16.2)\n\nf. Sulfonation (Section 16.2)\n\ng. Friedel-Crafts alkylation (Section 16.3)\n\nh. Friedel-Crafts acylation (Section 16.3)\n\n2. Reduction of aromatic nitro groups (Section 16.2)\n\n3. Nucleophilic aromatic substitution\na. By addition to activated aryl halides (Section 16.6)\n\nb. By formation of benzyne intermediate from unactivated aryl halide (Section 16.7)\n\n4. Oxidation of alkylbenzene side chain (Section 16.8)\n\n5. Benzylic bromination of alkylbenzene side chain (Section 16.8)\n\n6. Catalytic hydrogenation of aromatic ring (Section 16.9)\n\n7. Reduction of aryl alkyl ketones (Section 16.9)"}
{"id": 1040, "contents": "Visualizing Chemistry - \nPROBLEM Draw the product from reaction of each of the following substances with (1) $\\mathrm{Br}_{2}, \\mathrm{FeBr}_{3}$ and (2)\n16-24 $\\mathrm{CH}_{3} \\mathrm{COCl}, \\mathrm{AlCl}_{3}$.\n(a)\n\n(b)\n\n\nPROBLEM The following molecular model of a dimethyl-substituted biphenyl represents the lowest-energy\n$\\mathbf{1 6 - 2 5}$ conformation of the molecule. Why are the two benzene rings tilted at a $63^{\\circ}$ angle to each other\nrather than being in the same plane so that their $p$ orbitals overlap? Why doesn't complete rotation around the single bond joining the two rings occur?\n\n\nPROBLEM How would you synthesize the following compound starting from benzene? More than one step is 16-26 needed.\n\n\nPROBLEM The following compound can't be synthesized using the methods discussed in this chapter. Why 16-27 not?"}
{"id": 1041, "contents": "Mechanisms of Electrophilic Substitutions - \nPROBLEM Aromatic iodination can be carried out with a number of reagents, including iodine monochloride,\n16-28 ICl. What is the direction of polarization of ICl? Propose a mechanism for the iodination of an aromatic ring with ICl.\n\nPROBLEM The sulfonation of an aromatic ring with $\\mathrm{SO}_{3}$ and $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ is reversible. That is, heating\n16-29 benzenesulfonic acid with $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ yields benzene. Show the mechanism of the desulfonation reaction. What is the electrophile?\n\nPROBLEM The carbocation electrophile in a Friedel-Crafts reaction can be generated by an alternate means\n16-30 than reaction of an alkyl chloride with $\\mathrm{AlCl}_{3}$. For example, reaction of benzene with 2-methylpropene in the presence of $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ yields tert-butylbenzene. Propose a mechanism for this reaction.\n\nPROBLEM The $N, N, N$-trimethylammonium group, $-\\stackrel{+}{\\mathrm{N}}\\left(\\mathrm{CH}_{3}\\right)_{3}$, is one of the few groups that is a meta-directing\n16-31 deactivator yet has no electron- withdrawing resonance effect. Explain.\n\nPROBLEM The nitroso group, $-\\mathrm{N}=\\mathrm{O}$, is one of the few nonhalogens that is an ortho- and para-directing\n16-32 deactivator. Explain this behavior by drawing resonance structures of the carbocation intermediates in ortho, meta, and para electrophilic reaction on nitrosobenzene, $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{~N}=\\mathrm{O}$.\n\nPROBLEM Triphenylmethane can be prepared by reaction of benzene and chloroform in the presence of $\\mathrm{AlCl}_{3}$.\n16-33 Propose a mechanism for the reaction.\n\n\nPROBLEM Using resonance structures of the intermediates, explain why bromination of biphenyl occurs at\n16-34 ortho and para positions rather than at meta."}
{"id": 1042, "contents": "Mechanisms of Electrophilic Substitutions - \nPROBLEM Using resonance structures of the intermediates, explain why bromination of biphenyl occurs at\n16-34 ortho and para positions rather than at meta.\n\n\nBiphenyl\n\nPROBLEM Benzene and alkyl-substituted benzenes can be hydroxylated by reaction with $\\mathrm{H}_{2} \\mathrm{O}_{2}$ in the presence\n16-35 of an acidic catalyst. What is the structure of the reactive electrophile? Propose a mechanism for the reaction.\n\n\nAdditional Mechanism Practice\nPROBLEM Addition of HBr to 1-phenylpropene yields only (1-bromopropyl)benzene. Propose a mechanism for\n16-36 the reaction, and explain why none of the other regioisomer is produced.\n\n\nPROBLEM Hexachlorophene, a substance used in the manufacture of germicidal soaps, is prepared by\n16-37 reaction of 2,4,5-trichlorophenol with formaldehyde in the presence of concentrated sulfuric acid. Propose a mechanism for the reaction.\n\n\n\nHexachlorophene\nPROBLEM Benzenediazonium carboxylate decomposes when heated to yield $\\mathrm{N}_{2}, \\mathrm{CO}_{2}$, and a reactive 16-38 substance that can't be isolated. When benzene-diazonium carboxylate is heated in the presence of furan, the following reaction is observed:\n\n\nWhat intermediate is involved in this reaction? Propose a mechanism for its formation.\n\nPROBLEM 4-Chloropyridine undergoes reaction with dimethylamine to yield 4-dimethylaminopyridine.\n16-39 Propose a mechanism for the reaction.\n\n\nPROBLEM Propose a mechanism to account for the following reaction:\n16-40\n\n\nPROBLEM In the Gatterman-Koch reaction, a formyl group ( -CHO ) is introduced directly onto a benzene\n16-41 ring. For example, reaction of toluene with CO and HCl in the presence of mixed $\\mathrm{CuCl} / \\mathrm{AlCl}_{3}$ gives $p$-methylbenzaldehyde. Propose a mechanism."}
{"id": 1043, "contents": "Mechanisms of Electrophilic Substitutions - \nPROBLEM Treatment of p-tert-butylphenol with a strong acid such as $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ yields phenol and\n16-42 2-methylpropene. Propose a mechanism.\nPROBLEM Benzyl bromide is converted into benzaldehyde by heating in dimethyl sulfoxide. Propose a\n16-43 structure for the intermediate, and show the mechanisms of the two steps in the reaction.\n\n\nPROBLEM Propose a mechanism for the Smiles rearrangement below.\n\n16-44\n\n\n\nPROBLEM Because of their conjugation, azo dyes are highly colored compounds and the major artificial color\n16-45 source for textiles and food. Azo dyes are produced by the reaction of aryl diazonium salts with a second aromatic compound. In the product, the aromatic rings are linked by a diazo bridge ( $-\\mathrm{N}=\\mathrm{N}-$ ). From the reactants provided, propose a structure for each azo dye and draw the electron-pushing mechanism.\n(a)\n\n(b)\n\n(c)\n\n$+$\n\n\nReactivity and Orientation of Electrophilic Substitutions\nPROBLEM Identify each of the following groups as an activator or deactivator and as an o,p-director or 16-46 m -director:\n(a) $\\stackrel{\\gtrless}{<}\\left(\\mathrm{CH}_{3}\\right)_{2}$\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM Predict the major product(s) of nitration of the following substances. Which react faster than 16-47 benzene, and which slower?\n(a) Bromobenzene\n(b) Benzonitrile\n(c) Benzoic acid\n(d) Nitrobenzene\n(e) Benzenesulfonic acid (f) ethoxybenzene"}
{"id": 1044, "contents": "Mechanisms of Electrophilic Substitutions - \nPROBLEM Rank the compounds in each group according to their reactivity toward electrophilic substitution.\n16-48 (a) Chlorobenzene, o-dichlorobenzene, benzene\n(b) $p$-Bromonitrobenzene, nitrobenzene, phenol (c) Fluorobenzene, benzaldehyde, $o$-xylene\n(d) Benzonitrile, $p$-methylbenzonitrile, $p$-methoxybenzonitrile\n\nPROBLEM Predict the major monoalkylation products you would expect to obtain from reaction of the\n16-49 following substances with chloromethane and $\\mathrm{AlCl}_{3}$ :\n(a) Bromobenzene\n(b) $m$-Bromophenol\n(c) $p$-Chloroaniline\n(d) 2,4-Dichloronitrobenzene\n(e) 2,4-Dichlorophenol (f) Benzoic acid\n(g) $p$-Methylbenzenesulfonic acid\n(h) 2,5-Dibromotoluene\n\nPROBLEM Name and draw the major product(s) of electrophilic chlorination of the following compounds:\n16-50 (a) m-Nitrophenol\n(b) o-Xylene\n(c) p-Nitrobenzoic acid\n(d) $p$-Bromobenzenesulfonic acid\n\nPROBLEM Predict the major product(s) you would obtain from sulfonation of the following compounds:\n16-51 (a) Fluorobenzene (b) $m$-Bromophenol (c) $m$-Dichlorobenzene (d) 2,4-Dibromophenol\nPROBLEM Rank the following aromatic compounds in the expected order of their reactivity toward\n16-52 Friedel-Crafts alkylation. Which compounds are unreactive?\n(a) Bromobenzene\n(b) Toluene\n(c) Phenol\n(d) Aniline (e) Nitrobenzene\n(f) $p$-Bromotoluene\n\nPROBLEM What product(s) would you expect to obtain from the following reactions:\n16-53\n(a)\n\n(b)\n\n(c)\n\n(d)"}
{"id": 1045, "contents": "Mechanisms of Electrophilic Substitutions - \nPROBLEM What product(s) would you expect to obtain from the following reactions:\n16-53\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM Predict the major product(s) of the following reactions:\n16-54 (a)\n\n(c)\n\n\nOrganic Synthesis\nPROBLEM How would you synthesize the following substances starting from benzene or phenol? Assume that\n16-55 ortho- and para-substitution products can be separated.\n(a) o-Bromobenzoic acid\n(b) $p$-Methoxytoluene\n(c) 2,4,6-Trinitrobenzoic acid\n(d) $m$-Bromoaniline\n\nPROBLEM Starting with benzene as your only source of aromatic compounds, how would you synthesize the\n16-56 following substances? Assume that you can separate ortho and para isomers if necessary.\n(a) p-Chloroacetophenone\n(b) $m$-Bromonitrobenzene\n(c) $o$-Bromobenzenesulfonic acid\n(d) $m$-Chlorobenzenesulfonic acid\n\nPROBLEM Starting with either benzene or toluene, how would you synthesize the following substances:\n16-57 Assume that ortho and para isomers can be separated.\n(a) 2-Bromo-4-nitrotoluene\n(b) 1,3,5-Trinitrobenzene\n(c) 2,4,6-Tribromoaniline\n(d) m -Fluorobenzoic acid\n\nPROBLEM As written, the following syntheses have flaws. What is wrong with each?\n16-58 (a)\n\n(b)\n\n(c)"}
{"id": 1046, "contents": "General Problems - \nPROBLEM At what position and on what ring do you expect nitration of 4-bromo-biphenyl to occur? Explain,\n16-59 using resonance structures of the potential intermediates."}
{"id": 1047, "contents": "4-Bromobiphenyl - \nPROBLEM Electrophilic substitution on 3-phenylpropanenitrile occurs at the ortho and para positions, but\n16-60 reaction with 3 -phenylpropenenitrile occurs at the meta position. Explain, using resonance structures of the intermediates.\n\n\n3-Phenylpropanenitrile\n\n\n3-Phenylpropenenitrile\n\nPROBLEM At what position, and on what ring, would you expect the following substances to undergo\n16-61 electrophilic substitution?\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM At what position, and on what ring, would you expect bromination of benzanilide to occur? Explain\n16-62 by drawing resonance structures of the intermediates."}
{"id": 1048, "contents": "Benzanilide - \nPROBLEM Would you expect the Friedel-Crafts reaction of benzene with ( $R$ )-2-chlorobutane to yield optically\n16-63 active or racemic product? Explain.\n\nPROBLEM How would you synthesize the following substances starting from benzene?"}
{"id": 1049, "contents": "16-64 (a) - \n(b)\n\n(c)\n\n\nPROBLEM The compound MON-0585 is a nontoxic, biodegradable larvicide that is highly selective against\n16-65 mosquito larvae. Synthesize MON-0585 using either benzene or phenol as a source of the aromatic rings.\n\n\nMON-0585\n\nPROBLEM Phenylboronic acid, $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{~B}(\\mathrm{OH})_{2}$, is nitrated to give $15 \\%$ ortho- substitution product and $85 \\%$ meta.\n16-66 Explain the meta-directing effect of the $-\\mathrm{B}(\\mathrm{OH})_{2}$ group.\n\nPROBLEM Draw resonance structures of the intermediate carbocations in the bromination of naphthalene,\n16-67 and account for the fact that naphthalene undergoes electrophilic substitution at C1 rather than C2.\n\n\nPROBLEM Propose a mechanism for the reaction of 1-chloroanthraquinone with methoxide ion to give the\n16-68 substitution product 1-methoxyanthraquinone. Use curved arrows to show the electron flow in each step.\n\n\nPROBLEM $p$-Bromotoluene reacts with potassium amide to give a mixture of $m$ - and $p$-methylaniline. Explain. 16-69\n\nPROBLEM Propose a mechanism to account for the reaction of benzene with\n16-70 2,2,5,5-tetramethyltetrahydrofuran.\n\n\nPROBLEM How would you synthesize the following compounds from benzene? Assume that ortho and para\n16-71 isomers can be separated.\n(a)\n\n(b)\n\n\nPROBLEM You know the mechanism of HBr addition to alkenes, and you know the effects of various\n16-72 substituent groups on aromatic substitution. Use this knowledge to predict which of the following two alkenes reacts faster with HBr. Explain your answer by drawing resonance structures of the carbocation intermediates.\n\n\nPROBLEM Use your knowledge of directing effects, along with the following data, to deduce the directions of\n16-73 the dipole moments in aniline, bromobenzene, and bromoaniline.\n\n$\\mu=1.53 \\mathrm{D}$"}
{"id": 1050, "contents": "16-64 (a) - \n$\\mu=1.53 \\mathrm{D}$\n\n$\\mu=1.52 \\mathrm{D}$\n\n$\\mu=2.91 \\mathrm{D}$\n\nPROBLEM Identify the reagents represented by the letters a-e in the following scheme:\n16-74\n\n\n\nPROBLEM Phenols (ArOH) are relatively acidic, and the presence of a substituent group on the aromatic ring\n16-75 has a large effect. The $p K_{\\mathrm{a}}$ of unsubstituted phenol, for example, is 9.89 , while that of $p$-nitrophenol is 7.15. Draw resonance structures of the corresponding phenoxide anions and explain the data.\n\nPROBLEM Would you expect $p$-methylphenol to be more acidic or less acidic than unsubstituted phenol?\n16-76 Explain. (See Problem 16-75.)\nPROBLEM Predict the product(s) for each of the following reactions. In each case, draw the resonance forms of 16-77 the intermediate to explain the observed regiochemistry.\n(a)\n\n(b)\n\n(c)\n\n(d)"}
{"id": 1051, "contents": "Alcohols and Phenols - \nFIGURE 17.1 The phenol resveratrol, found in the skin of red grapes, continues to be studied for its potential anti-cancer, antiarthritic, and hypoglycemic properties. (credit: modification of work \"Weinreben-\u00d6tlingen\" by Pierre Likissas/Wikimedia Commons, CC BY 3.0)"}
{"id": 1052, "contents": "CHAPTER CONTENTS - 17.1 Naming Alcohols and Phenols\n17.2 Properties of Alcohols and Phenols\n17.3 Preparation of Alcohols: A Review\n17.4 Alcohols from Carbonyl Compounds: Reduction\n17.5 Alcohols from Carbonyl Compounds: Grignard Reaction\n17.6 Reactions of Alcohols\n17.7 Oxidation of Alcohols\n17.8 Protection of Alcohols\n17.9 Phenols and Their Uses\n17.10 Reactions of Phenols\n17.11 Spectroscopy of Alcohols and Phenols\n\nWHY THIS CHAPTER? Up to this point, we've focused on developing some general ideas of organic reactivity, looking at the chemistry of hydrocarbons and alkyl halides, and examining some of the tools used in structural studies. With that background, it's now time to begin a study of the oxygen-containing functional groups that lie at the heart of organic and biological chemistry. We'll look at alcohols in this chapter and move on to carbonyl compounds in Chapters 19 through 23.\n\nAlcohols and phenols can be thought of as organic derivatives of water in which one of water's hydrogens is replaced by an organic group: $\\mathrm{H}-\\mathrm{O}-\\mathrm{H}$ versus $\\mathrm{R}-\\mathrm{O}-\\mathrm{H}$ and $\\mathrm{Ar}-\\mathrm{O}-\\mathrm{H}$. In practice, the name alcohol is restricted to\ncompounds that have their - OH group bonded to a saturated, $s p^{3}$-hybridized carbon atom, while compounds with their -OH group bonded to a vinylic, $s p^{2}$-hybridized carbon are called enols. We'll look at enols in Chapter 22.\n\n\nAn alcohol\n\n\nA phenol"}
{"id": 1053, "contents": "CHAPTER CONTENTS - 17.1 Naming Alcohols and Phenols\nAn alcohol\n\n\nA phenol\n\n\nAlcohols occur widely in nature and have many industrial and pharmaceutical applications. Methanol, for instance, is one of the most important of all industrial chemicals. Historically, methanol was prepared by heating wood in the absence of air and thus came to be called wood alcohol. Today, approximately 173 million tons ( 50 billion gallons) of methanol is manufactured worldwide each year, most of it by catalytic reduction of carbon monoxide with hydrogen gas. Methanol is toxic to humans, causing blindness in small doses ( 15 mL ) and death in larger amounts ( $100-250 \\mathrm{~mL}$ ). Industrially, it is used both as a solvent and as a starting material for production of formaldehyde $\\left(\\mathrm{CH}_{2} \\mathrm{O}\\right)$ and acetic acid $\\left(\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right)$.\n\n\nEthanol was one of the first organic chemicals to be prepared and purified. Its production by fermentation of grains and sugars has been carried out for perhaps 9000 years, and its purification by distillation goes back at least as far as the 12 th century. Today, approximately 88 million tons ( 26 billion gallons) of ethanol are produced worldwide each year, most of it by fermentation of corn, barley, sorghum, and other plant sources. Almost all of this ethanol is used for bus and automobile fuel.\n\nEthanol for industrial use as a solvent or chemical intermediate is largely obtained by acid-catalyzed hydration of ethylene at high temperature.\n\n\nPhenols occur widely throughout nature and also serve as intermediates in the industrial synthesis of products as diverse as adhesives and antiseptics. Phenol itself is a general disinfectant found in coal tar; methyl salicylate is a flavoring agent found in oil of wintergreen; and urushiols are the allergenic constituents of poison oak and poison ivy. Note that the word phenol is the name both of the specific compound (hydroxybenzene) and of the class of compounds.\n\n\nPhenol (also known as carbolic acid)\n\n\nMethyl salicylate\n\n\nUrushiols ( $\\mathrm{R}=$ different $\\mathrm{C}_{15}$ alkyl and alkenyl chains)"}
{"id": 1054, "contents": "CHAPTER CONTENTS - 17.1 Naming Alcohols and Phenols\nAlcohols are classified as primary $\\left(1^{\\circ}\\right)$, secondary $\\left(2^{\\circ}\\right)$, or tertiary $\\left(3^{\\circ}\\right)$, depending on the number of organic groups bonded to the hydroxyl-bearing carbon."}
{"id": 1055, "contents": "A primary $\\left(\\mathbf{1}^{\\circ}\\right)$ alcohol A secondary (2 ${ }^{\\circ}$ ) alcohol A tertiary ( $\\mathbf{3}^{\\circ}$ ) alcohol - \nSimple alcohols are named in the IUPAC system as derivatives of the parent alkane, using the suffix -ol."}
{"id": 1056, "contents": "RULE 1 - \nSelect the longest carbon chain containing the hydroxyl group, and derive the parent name by replacing the $-e$ ending of the corresponding alkane with $-o l$. The $-e$ is deleted to prevent the occurrence of two adjacent vowels: propanol rather than propaneol, for example.\n\nRULE 2\nNumber the alkane chain beginning at the end nearer the hydroxyl group.\nRULE 3\nNumber the substituents according to their position on the chain, and write the name, listing the substituents in alphabetical order and identifying the position to which the -OH is bonded. Note that in naming cis-1,4-cyclohexanediol, the final -e of cyclohexane is not deleted because the next letter, $d$, is not a vowel; that is, cyclohexanediol rather than cyclohexandiol. Also, as with alkenes (Section 7.3), newer IUPAC naming recommendations place the locant immediately before the suffix rather than before the parent.\n\n\n2-Methyl-2-pentanol\n(New: 2-Methylpentan-2-ol) (New: cis-Cyclohexane-1,4-diol)\n\n\n3-Phenyl-2-butanol\n\nSome simple and widely occurring alcohols have common names that are accepted by IUPAC. For example:"}
{"id": 1057, "contents": "Allyl alcohol (2-propen-1-ol) - \ntert-Butyl alcohol (2-methyl-2-propanol)\n\n$$\n\\mathrm{HOCH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}\n$$\n\nEthylene glycol (1,2-ethanediol)\n\n\nGlycerol (1,2,3-propanetriol)\n\nPhenols are named as described previously for aromatic compounds according to the rules discussed in Section 15.1, with -phenol used as the parent name rather than -benzene.\n\n\nPROBLEM Give IUPAC names for the following compounds:\n17-1 (a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n\nPROBLEM Draw structures corresponding to the following IUPAC names:\n17-2 (a) (Z)-2-Ethyl-2-buten-1-ol\n(b) 3-Cyclohexen-1-ol\n(c) trans-3-Chlorocycloheptanol\n(d) 1,4-Pentanediol\n(e) 2,6-Dimethylphenol\n(f)\n-(2-Hydroxyethyl)phenol"}
{"id": 1058, "contents": "Allyl alcohol (2-propen-1-ol) - 17.2 Properties of Alcohols and Phenols\nAlcohols and phenols have nearly the same geometry around the oxygen atom as water. The $\\mathrm{R}-\\mathrm{O}-\\mathrm{H}$ bond angle has an approximately tetrahedral value ( $108.5^{\\circ}$ in methanol, for instance), and the oxygen atom is $s p^{3}$-hybridized.\n\nAlso like water, alcohols and phenols have higher boiling points than might be expected, because of hydrogenbonding (Section 2.12). A positively polarized -OH hydrogen atom from one molecule is attracted to a lone pair of electrons on the electronegative oxygen atom of another molecule, resulting in a weak force that holds the molecules together (FIGURE 17.2). These intermolecular attractions must be overcome for a molecule to break free from the liquid and enter the vapor state, so the boiling temperature is raised. For example, 1-propanol ( $\\mathrm{MW}=60$ ), butane $(\\mathrm{MW}=58)$, and chloroethane $(\\mathrm{MW}=65)$ have similar molecular weights, yet 1-propanol boils at $97^{\\circ} \\mathrm{C}$, compared with $-0.5^{\\circ} \\mathrm{C}$ for the alkane and $12.5^{\\circ} \\mathrm{C}$ for the chloroalkane.\n\n\nFIGURE 17.2 Hydrogen-bonding in alcohols and phenols. Attraction between a positively polarized -OH hydrogen and a negatively polarized oxygen holds molecules together. The electrostatic potential map of methanol shows the positively polarized -OH hydrogen and the negatively polarized oxygen.\n\nAnother similarity with water is that alcohols and phenols are both weakly basic and weakly acidic. As weak bases, they are reversibly protonated by strong acids to yield oxonium ions, $\\mathrm{ROH}_{2}{ }^{+}$.\n\n\nAn alcohol\nAn oxonium ion\n\n$$\n\\left[\\text { or } \\mathrm{ArOH}+\\mathrm{HX} \\rightleftarrows \\stackrel{+}{\\mathrm{H}}_{2} \\mathrm{ArO}^{-}\\right]\n$$"}
{"id": 1059, "contents": "Allyl alcohol (2-propen-1-ol) - 17.2 Properties of Alcohols and Phenols\nAs weak acids, they dissociate slightly in dilute aqueous solution by donating a proton to water, generating $\\mathrm{H}_{3} \\mathrm{O}^{+}$ and an alkoxide ion, $\\mathrm{RO}^{-}$, or a phenoxide ion, $\\mathrm{ArO}^{-}$.\n\n\nAn alcohol\nAn alkoxide ion\n\n\nRecall from the earlier discussion of acidity in Section 2.7 to Section 2.11 that the strength of any acid HA in water can be expressed by an acidity constant, $K_{\\mathrm{a}}$.\n\n$$\nK_{\\mathrm{a}}=\\frac{\\left[\\mathrm{A}^{-}\\right]\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]}{[\\mathrm{HA}]} \\quad \\mathrm{p} K_{\\mathrm{a}}=-\\log K_{\\mathrm{a}}\n$$\n\nCompounds with a smaller $K_{\\mathrm{a}}$ and larger $\\mathrm{p} K_{\\mathrm{a}}$ are less acidic, whereas compounds with a larger $K_{\\mathrm{a}}$ and smaller $\\mathrm{p} K_{\\mathrm{a}}$ are more acidic. As shown in TABLE 17.1, simple alcohols like methanol and ethanol are about as acidic as water, but the more highly substituted tert-butyl alcohol is somewhat weaker. Substituent groups also have a significant effect: 2,2,2-trifluoroethanol is approximately 3700 times stronger than ethanol, for instance. Phenols and thiols, the sulfur analogs of alcohols, are substantially more acidic than water."}
{"id": 1060, "contents": "Allyl alcohol (2-propen-1-ol) - 17.2 Properties of Alcohols and Phenols\n| Compound | $\\mathrm{p} K_{\\mathrm{a}}$ |  |\n| :---: | :---: | :---: |\n| $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{COH}$ | 18 | Weaker acid |\n| $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$ | 16 |  |\n| $\\mathrm{H}_{2} \\mathrm{O}$ | 15.74 |  |\n| $\\mathrm{CH}_{3} \\mathrm{OH}$ | 15.54 |  |\n| $\\mathrm{CF}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$ | 12.43 |  |\n| p-Aminophenol | 10.46 |  |\n| $\\mathrm{CH}_{3} \\mathrm{SH}$ | 10.3 |  |\n| $p$-Methylphenol | 10.17 |  |\n| Phenol | 9.89 |  |\n| $p$-Chlorophenol | 9.38 | acid |\n| p-Nitrophenol | 7.15 |  |\n\nThe effect of alkyl substitution on alcohol acidity is due primarily to solvation of the alkoxide ion formed on acid dissociation. The more readily the alkoxide ion is solvated by water, the more stable it is, the more its formation is energetically favored, and the greater the acidity of the parent alcohol. For example, the oxygen atom of an unhindered alkoxide ion, such as that from methanol, is sterically accessible and is easily solvated by water. The oxygen atom of a hindered alkoxide ion, however, such as that from tert-butyl alcohol, is less easily solvated and is therefore less stable."}
{"id": 1061, "contents": "Allyl alcohol (2-propen-1-ol) - 17.2 Properties of Alcohols and Phenols\nInductive effects (Section 16.4) are also important in determining alcohol acidities. Electron-withdrawing halogen substituents, for instance, stabilize an alkoxide ion by spreading the charge over a larger volume, thus making the alcohol more acidic. Compare, for instance, the acidities of ethanol ( $\\mathrm{p} K_{\\mathrm{a}}=16$ ) and 2,2,2-trifluoroethanol ( $\\mathrm{p} K_{\\mathrm{a}}=12.43$ ), or of tert-butyl alcohol ( $\\mathrm{p} K_{\\mathrm{a}}=18$ ) and nonafluoro-tert-butyl alcohol $\\left(\\mathrm{p} K_{\\mathrm{a}}=\\right.$ 5.4).\n\nElectron-withdrawing groups stabilize the alkoxide ion and lower the $\\mathrm{p} K_{\\mathrm{a}}$ of the alcohol.\n\n\nBecause alcohols are weak acids, they don't react with weak bases, such as amines or bicarbonate ion, and they only react to a limited extent with metal hydroxides such as NaOH . Alcohols do, however, react with alkali metals and with strong bases such as sodium hydride ( NaH ), sodium amide ( $\\mathrm{NaNH}_{2}$ ), and Grignard reagents ( RMgX ). Alkoxides are themselves bases that are frequently used as reagents in organic chemistry. They are named systematically by adding the -ate suffix to the name of the alcohol. Methanol becomes methanolate, for instance.\n\n\nPhenols are about a million times more acidic than alcohols (TABLE 17.1). They are therefore soluble in dilute aqueous NaOH and can often be separated from a mixture simply by basic extraction into aqueous solution, followed by reacidification."}
{"id": 1062, "contents": "Allyl alcohol (2-propen-1-ol) - 17.2 Properties of Alcohols and Phenols\nPhenols are more acidic than alcohols because the phenoxide anion is resonance-stabilized. Delocalization of the negative charge over the ortho and para positions of the aromatic ring results in increased stability of the phenoxide anion relative to undissociated phenol and in a consequently lower $\\Delta G^{\\circ}$ for dissociation. FIGURE 17.3 compares electrostatic potential maps of an alkoxide ion $\\left(\\mathrm{CH}_{3} \\mathrm{O}^{-}\\right)$with phenoxide ion to show how the negative charge in phenoxide ion is delocalized from oxygen to the ring.\n\n\nFIGURE 17.3 The resonance-stabilized phenoxide ion is more stable than an alkoxide ion. Electrostatic potential maps show how the negative charge is concentrated on oxygen in the methoxide ion but is spread over the aromatic ring in the phenoxide ion.\n\nSubstituted phenols can be either more acidic or less acidic than phenol itself, depending on whether the substituent is electron-withdrawing or electron-donating (Section 16.4). Phenols with an electron-withdrawing substituent are more acidic because these substituents delocalize the negative charge; phenols with an electron-donating substituent are less acidic because these substituents concentrate the charge. The acidifying effect of an electron-withdrawing substituent is particularly noticeable in phenols with a nitro group at the ortho or para position."}
{"id": 1063, "contents": "(8) WORKED EXAMPLE 17.1 - \nPredicting the Relative Acidity of a Substituted Phenol\nIs $p$-hydroxybenzaldehyde more acidic or less acidic than phenol?"}
{"id": 1064, "contents": "Strategy - \nIdentify the substituent on the aromatic ring, and decide whether it is electron-donating or electronwithdrawing. Electron-withdrawing substituents make the phenol more acidic by stabilizing the phenoxide anion, and electron-donating substituents make the phenol less acidic by destabilizing the anion."}
{"id": 1065, "contents": "Solution - \nWe saw in Section 16.4 that a carbonyl group is electron-withdrawing. Thus, $p$-hydroxybenzaldehyde is more\nacidic $\\left(\\mathrm{p} K_{\\mathrm{a}}=7.9\\right)$ than phenol $\\left(\\mathrm{p} K_{\\mathrm{a}}=9.89\\right)$.\n\np-Hydroxybenzaldehyde\n( $\\mathrm{p} K_{\\mathrm{a}}=7.9$ )\n\nPROBLEM The following data for isomeric four-carbon alcohols show that there is a decrease in boiling point\n17-3 with increasing substitution of the OH -bearing carbon. How might you account for this trend?\n\n$$\n\\begin{aligned}\n& \\text { 1-Butanol, bp } 117.5^{\\circ} \\mathrm{C} \\\\\n& \\text { 2-Butanol, bp } 99.5^{\\circ} \\mathrm{C} \\\\\n& \\text { 2-Methyl-2-propanol, bp } 82.2^{\\circ} \\mathrm{C}\n\\end{aligned}\n$$\n\nPROBLEM Rank the following substances in order of increasing acidity:\n17-4 (a) $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CHOH}, \\mathrm{HC} \\equiv \\mathrm{CH},\\left(\\mathrm{CF}_{3}\\right)_{2} \\mathrm{CHOH}, \\mathrm{CH}_{3} \\mathrm{OH}$\n(b) Phenol, $p$-methylphenol, $p$-(trifluoromethyl)phenol\n(c) Benzyl alcohol, phenol, $p$-hydroxybenzoic acid\n\nPROBLEM $p$-Nitrobenzyl alcohol is more acidic than benzyl alcohol, but p-methoxybenzyl alcohol is less 17-5 acidic. Explain."}
{"id": 1066, "contents": "Solution - 17.3 Preparation of Alcohols: A Review\nAlcohols occupy a central position in organic chemistry. They can be prepared from many other kinds of compounds (alkenes, alkyl halides, ketones, esters, and aldehydes, among others), and they can be transformed into an equally wide assortment of compounds (FIGURE 17.4).\n\n\nFIGURE 17.4 The central position of alcohols in organic chemistry. Alcohols can be prepared from, and converted into, many other kinds of compounds.\n\nWe've already seen several methods of alcohol synthesis:\n\n- Alcohols can be prepared by hydration of alkenes. Because the direct hydration of alkenes with aqueous acid is generally a poor reaction in the laboratory, two indirect methods are commonly used. Hydroboration-oxidation yields the syn, non-Markovnikov hydration product (Section 8.5), whereas oxymercuration-demercuration yields the Markovnikov hydration product (Section 8.4).\n\n- 1,2-Diols can be prepared either by direct hydroxylation of an alkene with $\\mathrm{OsO}_{4}$ followed by reduction with $\\mathrm{NaHSO}_{3}$ or by acid-catalyzed hydrolysis of an epoxide (Section 8.7). The $\\mathrm{OsO}_{4}$ reaction occurs with syn stereochemistry to give a cis diol, and epoxide opening occurs with anti stereochemistry to give a trans diol.\n\n\nPROBLEM Predict the products of the following reactions:\n17-6 (a)\n\n(c)"}
{"id": 1067, "contents": "Solution - 17.4 Alcohols from Carbonyl Compounds: Reduction\nThe most general method for preparing alcohols, both in the laboratory and in living organisms, is by reduction of a carbonyl compound. Just as reduction of an alkene adds hydrogen to a $\\mathrm{C}=\\mathrm{C}$ bond to give an alkane (Section 8.6), reduction of a carbonyl compound adds hydrogen to a $\\mathrm{C}=\\mathrm{O}$ bond to give an alcohol. All kinds of carbonyl compounds can be reduced, including aldehydes, ketones, carboxylic acids, and esters."}
{"id": 1068, "contents": "A carbonyl compound - \nAn alcohol"}
{"id": 1069, "contents": "Reduction of Aldehydes and Ketones - \nAldehydes are easily reduced to give primary alcohols, and ketones are reduced to give secondary alcohols.\n\n\nDozens of reagents are used in the laboratory to reduce aldehydes and ketones, depending on the circumstances, but sodium borohydride, $\\mathrm{NaBH}_{4}$, is usually chosen because of its safety and ease of handling. Sodium borohydride is a white, crystalline solid that can be weighed in the open atmosphere and used in either water or alcohol solution."}
{"id": 1070, "contents": "Butanal - \n1-Butanol (85\\%)\n(a $1^{\\circ}$ alcohol)"}
{"id": 1071, "contents": "Ketone reduction - \nLithium aluminum hydride, $\\mathrm{LiAlH}_{4}$, is another reducing agent often used for reduction of aldehydes and ketones. A grayish powder that is soluble in ether and tetrahydrofuran, $\\mathrm{LiAlH}_{4}$ is much more reactive than $\\mathrm{NaBH}_{4}$ but also much more dangerous. It reacts violently with water and decomposes explosively when heated above $120^{\\circ} \\mathrm{C}$.\n\n\n2-Cyclohexenone\n2-Cyclohexenol (94\\%)\nWe'll defer a detailed discussion of these reductions until Chapter 19. For the moment, we'll simply note that they involve the addition of a nucleophilic hydride ion $\\left(: \\mathrm{H}^{-}\\right)$to the positively polarized, electrophilic carbon atom of the carbonyl group. The initial product is an alkoxide ion, which is protonated by addition of $\\mathrm{H}_{3} \\mathrm{O}^{+}$in a second step to yield the alcohol product.\n\n\nIn living organisms, aldehyde and ketone reductions are carried out by either of the coenzymes NADH (reduced nicotinamide adenine dinucleotide) or NADPH (reduced nicotinamide adenine dinucleotide phosphate). Although these biological \"reagents\" are much more complex structurally than $\\mathrm{NaBH}_{4}$ or $\\mathrm{LiAlH}_{4}$, the mechanisms of laboratory and biological reactions are similar. The coenzyme acts as a hydride-ion donor to give an alkoxide anion, and the intermediate anion is then protonated by acid. An example is the reduction of acetoacetyl ACP to $\\beta$-hydroxybutyryl ACP, a step in the biological synthesis of fats (FIGURE 17.5). Note that the pro- $R$ hydrogen of NADPH is the one transferred in this example. Enzyme-catalyzed reactions usually occur with high specificity, although it's not usually possible to predict the stereochemical result before the fact.\n\n\nFIGURE 17.5 The biological reduction of a ketone (acetoacetyl ACP) to an alcohol ( $\\beta$-hydroxybutyryl ACP) by NADPH."}
{"id": 1072, "contents": "Reduction of Carboxylic Acids and Esters - \nCarboxylic acids and esters are reduced to give primary alcohols.\n\n\nA carboxylic acid\nAn ester\nA primary alcohol\nThese reactions aren't as rapid as the reductions of aldehydes and ketones. $\\mathrm{NaBH}_{4}$ reduces esters very slowly and does not reduce carboxylic acids at all. Instead, carboxylic acid and ester reductions are usually carried out with the more reactive reducing agent $\\mathrm{LiAlH}_{4}$. All carbonyl groups, including acids, esters, ketones, and aldehydes, are reduced by $\\mathrm{LiAlH}_{4}$. Note that one hydrogen atom is delivered to the carbonyl carbon atom during aldehyde and ketone reductions but that two hydrogens become bonded to the former carbonyl carbon during carboxylic acid and ester reductions. We'll defer a discussion of the mechanisms of these reactions until Chapter 21.\n\nCarboxylic acid reduction\n\n\nEster reduction\n\n\nMethyl 2-pentenoate 2-Penten-1-ol (91\\%)"}
{"id": 1073, "contents": "Identifying a Reactant, Given the Product - \nWhat carbonyl compounds would you reduce to obtain the following alcohols?\n(a)\n\n(b)"}
{"id": 1074, "contents": "Strategy - \nIdentify the target alcohol as primary, secondary, or tertiary. A primary alcohol can be prepared by reduction of an aldehyde, an ester, or a carboxylic acid; a secondary alcohol can be prepared by reduction of a ketone; and a tertiary alcohol can't be prepared by reduction."}
{"id": 1075, "contents": "Solution - \n(a) The target molecule is a secondary alcohol, which can be prepared only by reduction of a ketone. Either $\\mathrm{NaBH}_{4}$ or $\\mathrm{LiAlH}_{4}$ can be used.\n\n(b) The target molecule is a primary alcohol, which can be prepared by reduction of an aldehyde, an ester, or a carboxylic acid. $\\mathrm{LiAlH}_{4}$ is needed for the ester and carboxylic acid reductions.\n\n\nPROBLEM What reagent would you use to accomplish each of the following reactions?\n17-7 (a)\n\n\n(b)\n\n\n\nPROBLEM What carbonyl compounds give the following alcohols on reduction with $\\mathrm{LiAlH}_{4}$ ? Show all 17-8 possibilities.\n(a)\n\n(b)\n\n(c)\n\n(d) $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CHCH}_{2} \\mathrm{OH}$"}
{"id": 1076, "contents": "Solution - 17.5 Alcohols from Carbonyl Compounds: Grignard Reaction\nGrignard reagents (RMgX), prepared by reaction of organohalides with magnesium (Section 10.6), react with carbonyl compounds to yield alcohols in much the same way that hydride reducing agents do. Just as carbonyl reduction involves addition of a hydride ion nucleophile to the $\\mathrm{C}=\\mathrm{O}$ bond, Grignard reaction involves addition of a carbanion nucleophile ( $\\mathrm{R} \\mathrm{P}^{-+} \\mathrm{MgX}$ ).\n\n$$\n\\left[\\mathrm{R}-\\mathrm{X}+\\mathrm{Mg} \\longrightarrow \\underset{\\begin{array}{c}\n\\text { A Grignard } \\\\\n\\text { reagent }\n\\end{array}}{\\stackrel{\\delta^{-}}{\\mathrm{R}-\\mathrm{ingX}} \\mathrm{\\delta}^{+}}\\left\\{\\begin{array}{l}\n\\mathrm{R}=1^{\\circ}, 2^{\\circ}, \\text { or } 3^{\\circ} \\text { alkyl, aryl, or vinylic } \\\\\n\\mathrm{X}=\\mathrm{Cl}, \\mathrm{Br}, \\mathrm{I}\n\\end{array}\\right]\\right.\n$$\n\nThe nucleophilic addition reaction of Grignard reagents to carbonyl compounds has no direct counterpart in biological chemistry because organomagnesium compounds are too strongly basic to exist in an aqueous medium. Nevertheless, the reaction is worth understanding for two reasons. First, the reaction is an unusually broad and useful method of alcohol synthesis and demonstrates again the relative freedom with which chemists can operate in the laboratory. Second, the reaction does have an indirect biological counterpart, for we'll see in Chapter 23 that the addition of stabilized carbon nucleophiles to carbonyl compounds is used in almost all metabolic pathways as the major process for forming carbon-carbon bonds.\n\nAs examples of their addition to carbonyl compounds, Grignard reagents react with formaldehyde, $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{O}$, to give primary alcohols, with aldehydes to give secondary alcohols, and with ketones to give tertiary alcohols."}
{"id": 1077, "contents": "Aldehyde reaction - \nKetone reaction\n\n\nEsters react with Grignard reagents to yield tertiary alcohols in which two of the substituents bonded to the hydroxyl-bearing carbon have come from the Grignard reagent, just as $\\mathrm{LiAlH}_{4}$ reduction of an ester adds two hydrogens.\n\n\nCarboxylic acids don't give addition products when treated directly with Grignard reagents because the acidic carboxyl hydrogen reacts with the basic Grignard reagent to yield a hydrocarbon and the magnesium salt of the acid.\n\n$$\n[R-X+M g \\quad R-M g X]\n$$"}
{"id": 1078, "contents": "A carboxylic acid - \nA carboxylic acid salt\nCarboxylic acids do, however, react with Grignard reagents to give ketones if they are first treated with $i-\\mathrm{Pr}_{2} \\mathrm{NMgCl}-\\mathrm{LiCl}$, called the turbo-Hauser base, to form a complex and increase the electrophilicity of the carboxylate anion toward nucleophilic addition with a Grignard reagent.\n\n\nThe Grignard reaction, although useful, does have limitations. One major problem is that a Grignard reagent can't be prepared from an organohalide if other reactive functional groups are present in the same molecule. For example, a compound that is both an alkyl halide and a ketone can't form a Grignard reagent because it would react with itself. Similarly, a compound that is both an alkyl halide and a carboxylic acid, alcohol, or amine can't form a Grignard reagent because the acidic $\\mathrm{RCO}_{2} \\mathrm{H}, \\mathrm{ROH}$, or $\\mathrm{RNH}_{2}$ hydrogen present in the same molecule would react with the basic Grignard reagent as rapidly as it forms. In general, Grignard reagents can't be prepared from alkyl halides that contain the following functional groups (FG):\n\n\nAs with the reduction of carbonyl compounds discussed in the previous section, we'll defer a detailed treatment of the Grignard reactions until Chapter 19. For the moment, it's sufficient to note that Grignard reagents act as nucleophilic carbanions ( $: \\mathrm{R}^{-}$) and that their addition to a carbonyl compound is analogous to the addition of hydride ion. The intermediate is an alkoxide ion, which is protonated by addition of $\\mathrm{H}_{3} \\mathrm{O}^{+}$in a second step."}
{"id": 1079, "contents": "Using a Grignard Reaction to Synthesize an Alcohol - \nHow could you use the addition of a Grignard reagent to a ketone to synthesize 2-phenyl-2-butanol?"}
{"id": 1080, "contents": "Strategy - \nDraw the product, and identify the three groups bonded to the alcohol carbon atom. One of the three will have come from the Grignard reagent, and the remaining two will have come from the ketone."}
{"id": 1081, "contents": "Solution - \n2-Phenyl-2-butanol has a methyl group, an ethyl group, and a phenyl group ( $-\\mathrm{C}_{6} \\mathrm{H}_{5}$ ) attached to the alcohol carbon atom. Thus, the possibilities are addition of ethylmagnesium bromide to acetophenone, addition of methylmagnesium bromide to propiophenone, and addition of phenylmagnesium bromide to 2-butanone."}
{"id": 1082, "contents": "Using a Grignard Reaction to Synthesize an Alcohol - \nHow could you use the reaction of a Grignard reagent with a carbonyl compound to synthesize 2-methyl-2-pentanol?"}
{"id": 1083, "contents": "Strategy - \nDraw the product, and identify the three groups bonded to the alcohol carbon atom. If the three groups are all different, the starting carbonyl compound must be a ketone. If two of the three groups are identical, the starting carbonyl compound could be either a ketone or an ester."}
{"id": 1084, "contents": "Solution - \nIn the present instance, the product is a tertiary alcohol with two methyl groups and one propyl group. Starting from a ketone, the possibilities are addition of methylmagnesium bromide to 2 -pentanone and addition of propylmagnesium bromide to acetone."}
{"id": 1085, "contents": "Acetone - \nStarting from an ester, the only possibility is addition of methylmagnesium bromide to an ester of butanoic acid, such as methyl butanoate.\n\n\nPROBLEM Show the products obtained from addition of methylmagnesium bromide to the following 17-9 compounds:\n(a) Cyclopentanone\n(b) Benzophenone (diphenyl ketone)\n(c) 3-Hexanone\n\nPROBLEM Use a Grignard reaction to prepare the following alcohols:\n17-10 (a) 2-Methyl-2-propanol (b) 1-Methylcyclohexanol (c) 3-Methyl-3-pentanol"}
{"id": 1086, "contents": "(d) 2-Phenyl-2-butanol (e) Benzyl alcohol (f) 4-Methyl-1-pentanol - \nPROBLEM Use the reaction of a Grignard reagent with a carbonyl compound to synthesize the following 17-11 compound:"}
{"id": 1087, "contents": "(d) 2-Phenyl-2-butanol (e) Benzyl alcohol (f) 4-Methyl-1-pentanol - 17.6 Reactions of Alcohols\nWe've already seen several reactions of alcohols-their conversion into alkyl halides and tosylates in Section 10.5 and their dehydration to give alkenes in Section 8.1-albeit without mechanistic details. Let's now look at those details."}
{"id": 1088, "contents": "Conversion of Alcohols into Alkyl Halides - \nTertiary alcohols react with either HCl or HBr at $0^{\\circ} \\mathrm{C}$ by an $\\mathrm{S}_{\\mathrm{N}} 1$ mechanism through a carbocation intermediate. Primary and secondary alcohols are much more resistant to acid, however, and are best converted into halides by treatment with either $\\mathrm{SOCl}_{2}$ or $\\mathrm{PBr}_{3}$ through an $\\mathrm{S}_{\\mathrm{N}} 2$ mechanism.\n\nThe reaction of a tertiary alcohol with HX takes place by an $\\mathrm{S}_{\\mathrm{N}} 1$ mechanism when acid protonates the hydroxyl oxygen atom. Water is expelled to generate a carbocation, and the cation reacts with nucleophilic halide ion to give the alkyl halide product.\n\n\nThe reactions of primary and secondary alcohols with $\\mathrm{SOCl}_{2}$ and $\\mathrm{PBr}_{3}$ take place by $\\mathrm{S}_{\\mathrm{N}} 2$ mechanisms. Hydroxide ion itself is too poor a leaving group to be displaced by nucleophiles in $\\mathrm{S}_{\\mathrm{N}} 2$ reactions, but reaction of an alcohol with $\\mathrm{SOCl}_{2}$ or $\\mathrm{PBr}_{3}$ converts the -OH into a much better leaving group, either a chlorosulfite (-OSOCl) or a dibromophosphite $\\left(-\\mathrm{OPBr}_{2}\\right)$, which is readily expelled by backside nucleophilic substitution."}
{"id": 1089, "contents": "Conversion of Alcohols into Tosylates - \nAlcohols react with $p$-toluenesulfonyl chloride (tosyl chloride, $p$ - TosCl ) in pyridine solution to yield alkyl\ntosylates, ROTos (Section 11.1). Only the $\\mathrm{O}-\\mathrm{H}$ bond of the alcohol is broken in this reaction; the $\\mathrm{C}-\\mathrm{O}$ bond remains intact, so no change of configuration occurs if the oxygen is attached to a chirality center. The resultant alkyl tosylates behave much like alkyl halides, undergoing both $\\mathrm{S}_{\\mathrm{N}} 1$ and $\\mathrm{S}_{\\mathrm{N}} 2$ substitution reactions.\n\n\nOne of the most important reasons for using tosylates in $S_{N} 2$ reactions is stereochemical. The $S_{N} 2$ reaction of an alcohol with an alkyl halide proceeds with two inversions of configuration-one to make the halide from the alcohol and one to substitute the halide-and yields a product with the same stereochemistry as the starting alcohol. The $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of an alcohol with a tosylate, however, proceeds with only one inversion and yields a product of opposite stereochemistry to the starting alcohol. FIGURE 17.6 shows a series of reactions on the $R$ enantiomer of 2 -octanol that illustrates these stereochemical relationships.\n\n\nFIGURE 17.6 Stereochemical consequences of $\\mathbf{S}_{\\mathbf{N}} \\mathbf{2}$ reactions on derivatives of ( $\\boldsymbol{R}$ )-2-octanol. Substitution through the halide gives a product with the same stereochemistry as the starting alcohol; substitution through the tosylate gives a product with opposite stereochemistry to the starting alcohol.\n\nPROBLEM How would you carry out the following transformation, a step used in the commercial synthesis of 17-12 (S)-ibuprofen?"}
{"id": 1090, "contents": "Dehydration of Alcohols to Yield Alkenes - \nA third important reaction of alcohols, both in the laboratory and in biological pathways, is their dehydration to give alkenes. Because of the usefulness of the reaction, a number of methods have been devised for carrying out dehydrations. One method that works particularly well for tertiary alcohols is the acid-catalyzed reaction discussed in Section 8.1. For example, treatment of 1-methylcyclohexanol with warm, aqueous sulfuric acid in a solvent such as tetrahydrofuran results in loss of water and formation of 1-methylcyclohexene."}
{"id": 1091, "contents": "1-Methylcyclohexanol 1-Methylcyclohexene (91\\%) - \nAcid-catalyzed dehydrations usually follow Zaitsev's rule (Section 11.7) and yield the more stable alkene as the major product. Thus, 2-methyl-2-butanol gives primarily 2-methyl-2-butene (trisubstituted double bond)\nrather than 2-methyl-1-butene (disubstituted double bond).\n\n\n2-Methyl-2-butanol\n2-Methyl-2-butene 2-Methyl-1-butene (trisubstituted) (disubstituted) Major product Minor product\nThis reaction is an E1 process (Section 11.10) and occurs by the three-step mechanism shown in FIGURE 17.7 on the next page. Protonation of the alcohol oxygen is followed by unimolecular loss of water to generate a carbocation intermediate and final loss of a proton from the neighboring carbon atom to complete the process. As with most E1 reactions, tertiary alcohols react fastest because they lead to stabilized, tertiary carbocation intermediates. Secondary alcohols can be made to react, but the conditions are severe $\\left(75 \\% \\mathrm{H}_{2} \\mathrm{SO}_{4}, 100^{\\circ} \\mathrm{C}\\right)$ and sensitive molecules don't survive.\n\nFIGURE 17.7 MECHANISM\nMechanism for the acid-catalyzed dehydration of a tertiary alcohol to yield an\nalkene. The process is an E1 reaction and involves a carbocation intermediate.\n(1) Two electrons from the oxygen atom bond to $\\mathrm{H}^{+}$, yielding a protonated alcohol intermediate.\n\n(2) The carbon-oxygen bond breaks, and the two electrons from the bond stay with oxygen, leaving a carbocation intermediate.\n\n\nCarbocation\n(3) Two electrons from a neighboring carbon-hydrogen bond form the alkene $\\pi$ bond, and $\\mathrm{H}^{+}$(a proton) is eliminated."}
{"id": 1092, "contents": "1-Methylcyclohexanol 1-Methylcyclohexene (91\\%) - \nCarbocation\n(3) Two electrons from a neighboring carbon-hydrogen bond form the alkene $\\pi$ bond, and $\\mathrm{H}^{+}$(a proton) is eliminated.\n\n\n\nTo circumvent the need for strong acid and allow the dehydration of secondary alcohols in a gentler way, reagents have been developed that are effective under mild, basic conditions. One such reagent, phosphorus oxychloride $\\left(\\mathrm{POCl}_{3}\\right)$ in the basic amine solvent pyridine, is often able to effect the dehydration of secondary and\ntertiary alcohols at $0^{\\circ} \\mathrm{C}$.\n\n\n1-Methylcyclohexanol\n1-Methylcyclohexene (96\\%)\nAlcohol dehydrations carried out with $\\mathrm{POCl}_{3}$ in pyridine take place by an E2 mechanism, as shown in FIGURE 17.8. Because hydroxide ion is a poor leaving group (Section 11.3), direct E 2 elimination of water from an alcohol does not occur. On reaction with $\\mathrm{POCl}_{3}$, however, the -OH group is converted into a dichlorophosphate $\\left(-\\mathrm{OPOCl}_{2}\\right)$, which is a good leaving group and is readily eliminated. Pyridine is both the reaction solvent and the base that removes a neighboring proton in the E2 elimination step.\n\nFIGURE 17.8 MECHANISM\n\nMechanism for the dehydration of secondary and tertiary alcohols by reaction with $\\mathrm{POCl}_{3}$ in pyridine. The reaction is an E2 process.\n(1) The alcohol hydroxyl group reacts with $\\mathrm{POCl}_{3}$ to form a dichlorophosphate intermediate.\n\n2 E2 elimination then occurs by the usual one-step mechanism as the amine base pyridine abstracts a proton from the neighboring carbon at the same time that the\ndichlorophosphate group is leaving."}
{"id": 1093, "contents": "1-Methylcyclohexanol 1-Methylcyclohexene (91\\%) - \n2 E2 elimination then occurs by the usual one-step mechanism as the amine base pyridine abstracts a proton from the neighboring carbon at the same time that the\ndichlorophosphate group is leaving.\n\n\n\n\nAs noted in Section 11.11, biological dehydrations are also common and usually occur by an E1cB mechanism on a substrate in which the - OH group is two carbons away from a carbonyl group. One example occurs in the biosynthesis of the aromatic amino acid tyrosine. A base (:B) first abstracts a proton from the carbon adjacent to the carbonyl group, and the anion intermediate then expels the - OH group with simultaneous protonation by an acid (HA) to form water.\n\n\nPROBLEM What product(s) would you expect from dehydration of the following alcohols with $\\mathrm{POCl}_{3}$ in 17-13 pyridine? Indicate the major product in each case.\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)"}
{"id": 1094, "contents": "Conversion of Alcohols into Esters - \nAlcohols react with carboxylic acids to give esters, a reaction that is common in both the laboratory and living organisms. In the laboratory, the reaction can be carried out in a single step if a strong acid is used as catalyst. More frequently, though, the reactivity of the carboxylic acid is enhanced by first converting it into a carboxylic acid chloride, which then reacts with the alcohol.\n\n\nIn living organisms, a similar process occurs, though a thioester or acyl adenosyl phosphate acts as substrate rather than a carboxylic acid chloride. We'll look at the mechanisms of these reactions in Chapter 21.\n\n\nAn acyl adenosyl phosphate"}
{"id": 1095, "contents": "Conversion of Alcohols into Esters - 17.7 Oxidation of Alcohols\nPerhaps the most valuable reaction of alcohols is their oxidation to give carbonyl compounds-the opposite of the reduction of carbonyl compounds to give alcohols. Primary alcohols are oxidized either to aldehydes or carboxylic acids, and secondary alcohols are oxidized to ketones, but tertiary alcohols don't normally react with most oxidizing agents.\n\nPrimary alcohol\n\n\nSecondary alcohol\n\n\nA ketone\n\nTertiary alcohol\n\n\nPrimary and secondary alcohols can be oxidized by any of a number of reagents, including $\\mathrm{CrO}_{3}$ in aqueous acetic acid and $\\mathrm{KMnO}_{4}$ in aqueous NaOH , but chromium-based reagents are rarely used today because of their toxicity and fire danger. Today, primary and secondary alcohols are oxidized to aldehydes and ketones, respectively, using the iodine-containing Dess-Martin periodinane in dichloromethane solution.\n\n\nPrimary alcohols are oxidized to carboxylic acids by heating with $\\mathrm{KMnO}_{4}$ in a basic aqueous solution. An aldehyde is involved as an intermediate in the $\\mathrm{KMnO}_{4}$ reaction but can't usually be isolated because it is further oxidized too rapidly.\n\n\nAll these oxidations occur by a mechanism that is closely related to the E2 reaction (Section 11.8). In the Dess-Martin oxidation, for instance, the first step involves a substitution reaction between the alcohol and the $\\mathrm{I}(\\mathrm{V})$ reagent to form a new periodinane intermediate, followed by expulsion of reduced $\\mathrm{I}(\\mathrm{III})$ as the leaving group. Similarly, when a $\\mathrm{Cr}(\\mathrm{VI})$ reagent, such as $\\mathrm{CrO}_{3}$, is the oxidant, reaction with the alcohol gives a chromate\nintermediate followed by expulsion of a reduced $\\mathrm{Cr}(\\mathrm{VI})$ species."}
{"id": 1096, "contents": "Conversion of Alcohols into Esters - 17.7 Oxidation of Alcohols\nBiological alcohol oxidations are the opposite of biological carbonyl reductions and are facilitated by the coenzymes $\\mathrm{NAD}^{+}$and $\\mathrm{NADP}^{+}$. A base removes the -OH proton, and the alkoxide ion transfers a hydride ion to the coenzyme. An example is the oxidation of sn-glycerol 3-phosphate to dihydroxyacetone phosphate, a step in the biological metabolism of fats (FIGURE 17.9). Note that addition occurs exclusively on the Re face of the NAD ${ }^{+}$ ring, adding a hydrogen with pro- $R$ stereochemistry.\n\n\n\nFIGURE 17.9 The biological oxidation of an alcohol (sn-glycerol 3-phosphate) to give a ketone (dihydroxyacetone phosphate). This mechanism is the exact opposite of the ketone reduction shown previously in FIGURE 17.5.\n\nPROBLEM What alcohols would give the following products on oxidation?\n17-14 (a)\n\n(b)\n\n(c)\n\n\nPROBLEM What products would you expect from oxidation of the following compounds with the Dess-Martin 17-15 periodinane?\n(a) 1-Hexanol\n(b) 2-Hexanol\n(c) Hexanal"}
{"id": 1097, "contents": "Conversion of Alcohols into Esters - 17.8 Protection of Alcohols\nIt often happens, particularly during the synthesis of complex molecules, that one functional group in a molecule interferes with an intended reaction on another functional group elsewhere in the same molecule. We saw earlier in this chapter, for instance, that a Grignard reagent can't be prepared from an alcohol-containing halidel because the $\\mathrm{C}-\\mathrm{Mg}$ bond is not compatible with the presence of an acidic -OH group in the same molecule.\n\n\nWhen this kind of incompatibility arises, it's sometimes possible to circumvent the problem by protecting the interfering functional group. Protection involves three steps: (1) introducing a protecting group to block the interfering function, (2) carrying out the desired reaction, and (3) removing the protecting group.\n\nOne of the more common methods of alcohol protection involves reaction with a chlorotrialkylsilane, $\\mathrm{Cl}^{2}-\\mathrm{SiR}_{3}$, to yield a trialkylsilyl ether, $\\mathrm{R}^{\\prime}-\\mathrm{O}-\\mathrm{SiR}_{3}$. (Chloro-tert-butyldimethylsilane), usually abbreviated either TBS or TBDMS is often used, as is chlorotrimethylsilane (TMS), and the reaction is carried out in the presence of a base, such as triethylamine, to help form the alkoxide anion from the alcohol and to remove the HCl by-product from the reaction.\n\n\nFor example:\n\n\nThe ether-forming step is an $\\mathrm{S}_{\\mathrm{N}} 2$-like reaction of the alkoxide ion on the silicon atom, with concurrent loss of the leaving chloride anion. Unlike most $\\mathrm{S}_{\\mathrm{N}} 2$ reactions, though, this reaction takes place at a tertiary center-a trialkyl-substituted silicon atom. The reaction occurs because silicon, a third-row atom, is larger than carbon and forms longer bonds. In addition, three alkyl substituents attached to silicon offer little steric hindrance to ether formation."}
{"id": 1098, "contents": "Conversion of Alcohols into Esters - 17.8 Protection of Alcohols\nLike most other ethers, which we'll study in the next chapter, trialkylsilylethers are relatively unreactive. They have no acidic hydrogens and don't react with oxidizing agents, reducing agents, or Grignard reagents. They do, however, react with aqueous acid or with fluoride ion to regenerate the alcohol. Tetrabutylammonium fluoride is often used.\n\n\nTo solve the problem posed at the beginning of this section, note that it's possible to use an alcohol-containing halide in a Grignard reaction by employing a protection sequence. For example, we can add 3-bromo-1-propanol to acetaldehyde by the route shown in FIGURE 17.10.\n\nStep 1 Protect alcohol:\n\n\nStep 2a Form Grignard reagent:\n\n\nStep 2b Do Grignard reaction:\n\n\nStep 3 Remove protecting group:\nFIGURE 17.10 Use of a trialkylsilyl-protected alcohol during a Grignard reaction.\nPROBLEM Propose a mechanism for the reaction of cyclohexyl TMS ether with LiF to deprotect the silyl ether. 17-16"}
{"id": 1099, "contents": "Conversion of Alcohols into Esters - 17.9 Phenols and Their Uses\nThe outbreak of World War I provided a stimulus for the industrial preparation of large amounts of synthetic phenol, which was needed as a raw material to manufacture the explosive, picric acid (2,4,6-trinitrophenol). Today, approximately 132 million tons of phenol is manufactured worldwide each year for use in such products as Bakelite resin and adhesives for binding plywood.\n\nPhenol was manufactured for many years by the Dow process, in which chlorobenzene reacts with NaOH at high temperature and pressure (Section 16.7). Now, however, an alternative synthesis uses isopropylbenzene, commonly called cumene. Cumene reacts with air at high temperature by benzylic oxidation through a radical mechanism to form cumene hydroperoxide, which is converted into phenol and acetone by treatment with acid. This is a particularly efficient process because two valuable chemicals are prepared at the same time.\n\n\nAs shown in FIGURE 17.11, the reaction occurs by protonation of oxygen followed by shift of the phenyl group from carbon to oxygen with simultaneous loss of water. Replacement of the water then yields an intermediate called a hemiacetal-a compound that contains one -OR group and one -OH group bonded to the same carbon atom-which breaks down to phenol and acetone."}
{"id": 1100, "contents": "FIGURE 17.11 MECHANISM - \nMechanism for the formation of phenol by acid-catalyzed rearrangement of cumene hydroperoxide.\n\n\nProtonation of the hydroperoxy group\non the terminal oxygen atom gives an oxonium ion...\n(1) $\\downarrow \\mathrm{H}_{3} \\mathrm{O}^{+}$\n\n(2 $\\downarrow-\\mathrm{H}_{2} \\mathrm{O}$\n\n(3) Nucleophilic addition of water to the carbocation yields another oxonium ion...\n\n\n(4)... which rearranges by a proton shift from one oxygen to another.\n(4) $\\downarrow$\n\n(5) Elimination of phenol gives acetone as co-product and regenerates the acid catalyst.\n\n$$\n\\text { (5 } \\|-\\mathrm{H}_{3} \\mathrm{O}^{+}\n$$\n\n\n\nIn addition to its use in making resins and adhesives, phenol is also the starting material for the synthesis of chlorinated phenols and the food preservatives BHT (butylated hydroxytoluene) and BHA (butylated hydroxyanisole). Pentachlorophenol, a widely used wood preservative, is prepared by reaction of phenol with excess $\\mathrm{Cl}_{2}$. The herbicide 2,4-D (2,4-dichlorophenoxyacetic acid) is prepared from 2,4-dichlorophenol, and the hospital antiseptic agent hexachlorophene is prepared from 2,4,5-trichlorophenol.\n\n\nPentachlorophenol (wood preservative)\n\n\n2,4-Dichlorophenoxyacetic acid, 2,4-D (herbicide)\n\n\nHexachlorophene (antiseptic)\n\nThe food preservative BHT is prepared by Friedel-Crafts alkylation of $p$-methylphenol ( $p$-cresol) with 2 -methylpropene in the presence of acid; BHA is prepared similarly by alkylation of $p$-methoxyphenol.\n\n\nBHT\n\n\nBHA"}
{"id": 1101, "contents": "FIGURE 17.11 MECHANISM - \nBHT\n\n\nBHA\n\nPROBLEM Show the mechanism for the reaction of $p$-methylphenol with 2-methylpropene and $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ catalyst 17-17 to yield the food additive BHT."}
{"id": 1102, "contents": "Electrophilic Aromatic Substitution Reactions - \nThe hydroxyl group is a strongly activating, ortho- and para-directing substituent in electrophilic aromatic substitution reactions (Section 16.4). As a result, phenols are highly reactive substrates for electrophilic halogenation, nitration, sulfonation, and Friedel-Crafts reactions.\n\n\n$+$"}
{"id": 1103, "contents": "Oxidation of Phenols: Quinones - \nPhenols don't undergo oxidation in the same way as alcohols because they don't have a hydrogen atom on the hydroxyl-bearing carbon. Instead, oxidation of a phenol yields a 2,5-cyclohexadiene-1,4-dione, or quinone. Many different oxidizing agents will accomplish the transformation, but potassium nitrosodisulfonate [ $\\left(\\mathrm{KSO}_{3}\\right)_{2} \\mathrm{NO}$ ], called Fremy's salt, is often used.\n\n\nQuinones are a valuable class of compounds because of their oxidation-reduction, or redox, properties. They can be easily reduced to hydroquinones (p-dihydroxybenzenes) by reagents such as $\\mathrm{NaBH}_{4}$ and $\\mathrm{SnCl}_{2}$, and hydroquinones can be easily reoxidized back to quinones.\n\n\nBenzoquinone\nHydroquinone\nThe redox properties of quinones are crucial to the functioning of living cells, where compounds called ubiquinones act as biochemical oxidizing agents to mediate the electron-transfer processes involved in energy production. Ubiquinones, also called coenzymes $Q$, are components of the cells in all aerobic organisms, from the simplest bacterium to humans. They are so named because of their ubiquitous occurrence throughout nature.\n\n\nUbiquinones ( $\\boldsymbol{n}=\\mathbf{1 - 1 0}$ )\nUbiquinones function within the mitochondria of cells to mediate the respiration process in which electrons are transported from the biological reducing agent NADH to molecular oxygen. Through a complex series of steps, the ultimate result is a cycle whereby NADH is oxidized to $\\mathrm{NAD}^{+}, \\mathrm{O}_{2}$ is reduced to water, and energy is produced. Ubiquinone acts only as an intermediary and is itself unchanged."}
{"id": 1104, "contents": "Step 2 - \nNet change: NADH $+\\frac{1}{2} \\mathrm{O}_{2}+\\mathrm{H}^{+} \\longrightarrow \\mathrm{NAD}^{+}+\\mathrm{H}_{2} \\mathrm{O}$"}
{"id": 1105, "contents": "Infrared Spectroscopy - \nAlcohols have a strong C-O stretching absorption near $1050 \\mathrm{~cm}^{-1}$ and a characteristic $\\mathrm{O}-\\mathrm{H}$ stretching absorption at 3300 to $3600 \\mathrm{~cm}^{-1}$. The exact position of the $\\mathrm{O}-\\mathrm{H}$ stretch depends on the extent of hydrogenbonding in the molecule. Unassociated alcohols show a fairly sharp absorption near $3600 \\mathrm{~cm}^{-1}$, whereas hydrogen-bonded alcohols show a broader absorption in the 3300 to $3400 \\mathrm{~cm}^{-1}$ range. The hydrogen-bonded hydroxyl absorption appears at $3350 \\mathrm{~cm}^{-1}$ in the IR spectrum of cyclohexanol (FIGURE 17.12).\n\n\nFIGURE 17.12 IR spectrum of cyclohexanol. Characteristic $\\mathrm{O}-\\mathrm{H}$ and $\\mathrm{C}-\\mathrm{O}$ stretching absorptions are indicated.\nPhenols also show a characteristic broad IR absorption at $3500 \\mathrm{~cm}^{-1}$ due to the -OH group, as well as the usual 1500 and $1600 \\mathrm{~cm}^{-1}$ aromatic bands (FIGURE 17.13). In phenol itself, monosubstituted aromatic-ring peaks are visible at 690 and $760 \\mathrm{~cm}^{-1}$.\n\n\nPROBLEM Assume that you need to prepare 5-cholesten-3-one from cholesterol. How could you use IR\n17-18 spectroscopy to tell whether the reaction was successful? What differences would you look for in the IR spectra of starting material and product?\n\n\nCholesterol\n\n\n5-Cholestene-3-one"}
{"id": 1106, "contents": "Nuclear Magnetic Resonance Spectroscopy - \nCarbon atoms bonded to electron-withdrawing - OH groups are deshielded and absorb at a lower field in the ${ }^{13} \\mathrm{C}$ NMR spectrum than do typical alkane carbons. Most alcohol carbon absorptions fall in the range 50 to $80 \\delta$, as shown in the following drawing for cyclohexanol:\n\n\nAlcohols also show characteristic absorptions in the ${ }^{1} \\mathrm{H}$ NMR spectrum. Hydrogens on the oxygen-bearing carbon atom are deshielded by the electron-withdrawing effect of the nearby oxygen, and their absorptions\noccur in the range 3.4 to $4.5 \\delta$. Spin-spin splitting, however, is not usually observed between the $\\mathrm{O}-\\mathrm{H}$ proton of an alcohol and the neighboring protons on carbon. Most samples contain small amounts of acidic impurities, which catalyze an exchange of the $\\mathrm{O}-\\mathrm{H}$ proton on a timescale so rapid that the effect of spin-spin splitting is removed. It's often possible to take advantage of this rapid proton exchange to identify the position of the $\\mathrm{O}-\\mathrm{H}$ absorption. If a small amount of deuterated water, $\\mathrm{D}_{2} \\mathrm{O}$, is added to an NMR sample tube, the $\\mathrm{O}-\\mathrm{H}$ proton is rapidly exchanged for deuterium and the hydroxyl absorption disappears from the spectrum.\n\n\nTypical spin-spin splitting is observed between protons on the oxygen-bearing carbon and other neighbors. For example, the signal of the two $-\\mathrm{CH}_{2} \\mathrm{O}$ - protons in 1-propanol is split into a triplet by coupling with the neighboring $-\\mathrm{CH}_{2}-$ protons (FIGURE 17.14)."}
{"id": 1107, "contents": "Nuclear Magnetic Resonance Spectroscopy - \nFIGURE 17.14 ${ }^{\\mathbf{1}} \\mathbf{H}$ NMR spectrum of 1-propanol. The protons on the oxygen-bearing carbon are split into a triplet at $3.58 \\delta$.\nPhenols, like all aromatic compounds, show ${ }^{1} \\mathrm{H}$ NMR absorptions near 7 to $8 \\delta$, the expected position for aromatic-ring protons (Section 15.7). In addition, phenol $0-H$ protons absorb at 3 to $8 \\delta$. In neither case are these absorptions uniquely diagnostic for phenols, since other kinds of protons absorb in the same range.\n\nPROBLEM When the ${ }^{1} \\mathrm{H}$ NMR spectrum of an alcohol is run in dimethyl sulfoxide (DMSO) solvent rather than\n17-19 in chloroform, exchange of the $\\mathrm{O}-\\mathrm{H}$ proton is slow and spin-spin splitting is seen between the $\\mathrm{O}-\\mathrm{H}$ proton and $\\mathrm{C}-\\mathrm{H}$ protons on the adjacent carbon. What spin multiplicities would you expect for the hydroxyl protons in the following alcohols?\n(a) 2-Methyl-2-propanol\n(b) Cyclohexanol\n(c) Ethanol\n(d) 2-Propanol\n(e) Cholesterol\n(f) 1-Methylcyclohexanol"}
{"id": 1108, "contents": "Mass Spectrometry - \nAs noted in Section 12.3, alcohols undergo fragmentation in the mass spectrometer by two characteristic pathways, alpha cleavage and dehydration. In the alpha-cleavage pathway, a $\\mathrm{C}-\\mathrm{C}$ bond nearest the hydroxyl group is broken, yielding a neutral radical plus a resonance-stabilized, oxygen-containing cation.\n\n\nIn the dehydration pathway, water is eliminated, yielding an alkene radical cation.\n\n\nBoth fragmentation modes are apparent in the mass spectrum of 1-butanol (FIGURE 17.15). The peak at $\\mathrm{m} / \\mathrm{z}=$ 56 is due to loss of water from the molecular ion, and the peak at $m / z=31$ is due to an alpha cleavage.\n\n\nFIGURE 17.15 Mass spectrum of 1-butanol ( $\\mathbf{M}^{+}=\\mathbf{7 4}$ ). Dehydration gives a peak at $m / z=56$, and fragmentation by alpha cleavage gives a peak at $m / z=31$."}
{"id": 1109, "contents": "Ethanol: Chemical, Drug, and Poison - \nThe production of ethanol by fermentation of grains and sugars is one of the oldest known organic reactions, going back at least 8000 years in the Middle East and perhaps as many as 9000 years in China. Fermentation is carried out by adding yeast to an aqueous sugar solution, where enzymes break down carbohydrates into ethanol and $\\mathrm{CO}_{2}$. As noted in the chapter introduction, approximately 26 billion gallons of ethanol is produced each year by fermentation of corn and sugar, with essentially the entire amount used to make bus and automobile fuel.\n\n$$\n\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6} \\xrightarrow{\\text { Yeast }} 2 \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}+2 \\mathrm{CO}_{2}\n$$"}
{"id": 1110, "contents": "A carbohydrate - \nEthanol is classified medically as a central nervous system (CNS) depressant. Its effects-that is, being drunk-resemble the human response to anesthetics. There is an initial excitability and increase in sociable behavior, but this results from depression of inhibition rather than from stimulation. At a blood alcohol concentration of $0.1 \\%$ to $0.3 \\%$, motor coordination is affected, accompanied by loss of balance, slurred speech, and amnesia. When blood alcohol concentration rises to between $0.3 \\%$ and $0.4 \\%$, nausea and loss of consciousness occur. Above $0.6 \\%$, spontaneous respiration and cardiovascular regulation are affected, ultimately leading to death. The $\\mathrm{LD}_{50}$ of ethanol is $10.6 \\mathrm{~g} / \\mathrm{kg}$ (Chapter 1 Chemistry Matters).\n\nThe passage of ethanol through the body begins with its absorption in the stomach and small intestine, followed by rapid distribution to all body fluids and organs. In the pituitary gland, ethanol inhibits the production of a hormone that regulates urine flow, causing increased urine production and dehydration. In the stomach, ethanol stimulates production of acid. Throughout the body, ethanol causes blood vessels to dilate, resulting in flushing of the skin and a sensation of warmth as blood moves into capillaries beneath the surface. The result is not a warming of the body, but an increased loss of heat at the surface.\n\n\nFIGURE 17.16 The Harger Drunkometer was the first breath analyzer, introduced in 1938 to help convict drunk drivers. (credit: \"Unidentified police officer with a Harger \"Drunkometer\" breathalyzer\" by Florida Memory, State Library and Archives of Florida/Wikimedia Commons, Public Doman)\n\nEthanol metabolism occurs mainly in the liver and proceeds by oxidation in two steps, first to acetaldehyde $\\left(\\mathrm{CH}_{3} \\mathrm{CHO}\\right)$ and then to acetic acid $\\left(\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right)$. When continuously present in the body, ethanol and acetaldehyde are toxic, leading to the devastating physical and metabolic deterioration seen in people with chronic alcohol use disorder. The liver usually suffers the worst damage since it is the major site of alcohol metabolism."}
{"id": 1111, "contents": "A carbohydrate - \nApproximately 17,000 people are killed each year in the United States in alcohol-related automobile accidents. Thus, all 50 states have made it illegal to drive with a blood alcohol concentration (BAC) above $0.08 \\%$.\n\nFortunately, simple tests have been devised for measuring blood alcohol concentration. The original breath analyzer test measured alcohol concentration in expired air by the color change occurring when the brightorange oxidizing agent potassium dichromate $\\left(\\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}\\right)$ reduced to blue-green chromium(III). Current consumer devices use a conductivity sensor, and tests used by law-enforcement agencies use IR spectroscopy to measure blood-alcohol levels in expired air. Just breathe into the machine, and let the spectrum tell the tale."}
{"id": 1112, "contents": "Key Terms - \n- alcohol\n- phenoxide ion $\\mathrm{ArO}^{-}$\n- alkoxide ion, $\\mathrm{RO}^{-}$\n- protecting group\n- hydroquinone\n- quinone\n- phenol"}
{"id": 1113, "contents": "Summary - \nIn previous chapters, we focused on developing general ideas of organic reactivity, looking at the chemistry of hydrocarbons and alkyl halides and seeing some of the tools used in structural studies. With that accomplished, we have now begun to study the oxygen-containing functional groups that lie at the heart of organic and biological chemistry.\n\nAlcohols are among the most versatile of all organic compounds. They occur widely in nature, are important industrially, and have an unusually rich chemistry. The most widely used methods of alcohol synthesis start with carbonyl compounds. Aldehydes, esters, and carboxylic acids are reduced by reaction with $\\mathrm{LiAlH}_{4}$ to give primary alcohols $\\left(\\mathrm{RCH}_{2} \\mathrm{OH}\\right)$; ketones are reduced to yield secondary alcohols ( $\\mathrm{R}_{2} \\mathrm{CHOH}$ ).\n\nAlcohols are also prepared by reaction of carbonyl compounds with Grignard reagents, RMgX. Addition of a Grignard reagent to formaldehyde yields a primary alcohol, addition to an aldehyde yields a secondary alcohol, and addition to a ketone or an ester yields a tertiary alcohol. The Grignard reaction is limited by the fact that Grignard reagents can't be prepared from alkyl halides that contain reactive functional groups in the same molecule. This problem can sometimes be avoided by protecting the interfering functional group. Alcohols are often protected by formation of trialkylsilyl ethers.\n\nAlcohols undergo many reactions and can be converted into many other functional groups. They can be dehydrated to give alkenes by treatment with $\\mathrm{POCl}_{3}$ and can be transformed into alkyl halides by treatment with $\\mathrm{PBr}_{3}$ or $\\mathrm{SOCl}_{2}$. Furthermore, alcohols are weakly acidic ( $\\mathrm{p} K_{\\mathrm{a}} \\approx 16-18$ ) and react with strong bases and with alkali metals to form alkoxide anions, which are used frequently in organic synthesis. Perhaps the most important reaction of alcohols is their oxidation to carbonyl compounds. Primary alcohols yield either aldehydes or carboxylic acids, secondary alcohols yield ketones, but tertiary alcohols are not normally oxidized."}
{"id": 1114, "contents": "Summary - \nPhenols are aromatic counterparts of alcohols but are more acidic ( $\\mathrm{p} K_{\\mathrm{a}} \\approx 10$ ) because their corresponding phenoxide anions are resonance stabilized by delocalization of the negative charge into the aromatic ring. Substitution of the aromatic ring by an electron-withdrawing group increases phenol acidity, and substitution by an electron-donating group decreases acidity. Phenols can be oxidized to quinones, and quinones can be reduced back to hydroquinones."}
{"id": 1115, "contents": "Summary of Reactions - \n1. Synthesis of alcohols\na. Reduction of carbonyl compounds (Section 17.4)\n(1) Aldehydes\n\n\nPrimary alcohol\n(2) Ketones"}
{"id": 1116, "contents": "Secondary alcohol - \n(3) Esters\n\n\nPrimary alcohol\n(4) Carboxylic acids\n\n\nPrimary alcohol\nb. Grignard addition to carbonyl compounds (Section 17.5)\n(1) Formaldehyde\n\n\nPrimary alcohol\n(2) Aldehydes\n\n\nSecondary alcohol\n(3) Ketones\n\n\nTertiary alcohol\n(4) Esters\n\n\nTertiary alcohol\n2. Reactions of alcohols\na. Dehydration (Section 17.6)\n(1) Tertiary alcohols\n\n(2) Secondary and tertiary alcohols\nb. Oxidation (Section 17.7)\n(1) Primary alcohols\n\n\n\n\nCarboxylic acid\n(2) Secondary alcohols\n\n\nKetone\n3. Oxidation of phenols to quinones (Section 17.10)"}
{"id": 1117, "contents": "Visualizing Chemistry - \nPROBLEM Give IUPAC names for the following compounds:"}
{"id": 1118, "contents": "17-20 (a) - \n(c)\n\n(b)\n\n(d)\n\n\nPROBLEM Draw the structure of the carbonyl compound(s) from which each of the following alcohols might\n17-21 have been prepared, and show the products you would obtain by treatment of each alcohol with (1) Na metal, (2) $\\mathrm{SOCl}_{2}$, and (3) Dess-Martin periodinane.\n(a)\n\n(b)\n\n\nPROBLEM Predict the product from reaction of the following substance (reddish brown $=\\mathrm{Br}$ ) with:"}
{"id": 1119, "contents": "17-22 - \n(a) $\\mathrm{PBr}_{3}$\n(b) Aqueous $\\mathrm{H}_{2} \\mathrm{SO}_{4}$\n(c) $\\mathrm{SOCl}_{2}$\n(d) Dess-Martin periodinane (e) $\\mathrm{Br}_{2}, \\mathrm{FeBr}_{3}$\n\nPROBLEM Predict the product from reaction of the following substance with:\n17-23\n\n(a) $\\mathrm{NaBH}_{4}$; then $\\mathrm{H}_{3} \\mathrm{O}^{+}$(b) $\\mathrm{LiAlH}_{4}$; then $\\mathrm{H}_{3} \\mathrm{O}^{+}$(c) $2 \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{MgBr}$; then $\\mathrm{H}_{3} \\mathrm{O}^{+}$\n\nPROBLEM Name and assign $R$ or $S$ stereochemistry to the product(s) you would obtain by reaction of the\n17-24 following substance with ethylmagnesium bromide. Is the product chiral? Is it optically active? Explain."}
{"id": 1120, "contents": "Mechanism Problems - \nPROBLEM Evidence for the intermediate carbocations in the acid-catalyzed dehydration of alcohols comes\n17-25 from the observation that rearrangements sometimes occur. Propose a mechanism to account for the formation of 2,3-dimethyl-2-butene from 3,3-dimethyl-2-butanol.\n\n\nPROBLEM Acid-catalyzed dehydration of 2,2-dimethylcyclohexanol yields a mixture of\n17-26 1,2-dimethylcyclohexene and isopropylidenecyclopentane. Propose a mechanism to account for the formation of both products."}
{"id": 1121, "contents": "Isopropylidenecyclopentane - \nPROBLEM Epoxides react with Grignard reagents to yield alcohols. Propose a mechanism.\n17-27\n\n\nPROBLEM Treatment of the following epoxide with aqueous acid produces a carbocation intermediate that\n$\\mathbf{1 7 - 2 8}$ reacts with water to give a diol product. Show the structure of the carbocation, and propose a mechanism for the second step.\n\n\nPROBLEM Reduction of 2-butanone with $\\mathrm{NaBH}_{4}$ yields 2-butanol. Is the product chiral? Is it optically active?\n17-29 Explain.\nPROBLEM The conversion of $3^{\\circ}$ alcohols into $3^{\\circ}$ alkyl halides under acidic conditions involves two cationic\n17-30 intermediates. For each reaction, draw the complete mechanism using curved arrows.\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Identify the type of substitution mechanism $\\left(\\mathrm{S}_{\\mathrm{N}} 1, \\mathrm{~S}_{\\mathrm{N}} 2\\right)$ involved in the conversion of the following\n17-31 alcohols into the corresponding alkyl halide.\n(a)\n\n\n\n(b)\n\n(c)\n\n\nPROBLEM The conversion of $3^{\\circ}$ alcohols into alkenes under acidic conditions involves two cationic 17-32 intermediates. For each reaction, draw the complete mechanism using curved arrows.\n(a)\n\n(b)\n\n\n(c)\n\n\nPROBLEM For each reaction, write the mechanism using curved arrows for the conversion of the alcohol into\n17-33 the corresponding alkene with $\\mathrm{POCl}_{3}$. In each case, explain the regiochemistry of the elimination.\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM The trimethylsilyl (TMS) protecting group is one of several silicon protecting groups for alcohols.\n17-34 For each reaction, draw the mechanism for the protection of ( $R$ )-3-bromo-1-butanol with the following silyl chlorides, using triethylamine as the base:\n(a) tert-butyldimethylsilyl chloride (TBS-Cl)\n(b) triisopropylsilyl chloride (TIPS-Cl)\n(c) triethylsilyl chloride (TES-Cl)"}
{"id": 1122, "contents": "Isopropylidenecyclopentane - \nPROBLEM When the following alcohol is treated with $\\mathrm{POCl}_{3}$ and pyridine, the expected elimination product\n17-35 is formed. However, when the same alcohol is treated with $\\mathrm{H}_{2} \\mathrm{SO}_{4}$, the elimination product is 1,2-dimethylcyclopentene. Propose a mechanism for each pathway to account for these differences.\n\n\nPROBLEM Phenols generally have lower $\\mathrm{p} K_{\\mathrm{a}}$ 's than alcohols because of resonance stabilization with the\n17-36 aromatic ring. Draw all of the resonance contributors for the following phenolate ions.\n(a)\n\n(b)\n\n(c)\n\n\nNaming Alcohols\nPROBLEM Give IUPAC names for the following compounds:\n17-37 (a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n\n\nPROBLEM Draw and name the eight isomeric alcohols with formula $\\mathrm{C}_{5} \\mathrm{H}_{12} \\mathrm{O}$. 17-38\n\nPROBLEM Draw structures corresponding to the following IUPAC names:\n17-39 (a) Trans-3-Chlorocycloheptanol\n(b) 2-Ethyl-2-buten-1-ol\n(c) o-(2-Hydroxyethyl)phenol\n(d) 3-Methyl-1-phenyl-1-butanol\n\nPROBLEM Bombykol, the sex pheromone secreted by the female silkworm moth has the formula $\\mathrm{C}_{16} \\mathrm{H}_{28} \\mathrm{O}$\n17-40 and the systematic name ( $10 E, 12 Z$ )-10,12-hexadecadien-1-ol. Draw bombykol, showing the correct geometry for the two double bonds.\n\nPROBLEM Carvacrol is a naturally occurring substance isolated from oregano, thyme, and marjoram. What is 17-41 its IUPAC name?\n\n\nCarvacrol\n\nSynthesizing Alcohols\nPROBLEM What Grignard reagent and what carbonyl compound might you start with to prepare the following\n17-42 alcohols:\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)"}
{"id": 1123, "contents": "Isopropylidenecyclopentane - \n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n\nPROBLEM What carbonyl compounds would you reduce to prepare the following alcohols: List all possibilities.\n17-43 (a)\n\n\n(b)\n\n(c)\n\n\nPROBLEM What carbonyl compounds might you start with to prepare the following compounds by Grignard\n17-44 reaction? List all possibilities.\n(a) 2-Methyl-2-propanol\n(b) 1-Ethylcyclohexanol\n(c) 3-Phenyl-3-pentanol\n(d) 2-Phenyl-2-pentanol\n(e)\n\n(f)\n\n\nPROBLEM How would you synthesize the following alcohols, starting with benzene and other alcohols of six or 17-45 fewer carbons as your only organic reagents?\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n\nReactions of Alcohols\nPROBLEM What products would you obtain from reaction of 1-pentanol with the following reagents:\n17-46\n(a) $\\mathrm{PBr}_{3}$\n(b) $\\mathrm{SOCl}_{2}$\n(c) Dess-Martin periodinane\n\nPROBLEM How would you prepare the following compounds from 2-phenylethanol: More than one step may 17-47 be required.\n(a) Styrene $\\left(\\mathrm{PhCH}=\\mathrm{CH}_{2}\\right)$ (b) Phenylacetaldehyde $\\left(\\mathrm{PhCH}_{2} \\mathrm{CHO}\\right)$\n(c) Phenylacetic acid $\\left(\\mathrm{PhCH}_{2} \\mathrm{CO}_{2} \\mathrm{H}\\right)$ (d) Benzoic acid (e) Ethylbenzene (f) Benzaldehyde\n(g) 1-Phenylethanol (h) 1-Bromo-2-phenylethane\n\nPROBLEM How would you prepare the following compounds from 1-phenylethanol: More than one step may 17-48 be required.\n(a) Acetophenone $\\left(\\mathrm{PhCOCH}_{3}\\right)$\n(b) Benzyl alcohol\n(c) m-Bromobenzoic acid\n(d) 2-Phenyl-2-propanol"}
{"id": 1124, "contents": "Isopropylidenecyclopentane - \nPROBLEM How would you prepare the following substances from cyclopentanol: More than one step may be 17-49 required.\n(a) Cyclopentanone\n(b) Cyclopentene\n(c) 1-Methylcyclopentanol\n(d) trans-2-Methylcyclopentanol\n\nPROBLEM What products would you expect to obtain from reaction of 1-methylcyclohexanol with the 17-50 following reagents?\n(a) HBr (b)\n(b) NaH (c) $\\mathrm{H}_{2} \\mathrm{SO}_{4}$"}
{"id": 1125, "contents": "Spectroscopy - \nPROBLEM The following ${ }^{1} \\mathrm{H}$ NMR spectrum is that of an alcohol, $\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{O}$. Propose a structure."}
{"id": 1126, "contents": "17-51 - \nPROBLEM Propose structures for alcohols that have the following ${ }^{1}$ H NMR spectra:\n17-52\n(a) $\\mathrm{C}_{5} \\mathrm{H}_{12} \\mathrm{O}$\n\n(b) $\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{O}$\n\n\nPROBLEM Propose a structure consistent with the following spectral data for a compound $\\mathrm{C}_{8} \\mathrm{H}_{18} \\mathrm{O}_{2}$ :\n17-53\n\n> IR: $3350 \\mathrm{~cm}^{-1}$\n> ${ }^{1} \\mathrm{H}$ NMR: $1.24 \\delta(12 \\mathrm{H}$, singlet $) ; 1.56 \\delta(4 \\mathrm{H}$, singlet $) ; 1.95 \\delta(2 \\mathrm{H}$, singlet $)$\n\nPROBLEM The ${ }^{1} \\mathrm{H}$ NMR spectrum shown is that of 3-methyl-3-buten-1-ol. Assign all the observed resonance\n17-54 peaks to specific protons, and account for the splitting patterns."}
{"id": 1127, "contents": "17-51 - \nPROBLEM A compound of unknown structure gave the following spectroscopic data:\n17-55\nMass spectrum: $\\mathrm{M}^{+}=88.1$\nIR: $3600 \\mathrm{~cm}^{-1}$\n${ }^{1} \\mathrm{H}$ NMR: $1.4 \\delta(2 \\mathrm{H}$, quartet, $J=7 \\mathrm{~Hz}) ; 1.2 \\delta(6 \\mathrm{H}$, singlet); $1.0 \\delta(1 \\mathrm{H}$, singlet); $0.9 \\delta(3 \\mathrm{H}$, triplet, $J=7 \\mathrm{~Hz})$\n${ }^{13}$ C NMR: 74, 35, 27, $25 \\delta$\n(a) Assuming that the compound contains C and H but may or may not contain O , give three possible molecular formulas.\n(b) How many hydrogens does the compound contain?\n(c) What functional group(s) does the compound contain?\n(d) How many carbons does the compound contain?\n(e) What is the molecular formula of the compound?\n(f) What is the structure of the compound?\n(g) Assign peaks in the molecule's ${ }^{1} \\mathrm{H}$ NMR spectrum corresponding to specific protons.\n\nPROBLEM Propose a structure for a compound $\\mathrm{C}_{15} \\mathrm{H}_{24} \\mathrm{O}$ that has the following ${ }^{1} \\mathrm{H}$ NMR spectrum. The peak 17-56 marked by an asterisk disappears when $\\mathrm{D}_{2} \\mathrm{O}$ is added to the sample."}
{"id": 1128, "contents": "General Problems - \nPROBLEM How would you carry out the following transformations?\n17-57 (a)\n\n(b)\n\n(c)\n\n\nPROBLEM Benzoquinone is an excellent dienophile in the Diels-Alder reaction. What product would you\n17-58 expect from reaction of benzoquinone with 1 equivalent of 1,3-butadiene? From reaction with 2 equivalents of 1,3 -butadiene?\n\nPROBLEM Rank the following substituted phenols in order of increasing acidity, and explain your answer:\n17-59\n\n\n\n\n\nPROBLEM Benzyl chloride can be converted into benzaldehyde by treatment with nitromethane and base.\n17-60 The reaction involves initial conversion of nitromethane into its anion, followed by $S_{N} 2$ reaction of the anion with benzyl chloride and subsequent E2 reaction. Write the mechanism in detail, using curved arrows to indicate the electron flow in each step.\n\n\nPROBLEM Reaction of (S)-3-methyl-2-pentanone with methylmagnesium bromide followed by acidification\n17-61 yields 2,3-dimethyl-2-pentanol. What is the stereochemistry of the product? Is the product optically active?\n\n\n3-Methyl-2-pentanone\n\nPROBLEM Testosterone is one of the most important male steroid hormones. When testosterone is dehydrated\n17-62 by treatment with acid, rearrangement occurs to yield the product shown. Propose a mechanism to account for this reaction.\n\n\nPROBLEM Starting from testosterone (Problem 17-62), how would you prepare the following substances?\n\n17-63 (a)\n\n(c)\n\n(b)\n\n(d)\n\n\nPROBLEM $p$-Nitrophenol and 2,6-dimethyl-4-nitrophenol both have $\\mathrm{p} K_{\\mathrm{a}}=7.15$, but 17-64 3,5-dimethyl-4-nitrophenol has $\\mathrm{p} K_{\\mathrm{a}}=8.25$. Why is 3,5-dimethyl-4-nitrophenol so much less acidic?\n\n$\\mathrm{p} K_{\\mathrm{a}}=7.15$\n\n$\\mathrm{p} K_{\\mathrm{a}}=7.15$"}
{"id": 1129, "contents": "General Problems - \n$\\mathrm{p} K_{\\mathrm{a}}=7.15$\n\n$\\mathrm{p} K_{\\mathrm{a}}=7.15$\n\n$\\mathrm{p} K_{\\mathrm{a}}=8.25$\n\nPROBLEM Compound $\\mathbf{A}, \\mathrm{C}_{10} \\mathrm{H}_{18} \\mathrm{O}$, undergoes reaction with dilute $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ at $25{ }^{\\circ} \\mathrm{C}$ to yield a mixture of two 17-65 alkenes, $\\mathrm{C}_{10} \\mathrm{H}_{16}$. The major alkene product, $\\mathbf{B}$, gives only cyclopentanone after ozone treatment followed by reduction with zinc in acetic acid. Write the reactions involved, and identify $\\mathbf{A}$ and $\\mathbf{B}$.\n\nPROBLEM Compound $\\mathbf{A}, \\mathrm{C}_{5} \\mathrm{H}_{10} \\mathrm{O}$, is one of the basic building blocks of nature. All steroids and many other 17-66 naturally occurring compounds are built from compound $\\mathbf{A}$. Spectroscopic analysis of $\\mathbf{A}$ yields the following information:\n\nIR: $3400 \\mathrm{~cm}^{-1} ; 1640 \\mathrm{~cm}^{-1}$\n${ }^{1} \\mathrm{H}$ NMR: $1.63 \\delta(3 \\mathrm{H}$, singlet $) ; 1.70 \\delta(3 \\mathrm{H}$, singlet $) ; 3.83 \\delta(1 \\mathrm{H}$, broad singlet); $4.15 \\delta(2 \\mathrm{H}$, doublet, $J=7 \\mathrm{~Hz}) ; 5.70 \\delta(1 \\mathrm{H}$, triplet, $J=7 \\mathrm{~Hz})$\n(a) How many double bonds and/or rings does $\\mathbf{A}$ have?\n(b) From the IR spectrum, what is the identity of the oxygen-containing functional group?\n(c) What kinds of hydrogens are responsible for the NMR absorptions listed?\n(d) Propose a structure for $\\mathbf{A}$."}
{"id": 1130, "contents": "General Problems - \nPROBLEM Dehydration of trans-2-methylcyclopentanol with $\\mathrm{POCl}_{3}$ in pyridine yields predominantly 17-67 3-methylcyclopentene. Is the stereochemistry of this dehydration syn or anti?\n\nPROBLEM 2,3-Dimethyl-2,3-butanediol has the common name pinacol. On heating with aqueous acid, pinacol\n17-68 rearranges to pinacolone, 3,3-dimethyl-2-butanone. Suggest a mechanism for this reaction.\n\n\nPROBLEM As a rule, axial alcohols oxidize somewhat faster than equatorial alcohols. Which would you expect\n17-69 to oxidize faster, cis-4-tert-butylcyclohexanol or trans-4-tert-butylcyclohexanol? Draw the more stable chair conformation of each molecule.\n\nPROBLEM Propose a synthesis of bicyclohexylidene, starting from cyclohexanone as the only source of carbon.\n17-70"}
{"id": 1131, "contents": "Bicyclohexylidene - \nPROBLEM A problem often encountered in the oxidation of primary alcohols to carboxylic acids is that esters\n17-71 are sometimes produced as by-products. For example, oxidation of ethanol yields acetic acid and ethyl acetate:\n\n\nPropose a mechanism to account for the formation of ethyl acetate. Take into account the reversible reaction between aldehydes and alcohols:\n\n\nPROBLEM Identify the reagents a-f in the following scheme:\n17-72\n\n\n\nPROBLEM Galactose, a constituent of the disaccharide lactose found in dairy products, is metabolized by a\n17-73 pathway that includes the isomerization of UDP-galactose to UDP-glucose, where UDP = uridylyl diphosphate. The enzyme responsible for the transformation uses $\\mathrm{NAD}^{+}$as cofactor. Propose a mechanism.\n\n\nPROBLEM Propose structures for alcohols that have the following ${ }^{1} \\mathrm{H}$ NMR spectra:\n17-74 (a) $\\mathrm{C}_{9} \\mathrm{H}_{12} \\mathrm{O}$\n\n(b) $\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{O}_{2}$\n\n\nPROBLEM Compound $\\mathbf{A}, \\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{O}$, has the IR and ${ }^{1} \\mathrm{H}$ NMR spectra shown. Propose a structure consistent with\n17-75 the observed spectra, and label each peak in the NMR spectrum. Note that the absorption at $5.5 \\delta$ disappears when $\\mathrm{D}_{2} \\mathrm{O}$ is added."}
{"id": 1132, "contents": "CHAPTER 18 <br> Ethers and Epoxides; Thiols and Sulfides - \nFIGURE 18.1 The appalling and unforgettable odor of skunks is due to a mixture of several simple thiols. (credit: modification of work \"Striped Skunk\" by Tom Friedel/Wikimedia Commons, CC BY 3.0)"}
{"id": 1133, "contents": "CHAPTER CONTENTS - \n18.1 Names and Properties of Ethers\n18.2 Preparing Ethers\n18.3 Reactions of Ethers: Acidic Cleavage\n18.4 Cyclic Ethers: Epoxides\n18.5 Reactions of Epoxides: Ring-Opening\n18.6 Crown Ethers\n18.7 Thiols and Sulfides\n18.8 Spectroscopy of Ethers\n\nWHY THIS CHAPTER? This chapter finishes the coverage of functional groups with $\\mathrm{C}-\\mathrm{O}$ and $\\mathrm{C}-\\mathrm{S}$ single bonds that was begun in the chapter on Alcohols and Phenols. We'll focus primarily on ethers and take only a brief look at thiols and sulfides before going on to an extensive coverage of compounds with $\\mathrm{C}=\\mathrm{O}$ double bonds in Chapters 19 through 23.\n\nEthers ( $\\mathbf{R} \\mathbf{- 0}-\\mathbf{R}^{\\prime}$ ), like the alcohols we saw in the preceding chapter, are organic derivatives of water, but they have two organic groups bonded to the same oxygen atom rather than one. The organic groups might be alkyl, aryl, or vinylic, and the oxygen atom might be in an open chain or a ring.\n\nPerhaps the most well-known ether is diethyl ether, which has a long history of medicinal use as an anesthetic and industrial use as a solvent. Other useful ethers include anisole, a pleasant-smelling aromatic ether used in perfumery, and tetrahydrofuran (THF), a cyclic ether often used as a solvent.\n\n$\\mathrm{CH}_{3} \\mathrm{CH}_{2}, \\mathrm{O}-\\mathrm{CH}_{2} \\mathrm{CH}_{3}$\n\nDiethyl ether\n\n\n\nAnisole (methyl phenyl ether)\n\n\n\nTetrahydrofuran\n\nThiols ( $\\mathrm{R}-\\mathrm{S}-\\mathrm{H}$ ) and sulfides ( $\\mathrm{R}-\\mathrm{S}-\\mathrm{R}^{\\prime}$ ) are sulfur analogs of alcohols and ethers, respectively. Both functional groups are found in various biomolecules, although not as commonly as their oxygen-containing relatives."}
{"id": 1134, "contents": "CHAPTER CONTENTS - 18.1 Names and Properties of Ethers\nSimple ethers with no other functional groups are named by identifying the two organic substituents and adding the word ether.\n\n\nIsopropyl methyl ether\n\n\nEthyl phenyl ether\n\nIf other functional groups are present, the ether part is considered an alkoxy substituent. For example:\n\n$p$-Dimethoxybenzene\n\n\n4-tert-Butoxy-1-cyclohexene\n\nLike alcohols, ethers have nearly the same geometry as water. The $\\mathrm{R}-\\mathrm{O}-\\mathrm{R}$ bonds have an approximately tetrahedral bond angle ( $112^{\\circ}$ in dimethyl ether), and the oxygen atom is $s p^{3}$-hybridized.\n\n\nThe electronegative oxygen atom gives ethers a slight dipole moment, and the boiling points of ethers are often slightly higher than the boiling points of comparable alkanes. TABLE 18.1 compares the boiling points of some common ethers and their corresponding hydrocarbons.\n\nTABLE 18.1 Comparison of Boiling Points of Ethers and Hydrocarbons\n\n| Ether | Boiling point ${ }^{\\circ} \\mathrm{C}$ | Hydrocarbon | Boiling point ${ }^{\\circ} \\mathrm{C}$ |\n| :--- | :---: | :--- | :---: |\n| $\\mathrm{CH}_{3} \\mathrm{OCH}_{3}$ | -25 | $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$ | -45 |\n| $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OCH}_{2} \\mathrm{CH}_{3}$ | 34.6 | $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$ | 36 |\n\nTABLE 18.1 Comparison of Boiling Points of Ethers and Hydrocarbons\n\n| Ether | Boiling point ${ }^{\\circ} \\mathrm{C}$ | Hydrocarbon | Boiling point ${ }^{\\circ} \\mathrm{C}$ |\n| :---: | :---: | :---: | :---: |\n|  | 65 | 49 |  |\n|  |  |  |  |"}
{"id": 1135, "contents": "CHAPTER CONTENTS - 18.1 Names and Properties of Ethers\nEthers are relatively stable and unreactive in many respects, but some ethers react slowly with the oxygen in air to give peroxides, compounds that contain an $\\mathrm{O}-\\mathrm{O}$ bond. The peroxides from low-molecular-weight ethers such as diisopropyl ether and tetrahydrofuran are explosive and extremely dangerous, even in small amounts. Ethers are very useful as solvents in the laboratory, but they must always be used cautiously and should not be stored for long periods of time.\n\nPROBLEM Name the following ethers:\n\n18-1 (a)\n\n(e) $\\underset{\\substack{\\mathrm{CH} \\\\ \\mathrm{CH}_{3} \\mathrm{CHCH}_{2} \\mathrm{OCH}_{2} \\mathrm{CH}_{3}}}{\\text { (e) }}$\n(b)\n\n(f) $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}_{2} \\mathrm{OCH}=\\mathrm{CH}_{2}$\n(c)\n\n(d)"}
{"id": 1136, "contents": "CHAPTER CONTENTS - 18.2 Preparing Ethers\nDiethyl ether and other simple symmetrical ethers are prepared industrially by the sulfuric-acid-catalyzed reaction of alcohols. The reaction occurs by $\\mathrm{S}_{\\mathrm{N}} 2$ displacement of water from a protonated ethanol molecule by the oxygen atom of a second ethanol. Unfortunately, this method is limited to use with primary alcohols because secondary and tertiary alcohols dehydrate by an E1 mechanism to yield alkenes (Section 17.6)."}
{"id": 1137, "contents": "The Williamson Ether Synthesis - \nThe most generally useful method of preparing ethers is the Williamson ether synthesis, in which an alkoxide ion reacts with a primary alkyl halide or tosylate in an $\\mathrm{S}_{\\mathrm{N}} 2$ reaction. As we saw in Section 17.2, the alkoxide ion is normally prepared by reaction of an alcohol with a strong base such as sodium hydride, NaH .\n\n\nCyclopentanol\n\n\nAlkoxide ion\n\n\nCyclopentyl methyl ether (74\\%)\n\nA useful variation of the Williamson synthesis involves silver oxide, $\\mathrm{Ag}_{2} \\mathrm{O}$, as a mild base rather than NaH . Under these conditions, the free alcohol reacts directly with the alkyl halide, so there is no need to preform the metal\nalkoxide intermediate. Sugars react particularly well; glucose, for example, reacts with excess iodomethane in the presence of $\\mathrm{Ag}_{2} \\mathrm{O}$ to generate a pentaether in $85 \\%$ yield.\n\n\nBecause the Williamson synthesis is an $\\mathrm{S}_{\\mathrm{N}} 2$ reaction, it is subject to all the usual constraints, as discussed in Section 11.3. Primary halides and tosylates work best because competitive E2 elimination can occur with more hindered substrates. Unsymmetrical ethers should therefore be synthesized by reaction between the more hindered alkoxide partner and less hindered halide partner rather than vice versa. For example, tert-butyl methyl ether, a substance used in the 1990s as an octane booster in gasoline, is best prepared by reaction of tert-butoxide ion with iodomethane rather than by reaction of methoxide ion with 2-chloro-2-methylpropane.\n\ntert-Butoxide Iodomethane tert-Butyl methyl ether\n\n\nPROBLEM Why do you suppose only symmetrical ethers are prepared by the sulfuric-acid-catalyzed 18-2 dehydration procedure? What product(s) would you expect if ethanol and 1-propanol were allowed to react together? In what ratio would the products be formed if the two alcohols were of equal reactivity?"}
{"id": 1138, "contents": "The Williamson Ether Synthesis - \nPROBLEM How would you prepare the following ethers using a Williamson synthesis?\n18-3 (a) Methyl propyl ether (b) Anisole (methyl phenyl ether) (c) Benzyl isopropyl ether\n(d) Ethyl 2,2-dimethylpropyl ether"}
{"id": 1139, "contents": "Alkoxymercuration of Alkenes - \nWe saw in Section 8.4 that alkenes react with water in the presence of mercuric acetate to yield a hydroxymercuration product. Subsequent treatment with $\\mathrm{NaBH}_{4}$ breaks the $\\mathrm{C}-\\mathrm{Hg}$ bond and yields the alcohol. A similar alkoxymercuration reaction occurs when an alkene is treated with an alcohol in the presence of mercuric acetate or, even better, mercuric trifluoroacetate, $\\left(\\mathrm{CF}_{3} \\mathrm{CO}_{2}\\right)_{2} \\mathrm{Hg}$. Demercuration by reaction with $\\mathrm{NaBH}_{4}$ then yields an ether. The net result is Markovnikov addition of the alcohol to the alkene.\n\n\n\nThe mechanism of the alkoxymercuration reaction is similar to that described in Section 8.4 for hydroxymercuration. The reaction is initiated by electrophilic addition of $\\mathrm{Hg}^{2+}$ to the alkene, followed by reaction of the intermediate cation with alcohol and reduction of the $\\mathrm{C}-\\mathrm{Hg}$ bond by $\\mathrm{NaBH}_{4}$. A variety of alcohols and alkenes can be used in alkoxymercuration. Primary, secondary, and even tertiary alcohols react well, but ditertiary ethers can't be prepared because of steric hindrance."}
{"id": 1140, "contents": "Synthesizing an Ether - \nHow would you prepare ethyl phenyl ether? Use whichever method you think is more appropriate, Williamson synthesis or the alkoxymercuration reaction."}
{"id": 1141, "contents": "Strategy - \nDraw the target ether, identify the two groups attached to oxygen, and recall the limitations of the two methods for preparing ethers. Williamson synthesis uses an $S_{N} 2$ reaction and requires that one of the two groups attached to oxygen be either secondary or (preferably) primary. The alkoxymercuration reaction requires that one of the two groups come from an alkene precursor. Ethyl phenyl ether could be made by either method.\n\nEthyl phenyl ether\n\n\nSolution\n\n\nPhenol\n\n\n\nEthyl phenyl ether\n\nPROBLEM Review the mechanism of oxymercuration shown in Figure 8.4, and then write the mechanism of 18-4 the alkoxymercuration reaction of 1-methylcyclopentene with ethanol. Use curved arrows to show the electron flow in each step.\n\nPROBLEM How would you prepare the following ethers? Use whichever method you think is more appropriate, 18-5 Williamson synthesis or the alkoxymercuration reaction.\n(a) Butyl cyclohexyl ether (b) Benzyl ethyl ether $\\left(\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CH}_{2} \\mathrm{OCH}_{2} \\mathrm{CH}_{3}\\right)$\n(c) sec-Butyl tert-butyl ether (d) Tetrahydrofuran\n\nPROBLEM Rank the following halides in order of their reactivity in the Williamson synthesis:\n18-6 (a) Bromoethane, 2-bromopropane, bromobenzene\n(b) Chloroethane, bromoethane, 1-Iodopropene"}
{"id": 1142, "contents": "Strategy - 18.3 Reactions of Ethers: Acidic Cleavage\nEthers are unreactive to many reagents used in organic chemistry, a property that accounts for their wide use as reaction solvents. Halogens, dilute acids, bases, and nucleophiles have no effect on most ethers. In fact, ethers undergo only one truly general reaction-they are cleaved by strong acids. Aqueous HBr and HI both work well, but HCl does not cleave ethers.\n\n\nAcidic ether cleavages are typical nucleophilic substitution reactions and take place by either $\\mathrm{S}_{\\mathrm{N}} 1$ or $\\mathrm{S}_{\\mathrm{N}} 2$ mechanisms, depending on the structure of the substrate. Ethers with only primary and secondary alkyl groups react by an $\\mathrm{S}_{\\mathrm{N}} 2$ mechanism, in which $\\mathrm{I}^{-}$or $\\mathrm{Br}^{-}$reacts with the protonated ether at the less hindered site. This usually results in a selective cleavage into a single alcohol and a single alkyl halide. For example, ethyl isopropyl ether exclusively yields isopropyl alcohol and iodoethane on cleavage by HI because nucleophilic attack by iodide ion occurs at the less hindered primary site rather than at the more hindered secondary site.\n\n\nEthers with a tertiary, benzylic, or allylic group cleave by either an $S_{N} 1$ or E1 mechanism because these substrates can produce stable intermediate carbocations. These reactions are often fast and take place at moderate temperatures. tert-Butyl ethers, for example, react by an E1 mechanism on treatment with trifluoroacetic acid at $0^{\\circ} \\mathrm{C}$. We'll see in Section 26.7 that this reaction is often used in the laboratory synthesis of peptides."}
{"id": 1143, "contents": "Predicting the Product of an Ether Cleavage Reaction - \nPredict the products of the following reaction:"}
{"id": 1144, "contents": "Strategy - \nIdentify the substitution pattern of the two groups attached to oxygen-in this case a tertiary alkyl group and a primary alkyl group. Then recall the guidelines for ether cleavages. An ether with only primary and secondary alkyl groups usually undergoes cleavage by $\\mathrm{S}_{\\mathrm{N}} 2$ attack of a nucleophile on the less hindered alkyl group, but an ether with a tertiary alkyl group usually undergoes cleavage by an $\\mathrm{S}_{\\mathrm{N}} 1$ mechanism. In this case, an $\\mathrm{S}_{\\mathrm{N}} 1$ cleavage\nof the tertiary $\\mathrm{C}-\\mathrm{O}$ bond will occur, giving 1-propanol and a tertiary alkyl bromide. In addition, a competitive E1 reaction leading to alkene might occur."}
{"id": 1145, "contents": "Solution - \nPROBLEM Predict the products of the following reactions:\n\n18-7 (a)\n\n(b)\n\n\nPROBLEM Write the mechanism of the acid-induced cleavage of tert-butyl cyclohexyl ether to yield\n18-8 cyclohexanol and 2-methylpropene.\nPROBLEM Why are HI and HBr more effective than HCl in cleaving ethers? (See Section 11.3.)\n18-9"}
{"id": 1146, "contents": "Solution - 18.4 Cyclic Ethers: Epoxides\nFor the most part, cyclic ethers behave like acyclic ethers. The chemistry of the ether functional group is the same, whether it's in an open chain or in a ring. Common cyclic ethers such as tetrahydrofuran and dioxane are often used as solvents because of their inertness, yet they can be cleaved by strong acids.\n\n\n1,4-Dioxane\n\n\nTetrahydrofuran\n\nThe one group of cyclic ethers that behaves differently from open-chain ethers are the three-membered-ring compounds called epoxides, or oxiranes, which we saw in Section 8.7. The strain of the three-membered ring gives epoxides unique chemical reactivity.\n\nEthylene oxide, the simplest epoxide, is an intermediate in the manufacture of both ethylene glycol, used for automobile antifreeze, and polyester polymers. Approximately 37 million tons of ethylene oxide is produced worldwide each year, most of it by air oxidation of ethylene over a silver oxide catalyst at $300^{\\circ} \\mathrm{C}$. This process is not useful for other epoxides, however, and is of little value in the laboratory.\n\nNote that the name ethylene oxide is not a systematic one because the -ene ending implies the presence of a double bond in the molecule. The name is frequently used, however, because ethylene oxide is derived from ethylene by addition of an oxygen atom. Other simple epoxides are named similarly. The systematic name for ethylene oxide is 1,2-epoxyethane.\n\n\nIn the laboratory, as we saw in Section 8.7, epoxides are usually prepared by treatment of an alkene with a\nperoxyacid $\\left(\\mathrm{RCO}_{3} \\mathrm{H}\\right)$, typically $m$-chloroperoxybenzoic acid.\n\n\nEpoxides can also be prepared from halohydrins, themselves produced by electrophilic addition of $\\mathrm{Cl}_{2}$ or Br 2 to an alkene in the presence of water (Section 8.3). When a halohydrin is treated with base, HX is eliminated and an epoxide is produced by an intramolecular Williamson ether synthesis. That is, the nucleophilic alkoxide ion and the electrophilic alkyl halide are in the same molecule."}
{"id": 1147, "contents": "Solution - 18.4 Cyclic Ethers: Epoxides\nPROBLEM What product would you expect from reaction of trans-2-butene with $m$-chloroperoxybenzoic acid?\n18-10 Show the stereochemistry.\nPROBLEM Reaction of cis-2-butene with m-chloroperoxybenzoic acid yields an epoxide different from that\n18-11 obtained by reaction of the trans isomer. Explain."}
{"id": 1148, "contents": "Acid-Catalyzed Epoxide Opening - \nEpoxides are cleaved by treatment with acid just as other ethers are, but under much milder conditions because of ring strain. As we saw in Section 8.7, dilute aqueous acid at room temperature is sufficient for facilitating the hydrolysis of epoxides to give 1,2-diols, also called vicinal glycols. (The word vicinal means \"adjacent,\" and a glycol is a diol.) The epoxide cleavage takes place by $\\mathrm{S}_{\\mathrm{N}} 2$-like backside attack of a nucleophile on the protonated epoxide, giving a trans-1,2-diol as product.\n\n\n1,2-Epoxycyclohexane\n\n\n\ntrans-1,2-Cyclohexanediol\n(86\\%)\n\n\nEpoxides can also be opened by reaction with acids other than $\\mathrm{H}_{3} \\mathrm{O}^{+}$. If anhydrous HX is used, for instance, an epoxide is converted into a trans halohydrin.\n\n\nA trans 2-halocyclohexanol\nwhere $X=\\mathrm{F}, \\mathrm{Br}, \\mathrm{Cl}$, or I\nThe regiochemistry of acid-catalyzed ring-opening depends on the epoxide's structure, and a mixture of products is often formed. When both epoxide carbon atoms are either primary or secondary, attack of the nucleophile occurs primarily at the less highly substituted site: an $\\mathrm{S}_{\\mathrm{N}} 2$-like result. When one of the epoxide carbon atoms is tertiary, however, nucleophilic attack occurs primarily at the more highly substituted site-an $\\mathrm{S}_{\\mathrm{N}} 1$-like result. Thus, 1,2-epoxypropane reacts with HCl to give primarily 1-chloro-2-propanol, but 2-methyl-1,2-epoxypropane gives 2-chloro-2-methyl-1- propanol as the major product.\n\n\n1,2-Epoxypropane\n\n\n2-Methyl-1,2-epoxypropane\n\n\n$+$"}
{"id": 1149, "contents": "Acid-Catalyzed Epoxide Opening - \n1,2-Epoxypropane\n\n\n2-Methyl-1,2-epoxypropane\n\n\n$+$\n\n1-Chloro-2-propanol 2-Chloro-1-propanol (90\\%)\n2-Chloro-2-methyl- 1-Chloro-2-methyl-1-propanol (60\\%) 2-propanol (40\\%)\n\nThe mechanisms of these acid-catalyzed epoxide openings are more complex than they initially appear. They seem to be neither purely $\\mathrm{S}_{\\mathrm{N}} 1$ nor $\\mathrm{S}_{\\mathrm{N}} 2$ but instead to be midway between the two extremes and to have characteristics of both. For instance, take the reaction of 1,2-epoxy-1-methylcyclohexane with HBr, shown in FIGURE 18.2. The reaction yields only a single stereoisomer of 2-bromo-2-methylcyclohexanol in which the -Br and -OH groups are trans, an $\\mathrm{S}_{\\mathrm{N}} 2$-like result caused by backside displacement of the epoxide oxygen. But the fact that $\\mathrm{Br}^{-}$attacks the more hindered tertiary side of the epoxide rather than the less hindered secondary side is an $\\mathrm{S}_{\\mathrm{N}} 1$-like result in which the more stable, tertiary carbocation is involved.\n\nEvidently, the transition state for acid-catalyzed epoxide opening has an $\\mathrm{S}_{\\mathrm{N}}$ 2-like geometry but also has a high degree of $\\mathrm{S}_{\\mathrm{N}} 1$-like carbocationic character. Because the positive charge in the protonated epoxide is shared by the more highly substituted carbon atom, backside attack of $\\mathrm{Br}^{-}$occurs at the more highly substituted site."}
{"id": 1150, "contents": "Acid-Catalyzed Epoxide Opening - \nFIGURE 18.2 Ring-opening of 1,2-epoxy-1-methylcyclohexane with $\\mathbf{H B r}$. There is a high degree of $\\mathrm{S}_{\\mathrm{N}} 1$-like carbocation character in the transition state, which leads to backside attack of the nucleophile at the tertiary center and to formation of a product isomer that has -Br and -OH groups trans."}
{"id": 1151, "contents": "Predicting the Product of Epoxide Ring-Opening - \nPredict the major product of the following reaction:"}
{"id": 1152, "contents": "Strategy - \nIdentify the substitution pattern of the two epoxide carbon atoms. In this case, one carbon is secondary and one is primary. Then recall the guidelines for epoxide cleavages. An epoxide with only primary and secondary carbons usually undergoes cleavage by $\\mathrm{S}_{\\mathrm{N}} 2$-like attack of a nucleophile on the less hindered carbon, but an epoxide with a tertiary carbon atom usually undergoes cleavage by backside attack on the more hindered carbon. In this case, an $\\mathrm{S}_{\\mathrm{N}} 2$ cleavage of the primary $\\mathrm{C}-\\mathrm{O}$ epoxide bond will occur."}
{"id": 1153, "contents": "Solution - \nPROBLEM Predict the major product of each of the following reactions:\n18-12 (a)\n\n(b)\n\n\nPROBLEM How would you prepare the following diols?\n18-13"}
{"id": 1154, "contents": "Base-Catalyzed Epoxide Opening - \nUnlike other ethers, epoxide rings can be cleaved by bases and nucleophiles as well as by acid. Although an ether oxygen is normally a poor leaving group in an $\\mathrm{S}_{\\mathrm{N}} 2$ reaction (Section 11.3), the strain of the threemembered ring causes epoxides to react with hydroxide ion at elevated temperatures."}
{"id": 1155, "contents": "Methylenecyclohexane oxide - \n1-Hydroxymethylcyclohexanol (70\\%)\nBase-catalyzed epoxide opening is a typical $S_{N} 2$ reaction in which attack of the nucleophile takes place at the less hindered epoxide carbon. For example, 1,2-epoxypropane reacts with ethoxide ion exclusively at the less highly substituted, primary carbon to give 1-ethoxy-2-propanol.\n\n\n\n1-Ethoxy-2-propanol (83\\%)\nhere ( $2^{\\circ}$ )\nMany different nucleophiles can be used for epoxide opening, including amines ( $\\mathrm{RNH}_{2}$ or $\\mathrm{R}_{2} \\mathrm{NH}$ ) and Grignard reagents ( RMgX ). An example of an amine reacting with an epoxide occurs in the commercial synthesis of metoprolol, a so-called $\\beta$-blocker that is used for treatment of cardiac arrhythmias, hypertension, and heart attacks. $\\beta$-Blockers are among the most widely prescribed drugs in the world.\n\n\n\nMetoprolol\nA similar nucleophilic ring-opening occurs when epoxides are treated with Grignard reagents. Ethylene oxide is frequently used, thereby allowing the conversion of a Grignard reagent into a primary alcohol having two more carbons than the starting alkyl halide. 1-Bromobutane, for example, is converted into 1-hexanol by reaction of its Grignard reagent with ethylene oxide.\n\n\nPROBLEM Predict the major product of the following reactions:"}
{"id": 1156, "contents": "Methylenecyclohexane oxide - 18.6 Crown Ethers\nCrown ethers, discovered in the early 1960s by Charles Pedersen at the DuPont Company, are a relatively recent addition to the ether family. They are named according to the general format $x$-crown- $y$, where $x$ is the total number of atoms in the ring and $y$ is the number of oxygen atoms. Thus, 18-crown-6 ether is an 18-membered ring containing 6 ether oxygen atoms. Note the size and negative (red) character of the crown ether cavity in the following electrostatic potential map.\n\n\n18-Crown-6 ether\nThe importance of crown ethers stems from their ability to sequester specific metal cations in the center of the polyether cavity. 18-Crown-6, for example, binds strongly with potassium ion. As a result, a solution of 18 -crown-6 in a nonpolar organic solvent will dissolve many potassium salts. Potassium permanganate, $\\mathrm{KMnO}_{4}$, dissolves in toluene in the presence of 18 -crown- 6 , for instance, and the resulting solution is a valuable reagent for oxidizing alkenes.\n\nThe effect of using a crown ether to dissolve an inorganic salt in a hydrocarbon or ether solvent is similar to the effect of dissolving the salt in a polar aprotic solvent such as DMSO, DMF, or HMPA (Section 11.3). In both cases, the metal cation is strongly solvated, leaving the anion bare. Thus, the $\\mathrm{S}_{\\mathrm{N}} 2$ reactivity of an anion is tremendously enhanced in the presence of a crown ether.\n\nAlthough crown ethers do not occur naturally, a group of compounds called ionophores have similar ionbinding properties. Produced by various microorganisms, ionophores are fat-soluble molecules that bind to specific ions and facilitate transport of the ions through biological membranes. The antibiotic valinomycin, for instance, binds specifically to $\\mathrm{K}^{+}$ions with a ten-thousandfold selectivity over $\\mathrm{Na}^{+}$."}
{"id": 1157, "contents": "Methylenecyclohexane oxide - 18.6 Crown Ethers\nValinomycin\nPROBLEM 15-Crown-5 and 12-crown-4 ethers complex $\\mathrm{Na}^{+}$and $\\mathrm{Li}^{+}$, respectively. Make models of these crown 18-15 ethers, and compare the sizes of the cavities."}
{"id": 1158, "contents": "Thiols - \nThiols, also called mercaptans, are sulfur analogs of alcohols. They are named by the same system used for alcohols, with the suffix -thiol used in place of -ol. The -SH group itself is referred to as a mercapto group. Like alcohols, thiols are weakly acidic; the $\\mathrm{p}_{\\mathrm{a}}$ of $\\mathrm{CH}_{3} \\mathrm{SH}$, for instance, is 10.3 . Unlike alcohols, however, thiols don't typically form hydrogen bonds because the sulfur atom is not sufficiently electronegative.\n\n\nEthanethiol\n\n\nCyclohexanethiol\n\n$m$-Mercaptobenzoic acid\n\nThe most striking characteristic of thiols is their appalling odor. Skunk scent, for instance, is caused primarily by the simple thiols 3-methyl-1-butanethiol and 2-butene-1-thiol. Volatile thiols such as ethanethiol are also added to natural gas and liquefied propane to serve as an easily detectable warning in case of leaks.\n\nThiols are usually prepared from alkyl halides by $\\mathrm{S}_{\\mathrm{N}} 2$ displacement with a sulfur nucleophile such as hydrosulfide anion, ${ }^{-}$SH.\n\n\nThe reaction often works poorly unless an excess of the nucleophile is used because the product thiol can undergo a second $S_{N} 2$ reaction with alkyl halide to give a sulfide ( $\\mathbf{R}-\\mathbf{S}-\\mathbf{R}$ ') as by-product. To circumvent this problem, thiourea, $\\left(\\mathrm{NH}_{2}\\right)_{2} \\mathrm{C}=\\mathrm{S}$, is often used as the nucleophile in the preparation of a thiol from an alkyl halide. The reaction occurs by displacement of the halide ion to yield an intermediate alkyl isothiourea salt, which is hydrolyzed by subsequent reaction with aqueous base."}
{"id": 1159, "contents": "Thiols - \nThiols can be oxidized by $\\mathrm{Br}_{2}$ or $\\mathrm{I}_{2}$ to yield disulfides ( $\\mathbf{R}-\\mathbf{S}-\\mathbf{S}-\\mathbf{R}^{\\prime}$ ). The reaction is easily reversed, and a disulfide can be reduced back to a thiol by treatment with zinc and acid.\n\n$$\n2 \\mathrm{R}-\\mathrm{SH} \\underset{\\mathrm{Zn}, \\mathrm{H}^{+}}{\\stackrel{\\mathrm{I}_{2}}{\\rightleftarrows}} \\mathrm{R}-\\mathrm{S}-\\mathrm{S}-\\mathrm{R}+2 \\mathrm{HI}\n$$\n\nA thiol\nA disulfide\nThis thiol-disulfide interconversion is a key part of numerous biological processes. We'll see in Chapter 26, for instance, that disulfide formation is involved in defining the structure and three-dimensional conformations of proteins, where disulfide \"bridges\" often form cross-links between cysteine amino-acid units in the protein chains. Disulfide formation is also involved in the process by which cells protect themselves from oxidative degradation. A cellular component called glutathione removes potentially harmful oxidants and is itself oxidized to glutathione disulfide in the process. Reduction back to the thiol requires the coenzyme reduced flavin adenine dinucleotide, abbreviated $\\mathrm{FADH}_{2}$."}
{"id": 1160, "contents": "Sulfides - \nSulfides are the sulfur analogs of ethers just as thiols are the sulfur analogs of alcohols. Sulfides are named by following the same rules used for ethers, with sulfide used in place of ether for simple compounds and alkylthio used in place of alkoxy for more complex substances."}
{"id": 1161, "contents": "Dimethyl sulfide - \nMethyl phenyl sulfide\n\n\n3-(Methylthio)cyclohexene\n\nTreatment of a thiol with a base, such as NaH , gives the corresponding thiolate ion ( $\\mathrm{RS}^{-}$), which undergoes reaction with a primary or secondary alkyl halide to give a sulfide. The reaction occurs by an $S_{N} 2$ mechanism, analogous to the Williamson synthesis of ethers (Section 18.2).\n\n\nDespite their close structural similarity, sulfides and ethers differ substantially in their chemistry. Because the valence electrons on sulfur are farther from the nucleus and are less tightly held than those on oxygen ( $3 p$ electrons versus $2 p$ electrons), sulfur compounds are more nucleophilic than their oxygen analogs. Unlike dialkyl ethers, dialkyl sulfides react rapidly with primary alkyl halides by an $\\mathrm{S}_{\\mathrm{N}} 2$ mechanism to give sulfonium ions $\\left(\\mathrm{R}_{3} \\mathrm{~S}^{+}\\right)$.\n\n\nDimethyl sulfide Iodomethane Trimethylsulfonium iodide\nThe most common example of this process in living organisms is the reaction of the amino acid methionine with adenosine triphosphate (ATP; Section 6.8) to give $S$-adenosylmethionine. The reaction is somewhat unusual in that the biological leaving group in this $\\mathrm{S}_{\\mathrm{N}} 2$ process is the triphosphate ion rather than the more frequently seen diphosphate ion (Section 11.6).\n\n\nAdenosine triphosphate (ATP)\n$S$-Adenosylmethionine\nSulfonium ions are themselves useful alkylating agents because a nucleophile can attack one of the groups bonded to the positively charged sulfur, displacing a neutral sulfide as leaving group. We saw an example of this in Section 11.6 (FIGURE 11.17), in which $S$-adenosylmethionine transferred a methyl group to norepinephrine to give adrenaline."}
{"id": 1162, "contents": "Dimethyl sulfide - \nAnother difference between sulfides and ethers is that sulfides are easily oxidized. Treatment of a sulfide with hydrogen peroxide, $\\mathrm{H}_{2} \\mathrm{O}_{2}$, at room temperature yields the corresponding sulfoxide ( $\\mathbf{R}_{\\mathbf{2}} \\mathbf{S O}$ ), and further oxidation of the sulfoxide with a peroxyacid yields a sulfone ( $\\mathbf{R}_{\\mathbf{2}} \\mathbf{S O}_{\\mathbf{2}} \\mathbf{)}$.\n\n\nDimethyl sulfoxide (DMSO) is a particularly well-known sulfoxide that is often used as a polar aprotic solvent. It must be handled with care, however, because it has a remarkable ability to penetrate the skin, carrying along whatever is dissolved in it.\n\n\nDimethyl sulfoxide (a polar aprotic solvent)\n\nPROBLEM Name the following compounds:\n18-16 (a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n\nPROBLEM 2-Butene-1-thiol is one component of skunk spray. How would you synthesize this substance 18-17 (either cis or trans) from methyl 2-butenoate? From 1,3-butadiene?\n\n$\\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{CHCH}_{2} \\mathrm{SH}$\nMethyl 2-butenoate\n2-Butene-1-thiol"}
{"id": 1163, "contents": "Infrared Spectroscopy - \nEthers are difficult to identify by IR spectroscopy. Although they show an absorption due to C-O single-bond stretching in the range 1050 to $1150 \\mathrm{~cm}^{-1}$, many other kinds of absorptions occur in the same range. FIGURE 18.3 shows the IR spectrum of diethyl ether and identifies the $\\mathrm{C}-\\mathrm{O}$ stretch. Phenyl alkyl ethers show two strong absorbances for $\\mathrm{C}-\\mathrm{O}$ stretching at 1050 and $1250 \\mathrm{~cm}^{-1}$. FIGURE 18.4 shows the IR spectrum of anisole.\n\n\nFIGURE 18.3 The infrared spectrum of diethyl ether, $\\mathrm{CH}_{3} \\mathbf{C H}_{2} \\mathbf{O C H}_{2} \\mathbf{C H}_{3}$.\n\n\nFIGURE 18.4 The infrared spectrum of anisole."}
{"id": 1164, "contents": "Nuclear Magnetic Resonance Spectroscopy - \nHydrogens on carbon next to an ether oxygen are shifted downfield from the normal alkane resonance and show ${ }^{1}$ H NMR absorptions in the region 3.4 to $4.5 \\delta$. This downfield shift is clearly seen in the spectrum of dipropyl ether shown in FIGURE 18.5.\n\n\nFIGURE 18.5 The ${ }^{\\mathbf{1}} \\mathbf{H}$ NMR spectrum of dipropyl ether. Protons on carbon next to oxygen are shifted downfield to $3.4 \\delta$.\nEpoxides absorb at a slightly higher field than other ethers and show characteristic resonances at 2.5 to 3.5 $\\delta$ in their ${ }^{1}$ H NMR spectra, as indicated for 1,2-epoxypropane in FIGURE 18.6. The methylene protons of this epoxide are diastereotopic, and display complex splitting (see Section 13.7 and Section 13.8).\n\n\nEther carbon atoms also exhibit a downfield shift in the ${ }^{13}$ C NMR spectrum, where they usually absorb in the 50 to $80 \\delta$ range. For example, the carbon atoms next to oxygen in methyl propyl ether absorb at 58.5 and $74.8 \\delta$. Similarly, the methyl carbon in anisole absorbs at $54.8 \\delta$.\n\n\n\nPROBLEM The ${ }^{1} \\mathrm{H}$ NMR spectrum shown is that of a cyclic ether with the formula $\\mathrm{C}_{4} \\mathrm{H}_{8} \\mathrm{O}$. Propose a structure. 18-18"}
{"id": 1165, "contents": "(30) CHEMISTRY MATTERS - \nEpoxy Resins and Adhesives\nFew nonchemists know exactly what an epoxide is, but practically everyone has used an \"epoxy glue\" for household repairs or an epoxy resin for a protective coating. Worldwide, approximately 14 billion dollars' worth of epoxy resins and adhesives are used annually for a vast number of applications, including many in the aerospace industry. Much of the Boeing 787 Dreamliner, for instance, is held together with epoxy-based adhesives.\n\nEpoxy resins and adhesives generally consist of two components that are mixed just prior to use. One component is a liquid \"prepolymer,\" and the second is a \"curing agent\" that reacts with the prepolymer and causes it to solidify.\n\nThe most widely used epoxy resins and adhesives are based on a prepolymer made from bisphenol A and epichlorohydrin. On treatment with base, bisphenol A is converted into its anion, which acts as a nucleophile in an $\\mathrm{S}_{\\mathrm{N}} 2$ reaction with epichlorohydrin. Each epichlorohydrin molecule can react with two molecules of bisphenol $A$, once by $S_{N} 2$ displacement of chloride ion and once by nucleophilic opening of the epoxide ring. At the same time, each bisphenol A molecule can react with two epichlorohydrins, leading to a long polymer chain. Each end of a prepolymer chain has an unreacted epoxy group, and each chain has numerous secondary alcohol groups spaced regularly along its midsection.\n\n\nFIGURE 18.7 Kayaks are often made of a high-strength polymer coated with epoxy resin. (credit: \"Potomac River Festival Virginia 2017 Great Falls Kayak Race\" by Watts/Flickr, CC BY 2.0)\n\n\nBisphenol A\nEpichlorohydrin"}
{"id": 1166, "contents": "(30) CHEMISTRY MATTERS - \nBisphenol A\nEpichlorohydrin\n\n\nWhen an epoxide is to be used, a basic curing agent such as a tertiary amine, $\\mathrm{R}_{3} \\mathrm{~N}$, is added to cause the individual prepolymer chains to link together. This cross-linking of chains is simply a base-catalyzed, $\\mathrm{S}_{\\mathrm{N}} 2$ epoxide ring-opening of an -OH group in the middle of one chain with an epoxide group on the end of another chain. The result of such cross-linking is formation of a vast, three-dimensional tangle that has enormous strength and chemical resistance."}
{"id": 1167, "contents": "Key Terms - \n```\n\u2022 alkoxymercuration \u2022 sulfone\n- crown ether\n- sulfonium ion\n\u2022 disulfide (R-S-S-R') . sulfoxide\n- ether (R-0-R')\n- thiol\n- mercapto group\n- thiolate ion (RS`)\n- sulfide (R-S-R')\n- Williamson ether synthesis\n```"}
{"id": 1168, "contents": "Summary - \nThis chapter has finished the coverage of functional groups with $\\mathrm{C}-\\mathrm{O}$ and $\\mathrm{C}-\\mathrm{S}$ single bonds, focusing primarily on ethers, epoxides, thiols, and sulfides. Ethers are compounds that have two organic groups bonded to the same oxygen atom, ROR'. The organic groups can be alkyl, vinylic, or aryl, and the oxygen atom can be in a ring or in an open chain. Ethers are prepared by either Williamson ether synthesis, which involves $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of an alkoxide ion with a primary alkyl halide, or the alkoxymercuration reaction, which involves Markovnikov addition of an alcohol to an alkene.\n\nEthers are inert to most reagents but react with strong acids to give cleavage products. Both HI and HBr are often used. The cleavage reaction takes place by an $\\mathrm{S}_{\\mathrm{N}} 2$ mechanism at the less highly substituted site if only primary and secondary alkyl groups are bonded to the ether oxygen, but by an $\\mathrm{S}_{\\mathrm{N}} 1$ or E1 mechanism if one of the alkyl groups bonded to oxygen is tertiary.\n\nEpoxides are cyclic ethers with a three-membered, oxygen-containing ring. Because of the strain in the ring, epoxides undergo a cleavage reaction with both acids and bases. Acid-catalyzed ring-opening occurs with a regiochemistry that depends on the structure of the epoxide. Cleavage of the $\\mathrm{C}-\\mathrm{O}$ bond at the less highly substituted site occurs if both epoxide carbons are primary or secondary, but cleavage of the $\\mathrm{C}-\\mathrm{O}$ bond to the more highly substituted site occurs if one of the epoxide carbons is tertiary. Base-catalyzed epoxide ringopening occurs by $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of a nucleophile at the less hindered epoxide carbon."}
{"id": 1169, "contents": "Summary - \nThiols, the sulfur analogs of alcohols, are usually prepared by $S_{N} 2$ reaction of an alkyl halide with thiourea. Mild oxidation of a thiol yields a disulfide, and mild reduction of a disulfide returns the thiol. Sulfides, the sulfur analogs of ethers, are prepared by an $\\mathrm{S}_{\\mathrm{N}} 2$ reaction between a thiolate anion and a primary or secondary alkyl halide. Sulfides are more nucleophilic than ethers and can be alkylated by reaction with a primary alkyl halide to yield a sulfonium ion. Sulfides can also be oxidized to sulfoxides and to sulfones."}
{"id": 1170, "contents": "Summary of Reactions - \n1. Synthesis of ethers (Section 18.2)\na. Williamson ether synthesis\n\n$$\n\\mathrm{RO}^{-}+\\mathrm{R}^{\\prime} \\mathrm{CH}_{2} \\mathrm{X} \\longrightarrow \\mathrm{ROCH}_{2} \\mathrm{R}^{\\prime}+\\mathrm{X}^{-}\n$$\n\nb. Alkoxymercuration/demercuration\n\n2. Reactions of ethers\na. Cleavage by HBr or HI (Section 18.3)\n\n$$\n\\mathrm{R}-\\mathrm{O}-\\mathrm{R}^{\\prime} \\xrightarrow[\\mathrm{H}_{2} \\mathrm{O}]{\\mathrm{HX}} \\mathrm{RX}+\\mathrm{R}^{\\prime} \\mathrm{OH}\n$$\n\nb. Acid-catalyzed epoxide opening (Section 18.5)\n\nc. Base-catalyzed epoxide opening (Section 18.5)\n\n\n3. Synthesis of thiols (Section 18.7)\n\n$$\n\\mathrm{RCH}_{2} \\mathrm{Br} \\xrightarrow[\\text { 2. } \\mathrm{H}_{2} \\mathrm{O}, \\mathrm{NaOH}]{\\text { 1. } \\left.\\mathrm{H}_{2} \\mathrm{~N}\\right)_{2} \\mathrm{C}=\\mathrm{S}} \\quad \\mathrm{RCH}_{2} \\mathrm{SH}\n$$\n\n4. Oxidation of thiols to disulfides (Section 18.7)\n\n$$\n2 \\mathrm{RSH} \\xrightarrow{\\mathrm{I}_{2}, \\mathrm{H}_{2} \\mathrm{O}} \\quad \\mathrm{RS}-\\mathrm{SR}\n$$\n\n5. Synthesis of sulfides (Section 18.7)\n\n$$\n\\mathrm{RS}^{-}+\\mathrm{R}^{\\prime} \\mathrm{CH}_{2} \\mathrm{Br} \\longrightarrow \\mathrm{RSCH}_{2} \\mathrm{R}^{\\prime}+\\mathrm{Br}^{-}\n$$\n\n6. Oxidation of sulfides to sulfoxides and sulfones (Section 18.7)"}
{"id": 1171, "contents": "Visualizing Chemistry - \nPROBLEM Give IUPAC names for the following compounds (red = O; reddish brown = Br; yellow $=\\mathrm{S}$ ):\n\n18-19 (a)\n\n(b)\n\n(c)\n\n\nPROBLEM Show the product, including stereochemistry, that would result from reaction of the following\n18-20 epoxide with HBr :\n\n\nPROBLEM Show the product, including stereochemistry, of the following reaction:\n18-21\n\n\nPROBLEM Treatment of the following alkene with a peroxyacid yields an epoxide different from that obtained\n18-22 by reaction with aqueous $\\mathrm{Br}_{2}$ followed by base treatment. Propose structures for the two epoxides, and explain the result."}
{"id": 1172, "contents": "Mechanism Problems - \nPROBLEM Predict the product(s) and provide the mechanism for each of the following reactions.\n\n18-23 (a)\n\n(b)\n\n\nPROBLEM Predict the product(s) and provide the mechanism for each of the following reactions.\n18-24\n(a)\n\n$\\xrightarrow{\\mathrm{HBr}}$ ?\n(b)\n\n\nPROBLEM Predict the product(s) and provide the mechanism for each of the following two-step processes.\n\n18-25 (a)\n\n(b)\n\n\nPROBLEM The alkoxymercuration of alkenes involves the formation of an organomercury intermediate (I),\n18-26 which is reduced with $\\mathrm{NaBH}_{4}$ to give an ether product. Predict the ether product and provide the mechanism for the following reaction.\n\n\nPROBLEM Predict the product(s) and provide the mechanism for the following reactions:\n\n18-27 (a)\n\n\n(b)\n\n\nPROBLEM Predict the product(s) and provide the mechanism for each of the following reactions.\n\n18-28 (a)\n\n(b)\n\n\nPROBLEM In the formation of the prepolymer used to make epoxy resins, a bisphenol reacts with\n18-29 epichlorohydrin in the presence of a base. Show the product and mechanism when two moles of phenol react with epichlorohydrin."}
{"id": 1173, "contents": "Epichlorohydrin - \nPROBLEM Ethers undergo an acid-catalyzed cleavage reaction when treated with the Lewis acid $\\mathrm{BBr}_{3}$ at room 18-30 temperature. Propose a mechanism for the reaction.\n\n\nPROBLEM Treatment of 1,1-diphenyl-1,2-epoxyethane with aqueous acid yields diphenyl acetaldehyde as the\n18-31 major product. Propose a mechanism for the reaction.\n\n\nPROBLEM Fluoxetine, a heavily prescribed antidepressant marketed under the name Prozac, can be prepared\n18-32 by a route that begins with reaction between a phenol and an alkyl chloride.\n\n(a) The rate of the reaction depends on both phenol and alkyl halide. Is this an $\\mathrm{S}_{\\mathrm{N}} 1$ or an $\\mathrm{S}_{\\mathrm{N}} 2$ reaction? Show the mechanism.\n(b) The physiologically active enantiomer of fluoxetine has $(S)$ stereochemistry. Based on your answer in part (a), draw the structure of the alkyl chloride you would need, showing the correct stereochemistry.\n\nPROBLEM When 2-methyl-2,5-pentanediol is treated with sulfuric acid, dehydration occurs and\n18-33 2,2-dimethyltetrahydrofuran is formed. Suggest a mechanism for this reaction. Which of the two oxygen atoms is most likely to be eliminated, and why?"}
{"id": 1174, "contents": "2,2-Dimethyltetrahydrofuran - \nPROBLEM Methyl aryl ethers, such as anisole, are cleaved to iodomethane and a phenoxide ion by treatment 18-34 with LiI in hot DMF. Propose a mechanism for this reaction.\n\nPROBLEM The herbicide acifluorfen can be prepared by a route that begins with reaction between a phenol 18-35 and an aryl fluoride. Propose a mechanism.\n\n\nPROBLEM Aldehydes and ketones undergo acid-catalyzed reaction with alcohols to yield hemiacetals, from 18-36 aldehydes or ketals with ketones compounds that have one alcohol-like oxygen and one ether-like oxygen bonded to the same carbon. Further reaction of a hemiacetal with alcohol then yields an acetal, a compound that has two ether-like oxygens bonded to the same carbon.\n\n(a) Show the structures of the hemiketal and ketal you would obtain by reaction of cyclohexanone with ethanol.\n(b) Propose a mechanism for the conversion of a hemiacetal into a ketal.\n\nPROBLEM Propose a mechanism to account for the following transformation. What two kinds of reactions are 18-37 occurring?"}
{"id": 1175, "contents": "Naming Ethers - \nPROBLEM Draw structures corresponding to the following IUPAC names:\n18-38 (a) Ethyl 1-ethylpropyl ether (b) Di(p-chlorophenyl) ether (c) 3,4-Dimethoxybenzoic acid\n(d) Cyclopentyloxycyclohexane (e) 4-Allyl-2-methoxyphenol (eugenol; from oil of cloves)\n\nPROBLEM Give IUPAC names for the following structures:\n18-39\n\n\nSynthesizing Ethers\nPROBLEM How would you prepare the following ethers?\n18-40 (a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n\nPROBLEM How would you prepare the following compounds from 1-phenylethanol?\n18-41\n(a) Methyl 1-phenylethyl ether\n(b) Phenylepoxyethane\n(c) tert-Butyl 1-phenylethyl ether\n(d) 1-Phenylethanethiol\n\nPROBLEM tert-Butyl ethers can be prepared by the reaction of an alcohol with 2 -methylpropene in the 18-42 presence of an acid catalyst. Propose a mechanism for this reaction.\n\nPROBLEM Treatment of trans-2-chlorocyclohexanol with NaOH yields 1,2-epoxycyclohexane, but reaction of\n18-43 the cis isomer under the same conditions yields cyclohexanone. Propose mechanisms for both reactions, and explain why the different results are obtained.\n\n\n\n$\\xrightarrow[\\mathrm{H}_{2} \\mathrm{O}]{\\mathrm{NaOH}}$\n\n\nReactions of Ethers and Epoxides\nPROBLEM Predict the products of the following ether cleavage reactions:\n\n18-44 (a)\n\n(c)\n\n(b)\n\n(d)\n\n\nPROBLEM How would you carry out the following transformations? More than one step may be required.\n18-45 (a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n\nPROBLEM What product would you expect from cleavage of tetrahydrofuran with HI ? 18-46"}
{"id": 1176, "contents": "Naming Ethers - \n(b)\n\n(c)\n\n(d)\n\n(e)\n\n\nPROBLEM What product would you expect from cleavage of tetrahydrofuran with HI ? 18-46\n\nPROBLEM Write the mechanism of the hydrolysis of cis-5,6-epoxydecane by reaction with aqueous acid. What 18-47 is the stereochemistry of the product, assuming normal backside $\\mathrm{S}_{\\mathrm{N}} 2$ attack?\n\nPROBLEM What is the stereochemistry of the product from acid-catalyzed hydrolysis of 18-48 trans-5,6-epoxydecane? How does the product differ from that formed in Problem 18-47?\n\nPROBLEM Acid-catalyzed hydrolysis of a 1,2-epoxycyclohexane produces a trans-diaxial 1,2-diol. What 18-49 product would you expect to obtain from acidic hydrolysis of cis-3-tert-butyl-1,2-epoxycyclohexane? (Recall that the bulky tert-butyl group locks the\ncyclohexane ring into a specific conformation.)\nPROBLEM Imagine that you have treated ( $2 R, 3 R$ )-2,3-epoxy-3-methylpentane with aqueous acid to carry out a 18-50 ring-opening reaction."}
{"id": 1177, "contents": "2,3-Epoxy-3-methylpentane (no stereochemistry implied) - \n(a) Draw the epoxide, showing stereochemistry.\n(b) Draw and name the product, showing stereochemistry. (c) Is the product chiral? Explain.\n(d) Is the product optically active? Explain.\n\nPROBLEM Epoxides are reduced by treatment with lithium aluminum hydride to yield alcohols. Propose a\n18-51 mechanism for this reaction.\n\n\nPROBLEM Show the structure and stereochemistry of the alcohol that would result if 1,2-epoxycyclohexane\n18-52 were reduced with lithium aluminum deuteride, $\\mathrm{LiAlD}_{4}$ (Problem 18-51)."}
{"id": 1178, "contents": "Spectroscopy - \nPROBLEM The red fox (Vulpes vulpes) uses a chemical communication system based on scent marks in urine.\n18-53 One component of fox urine is a sulfide whose mass spectrum has $\\mathrm{M}^{+}=116$. IR spectroscopy shows an intense band at $890 \\mathrm{~cm}^{-1}$, and ${ }^{1} \\mathrm{H}$ NMR spectroscopy reveals the following peaks:\n$1.74 \\delta(3 \\mathrm{H}$, singlet $) ; 2.11 \\delta(3 \\mathrm{H}$, singlet $) ; 2.27 \\delta(2 \\mathrm{H}$, triplet, $J=4.2 \\mathrm{~Hz}) ; 2.57 \\delta(2 \\mathrm{H}$, triplet, $J=$ $4.2 \\mathrm{~Hz}) ; 4.73 \\delta(2 \\mathrm{H}$, broad)\n\nPropose a structure consistent with these data. [Note: $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{~S}$ absorbs at $2.1 \\mathrm{\\delta}$ ].\nPROBLEM Anethole, $\\mathrm{C}_{10} \\mathrm{H}_{12} \\mathrm{O}$, a major constituent of the oil of anise, has the ${ }^{1} \\mathrm{H}$ NMR spectrum shown. On\n18-54 careful oxidation with $\\mathrm{Na}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$, anethole yields $p$-methoxybenzoic acid. What is the structure of anethole? Assign all peaks in the NMR spectrum, and account for the observed splitting patterns.\n\n\nPROBLEM Propose structures for compounds that have the following ${ }^{1} \\mathrm{H}$ NMR spectra:\n18-55\n(a) $\\mathrm{C}_{5} \\mathrm{H}_{12} \\mathrm{~S}(\\mathrm{An}-\\mathrm{SH}$ proton absorbs near 1.6 \u05e1.)\n\n(b) $\\mathrm{C}_{9} \\mathrm{H}_{11} \\mathrm{BrO}$\n\n(c) $\\mathrm{C}_{5} \\mathrm{H}_{12} \\mathrm{O}_{2}$"}
{"id": 1179, "contents": "General Problems - \nPROBLEM Predict the products of the following reactions:"}
{"id": 1180, "contents": "18-56 (a) - \n$\\xrightarrow{\\mathrm{HBr}}$\n(b)\n\n?\n\nc)\n\n(d)\n?\n\nPROBLEM How would you synthesize anethole (Problem 18-54) from phenol?\n18-57\nPROBLEM How could you prepare benzyl phenyl ether from benzene and phenol? More than one step is\n18-58 required.\nPROBLEM Meerwein's reagent, triethyloxonium tetrafluoroborate, is a powerful ethylating agent that converts\n18-59 alcohols into ethyl ethers at neutral pH . Show the reaction of Meerwein's reagent with cyclohexanol, and account for the fact that trialkyloxonium salts are much more reactive alkylating agents than alkyl iodides."}
{"id": 1181, "contents": "$\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2}\\right)_{3} \\mathrm{O}^{+} \\mathrm{BF}_{4}^{-} \\quad$ Meerwein's reagent - \nPROBLEM Safrole, a substance isolated from oil of sassafras, is used as a perfumery agent. Propose a synthesis\n18-60 of safrole from catechol (1,2-benzenediol)."}
{"id": 1182, "contents": "Safrole - \nPROBLEM Grignard reagents react with oxetane, a four-membered cyclic ether, to yield primary alcohols, but\n18-61 the reaction is much slower than the corresponding reaction with ethylene oxide. Suggest a reason for the difference in reactivity between oxetane and ethylene oxide.\n\n$$\n\\square \\mathrm{O} \\xrightarrow[\\text { 2. } \\mathrm{H}_{3} \\mathrm{O}^{+}]{\\text {1. } \\mathrm{RMgX}} \\mathrm{RCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}\n$$\n\nOxetane\nPROBLEM The Zeisel method is an old analytical procedure for determining the number of methoxyl groups\n18-62 in a compound. A weighed amount of the compound is heated with concentrated HI, ether cleavage occurs, and the iodomethane product is distilled off and passed into an alcohol solution of $\\mathrm{AgNO}_{3}$, where it reacts to form a precipitate of silver iodide. The AgI is then collected and weighed, and the percentage of methoxyl groups in the sample is thereby determined. For example, 1.06 g of vanillin, the material responsible for the characteristic odor of vanilla, yields 1.60 g of AgI. If vanillin has a molecular weight of 152, how many methoxyl groups does it contain?"}
{"id": 1183, "contents": "Safrole - \nPROBLEM Disparlure, $\\mathrm{C}_{19} \\mathrm{H}_{38} \\mathrm{O}$, is a sex attractant released by the female spongy moth, Lymantria dispar.\n18-63 The ${ }^{1} \\mathrm{H}$ NMR spectrum of disparlure shows a large absorption in the alkane region, 1 to $2 \\delta$, and a triplet at $2.8 \\delta$. Treatment of disparlure, first with aqueous acid and then with $\\mathrm{KMnO}_{4}$, yields two carboxylic acids identified as undecanoic acid and 6-methylheptanoic acid. ( $\\mathrm{KMnO}_{4}$ cleaves 1,2-diols to yield carboxylic acids.) Neglecting stereochemistry, propose a structure for disparlure. The actual compound is a chiral molecule with $7 R, 8 S$ stereochemistry. Draw disparlure, showing the correct stereochemistry.\n\nPROBLEM How would you synthesize racemic disparlure (Problem 18-63) from compounds having ten or\n18-64 fewer carbons?\nPROBLEM How would you prepare o-hydroxyphenylacetaldehyde from phenol? More than one step is\n18-65 required.\n\no-Hydroxyphenylacetaldehyde\n\nPROBLEM Identify the reagents a-e in the following scheme:"}
{"id": 1184, "contents": "18-66 - \nPROBLEM Propose structures for compounds that have the following ${ }^{1} \\mathrm{H}$ NMR spectra:\n18-67 (a) $\\mathrm{C}_{4} \\mathrm{H}_{10} \\mathrm{O}_{2}$\n\n(b) $\\mathrm{C}_{9} \\mathrm{H}_{10} \\mathrm{O}$\n\n\nPROBLEM We saw in Section 17.4 that ketones react with $\\mathrm{NaBH}_{4}$ to yield alcohols. We'll also see in Section\n18-68 22.3 that ketones react with $\\mathrm{Br}_{2}$ to yield $\\alpha$-bromo ketones. Perhaps surprisingly, treatment with $\\mathrm{NaBH}_{4}$ of the $\\alpha$-bromo ketone from acetophenone yields an epoxide rather than a bromo alcohol. Show the structure of the epoxide, and explain its formation."}
{"id": 1185, "contents": "Preview of Carbonyl Chemistry - \nCarbonyl compounds are everywhere. Most biological molecules contain carbonyl groups, as do most pharmaceutical agents and many of the synthetic chemicals that affect our everyday lives. Citric acid, found in lemons and oranges; acetaminophen, the active ingredient in many over-the-counter headache remedies; and Dacron, the polyester material used in clothing, all contain different kinds of carbonyl groups.\n\n\nCitric acid\n(a carboxylic acid)\n\n\nAcetaminophen (an amide)\n\n\nDacron\n(a polyester)\n\nTo a great extent, the chemistry of living organisms is the chemistry of carbonyl compounds. Thus, we'll spend the next five chapters discussing the chemistry of the carbonyl group, $\\mathbf{C}=\\mathbf{O}$ (pronounced car-bo-neel). There are many different kinds of carbonyl compounds and many different reactions, but there are only a few fundamental principles that tie the entire field together. The purpose of this brief preview is not to show details of specific reactions but rather to provide a framework for learning carbonyl-group chemistry. Read through this preview now, and return to it on occasion to remind yourself of the larger picture."}
{"id": 1186, "contents": "I Kinds of Carbonyl Compounds - \nTABLE 18.2 shows some of the many different kinds of carbonyl compounds. All contain an acyl group ( $\\mathrm{R}-\\mathrm{C}=\\mathrm{O}$ ) bonded to another substituent. The R part of the acyl group can be practically any organic part/ structure, and the other substituent to which the acyl group is bonded might be a carbon, hydrogen, oxygen, halogen, nitrogen, or sulfur.\n\n| Name | General formula | Name ending |\n| :---: | :---: | :---: |\n| Aldehyde | <smiles>[R]C=O</smiles> | -al |\n| Ketone | <smiles>[R]C([R])=O</smiles> | -one |\n| Carboxylic acid | <smiles>[R]C(=O)O</smiles> | -oic acid |\n| Acid halide | <smiles>[R]C([X])=O</smiles> | -yl or -oyl halide |\n| Acid anhydride | <smiles>[R]C([R])=O</smiles> | -oic anhydride |\n| Acyl phosphate | <smiles>[R]C(=O)OP(=O)([O-])[O-]</smiles> | -yl phosphate |\n\nTABLE 18.2 Some Types of Carbonyl Compounds\n\n| Name | General formula | Name ending |\n| :---: | :---: | :---: |\n| Ester | <smiles>[R]OC([R])=O</smiles> | -oate |\n| Lactone (cyclic ester) | <smiles>CC1(C)CCCOC1=O</smiles> | None |\n| Thioester | <smiles>[R]SC([R])=O</smiles> | -thioate |\n| Amide | <smiles>[R]C(=O)N(C)C</smiles> | - amide |\n| Lactam (cyclic amide) | <smiles>CN1CCCC(C)(C)C1=O</smiles> | None |"}
{"id": 1187, "contents": "I Kinds of Carbonyl Compounds - \nIt's useful to classify carbonyl compounds into two categories based on the kinds of chemistry they undergo. In one category are aldehydes and ketones; in the other are carboxylic acids and their derivatives. The acyl group in an aldehyde or ketone is bonded to an atom (H or C, respectively) that can't stabilize a negative charge and therefore can't act as a leaving group in a nucleophilic substitution reaction. The acyl group in a carboxylic acid or its derivative, however, is bonded to an atom (oxygen, halogen, sulfur, nitrogen) that can stabilize a negative charge and therefore can act as a leaving group in a nucleophilic substitution reaction.\n\n\nAldehyde\n\n\nKetone\n\nThe $-\\mathrm{R}^{\\prime}$ and -H in these compounds can't act as leaving groups in nucleophilic substitution reactions.\n\n\nCarboxylic acid\n\n\nAmide\n\n\nAcid halide\n\n\nEster\n\n\nAcid\nanhydride\n\n\nAcyl phosphate\n\n\nThioester\nThe $-\\mathrm{OH},-\\mathrm{X},-\\mathrm{OR}^{\\prime},-\\mathrm{SR},-\\mathrm{NH}_{2}$, $-\\mathrm{OCOR}^{\\prime}$, and $-\\mathrm{OPO}_{3}{ }^{2-}$ in these compounds can act as leaving groups in nucleophilic substitution reactions."}
{"id": 1188, "contents": "II Nature of the Carbonyl Group - \nThe carbon-oxygen double bond of a carbonyl group is similar in many respects to the carbon-carbon double bond of an alkene. The carbonyl carbon atom is $s p^{2}$-hybridized and forms three $\\sigma$ bonds. The fourth valence electron remains in a carbon $p$ orbital and forms a $\\pi$ bond to oxygen by overlapping with an oxygen $p$ orbital.\n\nThe oxygen atom also has two nonbonding pairs of electrons, which occupy its remaining two orbitals.\n\n\nLike alkenes, carbonyl compounds are planar about the double bond and have bond angles of approximately $120^{\\circ}$. FIGURE 18.8 shows the structure of acetaldehyde and indicates its bond lengths and angles. As you might expect, the carbon-oxygen double bond is both shorter ( 122 pm versus 143 pm ) and stronger [ $732 \\mathrm{~kJ} / \\mathrm{mol}$ (175 $\\mathrm{kcal} / \\mathrm{mol}$ ) versus $385 \\mathrm{~kJ} / \\mathrm{mol}$ ( $92 \\mathrm{kcal} / \\mathrm{mol}$ )] than a C-O single bond.\n\n| Bond angle | $\\left.\\mathbf{(}^{\\circ}\\right)$ | Bond length | (pm) |\n| :--- | :---: | :---: | :---: |\n| $\\mathrm{H}-\\mathrm{C}-\\mathrm{C}$ | 118 | $\\mathrm{C}=\\mathrm{O}$ | 122 |\n| $\\mathrm{C}-\\mathrm{C}=\\mathrm{O}$ | 121 | $\\mathrm{C}-\\mathrm{C}$ | 150 |\n| $\\mathrm{H}-\\mathrm{C}=\\mathrm{O}$ | 121 | $\\mathrm{OC}-\\mathrm{H}$ | 109 |"}
{"id": 1189, "contents": "II Nature of the Carbonyl Group - \nFIGURE 18.8 Structure of acetaldehyde.\nAs indicated by the electrostatic potential map in FIGURE 18.8, the carbon-oxygen double bond is strongly polarized because of the high electronegativity of oxygen relative to carbon. Thus, the carbonyl carbon atom carries a partial positive charge, is an electrophilic (Lewis acidic) site, and reacts with nucleophiles. Conversely, the carbonyl oxygen atom carries a partial negative charge, is a nucleophilic (Lewis basic) site, and reacts with electrophiles. We'll see in the next five chapters that the majority of carbonyl-group reactions can be rationalized by simple polarity arguments."}
{"id": 1190, "contents": "III General Reactions of Carbonyl Compounds - \nBoth in the laboratory and in living organisms, most reactions of carbonyl compounds take place by one of four general mechanisms: nucleophilic addition, nucleophilic acyl substitution, alpha substitution, and carbonyl condensation. These mechanisms have many variations, just as alkene electrophilic addition reactions and $\\mathrm{S}_{\\mathrm{N}} 2$ reactions do, but the variations are much easier to learn when the fundamental features of the mechanisms are made clear. Let's see what the four mechanisms are and what kinds of chemistry carbonyl compounds undergo."}
{"id": 1191, "contents": "Nucleophilic Addition Reactions of Aldehydes and Ketones (Chapter 19) - \nThe most common reaction of aldehydes and ketones is the nucleophilic addition reaction, in which a nucleophile, : $\\mathrm{Nu}^{-}$, adds to the electrophilic carbon of the carbonyl group. Because the nucleophile uses an electron pair to form a new bond to carbon, two electrons from the carbon-oxygen double bond must move toward the electronegative oxygen atom to give an alkoxide anion. The carbonyl carbon rehybridizes from $s p^{2}$ to $s p^{3}$ during the reaction, and the alkoxide ion product therefore has tetrahedral geometry.\n\n\nA carbonyl compound\nA tetrahedral intermediate\n( $s p^{3}$-hybridized carbon)\n\nOnce formed, and depending on the nature of the nucleophile, the tetrahedral alkoxide intermediate can undergo one of two further reactions, as shown in FIGURE 18.9. Often, the tetrahedral alkoxide intermediate is simply protonated by water or acid to form an alcohol product. Alternatively, the tetrahedral intermediate\ncan be protonated and expel the oxygen to form a new double bond between the carbonyl carbon and the nucleophile. We'll study both processes in detail in Chapter 19.\n\n\nFIGURE 18.9 The addition reaction of an aldehyde or a ketone with a nucleophile. Depending on the nucleophile, either an alcohol or a compound with a $\\mathrm{C}=\\mathrm{Nu}$ double bond is formed."}
{"id": 1192, "contents": "FORMATION OF AN ALCOHOL - \nThe simplest reaction of a tetrahedral alkoxide intermediate is protonation to yield an alcohol. We've already seen two examples of this kind of process during reduction of aldehydes and ketones with hydride reagents such as $\\mathrm{NaBH}_{4}$ and $\\mathrm{LiAlH}_{4}$ (Section 17.4) and during Grignard reactions (Section 17.5). During a reduction, the nucleophile that adds to the carbonyl group is a hydride ion, $\\mathrm{H}:^{-}$, while during a Grignard reaction, the nucleophile is a carbanion, $\\mathrm{R}_{3} \\mathrm{C}:^{-}$.\n\n\n\nKetone/ aldehyde\n\nTetrahedral intermediate\n\nAlcohol"}
{"id": 1193, "contents": "FORMATION OF $\\mathbf{C = N u}$ - \nThe second mode of nucleophilic addition, which often occurs with amine nucleophiles, involves elimination of oxygen and formation of a $\\mathrm{C}=\\mathrm{Nu}$ double bond. For example, aldehydes and ketones react with primary amines, $\\mathrm{RNH}_{2}$, to form imines, $\\mathrm{R}_{2} \\mathrm{C}=\\mathrm{NR}^{\\prime}$. These reactions use the same kind of tetrahedral intermediate as that formed during hydride reduction and Grignard reaction, but the initially formed alkoxide ion is not isolated. Instead, it is protonated and then loses water to form an imine, as shown in FIGURE 18.10.\n\nFIGURE 18.10 MECHANISM\nFormation of an imine, $\\mathrm{R}_{\\mathbf{2}} \\mathrm{C}=\\mathrm{NR}^{\\prime}$, by reaction of an amine with an aldehyde or a ketone.\n(1) Addition to the ketone or aldehyde carbonyl group by the neutral amine nucleophile gives a dipolar tetrahedral intermediate.\n\n(1)\n\n(2) Transfer of a proton from nitrogen to oxygen then yields an amino alcohol intermediate.\n\n\n(3) Dehydration of the amino alcohol intermediate gives neutral imine and water as final products."}
{"id": 1194, "contents": "Nucleophilic Acyl Substitution Reactions of Carboxylic Acid Derivatives (Chapter 21) - \nThe second fundamental reaction of carbonyl compounds, nucleophilic acyl substitution, is related to the nucleophilic addition reaction just discussed but occurs only with carboxylic acid derivatives rather than with aldehydes and ketones. When the carbonyl group of a carboxylic acid derivative reacts with a nucleophile, addition occurs in the usual way, but the initially formed tetrahedral alkoxide intermediate is not isolated. Because carboxylic acid derivatives have a leaving group bonded to the carbonyl-group carbon, the tetrahedral intermediate can react further by expelling the leaving group and forming a new carbonyl compound:\n\n\nThe net effect of nucleophilic acyl substitution is the replacement of the leaving group by the entering nucleophile. We'll see in Chapter 21, for instance, that acid chlorides are rapidly converted into esters by treatment with alkoxide ions (FIGURE 18.11)."}
{"id": 1195, "contents": "FIGURE 18.11 MECHANISM - \nNucleophilic acyl substitution of an acid chloride with an alkoxide ion yields an ester.\n\n1 Nucleophilic addition of alkoxide ion to an acid chloride yields a tetrahedral intermediate."}
{"id": 1196, "contents": "\u00a9 - \n(2) An electron pair from oxygen expels chloride ion and yields the substitution product, an ester."}
{"id": 1197, "contents": "Alpha-Substitution Reactions (Chapter 22) - \nThe third major reaction of carbonyl compounds, alpha substitution, occurs at the position next to the carbonyl group-the alpha ( $\\alpha$ ) position. This reaction results in the substitution of an $\\alpha$ hydrogen by an electrophile through the formation of an intermediate enol or enolate ion:\n\n\nAn enol\nFor reasons that we'll explore in Chapter 22, the presence of a carbonyl group renders the hydrogens on the $\\alpha$ carbon acidic. Carbonyl compounds therefore react with strong base to yield enolate ions.\n\n\nBecause they're negatively charged, enolate ions act as nucleophiles and undergo many of the reactions we've already studied. For example, enolates react with primary alkyl halides in the $\\mathrm{S}_{\\mathrm{N}} 2$ reaction. The nucleophilic enolate ion displaces halide ion, and a new $\\mathrm{C}-\\mathrm{C}$ bond forms:\n\n\nThe $\\mathrm{S}_{\\mathrm{N}} 2$ alkylation reaction between an enolate ion and an alkyl halide is a powerful method for making $\\mathrm{C}-\\mathrm{C}$ bonds, thereby building up larger molecules from smaller precursors. We'll study the alkylation of many kinds of carbonyl compounds in Chapter 22."}
{"id": 1198, "contents": "Carbonyl Condensation Reactions (Chapter 23) - \nThe fourth and last fundamental reaction of carbonyl groups, carbonyl condensation, takes place when two carbonyl compounds react with each other. When acetaldehyde is treated with base, for instance, two molecules combine to yield the hydroxy aldehyde product known as aldol (aldehyde + alcohol):\n\n\nTwo acetaldehydes\nAldol\nAlthough carbonyl condensation appears to be different from the three processes already discussed, it's actually quite similar. A carbonyl condensation reaction is simply a combination of a nucleophilic addition step and an $\\alpha$-substitution step. The initially formed enolate ion of one acetaldehyde molecule acts as a nucleophile and adds to the carbonyl group of another acetaldehyde molecule, as shown in FIGURE 18.12."}
{"id": 1199, "contents": "FIGURE 18.12 MECHANISM - \nA carbonyl condensation reaction between two molecules of acetaldehyde yields a hydroxy aldehyde product.\n(1) Base abstracts an acidic alpha hydrogen from one acetaldehyde molecule, yielding a resonancestabilized enolate ion.\n\nThe enolate ion adds as a nucleophile to the carbonyl group of a second acetaldehyde, producing a tetrahedral alkoxide ion.\n\n\n\n(2) $\\|$\n\n\nTetrahedral intermediate\n3 The tetrahedral intermediate is protonated by solvent to yield the neutral aldol product and regenerate the base catalyst."}
{"id": 1200, "contents": "IV Summary - \nTo a great extent, the chemistry of living organisms is the chemistry of carbonyl compounds. We have not looked at the details of specific carbonyl reactions in this short preview but rather have laid the groundwork for the next five chapters. All the carbonyl-group reactions we'll be studying in Chapters 19 through 23 fall into one of the four fundamental categories discussed in this preview. Knowing where we'll be heading should help you keep matters straight in understanding this most important of all functional groups.\n\nPROBLEM Judging from the following electrostatic potential maps, which kind of carbonyl compound has\n18-69 the more electrophilic carbonyl carbon atom, a ketone or an acid chloride? Which has the more nucleophilic carbonyl oxygen atom? Explain.\n\n\nPROBLEM Predict the product formed by nucleophilic addition of cyanide ion ( $\\mathrm{CN}^{-}$) to the carbonyl group of 18-70 acetone, followed by protonation to give an alcohol:\n\n\nPROBLEM Identify each of the following reactions as a nucleophilic addition, nucleophilic acyl substitution, an 18-71 $\\alpha$ substitution, or a carbonyl condensation:\n(a)\n\n(b)\n\n\n(c)"}
{"id": 1201, "contents": "CHAPTER 19 - \nAldehydes and Ketones: Nucleophilic Addition Reactions\n\n\nFIGURE 19.1 Few flowers are more beautiful or more fragrant than roses. Their perfumed odor is due to several simple organic compounds, including the ketone $\\beta$-damascenone. (credit: modification of work \"Rosa 'Precious Platinum'\" by Bernard Spragg/Wikimedia Commons, CC BY 1.0)"}
{"id": 1202, "contents": "CHAPTER CONTENTS - 19.1 Naming Aldehydes and Ketones\n19.2 Preparing Aldehydes and Ketones\n19.3 Oxidation of Aldehydes and Ketones\n19.4 Nucleophilic Addition Reactions of Aldehydes and Ketones\n19.5 Nucleophilic Addition of $\\mathrm{H}_{2} \\mathrm{O}$ : Hydration\n19.6 Nucleophilic Addition of HCN: Cyanohydrin Formation\n19.7 Nucleophilic Addition of Hydride and Grignard Reagents: Alcohol Formation\n19.8 Nucleophilic Addition of Amines: Imine and Enamine Formation\n19.9 Nucleophilic Addition of Hydrazine: The Wolff-Kishner Reaction\n19.10 Nucleophilic Addition of Alcohols: Acetal Formation\n19.11 Nucleophilic Addition of Phosphorus Ylides: The Wittig Reaction\n19.12 Biological Reductions\n19.13 Conjugate Nucleophilic Addition to $\\alpha, \\beta$-Unsaturated Aldehydes and Ketones\n19.14 Spectroscopy of Aldehydes and Ketones\n\nWHY THIS CHAPTER? Much of organic chemistry is the chemistry of carbonyl compounds. Aldehydes and ketones, in particular, are intermediates in the synthesis of many pharmaceutical agents, in almost all biological\npathways, and in numerous industrial processes, so an understanding of their properties and reactions is essential. In this chapter, we'll look at some of their most important reactions.\n\nAldehydes (RCHO) and ketones ( $\\mathbf{R}_{\\mathbf{2}} \\mathbf{C O}$ ) are among the most widely occurring of all compounds. In nature, many substances required by living organisms are aldehydes or ketones. The aldehyde pyridoxal phosphate, for instance, is a coenzyme involved in a large number of metabolic reactions; the ketone hydrocortisone is a steroid hormone secreted by the adrenal glands to regulate fat, protein, and carbohydrate metabolism.\n\n\nPyridoxal phosphate (PLP)\n\n\nHydrocortisone"}
{"id": 1203, "contents": "CHAPTER CONTENTS - 19.1 Naming Aldehydes and Ketones\nPyridoxal phosphate (PLP)\n\n\nHydrocortisone\n\nIn the chemical industry, simple aldehydes and ketones are produced in large quantities for use as solvents and as starting materials to prepare a host of other compounds. For example, more than 50 million tons per year of formaldehyde, $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{O}$, is produced worldwide for use in building insulation materials and in the adhesive resins that bind particle board and plywood. Acetone, $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{C}=\\mathrm{O}$, is widely used as an industrial solvent, with approximately 8 million tons per year produced worldwide. Formaldehyde is synthesized industrially by catalytic oxidation of methanol, and one method of acetone preparation involves oxidation of 2-propanol produced by hydration of propylene.\n\n\n\n\nAcetone"}
{"id": 1204, "contents": "CHAPTER CONTENTS - 19.1 Naming Aldehydes and Ketones\nAldehydes are named by replacing the terminal -e of the corresponding alkane name with -al. The parent chain must contain the - CHO group, and the - CHO carbon is numbered as carbon 1 . Note in the following examples that the longest chain in 2-ethyl-4-methylpentanal is actually a hexane, but this chain does not include the -CHO group and thus is not the parent.\n\n\nFor cyclic aldehydes in which the -CHO group is directly attached to a ring, the ring name followed by carbaldehyde is used.\n\n\nCyclohexanecarbaldehyde\n\n\n2-Naphthalenecarbaldehyde\n\nA few simple and well-known aldehydes have common names that are recognized by IUPAC. Several that you might encounter are listed in TABLE 19.1.\n\nTABLE 19.1 Common Names of Some Simple Aldehydes\n\n| Formula | Common name | Systematic name |\n| :--- | :--- | :--- |\n| HCHO | Formaldehyde | Methanal |\n| $\\mathrm{CH}_{3} \\mathrm{CHO}$ | Acetaldehyde | Ethanal |\n| $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCHO}$ | Acrolein | Propenal |\n| $\\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{CHCHO}$ | Crotonaldehyde | 2-Butenal |\n|  | Benzaldehyde | Benzenecarbaldehyde |\n\nKetones are named by replacing the terminal -e of the corresponding alkane name with -one. The parent chain is the longest one that includes the ketone group, and the numbering begins at the end nearer the carbonyl carbon. As with alkenes (Section 7.3) and alcohols (Section 17.1), the locant is placed before the parent name using older rules but before the suffix with the newer IUPAC guidelines. For example:\n\n\nA few ketones are allowed by IUPAC to retain their common names.\n\n\n\nAcetophenone\n\n\nBenzophenone"}
{"id": 1205, "contents": "CHAPTER CONTENTS - 19.1 Naming Aldehydes and Ketones\nA few ketones are allowed by IUPAC to retain their common names.\n\n\n\nAcetophenone\n\n\nBenzophenone\n\nWhen it's necessary to refer to the $\\mathrm{R}-\\mathrm{C}=\\mathrm{O}$ group as a substituent, the name acyl (a-sil) group is used and the name ending $-y l$ is attached. Thus, $-\\mathrm{COCH}_{3}$ is an acetyl group, -CHO is a formyl group, -COAr is an aroyl group, and $-\\mathrm{COC}_{6} \\mathrm{H}_{5}$ is a benzoyl group.\n\n\nAn acyl group\n\n\nAcetyl\n\n\nFormyl\n\n\nBenzoyl\n\nIf other functional groups are present and the double-bonded oxygen is considered a substituent on a parent chain, the prefix oxo- is used. For example:\n\n\nPROBLEM Name the following aldehydes and ketones:\n19-1 (a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\nf)\n\n\nPROBLEM Draw structures corresponding to the following names:\n19-2 (a) 3-Methylbutanal (b) 4-Chloro-2-pentanone (c) Phenylacetaldehyde\n(d) cis-3-tert-Butylcyclohexanecarbaldehyde\n(e) 3-Methyl-3-butenal\n(f) 2-(1-Chloroethyl)-5-methylheptanal"}
{"id": 1206, "contents": "Preparing Aldehydes - \nPerhaps the best method of aldehyde synthesis is by oxidation of a primary alcohol, as we saw in Section 17.7. The reaction is usually carried out using the Dess-Martin periodinane reagent in dichloromethane solvent at room temperature:\n\n\nAnother method of aldehyde synthesis is one that we'll mention here just briefly and then return to in Section 21.6. Certain carboxylic acid derivatives can be partially reduced to yield aldehydes. The partial reduction of an ester by diisobutylaluminum hydride (DIBAH, or DIBAL-H), for instance, is an important laboratory-scale method of aldehyde synthesis, and mechanistically related processes also occur in biological pathways. The reaction is normally carried out at $-78^{\\circ} \\mathrm{C}$ (dry-ice temperature) in toluene solution."}
{"id": 1207, "contents": "Methyl dodecanoate - \nDodecanal (88\\%)\n\n\nPROBLEM How would you prepare pentanal from the following starting materials?\n19-3 (a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}$\n(b) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}=\\mathrm{CH}_{2}$\n(c) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{CH}_{3}$\n(d) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}=\\mathrm{CH}_{2}$"}
{"id": 1208, "contents": "Preparing Ketones - \nFor the most part, methods of ketone synthesis are similar to those for aldehydes. Secondary alcohols are\noxidized by a variety of reagents to give ketones (Section 17.7). The choice of oxidant depends on such factors as reaction scale, cost, and acid or base sensitivity of the alcohol. The Dess-Martin periodinane is a common choice although chromic acid $\\left(\\mathrm{H}_{2} \\mathrm{CrO}_{4}\\right)$ was frequently used in older work.\n\n\nOther methods include the ozonolysis of alkenes in which one of the unsaturated carbon atoms is disubstituted (Section 8.8) and Friedel-Crafts acylation of an aromatic ring with an acid chloride in the presence of $\\mathrm{AlCl}_{3}$ catalyst (Section 16.3).\n\n\nIn addition to those methods already discussed, ketones can also be prepared from certain carboxylic acid derivatives, just as aldehydes can. Among the most useful reactions of this sort is that between an acid chloride and a lithium diorganocopper reagent, as we saw in Section 10.7. We'll discuss this reaction in more detail in Section 21.4.\n\n\nHexanoyl chloride\n2-Heptanone (81\\%)\nPROBLEM How would you carry out the following reactions? More than one step may be needed.\n19-4 (a) 3-Hexyne $\\rightarrow 3$-Hexanone (b) Benzene $\\rightarrow$ m-Bromoacetophenone\n(c) Bromobenzene $\\rightarrow$ Acetophenone (d) 1-Methylcyclohexene $\\rightarrow$ 2-Methylcyclohexanone"}
{"id": 1209, "contents": "Preparing Ketones - 19.3 Oxidation of Aldehydes and Ketones\nAldehydes are easily oxidized to yield carboxylic acids, but ketones are generally inert toward oxidation. The difference is a consequence of structure: aldehydes have a -CHO proton that can be abstracted during oxidation, but ketones do not.\n\n\nMany oxidizing agents, including alkaline $\\mathrm{KMnO}_{4}$ and hot $\\mathrm{HNO}_{3}$, convert aldehydes into carboxylic acids. The oxidation occurs rapidly at room temperature and generally works well.\n\n\nHexanal\nHexanoic acid (85\\%)\nAldehyde oxidations occur through intermediate 1,1-diols, or hydrates, which are formed by a reversible nucleophilic addition of water to the carbonyl group. Even though it's formed to only a small extent at equilibrium, the hydrate reacts like any typical primary or secondary alcohol and is oxidized to a carbonyl compound (Section 17.7).\n\n\nAn aldehyde\n\n\nA hydrate\n\n\nA carboxylic acid\n\nKetones are inert to most oxidizing agents but undergo a slow cleavage reaction of the $\\mathrm{C}-\\mathrm{C}$ bond next to the carbonyl group when treated with hot alkaline $\\mathrm{KMnO}_{4}$. The reaction is not often used, however, and is mentioned here only for completeness.\n\n\nCyclohexanone Hexanedioic acid (79\\%)"}
{"id": 1210, "contents": "Preparing Ketones - 19.4 Nucleophilic Addition Reactions of Aldehydes and Ketones\nAs we saw in the Preview of Carbonyl Chemistry, the most general reaction of aldehydes and ketones is the nucleophilic addition reaction. As shown in FIGURE 19.2, a nucleophile, : $\\mathrm{Nu}^{-}$, approaches the carbonyl group from an angle of about $105^{\\circ}$ opposite the carbonyl oxygen and forms a bond to the electrophilic $\\mathrm{C}-\\mathrm{O}$ carbon atom. At the same time, rehybridization of the carbonyl carbon from $s p^{2}$ to $s p^{3}$ occurs, an electron pair from the $\\mathrm{C}=\\mathrm{O}$ bond moves toward the electronegative oxygen atom, and a tetrahedral alkoxide ion intermediate is produced. Protonation of the alkoxide by addition of acid then gives an alcohol."}
{"id": 1211, "contents": "FIGURE 19.2 MECHANISM - \nA nucleophilic addition reaction to an aldehyde or ketone. The nucleophile approaches the carbonyl group from an angle of approximately $75^{\\circ}$ to the plane of the $s p^{2}$ orbitals, the carbonyl carbon rehybridizes from $s p^{2}$ to $s p^{3}$, and an alkoxide ion is formed. Protonation by addition of acid then gives an alcohol.\n(1) An electron pair from the nucleophile adds to the electrophilic carbon of the carbonyl group, pushing an electron pair from the $\\mathrm{C}=\\mathrm{O}$ bond onto oxygen and giving an alkoxide ion intermediate. The carbonyl carbon rehybridizes from $s p^{2}$ to $s p^{3}$.\n\n\nAldehyde or ketone\n\n\nAlkoxide ion\n(2) $\\mathrm{H}_{3} \\mathrm{O}^{+}$\n\n\nAlcohol\n\nThe nucleophile can be either negatively charged (: $\\mathrm{Nu}^{-}$) or neutral (: Nu ). If it's neutral, however, it usually carries a hydrogen atom that can subsequently be eliminated, : Nu-H. For example:\n\n\nNucleophilic additions to aldehydes and ketones have two general variations, as shown in FIGURE 19.3. In one variation, the tetrahedral intermediate is protonated by water or acid to give an alcohol as the final product. In the second variation, the carbonyl oxygen atom is protonated and then eliminated as $\\mathrm{HO}^{-}$or $\\mathrm{H}_{2} \\mathrm{O}$ to give a product with a $\\mathrm{C}=\\mathrm{Nu}$ double bond.\n\n\nFIGURE 19.3 Two general reaction pathways following addition of a nucleophile to an aldehyde or ketone. The top pathway leads to an alcohol product; the bottom pathway leads to a product with a $\\mathrm{C}=\\mathrm{Nu}$ double bond."}
{"id": 1212, "contents": "FIGURE 19.2 MECHANISM - \nAldehydes are generally more reactive than ketones in nucleophilic addition reactions for both steric and electronic reasons. Sterically, the presence of only one large substituent bonded to the $\\mathrm{C}=\\mathrm{O}$ carbon in an aldehyde versus two large substituents in a ketone means that a nucleophile is able to approach an aldehyde more readily. Thus, the transition state leading to the tetrahedral intermediate is less crowded and lower in energy for an aldehyde than for a ketone (FIGURE 19.4).\n\n\nFIGURE 19.4 Steric hindrance in nucleophilic addition reactions. (a) Nucleophilic addition to an aldehyde is sterically less hindered because only one relatively large substituent is attached to the carbonyl-group carbon. (b) A ketone, however, has two large substituents and is more hindered. The approach of the nucleophile is along the $\\mathrm{C}=\\mathrm{O}$ bond at an angle of about $75^{\\circ}$ to the plane of the carbon $s p^{2}$ orbitals.\n\nElectronically, aldehydes are more reactive than ketones because of the greater polarization of aldehyde carbonyl groups. To see this polarity difference, recall the stability order of carbocations (Section 7.9). A primary carbocation is higher in energy and thus more reactive than a secondary carbocation since it has only one alkyl group inductively stabilizing the positive charge rather than two. In the same way, an aldehyde has only one alkyl group inductively stabilizing the partial positive charge on the carbonyl carbon rather than two, and is a bit more electrophilic, and, therefore, more reactive than a ketone.\n\n\n\nAldehyde\n(less stabilization of $\\boldsymbol{\\delta}+$, more reactive)\n\n$2^{\\circ}$ carbocation (more stable, less reactive)\n\n\nKetone\n(more stabilization of $\\boldsymbol{\\delta}+$, less reactive)"}
{"id": 1213, "contents": "FIGURE 19.2 MECHANISM - \n$2^{\\circ}$ carbocation (more stable, less reactive)\n\n\nKetone\n(more stabilization of $\\boldsymbol{\\delta}+$, less reactive)\n\nOne further comparison: aromatic aldehydes, such as benzaldehyde, are less reactive in nucleophilic addition reactions than aliphatic aldehydes because the electron-donating resonance effect of the aromatic ring makes the carbonyl group less electrophilic. Comparing electrostatic potential maps of formaldehyde and benzaldehyde, for example, shows that the carbonyl carbon atom is less positive (less blue) in the aromatic\naldehyde.\n\n\n\nFormaldehyde\n\n\nBenzaldehyde\n\nPROBLEM Treatment of an aldehyde or ketone with cyanide ion ( ${ }^{-}: \\mathrm{C} \\equiv \\mathrm{N}$ ), followed by protonation of the 19-5 tetrahedral alkoxide ion intermediate, gives a cyanohydrin. Show the structure of the cyanohydrin obtained from cyclohexanone.\n\nPROBLEM $p$-Nitrobenzaldehyde is more reactive toward nucleophilic additions than $p$-methoxybenzaldehyde. 19-6 Explain."}
{"id": 1214, "contents": "FIGURE 19.2 MECHANISM - 19.5 Nucleophilic Addition of $\\mathrm{H}_{2} \\mathrm{O}$ : Hydration\nAldehydes and ketones react with water to yield 1,1-diols, or geminal (gem) diols. The hydration reaction is reversible, and a gem diol can eliminate water to regenerate an aldehyde or ketone.\n\n\nAcetone (99.9\\%)\nAcetone hydrate (0.1\\%)\nThe position of the equilibrium between a gem diol and an aldehyde or ketone depends on the structure of the carbonyl compound. Equilibrium generally favors the carbonyl compound for steric reasons, but the gem diol is favored for a few simple aldehydes. For example, an aqueous solution of formaldehyde consists of $99.9 \\%$ gem diol and $0.1 \\%$ aldehyde at equilibrium, whereas an aqueous solution of acetone consists of only about $0.1 \\%$ gem diol and 99.9\\% ketone.\n\n\nFormaldehyde (0.1\\%) Formaldehyde hydrate (99.9\\%)\nThe hydrate is also favored for $\\alpha$-keto carboxylic acids, which have adjacent positively charged carbons that destabilize the keto form and favor the hydrate. Such compounds occur commonly in many biological pathways. Pyruvic acid, which is $60 \\%$ hydrate at equilibrium, and $\\alpha$-ketoglutaric acid, which is $50 \\%$ hydrate, are examples.\n\n\nThe nucleophilic addition of water to an aldehyde or ketone is slow under neutral conditions but is catalyzed by both base and acid. Under basic conditions (FIGURE 19.5a), the nucleophile is negatively charged ( $\\mathrm{OH}^{-}$) and uses a pair of its electrons to form a bond to the electrophilic carbon atom of the $\\mathrm{C}=\\mathrm{O}$ group. At the same time, the $\\mathrm{C}=\\mathrm{O}$ carbon atom rehybridizes from $s p^{2}$ to $s p^{3}$ and two electrons from the $\\mathrm{C}=\\mathrm{O} \\pi$ bond are pushed onto the oxygen atom, giving an alkoxide ion. Protonation of the alkoxide ion by water then yields a neutral addition product plus regenerated $\\mathrm{OH}^{-}$."}
{"id": 1215, "contents": "FIGURE 19.2 MECHANISM - 19.5 Nucleophilic Addition of $\\mathrm{H}_{2} \\mathrm{O}$ : Hydration\nUnder acidic conditions (FIGURE 19.5b), the carbonyl oxygen atom is first protonated by $\\mathrm{H}_{3} \\mathrm{O}^{+}$to make the carbonyl group more strongly electrophilic. A neutral nucleophile, $\\mathrm{H}_{2} \\mathrm{O}$, then uses a pair of electrons to bond to the carbon atom of the $\\mathrm{C}=\\mathrm{O}$ group, and two electrons from the $\\mathrm{C}=\\mathrm{O} \\pi$ bond move onto the oxygen atom. The positive charge on oxygen is thereby neutralized, while the nucleophile gains a positive charge. Finally, deprotonation by water gives the neutral addition product and regenerates the $\\mathrm{H}_{3} \\mathrm{O}^{+}$catalyst.\n\nNote the key difference between the base-catalyzed and acid-catalyzed reactions. The base-catalyzed reaction takes place rapidly because water is converted into hydroxide ion, a much better nucleophile. The acidcatalyzed reaction takes place rapidly because the carbonyl compound is converted by protonation into a much better electrophile."}
{"id": 1216, "contents": "FIGURE 19.5 MECHANISM - \nThe mechanism for a nucleophilic addition reaction of aldehydes and ketones under both basic and acidic conditions. (a) Under basic conditions, a negatively charged nucleophile adds to the carbonyl group to give an alkoxide ion intermediate, which is subsequently protonated. (b) Under acidic conditions, protonation of the carbonyl group occurs first, followed by addition of a neutral nucleophile and subsequent deprotonation."}
{"id": 1217, "contents": "(a) Basic conditions - \n(1) The negatively charged nucleophile $\\mathrm{OH}^{-}$adds to the electrophilic carbon and pushes $\\pi$ electrons from the $\\mathrm{C}=0$ bond onto oxygen, giving an alkoxide ion.\n(2) The alkoxide ion is protonated by water to give the neutral hydrate as the addition product and regenerating $\\mathrm{OH}^{-}$.\n\n\nHydrate (gem diol)\n(b) Acidic conditions\n(1) The carbonyl oxygen is protonated by acid $\\mathrm{H}_{3} \\mathrm{O}^{+}$, making the carbon more strongly electrophilic.\noxygen. The oxygen becomes neutral, and the nucleophile gains the + charge.\n(3) Water deprotonates the intermediate, giving the neutral hydrate addition product and regenerating the acid catalyst $\\mathrm{H}_{3} \\mathrm{O}^{+}$.\n\n\nHydrate (gem diol)\n\nThe hydration reaction just described is typical of what happens when an aldehyde or ketone is treated with a nucleophile of the type $\\mathrm{H}-\\mathrm{Y}$, where the Y atom is electronegative and can stabilize a negative charge (oxygen, halogen, or sulfur, for instance). In such reactions, the nucleophilic addition is reversible, with the equilibrium generally favoring the carbonyl reactant rather than the tetrahedral addition product. In other words, treatment of an aldehyde or ketone with $\\mathrm{CH}_{3} \\mathrm{OH}, \\mathrm{H}_{2} \\mathrm{O}, \\mathrm{HCl}, \\mathrm{HBr}$, or $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ does not normally lead to a stable alcohol addition product.\n\n\nPROBLEM When dissolved in water, trichloroacetaldehyde exists primarily as its hydrate, called chloral 19-7 hydrate. Show the structure of chloral hydrate."}
{"id": 1218, "contents": "(a) Basic conditions - \nPROBLEM When dissolved in water, trichloroacetaldehyde exists primarily as its hydrate, called chloral 19-7 hydrate. Show the structure of chloral hydrate.\n\nPROBLEM The oxygen in water is primarily (99.8\\%) ${ }^{16} \\mathrm{O}$, but water enriched with the heavy isotope ${ }^{18} \\mathrm{O}$ is 19-8 also available. When an aldehyde or ketone is dissolved in ${ }^{18} \\mathrm{O}$-enriched water, the isotopic label becomes incorporated into the carbonyl group. Explain.\n\n$$\n\\mathrm{R}_{2} \\mathrm{C}=\\mathrm{O}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathrm{R}_{2} \\mathrm{C}=\\mathrm{O}+\\mathrm{H}_{2} \\mathrm{O} \\quad \\text { where } \\mathrm{O}={ }^{18} \\mathrm{O}\n$$"}
{"id": 1219, "contents": "(a) Basic conditions - 19.6 Nucleophilic Addition of HCN: Cyanohydrin Formation\nAldehydes and unhindered ketones undergo a nucleophilic addition reaction with HCN to yield cyanohydrins, $\\mathrm{RCH}(\\mathrm{OH}) \\mathrm{C} \\equiv \\mathrm{N}$. Studies carried out in the early 1900s by Arthur Lapworth showed that cyanohydrin formation is reversible and base-catalyzed. Reaction occurs slowly when pure HCN is used but rapidly when a small amount of base is added to generate the nucleophilic cyanide ion, $\\mathrm{CN}^{-}$. Addition of $\\mathrm{CN}^{-}$takes place by a typical nucleophilic addition pathway, yielding a tetrahedral intermediate that is protonated by HCN to give cyanohydrin product plus regenerated $\\mathrm{CN}^{-}$.\n\n\nCyanohydrin formation is somewhat unusual because it is one of the few examples of the addition of a protic acid ( $\\mathrm{H}-\\mathrm{Y}$ ) to a carbonyl group. As noted in the previous section, protic acids such as $\\mathrm{H}_{2} \\mathrm{O}, \\mathrm{HBr}, \\mathrm{HCl}$, and $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ don't normally yield carbonyl addition products because the equilibrium constants are unfavorable. With HCN, however, equilibrium favors the cyanohydrin adduct.\n\nCyanohydrin formation is useful because of the further chemistry that can be carried out on the product. For example, a nitrile ( $\\mathrm{R}-\\mathrm{C} \\equiv \\mathrm{N}$ ) can be reduced with $\\mathrm{LiAlH}_{4}$ to yield a primary amine $\\left(\\mathrm{RCH}_{2} \\mathrm{NH}_{2}\\right)$ and can be hydrolyzed by hot aqueous acid to yield a carboxylic acid. Thus, cyanohydrin formation provides a method for transforming an aldehyde or ketone into a different functional group.\n\n\nPROBLEM Cyclohexanone forms a cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not. 19-9 Explain."}
{"id": 1220, "contents": "Addition of Hydride Reagents: Reduction - \nWe saw in Section 17.4 that the most common method for preparing alcohols, both in the laboratory and in living organisms, is by the reduction of carbonyl compounds. Aldehydes are reduced with sodium borohydride $\\left(\\mathrm{NaBH}_{4}\\right)$ to give primary alcohols, and ketones are reduced similarly to give secondary alcohols.\n\n\nCarbonyl reduction occurs by a typical nucleophilic addition mechanism under basic conditions, as shown earlier in FIGURE 19.5a. Although the details of carbonyl-group reductions are complex, $\\mathrm{LiAlH}_{4}$ and $\\mathrm{NaBH}_{4}$ act as if they were donors of hydride ion nucleophile, $: \\mathrm{H}^{-}$, and the initially formed alkoxide ion intermediate is then protonated by addition of aqueous acid. The reaction is effectively irreversible because the reverse process would require expulsion of a very poor leaving group."}
{"id": 1221, "contents": "FIGURE 19.6 MECHANISM - \nMechanism of the Grignard reaction. Complexation of the carbonyl oxygen with the Lewis acid $\\mathrm{Mg}^{2+}$ and subsequent nucleophilic addition of a carbanion to an aldehyde or ketone is followed by protonation of the alkoxide intermediate to yield an alcohol.\n(1) The Lewis acid $\\mathrm{Mg}^{2+}$ first forms an acid-base complex with the basic oxygen atom of the aldehyde or ketone, thereby making the carbonyl group a better acceptor.\n(2) Nucleophilic addition of an alkyl group : $\\mathrm{R}^{-}$to the aldehyde or ketone produces a tetrahedral magnesium alkoxide intermediate...\n(2)\n\n\n\nA tetrahedral intermediate\n(3) . . . which undergoes hydrolysis when water is added in a separate step. The final product is a neutral alcohol.\n(3) $\\mathrm{H}_{2} \\mathrm{O}$\n\n\nAn alcohol"}
{"id": 1222, "contents": "Addition of Grignard Reagents, RMgX - \nJust as aldehydes and ketones undergo nucleophilic addition with hydride ion to give alcohols, they undergo a similar addition with Grignard reagent nucleophiles, $\\mathrm{R}:^{-+} \\mathrm{MgX}$. Aldehydes give secondary alcohols on reaction with Grignard reagents in ether solution, and ketones give tertiary alcohols.\n\n\nAs shown in FIGURE 19.6, a Grignard reaction begins with an acid-base complexation of $\\mathrm{Mg}^{2+}$ to the carbonyl oxygen atom of the aldehyde or ketone, thereby making the carbonyl group a better electrophile. Nucleophilic addition of R:- then produces a tetrahedral magnesium alkoxide intermediate, and protonation by addition of water or dilute aqueous acid in a separate step yields the neutral alcohol. Like reduction, Grignard additions are effectively irreversible because a carbanion is too poor a leaving group to be expelled in a reversal step."}
{"id": 1223, "contents": "Addition of Grignard Reagents, RMgX - 19.8 Nucleophilic Addition of Amines: Imine and Enamine Formation\nPrimary amines, $\\mathrm{RNH}_{2}$, add to aldehydes and ketones to yield imines, $\\mathrm{R}_{2} \\mathrm{C}=\\mathrm{NR}$. Secondary amines, $\\mathrm{R}_{2} \\mathrm{NH}$, add similarly to yield enamines, $\\mathrm{R}_{2} \\mathrm{~N}-\\mathrm{CR}=\\mathrm{CR}_{2}$ (ene + amine $=$ unsaturated amine).\n\n\nImines are particularly common as intermediates in biological pathways, where they are often called Schiff bases. The amino acid alanine, for instance, is metabolized in the body by reaction with the aldehyde pyridoxal phosphate (PLP), a derivative of vitamin $B_{6}$, to yield a Schiff base that is further degraded.\n\n\nImine formation and enamine formation seem different because one leads to a product with a $\\mathrm{C}=\\mathrm{N}$ bond and the other leads to a product with a $\\mathrm{C}=\\mathrm{C}$ bond. Actually, though, the reactions are quite similar. Both are typical examples of nucleophilic addition reactions in which water is eliminated from the initially formed tetrahedral intermediate and a new $\\mathrm{C}=\\mathrm{Nu}$ double bond is formed.\n\nImines are formed in a reversible, acid-catalyzed process (FIGURE 19.7) that begins with nucleophilic addition of the primary amine to the carbonyl group, followed by transfer of a proton from nitrogen to oxygen to yield a neutral amino alcohol, or carbinolamine. Protonation of the carbinolamine oxygen by an acid catalyst then converts the -OH into a better leaving group $\\left(-\\mathrm{OH}_{2}{ }^{+}\\right)$, and E1-like loss of water produces an iminium ion. Loss of a proton from nitrogen gives the final product and regenerates the acid catalyst."}
{"id": 1224, "contents": "FIGURE 19.7 MECHANISM - \nMechanism of imine formation by reaction of an aldehyde or ketone with a primary amine. The key step is the initial nucleophilic addition to yield a carbinolamine intermediate, which then loses water to give the imine.\n\n1 Nucleophilic attack on the ketone or aldehyde by the lone-pair electrons of an amine leads to a dipolar tetrahedral intermediate.\n\n(2) A proton is then transferred from nitrogen to oxygen, yielding a neutral carbinolamine.\n(2) $\\downarrow$ Proton transfer\n\n\nCarbinolamine\n3 Acid catalyst protonates the hydroxyl oxygen.\n(3) $\\downarrow \\mathrm{H}_{3} \\mathrm{O}^{+}$\n\n\n4 The nitrogen lone-pair electrons expel water, giving an iminium ion.\n\n\n\nIminium ion\nLoss of $\\mathrm{H}^{+}$from nitrogen then gives the neutral imine product.\n(5)\n\n\nImine\n\nImine formation with such reagents as hydroxylamine and 2,4-dinitrophenylhydrazine is sometimes useful because the products of these reactions-oximes and 2,4-dinitrophenylhydrazones (2,4-DNPS), respectively-are often crystalline and easy to handle. Such crystalline derivatives are occasionally prepared as a means of purifying and characterizing liquid ketones or aldehydes."}
{"id": 1225, "contents": "2,4-Dinitrophenylhydrazone - \nReaction of an aldehyde or ketone with a secondary amine, $\\mathrm{R}_{2} \\mathrm{NH}$, rather than a primary amine yields an enamine. As shown in FIGURE 19.8, the process is identical to imine formation up to the iminium ion stage, but at this point there is no proton on nitrogen that can be lost to form a neutral imine product. Instead, a proton is lost from the neighboring carbon (the $\\alpha$ carbon), yielding an enamine."}
{"id": 1226, "contents": "FIGURE 19.8 MECHANISM - \nMechanism for enamine formation by reaction of an aldehyde or ketone with a secondary amine, $\\mathbf{R}_{\\mathbf{2}} \\mathbf{N H}$. The iminium ion intermediate formed in step 3 has no hydrogen attached to N and so must lose $\\mathrm{H}^{+}$from the carbon two atoms away.\n\n1. Nucleophilic addition of a secondary amine to the ketone or aldehyde, followed by proton transfer from (1) $\\downarrow \\mathrm{R}_{2} \\mathrm{NH}$ nitrogen to oxygen, yields an intermediate carbinolamine in the normal way.\n\n\n2 Protonation of the hydroxyl by acid catalyst converts it into a better (2 $\\downarrow{ } \\mathrm{H}_{3} \\mathrm{O}^{+}$ leaving group.\n\n(3) Elimination of water by the lone-pair electrons on nitrogen then yields an intermediate iminium ion.\n\n$$\n\\text { (3 } \\downarrow \\|_{-\\mathrm{H}_{2} \\mathrm{O}}\n$$\n\n\n\n4 Loss of a proton from the alpha carbon atom yields the enamine product and regenerates the acid catalyst.\n(4) $\\downarrow$\n\n\nEnamine\n\nImine and enamine formation are slow at both high pH and low pH but reach a maximum rate at a weakly acidic pH around 4 to 5 . For example, the profile of pH versus rate shown in FIGURE 19.9 for the reaction between acetone and hydroxylamine, $\\mathrm{NH}_{2} \\mathrm{OH}$, indicates that the maximum reaction rate occurs at pH 4.5 .\n\nWe can explain the observed pH dependence of imine formation by looking at the individual steps in the mechanism. As indicated in FIGURE 19.8, an acid catalyst is required in step 3 to protonate the intermediate carbinolamine, thereby converting the - OH into a better leaving group. Thus, reaction will be slow if not enough acid is present (that is, at high pH ). On the other hand, if too much acid is present (low pH ), the basic amine nucleophile is completely protonated, so the initial nucleophilic addition step can't occur."}
{"id": 1227, "contents": "FIGURE 19.8 MECHANISM - \nFIGURE 19.9 Dependence on $\\mathbf{p H}$ of the rate of reaction between acetone and hydroxylamine: $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{C}=\\mathrm{O}+\\mathrm{NH}_{2} \\mathrm{OH} \\rightarrow\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{C}=\\mathrm{NOH}+\\mathrm{H}_{2} \\mathrm{O}$.\n\nEvidently, a pH of 4.5 represents a compromise between the need for some acid to catalyze the rate-limiting dehydration step but not too much acid so as to avoid complete protonation of the amine. Each nucleophilic addition reaction has its own requirements, and conditions must be optimized to obtain maximum reaction rates."}
{"id": 1228, "contents": "Predicting the Product of Reaction between a Ketone and an Amine - \nShow the products you would obtain by acid-catalyzed reaction of 3-pentanone with methylamine, $\\mathrm{CH}_{3} \\mathrm{NH}_{2}$, and with dimethylamine, $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NH}$."}
{"id": 1229, "contents": "Strategy - \nAn aldehyde or ketone reacts with a primary amine, $\\mathrm{RNH}_{2}$, to yield an imine, in which the carbonyl oxygen atom has been replaced by the $=\\mathrm{N}-\\mathrm{R}$ group of the amine. Reaction of the same aldehyde or ketone with a secondary amine, $\\mathrm{R}_{2} \\mathrm{NH}$, yields an enamine, in which the oxygen atom has been replaced by the $-\\mathrm{NR}_{2}$ group of the amine and the double bond has moved to a position between the former carbonyl carbon and the neighboring carbon."}
{"id": 1230, "contents": "Solution - \nAn enamine\n\nPROBLEM Show the products you would obtain by acid-catalyzed reaction of cyclohexanone with ethylamine, 19-10 $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{NH}_{2}$, and with diethylamine, $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2}\\right)_{2} \\mathrm{NH}$.\n\nPROBLEM Imine formation is reversible. Show all the steps involved in the acid-catalyzed reaction of an imine 19-11 with water (hydrolysis) to yield an aldehyde or ketone plus primary amine.\n\nPROBLEM Draw the following molecule as a skeletal structure, and show how it can be prepared from a ketone 19-12\nand an amine."}
{"id": 1231, "contents": "Solution - 19.9 Nucleophilic Addition of Hydrazine: The Wolff-Kishner Reaction\nA useful variant of the imine-forming reaction just discussed involves the treatment of an aldehyde or ketone with hydrazine, $\\mathrm{H}_{2} \\mathrm{~N}-\\mathrm{NH}_{2}$, in the presence of KOH . Called the Wolff-Kishner reaction, the process is a useful and general method for converting an aldehyde or ketone into an alkane, $\\mathrm{R}_{2} \\mathrm{C}=\\mathrm{O} \\rightarrow \\mathrm{R}_{2} \\mathrm{CH}_{2}$.\n\n\n\nAs shown in FIGURE 19.10, the Wolff-Kishner reaction involves formation of a hydrazone intermediate, $\\mathrm{R}_{2} \\mathrm{C}=\\mathrm{N}-\\mathrm{NH}_{2}$, followed by base-catalyzed double-bond migration, loss of $\\mathrm{N}_{2}$ gas to give a carbanion, and protonation to give the alkane product. The double-bond migration takes place when a base removes one of the weakly acidic NH protons in step 2 to generate a hydrazone anion, which has an allylic resonance structure that places the double bond between nitrogens and the negative charge on carbon. Reprotonation then occurs on carbon to generate the double-bond rearrangement product. The next step-loss of nitrogen and formation of an alkyl anion-is driven by the large thermodynamic stability of the $\\mathrm{N}_{2}$ molecule."}
{"id": 1232, "contents": "FIGURE 19.10 MECHANISM - \nMechanism for the Wolff-Kishner reduction of an aldehyde or ketone to yield an alkane.\n\n(1) Reaction of the aldehyde or ketone with hydrazine yields a hydrazone in the normal way.\n\n2 Base abstracts a weakly acidic $\\mathrm{N}-\\mathrm{H}$ proton, yielding a hydrazone anion. This anion has a resonance form that places the negative charge on carbon and the double bond between nitrogens.\n(3) Protonation of the hydrazone anion takes place on carbon to yield a neutral intermediate.\n(4) Deprotonation of the remaining weakly acidic N-H occurs with simultaneous loss of nitrogen to give a carbanion...\n5. . . which is protonated to give the alkane product\n\n(1) $\\downarrow \\downarrow$\n\n(5) $\\mathrm{H}_{2} \\mathrm{O}$\n\n\nNote that the Wolff-Kishner reduction accomplishes the same overall transformation as the catalytic hydrogenation of an acylbenzene to yield an alkylbenzene (Section 16.10). The Wolff-Kishner reduction is more general and more useful than catalytic hydrogenation, however, because it works well with both alkyl and aryl ketones.\n\nPROBLEM Show how you could prepare the following compounds from 4-methyl-3-penten-2-one, $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{C}=$ 19-13 $\\mathrm{CHCOCH}_{3}$.\n(a)\n\n(b)\n\n(c)"}
{"id": 1233, "contents": "FIGURE 19.10 MECHANISM - 19.10 Nucleophilic Addition of Alcohols: Acetal Formation\nAldehydes and ketones react reversibly with 2 equivalents of an alcohol in the presence of an acid catalyst to yield acetals, $\\mathbf{R}_{\\mathbf{2}} \\mathbf{C}\\left(\\mathbf{O R}^{\\prime}\\right)_{\\mathbf{2}}$, which are often called ketals if derived from a ketone. Cyclohexanone, for instance, reacts with methanol in the presence of HCl to give the corresponding dimethyl acetal.\n\n\nAcetal formation is similar to the hydration reaction discussed in Section 19.5. Like water, alcohols are weak nucleophiles that add to aldehydes and ketones slowly under neutral conditions. Under acidic conditions, however, the reactivity of the carbonyl group is increased by protonation, so addition of an alcohol occurs rapidly.\n\n\nA neutral carbonyl group is moderately electrophilic because of the polarity of the $\\mathrm{C}-\\mathrm{O}$ bond.\n\n\nA protonated carbonyl group is strongly electrophilic because of the positive charge on carbon.\n\nAs shown in FIGURE 19.11, nucleophilic addition of an alcohol to the carbonyl group initially yields a hydroxy ether called a hemiacetal, analogous to the gem diol formed by addition of water. Hemiacetals are formed reversibly, with equilibrium normally favoring the carbonyl compound. In the presence of acid, however, a further reaction occurs. Protonation of the - OH group, followed by an E1-like loss of water, leads to an oxonium ion, $\\mathrm{R}_{2} \\mathrm{C}=\\mathrm{O}^{+} \\mathrm{R}$, which undergoes a second nucleophilic addition of alcohol to yield the protonated acetal. Loss of a proton completes the reaction."}
{"id": 1234, "contents": "FIGURE 19.11 MECHANISM - \nMechanism of acid-catalyzed acetal formation by reaction of an aldehyde or ketone with an alcohol.\n\n1 Protonation of the carbonyl oxygen strongly polarizes the carbonyl group and...\n\n2 . . . activates the carbonyl group for nucleophilic attack by oxygen lone-pair electrons from the alcohol.\n\n3 Loss of a proton yields a neutral hemiacetal tetrahedral intermediate.\n\n(1)\n\n\n(3)\n\nHemiacetal\n\n(4) Protonation of the hemiacetal hydroxyl converts it into a good leaving group.\n\n(5) Dehydration yields an intermediate oxonium ion.\n(6) Addition of a second equivalent of alcohol gives a protonated acetal.\n(5)\n\n(6) $\\| R-\\ddot{o}-H$\n\n\nLoss of a proton yields the neutral acetal product.\n\n\nAcetal\n\n\nBecause all the steps in acetal formation are reversible, the reaction can be driven either forward (from carbonyl\ncompound to acetal) or backward (from acetal to carbonyl compound), depending on the conditions. The forward reaction is favored by conditions that remove water from the medium and thus drive the equilibrium to the right. In practice, this is often done by distilling off water as it forms. The reverse reaction is favored by treating the acetal with a large excess of aqueous acid to drive the equilibrium to the left.\n\nAcetals are useful because they can act as protecting groups for aldehydes and ketones in the same way that trimethylsilyl ethers act as protecting groups for alcohols (Section 17.8). As we saw previously, it sometimes happens that one functional group interferes with intended chemistry elsewhere in a complex molecule. For example, if we wanted to reduce only the ester group of ethyl 4 -oxopentanoate, the ketone would interfere. Treatment of the starting keto ester with $\\mathrm{LiAlH}_{4}$ would reduce both the keto group and the ester group to give a diol product."}
{"id": 1235, "contents": "FIGURE 19.11 MECHANISM - \nBy protecting the keto group as an acetal, however, the problem can be circumvented. Like other ethers, acetals are unreactive to bases, hydride reducing agents, Grignard reagents, and catalytic hydrogenation conditions, but they are cleaved by acid. Thus, we can accomplish the selective reduction of the ester group in ethyl 4 -oxopentanoate by first converting the keto group to an acetal, then reducing the ester with $\\mathrm{LiAlH}_{4}$, and then removing the acetal by treatment with aqueous acid. (In practice, it's often convenient to use 1 equivalent of a diol such as ethylene glycol as the alcohol to form a cyclic acetal. The mechanism of cyclic acetal formation using 1 equivalent of ethylene glycol is exactly the same as that using 2 equivalents of methanol or other monoalcohols.)\n\n\nAcetal and hemiacetal groups are particularly common in carbohydrate chemistry. Glucose, for instance, is a polyhydroxy aldehyde that undergoes an internal nucleophilic addition reaction and exists primarily as a cyclic hemiacetal."}
{"id": 1236, "contents": "Predicting the Product of Reaction between a Ketone and an Alcohol - \nShow the structure of the acetal you would obtain by acid-catalyzed reaction of 2-pentanone with"}
{"id": 1237, "contents": "Strategy - \nAcid-catalyzed reaction of an aldehyde or ketone with 2 equivalents of a monoalcohol or 1 equivalent of a diol yields an acetal, in which the carbonyl oxygen atom is replaced by two -OR groups from the alcohol."}
{"id": 1238, "contents": "Solution - \nPROBLEM Show all the steps in the acid-catalyzed formation of a cyclic acetal from ethylene glycol and an 19-14 aldehyde or ketone.\n\nPROBLEM Identify the carbonyl compound and the alcohol that were used to prepare the following acetal:\n19-15"}
{"id": 1239, "contents": "Solution - 19.11 Nucleophilic Addition of Phosphorus Ylides: The Wittig Reaction\nAldehydes and ketones are converted into alkenes by means of a nucleophilic addition called the Wittig reaction, after the German chemist Georg Wittig who won the 1979 Nobel Prize in Chemistry. The reaction has no direct biological counterpart but is important both because of its wide use in drug manufacture and because of its mechanistic similarity to reactions of the coenzyme thiamin diphosphate, which we'll see in Section 29.6.\n\nIn the Wittig reaction, a triphenylphosphorus ylide, $\\mathrm{R}_{2} \\stackrel{-}{\\mathrm{C}}-\\stackrel{+}{\\mathrm{P}} \\mathrm{Ph}_{3}$ also called a phosphorane and sometimes written in the resonance form $\\mathrm{R}_{2} \\mathrm{C}=\\mathrm{P}(\\mathrm{Ph})_{3}$, adds to an aldehyde or ketone to yield a four-membered cyclic intermediate called an oxaphosphetane. The oxaphosphetane is not isolated, but instead spontaneously decomposes to give an alkene plus triphenylphosphine oxide, $\\mathrm{O}=\\mathrm{PPh}_{3}$. In effect, the oxygen atom of the aldehyde or ketone and the $\\mathrm{R}_{2} \\mathrm{C}=$ bonded to phosphorus exchange places. (An ylide-pronounced ill-id-is a neutral, dipolar compound with adjacent positive and negative charges.)\n\n\nThe initial addition takes place by different pathways depending on the structure of the reactants and the exact experimental conditions. One pathway involves a one-step cycloaddition process analogous to the Diels-Alder cycloaddition reaction (Section 14.4). The other pathway involves a nucleophilic addition reaction to give a dipolar intermediate called a betaine (bay-ta-een), which undergoes ring closure."}
{"id": 1240, "contents": "Oxaphosphetane - \nThe phosphorus ylides necessary for Wittig reaction are easily prepared by $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of primary (and some secondary) alkyl halides with triphenylphosphine, ( Ph$)_{3} \\mathrm{P}$, followed by treatment with base. Triphenylphosphine is a good nucleophile in $\\mathrm{S}_{\\mathrm{N}} 2$ reactions, and yields of the resultant alkyltriphenylphosphonium salts are high. Because of the positive charge on phosphorus, the hydrogen on the neighboring carbon is weakly acidic and can be removed by a strong base such as butyllithium (BuLi) to generate the neutral ylide. For example:\n\n\nTriphenylphosphine\n\n\nMethyltriphenylphosphonium bromide\n\n\nMethylenetriphenylphosphorane\n\nThe Wittig reaction is extremely general, and a great many monosubstituted, disubstituted, and trisubstituted alkenes can be prepared from the appropriate combination of phosphorane and aldehyde or ketone. Tetrasubstituted alkenes, however, can't be prepared this way because of steric hindrance.\n\nThe real value of the Wittig reaction is that it yields a pure alkene of predictable structure. The $\\mathrm{C}=\\mathrm{C}$ bond in the product is always exactly where the $\\mathrm{C}=\\mathrm{O}$ group was in the reactant, and no alkene isomers (except $E, Z$ isomers) are formed. For example, Wittig reaction of cyclohexanone with methylenetriphenylphosphorane yields only the single alkene product methylenecyclohexane. By contrast, addition of methylmagnesium bromide to cyclohexanone, followed by dehydration with $\\mathrm{POCl}_{3}$, yields a roughly $9: 1$ mixture of two alkenes.\n\n\nWittig reactions are used commercially in the synthesis of numerous pharmaceutical products. For example, the German chemical company BASF prepares vitamin A by using a Wittig reaction between a 15 -carbon ylide and a 5-carbon aldehyde."}
{"id": 1241, "contents": "Oxaphosphetane - \n$$\n\\downarrow \\begin{aligned}\n& \\mathrm{Na}^{+-} \\mathrm{OCH}_{3} \\\\\n& \\mathrm{CH}_{3} \\mathrm{OH}\n\\end{aligned}\n$$\n\n\n\nVitamin A acetate"}
{"id": 1242, "contents": "Synthesizing an Alkene Using a Wittig Reaction - \nWhat carbonyl compound and what phosphorus ylide might you use to prepare 3-ethyl-2-pentene?"}
{"id": 1243, "contents": "Strategy - \nAn aldehyde or ketone reacts with a phosphorus ylide to yield an alkene in which the oxygen atom of the carbonyl reactant is replaced by the $=\\mathrm{CR}_{2}$ of the ylide. Preparation of the phosphorus ylide itself usually involves $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of a primary alkyl halide with triphenylphosphine, so the ylide is typically primary, $\\mathrm{RCH}=\\mathrm{P}(\\mathrm{Ph})_{3}$. This means that the disubstituted alkene carbon in the product comes from the carbonyl reactant, while the monosubstituted alkene carbon comes from the ylide.\n\nSolution\n\n\nPROBLEM What carbonyl compound and what phosphorus ylide might you use to prepare each of the 19-16 following compounds?\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n\nPROBLEM $\\beta$-Carotene, a yellow food-coloring agent and dietary source of vitamin A can be prepared by a\n19-17 double Wittig reaction between 2 equivalents of $\\beta$-ionylideneacetaldehyde and a diylide. Show the structure of the $\\beta$-carotene product."}
{"id": 1244, "contents": "Strategy - 19.12 Biological Reductions\nAs a general rule, nucleophilic addition reactions are characteristic only of aldehydes and ketones, not of carboxylic acid derivatives. The reason for the difference is structural. As discussed previously in the Preview of Carbonyl Chemistry, and shown in FIGURE 19.12, the tetrahedral intermediate produced by addition of a nucleophile to a carboxylic acid derivative can eliminate a leaving group, leading to a net nucleophilic acyl substitution reaction. The tetrahedral intermediate produced by addition of a nucleophile to an aldehyde or ketone, however, has only alkyl or hydrogen substituents and thus can't usually expel a leaving group. One exception to this rule is the Cannizzaro reaction, discovered in 1853.\n\n\n$$\n\\begin{array}{ll}\n\\text { Reaction occurs when: } & \\mathrm{Y}=-\\mathrm{Br},-\\mathrm{Cl},-\\mathrm{OR},-\\mathrm{NR}_{2} \\\\\n\\text { Reaction does NOT occur when: } & \\mathrm{Y}=-\\mathrm{H},-\\mathrm{R}\n\\end{array}\n$$\n\nFIGURE 19.12 Carboxylic acid derivatives have an electronegative substituent $\\mathbf{Y}=-\\mathrm{Br},-\\mathrm{Cl},-\\mathrm{OR},-\\mathrm{NR}_{\\mathbf{2}}$ that can be expelled as a leaving group from the tetrahedral intermediate formed by nucleophilic addition. Aldehydes and ketones have no such leaving group and thus do not usually undergo this reaction.\n\nThe Cannizzaro reaction takes place by nucleophilic addition of $\\mathrm{OH}^{-}$to an aldehyde to give a tetrahedral intermediate, which expels hydride ion as a leaving group and is thereby oxidized. A second aldehyde molecule accepts the hydride ion in another nucleophilic addition step and is thereby reduced. Benzaldehyde, for instance, yields benzyl alcohol plus benzoic acid when heated with aqueous NaOH ."}
{"id": 1245, "contents": "Strategy - 19.12 Biological Reductions\nThe Cannizzaro reaction is rarely used today but is interesting mechanistically because it is a simple laboratory analogy for the primary biological pathway by which carbonyl reductions occur in living organisms. In nature, as we saw in Section 17.4, one of the most important reducing agents is NADH, reduced nicotinamide adenine dinucleotide. NADH donates $\\mathrm{H}^{-}$to aldehydes and ketones, thereby reducing them, in much the same way that the tetrahedral alkoxide intermediate in a Cannizzaro reaction does. The electron lone pair on a nitrogen atom of NADH expels $\\mathrm{H}^{-}$as leaving group, which adds to a carbonyl group in another molecule (FIGURE 19.13). As an example, pyruvate is converted during intense muscle activity to (S)-lactate, a reaction catalyzed by lactate dehydrogenase.\n\n\nNADH\n\n\nFIGURE 19.13 Mechanism of biological aldehyde and ketone reductions by the coenzyme NADH. The key step is an expulsion of hydride ion from NADH and donation to the carbonyl group.\n\nPROBLEM When $o$-phthalaldehyde is treated with base, $o$-(hydroxymethyl)benzoic acid is formed. Show the\n19-18 mechanism of this reaction.\n\n\nPROBLEM What is the stereochemistry of the pyruvate reduction shown in Figure 19.13? Does NADH lose its\n19-19 pro- $R$ or pro- $S$ hydrogen? Does addition occur to the Si face or Re face of pyruvate? (Review Section 5.11.)"}
{"id": 1246, "contents": "Strategy - 19.13 Conjugate Nucleophilic Addition to $\\boldsymbol{\\alpha}, \\boldsymbol{\\beta}$-Unsaturated Aldehydes and Ketones\nAll the reactions we've been discussing to this point have involved the addition of a nucleophile directly to the carbonyl group, a so-called $\\mathbf{1 , 2}$-addition. Closely related to this direct addition is the conjugate addition, or 1,4-addition, of a nucleophile to the $\\mathrm{C}=\\mathrm{C}$ bond of an $\\alpha, \\beta$-unsaturated aldehyde or ketone. (The carbon atom next to a carbonyl group is often called the $\\alpha$ carbon, the next carbon is the $\\beta$ carbon, and so on. Thus, an $\\alpha, \\beta$-unsaturated aldehyde or ketone has a double bond conjugated with the carbonyl group.) The initial product of conjugate addition is a resonance-stabilized enolate ion, which typically undergoes protonation on the $\\alpha$ carbon to give a saturated aldehyde or ketone product (FIGURE 19.14)."}
{"id": 1247, "contents": "Conjugate (1,4) addition - \nFIGURE 19.14 A comparison of direct $(1,2)$ and conjugate $(1,4)$ nucleophilic addition reactions. In conjugate addition, a nucleophile adds to the $\\beta$ carbon of an $\\alpha, \\beta$-unsaturated aldehyde or ketone and protonation occurs on the $\\alpha$ carbon.\n\nThe conjugate addition of a nucleophile to an $\\alpha, \\beta$-unsaturated aldehyde or ketone is caused by the same electronic factors that are responsible for direct addition. The electronegative oxygen atom of the $\\alpha, \\beta$-unsaturated carbonyl compound withdraws electrons from the $\\beta$ carbon, thereby making it electron-poor and more electrophilic than a typical alkene carbon.\n\n\nAs noted, conjugate addition of a nucleophile to the $\\beta$ carbon of an $\\alpha, \\beta$-unsaturated aldehyde or ketone leads to an enolate ion intermediate, which is protonated on the $\\alpha$ carbon to give the saturated product (FIGURE 19.14). The net effect is addition of the nucleophile to the $\\mathrm{C}=\\mathrm{C}$ bond, with the carbonyl group itself unchanged. In fact, of course, the carbonyl group is crucial to the success of the reaction. Without the carbonyl group, the $\\mathrm{C}=\\mathrm{C}$ bond would not be activated for addition, and no reaction would occur.\n\nActivated double bond\n\n\nUnactivated double bond"}
{"id": 1248, "contents": "Conjugate Addition of Amines - \nBoth primary and secondary amines add to $\\alpha, \\beta$-unsaturated aldehydes and ketones to yield $\\beta$-amino aldehydes and ketones rather than the alternative imines. Under typical reaction conditions, both modes of addition occur\nrapidly. But because the reactions are reversible, they generally proceed with thermodynamic control rather than kinetic control (Section 14.3), so the more stable conjugate addition product is often obtained with the complete exclusion of the less stable direct addition product."}
{"id": 1249, "contents": "Conjugate Addition of Water - \nWater can add reversibly to $\\alpha, \\beta$-unsaturated aldehydes and ketones to yield $\\beta$-hydroxy aldehydes and ketones, although the position of the equilibrium generally favors unsaturated reactant rather than saturated adduct. Related additions to $\\alpha, \\beta$-unsaturated carboxylic acids occur in numerous biological pathways, such as the citric acid cycle of food metabolism in which cis-aconitate is converted into isocitrate by conjugate addition of water to a double bond.\n\n\nPROBLEM Assign $R$ or $S$ stereochemistry to the two chirality centers in isocitrate, and tell whether OH and H 19-20 add to the Si face or the Re face of the double bond."}
{"id": 1250, "contents": "Conjugate Addition of Alkyl Groups: Organocopper Reactions - \nThe conjugate addition of an alkyl or other organic group to an $\\alpha, \\beta$-unsaturated ketone (but not aldehyde) is one of the more useful 1,4-addition reactions, just as direct addition of a Grignard reagent is one of the more useful 1,2-additions.\n\n$\\alpha, \\beta$-Unsaturated ketone\nConjugate addition of an organic group is carried out by treating the $\\alpha, \\beta$-unsaturated ketone with a lithium diorganocopper reagent, $\\mathrm{R}_{2} \\mathrm{CuLi}$. As we saw in Section 10.7, lithium diorganocopper (Gilman) reagents are prepared by reaction between 1 equivalent of copper(I) iodide and 2 equivalents of an organolithium regent, RLi. The organolithium reagent, in turn, is formed by reaction of lithium metal with an organohalide in the same way that a Grignard reagent is prepared by reaction of magnesium metal with an organohalide.\n\n# $\\mathrm{RX} \\xrightarrow[\\text { Pentane }]{2 \\mathrm{Li}} \\mathrm{RLi}+\\mathrm{Li}^{+} \\mathrm{X}^{-}$ <br> $2 \\mathrm{RLi} \\xrightarrow[\\text { Ether }]{\\mathrm{CuI}} \\mathrm{Li}^{+}(\\mathrm{RCuR})+\\mathrm{Li}^{-} \\mathrm{I}^{-}$"}
{"id": 1251, "contents": "A lithium <br> diorganocopper <br> (Gilman reagent) - \nPrimary, secondary, and even tertiary alkyl groups undergo the conjugate addition reaction, as do aryl and alkenyl groups. Alkynyl groups, however, react poorly in the conjugate addition process. Diorganocopper reagents are unique in their ability to give conjugate addition products. Other organometallic reagents, such as Grignard reagents and organolithiums, normally result in direct carbonyl addition on reaction with $\\alpha, \\beta$-unsaturated ketones.\n\n\nThe mechanism of this reaction is thought to involve conjugate nucleophilic addition of the diorganocopper anion, $\\mathrm{R}_{2} \\mathrm{Cu}^{-}$, to the unsaturated ketone to give a copper-containing intermediate. Transfer of an R group from copper to carbon, followed by elimination of a neutral organocopper species, RCu, gives the final product."}
{"id": 1252, "contents": "Using a Conjugate Addition Reaction - \nHow might you use a conjugate addition reaction to prepare 2-methyl-3-propylcyclopentanone?\n\n\n2-Methyl-3-propylcyclopentanone"}
{"id": 1253, "contents": "Strategy - \nA ketone with a substituent group in its $\\beta$ position might be prepared by a conjugate addition of that group to an $\\alpha, \\beta$-unsaturated ketone. In the present instance, the target molecule has a propyl substituent on the $\\beta$ carbon and might therefore be prepared from 2-methyl-2-cyclopentenone by reaction with lithium dipropylcopper."}
{"id": 1254, "contents": "Solution - \nPROBLEM Treatment of 2-cyclohexenone with HCN/KCN yields a saturated keto nitrile rather than an\n19-21 unsaturated cyanohydrin. Show the structure of the product, and propose a mechanism for the reaction.\n\nPROBLEM How might conjugate addition reactions of lithium diorganocopper reagents be used to synthesize 19-22 the following compounds?\n(a)\n(b)\n\n(c)\n\n(d)"}
{"id": 1255, "contents": "Infrared Spectroscopy - \nAldehydes and ketones show a strong $\\mathrm{C}=\\mathrm{O}$ bond absorption in the IR region from 1660 to $1770 \\mathrm{~cm}^{-1}$, as the spectra of benzaldehyde and cyclohexanone demonstrate (FIGURE 19.15). In addition, aldehydes show two characteristic C-H absorptions between $2700-2760$ and $2800-2860 \\mathrm{~cm}^{-1}$. These absorbances are important for distinguishing between aldehydes and ketones. The higher-frequency absorbance is sometimes obscured in cases where the compound has numerous saturated $\\mathrm{C}-\\mathrm{H}$ groups, but the lower-frequency peak is almost always visible.\n\n\nFIGURE 19.15 Infrared spectra of (a) benzaldehyde and (b) cyclohexanone.\nThe exact position of the $\\mathrm{C}=\\mathrm{O}$ absorption is diagnostic of the nature of the carbonyl group. As the data in TABLE 19.2 indicate, saturated aldehydes usually show carbonyl absorptions near $1730 \\mathrm{~cm}^{-1}$ in the IR spectrum, but conjugation of the aldehyde to an aromatic ring or a double bond lowers the absorption by $25 \\mathrm{~cm}^{-1}$ to near $1705 \\mathrm{~cm}^{-1}$. Saturated aliphatic ketones and cyclohexanones both absorb near $1715 \\mathrm{~cm}^{-1}$, and conjugation with a double bond or an aromatic ring again lowers the absorption by $30 \\mathrm{~cm}^{-1}$ to $1685-1690 \\mathrm{~cm}^{-1}$. Angle strain in the carbonyl group, caused by reducing the ring size of cyclic ketones to four or five, raises the absorption position. Cyclohexanone has its $\\mathrm{C}=\\mathrm{O}$ stretch absorbance at $1715 \\mathrm{~cm}^{-1}$ while cyclopentanone's carbonyl stretch is at $1750 \\mathrm{~cm}^{-1}$ and cyclobutanone's is at $1785 \\mathrm{~cm}^{-1}$.\n\nTABLE 19.2 Infrared Absorptions of Some Aldehydes and Ketones"}
{"id": 1256, "contents": "Infrared Spectroscopy - \nTABLE 19.2 Infrared Absorptions of Some Aldehydes and Ketones\n\n| Carbonyl type | Example | Absorption (cm $\\left.{ }^{\\mathbf{- 1}}\\right)$ |\n| :--- | :--- | :---: |\n| Saturated aldehyde | $\\mathrm{CH}_{3} \\mathrm{CHO}$ | 1730 |\n| Aromatic aldehyde | PhCHO | 1705 |\n| $\\alpha, \\beta$-Unsaturated aldehyde | $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCHO}$ | 1705 |\n| Saturated ketone | $\\mathrm{CH}_{3} \\mathrm{COCH}_{3}$ | 1715 |\n| Cyclohexanone |  | 1715 |\n| Cyclopentanone |  | 1750 |\n| Cyclobutanone | $\\mathrm{PhCOCH}_{3}$ | 1785 |\n| Aromatic ketone | $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCOCH}_{3}$ | 1685 |\n| $\\alpha, \\beta$-Unsaturated ketone |  |  |\n\nThe values given in TABLE 19.2 are remarkably constant from one aldehyde or ketone to another. As a result, IR spectroscopy is a powerful tool for identifying the kind of a carbonyl group in a molecule of unknown structure. An unknown that shows an IR absorption at $1730 \\mathrm{~cm}^{-1}$ is almost certainly an aldehyde rather than a ketone; an unknown that shows an IR absorption at $1750 \\mathrm{~cm}^{-1}$ is almost certainly a cyclopentanone, and so on.\n\nPROBLEM How might you use IR spectroscopy to determine whether reaction between 2-cyclohexenone and 19-23 lithium dimethylcopper gives the direct addition product or the conjugate addition product?"}
{"id": 1257, "contents": "Infrared Spectroscopy - \nPROBLEM How might you use IR spectroscopy to determine whether reaction between 2-cyclohexenone and 19-23 lithium dimethylcopper gives the direct addition product or the conjugate addition product?\n\nPROBLEM Where would you expect each of the following compounds to absorb in the IR spectrum?\n19-24 (a) 4-Penten-2-one\n(b) 3-Penten-2-one\n(c) 2,2-Dimethylcyclopentanone\n(d) m-Chlorobenzaldehyde\n(e) 3-Cyclohexenone\n(f) 2-Hexenal"}
{"id": 1258, "contents": "Nuclear Magnetic Resonance Spectroscopy - \nAldehyde protons (RCHO) absorb near $10 \\delta$ in the ${ }^{1} \\mathrm{H}$ NMR spectrum and are very distinctive because no other absorptions occur in this region. The aldehyde proton shows spin-spin coupling with protons on the neighboring carbon, with coupling constant $J \\approx 3 \\mathrm{~Hz}$. Acetaldehyde, for example, shows a quartet at $9.79 \\delta$ for the aldehyde proton, indicating that there are three protons neighboring the - CHO group (FIGURE 19.16).\n\n\nFIGURE $19.16{ }^{\\mathbf{1}} \\mathbf{H}$ NMR spectrum of acetaldehyde. The absorption of the aldehyde proton appears at $9.79 \\delta$ and is split into a quartet.\nHydrogens on the carbon next to a carbonyl group are slightly deshielded and usually absorb near 2.0 to $2.3 \\delta$. The acetaldehyde methyl group in FIGURE 19.16, for instance, absorbs at $2.20 \\delta$. Methyl ketones are particularly distinctive because they always show a sharp three-proton singlet near 2.1 $\\delta$.\n\nThe carbonyl-group carbon atoms of aldehydes and ketones have characteristic ${ }^{13} \\mathrm{C}$ NMR resonances in the range 190 to $215 \\delta$. Since no other kinds of carbons absorb in this range, the presence of an NMR absorption near $200 \\delta$ is clear evidence for a carbonyl group. Saturated aldehyde or ketone carbons usually absorb in the region from 200 to $215 \\delta$, while aromatic and $\\alpha, \\beta$-unsaturated carbonyl carbons absorb in the 190 to $200 \\delta$ region."}
{"id": 1259, "contents": "Mass Spectrometry - \nAs discussed in Section 12.3, aliphatic aldehydes and ketones that have hydrogens on their gamma ( $\\gamma$ ) carbon atoms undergo a characteristic mass spectral cleavage called the McLafferty rearrangement. A hydrogen atom is transferred from the $\\gamma$ carbon to the carbonyl oxygen, the bond between the $\\alpha$ and $\\beta$ carbons is broken, and a neutral alkene fragment is produced. The charge remains with the oxygen-containing fragment.\n\n\nIn addition to fragmentation by the McLafferty rearrangement, aldehydes and ketones also undergo cleavage of the bond between the carbonyl group and the $\\alpha$ carbon, called an $\\alpha$ cleavage. Alpha cleavage yields a neutral radical and a resonance-stabilized acyl cation.\n\n\nFragment ions from both McLafferty rearrangement and $\\alpha$ cleavage are visible in the mass spectrum of 5-methyl-2-hexanone shown in FIGURE 19.17. McLafferty rearrangement with loss of 2-methylpropene yields a fragment with $m / z=58$. Alpha cleavage occurs primarily at the more substituted side of the carbonyl group, leading to a $\\left[\\mathrm{CH}_{3} \\mathrm{CO}\\right]^{+}$fragment with $m / z=43$.\n\n\n\nFIGURE 19.17 Mass spectrum and the related reactions of 5-methyl-2-hexanone. The peak at $m / z=58$ is due to McLafferty rearrangement. The abundant peak at $m / z=43$ is due to $\\alpha$ cleavage at the more highly substituted side of the carbonyl group. Note that the peak due to the molecular ion is very small.\n\nPROBLEM How might you use mass spectrometry to distinguish between the following pairs of isomers?\n19-25 (a) 3-Methyl-2-hexanone and 4-methyl-2-hexanone (b) 3-Heptanone and 4-heptanone\n(c) 2-Methylpentanal and 3-methylpentanal\n\nPROBLEM Describe the prominent IR absorptions and mass spectral peaks you would expect for the following 19-26 compound:"}
{"id": 1260, "contents": "Enantioselective Synthesis - \nWhenever a chiral product is formed by reaction between achiral reagents, the product is racemic; that is, both enantiomers of the product are formed in equal amounts. The epoxidation reaction of geraniol with $m$-chloroperoxybenzoic acid, for instance, gives a racemic mixture of $(2 S, 3 S)$ and $(2 R, 3 R)$ epoxides.\n\n\nUnfortunately, it's usually the case that only one enantiomer of a given drug or other important substance has the desired biological properties. The other enantiomer might be inactive or even dangerous. Thus, much work is currently being done on developing enantioselective methods of synthesis, which yield only one of the two possible enantiomers. So important has enantioselective synthesis become that the 2001 Nobel Prize in Chemistry was awarded to three pioneers in the field: William S. Knowles of the Monsanto Co., K. Barry Sharpless of the Scripps Research Institute, and Ryoji Noyori of Nagoya University.\nSeveral approaches to enantioselective synthesis have been developed, but the most efficient are those that use chiral catalysts to temporarily hold a substrate molecule in an unsymmetrical environment-the same strategy that nature uses when catalyzing reactions with chiral enzymes. While in that unsymmetrical environment, the substrate may be more open to reaction on one side than on another, leading to an excess of one enantiomeric product over another. As an analogy, think about picking up a coffee mug in your right hand to take a drink. The mug by itself is achiral, but as soon as you pick it up by the handle, it becomes chiral. One side of the mug now faces toward you so you can drink from it, but the other side faces away. The two sides are different, with one side much more accessible to you than the other.\n\n\nFIGURE 19.18 A substance made from the tartaric acid found at the bottom of this wine vat catalyzes enantioselective reactions. (credit: \"Pinot Noir fermentation in open vat\" by Pipers Brook Vineyard Media/Flickr, CC BY 2.0)"}
{"id": 1261, "contents": "Enantioselective Synthesis - \nAmong the thousands of enantioselective reactions now known, one of the most useful is the so-called Sharpless epoxidation, in which an allylic alcohol, such as geraniol, is treated with tert-butyl hydroperoxide,\n$\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{C}-\\mathrm{OOH}$, in the presence of titanium tetraisopropoxide and diethyl tartrate (DET) as a chiral auxiliary reagent. When $(R, R)$ tartrate is used, geraniol is converted into its $2 S, 3 S$ epoxide with $98 \\%$ selectivity, whereas the use of $(S, S)$ tartrate gives the $2 R, 3 R$ epoxide enantiomer. We say that the major product in each case is formed with an enantiomeric excess of $96 \\%$, meaning that $4 \\%$ of the product is racemic ( $2 \\% 2 S, 3 S$ plus $2 \\%$ $2 R, 3 R$ ) and an extra $96 \\%$ of a single enantiomer is formed. The mechanistic details by which the chiral catalyst works are a bit complex, although it appears that a chiral complex of two tartrate molecules with one titanium is involved.\n\n\n\n\n2S,3S isomer-98\\%\n\nGeraniol\n\n$2 R, 3 R$ isomer-98\\%"}
{"id": 1262, "contents": "Key Terms - \n```\n- acetal, R2C(OR')2\n- imine ( }\\mp@subsup{\\textrm{R}}{2}{}\\textrm{C}=\\textrm{NR}\n- acyl group\n- ketal\n- 1,2-addition\n- ketone (R2CO)\n- 1,4-addition\n- McLafferty rearrangement\n- aldehyde (RCHO)\n- nucleophilic addition reaction\n- Cannizzaro reaction\n- phosphorane\n- carbinolamine\n- Schiff bases\n- conjugate addition\n- Wittig reaction\n- cyanohydrin\n- Wolff-Kishner reaction\n- enamine ( }\\mp@subsup{\\textrm{R}}{2}{}\\textrm{N}-\\textrm{CR}=\\mp@subsup{\\textrm{CR}}{2}{}\n- ylide\n- hemiacetal\n```"}
{"id": 1263, "contents": "Summary - \nAldehydes and ketones are among the most important of all functional groups, both in the chemical industry and in biological pathways. In this chapter, we've looked at some of their typical reactions. Aldehydes are normally prepared in the laboratory by oxidation of primary alcohols or by partial reduction of esters. Ketones are similarly prepared by oxidation of secondary alcohols.\n\nThe nucleophilic addition reaction is the most common general reaction type for aldehydes and ketones. Many different kinds of products can be prepared by nucleophilic additions. Aldehydes and ketones are reduced by $\\mathrm{NaBH}_{4}$ or $\\mathrm{LiAlH}_{4}$ to yield primary and secondary alcohols, respectively. Addition of Grignard reagents to aldehydes and ketones also gives alcohols (secondary and tertiary, respectively), and addition of HCN yields cyanohydrins. Primary amines add to carbonyl compounds yielding imines, or Schiff bases, and secondary\namines yield enamines. Reaction of an aldehyde or ketone with hydrazine and base gives an alkane (the Wolff-Kishner reaction). Alcohols add to carbonyl groups to yield acetals, which are valuable as protecting groups. Phosphorus ylides add to aldehydes and ketones in the Wittig reaction to give alkenes.\n$\\alpha, \\beta$-Unsaturated aldehydes and ketones often react with nucleophiles to give the product of conjugate addition, or 1,4-addition. Particularly useful are the conjugate addition of an amine and the conjugate addition of an organic group by reaction with a diorganocopper reagent."}
{"id": 1264, "contents": "Summary - \nIR spectroscopy is helpful for identifying aldehydes and ketones. Carbonyl groups absorb in the IR range 1660 to $1770 \\mathrm{~cm}^{-1}$, with the exact position highly diagnostic of the kind of carbonyl group present in the molecule. ${ }^{13} \\mathrm{C}$ NMR spectroscopy is also useful for aldehydes and ketones because their carbonyl carbons show resonances in the 190 to $215 \\delta$ range. ${ }^{1} \\mathrm{H}$ NMR is useful for aldehyde - CHO protons, which absorb near $10 \\delta$. Aldehydes and ketones undergo two characteristic kinds of fragmentation in the mass spectrometer: $\\alpha$ cleavage and McLafferty rearrangement."}
{"id": 1265, "contents": "Summary of Reactions - \n1. Preparation of aldehydes (Section 19.2)\na. Oxidation of primary alcohols (Section 17.7)\n\nb. Partial reduction of esters (Section 19.2)\n\n2. Preparation of ketones\na. Oxidation of secondary alcohols (Section 17.7)\n\nb. Diorganocopper reaction with acid chlorides (Section 19.2)\n\n3. Oxidation of aldehydes (Section 19.3)\n\n4. Nucleophilic addition reactions of aldehydes and ketones\na. Addition of hydride to give alcohols: reduction (Section 19.7)\n\nb. Addition of Grignard reagents to give alcohols (Section 19.7)\n\n\n\nc. Addition of HCN to give cyanohydrins (Section 19.6)\n\nd. Addition of primary amines to give imines (Section 19.8)\n\ne. Addition of secondary amines to give enamines (Section 19.8)\n\nf. Wolff-Kishner reaction to give alkanes (Section 19.9)\n\ng. Addition of alcohols to give acetals (Section 19.10)\n\nh. Addition of phosphorus ylides to give alkenes: Wittig reaction (Section 19.11)\n\n5. Conjugate additions to $\\alpha, \\beta$-unsaturated aldehydes and ketones (Section 19.13)\na. Conjugate addition of amines\n\nb. Conjugate addition of water\n\nc. Conjugate addition of alkyl groups by diorganocopper reaction"}
{"id": 1266, "contents": "Visualizing Chemistry - \nPROBLEM Each of the following substances can be prepared by a nucleophilic addition reaction between an\n19-27 aldehyde or ketone and a nucleophile. Identify the reactants from which each was prepared. If the substance is an acetal, identify the carbonyl compound and the alcohol; if it is an imine, identify the carbonyl compound and the amine; and so forth.\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM The following molecular model represents a tetrahedral intermediate resulting from addition of a\n19-28 nucleophile to an aldehyde or ketone. Identify the reactants, and write the structure of the final product when the nucleophilic addition reaction is complete.\n\n\nPROBLEM The enamine prepared from acetone and dimethylamine is shown in its lowest-energy form.\n19-29 (a) What is the geometry and hybridization of the nitrogen atom?\n(b) What orbital on nitrogen holds the lone pair of electrons?\n(c) What is the geometric relationship between the $p$ orbitals of the double bond and the nitrogen orbital that holds the lone pair? Why do you think this geometry represents the minimum energy?"}
{"id": 1267, "contents": "Mechanism Problems - \nPROBLEM Predict the product(s) and propose a mechanism for each of the following reactions:\n19-30 (a)\n\n(b)\n\n$+$\n\n\nPROBLEM Predict the product(s) and propose a mechanism for each of the following reactions:\n19-31 (a)\n\n(b)\n\n\nPROBLEM Predict the product(s) and propose a mechanism for each of the following reactions:\n19-32\n\n\nb)\n\n\nPROBLEM Predict the product(s) and propose a mechanism for each of the following reactions:\n19-33 (a)\n\n(b)\n\n\nPROBLEM Predict the product(s) and propose mechanisms for the following reactions:\n\n19-34 (a)\n\n(b)\n\n\nPROBLEM The following reaction begins with an acetal and converts it into a different acetal. Predict the\n19-35 product(s) and propose a mechanism.\n(a)\n\n(b)\n\n\nPROBLEM When $\\alpha$-glucose is treated with an acid catalyst in the presence of an alcohol, an acetal is formed.\n19-36 Propose a mechanism for this process and give the structure of the stereoisomeric acetal that you would also expect as a product.\n\n\nPROBLEM Predict the products of the following Wolff-Kishner reactions. Write the mechanism for each,\n19-37 beginning from the hydrazone intermediate.\n(a)\n\n(b)\n\n\nPROBLEM Aldehydes can be prepared by the Wittig reaction using (methoxymethylene)triphenylphosphorane 19-38 as the Wittig reagent and then hydrolyzing the product with acid. For example,\n\n(a) How would you prepare the necessary phosphorane?\n(b) Propose a mechanism for the hydrolysis step.\n\nPROBLEM One of the steps in the metabolism of fats is the reaction of an unsaturated acyl CoA with water to 19-39 give a $\\beta$-hydroxyacyl CoA. Propose a mechanism.\n\n\nUnsaturated acyl CoA\n$\\beta$-Hydroxyacyl CoA\nPROBLEM Aldehydes and ketones react with thiols to yield thioacetals just as they react with alcohols to yield 19-40 acetals. Predict the product of the following reaction, and propose a mechanism:\n\n\nPROBLEM Ketones react with dimethylsulfonium methylide to yield epoxides. Suggest a mechanism for the\n19-41 reaction."}
{"id": 1268, "contents": "Mechanism Problems - \nPROBLEM Ketones react with dimethylsulfonium methylide to yield epoxides. Suggest a mechanism for the\n19-41 reaction.\n\n\nPROBLEM Propose a mechanism for the following reaction.\n19-42\n\n\nPROBLEM Paraldehyde, a sedative and hypnotic agent, is prepared by treatment of acetaldehyde with an acidic\n19-43 catalyst. Propose a mechanism for the reaction."}
{"id": 1269, "contents": "Paraldehyde - \nPROBLEM The Meerwein-Ponndorf-Verley reaction involves reduction of a ketone by treatment with an\n19-44 excess of aluminum triisopropoxide, $\\left[\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CHO}\\right]_{3} \\mathrm{Al}$. The mechanism of the process is closely related to the Cannizzaro reaction in that a hydride ion acts as a leaving group. Propose a mechanism.\n\n\nPROBLEM Propose a mechanism to account for the formation of 3,5-dimethylpyrazole from hydrazine and\n19-45 2,4-pentanedione. What has happened to each carbonyl carbon in going from starting material to product.\n\n\nPROBLEM In light of your answer to Problem 19-45, propose a mechanism for the formation of 19-46 3,5-dimethylisoxazole from hydroxylamine and 2,4-pentanedione.\n\n\n3,5-Dimethylisoxazole\n\nPROBLEM Trans alkenes are converted into their cis isomers and vice versa on epoxidation followed by 19-47 treatment of the epoxide with triphenylphosphine. Propose a mechanism for the reaction.\n\n\nPROBLEM Treatment of an $\\alpha, \\beta$-unsaturated ketone with basic aqueous hydrogen peroxide yields an epoxy\n19-48 ketone. The reaction is specific to unsaturated ketones; isolated alkene double bonds do not react. Propose a mechanism.\n\n\nPROBLEM One of the biological pathways by which an amine is converted to a ketone involves two steps: (1)\n19-49 oxidation of the amine by $\\mathrm{NAD}^{+}$to give an imine and (2) hydrolysis of the imine to give a ketone plus ammonia. Glutamate, for instance, is converted by this process into $\\alpha$-ketoglutarate. Show the structure of the imine intermediate, and propose mechanisms for both steps."}
{"id": 1270, "contents": "Paraldehyde - \nPROBLEM Primary amines react with esters to yield amides: $\\mathrm{RCO}_{2} \\mathrm{R}^{\\prime}+\\mathrm{R}^{\\prime \\prime} \\mathrm{NH}_{2} \\rightarrow \\mathrm{RCONHR}{ }^{\\prime \\prime}+\\mathrm{R}^{\\prime} \\mathrm{OH}$. Propose a 19-50 mechanism for the following reaction of an $\\alpha, \\beta$-unsaturated ester.\n\n\nPROBLEM When crystals of pure $\\alpha$-glucose are dissolved in water, isomerization occurs slowly to produce 19-51 $\\beta$-glucose. Propose a mechanism for the isomerization.\n\n\nPROBLEM The Wharton reaction converts an epoxy ketone to an allylic alcohol by reaction with hydrazine.\n19-52 Review the Wolff-Kishner reaction in Section 19.9 and then propose a mechanism.\n\n\nNaming Aldehydes and Ketones\nPROBLEM Draw structures corresponding to the following names:\n19-53\n(a) Bromoacetone\n(b) (S)-2-Hydroxypropanal\n(c) 2-Methyl-3-heptanone\n(d) $(2 S, 3 R)-2,3,4$-Trihydroxybutanal (e) 2,2,4,4-Tetramethyl-3-pentanone\n(f) 4-Methyl-3-penten-2-one (g) Butanedial (h) 3-Phenyl-2-propenal\n(i) 6,6-Dimethyl-2,4-cyclohexadienone (j) p-Nitroacetophenone\n\nPROBLEM Draw and name the seven aldehydes and ketones with the formula $\\mathrm{C}_{5} \\mathrm{H}_{10} \\mathrm{O}$. Which are chiral? 19-54\n\nPROBLEM Give IUPAC names for the following compounds:\n19-55 (a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)"}
{"id": 1271, "contents": "Paraldehyde - \nPROBLEM Give IUPAC names for the following compounds:\n19-55 (a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n\nPROBLEM Draw structures of compounds that fit the following descriptions:\n19-56 (a) An $\\alpha, \\beta$-unsaturated ketone, $\\mathrm{C}_{6} \\mathrm{H}_{8} \\mathrm{O}$\n(b) An $\\alpha$-diketone (c) An aromatic ketone, $\\mathrm{C}_{9} \\mathrm{H}_{10} \\mathrm{O}$\n(d) A diene aldehyde, $\\mathrm{C}_{7} \\mathrm{H}_{8} \\mathrm{O}$\n\nReactions of Aldehydes and Ketones\nPROBLEM Predict the products of the reaction of (1) phenylacetaldehyde and (2) acetophenone with the\n19-57 following reagents:\n(a) $\\mathrm{NaBH}_{4}$, then $\\mathrm{H}_{3} \\mathrm{O}^{+}$\n(b) Dess-Martin reagent\n(c) $\\mathrm{NH}_{2} \\mathrm{OH}, \\mathrm{HCl}$ catalyst\n(d) $\\mathrm{CH}_{3} \\mathrm{MgBr}$, then $\\mathrm{H}_{3} \\mathrm{O}^{+}$\n(e) $2 \\mathrm{CH}_{3} \\mathrm{OH}, \\mathrm{HCl}$ catalyst (f) $\\mathrm{H}_{2} \\mathrm{NNH}_{2}, \\mathrm{KOH}$\n(g) $\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{3} \\mathrm{P}=\\mathrm{CH}_{2}$\n(h) $\\mathrm{HCN}, \\mathrm{KCN}$\n\nPROBLEM Show how you might use a Wittig reaction to prepare the following alkenes. Identify the alkyl halide 19-58 and the carbonyl components.\n(a)\n\n(b)\n\n\nPROBLEM How would you use a Grignard reaction on an aldehyde or ketone to synthesize the following 19-59 compounds?\n(a) 2-Pentanol\n(b) 1-Butanol\n(c) 1-Phenylcyclohexanol\n(d) Diphenylmethanol"}
{"id": 1272, "contents": "Paraldehyde - \nPROBLEM How might you carry out the following selective transformations? One of the two schemes requires\n19-60 a protection step. (Recall from Section 19.4 that aldehydes are more reactive than ketones toward nucleophilic addition.)\n(a)\n\n(b)\n\n\nPROBLEM How would you prepare the following substances from 2-cyclohexenone? More than one step may\n19-61 be needed.\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n(Two ways)\n\nPROBLEM How would you synthesize the following substances from benzaldehyde and any other reagents 19-62 needed?\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Carvone is the major constituent of spearmint oil. What products would you expect from reaction 19-63 of carvone with the following reagents?\n\n\nCarvone\n(a) $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{Cu}^{-} \\mathrm{Li}^{+}$, then $\\mathrm{H}_{3} \\mathrm{O}^{+}$\n(b) $\\mathrm{LiAlH}_{4}$, then $\\mathrm{H}_{3} \\mathrm{O}^{+}$\n(c) $\\mathrm{CH}_{3} \\mathrm{NH}_{2}$\n(d) $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{MgBr}$, then $\\mathrm{H}_{3} \\mathrm{O}^{+}$\n(e) $\\mathrm{H}_{2} / \\mathrm{Pd}$\n(f) $\\mathrm{HOCH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}, \\mathrm{HCl}$\n(g) $\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{3} \\stackrel{+-}{\\mathrm{P}} \\mathrm{HCH}_{3}$\n\nPROBLEM How would you synthesize the following compounds from cyclohexanone?\n19-64 (a) 1-Methylcyclohexene (b) 2-Phenylcyclohexanone (c) cis-1,2-Cyclohexanediol\n(d) 1-Cyclohexylcyclohexanol"}
{"id": 1273, "contents": "Spectroscopy - \nPROBLEM At what position would you expect to observe IR absorptions for the following molecules?"}
{"id": 1274, "contents": "19-65 (a) - \n(b)\n\n\n1-Indanone\n(c)\n\n\nPROBLEM Acid-catalyzed dehydration of 3-hydroxy-3-phenylcyclohexanone leads to an unsaturated ketone.\n19-66 What possible structures are there for the product? At what position in the IR spectrum would you expect each to absorb? If the actual product has an absorption at $1670 \\mathrm{~cm}^{-1}$, what is its structure?\n\nPROBLEM Choose the structure that best fits the IR spectrum shown.\n19-67\n\n(a)\n\n(b)\n\n(c)\n\n(d)"}
{"id": 1275, "contents": "19-65 (a) - \nPROBLEM Choose the structure that best fits the IR spectrum shown.\n19-67\n\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM Propose structures for molecules that meet the following descriptions. Assume that the kinds of 19-68 carbons ( $1^{\\circ}, 2^{\\circ}, 3^{\\circ}$, or $4^{\\circ}$ ) have been assigned by DEPT-NMR.\n(a) $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}$; IR: $1715 \\mathrm{~cm}^{-1}$; ${ }^{13} \\mathrm{C}$ NMR: $8.0 \\delta\\left(1^{\\circ}\\right)$, $18.5 \\delta\\left(1^{\\circ}\\right), 33.5 \\delta\\left(2^{\\circ}\\right), 40.6 \\delta\\left(3^{\\circ}\\right), 214.0 \\delta\\left(4^{\\circ}\\right)$\n(b) $\\mathrm{C}_{5} \\mathrm{H}_{10} \\mathrm{O}$; IR: $1730 \\mathrm{~cm}^{-1}$; ${ }^{13} \\mathrm{C}$ NMR: $22.6 \\delta\\left(1^{\\circ}\\right), 23.6 \\delta\\left(3^{\\circ}\\right), 52.8 \\delta\\left(2^{\\circ}\\right), 202.4 \\delta\\left(3^{\\circ}\\right)$\n(c) $\\mathrm{C}_{6} \\mathrm{H}_{8} \\mathrm{O}$; IR: $1680 \\mathrm{~cm}^{-1}$; ${ }^{13} \\mathrm{C}$ NMR: $22.9 \\delta\\left(2^{\\circ}\\right), 25.8 \\delta\\left(2^{\\circ}\\right), 38.2 \\delta\\left(2^{\\circ}\\right), 129.8 \\delta\\left(3^{\\circ}\\right), 150.6 \\delta\\left(3^{\\circ}\\right)$, 198.7 ( $4^{\\circ}$ )"}
{"id": 1276, "contents": "19-65 (a) - \nPROBLEM Compound $\\mathbf{A}, \\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{O}_{2}$, has an intense IR absorption at $1750 \\mathrm{~cm}^{-1}$ and gives the ${ }^{13} \\mathrm{C}$ NMR spectrum 19-69 shown. Propose a structure for A.\n\n\nPROBLEM Propose structures for ketones or aldehydes that have the following ${ }^{1} \\mathrm{H}$ NMR spectra:\n19-70 (a) $\\mathrm{C}_{4} \\mathrm{H}_{7} \\mathrm{ClO}$\nIR: $1715 \\mathrm{~cm}^{-1}$\n\n(b) $\\mathrm{C}_{7} \\mathrm{H}_{14} \\mathrm{O}$\n\nIR: $1710 \\mathrm{~cm}^{-1}$"}
{"id": 1277, "contents": "General Problems - \nPROBLEM When 4-hydroxybutanal is treated with methanol in the presence of an acid catalyst, 19-71 2-methoxytetrahydrofuran is formed. Explain.\n\n\nPROBLEM The $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of (dibromomethyl)benzene, $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CHBr}_{2}$, with NaOH yields benzaldehyde rather 19-72 than (dihydroxymethyl)benzene, $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CH}(\\mathrm{OH})_{2}$. Explain.\n\nPROBLEM Reaction of 2-butanone with HCN yields a chiral product. What stereochemistry does the product 19-73 have? Is it optically active?\n\nPROBLEM The amino acid methionine is biosynthesized by a multistep route that includes reaction of an 19-74 imine of pyridoxal phosphate (PLP) to give an unsaturated imine, which then reacts with cysteine. What kinds of reactions are occurring in the two steps?"}
{"id": 1278, "contents": "O-SuccinylhomoserinePLP imine <br> Unsaturated imine - \nPROBLEM Each of the following reaction schemes has one or more flaws. What is wrong in each case? How 19-75 would you correct each scheme?\n(a)\n\n(b)\n\n\n\nPROBLEM 6-Methyl-5-hepten-2-one is a constituent of lemongrass oil. How could you synthesize this 19-76 substance from methyl 4-oxopentanoate?\nMethyl 4-oxopentanoate\n\nPROBLEM Tamoxifen is a drug used in the treatment of breast cancer. How would you prepare tamoxifen from 19-77 benzene, the following ketone, and any other reagents needed?\n\n\nPROBLEM Compound $\\mathbf{A}, \\mathrm{MW}=86$, shows an IR absorption at $1730 \\mathrm{~cm}^{-1}$ and a very simple ${ }^{1} \\mathrm{H}$ NMR spectrum 19-78 with peaks at $9.7 \\delta(1 \\mathrm{H}$, singlet) and $1.2 \\delta(9 \\mathrm{H}$, singlet). Propose a structure for A.\nPROBLEM Compound $\\mathbf{B}$ is isomeric with $\\mathbf{A}$ (Problem 19-78) and shows an IR peak at $1715 \\mathrm{~cm}^{-1}$. The ${ }^{1} \\mathrm{H}$ NMR 19-79 spectrum of $\\mathbf{B}$ has peaks at $2.4 \\delta(1 \\mathrm{H}$, septet, $J=7 \\mathrm{~Hz}), 2.1 \\delta(3 \\mathrm{H}$, singlet), and $1.2 \\delta(6 \\mathrm{H}$, doublet, $J$\n$=7 \\mathrm{~Hz})$. What is the structure of $\\mathbf{B}$ ?\nPROBLEM The ${ }^{1} \\mathrm{H}$ NMR spectrum shown is that of a compound with the formula $\\mathrm{C}_{9} \\mathrm{H}_{10} \\mathrm{O}$. How many double 19-80 bonds and/or rings does this compound contain? If the unknown compound has an IR absorption at $1690 \\mathrm{~cm}^{-1}$, what is a likely structure?"}
{"id": 1279, "contents": "O-SuccinylhomoserinePLP imine <br> Unsaturated imine - \nPROBLEM The ${ }^{1} \\mathrm{H}$ NMR spectrum shown is that of a compound isomeric with the one in Problem 19-80. This 19-81 isomer has an IR absorption at $1730 \\mathrm{~cm}^{-1}$. Propose a structure. [Note: Aldehyde protons (CHO) often show low coupling constants to adjacent hydrogens, so the splitting of aldehyde signals is not always apparent.]\n\n\nPROBLEM Propose structures for ketones or aldehydes that have the following ${ }^{1} \\mathrm{H}$ NMR spectra:\n19-82 (a) $\\mathrm{C}_{9} \\mathrm{H}_{10} \\mathrm{O}_{2}$ : IR: $1695 \\mathrm{~cm}^{-1}$\n\n(b) $\\mathrm{C}_{4} \\mathrm{H}_{6} \\mathrm{O}:$ IR: $1690 \\mathrm{~cm}^{-1}$\n\n\nPROBLEM Propose structures for ketones or aldehydes that have the following ${ }^{1} \\mathrm{H}$ NMR spectra.\n19-83 (a) $\\mathrm{C}_{10} \\mathrm{H}_{12} \\mathrm{O}$ : IR: $1710 \\mathrm{~cm}^{-1}$\n\n(b) $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{3}$ : IR: $1715 \\mathrm{~cm}^{-1}$\n\n\nPROBLEM When glucose (Problem 19-51) is treated with $\\mathrm{NaBH}_{4}$, reaction occurs to yield sorbitol, a 19-84 polyalcohol commonly used as a food additive. Show how this reduction occurs.\n\n\nGlucose\nSorbitol\n\nPROBLEM The proton and carbon NMR spectra for each of three isomeric ketones with the formula $\\mathrm{C}_{7} \\mathrm{H}_{14} \\mathrm{O}$ are\n19-85 shown. Assign a structure to each pair of spectra."}
{"id": 1280, "contents": "O-SuccinylhomoserinePLP imine <br> Unsaturated imine - \nPROBLEM The proton NMR spectrum for a compound with formula $\\mathrm{C}_{10} \\mathrm{H}_{12} \\mathrm{O}_{2}$ is shown below. The infrared 19-86 spectrum has a strong band at $1711 \\mathrm{~cm}^{-1}$. The broadband-decoupled 13C NMR spectral results are tabulated along with the DEPT-135 and DEPT-90 information. Draw the structure of this compound."}
{"id": 1281, "contents": "CHAPTER 20 - \nCarboxylic Acids and Nitriles\n\n\nFIGURE 20.1 The burning sensation produced by touching or eating chili peppers is due to capsaicin, a carboxylic acid derivative called an amide. (credit: modification of \"Chillis\" by Lucas Cobb/Flickr, CC BY 2.0)"}
{"id": 1282, "contents": "CHAPTER CONTENTS - \n20.1 Naming Carboxylic Acids and Nitriles\n20.2 Structure and Properties of Carboxylic Acids\n20.3 Biological Acids and the Henderson-Hasselbalch Equation\n20.4 Substituent Effects on Acidity\n20.5 Preparing Carboxylic Acids\n20.6 Reactions of Carboxylic Acids: An Overview\n20.7 Chemistry of Nitriles\n20.8 Spectroscopy of Carboxylic Acids and Nitriles\n\nWHY THIS CHAPTER? Carboxylic acids are present in many industrial processes and most biological pathways and are the starting materials from which other acyl derivatives are made. Thus, an understanding of their properties and reactions is fundamental to understanding organic chemistry. We'll look both at acids and at their close relatives, nitriles ( $\\mathrm{RC} \\equiv \\mathrm{N}$ ), in this chapter and at carboxylic acid derivatives in the next chapter.\n\nCarboxylic acids, $\\mathbf{R C O}_{\\mathbf{2}} \\mathbf{H}$, occupy a central place among carbonyl compounds. Not only are they valuable in themselves, they also serve as starting materials for preparing numerous carboxylic acid derivatives such as acid chlorides, esters, amides, and thioesters. In addition, carboxylic acids are present in the majority of biological pathways.\n\nAn acid chloride $\\quad$ An ester"}
{"id": 1283, "contents": "CHAPTER CONTENTS - \nAn acid chloride $\\quad$ An ester\n\nA great many carboxylic acids are found in nature: acetic acid, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$, is the chief organic component of vinegar; butanoic acid, $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$, is responsible for the rancid odor of sour butter; and hexanoic acid (caproic acid), $\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{4} \\mathrm{CO}_{2} \\mathrm{H}$, is responsible for the unmistakable aroma of goats and dirty gym socks (the name comes from the Latin caper, meaning \"goat\"). Other examples are cholic acid, a major component of human bile, and long-chain aliphatic acids such as palmitic acid, $\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{14} \\mathrm{CO}_{2} \\mathrm{H}$, a biological precursor of fats and vegetable oils.\n\n\n\nCholic acid\nApproximately 20 million tons of acetic acid is produced worldwide each year for a variety of purposes, including preparation of the vinyl acetate polymer used in paints and adhesives. About $20 \\%$ of the acetic acid synthesized industrially is obtained by oxidation of acetaldehyde. Much of the remaining $80 \\%$ is prepared by the rhodium-catalyzed reaction of methanol with carbon monoxide."}
{"id": 1284, "contents": "Carboxylic Acids, $\\mathrm{RCO}_{2} \\mathrm{H}$ - \nSimple carboxylic acids derived from open-chain alkanes are systematically named by replacing the terminal $-e$ of the corresponding alkane name with -oic acid. The $-\\mathrm{CO}_{2} \\mathrm{H}$ carbon atom is numbered C 1 .\n\n\nPropanoic acid\n\n\n4-Methylpentanoic acid\n\n\n3-Ethyl-6-methyloctanedioic acid\n\nCompounds that have $\\mathrm{a}-\\mathrm{CO}_{2} \\mathrm{H}$ group bonded to a ring are named using the suffix -carboxylic acid. The $\\mathrm{CO}_{2} \\mathrm{H}$ carbon is attached to C 1 in this system and is not itself numbered. As a substituent, the $\\mathrm{CO}_{2} \\mathrm{H}$ group is called a carboxyl group.\n\ntrans-4-Hydroxycyclohexanecarboxylic acid\n\n\n1-Cyclopentenecarboxylic acid\n\nBecause many carboxylic acids were among the first organic compounds to be isolated and purified, quite a few common names exist (TABLE 20.1). Biological chemists, in particular, make frequent use of these names, so you may find yourself referring back to this list on occasion. We'll use systematic names in this book, with a few exceptions such as formic (methanoic) acid and acetic (ethanoic) acid, whose names are accepted by IUPAC and are so well known that it makes little sense to refer to them any other way.\nAlso listed in TABLE 20.1 are the names of acyl groups $\\binom{\\mathrm{O}}{\\mathrm{R}-\\mathrm{C}-}$ derived from the parent acids. Except for the eight entries at the top of TABLE 20.1, whose names have a -yl ending, all other acyl groups are named using an -oyl ending.\n\nTABLE 20.1 Common Names of Some Carboxylic Acids and Acyl Groups"}
{"id": 1285, "contents": "Carboxylic Acids, $\\mathrm{RCO}_{2} \\mathrm{H}$ - \n| Structure | Name | Acyl group |\n| :---: | :---: | :---: |\n| $\\mathrm{HCO}_{2} \\mathrm{H}$ | Formic | Formyl |\n| $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$ | Acetic | Acetyl |\n| $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ | Propionic | Propionyl |\n| $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ | Butyric | Butyryl |\n| $\\mathrm{HO}_{2} \\mathrm{CCO}_{2} \\mathrm{H}$ | Oxalic | Oxalyl |\n| $\\mathrm{HO}_{2} \\mathrm{CCH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ | Malonic | Malonyl |\n| $\\mathrm{HO}_{2} \\mathrm{CCH}_{2} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ | Succinic | Succinyl |\n| $\\mathrm{HO}_{2} \\mathrm{CCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ | Glutaric | Glutaryl |\n| $\\mathrm{HO}_{2} \\mathrm{CCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ | Adipic | Adipoyl |\n| $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCO}_{2} \\mathrm{H}$ | Acrylic | Acryloyl |\n| $\\mathrm{HO}_{2} \\mathrm{CCH}=\\mathrm{CHCO}_{2} \\mathrm{H}$ | Maleic (cis) Fumaric (trans) | Maleoyl Fumaroyl |\n| $\\mathrm{HOCH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ | Glycolic | Glycoloyl |"}
{"id": 1286, "contents": "Carboxylic Acids, $\\mathrm{RCO}_{2} \\mathrm{H}$ - \n| $\\mathrm{HOCH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ | Glycolic | Glycoloyl |\n| <smiles>CC(O)C(=O)O</smiles> | Lactic | Lactoyl |\n| <smiles>CC(=O)O</smiles> | Pyruvic | Pyruvoyl |\n| <smiles>O=C(O)C(O)CO</smiles> | Glyceric | Glyceroyl |"}
{"id": 1287, "contents": "Carboxylic Acids, $\\mathrm{RCO}_{2} \\mathrm{H}$ - \nTABLE 20.1 Common Names of Some Carboxylic Acids and Acyl Groups\n\n| Structure | Name | Acyl group |\n| :---: | :---: | :---: |\n| <smiles>O=C(O)CC(O)C(=O)O</smiles> | Malic | Maloyl |\n| <smiles>O=C(O)CC(=O)O</smiles> | Oxaloacetic | Oxaloacetyl |\n| <smiles>O=C(O)c1ccccc1</smiles> | Benzoic | Benzoyl |\n| <smiles>O=C(O)c1ccccc1C(=O)O</smiles> | Phthalic | Phthaloyl |"}
{"id": 1288, "contents": "Nitriles, $\\mathrm{RC} \\equiv \\mathrm{N}$ - \nCompounds containing the $-\\mathrm{C} \\equiv \\mathrm{N}$ functional group are called nitriles and can undergo some chemistry similar to that of carboxylic acids. Simple open-chain nitriles are named by adding -nitrile as a suffix to the alkane name, with the nitrile carbon numbered C 1 .\n\n$$\n\\underset{5}{\\mathrm{CH}_{3}} \\stackrel{\\mathrm{C}_{4}^{\\mathrm{CH}} \\mathrm{CH}_{3}}{\\mathrm{CH}_{3}}\n$$\n\nNitriles can also be named as derivatives of carboxylic acids by replacing the -ic acid or -oic acid ending with -onitrile, or by replacing the -carboxylic acid ending with -carbonitrile. The nitrile carbon atom is attached to C 1 but is not itself numbered.\n\n$$\n\\mathrm{CH}_{3} \\mathrm{C} \\equiv \\mathrm{~N}\n$$"}
{"id": 1289, "contents": "Acetonitrile (from acetic acid) <br> Benzonitrile <br> (from benzoic acid) - \n2,2-Dimethylcyclohexanecarbonitrile (from 2,2-dimethylcyclohexanecarboxylic acid)\n\nIf another carboxylic acid derivative is present in the same molecule, the prefix cyano- is used for the $\\mathrm{C} \\equiv \\mathrm{N}$ group."}
{"id": 1290, "contents": "Methyl 4-cyanopentanoate - \nPROBLEM Give IUPAC names for the following compounds:\n20-1 (a)\n\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n\nPROBLEM Draw structures corresponding to the following IUPAC names:\n20-2 (a) 2,3-Dimethylhexanoic acid (b) 4-Methylpentanoic acid\n(c) trans-1,2-Cyclobutanedicarboxylic acid (d) o-Hydroxybenzoic acid\n(e) (9Z,12Z)-9,12-Octadecadienoic acid (f) 2-Pentenenitrile"}
{"id": 1291, "contents": "Methyl 4-cyanopentanoate - 20.2 Structure and Properties of Carboxylic Acids\nCarboxylic acids are similar in some respects to both ketones and alcohols. Like ketones, the carboxyl carbon is $s p^{2}$-hybridized, and carboxylic acid groups are therefore planar with $\\mathrm{C}-\\mathrm{C}=\\mathrm{O}$ and $\\mathrm{O}=\\mathrm{C}-\\mathrm{O}$ bond angles of approximately $120^{\\circ}$ (TABLE 20.2).\n\n| TABLE 20.2 Physical Parameters for Acetic Acid |  |  |  |  |\n| :--- | :--- | :--- | :--- | :---: |\n|  |  |  |  |  |\n|  |  |  |  |  |\n| Bond angle | (degrees) | Bond length | (pm) |  |\n| $\\mathrm{C}-\\mathrm{C}=\\mathrm{O}$ | 119 | $\\mathrm{C}-\\mathrm{C}$ | 152 |  |\n| $\\mathrm{C}-\\mathrm{C}-\\mathrm{OH}$ | 119 | $\\mathrm{C}=\\mathrm{O}$ | 125 |  |\n| $\\mathrm{O}=\\mathrm{C}-\\mathrm{OH}$ | 122 | $\\mathrm{C}-\\mathrm{OH}$ | 131 |  |\n\nLike alcohols, carboxylic acids are strongly associated because of hydrogen-bonding. Most carboxylic acids exist as cyclic dimers held together by two hydrogen bonds. This strong hydrogen-bonding has a noticeable effect on boiling points, making carboxylic acids boil far less easily than their corresponding alcohols. Acetic acid, for instance, has a boiling point of $117.9^{\\circ} \\mathrm{C}$, versus $78.3^{\\circ} \\mathrm{C}$ for ethanol, even though both compounds have two carbons.\n\n\nAcetic acid dimer"}
{"id": 1292, "contents": "Methyl 4-cyanopentanoate - 20.2 Structure and Properties of Carboxylic Acids\nAcetic acid dimer\n\n\nThe most obvious property of carboxylic acids is implied by their name: carboxylic acids are acidic. They therefore react with bases such as NaOH and $\\mathrm{NaHCO}_{3}$ to give metal carboxylate salts, $\\mathrm{RCO}_{2}{ }^{-} \\mathrm{M}^{+}$. Carboxylic acids with more than six carbons are only slightly soluble in water, but the alkali metal salts of carboxylic acids are often highly water-soluble. In fact, it's often possible to purify an acid by extracting its salt into aqueous base, then reacidifying and extracting the pure acid back into an organic solvent.\n\n\nLike other Br\u00f8nsted-Lowry acids discussed in Section 2.7, carboxylic acids dissociate slightly in dilute aqueous solution to give $\\mathrm{H}_{3} \\mathrm{O}^{+}$and the corresponding carboxylate anions, $\\mathrm{RCO}_{2}{ }^{-}$. The extent of dissociation is given by an acidity constant, $K_{\\mathrm{a}}$.\n\n\nA list of $K_{\\mathrm{a}}$ values for various carboxylic acids is given in TABLE 20.3. For most, $K_{\\mathrm{a}}$ is approximately $10^{-4}$ to $10^{-5}$. Acetic acid, for instance, has $K_{\\mathrm{a}}=1.75 \\times 10^{-5}$ at $25^{\\circ} \\mathrm{C}$, which corresponds to a $\\mathrm{p} K_{\\mathrm{a}}$ of 4.76 . In practical terms, a $K_{\\mathrm{a}}$ value near $10^{-5}$ means that only about $0.1 \\%$ of the molecules in a 0.1 M solution are dissociated, as opposed to the $100 \\%$ dissociation found with strong mineral acids like HCl ."}
{"id": 1293, "contents": "Methyl 4-cyanopentanoate - 20.2 Structure and Properties of Carboxylic Acids\n| Structure | $K_{\\text {a }}$ | $\\mathrm{p} K_{\\mathrm{a}}$ |  |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{CF}_{3} \\mathrm{CO}_{2} \\mathrm{H}$ | 0.59 | 0.23 | Stronger acid |\n| $\\mathrm{HCO}_{2} \\mathrm{H}$ | $1.77 \\times 10^{-4}$ | 3.75 |  |\n| $\\mathrm{HOCH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ | $1.5 \\times 10^{-4}$ | 3.84 |  |\n| $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CO}_{2} \\mathrm{H}$ | $6.46 \\times 10^{-5}$ | 4.19 |  |\n| $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCO}_{2} \\mathrm{H}$ | $5.6 \\times 10^{-5}$ | 4.25 |  |\n| $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$ | $1.75 \\times 10^{-5}$ | 4.76 |  |\n| $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ | $1.34 \\times 10^{-5}$ | 4.87 | acid |\n| $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$ (ethanol) | $\\left(1 \\times 10^{-16}\\right)$ | (16) |  |\n\nAlthough much weaker than mineral acids, carboxylic acids are nevertheless much stronger acids than alcohols and phenols. The $K_{\\mathrm{a}}$ of ethanol, for example, is approximately $10^{-16}$, making it a weaker acid than acetic acid by a factor of $10^{11}$."}
{"id": 1294, "contents": "Methyl 4-cyanopentanoate - 20.2 Structure and Properties of Carboxylic Acids\nWhy are carboxylic acids so much more acidic than alcohols, even though both contain - OH groups? An alcohol dissociates to give an alkoxide ion, in which the negative charge is localized on a single electronegative atom. A carboxylic acid, however, gives a carboxylate ion, in which the negative charge is delocalized over two equivalent oxygen atoms (FIGURE 20.2). In resonance terms (Section 2.4), a carboxylate ion is a stabilized resonance hybrid of two equivalent structures. Since a carboxylate ion is more stable than an alkoxide ion, it is lower in energy and more favored in the dissociation equilibrium.\n\n\nEthanol\n\n\n$+$\n\n\nEthoxide ion (localized charge)\n\n\n\nAcetate ion\n(delocalized charge)\n\nFIGURE 20.2 An alkoxide ion has its charge localized on one oxygen atom and is less stable, while a carboxylate ion has the charge spread equally over both oxygens and is therefore more stable.\n\nExperimental evidence for the equivalence of the two carboxylate oxygens comes from X-ray crystallographic studies on sodium formate. Both carbon-oxygen bonds are 127 pm in length, midway between the $\\mathrm{C}=\\mathrm{O}$ double bond ( 120 pm ) and the $\\mathrm{C}-\\mathrm{O}$ single bond ( 134 pm ) of formic acid. An electrostatic potential map of the formate ion also shows how the negative charge (red) is spread equally over both oxygens.\n\n\nPROBLEM Assume you have a mixture of naphthalene and benzoic acid that you want to separate. How might 20-3 you take advantage of the acidity of one component in the mixture to effect a separation?\n\nPROBLEM The $K_{\\mathrm{a}}$ for dichloroacetic acid is $3.32 \\times 10^{-2}$. Approximately what percentage of the acid is 20-4 dissociated in a 0.10 M aqueous solution?"}
{"id": 1295, "contents": "Methyl 4-cyanopentanoate - 20.3 Biological Acids and the Henderson-Hasselbalch Equation\nIn acidic solution, at low pH , a carboxylic acid is completely undissociated and exists entirely as RCOH. In basic solution, at high pH , a carboxylic acid is completely dissociated and exists entirely as $\\mathrm{RCO}_{2}{ }^{-}$. Inside living cells, however, the pH is neither acidic nor basic but is instead buffered to a nearly neutral pH of $3.5-4.5$ in humans, a value often referred to as physiological $p H$. In what form, then, do carboxylic acids exist inside cells? The question is an important one for understanding the acid catalysts so often found in biological reactions.\n\nIf the $\\mathrm{p} K_{\\mathrm{a}}$ value of a given acid and the pH of the medium are known, the percentages of dissociated and\nundissociated forms can be calculated using the Henderson-Hasselbalch equation.\nFor any acid HA, we have\n\n$$\n\\begin{aligned}\n\\mathrm{p} K_{\\mathrm{a}} & =-\\log \\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]}=-\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]-\\log \\frac{\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]} \\\\\n& =\\mathrm{pH}-\\log \\frac{\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]}\n\\end{aligned}\n$$\n\nwhich can be rearranged to give\n\n$$\n\\mathrm{pH}=\\mathrm{p} K_{\\mathrm{a}}+\\log \\frac{\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]} \\quad \\text { Henderson-Hasselbalch equation }\n$$\n\nso\n\n$$\n\\log \\frac{\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]}=\\mathrm{pH}-\\mathrm{p} K_{\\mathrm{a}}\n$$"}
{"id": 1296, "contents": "Methyl 4-cyanopentanoate - 20.3 Biological Acids and the Henderson-Hasselbalch Equation\nso\n\n$$\n\\log \\frac{\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]}=\\mathrm{pH}-\\mathrm{p} K_{\\mathrm{a}}\n$$\n\nThis equation says that the logarithm of the concentration of dissociated acid [ $\\mathrm{A}^{-}$] divided by the concentration of undissociated acid [HA] is equal to the pH of the solution minus the $\\mathrm{p} K_{\\mathrm{a}}$ of the acid. Thus, if we know both the pH of the solution and the $\\mathrm{p} K_{\\mathrm{a}}$ of the acid, we can calculate the ratio of [ $\\mathrm{A}^{-}$] to [HA]. Furthermore, when $\\mathrm{pH}=\\mathrm{p} K_{\\mathrm{a}}$, then HA and $\\mathrm{A}^{-}$are present in equal amounts because $\\log 1=0$.\n\nAs an example of how to use the Henderson-Hasselbalch equation, let's find out what species are present in a 0.0010 M solution of acetic acid at $\\mathrm{pH}=7.3$. According to Table $20-3$, the $\\mathrm{p} K_{\\mathrm{a}}$ of acetic acid is 4.76 . From the Henderson-Hasselbalch equation, we have\n\n$$\n\\begin{aligned}\n\\log \\frac{\\left[\\mathrm{A}^{-}\\right]}{\\mathrm{HA}} & =\\mathrm{pH}-\\mathrm{p} K_{\\mathrm{a}}=7.3-4.76=2.54 \\\\\n\\frac{\\left[\\mathrm{~A}^{-}\\right]}{\\mathrm{HA}} & =\\operatorname{antilog}(2.54)=3.5 \\times 10^{2} \\quad \\text { so }\\left[\\mathrm{A}^{-}\\right]=\\left(3.5 \\times 10^{2}\\right)[\\mathrm{HA}]\n\\end{aligned}\n$$\n\nIn addition, we know that\n\n$$\n\\left[\\mathrm{A}^{-}\\right]+[\\mathrm{HA}]=0.0010 \\mathrm{M}\n$$"}
{"id": 1297, "contents": "Methyl 4-cyanopentanoate - 20.3 Biological Acids and the Henderson-Hasselbalch Equation\nIn addition, we know that\n\n$$\n\\left[\\mathrm{A}^{-}\\right]+[\\mathrm{HA}]=0.0010 \\mathrm{M}\n$$\n\nSolving the two simultaneous equations gives $\\left[\\mathrm{A}^{-}\\right]=0.0010 \\mathrm{M}$ and $[\\mathrm{HA}]=3 \\times 10^{-6} \\mathrm{M}$. In other words, at a physiological pH of 7.3 , essentially $100 \\%$ of acetic acid molecules in a 0.0010 M solution are dissociated to the acetate ion.\n\nWhat is true for acetic acid is also true for other carboxylic acids: At the physiological pH that occurs inside cells, carboxylic acids are almost entirely dissociated. To reflect this fact, we always refer to cellular carboxylic acids by the name of their anion-acetate, lactate, citrate, and so forth, rather than acetic acid, lactic acid, and citric acid.\n\nPROBLEM Calculate the percentages of dissociated and undissociated forms present in the following 20-5 solutions:\n(a) 0.0010 M glycolic acid $\\left(\\mathrm{HOCH}_{2} \\mathrm{CO}_{2} \\mathrm{H} ; \\mathrm{p} K_{\\mathrm{a}}=3.83\\right)$ at $\\mathrm{pH}=4.50$\n(b) 0.0020 M propanoic $\\operatorname{acid}\\left(\\mathrm{p} K_{\\mathrm{a}}=4.87\\right)$ at $\\mathrm{pH}=5.30$"}
{"id": 1298, "contents": "Methyl 4-cyanopentanoate - 20.4 Substituent Effects on Acidity\nThe listing of $\\mathrm{p} K_{\\mathrm{a}}$ values shown previously in TABLE 20.3 indicates that there are substantial differences in acidity from one carboxylic acid to another. For example, trifluoroacetic acid ( $K_{\\mathrm{a}}=0.59$ ) is 33,000 times as strong as acetic acid $\\left(K_{\\mathrm{a}}=1.75 \\times 10^{-5}\\right)$. How can we account for such differences?\n\nBecause the dissociation of a carboxylic acid is an equilibrium process, any factor that stabilizes the carboxylate anion relative to undissociated carboxylic acid will drive the equilibrium toward increased dissociation and result in increased acidity. For instance, three electron-withdrawing fluorine atoms delocalize the negative charge in the trifluoroacetate anion, thereby stabilizing the ion and increasing the acidity of $\\mathrm{CF}_{3} \\mathrm{CO}_{2} \\mathrm{H}$. In the same way, glycolic acid $\\left(\\mathrm{HOCH}_{2} \\mathrm{CO}_{2} \\mathrm{H} ; \\mathrm{p} K_{\\mathrm{a}}=3.83\\right)$ is stronger than acetic acid because of the electronwithdrawing effect of the electronegative oxygen atom.\n\n\nBecause inductive effects operate through $\\sigma$ bonds and are dependent on distance, the effect of halogen substitution decreases as the substituent moves farther from the carboxyl. Thus, 2-chlorobutanoic acid has $\\mathrm{p} K_{\\mathrm{a}}=2.86,3$-chlorobutanoic acid has $\\mathrm{p} K_{\\mathrm{a}}=4.05$, and 4-chlorobutanoic acid has $\\mathrm{p} K_{\\mathrm{a}}=4.52$, similar to that of butanoic acid itself.\n\n$\\mathrm{p} K_{\\mathrm{a}}=4.52$\n\n$\\mathrm{p} K_{\\mathrm{a}}=4.05$\nAcidity"}
{"id": 1299, "contents": "Methyl 4-cyanopentanoate - 20.4 Substituent Effects on Acidity\n$\\mathrm{p} K_{\\mathrm{a}}=4.52$\n\n$\\mathrm{p} K_{\\mathrm{a}}=4.05$\nAcidity\n\nSubstituent effects on acidity are also found in substituted benzoic acids. We said during the discussion of electrophilic aromatic substitution in Section 16.4 that substituents on the aromatic ring strongly affect reactivity. Aromatic rings with electron-donating groups are activated toward further electrophilic substitution, and aromatic rings with electron-withdrawing groups are deactivated. Exactly the same effects can be observed on the acidity of substituted benzoic acids (TABLE 20.4).\n\n|  | Y | <smiles>[Y]c1ccc(C(=O)O)cc1</smiles> |  |  |\n| :---: | :---: | :---: | :---: | :---: |\n|  |  | $K_{\\mathrm{a}} \\times 1$ | $\\mathrm{p} K_{\\mathrm{a}}$ |  |\n| Stronger acid | $-\\mathrm{NO}_{2}$ | 39 | 3.41 | Deactivating groups |\n|  | -CN | 28 | 3.55 |  |\n|  | - CHO | 18 | 3.75 |  |\n|  | -Br | 11 | 3.96 |  |\n|  | -Cl | 10 | 4.0 |  |\n|  | -H | 6.46 | 4.19 |  |\n| Weaker | $-\\mathrm{CH}_{3}$ | 4.3 | 4.34 | Activating groups |\n|  | $-\\mathrm{OCH}_{3}$ | 3.5 | 4.46 |  |\n|  | -OH | 3.3 | 4.48 |  |\n\nAs TABLE 20.4 shows, an electron-donating (activating) group such as methoxy decreases acidity by destabilizing the carboxylate anion, and an electron-withdrawing (deactivating) group such as nitro increases acidity by stabilizing the carboxylate anion."}
{"id": 1300, "contents": "Methyl 4-cyanopentanoate - 20.4 Substituent Effects on Acidity\n$p$-Methoxybenzoic acid ( $\\mathrm{p} K_{\\mathrm{a}}=4.46$ )\n\n\nBenzoic acid\n( $\\mathrm{p} K_{\\mathrm{a}}=4.19$ )\n\n$p$-Nitrobenzoic acid $\\left(\\mathrm{p} K_{\\mathrm{a}}=3.41\\right)$"}
{"id": 1301, "contents": "Acidity - \nBecause it's much easier to measure the acidity of a substituted benzoic acid than it is to determine the relative reactivity of an aromatic ring toward electrophilic substitution, the correlation between the two effects is useful for predicting reactivity. If we want to know the effect of a certain substituent on electrophilic reactivity, we can simply find the acidity of the corresponding benzoic acid. Worked Example 20.1 gives an illustration.\n\n... lets us predict the reactivity of this substituted benzene to electrophilic attack."}
{"id": 1302, "contents": "Predicting the Effect of a Substituent on the Reactivity of an Aromatic Ring toward Electrophilic Substitution - \nThe $\\mathrm{p} K_{\\mathrm{a}}$ of $p$-(trifluoromethyl)benzoic acid is 3.6 . Is the trifluoromethyl substituent an activating or deactivating group in electrophilic aromatic substitution?"}
{"id": 1303, "contents": "Strategy - \nDecide whether $p$-(trifluoromethyl)benzoic acid is stronger or weaker than benzoic acid. A substituent that strengthens the acid is a deactivating group because it withdraws electrons, and a substituent that weakens the acid is an activating group because it donates electrons."}
{"id": 1304, "contents": "Solution - \nA p $K_{\\mathrm{a}}$ of 3.6 means that $p$-(trifluoromethyl)benzoic acid is stronger than benzoic acid, whose $\\mathrm{p} K_{\\mathrm{a}}$ is 4.19. Thus, the trifluoromethyl substituent favors dissociation by helping stabilize the negative charge. Trifluoromethyl must therefore be an electron-withdrawing, deactivating group.\n\nPROBLEM Which would you expect to be a stronger acid, the lactic acid found in tired muscles or acetic acid? 20-6 Explain.\n\n\nLactic acid\n\nPROBLEM Dicarboxylic acids have two dissociation constants, one for the initial dissociation into a monoanion 20-7 and one for the second dissociation into a dianion. For oxalic acid, $\\mathrm{HO}_{2} \\mathrm{C}-\\mathrm{CO}_{2} \\mathrm{H}$, the first ionization constant is $\\mathrm{p} K_{\\mathrm{a} 1}=1.2$ and the second ionization constant is $\\mathrm{p} K_{\\mathrm{a} 2}=4.2$. Why is the second carboxyl group far less acidic than the first?\n\nPROBLEM The $\\mathrm{p} K_{\\mathrm{a}}$ of $p$-cyclopropylbenzoic acid is 4.45. Is cyclopropylbenzene likely to be more reactive or 20-8 less reactive than benzene toward electrophilic bromination? Explain.\n\nPROBLEM Rank the following compounds in order of increasing acidity. Don't look at a table of $\\mathrm{p} K_{\\mathrm{a}}$ data to help 20-9 with your answer.\n(a) Benzoic acid, $p$-methylbenzoic acid, $p$-chlorobenzoic acid\n(b) $p$-Nitrobenzoic acid, acetic acid, benzoic acid"}
{"id": 1305, "contents": "Solution - 20.5 Preparing Carboxylic Acids\nLet's review briefly some of the methods for preparing carboxylic acids that we've seen in previous chapters.\n\n- Oxidation of a substituted alkylbenzene with $\\mathrm{KMnO}_{4}$ gives a substituted benzoic acid (Section 16.8). Both primary and secondary alkyl groups can be oxidized, but tertiary groups are not affected.\n\n$p-$ Nitrotoluene\n$p$-Nitrobenzoic acid (88\\%)\n- Oxidation of a primary alcohol or an aldehyde yields a carboxylic acid (Section 17.7 and Section 19.3). Primary alcohols are often oxidized with the Dess Martin periodinane, and aldehydes are similarly oxidized with alkaline $\\mathrm{KMnO}_{4}$.\n\n\n4-Methyl-1-pentanol\n4-Methylpentanoic acid\n\n\nHexanal\nHexanoic acid"}
{"id": 1306, "contents": "Hydrolysis of Nitriles - \nCarboxylic acids can be prepared from nitriles on heating with aqueous acid or base by a mechanism that we'll discuss in Section 20.7. Since nitriles themselves are usually made by $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of a primary or secondary alkyl halide with $\\mathrm{CN}^{-}$, the two-step sequence of cyanide displacement followed by nitrile hydrolysis is a good way to make a carboxylic acid from an alkyl halide ( $\\mathrm{RBr} \\rightarrow \\mathrm{RC} \\equiv \\mathrm{N} \\rightarrow \\mathrm{RCO}_{2} \\mathrm{H}$ ). Note that the product acid has one more carbon than the starting alkyl halide. One example occurs in a commercial route for the synthesis of the nonsteroidal anti-inflammatory drug ibuprofen. (See Chapter 15 Chemistry Matters.)\n\n$\\downarrow$ 1. $\\mathrm{NaOH}, \\mathrm{H}_{2} \\mathrm{O}$\n\n\nIbuprofen"}
{"id": 1307, "contents": "Carboxylation of Grignard Reagents - \nAnother method for preparing carboxylic acids is by reaction of a Grignard reagent with $\\mathrm{CO}_{2}$ to yield a metal carboxylate, followed by protonation to give a carboxylic acid. This carboxylation reaction is usually carried out by bubbling a stream of dry $\\mathrm{CO}_{2}$ gas through a solution of the Grignard reagent. The organomagnesium halide adds to a $\\mathrm{C}=\\mathrm{O}$ bond of carbon dioxide in a typical nucleophilic carbonyl addition reaction, and protonation of\nthe carboxylate by addition of aqueous HCl in a separate step then gives the free carboxylic acid. For example:\n\n\nAs noted previously, there are no Grignard reagents inside living cells, but there are other types of stabilized carbanions that are often carboxylated. One of the initial steps in fatty-acid biosynthesis, for instance, involves the formation of a carbanion from acetyl CoA, followed by carboxylation to yield malonyl CoA."}
{"id": 1308, "contents": "Devising a Synthesis Route for a Carboxylic Acid - \nHow would you prepare phenylacetic acid $\\left(\\mathrm{PhCH}_{2} \\mathrm{CO}_{2} \\mathrm{H}\\right)$ from benzyl bromide $\\left(\\mathrm{PhCH}_{2} \\mathrm{Br}\\right)$ ?"}
{"id": 1309, "contents": "Strategy - \nWe've seen two methods for preparing carboxylic acids from alkyl halides: (1) cyanide ion displacement followed by hydrolysis and (2) formation of a Grignard reagent followed by carboxylation. The first method involves an $\\mathrm{S}_{\\mathrm{N}} 2$ reaction and is therefore limited to use with primary and some secondary alkyl halides. The second method involves formation of a Grignard reagent and is therefore limited to use with organic halides that have no acidic hydrogens or reactive functional groups elsewhere in the molecule. In the present instance, either method would work well.\n\nSolution\n\n\nPROBLEM How would you prepare the following carboxylic acids?\n20-10 (a) $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CCO}_{2} \\mathrm{H}$ from $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CCl}$\n(b) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ from $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{Br}$"}
{"id": 1310, "contents": "Strategy - 20.6 Reactions of Carboxylic Acids: An Overview\nWe commented earlier in this chapter that carboxylic acids are similar in some respects to both alcohols and\nketones. Like alcohols, carboxylic acids can be deprotonated to give anions, which are good nucleophiles in $\\mathrm{S}_{\\mathrm{N}} 2$ reactions. Like ketones, carboxylic acids undergo addition of nucleophiles to the carbonyl group. However, carboxylic acids also undergo other reactions characteristic of neither alcohols nor ketones. FIGURE 20.3 shows some of the general reactions of carboxylic acids.\n\n\nDeprotonation\n\n\nReduction\n\n\nFIGURE 20.3 Some general reactions of carboxylic acids.\nReactions of carboxylic acids can be grouped into the four categories indicated in FIGURE 20.3. Of the four, we've already discussed the acidic behavior of carboxylic acids in Section 20.2 through Section 20.4, and we mentioned reduction by treatment with $\\mathrm{LiAlH}_{4}$ in Section 17.4. The remaining two categories are examples of fundamental carbonyl-group reaction mechanisms-nucleophilic acyl substitution and $\\alpha$ substitution-that will be discussed in detail in Chapters 21 and 22.\n\nPROBLEM How might you prepare 2-phenylethanol from benzyl bromide? More than one step is needed.\n20-11\n\n\nPROBLEM How might you carry out the following transformation? More than one step is needed.\n20-12"}
{"id": 1311, "contents": "Strategy - 20.7 Chemistry of Nitriles\nNitriles are analogous to carboxylic acids in that both have a carbon atom with three bonds to an electronegative atom and both contain a $\\pi$ bond. Thus, some reactions of nitriles and carboxylic acids are similar. Both kinds of compounds are electrophiles, for instance, and both undergo nucleophilic addition reactions.\n\n$$\n\\mathrm{R}-\\mathrm{C} \\equiv \\mathrm{~N}\n$$"}
{"id": 1312, "contents": "A nitrile-three bonds to nitrogen - \nAn acid-three bonds to two oxygens\n\nNitriles occur infrequently in living organisms, although several hundred examples are known. Cyanocycline A, for instance, has been isolated from the bacterium Streptomyces lavendulae and was found to have both antimicrobial and antitumor activity. In addition, more than 1000 compounds called cyanogenic glycosides are known. Derived primarily from plants, cyanogenic glycosides contain a sugar with an acetal carbon, one oxygen of which is bonded to a nitrile-bearing carbon (sugar-O-C-CN). On hydrolysis with aqueous acid, the acetal is cleaved (Section 19.10), generating a cyanohydrin (HO-C-CN), which releases hydrogen cyanide. It's thought that the primary function of cyanogenic glycosides is to protect the plant by poisoning any animal foolish enough to eat it. Lotaustralin from the cassava plant is an example.\n\n\nCyanocycline A\n\n\nLotaustralin\n(a cyanogenic glycoside)"}
{"id": 1313, "contents": "Preparation of Nitriles - \nThe simplest method of nitrile preparation is the $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of $\\mathrm{CN}^{-}$with a primary or secondary alkyl halide, as discussed in Section 20.5. Another method for preparing nitriles is by dehydration of a primary amide, $\\mathrm{RCONH}_{2}$. Thionyl chloride is often used for this reaction, although other dehydrating agents such as $\\mathrm{POCl}_{3}$ also work.\n\n\nThe dehydration occurs by initial reaction of $\\mathrm{SOCl}_{2}$ on the nucleophilic amide oxygen atom, followed by deprotonation and a subsequent E2-like elimination reaction.\n\n\nBoth methods of nitrile synthesis- $\\mathrm{S}_{\\mathrm{N}} 2$ displacement by $\\mathrm{CN}^{-}$on an alkyl halide and amide dehydration-are useful, but the synthesis from amides is more general because it is not limited by steric hindrance."}
{"id": 1314, "contents": "Reactions of Nitriles - \nLike a carbonyl group, a nitrile group is strongly polarized and has an electrophilic carbon atom. Nitriles therefore react with nucleophiles to yield $s p^{2}$-hybridized imine anions in a reaction analogous to the formation of a $s p^{3}$-hybridized alkoxide ion by nucleophilic addition to a carbonyl group."}
{"id": 1315, "contents": "Carbonyl compound - \nNitrile\n\n\n\nSome general reactions of nitriles are shown in FIGURE 20.4.\n\n\nHydrolysis: Conversion of Nitriles into Carboxylic Acids\nAmong the most useful reactions of nitriles is their hydrolysis to yield first an amide and then a carboxylic acid plus ammonia or an amine. The reaction occurs in either basic or acidic aqueous solution:\n\n\nAs shown in FIGURE 20.5, base-catalyzed nitrile hydrolysis involves nucleophilic addition of hydroxide ion to the polar $\\mathrm{C} \\equiv \\mathrm{N}$ bond to give an imine anion in a process similar to the nucleophilic addition to a polar $\\mathrm{C}=\\mathrm{O}$ bond to give an alkoxide anion. Protonation then gives a hydroxy imine, which tautomerizes (Section 9.4) to an amide in a step similar to the tautomerization of an enol to a ketone. Further hydrolysis gives a carboxylate ion."}
{"id": 1316, "contents": "FIGURE 20.5 MECHANISM - \nMechanism for the basic hydrolysis of a nitrile to yield an amide, which is then hydrolyzed further to a carboxylic acid anion.\n(1) Nucleophilic addition of hydroxide ion to the CN triple bond gives an imine anion addition product.\n(2) Protonation of the imine anion by water yields a hydroxyimine and regenerates the base catalyst.\n\n(3) Tautomerization of the hydroxyimine yields an amide in a reaction analogous to the tautomerization of an enol to give a ketone.\n\n(4) Further hydrolysis of the amide gives the anion of a carboxylic acid by a mechanism we'll discuss in Section 21.7.\n\n\nCarboxylate ion\n\nThe further hydrolysis of the amide intermediate takes place by a nucleophilic addition of hydroxide ion to the amide carbonyl group, which yields a tetrahedral alkoxide ion. Expulsion of amide ion, $\\mathrm{NH}_{2}{ }^{-}$, as leaving group gives the carboxylate ion, thereby driving the reaction toward the products. Subsequent acidification in a separate step yields the carboxylic acid. We'll look at this process in more detail in Section 21.7."}
{"id": 1317, "contents": "Reduction: Conversion of Nitriles into Amines - \nReduction of a nitrile with $\\mathrm{LiAlH}_{4}$ gives a primary amine, $\\mathrm{RNH}_{2}$. The reaction occurs by nucleophilic addition of hydride ion to the polar $\\mathrm{C} \\equiv \\mathrm{N}$ bond, yielding an imine anion, which still contains a $\\mathrm{C}=\\mathrm{N}$ bond and therefore undergoes a second nucleophilic addition of hydride to give a dianion. Both monoanion and dianion intermediates are undoubtedly stabilized by Lewis acid-base complexation to an aluminum species, facilitating the second addition that would otherwise be difficult. Protonation of the dianion by addition of water in a subsequent step gives the amine.\n\n\nReaction of Nitriles with Grignard Reagents\nGrignard reagents add to a nitrile to give an intermediate imine anion that is hydrolyzed by addition of water to yield a ketone. The mechanism of the hydrolysis is the exact reverse of imine formation (see FIGURE 19.7).\n\n\nThis reaction is similar to the reduction of a nitrile to an amine, except that only one nucleophilic addition occurs rather than two and the attacking nucleophile is a carbanion ( $\\mathrm{R}:^{-}$) rather than a hydride ion. For example:"}
{"id": 1318, "contents": "Synthesizing a Ketone from a Nitrile - \nHow would you prepare 2-methyl-3-pentanone from a nitrile?\n\n\n2-Methyl-3-pentanone"}
{"id": 1319, "contents": "Strategy - \nA ketone results from the reaction between a Grignard reagent and a nitrile, with the $\\mathrm{C} \\equiv \\mathrm{N}$ carbon of the nitrile becoming the carbonyl carbon. Identify the two groups attached to the carbonyl carbon atom in the product. One will come from the Grignard reagent and the other will come from the nitrile."}
{"id": 1320, "contents": "Solution - \nThere are two possibilities.\n\n\nPROBLEM How would you prepare the following carbonyl compounds from a nitrile?\n20-13\n(a)\n\n(b)\n\n\nPROBLEM How would you prepare 1-phenyl-2-butanone, $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CH}_{2} \\mathrm{COCH}_{2} \\mathrm{CH}_{3}$, from benzyl bromide,\n20-14 $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CH}_{2} \\mathrm{Br}$ ? More than one step is needed."}
{"id": 1321, "contents": "Infrared Spectroscopy - \nCarboxylic acids have two characteristic IR absorptions that make the $-\\mathrm{CO}_{2} \\mathrm{H}$ group easily identifiable. The $\\mathrm{O}-\\mathrm{H}$ bond of the carboxyl group gives rise to a very broad absorption over the range 2500 to $3300 \\mathrm{~cm}^{-1}$. The $\\mathrm{C}=\\mathrm{O}$ bond shows an absorption between 1710 and $1760 \\mathrm{~cm}^{-1}$. The exact position of $\\mathrm{C}=\\mathrm{O}$ absorption depends both on the structure of the molecule and on whether the acid is free (monomeric) or hydrogen-bonded (dimeric). Free carboxyl groups absorb at $1760 \\mathrm{~cm}^{-1}$, but the more commonly encountered dimeric carboxyl groups absorb in a broad band centered around $1710 \\mathrm{~cm}^{-1}$. As with other carbonyl-containing functional groups, conjugation with an alkene or benzene ring lowers the frequency of the $\\mathrm{C}=\\mathrm{O}$ stretch by 20 to $30 \\mathrm{~cm}^{-1}$.\n\n\n\nAssociated carboxyl (usual case), $1710 \\mathrm{~cm}^{-1}$\n\n\nBoth the broad $\\mathrm{O}-\\mathrm{H}$ absorption and the $\\mathrm{C}=\\mathrm{O}$ absorption at $1710 \\mathrm{~cm}^{-1}$ (dimeric) are identified in the IR spectrum of butanoic acid shown in FIGURE 20.6.\n\n\nFIGURE 20.6 IR spectrum of butanoic acid, $\\mathbf{C H}_{3} \\mathbf{C H}_{2} \\mathbf{C H}_{2} \\mathbf{C O}_{2} \\mathbf{H}$.\nNitriles show an intense and easily recognizable $\\mathrm{C} \\equiv \\mathrm{N}$ bond absorption near $2250 \\mathrm{~cm}^{-1}$ for saturated compounds and $2230 \\mathrm{~cm}^{-1}$ for aromatic and conjugated molecules. Few other functional groups absorb in this region, so IR spectroscopy is highly diagnostic for nitriles."}
{"id": 1322, "contents": "Infrared Spectroscopy - \nPROBLEM Cyclopentanecarboxylic acid and 4-hydroxycyclohexanone have the same formula $\\left(\\mathrm{C}_{6} \\mathrm{H}_{10} \\mathrm{O}_{2}\\right)$, and\n20-15 both contain an -OH and a $\\mathrm{C}=\\mathrm{O}$ group. How could you distinguish between them using IR spectroscopy?"}
{"id": 1323, "contents": "Nuclear Magnetic Resonance Spectroscopy - \nCarboxyl carbon atoms absorb in the range 165 to $185 \\delta$ in the ${ }^{13} \\mathrm{C}$ NMR spectrum, with aromatic and $\\alpha, \\beta$-unsaturated acids near the upfield end of the range ( $\\sim 165 \\delta$ ) and saturated aliphatic acids near the downfield end ( $\\sim 185 \\delta$ ). Nitrile carbons absorb in the range 115 to $130 \\delta$.\n\n\nIn the ${ }^{1} \\mathrm{H}$ NMR spectrum, the acidic $-\\mathrm{CO}_{2} \\mathrm{H}$ proton normally absorbs as a singlet near $12 \\delta$. The chemical shift of the carboxyl proton is concentration and solvent dependent, as these variables can change the extent of hydrogen-bonding in the sample. In some cases, the carboxyl-proton resonance is broadened to the point of being nearly undetectable. Traces of water in the sample can exacerbate the situation. As with alcohols (Section 17.11), the $-\\mathrm{CO}_{2} \\mathrm{H}$ proton can be replaced by deuterium when $\\mathrm{D}_{2} \\mathrm{O}$ is added to the sample tube, causing the absorption to disappear from the NMR spectrum. FIGURE 20.7 shows the ${ }^{1} \\mathrm{H}$ NMR spectrum of phenylacetic acid. Note that the $-\\mathrm{CO}_{2} \\mathrm{H}$ absorption occurs at $12.0 \\delta$.\n\n\nFIGURE 20.7 Proton NMR spectrum of phenylacetic acid, $\\mathbf{C}_{\\mathbf{6}} \\mathbf{H}_{\\mathbf{5}} \\mathbf{C H}_{\\mathbf{2}} \\mathbf{C O}_{\\mathbf{2}} \\mathbf{H}$.\nPROBLEM How could you distinguish between the isomers cyclopentanecarboxylic acid and 20-16 4-hydroxycyclohexanone by ${ }^{1} \\mathrm{H}$ and ${ }^{13} \\mathrm{C}$ NMR spectroscopy? (See Problem 20-15.)"}
{"id": 1324, "contents": "Vitamin C - \nThe word vitamin, despite its common usage, is actually an imprecise term. Generally speaking, a vitamin is an organic substance that a given organism requires in small amounts to live and grow but is unable to synthesize and must obtain in its diet. Thus, to be considered a vitamin, only a small amount of the substance is needed-anywhere from a few micrograms to 100 mg or so per day. Dietary substances needed in larger amounts, such as some amino acids and unsaturated fats, are not considered vitamins.\n\n\nFIGURE 20.8 In addition to the hazards of weather, participants in early polar expeditions often suffered from scurvy, caused by a dietary vitamin C deficiency. (credit: \"Shackleton's expedition to the Antarctic last moments of the Endurance\" by Underwood \\& Underwood/ Wikimedia Commons, Public Domain)\n\nFurthermore, different organisms need different vitamins. More than 4000 species of mammals can synthesize ascorbic acid in their bodies, for instance, but humans are not among them. Ascorbic acid is therefore a human vitamin-what we all know as vitamin C -and must be provided by our diet. Small amounts of more than a dozen other substances are similarly required by humans: retinol (vitamin A), thiamine (vitamin $B_{1}$ ), and tocopherol (vitamin E), for instance.\n\nVitamin C is surely the best known of all human vitamins. It was the first to be discovered (1912), the first to be isolated (1928), the first to be structurally characterized (1933), and the first to be synthesized in the laboratory (1933). Over 121,000 tons of vitamin C is synthesized worldwide each year, more than the total amount of all other vitamins combined. In addition to its use as a supplement, vitamin C is used as a food preservative, a \"flour improver\" in bakeries, and an animal-food additive.\n\n\nVitamin C (ascorbic acid)\n\n\nVitamin C is perhaps most well-known for its antiscorbutic properties, meaning that it prevents the onset of scurvy, a bleeding disease affecting those with a deficiency of fresh vegetables and citrus fruits in their diet. Sailors in the Age of Exploration were particularly susceptible to scurvy, and the death toll was high. The"}
{"id": 1325, "contents": "Vitamin C - \nPortuguese explorer Vasco da Gama lost more than half his crew to scurvy during his two-year voyage around the Cape of Good Hope in 1497-1499.\n\nMore recently, large doses of vitamin $C$ have been claimed to prevent the common cold, cure infertility, delay the onset of symptoms in acquired immunodeficiency syndrome (AIDS), and inhibit the development of gastric and cervical cancers. None of these claims have been backed by medical evidence, however. In the largest study yet of the effect of vitamin $C$ on the common cold, a meta-analysis of more than 100 separate trials covering 40,000 people found no difference in the incidence of colds between those who took supplemental vitamin C regularly and those who did not. When taken during a cold, however, vitamin $C$ does appear to decrease the cold's duration by perhaps a day.\n\nThe industrial preparation of vitamin $C$ involves an unusual blend of biological and laboratory organic chemistry, beginning with glucose and following the five-step route shown in FIGURE 20.9. Glucose, a pentahydroxy aldehyde, is first reduced to sorbitol, which is then oxidized by the microorganism Acetobacter suboxydans. No chemical reagent is known that is selective enough to oxidize only one of the six alcohol groups in sorbitol, so an enzymatic reaction is used. Treatment with acetone and an acid catalyst then converts four of the other hydroxyl groups into acetal linkages, and the remaining hydroxyl group is chemically oxidized to a carboxylic acid by reaction with aqueous NaOCl (household bleach). Hydrolysis with acid then removes the two acetal groups and causes an internal ester-forming reaction to give ascorbic acid. Each of the five steps has a yield better than $90 \\%$.\n\n\nFIGURE 20.9 The industrial synthesis of ascorbic acid from glucose."}
{"id": 1326, "contents": "Key Terms - \n- carboxyl group - Henderson-Hasselbalch equation\n- carboxylation\n- nitrile\n- carboxylic acid, $\\mathrm{RCO}_{2} \\mathrm{H}$"}
{"id": 1327, "contents": "Summary - \nCarboxylic acids are among the most useful building blocks for synthesizing other molecules, both in nature and in the laboratory. Thus, an understanding of their properties and reactions is fundamental to understanding biological chemistry. In this chapter, we've looked both at acids and at their close relatives, nitriles ( $\\mathrm{RC} \\equiv \\mathrm{N}$ ).\n\nCarboxylic acids are named systematically by replacing the terminal -e of the corresponding alkane name with -oic acid. Like aldehydes and ketones, the carbonyl carbon atom is $s p^{2}$-hybridized; like alcohols, carboxylic acids are associated through hydrogen-bonding and therefore have high boiling points.\n\nThe distinguishing characteristic of carboxylic acids is their acidity. Although weaker than mineral acids such as HCl , carboxylic acids dissociate much more readily than alcohols because the resultant carboxylate ions are stabilized by resonance between two equivalent forms.\n\nMost carboxylic acids have $\\mathrm{p} K_{\\mathrm{a}}$ values near 5, but the exact $\\mathrm{p} K_{\\mathrm{a}}$ of a given acid depends on structure. Carboxylic acids substituted by electron-withdrawing groups are more acidic (have a lower $\\mathrm{p} K_{\\mathrm{a}}$ ) because their carboxylate ions are stabilized. Carboxylic acids substituted by electron-donating groups are less acidic (have a higher $\\mathrm{p} K_{\\mathrm{a}}$ ) because their carboxylate ions are destabilized. The extent of dissociation of a carboxylic acid in a buffered solution of a given pH can be calculated with the Henderson-Hasselbalch equation. Inside living cells, where the physiological $\\mathrm{pH}=7.3$, carboxylic acids are entirely dissociated and exist as their carboxylate anions."}
{"id": 1328, "contents": "Summary - \nMethods of synthesis for carboxylic acids include (1) oxidation of alkylbenzenes, (2) oxidation of primary alcohols or aldehydes, (3) reaction of Grignard reagents with $\\mathrm{CO}_{2}$ (carboxylation), and (4) hydrolysis of nitriles. General reactions of carboxylic acids include (1) loss of the acidic proton, (2) nucleophilic acyl substitution at the carbonyl group, (3) substitution on the $\\alpha$ carbon, and (4) reduction.\n\nNitriles are similar in some respects to carboxylic acids and are prepared either by $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of an alkyl halide with cyanide ion or by dehydration of an amide. Nitriles undergo nucleophilic addition to the polar $\\mathrm{C} \\equiv \\mathrm{N}$ bond in the same way that carbonyl compounds do. The most important reactions of nitriles are their hydrolysis to carboxylic acids, reduction to primary amines, and reaction with Grignard reagents to yield ketones.\n\nCarboxylic acids and nitriles are easily distinguished spectroscopically. Acids show a characteristic IR absorption at 2500 to $3300 \\mathrm{~cm}^{-1}$ due to the $\\mathrm{O}-\\mathrm{H}$ bond and another at 1710 to $1760 \\mathrm{~cm}^{-1}$ due to the $\\mathrm{C}=\\mathrm{O}$ bond; nitriles have an absorption at $2250 \\mathrm{~cm}^{-1}$. Acids also show ${ }^{13} \\mathrm{C}$ NMR absorptions at 165 to $185 \\delta$ and ${ }^{1} \\mathrm{H}$ NMR absorptions near $12 \\delta$. Nitriles have a ${ }^{13} \\mathrm{C}$ NMR absorption in the range 115 to $130 \\delta$."}
{"id": 1329, "contents": "Summary of Reactions - \n1. Preparation of carboxylic acids (Section 20.5)\na. Carboxylation of Grignard reagents\n\n$$\n\\mathrm{R}-\\mathrm{MgX} \\stackrel{\\text { 1. } \\mathrm{CO}_{2}}{\\text { 2. } \\mathrm{H}_{3} \\mathrm{O}^{+}} \\quad \\stackrel{\\stackrel{\\mathrm{O}}{\\|}}{\\mathrm{R}^{-}}{ }_{\\mathrm{OH}}\n$$\n\nb. Hydrolysis of nitriles\n\n$$\n\\mathrm{R}-\\mathrm{C} \\equiv \\mathrm{~N} \\stackrel{\\mathrm{or}^{\\mathrm{NaOH}, \\mathrm{H}_{2} \\mathrm{O}}}{\\mathrm{R}^{-} \\stackrel{\\mathrm{O}}{\\mathrm{C}} \\mathrm{C}_{\\mathrm{OH}}^{+}}\n$$\n\n2. Preparation of nitriles (Section 20.7)\na. $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of alkyl halides\n\n$$\n\\mathrm{RCH}_{2} \\mathrm{Br} \\quad \\xrightarrow{\\mathrm{NaCN}} \\mathrm{RCH}_{2} \\mathrm{C} \\equiv \\mathrm{~N}\n$$\n\nb. Dehydration of amides\n\n3. Reactions of nitriles (Section 20.7)\na. Hydrolysis to yield carboxylic acids\n\n$$\n\\mathrm{R}-\\mathrm{C} \\equiv \\mathrm{~N} \\xrightarrow[\\text { 2. } \\mathrm{H}_{3} \\mathrm{O}^{+}]{\\stackrel{\\text { 1. } \\mathrm{NaOH}, \\mathrm{H}_{2} \\mathrm{O}}{\\mathrm{R}^{-\\quad}} \\stackrel{\\mathrm{O}}{\\mathrm{C}} \\mathrm{OH}_{\\mathrm{OH}}}+\\mathrm{NH}_{3}\n$$\n\nb. Reduction to yield primary amines\n\nc. Reaction with Grignard reagents to yield ketones"}
{"id": 1330, "contents": "Summary of Reactions - \nb. Reduction to yield primary amines\n\nc. Reaction with Grignard reagents to yield ketones\n\n$$\n\\mathrm{R}-\\mathrm{C} \\equiv \\mathrm{~N} \\quad \\stackrel{\\text { 1. } \\mathrm{R}^{\\prime} \\mathrm{MgX} \\text {, ether }}{\\text { 2. } \\mathrm{H}_{3} \\mathrm{O}^{+}} \\quad \\stackrel{\\text { R }}{\\stackrel{\\mathrm{O}}{\\mathrm{C}}>_{\\mathrm{R}^{\\prime}}}+\\mathrm{NH}_{3}\n$$"}
{"id": 1331, "contents": "Visualizing Chemistry - \nPROBLEM Give IUPAC names for the following carboxylic acids (reddish brown $=\\mathrm{Br}$ ):\n\n20-17 (a)\n\n(c)\n\n(b)\n\n(d)\n\n\nPROBLEM Would you expect the following carboxylic acids to be more acidic or less acidic than benzoic acid?\n20-18 Explain. (Reddish brown $=\\mathrm{Br}$.)\n(a)\n\n(b)\n\n\nPROBLEM The following carboxylic acid can't be prepared from an alkyl halide by either the nitrile hydrolysis\n20-19 route or the Grignard carboxylation route. Explain.\n\n\nPROBLEM Electrostatic potential maps of anisole and thioanisole are shown. Which do you think is the 20-20 stronger acid, $p$-methoxybenzoic acid or $p$-(methylthio) benzoic acid? Explain.\n\n\nAnisole $\\left(\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{OCH}_{3}\\right)$\n\n\nThioanisole $\\left(\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{SCH}_{3}\\right)$"}
{"id": 1332, "contents": "Mechanism Problems - \nPROBLEM Predict the product(s) and write the mechanism of each of the following reactions:\n\n20-21 (a)\n\n\n1. Mg\n$\\xrightarrow[\\text { 3. } \\mathrm{H}_{3} \\mathrm{O}^{+}]{\\text {2. } \\mathrm{CO}_{2}}$ ?\n(b)\n\n\nPROBLEM Predict the product(s) and write the mechanism of each of the following reactions:"}
{"id": 1333, "contents": "20-22 (a) - \n(b)\n\n\nPROBLEM Predict the product(s) and write the mechanism of each of the following reactions:\n\n20-23 (a)\n\n(b)\n\n\nPROBLEM Predict the product(s) and write the complete mechanism of each of the following reactions:\n20-24 (a)\n\n(b)\n\n\nPROBLEM Acid-catalyzed hydrolysis of a nitrile to give a carboxylic acid occurs by initial protonation of the\n20-25 nitrogen atom, followed by nucleophilic addition of water. Review the mechanism of base-catalyzed nitrile hydrolysis in Section 20.7 and then predict the products for the following reactions. Write the steps involved in the acid-catalyzed reaction, using curved arrows.\n(a)\n\n(b)\n\n\nPROBLEM Nitriles can be converted directly to esters by the Pinner reaction, which first produces an\n$\\mathbf{2 0 - 2 6}$ iminoester salt that is isolated and then treated with water to give the final product. Propose a mechanism for the Pinner reaction using curved arrows to show the flow of electrons at each step.\n\n\nPROBLEM Naturally occurring compounds called cyanogenic glycosides, such as lotaustralin, release\n20-27 hydrogen cyanide, HCN, when treated with aqueous acid. The reaction occurs by hydrolysis of the acetal linkage to form a cyanohydrin, which then expels HCN and gives a carbonyl compound.\n(a) Show the mechanism of the acetal hydrolysis and the structure of the cyanohydrin that results.\n(b) Propose a mechanism for the loss of HCN, and show the structure of the carbonyl compound that forms."}
{"id": 1334, "contents": "Lotaustralin - \nPROBLEM 2-Bromo-6,6-dimethylcyclohexanone gives 2,2-dimethylcyclopentanecarboxylic acid on treatment\n20-28 with aqueous NaOH followed by acidification, a process called the Favorskii reaction. Propose a mechanism.\n\n\nPROBLEM Naturally occurring compounds called terpenoids, which we'll discuss in Section 27.5, are\n20-29 biosynthesized by a pathway that involves loss of $\\mathrm{CO}_{2}$ from 3-phosphomevalonate 5-diphosphate to yield isopentenyl diphosphate. Use curved arrows to show the mechanism of this reaction.\n\n\nPROBLEM In the Ritter reaction, an alkene reacts with a nitrile in the presence of strong aqueous sulfuric acid 20-30 to yield an amide. Propose a mechanism."}
{"id": 1335, "contents": "Naming Carboxylic Acids and Nitriles - \nPROBLEM Give IUPAC names for the following compounds:\n20-31 (a)\n\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n(g)\n\n(h)\n\n\nPROBLEM Draw structures corresponding to the following IUPAC names:\n20-32 (a) cis-1,2-Cyclohexanedicarboxylic acid (b) Heptanedioic acid (c) 2-Hexen-4-ynoic acid\n(d) 4-Ethyl-2-propyloctanoic acid (e) 3-Chlorophthalic acid (f) Triphenylacetic acid\n(g) 2-Cyclobutenecarbonitrile (h) m-Benzoylbenzonitrile\n\nPROBLEM Draw and name the following compounds:\n20-33 (a) The eight carboxylic acids with the formula $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{2}$\n(b) Three nitriles with the formula $\\mathrm{C}_{5} \\mathrm{H}_{7} \\mathrm{~N}$\n\nPROBLEM Pregabalin, marketed as Lyrica, is an anticonvulsant drug that is also effective in treating chronic\n20-34 pain. The IUPAC name of pregabalin is (S)-3-(aminomethyl)-5-methylhexanoic acid. (An aminomethyl group is $-\\mathrm{CH}_{2} \\mathrm{NH}_{2}$.) Draw the structure of pregabalin.\n\nPROBLEM Isocitric acid, an intermediate in the citric acid cycle of food metabolism, has the systematic name\n20-35 ( $2 R, 3 S$ )-3-carboxy-2-hydroxypentanedioic acid. Draw the structure."}
{"id": 1336, "contents": "Acidity of Carboxylic Acids - \nPROBLEM Order the compounds in each of the following sets with respect to increasing acidity:\n20-36 (a) Acetic acid, oxalic acid, formic acid\n(b) $p$-Bromobenzoic acid, $p$-nitrobenzoic acid, 2,4-dinitrobenzoic acid\n(c) Fluoroacetic acid, 3-fluoropropanoic acid, iodoacetic acid\n\nPROBLEM Arrange the compounds in each of the following sets in order of increasing basicity:\n20-37 (a) Magnesium acetate, magnesium hydroxide, methylmagnesium bromide\n(b) Sodium benzoate, sodium $p$-nitrobenzoate, sodium acetylide\n(c) Lithium hydroxide, lithium ethoxide, lithium formate"}
{"id": 1337, "contents": "Acidity of Carboxylic Acids - \nPROBLEM Calculate the $\\mathrm{p} K_{\\mathrm{a}}$ 's of the following acids:\n20-38 (a) Lactic acid, $K_{\\mathrm{a}}=8.4 \\times 10^{-4}$ (b) Acrylic acid, $K_{\\mathrm{a}}=5.6 \\times 10^{-6}$\nPROBLEM Calculate the $K_{\\mathrm{a}}$ 's of the following acids:\n20-39 (a) Citric acid, $\\mathrm{p} K_{\\mathrm{a}}=3.14$ (b) Tartaric acid, $\\mathrm{p} K_{\\mathrm{a}}=2.98$\nPROBLEM Thioglycolic acid, $\\mathrm{HSCH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$, a substance used in depilatory agents (hair removers) has $\\mathrm{p} K_{\\mathrm{a}}=$\n20-40 3.42. What is the percent dissociation of thioglycolic acid in a buffer solution at $\\mathrm{pH}=3.0$ ?\nPROBLEM In humans, the final product of purine degradation from DNA is uric acid, $\\mathrm{p} K_{\\mathrm{a}}=5.61$, which is 20-41 excreted in the urine. What is the percent dissociation of uric acid in urine at a typical $\\mathrm{pH}=6.0$ ? Why do you think uric acid is acidic even though it does not have a $\\mathrm{CO}_{2} \\mathrm{H}$ group?\n\n\nUric acid\n\nPROBLEM Some $\\mathrm{p} K_{\\mathrm{a}}$ data for simple dibasic acids is shown. How can you account for the fact that the difference\n20-42 between the first and second ionization constants decreases with increasing distance between the carboxyl groups?"}
{"id": 1338, "contents": "Acidity of Carboxylic Acids - \n| Name | Structure | $\\mathbf{p K}$ | $\\mathbf{p} \\mathbf{K}_{\\mathbf{2}}$ |\n| :---: | :---: | :---: | :---: |\n| Oxalic | $\\mathrm{HO}_{2} \\mathrm{CCO}_{2} \\mathrm{H}$ | 1.2 | 4.2 |\n\n\n| Succinic | $\\mathrm{HO}_{2} \\mathrm{CCH}_{2} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ | 4.2 | 5.6 |\n| :---: | :---: | :---: | :---: |\n| Adipic | $\\mathrm{HO}_{2} \\mathrm{C}\\left(\\mathrm{CH}_{2}\\right)_{4} \\mathrm{CO}_{2} \\mathrm{H}$ | 4.4 | 5.4 |\n\nReactions of Carboxylic Acids and Nitriles\nPROBLEM How could you convert butanoic acid into the following compounds? Write each step showing the 20-43 reagents needed.\n(a) 1-Butanol\n(b) 1-Bromobutane\n(c) Pentanoic acid\n(d) 1-Butene\n(e) Octane\n\nPROBLEM How could you convert each of the following compounds into butanoic acid? Write each step 20-44 showing all reagents.\n(a) 1-Butanol\n(b) 1-Bromobutane\n(c) 1-Butene\n(d) 1-Bromopropane\n(e) 4-Octene\n\nPROBLEM How could you convert butanenitrile into the following compounds? Write each step showing the 20-45 reagents needed.\n(a) 1-Butanol\n(b) Butylamine\n(c) 2-Methyl-3-hexanone"}
{"id": 1339, "contents": "Acidity of Carboxylic Acids - \nPROBLEM How would you prepare the following compounds from benzene? More than one step is needed in 20-46 each case.\n(a) $m$-Chlorobenzoic acid\n(b) p-Bromobenzoic acid\n(c) Phenylacetic acid, $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$\n\nPROBLEM Predict the product of the reaction of $p$-methylbenzoic acid with each of the following:\n20-47 (a) $\\mathrm{LiAlH}_{4}$, then $\\mathrm{H}_{3} \\mathrm{O}^{+}$\n(b) $N$-Bromosuccinimide in $\\mathrm{CCl}_{4}$\n(c) $\\mathrm{CH}_{3} \\mathrm{MgBr}$ in ether, then $\\mathrm{H}_{3} \\mathrm{O}^{+}$\n(d) $\\mathrm{KMnO}_{4}, \\mathrm{H}_{3} \\mathrm{O}^{+}$\n\nPROBLEM Using ${ }^{13} \\mathrm{CO}_{2}$ as your only source of labeled carbon, along with any other compounds needed, how\n20-48 would you synthesize the following compounds?\n(a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2}{ }^{13} \\mathrm{CO}_{2} \\mathrm{H}$ (b) $\\mathrm{PRODCH}_{3}{ }^{13} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$\n\nPROBLEM How would you carry out the following transformations?\n20-49\n\n\n\nPROBLEM Which method-Grignard carboxylation or nitrile hydrolysis-would you use for each of the 20-50 following reactions? Explain.\n(a)\n\n(b)\n\n(c)\n\n(d) $\\mathrm{HOCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{Br} \\longrightarrow \\mathrm{HOCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$"}
{"id": 1340, "contents": "Acidity of Carboxylic Acids - \nPROBLEM 1,6-Hexanediamine, a starting material needed for making nylon, can be made from 1,3-butadiene.\n20-51 How would you accomplish the synthesis?\n\n\nPROBLEM 3-Methyl-2-hexenoic acid (mixture of $E$ and $Z$ isomers) has been identified as the substance 20-52 responsible for the odor of human sweat. Synthesize the compound from starting materials having five or fewer carbons."}
{"id": 1341, "contents": "Spectroscopy - \nPROBLEM Propose a structure for a compound $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{2}$ that dissolves in dilute NaOH and shows the following 20-53 ${ }^{1} \\mathrm{H}$ NMR spectrum: $1.08 \\delta(9 \\mathrm{H}$, singlet), $2.2 \\delta(2 \\mathrm{H}$, singlet), and $11.2 \\delta(1 \\mathrm{H}$, singlet $)$.\n\nPROBLEM What spectroscopic method could you use to distinguish among the following three isomeric acids?\n20-54 Tell what characteristic features you would expect for each acid.\n$\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{3} \\mathrm{CO}_{2} \\mathrm{H}$\n$\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CHCH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$\n\nPentanoic acid\n3-Methylbutanoic acid\n2,2-Dimethylpropanoic acid\nPROBLEM How would you use NMR (either ${ }^{13} \\mathrm{C}$ or ${ }^{1} \\mathrm{H}$ ) to distinguish between the following pairs of isomers?\n20-55 (a)\n\nand\n\n(c) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ and\n$\\mathrm{HOCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CHO}$\n(d)\n$\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{C}=\\mathrm{CHCH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$\nand\n\n\nPROBLEM Compound $\\mathbf{A}, \\mathrm{C}_{4} \\mathrm{H}_{8} \\mathrm{O}_{3}$, has infrared absorptions at 1710 and 2500 to $3100 \\mathrm{~cm}^{-1}$ and has the ${ }^{1} \\mathrm{H}$ 20-56 NMR spectrum shown. Propose a structure for A."}
{"id": 1342, "contents": "General Problems - \nPROBLEM A chemist in need of 2,2-dimethylpentanoic acid decided to synthesize some by reaction of 20-57 2-chloro-2-methylpentane with NaCN , followed by hydrolysis of the product. After the reaction sequence was carried out, however, none of the desired product could be found. What do you suppose went wrong?\n\nPROBLEM Show how you might prepare the anti-inflammatory agent ibuprofen starting from 20-58 isobutylbenzene. More than one step is needed.\n\n\nPROBLEM The following synthetic schemes all have at least one flaw in them. What is wrong with each?\n20-59 (a)\n\n(b)\n\n(c)\n\n\nPROBLEM p-Aminobenzoic acid (PABA) was once widely used as a sunscreen agent. Propose a synthesis of\n20-60 PABA starting from toluene.\nPROBLEM Propose a synthesis of the anti-inflammatory drug fenclorac from phenylcyclohexane."}
{"id": 1343, "contents": "20-61 - \nPROBLEM The $\\mathrm{p} K_{\\mathrm{a}}$ 's of five $p$-substituted benzoic acids $\\left(\\mathrm{YC}_{6} \\mathrm{H}_{4} \\mathrm{CO}_{2} \\mathrm{H}\\right)$ are listed below. Rank the corresponding\n20-62 substituted benzenes $\\left(\\mathrm{YC}_{6} \\mathrm{H}_{5}\\right)$ in order of their increasing reactivity toward electrophilic aromatic substitution. If benzoic acid has $\\mathrm{p} K_{\\mathrm{a}}=4.19$, which of the substituents are activators and which are deactivators?\n\n| Substituent $\\mathbf{Y}$ |  |\n| :---: | :---: |\n| $-\\mathrm{Si}\\left(\\mathrm{CH}_{3}\\right)_{3}$ | 4.27 |\n| $-\\mathrm{CH}=\\mathrm{CHC} \\equiv \\mathrm{N}$ | 4.03 |\n| $-\\mathrm{HgCH}_{3}$ | 4.10 |\n| $-\\mathrm{OSO}_{2} \\mathrm{CH}_{3}$ | 3.84 |\n| $-\\mathrm{PCl}_{2}$ | 3.59 |\n\nPROBLEM How would you carry out the following transformations? More than one step is needed in each case.\n\n20-63 (a)\n\n(b)\n\n\nPROBLEM The following $\\mathrm{p} K_{\\mathrm{a}}$ values have been measured. Explain why a hydroxyl group in the para position 20-64 decreases the acidity while a hydroxyl group in the meta position increases the acidity.\n\n$\\mathrm{p} K_{\\mathrm{a}}=4.48$\n\n$\\mathrm{p} K_{\\mathrm{a}}=4.19$\n\n$\\mathrm{p} K_{\\mathrm{a}}=4.07$\n\nPROBLEM Identify the missing reagents a-f in the following scheme:"}
{"id": 1344, "contents": "20-65 - \nPROBLEM Propose a structure for a compound, $\\mathrm{C}_{4} \\mathrm{H}_{7} \\mathrm{~N}$, that has the following IR and ${ }^{1} \\mathrm{H}$ NMR spectra:\n20-66\n\n\nPROBLEM The two ${ }^{1} \\mathrm{H}$ NMR spectra shown here belong to crotonic acid (trans $-\\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{CHCO}_{2} \\mathrm{H}$ ) and 20-67 methacrylic acid $\\left[\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{C}\\left(\\mathrm{CH}_{3}\\right) \\mathrm{CO}_{2} \\mathrm{H}\\right]$. Which spectrum corresponds to which acid? Explain.\n(a)\n\n(b)\n\n\nPROBLEM The ${ }^{1} \\mathrm{H}$ and ${ }^{13} \\mathrm{C}$ NMR spectra below belong to a compound with formula $\\mathrm{C}_{6} \\mathrm{H}_{10} \\mathrm{O}_{2}$. Propose a 20-68 structure for this compound.\n(a)\n\n(b)"}
{"id": 1345, "contents": "20-65 - \n(b)\n\n\nPROBLEM Propose structures for carboxylic acids that show the following peaks in their ${ }^{13}$ C NMR spectra.\n20-69 Assume that the kinds of carbons $\\left(1^{\\circ}, 2^{\\circ}, 3^{\\circ}\\right.$, or $\\left.4^{\\circ}\\right)$ have been assigned by DEPT-NMR.\n(a) $\\mathrm{C}_{7} \\mathrm{H}_{12} \\mathrm{O}_{2}: 25.5 \\delta\\left(2^{\\circ}\\right), 25.9 \\delta\\left(2^{\\circ}\\right), 29.0 \\delta\\left(2^{\\circ}\\right), 43.1 \\delta\\left(3^{\\circ}\\right), 183.0 \\delta\\left(4^{\\circ}\\right)$\n(b) $\\mathrm{C}_{8} \\mathrm{H}_{8} \\mathrm{O}_{2}: 21.4 \\delta\\left(1^{\\circ}\\right), 128.3 \\delta\\left(4^{\\circ}\\right), 129.0 \\delta\\left(3^{\\circ}\\right), 129.7 \\delta\\left(3^{\\circ}\\right), 143.1 \\delta\\left(4^{\\circ}\\right), 168.2 \\delta\\left(4^{\\circ}\\right)$\n\nPROBLEM Carboxylic acids having a second carbonyl group two atoms away lose $\\mathrm{CO}_{2}$ (decarboxylate) through 20-70 an intermediate enolate ion when treated with base. Write the mechanism of this decarboxylation reaction using curved arrows to show the electron flow in each step.\n\n\nAn enolate ion"}
{"id": 1346, "contents": "CHAPTER 21 <br> Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Reactions - \nFIGURE 21.1 The lives of rock climbers depend on their ropes, typically made of a nylon polymer prepared by a nucleophilic acyl substitution reaction. (credit: modification of work \"110823-A-NR754-001\" by USASOC News Service/Flickr, CC BY 2.0)"}
{"id": 1347, "contents": "CHAPTER CONTENTS - 21.1 Naming Carboxylic Acid Derivatives\n21.2 Nucleophilic Acyl Substitution Reactions\n21.3 Reactions of Carboxylic Acids\n21.4 Chemistry of Acid Halides\n21.5 Chemistry of Acid Anhydrides\n21.6 Chemistry of Esters\n21.7 Chemistry of Amides\n21.8 Chemistry of Thioesters and Acyl Phosphates: Biological Carboxylic Acid Derivatives\n21.9 Polyamides and Polyesters: Step-Growth Polymers\n21.10 Spectroscopy of Carboxylic Acid Derivatives\n\nWHY THIS CHAPTER? Carboxylic acid derivatives are among the most widely occurring of all molecules, both in laboratory chemistry and in biological pathways. Thus, a study of them and their primary reaction-nucleophilic acyl substitution-is fundamental to understanding organic chemistry. We'll begin this chapter by first learning about carboxylic acid derivatives, and we'll then explore the chemistry of acyl substitution reactions.\n\nClosely related to the carboxylic acids and nitriles discussed in the previous chapter are the carboxylic acid\nderivatives, compounds in which an acyl group is bonded to an electronegative atom or substituent that can act as a leaving group in the nucleophilic acyl substitution reaction that we saw briefly in the Preview of Carbonyl Chemistry:\n\n\nMany kinds of acid derivatives are known, but we'll be concerned primarily with four of the most common ones: acid halides, acid anhydrides, esters, and amides. Acid halides and acid anhydrides are used only in the laboratory, while esters and amides are common in both laboratory and biological chemistry. In addition, carboxylic acid derivatives called thioesters and acyl phosphates are encountered primarily in biological chemistry. Note the structural similarity between acid anhydrides and acyl phosphates.\n\n\n\nAcid halide ( $\\mathrm{X}=\\mathrm{Cl}, \\mathrm{Br}$ )\n\n\nAcid anhydride\n\n\nEster\n\n\nAmide\n\n\nThioester\n\n\nAcyl phosphate"}
{"id": 1348, "contents": "Acid Halides, RCOX - \nAcid halides are named by identifying first the acyl group and then the halide. As described in Section 20.1 and shown in TABLE 20.1, the acyl group name is derived from the carboxylic acid name by replacing the -ic acid or -oic acid ending with -oyl, or the -carboxylic acid ending with -carbonyl. To keep things interesting, however, IUPAC recognizes eight exceptions for which a -yl rather than an -oyl ending is used: formic (formyl), acetic (acetyl), propionic (propionyl), butyric (butyryl), oxalic (oxalyl), malonic (malonyl), succinic (succinyl), and glutaric (glutaryl).\n\n\nAcetyl chloride\n\n\nBenzoyl bromide\n\n\nCyclohexanecarbonyl chloride"}
{"id": 1349, "contents": "Acid Anhydrides, $\\mathrm{RCO}_{2} \\mathrm{COR}^{\\prime}$ - \nSymmetrical anhydrides of unsubstituted monocarboxylic acids and cyclic anhydrides of dicarboxylic acids are named by replacing the word acid with anhydride.\n\n\nAcetic anhydride\n\n\nBenzoic anhydride\n\n\nSuccinic anhydride\n\nUnsymmetrical anhydrides-those prepared from two different carboxylic acids-are named by listing the two acids alphabetically and then adding anhydride.\n\n\nAcetic benzoic anhydride"}
{"id": 1350, "contents": "Esters, $\\mathrm{RCO}_{2} \\mathrm{R}^{\\prime}$ - \nEsters are named by first identifying the alkyl group attached to oxygen and then the carboxylic acid, with the -ic acid ending replaced by -ate.\n\n\nEthyl acetate\n\n\nDimethyl malonate\n\ntert-Butyl cyclohexanecarboxylate"}
{"id": 1351, "contents": "Amides, $\\mathrm{RCONH}_{2}$ - \nAmides with an unsubstituted $-\\mathrm{NH}_{2}$ group are named by replacing the -oic acid or -ic acid ending with -amide, or by replacing the -carboxylic acid ending with -carboxamide.\n\n\nAcetamide\n\n\nHexanamide\n\n\nCyclopentanecarboxamide\n\nIf the nitrogen atom is further substituted, the compound is named by first identifying the substituent groups and then the parent amide. The substituents are preceded by the letter $N$ to identify them as being directly attached to nitrogen.\n\n\nN -Methylpropanamide\n\n$N, N$-Diethylcyclohexanecarboxamide"}
{"id": 1352, "contents": "Thioesters, RCOSR' - \nThioesters are named like the corresponding esters. If the related ester has a common name, the prefix thiois added to the name of the carboxylate: acetate becomes thioacetate, for instance. If the related ester has a systematic name, the -oate or -carboxylate ending is replaced by -thioate or -carbothioate: butanoate becomes butanethioate and cyclohexanecarboxylate becomes cyclohexanecarbothioate, for instance.\n\n\nMethyl thioacetate\n\n\nEthyl butanethioate\n\n\nMethyl cyclohexanecarbothioate"}
{"id": 1353, "contents": "Acyl Phosphates, $\\mathrm{RCO}_{2} \\mathrm{PO}_{3}{ }^{2-}$ and $\\mathrm{RCO}_{2} \\mathrm{PO}_{3} \\mathrm{R}^{r}$ - \nAcyl phosphates are named by citing the acyl group and adding the word phosphate. If an alkyl group is attached to one of the phosphate oxygens, it is identified after the name of the acyl group. In biological chemistry, acyl\nadenosyl phosphates are particularly common.\n\n\nBenzoyl phosphate\n\n\nAcetyl adenosyl phosphate\n\nA summary of nomenclature rules for carboxylic acid derivatives is given in TABLE 21.1.\nTABLE 21.1 Nomenclature of Carboxylic Acid Derivatives\n\n| Functional group | Structure | Name ending |\n| :---: | :---: | :---: |\n| Carboxylic acid | <smiles>[R]C(=O)O</smiles> | -ic acid <br> (-carboxylic acid) |\n| Acid halide | <smiles>[R]C([X])=O</smiles> | -oyl halide <br> (-carbonyl halide) |\n| Acid anhydride | <smiles>[R]C(=O)OC([R])[R]</smiles> | anhydride |\n| Amide | <smiles>R16N</smiles> | -amide <br> (-carboxamide) |\n| Ester | <smiles>[R]OC([R])=O</smiles> | -oate <br> (-carboxylate) |\n| Thioester | <smiles>[R]C(=O)[Sn]</smiles> | -thioate <br> (-carbothioate) |\n| Acyl phosphate | <smiles></smiles> | -oyl phosphate |\n\nPROBLEM Give IUPAC names for the following substances:\n\n21-1 (a)\n\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n(g)\n\n(h)\n\n(i)"}
{"id": 1354, "contents": "Acyl Phosphates, $\\mathrm{RCO}_{2} \\mathrm{PO}_{3}{ }^{2-}$ and $\\mathrm{RCO}_{2} \\mathrm{PO}_{3} \\mathrm{R}^{r}$ - \nPROBLEM Give IUPAC names for the following substances:\n\n21-1 (a)\n\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n(g)\n\n(h)\n\n(i)\n\n\nPROBLEM Draw structures corresponding to the following names:\n21-2 (a) Phenyl benzoate (b) $N$-Ethyl- $N$-methylbutanamide (c) 2,4-Dimethylpentanoyl chloride\n(d) Methyl 1-methylcyclohexanecarboxylate (e) Ethyl 3-oxopentanoate\n(f) Methyl $p$-bromobenzenethioate (g) Formic propanoic anhydride\n(h) cis-2-Methylcyclopentanecarbonyl bromide"}
{"id": 1355, "contents": "Acyl Phosphates, $\\mathrm{RCO}_{2} \\mathrm{PO}_{3}{ }^{2-}$ and $\\mathrm{RCO}_{2} \\mathrm{PO}_{3} \\mathrm{R}^{r}$ - 21.2 Nucleophilic Acyl Substitution Reactions\nThe addition of a nucleophile to a polar $\\mathrm{C}=\\mathrm{O}$ bond is the key step in three of the four major carbonyl-group reactions. We saw in the chapter on Aldehydes and Ketones: Nucleophilic Addition Reactions that when a nucleophile adds to an aldehyde or ketone, the initially formed tetrahedral intermediate can be protonated to yield an alcohol. When a nucleophile adds to a carboxylic acid derivative, however, a different reaction path is taken. The initially formed tetrahedral intermediate eliminates one of the two substituents originally bonded to the carbonyl carbon, leading to a net nucleophilic acyl substitution reaction (FIGURE 21.2).\n(a) Aldehyde or ketone: nucleophilic addition\n\n(b) Carboxylic acid derivative: nucleophilic acyl substitution\n\n\nFIGURE 21.2 The general mechanisms of nucleophilic addition and nucleophilic acyl substitution reactions. Both reactions begin with addition of a nucleophile to a polar $\\mathrm{C}=\\mathrm{O}$ bond to give a tetrahedral, alkoxide ion intermediate. (a) The intermediate formed from an aldehyde or ketone is protonated to give an alcohol, but (b) the intermediate formed from a carboxylic acid derivative expels a leaving group to give a new carbonyl compound.\n\nThe difference in behavior between aldehydes/ketones and carboxylic acid derivatives is a consequence of structure. Carboxylic acid derivatives have an acyl carbon bonded to a group -Y that can act as a leaving group, often as a stable anion. As soon as the tetrahedral intermediate is formed, the leaving group is expelled to generate a new carbonyl compound. Aldehydes and ketones, however, have no such leaving group, however, and therefore don't undergo substitution.\n\n\nA carboxylic acid derivative\n\n\nAn aldehyde\n\n\nA ketone"}
{"id": 1356, "contents": "Acyl Phosphates, $\\mathrm{RCO}_{2} \\mathrm{PO}_{3}{ }^{2-}$ and $\\mathrm{RCO}_{2} \\mathrm{PO}_{3} \\mathrm{R}^{r}$ - 21.2 Nucleophilic Acyl Substitution Reactions\nA carboxylic acid derivative\n\n\nAn aldehyde\n\n\nA ketone\n\nThe net effect of the addition/elimination sequence is a substitution of the nucleophile for the -Y group that was originally bonded to the acyl carbon. Thus, the overall reaction is superficially similar to the kind of nucleophilic substitution that occurs during an $\\mathrm{S}_{\\mathrm{N}} 2$ reaction (Section 11.3), but the mechanisms of the two reactions are completely different. An $\\mathrm{S}_{\\mathrm{N}} 2$ reaction occurs in a single step by backside displacement of the leaving group, while a nucleophilic acyl substitution takes place in two steps and involves a tetrahedral intermediate.\n\nPROBLEM Show the mechanism of the following nucleophilic acyl substitution reaction, using curved arrows\n21-3 to indicate the electron flow in each step:"}
{"id": 1357, "contents": "Relative Reactivity of Carboxylic Acid Derivatives - \nBoth the initial addition step and the subsequent elimination step can affect the overall rate of a nucleophilic acyl substitution reaction, but the addition is usually the rate-limiting step. Thus, any factor that makes the carbonyl group more reactive toward nucleophiles favors the substitution process.\n\nSteric and electronic factors are both important in determining reactivity. Sterically, we find within a series of similar acid derivatives that unhindered, accessible carbonyl groups react with nucleophiles more readily than do sterically hindered carbonyl groups. The reactivity order is"}
{"id": 1358, "contents": "Reactivity - \nElectronically, we find that strongly polarized acyl compounds react more readily than less polar ones. Thus, acid chlorides are the most reactive because the electronegative chlorine atom withdraws electrons from the carbonyl carbon, whereas amides are the least reactive. Although subtle, electrostatic potential maps of various carboxylic acid derivatives indicate these differences by the relative blueness on the $\\mathrm{C}=\\mathrm{O}$ carbons. Acyl phosphates are hard to place on this scale because they are not often used in the laboratory, but in biological systems they appear to be somewhat more reactive than thioesters.\n\n\n\nAmide\n\n\n\nEster\n\n\n\nThioester\n\n\n\nAcid anhydride\n\n<\n\n\nAcid chloride"}
{"id": 1359, "contents": "Reactivity - \nThe way in which various substituents affect the polarization of a carbonyl group is similar to the way they affect the reactivity of an aromatic ring toward electrophilic substitution (Section 16.4). A chlorine substituent, for example, inductively withdraws electrons from an acyl group in the same way that it withdraws electrons from and thus deactivates an aromatic ring. Similarly, amino, methoxyl, and methylthio substituents donate electrons to acyl groups by resonance in the same way that they donate electrons to, and thus activate, aromatic rings.\n\nAs a consequence of these reactivity differences, it's usually possible to convert a more reactive acid derivative into a less reactive one. Acid chlorides, for instance, can be directly converted into anhydrides, thioesters, esters, and amides, but amides can't be directly converted into esters, thioesters, anhydrides, or acid chlorides. Remembering the reactivity order is therefore a way to keep track of a large number of reactions (FIGURE 21.3). Another consequence, as noted previously, is that only acyl phosphates, thioesters, esters, and amides are commonly found in nature. Acid halides and acid anhydrides react so rapidly with water that they can't exist for long in living organisms.\n\n\nFIGURE 21.3 Interconversions of carboxylic acid derivatives. A more reactive acid derivative can be converted into a less reactive one, but not vice versa.\n\nIn studying the chemistry of carboxylic acid derivatives in the next few sections, we'll be concerned largely with the reactions of just a few nucleophiles and will see that the same kinds of reactions tend to reoccur (FIGURE 21.4).\n\n- Hydrolysis\n- Alcoholysis\n- Aminolysis\n- Reduction\n- Grignard reaction\n\nReaction with water to yield a carboxylic acid\nReaction with an alcohol to yield an ester\nReaction with ammonia or an amine to yield an amide\nReaction with a hydride reducing agent to yield an aldehyde or an alcohol\nReaction with an organometallic reagent to yield a ketone or an alcohol\n\n\nFIGURE 21.4 Some general reactions of carboxylic acid derivatives."}
{"id": 1360, "contents": "Predicting the Product of a Nucleophilic Acyl Substitution Reaction - \nPredict the product of the following nucleophilic acyl substitution reaction of benzoyl chloride with 2-propanol:\n\n\nBenzoyl chloride"}
{"id": 1361, "contents": "Strategy - \nA nucleophilic acyl substitution reaction involves the substitution of a nucleophile for a leaving group in a carboxylic acid derivative. Identify the leaving group ( $\\mathrm{Cl}^{-}$in the case of an acid chloride) and the nucleophile (an alcohol in this case), and replace one by the other. The product is isopropyl benzoate.\n\nSolution\n\n\nPROBLEM Rank the compounds in each of the following sets in order of their expected reactivity toward 21-4 nucleophilic acyl substitution:\n(a)\n\n\n(b)\n\n\n\n\nPROBLEM Predict the products of the following nucleophilic acyl substitution reactions:\n21-5\n\n\nPROBLEM The following structure represents a tetrahedral alkoxide ion intermediate formed by addition of a 21-6 nucleophile to a carboxylic acid derivative. Identify the nucleophile, the leaving group, the starting acid derivative, and the ultimate product."}
{"id": 1362, "contents": "Strategy - 21.3 Reactions of Carboxylic Acids\nThe direct nucleophilic acyl substitution of a carboxylic acid is difficult because -OH is a poor leaving group (Section 11.3). Thus, it's usually necessary to enhance the reactivity of the acid, either by using a strong acid catalyst to protonate the carboxyl and make it a better acceptor or by converting the - OH into a better leaving group. Under the right circumstances, however, acid chlorides, anhydrides, esters, and amides can all be prepared from carboxylic acids by nucleophilic acyl substitution reactions."}
{"id": 1363, "contents": "Conversion of Carboxylic Acids into Acid Chlorides - \nIn the laboratory, carboxylic acids are converted into acid chlorides by treatment with thionyl chloride, $\\mathrm{SOCl}_{2}$.\n\n\n2,4,6-Trimethylbenzoic acid"}
{"id": 1364, "contents": "2,4,6-Trimethylbenzoyl - \nchloride (90\\%)This reaction occurs by a nucleophilic acyl substitution pathway in which the carboxylic acid is first converted into an acyl chlorosulfite intermediate, thereby replacing the -OH of the acid with a much better leaving group. The chlorosulfite then reacts with a nucleophilic chloride ion. You might recall from Section 17.6 that an analogous chlorosulfite is involved in the reaction of an alcohol with $\\mathrm{SOCl}_{2}$ to yield an alkyl chloride.\n\n\nCarboxylic acid\n\nA chlorosulfite\nAcid chloride"}
{"id": 1365, "contents": "Conversion of Carboxylic Acids into Acid Anhydrides - \nAcid anhydrides can be derived from two molecules of carboxylic acid by heating to remove 1 equivalent of water. Because of the high temperatures needed, however, only acetic anhydride is commonly prepared this\nway."}
{"id": 1366, "contents": "Conversion of Carboxylic Acids into Esters - \nPerhaps the most useful reaction of carboxylic acids is their conversion into esters. There are many methods for accomplishing this, including the $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of a carboxylate anion with a primary alkyl halide that we saw in Section 11.3."}
{"id": 1367, "contents": "Sodium butanoate - \nMethyl butanoate (97\\%)\nEsters can also be synthesized by an acid-catalyzed nucleophilic acyl substitution reaction of a carboxylic acid with an alcohol, a process called the Fischer esterification reaction. Unfortunately, the need for an excess of a liquid alcohol as solvent effectively limits the method to the synthesis of methyl, ethyl, propyl, and butyl esters.\n\n\nThe mechanism of the Fischer esterification reaction is shown in FIGURE 21.5. Carboxylic acids are not reactive enough to undergo nucleophilic addition directly, but their reactivity is greatly enhanced in the presence of a strong acid such as HCl or $\\mathrm{H}_{2} \\mathrm{SO}_{4}$. The mineral acid protonates the carbonyl-group oxygen atom, thereby giving the carboxylic acid a positive charge and rendering it much more reactive toward nucleophiles. Subsequent loss of water from the tetrahedral intermediate yields the ester product."}
{"id": 1368, "contents": "FIGURE 21.5 MECHANISM - \nMechanism of Fischer esterification. The reaction is an acid-catalyzed, nucleophilic acyl substitution of a carboxylic acid.\n\n\nThe net effect of Fischer esterification is substitution of an -OH group by -OR'. All steps are reversible, and the reaction typically has an equilibrium constant close to 1 . Thus, the reaction can be driven in either direction by the choice of reaction conditions. Ester formation is favored when a large excess of alcohol is used as solvent, but carboxylic acid formation is favored when a large excess of water is present.\n\nEvidence in support of the mechanism shown in FIGURE 21.5 comes from isotope-labeling experiments. When ${ }^{18} \\mathrm{O}$-labeled methanol reacts with benzoic acid, the methyl benzoate produced is found to be ${ }^{18} \\mathrm{O}$-labeled whereas the water produced is unlabeled. Thus, it is the $\\mathrm{C}-\\mathrm{OH}$ bond of the carboxylic acid that is broken during the reaction rather than the $\\mathrm{CO}-\\mathrm{H}$ bond and the $\\mathrm{RO}-\\mathrm{H}$ bond of the alcohol that is broken rather than the $\\mathrm{R}-\\mathrm{OH}$ bond."}
{"id": 1369, "contents": "Synthesizing an Ester from an Acid - \nHow might you prepare the following ester using a Fischer esterification reaction?"}
{"id": 1370, "contents": "Strategy - \nBegin by identifying the two parts of the ester. The acyl part comes from the carboxylic acid and the -OR part comes from the alcohol. In this case, the target molecule is propyl o-bromobenzoate, so it can be prepared by treating $o$-bromobenzoic acid with 1-propanol."}
{"id": 1371, "contents": "Solution - \nPROBLEM How might you prepare the following esters from the corresponding acids?\n\n21-7 (a)\n\n(b)\n\n(c)\n\n\nPROBLEM If the following molecule is treated with acid catalyst, an intramolecular esterification reaction\n21-8 occurs. What is the structure of the product? (Intramolecular means within the same molecule.)"}
{"id": 1372, "contents": "Conversion of Carboxylic Acids into Amides - \nAmides are difficult to prepare by direct reaction of carboxylic acids with amines because amines are bases that convert acidic carboxyl groups into their unreactive carboxylate anions. Thus, the - OH must be replaced by a better, nonacidic leaving group to carry out a nucleophilic acyl substitution. In practice, amides are often prepared by activating the carboxylic acid with a carbodiimide ( $\\mathrm{R}-\\mathrm{N}=\\mathrm{C}=\\mathrm{N}-\\mathrm{R}$ ), followed by addition of the amine. Dicyclohexylcarbodiimide (DCC) and 1-ethyl-3-(3-dimethylaminopropylcarbodiimide (EDEC) are commonly used.\n\n\nDicyclohexylcarbodiimide 1-Ethyl-3-(3-dimethylaminopropyl)-carbodiimide\n\nAs shown in FIGURE 21.6, the acid first adds to a $\\mathrm{C}=\\mathrm{N}$ double bond of DCC , and nucleophilic acyl substitution by amine then ensues. Alternatively, and depending on the reaction solvent, the reactive acyl intermediate might also react with a second equivalent of carboxylate ion to generate an acid anhydride that then reacts with the amine. The product from either pathway is the same."}
{"id": 1373, "contents": "FIGURE 21.6 MECHANISM - \nMechanism of amide formation by reaction of a carboxylic acid and an amine with dicyclohexylcarbodiimide (DCC).\n\nDicyclohexylcarbodiimide is first protonated by the carboxylic acid to make it a better acceptor.\n\n(2) The carboxylate then adds to the protonated carbodiimide to yield a reactive acylating agent.\n(3) Nucleophilic attack of the amine on the acylating agent gives a tetrahedral intermediate. dicyclohexylurea and gives the amide.\n(4) $\\downarrow$\n\n\n\nAmide\n\n\nDicyclohexylurea\n\nWe'll see in Section 26.7 that this carbodiimide method of amide formation is the key step in the laboratory synthesis of small proteins, or peptides. For instance, when one amino acid with its $\\mathrm{NH}_{2}$ rendered unreactive and a second amino acid with its $-\\mathrm{CO}_{2} \\mathrm{H}$ rendered unreactive are treated with carbodiimide, a dipeptide is formed.\n\n\nAmino acid 1\nAmino acid 2\nA dipeptide"}
{"id": 1374, "contents": "Conversion of Carboxylic Acids into Alcohols - \nWe said in Section 17.4 that carboxylic acids are reduced by $\\mathrm{LiAlH}_{4}$ to give primary alcohols, but we deferred a discussion of the reaction mechanism at that time. In fact, the reduction is a nucleophilic acyl substitution reaction in which -H replaces -OH to give an aldehyde that is further reduced by nucleophilic addition to produce a primary alcohol. The aldehyde intermediate is much more reactive than the starting acid, so it reacts instantly and is not isolated.\n\n\nBecause hydride ion is a base as well as a nucleophile, the actual nucleophilic acyl substitution step takes place on the carboxylate ion rather than on the free carboxylic acid and gives a high-energy dianion intermediate. In this intermediate, the two oxygens are complexed to a Lewis acidic aluminum species. Thus, the reaction is relatively difficult, and acid reductions require higher temperatures and extended reaction times.\n\n\nAlternatively, borane in tetrahydrofuran $\\left(\\mathrm{BH}_{3} / \\mathrm{THF}\\right)$ is a useful reagent for reducing carboxylic acids to primary alcohols. Reaction of an acid with $\\mathrm{BH}_{3} / \\mathrm{THF}$ occurs rapidly at room temperature, and the procedure is often preferred to reduction with $\\mathrm{LiAlH}_{4}$ because of its relative ease and safety. Borane reacts with carboxylic acids faster than with any other functional group, thereby allowing selective transformations such as that on p-nitrophenylacetic acid. If the reduction of $p$-nitrophenylacetic acid were done with $\\mathrm{LiAlH}_{4}$, both the nitro and carboxyl groups would be reduced."}
{"id": 1375, "contents": "Biological Conversions of Carboxylic Acids - \nThe direct conversion of a carboxylic acid to an acyl derivative by nucleophilic acyl substitution does not occur in biological chemistry. As in the laboratory, the acid must first be activated by converting the - OH into a better leaving group. This activation is often accomplished in living organisms by reaction of the acid with adenosine triphosphate (ATP) to give an acyl adenosyl phosphate, or acyl adenylate, a mixed anhydride combining a carboxylic acid and adenosine monophosphate (AMP, also known as adenylic acid). In the biosynthesis of fats, for example, a long-chain carboxylic acid reacts with ATP to give an acyl adenylate, followed by subsequent nucleophilic acyl substitution of a thiol group in coenzyme A to give the corresponding acyl CoA (FIGURE 21.7)."}
{"id": 1376, "contents": "FIGURE 21.7 MECHANISM - \nIn fatty-acid biosynthesis, a carboxylic acid is activated by reaction with ATP to give an acyl adenylate, which undergoes nucleophilic acyl substitution with the -SH group on coenzyme A. (ATP $=$ adenosine triphosphate; $A M P=$ adenosine monophosphate.)\n(1) ATP is activated by coordination to magnesium ion, and nucleophilic addition of a fatty acid carboxylate to phosphorus then yields a pentacoordinate intermediate...\n(2... which expels diphosphate ion ( $\\mathrm{PP}_{\\mathrm{i}}$ ) as leaving group and gives an acyl adenosyl phosphate in a process analogous to a nucleophilic acyl substitution reaction.\n(3) The -SH group of coenzyme A adds to the acyl adenosyl phosphate, giving a tetrahedral alkoxide intermediate . . .\n(4)... which expels adenosine monophosphate (AMP) as leaving group and yields the fatty acyl CoA.\n\n\nATP\n$\\mathrm{Mg}^{2+}$\n(1)\n\n\nPentacoordinate intermediate\n(2)\n\n\nAcetyl adenosyl phosphate"}
{"id": 1377, "contents": "(acyl adenylate) - \n(3)\n\n\u00a9\n\n\nNote that the first step in FIGURE 21.7-reaction of the carboxylate with ATP to give an acyl adenylate-is itself a nucleophilic acyl substitution on phosphorus. The carboxylate first adds to a $\\mathrm{P}=\\mathrm{O}$ double bond, giving a fivecoordinate phosphorus intermediate that expels diphosphate ion as a leaving group."}
{"id": 1378, "contents": "Preparation of Acid Halides - \nAcid chlorides are prepared from carboxylic acids by reaction with thionyl chloride $\\left(\\mathrm{SOCl}_{2}\\right)$, as we saw in the previous section. Similar reaction of a carboxylic acid with phosphorus tribromide ( $\\mathrm{PBr}_{3}$ ) yields the acid bromide."}
{"id": 1379, "contents": "Reactions of Acid Halides - \nAcid halides are among the most reactive of carboxylic acid derivatives and can be converted into many other kinds of compounds by nucleophilic acyl substitution mechanisms. The halogen can be replaced by -OH to yield an acid, by -OCOR to yield an anhydride, by -OR to yield an ester, by $-\\mathrm{NH}_{2}$ to yield an amide, or by R' to yield a ketone. In addition, the reduction of an acid halide yields a primary alcohol, and reaction with a Grignard reagent yields a tertiary alcohol. Although the reactions we'll be discussing in this section are illustrated only for acid chlorides, similar processes take place with other acid halides."}
{"id": 1380, "contents": "Acid anhydride - \nEster\n\n$\\mathrm{R}^{\\prime} \\mathrm{OH}$\n\n\n\nAmide\n\n\n\n\nKetone\n\nConversion of Acid Halides into Carboxylic Acids: Hydrolysis\nAcid chlorides react with water to yield carboxylic acids. This hydrolysis reaction is a typical nucleophilic acyl substitution process and is initiated by attack of water on the acid chloride carbonyl group. The tetrahedral intermediate undergoes elimination of $\\mathrm{Cl}^{-}$and loss of $\\mathrm{H}^{+}$to give the product carboxylic acid plus HCl .\n\n\nBecause HCl is formed during hydrolysis, the reaction is often carried out in the presence of a base such as pyridine or NaOH to remove the HCl and prevent it from causing side reactions."}
{"id": 1381, "contents": "Conversion of Acid Halides into Anhydrides - \nNucleophilic acyl substitution reaction of an acid chloride with a carboxylate anion gives an acid anhydride. Both symmetrical and unsymmetrical acid anhydrides can be prepared."}
{"id": 1382, "contents": "Conversion of Acid Halides into Esters: Alcoholysis - \nAcid chlorides react with alcohols to yield esters in a process analogous to their reaction with water to yield acids. In fact, this reaction is probably the most common method for preparing esters in the laboratory. As with hydrolysis, alcoholysis reactions are usually carried out in the presence of pyridine or NaOH to react with the HCl formed.\n\n\nThe reaction of an alcohol with an acid chloride is strongly affected by steric hindrance. Bulky groups on either partner slow down the reaction considerably, resulting in a reactivity order among alcohols of primary > secondary > tertiary. As a result, it's often possible to selectively esterify an unhindered alcohol in the presence of a more hindered one. This can be important in complex syntheses in which it's sometimes necessary to distinguish between similar functional groups. For example,\n\n\nPROBLEM How might you prepare the following esters using a nucleophilic acyl substitution reaction of an 21-9 acid chloride?\n(a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{CH}_{3}$\n(b) $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$\n(c) Ethyl benzoate\n\nPROBLEM Which method would you choose if you wanted to prepare cyclohexyl benzoate-Fischer\n21-10 esterification or reaction of an acid chloride with an alcohol? Explain."}
{"id": 1383, "contents": "Conversion of Acid Halides into Amides: Aminolysis - \nAcid chlorides react rapidly with ammonia and amines to give amides. As with the acid chloride-plus-alcohol method for preparing esters, this reaction of acid chlorides with amines is the most commonly used laboratory method for preparing amides. Both monosubstituted and disubstituted amines can be used, but not trisubstituted amines ( $\\mathrm{R}_{3} \\mathrm{~N}$ ).\n\n\nBecause HCl is formed during the reaction, two equivalents of the amine must be used. One equivalent reacts with the acid chloride, and one equivalent reacts with the HCl by-product to form an ammonium chloride salt. If, however, the amine component is valuable, amide synthesis is often carried out using one equivalent of the amine plus one equivalent of an inexpensive base such as NaOH . For example, the sedative trimetozine is prepared commercially by reaction of $3,4,5$-trimethoxybenzoyl chloride with the amine morpholine in the presence of one equivalent of NaOH ."}
{"id": 1384, "contents": "Synthesizing an Amide from an Acid Chloride - \nHow might you prepare $N$-methylpropanamide by reaction of an acid chloride with an amine?"}
{"id": 1385, "contents": "Strategy - \nAs its name implies, $N$-methylpropanamide can be made by reaction of methylamine with the acid chloride of propanoic acid."}
{"id": 1386, "contents": "Solution - \nPropanoyl Methylamine N-Methylpropanamide chloride\n\nPROBLEM Write the mechanism of the reaction just shown between 3,4,5-trimethoxybenzoyl chloride and 21-11 morpholine to form trimetozine. Use curved arrows to show the electron flow in each step.\n\nPROBLEM How could you prepare the following amides using an acid chloride and an amine or ammonia?\n21-12 (a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CONHCH}_{3}$\n(b) $N, N$-Diethylbenzamide\n(c) Propanamide"}
{"id": 1387, "contents": "Conversion of Acid Chlorides into Alcohols: Reduction and Grignard Reaction - \nAcid chlorides are reduced by $\\mathrm{LiAlH}_{4}$ to yield primary alcohols. The reaction is of little practical value, however, because the parent carboxylic acids are generally more readily available and can themselves be reduced by $\\mathrm{LiAlH}_{4}$ to yield alcohols.\n\nReduction occurs via a typical nucleophilic acyl substitution mechanism in which a hydride ion ( $\\mathrm{H}:^{-}$) adds to the carbonyl group, yielding a tetrahedral intermediate that expels $\\mathrm{Cl}^{-}$. The net effect is a substitution of -Cl by -H to yield an aldehyde, which is then further reduced by $\\mathrm{LiAlH}_{4}$ in a second step to yield the primary alcohol.\n\n\nBenzoyl chloride\n\n\n\nBenzyl alcohol (96\\%)\n\nGrignard reagents react with acid chlorides to yield tertiary alcohols with two identical substituents. The mechanism of the reaction is similar to that of $\\mathrm{LiAlH}_{4}$ reduction. The first equivalent of Grignard reagent adds to the acid chloride, loss of $\\mathrm{Cl}^{-}$from the tetrahedral intermediate yields a ketone, and a second equivalent of Grignard reagent immediately adds to the ketone to produce an alcohol.\n\n\nConversion of Acid Chlorides into Ketones: Diorganocopper Reaction\nThe ketone intermediate formed in the reaction of an acid chloride with a Grignard reagent can't usually be isolated because addition of the second equivalent of organomagnesium reagent occurs too rapidly. A ketone can, however, be isolated from the reaction of an acid chloride with a lithium diorganocopper (Gilman) reagent, $\\mathrm{Li}^{+} \\mathrm{R}_{2} \\mathrm{Cu}^{-}$. The reaction occurs by initial nucleophilic acyl substitution on the acid chloride by the diorganocopper anion to yield an acyl diorganocopper intermediate, followed by loss of $\\mathrm{R}^{\\prime} \\mathrm{Cu}$ and formation of the ketone."}
{"id": 1388, "contents": "An acid An acyl A ketone chloride <br> diorganocopper - \nThe reaction is generally carried out at $-78^{\\circ} \\mathrm{C}$ in ether solution, and yields are often excellent. For example, manicone, a substance secreted by male ants to coordinate ant pairing and mating, has been synthesized by reaction of lithium diethylcopper with (E)-2,4-dimethyl-2-hexenoyl chloride.\n\n\nNote that the diorganocopper reaction occurs only with acid chlorides. Carboxylic acids, esters, acid anhydrides, and amides do not react with lithium diorganocopper reagents.\n\nPROBLEM How could you prepare the following ketones by reaction of an acid chloride with a lithium 21-13 diorganocopper reagent?"}
{"id": 1389, "contents": "Preparation of Acid Anhydrides - \nAcid anhydrides are typically prepared by nucleophilic acyl substitution reaction of an acid chloride with a carboxylate anion, as we saw in the previous section. Both symmetrical and unsymmetrical acid anhydrides can be prepared in this way."}
{"id": 1390, "contents": "Reactions of Acid Anhydrides - \nThe chemistry of acid anhydrides is similar to that of acid chlorides, although anhydrides react more slowly. Thus, acid anhydrides react with water to form acids, with alcohols to form esters, with amines to form amides, and with $\\mathrm{LiAlH}_{4}$ to form primary alcohols. Only the ester- and amide-forming reactions are commonly used, however.\n\n\nAlcoholysis\nAminolysis"}
{"id": 1391, "contents": "Conversion of Acid Anhydrides into Esters - \nAcetic anhydride is often used to prepare acetate esters from alcohols. For example, aspirin (acetylsalicylic acid) is prepared commercially by the acetylation of $o$-hydroxybenzoic acid (salicylic acid) with acetic anhydride.\n\n\nAspirin (an ester)"}
{"id": 1392, "contents": "Conversion of Acid Anhydrides into Amides - \nAcetic anhydride is also commonly used to prepare $N$-substituted acetamides from amines. For example, acetaminophen, a drug used in over-the-counter analgesics such as Tylenol, is prepared by reaction of $p$-hydroxyaniline with acetic anhydride. Only the more nucleophilic $-\\mathrm{NH}_{2}$ group reacts rather than the less nucleophilic - OH group.\n\n\nNotice in both of the previous reactions that only \"half\" of the anhydride molecule is used, while the other half acts as the leaving group during the nucleophilic acyl substitution step and produces acetate ion as a by-product. Thus, anhydrides are inefficient, and acid chlorides are normally preferred for introducing acyl substituents other than acetyl groups.\n\nPROBLEM Write the mechanism of the reaction between $p$-hydroxyaniline and acetic anhydride to prepare\n21-14 acetaminophen.\nPROBLEM What product would you expect from reaction of one equivalent of methanol with a cyclic\n21-15 anhydride, such as phthalic anhydride (1,2-benzenedicarboxylic anhydride)? What is the fate of the second \"half\" of the anhydride?"}
{"id": 1393, "contents": "Phthalic anhydride - 21.6 Chemistry of Esters\nEsters are among the most widespread of all naturally occurring compounds. Many simple esters are pleasantsmelling liquids that are responsible for the fragrant odors of fruits and flowers. For example, methyl butanoate is found in pineapple oil, and isopentyl acetate is a constituent of banana oil. The ester linkage is also present in animal fats and in many biologically important molecules.\n\n$A$ fat\n$\\left(R=C_{11-17}\\right.$ chains $)$\n\nThe chemical industry uses esters for a variety of purposes. Ethyl acetate, for instance, is a commonly used solvent, and dialkyl phthalates are used as plasticizers to keep polymers from becoming brittle. You may be aware that there is current concern about the possible toxicity of phthalates at high concentrations, although a recent assessment by the U.S. Food and Drug Administration found the risk to be minimal for most people, with the possible exception of male infants.\n\n\nDibutyl phthalate (a plasticizer)"}
{"id": 1394, "contents": "Preparation of Esters - \nEsters are usually prepared from carboxylic acids by the methods already discussed. Thus, carboxylic acids are converted directly into esters by $S_{N} 2$ reaction of a carboxylate ion with a primary alkyl halide or by Fischer esterification of a carboxylic acid with an alcohol in the presence of a mineral acid catalyst. In addition, acid chlorides are converted into esters by treatment with an alcohol in the presence of base (Section 21.4)."}
{"id": 1395, "contents": "Reactions of Esters - \nEsters undergo the same kinds of reactions that we've seen for other carboxylic acid derivatives, but they are less reactive toward nucleophiles than either acid chlorides or anhydrides. All their reactions are applicable to both acyclic and cyclic esters, called lactones."}
{"id": 1396, "contents": "A lactone (cyclic ester) - \nConversion of Esters into Carboxylic Acids: Hydrolysis\nAn ester is hydrolyzed, either by aqueous base or aqueous acid, to yield a carboxylic acid plus an alcohol.\n\n\nEster hydrolysis in basic solution is called saponification, after the Latin word sapo, meaning \"soap.\" We'll see in Section 27.2 that soap is in fact made by boiling animal fat with aqueous base to hydrolyze the ester linkages.\n\nAs shown in FIGURE 21.8, ester hydrolysis occurs through a typical nucleophilic acyl substitution pathway in which hydroxide ion is the nucleophile that adds to the ester carbonyl group to give a tetrahedral intermediate. Loss of alkoxide ion then gives a carboxylic acid, which is deprotonated to give the carboxylate ion. Addition of aqueous HCl , in a separate step after the saponification is complete, protonates the carboxylate ion and gives the carboxylic acid."}
{"id": 1397, "contents": "FIGURE 21.8 MECHANISM - \nMechanism of base-induced ester hydrolysis (saponification).\nNucleophilic addition of hydroxide ion\nto the ester carbonyl group gives the\nusual tetrahedral alkoxide\nintermediate.\n2 Elimination of alkoxide ion then\ngenerates the carboxylic acid.\n3 Alkoxide ion abstracts the acidic\nproton from the carboxylic acid and\nyields a carboxylate ion.\n4 Protonation of the carboxylate ion\nby addition of aqueous mineral acid\nin a separate step then gives the\nfree carboxylic acid.\n\nThe mechanism shown in FIGURE 21.8 is supported by isotope-labeling studies. When ethyl propanoate labeled with ${ }^{18} \\mathrm{O}$ in the ether-like oxygen is hydrolyzed in aqueous NaOH , the ${ }^{18} \\mathrm{O}$ label shows up exclusively in the ethanol product. None of the label remains with the propanoic acid, indicating that saponification occurs by cleavage of the $\\mathrm{C}-\\mathrm{OR}^{\\prime}$ bond rather than the $\\mathrm{CO}-\\mathrm{R}^{\\prime}$ bond.\n\n\nAcid-catalyzed ester hydrolysis can occur by more than one mechanism, depending on the structure of the ester. The usual pathway, however, is just the reverse of a Fischer esterification reaction (Section 21.3). As shown in FIGURE 21.9, the ester is first activated toward nucleophilic attack by protonation of the carboxyl oxygen atom, and nucleophilic addition of water then occurs. Transfer of a proton and elimination of alcohol yields the carboxylic acid. Because this hydrolysis reaction is the reverse of a Fischer esterification reaction, FIGURE 21.9 is the reverse of FIGURE 21.5."}
{"id": 1398, "contents": "FIGURE 21.9 MECHANISM - \nMechanism of acid-catalyzed ester hydrolysis. The forward reaction is a hydrolysis; the back-reaction is a Fischer esterification and is thus the reverse of FIGURE 21.5.\n\n\nEster hydrolysis is common in biological chemistry, particularly in the digestion of dietary fats and oils. We'll save a complete discussion of the mechanistic details of fat hydrolysis until Section 29.2 but will note for now that the reaction is catalyzed by various lipase enzymes and involves two sequential nucleophilic acyl substitution reactions. The first is a transesterification reaction in which an alcohol group on the lipase adds to an ester linkage in the fat molecule to give a tetrahedral intermediate that expels alcohol and forms an acyl enzyme intermediate. The second is an addition of water to the acyl enzyme, followed by expulsion of the enzyme to give a hydrolyzed acid and a regenerated enzyme.\n\n\nPROBLEM Why is the saponification of an ester irreversible? In other words, why doesn't treatment of a\n21-16 carboxylic acid with an alkoxide ion yield an ester?"}
{"id": 1399, "contents": "Conversion of Esters into Amides: Aminolysis - \nEsters react with ammonia and amines to yield amides. The reaction is not often used, however, because it's usually easier to prepare an amide by starting with an acid chloride (Section 21.4)."}
{"id": 1400, "contents": "Conversion of Esters into Alcohols: Reduction - \nEsters are easily reduced by treatment with $\\mathrm{LiAlH}_{4}$ to yield primary alcohols (Section 17.4)."}
{"id": 1401, "contents": "A lactone - \nThe mechanism of ester reduction is similar to that of acid chloride reduction in that a hydride ion first adds to the carbonyl group, followed by elimination of alkoxide ion to yield an aldehyde. Further reduction of the aldehyde gives the primary alcohol.\n\n\nThe aldehyde intermediate can be isolated if 1 equivalent of diisobutylaluminum hydride (DIBAH, or DIBALH ) is used as the reducing agent instead of $\\mathrm{LiAlH}_{4}$. The reaction has to be carried out at $-78{ }^{\\circ} \\mathrm{C}$ to avoid further reduction to the alcohol. Such partial reductions of carboxylic acid derivatives to aldehydes also occur\nin numerous biological pathways, although the substrate is either a thioester or acyl phosphate rather than an ester.\n\n\nPROBLEM What product would you expect from the reaction of butyrolactone with $\\mathrm{LiAlH}_{4}$ ? With DIBAH?\n21-17\n\n\nPROBLEM Show the products you would obtain by reduction of the following esters with $\\mathrm{LiAlH}_{4}$ :\n\n21-18 (a)\n\n(b)\n\n\nConversion of Esters into Alcohols: Grignard Reaction\nEsters react with 2 equivalents of a Grignard reagent to yield a tertiary alcohol in which two of the substituents are identical (Section 17.5). The reaction occurs by the usual nucleophilic substitution mechanism to give an intermediate ketone, which reacts further with the Grignard reagent to yield a tertiary alcohol.\n\n\nPROBLEM What ester and what Grignard reagent might you start with to prepare the following alcohols?\n\n21-19 (a)\n\n(b)\n\n(c)"}
{"id": 1402, "contents": "A lactone - 21.7 Chemistry of Amides\nAmides, like esters, are abundant in all living organisms. Proteins, nucleic acids, and many pharmaceutical agents have amide functional groups. The reason for this abundance of amides is that they are stable in the aqueous conditions found in living organisms. Amides are the least reactive of the common acid derivatives and undergo relatively few nucleophilic acyl substitution reactions.\n\n\nA protein segment\n\n\nBenzylpenicillin (penicillin G)\n\n\nUridine 5 '-phosphate (a ribonucleotide)"}
{"id": 1403, "contents": "Preparation of Amides - \nAmides are usually prepared by reaction of an amine with an acid chloride (Section 21.4). Ammonia, monosubstituted amines, and disubstituted amines all undergo the reaction."}
{"id": 1404, "contents": "Conversion of Amides into Carboxylic Acids: Hydrolysis - \nAmides undergo hydrolysis to yield carboxylic acids plus ammonia or an amine upon heating in either aqueous acid or aqueous base. The conditions required for amide hydrolysis are more extreme than those required for the hydrolysis of acid chlorides or esters, but the mechanisms are similar. Acidic hydrolysis reaction occurs by nucleophilic addition of water to the protonated amide, followed by transfer of a proton from oxygen to nitrogen to make the nitrogen a better leaving group, and subsequent elimination. The steps are reversible, with the equilibrium shifted toward product by protonation of $\\mathrm{NH}_{3}$ in the final step.\n\n\nBasic hydrolysis occurs by nucleophilic addition of $\\mathrm{OH}^{-}$to the amide carbonyl group, followed by elimination of amide ion $\\left({ }^{-} \\mathrm{NH}_{2}\\right)$ and subsequent deprotonation of the initially formed carboxylic acid by ammonia. The steps are reversible, with the equilibrium shifted toward product by the final deprotonation of the carboxylic acid. Basic hydrolysis is substantially more difficult than the analogous acid-catalyzed reaction because amide ion is a very poor leaving group, making the elimination step difficult.\n\n\nAmide hydrolysis is common in biological chemistry. Just as the hydrolysis of esters is the initial step in the digestion of dietary fats, the hydrolysis of amides is the initial step in the digestion of dietary proteins. The reaction is catalyzed by protease enzymes and occurs by a mechanism almost identical to what we just saw for fat hydrolysis. That is, an initial nucleophilic acyl substitution of an alcohol group in the enzyme on an amide linkage in the protein gives an acyl enzyme intermediate, which then undergoes hydrolysis."}
{"id": 1405, "contents": "Conversion of Amides into Carboxylic Acids: Hydrolysis - \nConversion of Amides into Amines: Reduction\nLike other carboxylic acid derivatives, amides can be reduced by $\\mathrm{LiAlH}_{4}$. The product of the reduction, however, is an amine rather than an alcohol. The net effect of an amide reduction is thus the conversion of the amide carbonyl group into a methylene group $\\left(\\mathrm{C}=\\mathrm{O} \\longrightarrow \\mathrm{CH}_{2}\\right)$. This kind of reaction is specific to amides and does not occur with other carboxylic acid derivatives.\n\n$\\boldsymbol{N}$-Methyldodecanamide\nDodecylmethylamine (95\\%)\nAmide reduction occurs by nucleophilic addition of hydride ion to the amide carbonyl group, followed by expulsion of the oxygen atom as an aluminate anion leaving group to give an iminium ion intermediate. The intermediate iminium ion is further reduced by $\\mathrm{LiAlH}_{4}$ to yield the amine.\n\n\nThe reaction is effective with both acyclic and cyclic amides, or lactams, and is a good method for preparing cyclic amines."}
{"id": 1406, "contents": "Synthesizing an Amine from an Amide - \nHow could you prepare $N$-ethylaniline by reduction of an amide with $\\mathrm{LiAlH}_{4}$ ?\n\n$N$-Ethylaniline"}
{"id": 1407, "contents": "Strategy - \nReduction of an amide with $\\mathrm{LiAlH}_{4}$ yields an amine. To find the starting material for synthesis of N -ethylaniline, look for a $\\mathrm{CH}_{2}$ position next to the nitrogen atom and replace that $\\mathrm{CH}_{2}$ by $\\mathrm{C}=\\mathrm{O}$. In this case, the amide is $N$-phenylacetamide.\n\nSolution\n\n\nPROBLEM How would you convert $N$-ethylbenzamide to each of the following products?\n21-20 (a) Benzoic acid\n(b) Benzyl alcohol\n(c) $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CH}_{2} \\mathrm{NHCH}_{2} \\mathrm{CH}_{3}$\n\nPROBLEM How would you use the reaction of an amide with $\\mathrm{LiAlH}_{4}$ as the key step in going from\n21-21 bromocyclohexane to ( $N, N$-dimethylaminomethyl)cyclohexane? Write all the steps in the reaction sequence.\n\n(N,N-Dimethylaminomethyl)cyclohexane"}
{"id": 1408, "contents": "Strategy - 21.8 Chemistry of Thioesters and Acyl Phosphates: Biological Carboxylic Acid Derivatives\nAs mentioned in the chapter introduction, the substrate for a nucleophilic acyl substitution reaction in living organisms is generally either a thioester ( $\\mathrm{RCOSR}{ }^{\\prime}$ ) or an acyl phosphate ( $\\mathrm{RCO}_{2} \\mathrm{PO}_{3}{ }^{2-}$ or $\\mathrm{RCO}_{2} \\mathrm{PO}_{3} \\mathrm{R}^{\\prime-}$ ). Neither is as reactive as an acid chloride or acid anhydride, yet both are stable enough to exist in living organisms while still reactive enough to undergo acyl substitution.\n\nAcyl CoA's, such as acetyl CoA, are the most common thioesters in nature. Coenzyme A, abbreviated CoA, is a thiol formed by a phosphoric anhydride linkage ( $\\mathrm{O}=\\mathrm{P}-\\mathrm{O}-\\mathrm{P}=\\mathrm{O}$ ) between phosphopantetheine and adenosine $3^{\\prime}, 5^{\\prime}$-bisphosphate. (The prefix bis- means \"two\" and indicates that adenosine $3^{\\prime}, 5^{\\prime}$-bisphosphate has two phosphate groups, one on C3' and one on $\\mathrm{C}^{\\prime}$.) Reaction of coenzyme A with an acyl phosphate or acyl adenylate gives acyl CoA (FIGURE 21.10). As we saw in Section 21.3 (FIGURE 21.7), formation of the acyl adenylate occurs by reaction of a carboxylic acid with ATP and is itself a nucleophilic acyl substitution reaction that takes place on phosphorus.\n\nAdenosine $\\mathbf{3}^{\\prime}$, $\\mathbf{5}^{\\prime}$-bisphosphate"}
{"id": 1409, "contents": "Coenzyme A (CoA) - \nAcetyl adenylate"}
{"id": 1410, "contents": "Acetyl CoA - \nFIGURE 21.10 Formation of the thioester acetyl CoA by nucleophilic acyl substitution reaction of coenzyme $A$ (CoA) with acetyl adenylate.\n\nOnce formed, an acyl CoA is a substrate for further nucleophilic acyl substitution reactions. For example, N -acetylglucosamine, a component of cartilage and other connective tissues, is synthesized by an aminolysis reaction between glucosamine and acetyl CoA.\n\n\nAnother example of a nucleophilic acyl substitution reaction on a thioester-this one a substitution by hydride ion to effect the partial reduction of a thioester to an aldehyde-occurs in the biosynthesis of mevaldehyde, an intermediate in terpenoid synthesis, which we'll discuss in some detail in Section 27.5. In this reaction, (3S)-3-hydroxy-3-methylglutaryl CoA is reduced by hydride donation from NADPH.\n\n\nPROBLEM Write the mechanism of the reaction shown in Figure 21.10 between coenzyme A and acetyl\n21-22 adenylate to give acetyl CoA."}
{"id": 1411, "contents": "Acetyl CoA - 21.9 Polyamides and Polyesters: Step-Growth Polymers\nWhen an amine reacts with an acid chloride, an amide is formed. What would happen, though, if a diamine and a diacid chloride were allowed to react? Each partner would form two amide bonds, linking more and more molecules together until a giant polyamide resulted. In the same way, reaction of a diol with a diacid would lead to a polyester.\n\n\nA diamine A diacid chloride\n\nA polyamide (nylon)\n\n\nA diacid\nA polyester\nThere are two main classes of synthetic polymers: chain-growth polymers and step-growth polymers. The polyethylene and other alkene and diene polymers that we saw in Section 8.10 and Section 14.6 are called chain-growth polymers because they are produced in chain-reaction processes. An initiator adds to a $\\mathrm{C}=\\mathrm{C}$ bond to give a reactive intermediate, which adds to a second alkene molecule to produce a new intermediate, which adds to a third molecule, and so on. By contrast, polyamides and polyesters are step-growth polymers because each bond in the polymer is formed independently in a discrete step, often the nucleophilic acyl substitution reaction of a carboxylic acid derivative. Some commercially important step-growth polymers are shown in TABLE 21.2.\n\nTABLE 21.2 Some Common Step-Growth Polymers and Their Uses"}
{"id": 1412, "contents": "Acetyl CoA - 21.9 Polyamides and Polyesters: Step-Growth Polymers\nTABLE 21.2 Some Common Step-Growth Polymers and Their Uses\n\n| Monomers | Structure | Polymer | Uses |\n| :---: | :---: | :---: | :---: |\n| Adipic acid <br> Hexamethylenediamine | <smiles>O=CCCCCC[Ca]=O</smiles> $\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}$ | Nylon 66 | Fibers, clothing, tire cord |\n| Dimethyl terephthalate | <smiles>COC(=O)c1ccc(C(=O)OC)cc1</smiles> | Dacron, Mylar, Terylene | Fibers, clothing, films, tire cord |\n| Ethylene glycol | $\\mathrm{HOCH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}$ |  |  |\n\nTABLE 21.2 Some Common Step-Growth Polymers and Their Uses\nMonomers"}
{"id": 1413, "contents": "Polyamides (Nylons) - \nThe best known step-growth polymers are the polyamides, or nylons, first prepared in 1930 by Wallace Carothers at the DuPont Company by heating a diamine with a diacid. For instance, nylon 66 is prepared by reaction of adipic acid (hexanedioic acid) with hexamethylenediamine (1,6-hexanediamine) at $280{ }^{\\circ} \\mathrm{C}$. The designation \" 66 \" indicates the number of carbon atoms in the diamine (the first 6 ) and the diacid (the second 6).\n\n\nNylons are used both in engineering applications and in making fibers. A combination of high impact strength and abrasion resistance makes nylon an excellent metal substitute for bearings and gears. As fiber, nylon is used in a variety of applications, from clothing to tire cord to ropes."}
{"id": 1414, "contents": "Polyesters - \nThe most generally useful polyester is made by reaction between dimethyl terephthalate (dimethyl 1,4-benzenedicarboxylate) and ethylene glycol (1,2-ethanediol). The product is used under the trade name Dacron to make clothing fiber or tire cord and under the name Mylar to make recording tape. The tensile strength of poly(ethylene terephthalate) film is nearly equal to that of steel.\n\n\nLexan, a polycarbonate prepared from diphenyl carbonate and bisphenol A, is another commercially valuable polyester. Lexan has an unusually high impact strength, making it valuable for use in bicycle helmets and laptop cases."}
{"id": 1415, "contents": "Sutures and Biodegradable Polymers - \nBecause plastics are too often thrown away rather than recycled, much work has been carried out on developing biodegradable polymers, which can be broken down rapidly in landfills by soil microorganisms. Among the most common biodegradable polymers are poly(glycolic acid) (PGA), poly(lactic acid) (PLA), and poly(hydroxybutyrate) (PHB). All are polyesters and are therefore susceptible to hydrolysis of their ester links. Copolymers of PGA with PLA have found a particularly wide range of uses. A 90/10 copolymer of poly(glycolic acid) with poly(lactic acid) is used to make absorbable sutures, for instance. The sutures are entirely hydrolyzed and absorbed by the body within 90 days after surgery.\n\n\nGlycolic acid\n\n\nPoly(glycolic acid)\n\n\nLactic acid\n$\\downarrow$ Heat\n\n\nPoly(lactic acid)\n\n\n3-Hydroxybutyric acid\n$\\downarrow$ Heat\n\n\nPoly(hydroxybutyrate)\n\nIn Europe, interest has centered particularly on poly(hydroxybutyrate), which can be made into films for packaging as well as into molded items. The polymer degrades within four weeks in landfills, both by ester hydrolysis and by an E1cB elimination reaction of the oxygen atom $\\beta$ to the carbonyl group. The use of poly(hydroxybutyrate) is limited at present by its cost-about four times that of polypropylene.\n\nPROBLEM Draw structures of the step-growth polymers you would expect to obtain from the following\n21-23 reactions:\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Kevlar, a nylon polymer prepared by reaction of 1,4-benzenedicarboxylic acid (terephthalic acid)\n21-24 with 1,4 -benzenediamine ( $p$-phenylenediamine), is so strong that it's used to make bulletproof vests. Draw the structure of a segment of Kevlar."}
{"id": 1416, "contents": "Infrared Spectroscopy - \nAll carbonyl-containing compounds have intense IR absorptions in the range 1650 to $1850 \\mathrm{~cm}^{-1}$. As shown in TABLE 21.3 the exact position of the absorption provides information about the specific kind of carbonyl group. For comparison, the IR absorptions of aldehydes, ketones, and carboxylic acids are included in the table, along with values for carboxylic acid derivatives.\n\nTABLE 21.3 Infrared Absorptions of Some Carbonyl Compounds\n\n| Carbonyl type | Example | Absorption $\\left(\\mathrm{cm}^{\\mathbf{- 1}}\\right)$ |\n| :--- | :--- | :--- |\n| Saturated acid chloride | Acetyl chloride | 1810 |\n| Aromatic acid chloride | Benzoyl chloride | 1770 |\n| Saturated acid anhydride | Acetic anhydride | 1820,1760 |\n| Saturated ester | Ethyl acetate | 1735 |\n| Aromatic ester | Ethyl benzoate | 1720 |\n| Saturated amide | Acetamide | 1690 |\n| $N$-Substituted amide | Benzamide | 1675 |\n| $N, N$-Disubstituted amide | $N, N$-Dimethylacetamide | 1650 |\n| Saturated aldehyde | Acetaldehyde | 1730 |\n| Saturated ketone | Acetone | 1715 |\n| Saturated carboxylic acid | Acetic acid | 1710 |"}
{"id": 1417, "contents": "Infrared Spectroscopy - \nAcid chlorides are easily detected by their characteristic absorption near $1810 \\mathrm{~cm}^{-1}$. Acid anhydrides can be identified by a pair of absorptions in the carbonyl region, one at $1820 \\mathrm{~cm}^{-1}$ and another at $1760 \\mathrm{~cm}^{-1}$. Note that each of these functional groups has a strong electron-withdrawing group attached to the carbonyl. The inductive withdrawal of electron density shortens the $\\mathrm{C}=\\mathrm{O}$ bond, thereby raising its stretching frequency. Esters are detected by their absorption at $1735 \\mathrm{~cm}^{-1}$, a position somewhat higher than that for either aldehydes or ketones. Amides, by contrast, absorb near the low-wavenumber end of the carbonyl region, with the degree of substitution on nitrogen affecting the exact position of the IR band. That is, for amides, the delocalization of\nelectron density (resonance) from nitrogen into the carbonyl lengthens the $\\mathrm{C}=\\mathrm{O}$ bond and lowers its stretching frequency.\n\nPROBLEM What kinds of functional groups might compounds have if they show the following IR absorptions?\n21-25 (a) Absorption at $1735 \\mathrm{~cm}^{-1}$ (b) Absorption at $1810 \\mathrm{~cm}^{-1}$\n(c) Absorptions at 2500 to $3300 \\mathrm{~cm}^{-1}$ and $1710 \\mathrm{~cm}^{-1}$\n(d) Absorption at $1715 \\mathrm{~cm}^{-1}$\n\nPROBLEM Propose structures for compounds that have the following formulas and IR absorptions:\n21-26 (a) $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{2}, 1735 \\mathrm{~cm}^{-1}$\n(b) $\\mathrm{C}_{4} \\mathrm{H}_{9} \\mathrm{NO}, 1650 \\mathrm{~cm}^{-1}$\n(c) $\\mathrm{C}_{4} \\mathrm{H}_{5} \\mathrm{ClO}, 1780 \\mathrm{~cm}^{-1}$"}
{"id": 1418, "contents": "Nuclear Magnetic Resonance Spectroscopy - \nHydrogens on the carbon next to a carbonyl group are slightly deshielded and absorb near $2 \\delta$ in the ${ }^{1} \\mathrm{H}$ NMR spectrum, although the exact identity of the carbonyl group can't be determined. FIGURE 21.11 shows the ${ }^{1} \\mathrm{H}$ NMR spectrum of ethyl acetate.\n\n\nFIGURE 21.11 ${ }^{\\mathbf{1}} \\mathbf{H}$ NMR spectrum of ethyl acetate.\nAlthough ${ }^{13} \\mathrm{C}$ NMR is useful for determining the presence or absence of a carbonyl group in a molecule, the identity of the carbonyl group is difficult to determine. Aldehydes and ketones absorb near $200 \\delta$, while the carbonyl carbon atoms of various acid derivatives absorb in the range 160 to $180 \\delta$ (TABLE 21.4).\n\nTABLE $21.4^{13} \\mathrm{C}$ NMR Absorptions of Some Carbonyl Compounds\n\n| Compound | Absorption (\u05e1) |\n| :--- | :--- |\n| Acetic acid | 177.3 |\n| Ethyl acetate | 170.7 |\n| Acetyl chloride | 170.3 |\n| Acetamide | 172.6 |\n| Acetic anhydride | 166.9 |\n| Acetone | 205.6 |\n| Acetaldehyde | 201.0 |"}
{"id": 1419, "contents": "$\\boldsymbol{\\beta}$-Lactam Antibiotics - \nYou should never underestimate the value of hard work and logical thinking, but it's also true that blind luck often plays a role in most real scientific breakthroughs. What has been called \"the supreme example of luck in all scientific history\" occurred in the late summer of 1928, when the Scottish bacteriologist Alexander Fleming went on vacation, leaving in his lab a culture plate recently inoculated with the bacterium Staphylococcus aureus.\n\nWhile Fleming was away, an extraordinary chain of events occurred. First, a nine-day cold spell lowered the laboratory temperature to a point where the Staphylococcus in the culture plate could not grow. During this time, spores from a colony of the mold Penicillium notatum, being grown in a lab on the floor below, wafted up into Fleming's lab and landed in the culture plate. The temperature then rose, and both Staphylococcus and Penicillium began to grow. On returning from vacation, Fleming discarded the plate into a tray of antiseptic, intending to sterilize it. Evidently, though, the plate did not sink deeply enough into the antiseptic, because when Fleming happened to glance at it a few days later, what he saw changed the course of history. He noticed that the growing Penicillium mold appeared to dissolve the colonies of staphylococci.\n\nFleming realized that the Penicillium mold must be producing a chemical that killed the Staphylococcus bacteria, and he spent several years trying to isolate the substance. Finally, in 1939, the Australian pathologist Howard Florey and the German refugee Ernst Chain managed to isolate the active substance, called penicillin. The dramatic ability of penicillin to cure infections in mice was soon demonstrated, and successful tests in humans followed shortly thereafter. By 1943, penicillin was being produced on a large scale for military use in World War II, and by 1944 it was being used on civilians. Fleming, Florey, and Chain shared the 1945 Nobel Prize in Physiology or Medicine."}
{"id": 1420, "contents": "$\\boldsymbol{\\beta}$-Lactam Antibiotics - \nNow called benzylpenicillin, or penicillin G, the substance first discovered by Fleming is but one member of a large class of so-called $\\boldsymbol{\\beta}$-lactam antibiotics, compounds with a four-membered lactam (cyclic amide) ring. The four-membered lactam ring is fused to a five-membered, sulfur-containing ring, and the carbon atom next to the lactam carbonyl group is bonded to an acylamino substituent, RCONH-. This acylamino side chain can be varied in the laboratory to provide many hundreds of penicillin analogs with different biological activity profiles. Ampicillin, for instance, has an $\\alpha$-aminophenylacetamido substituent [ $\\left.\\mathrm{PhCH}\\left(\\mathrm{NH}_{2}\\right) \\mathrm{CONH}-\\right]$.\n\n\nClosely related to the penicillins are the cephalosporins, a group of $\\beta$-lactam antibiotics that contain an unsaturated, six-membered, sulfur-containing ring. Cephalexin, marketed under the trade name Keflex, is an example. Cephalosporins generally have much greater antibacterial activity than penicillins, particularly against resistant strains of bacteria.\n\n\nCephalexin (a cephalosporin)\n\nThe biological activity of penicillins and cephalosporins is due to the presence of the strained $\\beta$-lactam ring, which reacts with and deactivates the transpeptidase enzyme needed to synthesize and repair bacterial cell walls. With the wall either incomplete or weakened, the bacterial cell ruptures and dies.\n\n\n(inactive enzyme)"}
{"id": 1421, "contents": "Key Terms - \n- acid anhydride\n- acid halide\n- acyl phosphate\n- amide\n- carboxylic acid derivative\n- chain-growth polymer\n- ester\n- Fischer esterification reaction\n- lactam\n- lactone\n- nucleophilic acyl substitution reaction\n- saponification\n- step-growth polymer\n- thioester"}
{"id": 1422, "contents": "Summary - \nCarboxylic acid derivatives-compounds in which the -OH group of a carboxylic acid has been replaced by another substituent-are among the most widely occurring of all molecules and are involved in almost all biological pathways. In this chapter, we covered the chemistry necessary for understanding them and thus also necessary for understanding living organisms. Acid halides, acid anhydrides, esters, and amides are the most common such derivatives in the laboratory; thioesters and acyl phosphates are common in biological molecules.\n\nThe chemistry of carboxylic acid derivatives is dominated by the nucleophilic acyl substitution reaction. Mechanistically, these substitutions take place by addition of a nucleophile to the polar carbonyl group of the acid derivative to give a tetrahedral intermediate, followed by expulsion of a leaving group.\n\n\nThe reactivity of an acid derivative toward substitution depends both on the steric environment near the carbonyl group and on the electronic nature of the substituent, Y. The reactivity order is acid halide >acid anhydride $>$ thioester $>$ ester $>$ amide.\n\nThe most common reactions of carboxylic acid derivatives are substitution by water to yield an acid (hydrolysis), by an alcohol to yield an ester (alcoholysis), by an amine to yield an amide (aminolysis), by hydride ion to yield\nan alcohol (reduction), and by an organomagnesium halide to yield an alcohol (Grignard reaction).\nStep-growth polymers, such as polyamides and polyesters, are prepared by reactions between difunctional molecules. Polyamides (nylons) are formed by reaction between a diacid and a diamine; polyesters are formed from a diacid and a diol.\n\nIR spectroscopy is a valuable tool for the structural analysis of acid derivatives. Acid chlorides, anhydrides, esters, and amides all show characteristic IR absorptions that can be used to identify these functional groups."}
{"id": 1423, "contents": "Summary of Reactions - \n1. Reactions of carboxylic acids (Section 21.3)\na. Conversion into acid chlorides\n\nb. Conversion into esters\n\nc. Conversion into amides\n\nd. Reduction to yield primary alcohols\n\n2. Reactions of acid chlorides (Section 21.4)\na. Hydrolysis to yield acids\n\nb. Reaction with carboxylates to yield anhydrides\n\nc. Alcoholysis to yield esters\n\nd. Aminolysis to yield amides\n\ne. Reduction to yield primary alcohols\n\nf. Grignard reaction to yield tertiary alcohols\n\ng. Diorganocopper reaction to yield ketones\n\n3. Reactions of acid anhydrides (Section 21.5)\na. Hydrolysis to yield acids\n\nb. Alcoholysis to yield esters\n\nc. Aminolysis to yield amides\n\n4. Reactions of esters (Section 21.6)\na. Hydrolysis to yield acids\n\nb. Reduction to yield primary alcohols\n\nc. Partial reduction to yield aldehydes\n\nd. Grignard reaction to yield tertiary alcohols\n\n5. Reactions of amides (Section 21.7)\na. Hydrolysis to yield acids\n\nb. Reduction to yield amines"}
{"id": 1424, "contents": "Visualizing Chemistry - \nPROBLEM Name the following compounds:\n21-27\n(a)\n\n(b)\n\n\nPROBLEM How would you prepare the following compounds starting with an appropriate carboxylic acid and\n21-28 any other reagents needed? (reddish brown $=\\mathrm{Br}$.)\n(a)\n\n(b)\n\n\nPROBLEM The following structure represents a tetrahedral alkoxide-ion intermediate formed by addition of a\n21-29 nucleophile to a carboxylic acid derivative. Identify the nucleophile, the leaving group, the starting acid derivative, and the ultimate product (green $=\\mathrm{Cl}$ ).\n\n\nPROBLEM Electrostatic potential maps of a typical amide (acetamide) and an acyl azide (acetyl azide) are\n21-30 shown. Which of the two do you think is more reactive in nucleophilic acyl substitution reactions? Explain.\n\nAcetamide"}
{"id": 1425, "contents": "Mechanism Problems - \nPROBLEM Predict the product(s) and write the mechanism for the following reactions:\n21-31 (a)\n\n(b)\n\n\nPROBLEM Predict the product(s) and write the mechanism for the following reactions:"}
{"id": 1426, "contents": "21-32 (a) - \n(b)\n\n\nPROBLEM Predict the product(s) and write the mechanism for each of the following reactions:\n21-33 (a)\n\n\n\nPROBLEM Predict the product(s) and write the mechanism for the following reactions:\n21-34 (a)\n\n(b)\n\n\nPROBLEM Pivalic mixed anhydrides are often used to form amide bonds between amino acids. Unlike with a\n21-35 symmetrical anhydride, this reaction is highly regioselective, with the nucleophile adding only to the amino-acid carbonyl. Provide the complete mechanism for the following reaction and explain the regioselectivity.\n\n\nPROBLEM When 4-dimethylaminopyridine (DMAP) is added in catalytic amounts to acetic anhydride and an\n21-36 alcohol, it significantly increases the rate of ester formation. The process begins with a reaction between acetic anhydride and DMAP to form a highly reactive acetylpyridinium intermediate that is more reactive than acetic anhydride itself. Propose a mechanism for this process that includes the formation and reaction of the acetylpyridinium intermediate.\n\nPROBLEM Fats are biosynthesized from glycerol 3-phosphate and fatty-acyl CoA's by a reaction sequence that\n21-37 begins with the following step. Show the mechanism of the reaction.\n\n\nPROBLEM Treatment of an $\\alpha$-amino acid with DCC yields a 2,5-diketopiperazine. Propose a mechanism. 21-38\n\n\nPROBLEM Succinic anhydride yields the cyclic imide succinimide when heated with ammonium chloride\n21-39 at $200^{\\circ} \\mathrm{C}$. Propose a mechanism for this reaction. Why do you suppose such a high reaction temperature is required?"}
{"id": 1427, "contents": "21-32 (a) - \nPROBLEM The hydrolysis of a biological thioester to the corresponding carboxylate is often more complex\n21-40 than the overall result might suggest. The conversion of succinyl CoA to succinate in the citric acid cycle, for instance, occurs by initial formation of an acyl phosphate, followed by reaction with guanosine diphosphate (GDP, a relative of adenosine diphosphate [ADP]) to give succinate and guanosine triphosphate (GTP, a relative of ATP). Suggest mechanisms for both steps.\n\n\nSuccinyl CoA\n\n\nAcyl phosphate\n\n\n\nGTP\n\nPROBLEM One step in the gluconeogenesis pathway for the biosynthesis of glucose is the partial reduction\n21-41 of 3-phosphoglycerate to give glyceraldehyde 3-phosphate. The process occurs by phosphorylation with ATP to give 1,3-bisphosphoglycerate, reaction with a thiol group on the enzyme to give an enzyme-bound thioester, and reduction with NADH. Suggest mechanisms for all three reactions.\n\n\n3-Phosphoglycerate\n\n\n1,3-Bisphosphoglycerate\n\n\n(Enzyme-bound thioester)\n\n\nGlyceraldehyde 3-phosphate\n\n\nPROBLEM Bacteria typically develop a resistance to penicillins and other $\\beta$-lactam antibiotics (see Chemistry\n21-42 Matters at the end of this chapter) due to bacterial synthesis of $\\beta$-lactamase enzymes. Tazobactam, however, is able to inhibit the activity of the $\\beta$-lactamase by trapping it, thereby preventing a resistance from developing.\n\n(a) The first step in trapping is reaction of a hydroxyl group on the $\\beta$-lactamase to open the $\\beta$-lactam ring of tazobactam. Show the mechanism.\n(b) The second step is opening the sulfur-containing ring in tazobactam to give an acyclic imine intermediate. Show the mechanism.\n(c) Cyclization of the imine intermediate gives the trapped $\\beta$-lactamase product. Show the mechanism."}
{"id": 1428, "contents": "21-32 (a) - \nPROBLEM The following reaction, called the benzilic acid rearrangement, takes place by typical carbonyl-\n21-43 group reactions. Propose a mechanism ( $\\mathrm{Ph}=$ phenyl).\n\n\nPROBLEM In the iodoform reaction, a triiodomethyl ketone reacts with aqueous NaOH to yield a carboxylate\n21-44 ion and iodoform (triiodomethane). Propose a mechanism for this reaction."}
{"id": 1429, "contents": "Naming Carboxylic Acid Derivatives - \nPROBLEM Give IUPAC names for the following compounds:\n21-45\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n(g)\n\n(h)\n\n\nPROBLEM Draw structures corresponding to the following names:\n21-46 (a) $p$-Bromophenylacetamide (b) $m$-Benzoylbenzamide (c) 2,2-Dimethylhexanamide\n(d) Cyclohexyl cyclohexanecarboxylate (e) Ethyl 2-cyclobutenecarboxylate\n(f) Succinic anhydride\n\nPROBLEM Draw and name compounds that meet the following descriptions:\n21-47 (a) Three acid chlorides having the formula $\\mathrm{C}_{6} \\mathrm{H}_{9} \\mathrm{ClO}$\n(b) Three amides having the formula $\\mathrm{C}_{7} \\mathrm{H}_{11} \\mathrm{NO}$\n\nNucleophilic Acyl Substitution Reactions\nPROBLEM Predict the product, if any, of reaction between propanoyl chloride and the following reagents:\n21-48\n(a) $\\mathrm{Li}(\\mathrm{Ph})_{2} \\mathrm{Cu}$ in ether\n(b) $\\mathrm{LiAlH}_{4}$, then $\\mathrm{H}_{3} \\mathrm{O}^{+}$\n(c) $\\mathrm{CH}_{3} \\mathrm{MgBr}$, then $\\mathrm{H}_{3} \\mathrm{O}^{+}$\n(d) $\\mathrm{H}_{3} \\mathrm{O}^{+}$\n(e) Cyclohexanol (f) Aniline (g) $\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-} \\mathrm{Na}^{+}$"}
{"id": 1430, "contents": "Naming Carboxylic Acid Derivatives - \nPROBLEM Answer Problem 21-48 for reaction of the listed reagents with methyl propanoate.\n21-49\nPROBLEM Answer Problem 21-48 for reaction of the listed reagents with propanamide.\n21-50\nPROBLEM What product would you expect to obtain from Grignard reaction of an excess of phenylmagnesium\n21-51 bromide with dimethyl carbonate, $\\mathrm{CH}_{3} \\mathrm{OCO}_{2} \\mathrm{CH}_{3}$ ?\nPROBLEM How might you prepare the following compounds from butanoic acid?\n21-52 (a) 1-Butanol (b) Butanal (c) 1-Bromobutane (d) Pentanenitrile (e) 1-Butene\n(f) $N$-Methylpentanamide (g) 2-Hexanone (h) Butylbenzene (i) Butanenitrile\n\nPROBLEM Predict the product(s) of the following reactions:\n\n21-53 (a)\n\n\n(b)\n$\\xrightarrow[\\text { 2. } \\mathrm{H}_{3} \\mathrm{O}^{+}]{\\text {1. } \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{MgBr}}$ ?\n\n(d)\n\n\n(e)\n\n(h)\n\n\n\nPROBLEM The following reactivity order has been found for the saponification of alkyl acetates by aqueous\n21-54 NaOH. Explain.\n$\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{CH}_{3}>\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}>\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{CH}\\left(\\mathrm{CH}_{3}\\right)_{2}>\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{C}\\left(\\mathrm{CH}_{3}\\right)_{3}$\n\nPROBLEM Explain the observation that attempted Fischer esterification of 2,4,6-trimethylbenzoic acid with\n21-55 methanol and HCl is unsuccessful. No ester is obtained, and the acid is recovered unchanged. What alternative method of esterification might be successful?"}
{"id": 1431, "contents": "Naming Carboxylic Acid Derivatives - \nPROBLEM Outline methods for the preparation of acetophenone (phenyl methyl ketone) starting from the 21-56 following:\n(a) Benzene\n(b) Bromobenzene\n(c) Methyl benzoate\n(d) Benzonitrile\n(e) Styrene\n\nPROBLEM Treatment of 5 -aminopentanoic acid with DCC (dicyclohexylcarbodiimide) yields a lactam. Show 21-57 the structure of the product and the mechanism of the reaction.\n\nPROBLEM When ethyl benzoate is heated in methanol containing a small amount of HCl , methyl benzoate is 21-58 formed. Propose a mechanism for the reaction.\n\nPROBLEM tert-Butoxycarbonyl azide, a reagent used in protein synthesis, is prepared by treating 21-59 tert-butoxycarbonyl chloride with sodium azide. Propose a mechanism for this reaction."}
{"id": 1432, "contents": "Step-Growth Polymers - \nPROBLEM The step-growth polymer nylon 6 is prepared from caprolactam. The reaction involves initial\n21-60 reaction of caprolactam with water to give an intermediate open-chain amino acid, followed by heating to form the polymer. Propose mechanisms for both steps, and show the structure of nylon 6.\n\n\nCaprolactam\n\nPROBLEM Qiana, a polyamide fiber with a silky texture, has the following structure. What are the monomer\n21-61 units used in the synthesis of Qiana?\n\n\nQiana\n\nPROBLEM What is the structure of the polymer produced by treatment of $\\beta$-propiolactone with a small amount\n21-62 of hydroxide ion?\n\n$\\beta$-Propiolactone\n\nPROBLEM Polyimides with the structure shown are used as coatings on glass and plastics to improve scratch 21-63 resistance. How would you synthesize a polyimide? (See Problem 21-39.)"}
{"id": 1433, "contents": "A polyimide - \nSpectroscopy\nPROBLEM How would you distinguish spectroscopically between the following isomer pairs? Tell what 21-64 differences you would expect to see.\n(a) N -Methylpropanamide and $\\mathrm{N}, \\mathrm{N}$-dimethylacetamide\n(b) 5-Hydroxypentanenitrile and cyclobutanecarboxamide\n(c) 4-Chlorobutanoic acid and 3-methoxypropanoyl chloride\n(d) Ethyl propanoate and propyl acetate\n\nPROBLEM Propose a structure for a compound, $\\mathrm{C}_{4} \\mathrm{H}_{7} \\mathrm{ClO}_{2}$, that has the following IR and ${ }^{1} \\mathrm{H}$ NMR spectra:\n21-65\n\n\n\nPROBLEM Assign structures to compounds with the following ${ }^{1} \\mathrm{H}$ NMR spectra:\n21-66\n(a) $\\mathrm{C}_{4} \\mathrm{H}_{7} \\mathrm{ClO}$\n\nIR: $1810 \\mathrm{~cm}^{-1}$\n\n(b) $\\mathrm{C}_{5} \\mathrm{H}_{7} \\mathrm{NO}_{2}$\n\nIR: 2250, $1735 \\mathrm{~cm}^{-1}$"}
{"id": 1434, "contents": "General Problems - \nPROBLEM The following reactivity order has been found for the basic hydrolysis of $p$-substituted methyl\n21-67 benzoates:\n$\\mathrm{Y}=\\mathrm{NO}_{2}>\\mathrm{Br}>\\mathrm{H}>\\mathrm{CH}_{3}>\\mathrm{OCH}_{3}$\nHow can you explain this reactivity order? Where would you expect $\\mathrm{Y}=\\mathrm{C} \\equiv \\mathrm{N}, \\mathrm{Y}=\\mathrm{CHO}$, and $\\mathrm{Y}=\\mathrm{NH}_{2}$ to be in the reactivity list?\n\n\nPROBLEM When a carboxylic acid is dissolved in isotopically labeled water, the label rapidly becomes\n21-68 incorporated into both oxygen atoms of the carboxylic acid. Explain.\n\n\nPROBLEM We said in Section 21.6 that mechanistic studies on ester hydrolysis have been carried out using\n21-69 ethyl propanoate labeled with ${ }^{18} \\mathrm{O}$ in the ether-like oxygen. Assume that ${ }^{18} \\mathrm{O}$-labeled acetic acid is your only source of isotopic oxygen, and then propose a synthesis of the labeled ethyl propanoate.\n\nPROBLEM Treatment of a carboxylic acid with trifluoroacetic anhydride leads to an unsymmetrical anhydride\n21-70 that rapidly reacts with alcohol to give an ester.\n\n(a) Propose a mechanism for formation of the unsymmetrical anhydride.\n(b) Why is the unsymmetrical anhydride unusually reactive?\n(c) Why does the unsymmetrical anhydride react as indicated rather than giving a trifluoroacetate ester plus carboxylic acid?\n\nPROBLEM Butacetin is an analgesic (pain-killing) agent that is synthesized commercially from 21-71 p-fluoronitrobenzene. Propose a synthesis.\n\n\nButacetin\n\nPROBLEM Phenyl 4-aminosalicylate is a drug used in the treatment of tuberculosis. Propose a synthesis of this\n21-72 compound starting from 4-nitrosalicylic acid."}
{"id": 1435, "contents": "General Problems - \nButacetin\n\nPROBLEM Phenyl 4-aminosalicylate is a drug used in the treatment of tuberculosis. Propose a synthesis of this\n21-72 compound starting from 4-nitrosalicylic acid.\n\n\nPROBLEM $N, N$-Diethyl- $m$-toluamide (DEET) is the active ingredient in many insect-repellent preparations.\n21-73 How might you synthesize this substance from $m$-bromotoluene?\n\n$N, N$-Diethyl-m-toluamide\n\nPROBLEM Tranexamic acid, a drug useful against blood clotting, is prepared commercially from\n21-74 p-methylbenzonitrile. Formulate the steps likely to be used in the synthesis. (Don't worry about cis-trans isomers; heating to $300^{\\circ} \\mathrm{C}$ interconverts the isomers.)\n\n\nPROBLEM One frequently used method for preparing methyl esters is by reaction of carboxylic acids with\n21-75 diazomethane, $\\mathrm{CH}_{2} \\mathrm{~N}_{2}$.\n\n\nThe reaction occurs in two steps: (1) protonation of diazomethane by the carboxylic acid to yield methyldiazonium ion, $\\mathrm{CH}_{3} \\mathrm{~N}_{2}{ }^{+}$, plus a carboxylate ion; and (2) reaction of the carboxylate ion with $\\mathrm{CH}_{3} \\mathrm{~N}_{2}{ }^{+}$.\n(a) Draw two resonance structures of diazomethane, and account for step 1.\n(b) What kind of reaction occurs in step 2?\n\nPROBLEM Draw the structure of the polymer you would expect to obtain from reaction of dimethyl 21-76 terephthalate with a triol such as glycerol. What structural feature would this new polymer have that was not present in Dacron (Table 21.2)? How do you think this new feature might affect the properties of the polymer?"}
{"id": 1436, "contents": "General Problems - \nPROBLEM Assign structures to compounds with the following ${ }^{1} \\mathrm{H}$ NMR spectra:\n21-77 (a) $\\mathrm{C}_{5} \\mathrm{H}_{10} \\mathrm{O}_{2}$\nIR: $1735 \\mathrm{~cm}^{-1}$\n\n(b) $\\mathrm{C}_{11} \\mathrm{H}_{12} \\mathrm{O}_{2}$\n\nIR: $1710 \\mathrm{~cm}^{-1}$\n\n\nPROBLEM Propose structures for compounds with the following ${ }^{1} \\mathrm{H}$ NMR spectra:\n21-78 (a) $\\mathrm{C}_{5} \\mathrm{H}_{9} \\mathrm{ClO}_{2}$\nIR: $1735 \\mathrm{~cm}^{-1}$\n\n(b) $\\mathrm{C}_{7} \\mathrm{H}_{12} \\mathrm{O}_{4}$\n\nIR: $1735 \\mathrm{~cm}^{-1}$\n\n\nPROBLEM Propose a structure for the compound with the formula $\\mathrm{C}_{19} \\mathrm{H}_{9} \\mathrm{NO}_{2}$ and the following IR and NMR 21-79 spectra\n\n\n\n\nPROBLEM Draw the structure of the compound that produced the following spectra. The infrared spectrum has 21-80 strong bands at 1720 and $1738 \\mathrm{~cm}^{-1}$.\n\n\nPROBLEM When an amide is formed from an acid chloride or an anhydride, two equivalents of base are\n21-81 required. However, when an ester is used as the starting material, only one equivalent of base is needed. Explain this reactivity in terms of basicity of the leaving groups.\n\nPROBLEM Epoxy adhesives are prepared in two steps. $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of the disodium salt of bisphenol A with\n21-82 epichlorohydrin forms a \"prepolymer,\" which is then \"cured\" by treatment with a triamine such as $\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NHCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}$."}
{"id": 1437, "contents": "General Problems - \nDraw structures to show how addition of the triamine results in a strengthening of the polymer."}
{"id": 1438, "contents": "CHAPTER 22 - \nCarbonyl Alpha-Substitution Reactions\n\n\nFIGURE 22.1 Tear gas is a controversial riot-control device used by some military and police forces. It is a simple chloroketone made by a carbonyl $\\alpha$-substitution reaction. (credit: modification of work \"Tear Gas in CWB Hennessy Road View\" by The Stand News/Wikimedia Commons, Public Domain)"}
{"id": 1439, "contents": "CHAPTER CONTENTS - 22.1 Keto-Enol Tautomerism\n22.2 Reactivity of Enols: $\\alpha$-Substitution Reactions\n22.3 Alpha Halogenation of Aldehydes and Ketones\n22.4 Alpha Bromination of Carboxylic Acids\n22.5 Acidity of Alpha Hydrogen Atoms: Enolate Ion Formation\n22.6 Reactivity of Enolate Ions\n22.7 Alkylation of Enolate Ions\n\nWHY THIS CHAPTER? As with nucleophilic additions and nucleophilic acyl substitutions, many laboratory schemes, pharmaceutical syntheses, and biochemical pathways make frequent use of carbonyl $\\alpha$-substitution reactions. Their great value comes from the fact that they constitute one of the few general methods for forming carbon-carbon bonds, thereby making it possible to build larger molecules from smaller precursors. In this chapter, we'll see how and why these reactions occur.\n\nWe said in the Preview of Carbonyl Chemistry that much of the chemistry of carbonyl compounds can be explained by just four fundamental reaction types: nucleophilic additions, nucleophilic acyl substitutions, $\\alpha$ substitutions, and carbonyl condensations. Having studied the first two of these reactions in the past three chapters, let's now look in more detail at the third major carbonyl-group process-the $\\boldsymbol{a}$-substitution reaction.\n\nAlpha-substitution reactions occur at the position next to the carbonyl group-the $\\alpha$ position-and involve the substitution of an $\\alpha$ hydrogen atom by an electrophile, E, through either an enol or enolate ion intermediate. Let's begin by learning more about these two species."}
{"id": 1440, "contents": "CHAPTER CONTENTS - 22.1 Keto-Enol Tautomerism\nA carbonyl compound with a hydrogen atom on its $\\alpha$ carbon is in equilibrium with its corresponding enol isomer (Section 9.4). This spontaneous interconversion between two isomers, usually with the change in position of a hydrogen, is called tautomerism, from the Greek tauto, meaning \"the same,\" and meros, meaning \"part.\" The individual keto and enol isomers are called tautomers.\n\n\nNote the difference between tautomers and resonance forms. Tautomers are constitutional isomers-different compounds with different structures-while resonance forms are different representations of a single compound. Tautomers have their atoms arranged differently, while resonance forms differ only in the position of their $\\pi$ and nonbonding electrons.\n\nMost monocarbonyl compounds exist almost entirely in their keto form at equilibrium, and it's usually difficult to isolate the pure enol. Cyclohexanone, for example, contains only about $0.0001 \\%$ of its enol tautomer at room temperature. The percentage of enol tautomer is even lower for carboxylic acids, esters, and amides. The enol can only predominate when stabilized by conjugation or intramolecular hydrogen bonding. Thus, 2,4-pentanedione is about $76 \\%$ enol tautomer. Although enols are scarce at equilibrium, they are nevertheless responsible for much of the chemistry of carbonyl compounds because they are so reactive.\n\n\nCyclohexanone"}
{"id": 1441, "contents": "Acetone - \nKeto-enol tautomerism of carbonyl compounds is catalyzed by both acids and bases. Acid catalysis occurs by protonation of the carbonyl oxygen atom to give an intermediate cation that loses $\\mathrm{H}^{+}$from its $\\alpha$ carbon to yield a neutral enol (FIGURE 22.2a). This proton loss from the cation intermediate is similar to what occurs during an E1 reaction when a carbocation loses $\\mathrm{H}^{+}$to form an alkene (Section 11.10).\n\nBase-catalyzed enol formation occurs because the presence of a carbonyl group makes the hydrogens on the $\\alpha$ carbon weakly acidic. Thus, a carbonyl compound can act as an acid and donate one of its $\\alpha$ hydrogens\nto a sufficiently strong base. The resultant resonance-stabilized anion, an enolate ion, is then protonated to yield a neutral compound. If protonation of the enolate ion takes place on the $\\alpha$ carbon, the keto tautomer is regenerated and no net change occurs. If, however, protonation takes place on the oxygen atom, an enol tautomer is formed (FIGURE 22.2b)."}
{"id": 1442, "contents": "FIGURE 22.2 MECHANISM - \nMechanism of enol formation under both acid-catalyzed and base-catalyzed conditions. (a) Acid catalysis involves (1) initial protonation of the carbonyl oxygen followed by (2) removal of $\\mathrm{H}^{+}$from the $\\alpha$ position. (b) Base catalysis involves (1) initial deprotonation of the $\\alpha$ position to give an enolate ion, followed by (2) reprotonation on oxygen.\n\n\nNote that only the hydrogens on the $\\alpha$ position of carbonyl compounds are acidic. Hydrogens at $\\beta, \\gamma, \\delta$, and so on, aren't acidic and can't be removed by base because the resulting anions can't be resonance-stabilized by the carbonyl group.\n\n\nPROBLEM Draw structures for the enol tautomers of the following compounds:\n22-1 (a) Cyclopentanone (b) Methyl thioacetate (c) Ethyl acetate (d) Propanal (e) Acetic acid (f) Phenylacetone\n\nPROBLEM How many acidic hydrogens does each of the molecules listed in Problem 22-1 have? Identify them. 22-2\n\nPROBLEM Draw structures for all monoenol forms of the following molecule. Which would you expect to be 22-3 most stable? Explain."}
{"id": 1443, "contents": "FIGURE 22.2 MECHANISM - 22.2 Reactivity of Enols: $\\boldsymbol{\\alpha}$-Substitution Reactions\nWhat kind of chemistry do enols have? Because their double bonds are electron-rich, enols behave as nucleophiles and react with electrophiles in much the same way that alkenes do. But because of resonanceelectron donation of a lone pair of electrons on the neighboring oxygen, enols are more electron-rich and correspondingly more reactive than alkenes. Notice in the following electrostatic potential map of ethenol $\\left(\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHOH}\\right)$ how there is a substantial electron density (yellow-red) on the $\\alpha$ carbon.\n\n\nWhen an alkene reacts with an electrophile, $\\mathrm{E}^{+}$, initial addition gives an intermediate cation and subsequent reaction with a nucleophile, such as a halide ion, yields an addition product (Section 7.7). When an enol reacts with an electrophile, however, only the initial addition step is the same (FIGURE 22.3). Instead of reacting with a nucleophile to give an addition product, the intermediate cation loses the -OH proton to give an $\\alpha$-substituted carbonyl compound."}
{"id": 1444, "contents": "FIGURE 22.3 MECHANISM - \nGeneral mechanism of a carbonyl $\\boldsymbol{\\alpha}$-substitution reaction. In step 3, the initially formed cation loses $\\mathrm{H}^{+}$to regenerate a carbonyl compound.\n\nAcid-catalyzed enol formation occurs by the usual mechanism.\n\n\n2 An electron pair from the enol oxygen attacks an electrophile ( $\\mathrm{E}^{+}$), forming a new bond and leaving a cation intermediate that is stabilized by resonance between two forms.\n\nLoss of a proton from oxygen yields the neutral alpha-substitution product as a new $\\mathrm{C}=\\mathrm{O}$ bond is formed.\n(1) $\\downarrow$ Acid catalyst\n\n\u00a9"}
{"id": 1445, "contents": "FIGURE 22.3 MECHANISM - 22.3 Alpha Halogenation of Aldehydes and Ketones\nA particularly common $\\alpha$-substitution reaction in the laboratory is the halogenation of aldehydes and ketones at their $\\alpha$ positions by reaction with $\\mathrm{Cl}_{2}, \\mathrm{Br}_{2}$, or $\\mathrm{I}_{2}$ in acidic solution. Bromine in acetic acid solvent is often used.\n\n\nRemarkably, ketone halogenation also occurs in biological systems, particularly in marine alga, where dibromoacetaldehyde, bromoacetone, 1,1,1-tribromoacetone, and other related compounds have been found.\n\n\n\n\nFrom the Hawaiian alga Asparagopsis taxiformis\nThis form of halogenation is a typical $\\alpha$-substitution reaction that proceeds by acid-catalyzed formation of an enol intermediate, as shown in FIGURE 22.4.\n\n\nEvidence for the mechanism shown in FIGURE 22.4 includes the observation that acid-catalyzed halogenations show second-order kinetics and follow the rate law\n\n$$\n\\text { Reaction rate }=k[\\text { Ketone }]\\left[\\mathrm{H}^{+}\\right]\n$$\n\nIn other words, the rate of halogenation depends only on the concentrations of ketone and acid and is independent of halogen concentration. Since halogen is not involved in the rate-limiting step, chlorination, bromination, and iodination of a given substrate all occur at the same rate.\n\nFurthermore, if an aldehyde or ketone is treated with $\\mathrm{D}_{3} \\mathrm{O}^{+}$, the acidic $\\alpha$ hydrogens are replaced by deuterium. For a given ketone, the rate of deuterium exchange is identical to the rate of halogenation, implying that a common intermediate-presumably the enol-is involved in both processes."}
{"id": 1446, "contents": "FIGURE 22.3 MECHANISM - 22.3 Alpha Halogenation of Aldehydes and Ketones\n$\\alpha$-Bromo ketones are useful in the laboratory because they can be dehydrobrominated by base treatment to yield $\\alpha, \\beta$-unsaturated ketones. For example, 2-methylcyclohexanone gives 2-bromo-2-methylcyclohexanone on halogenation, and the $\\alpha$-bromo ketone gives 2-methyl-2-cyclohexenone when heated in pyridine. The reaction takes place by an E2 elimination pathway (Section 11.8) and is a good method for introducing a $\\mathrm{C}=\\mathrm{C}$ bond into a molecule. Note that bromination of 2-methylcyclohexanone occurs primarily on the more highly substituted $\\alpha$ position because the more highly substituted enol is favored over the less highly substituted one (Section 7.6).\n\n\nPROBLEM Write the complete mechanism for the deuteration of acetone on treatment with $\\mathrm{D}_{3} \\mathrm{O}^{+}$.\n22-4\n\n\nPROBLEM Show how you might prepare 1-penten-3-one from 3-pentanone.\n22-5"}
{"id": 1447, "contents": "FIGURE 22.3 MECHANISM - 22.4 Alpha Bromination of Carboxylic Acids\nThe $\\alpha$ bromination of carbonyl compounds by $\\mathrm{Br}_{2}$ in acetic acid is limited to aldehydes and ketones because acids, esters, and amides don't enolize to a sufficient extent. Carboxylic acids, however, can be $\\alpha$ brominated by a mixture of $\\mathrm{Br}_{2}$ and $\\mathrm{PBr}_{3}$ in the Hell-Volhard-Zelinskii (HVZ) reaction.\n\n\nThe Hell-Volhard-Zelinskii reaction is a bit more complex than it looks and actually involves $\\alpha$ substitution of an acid bromide enol rather than a carboxylic acid enol. The process begins by reaction of the carboxylic acid with $\\mathrm{PBr}_{3}$ to form an acid bromide plus HBr (Section 21.4). The HBr then catalyzes enolization of the acid bromide, and the resultant enol reacts with $\\mathrm{Br}_{2}$ in an $\\alpha$-substitution reaction to give an $\\alpha$-bromo acid bromide. Addition of water hydrolyzes the acid bromide in a nucleophilic acyl substitution reaction and yields the $\\alpha$-bromo carboxylic acid product.\n\n\nPROBLEM If methanol rather than water is added at the end of a Hell-Volhard-Zelinskii reaction, an ester 22-6 rather than an acid is produced. Show how you would carry out the following transformation, and propose a mechanism for the ester-forming step."}
{"id": 1448, "contents": "FIGURE 22.3 MECHANISM - 22.5 Acidity of Alpha Hydrogen Atoms: Enolate Ion Formation\nAs noted in Section 22.1, a hydrogen on the $\\alpha$ position of a carbonyl compound is weakly acidic and can be removed by a strong base to yield an enolate ion. In comparing acetone ( $\\mathrm{p} K_{\\mathrm{a}}=19.3$ ) with ethane ( $\\mathrm{p} K_{\\mathrm{a}} \\approx 60$ ), for instance, the presence of a neighboring carbonyl group increases the acidity of the ketone over the alkane by a factor of $10^{40}$.\n\n\nAcetone\n$\\left(p K_{a}=19.3\\right)$\n\n\nEthane\n( $\\mathrm{p} K_{\\mathrm{a}} \\approx 60$ )\n\nProton abstraction from a carbonyl compound occurs when the $\\alpha \\mathrm{C}-\\mathrm{H}$ bond is oriented roughly parallel to the $p$ orbitals of the carbonyl group. The $\\alpha$ carbon atom of the enolate ion is $s p^{2}$-hybridized and has a $p$ orbital that overlaps the neighboring carbonyl $p$ orbitals. Thus, the negative charge is shared by the electronegative oxygen atom, and the enolate ion is stabilized by resonance (FIGURE 22.5).\n\n\nFIGURE 22.5 Mechanism of enolate-ion formation by abstraction of an $\\alpha$ proton from a carbonyl compound. The enolate ion is stabilized by resonance, and the negative charge (red) is shared by the oxygen and the $\\alpha$ carbon atom, as indicated in the electrostatic potential map.\n\nBecause carbonyl compounds are only weakly acidic, a strong base is needed for the formation of an enolateion. If an alkoxide ion, such as sodium ethoxide, is used as base, deprotonation of acetone only takes place to the extent of about $0.1 \\%$ because acetone is a weaker acid than ethanol ( $\\mathrm{p} K_{\\mathrm{a}}=16$ ). If, however, a more powerful base is used, then a carbonyl compound is completely converted into its enolate ion."}
{"id": 1449, "contents": "FIGURE 22.3 MECHANISM - 22.5 Acidity of Alpha Hydrogen Atoms: Enolate Ion Formation\nIn practice, the strong base lithium diisopropylamide [ $\\mathrm{LiN}\\left(i-\\mathrm{C}_{3} \\mathrm{H}_{7}\\right)_{2}$; abbreviated LDA] is commonly used for making enolate ions. As the lithium salt of the weak acid diisopropylamine, $\\mathrm{p} K_{\\mathrm{a}}=36$, LDA can readily deprotonate most carbonyl compounds. It is easily prepared by reaction of butyllithium with diisopropylamine and is soluble in organic solvents because of its two alkyl groups.\n\n\nMany types of carbonyl compounds, including aldehydes, ketones, esters, thioesters, acids, and amides, can be converted into enolate ions by reaction with LDA. TABLE 22.1 lists the approximate $\\mathrm{p} K_{\\mathrm{a}}$ values of different types of carbonyl compounds and shows how these values compare to other acidic substances we've seen. Note that nitriles, too, are acidic and can be converted into enolate-like anions.\n\n| Functional group | Example | $\\mathrm{p} K_{\\mathrm{a}}$ |\n| :---: | :---: | :---: |\n| Carboxylic acid | <smiles>CC=O</smiles> | 5 |\n| 1,3-Diketone | <smiles>CC(=O)C=O</smiles> | 9 |\n| 3-Keto ester | <smiles>CC(=O)C=O</smiles> | 11 |\n| 1,3-Diester | <smiles>CC(=O)CC(C)=O</smiles> | 13 |\n| Alcohol | $\\mathrm{CH}_{3} \\mathrm{OH}$ | 16 |\n| Acid chloride | <smiles>CC(Cl)Cl</smiles> | 16 |\n| Aldehyde | <smiles>C=O</smiles> | 17 |\n| Ketone | <smiles>CC(C)=O</smiles> | 19 |"}
{"id": 1450, "contents": "FIGURE 22.3 MECHANISM - 22.5 Acidity of Alpha Hydrogen Atoms: Enolate Ion Formation\n| Functional group | Example | $\\mathrm{p} K_{\\mathrm{a}}$ |\n| :---: | :---: | :---: |\n| Thioester | <smiles>C[AsH2]=O</smiles> | 21 |\n| Ester | <smiles>CC(C)=O</smiles> | 25 |\n| Nitrile | $\\mathrm{CH}_{3} \\mathrm{C}=\\mathrm{N}$ | 25 |\n| $N, N$-Dialkylamide | <smiles>CCCCCC(C)=O</smiles> | 30 |\n| Dialkylamine | $\\mathrm{HN}\\left(\\mathrm{i}-\\mathrm{C}_{3} \\mathrm{H}_{7}\\right)_{2}$ | 36 |\n\nWhen a hydrogen atom is flanked by two carbonyl groups, its acidity is enhanced further. TABLE 22.1 thus shows that 1,3 -diketones ( $\\beta$-diketones), 3 -oxo esters ( $\\beta$-keto esters), and 1,3 -diesters (malonic esters) are more acidic than water. The enolate ions derived from these $\\beta$-dicarbonyl compounds are stabilized by sharing the negative charge of the two neighboring carbonyl oxygens. The enolate ion of 2,4 -pentanedione, for instance, has three resonance forms. Similar resonance forms can be drawn for other doubly stabilized enolate ions."}
{"id": 1451, "contents": "Identifying the Acidic Hydrogens in a Compound - \nIdentify the most acidic hydrogens in each of the following compounds, and rank the compounds in order of increasing acidity:\n(a)\n\n(b)\n\n(c)"}
{"id": 1452, "contents": "Strategy - \nHydrogens on carbon next to a carbonyl group are acidic. In general, a $\\beta$-dicarbonyl compound is most acidic, a ketone or aldehyde is next most acidic, and a carboxylic acid derivative is least acidic. Remember that alcohols, phenols, and carboxylic acids are also acidic because of their -OH hydrogens."}
{"id": 1453, "contents": "Solution - \nThe acidity order is $(\\mathbf{a})>(\\mathbf{c})>(\\mathbf{b})$. Acidic hydrogens are shown in red.\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Identify the most acidic hydrogens in each of the following molecules:\n22-7 (a)\n$\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CHO}$\n(b) $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{C}-\\mathrm{COCH}_{3}$\n(c) $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$\n(d) Benzamide\n(e) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CN}$\n(f) $\\mathrm{CH}_{3} \\mathrm{CON}\\left(\\mathrm{CH}_{3}\\right)_{2}$\n\nPROBLEM Draw a resonance structure of the acetonitrile anion, ${ }^{-}: \\mathrm{CH}_{2} \\mathrm{C} \\equiv \\mathrm{N}$, and account for the acidity of 22-8 nitriles."}
{"id": 1454, "contents": "Solution - 22.6 Reactivity of Enolate Ions\nEnolate ions are more useful than enols for two reasons. First, pure enols can't normally be isolated but are instead generated only as short-lived intermediates in low concentration. By contrast, stable solutions of pure enolate ions are easily prepared from most carbonyl compounds by reaction with a strong base. Second, enolate ions are more reactive than enols and undergo many reactions that enols don't. Whereas enols are neutral, enolate ions are negatively charged, making them much better nucleophiles.\n\nBecause they are resonance hybrids of two nonequivalent forms, enolate ions can be looked at either as vinylic alkoxides ( $\\mathrm{C}=\\mathrm{C}-\\mathrm{O}^{-}$) or as $\\alpha$-keto carbanions ( ${ }^{-} \\mathrm{C}-\\mathrm{C}=\\mathrm{O}$ ). Thus, enolate ions can react with electrophiles either on oxygen or on carbon. Reaction on oxygen yields an enol derivative, while reaction on carbon yields an $\\alpha$-substituted carbonyl compound (FIGURE 22.6). Both kinds of reactivity are known, but reaction on carbon is more common.\n\n\n\n\n\nAn enol derivative\n\n\nAn $\\alpha$-substituted\ncarbonyl compound\n\nFIGURE 22.6 The electrostatic potential map of acetone enolate ion shows how the negative charge is delocalized over both the oxygen and the $\\boldsymbol{\\alpha}$ carbon. As a result, two modes of reaction of an enolate ion with an electrophile $\\mathrm{E}^{+}$are possible. Reaction on carbon to yield an $\\alpha$-substituted carbonyl product is more common.\n\nAs an example of enolate ion reactivity, aldehydes and ketones undergo base-promoted $\\alpha$ halogenation. Even relatively weak bases such as hydroxide ion are effective for halogenation because it's not necessary to convert the ketone completely into its enolate ion. As soon as a small amount of enolate is generated, halogen reacts with it immediately, removing it from the reaction and driving the equilibrium toward further enolate ion\nformation."}
{"id": 1455, "contents": "Solution - 22.6 Reactivity of Enolate Ions\nBase-promoted halogenation of aldehydes and ketones is seldom used in practice because it's difficult to stop the reaction at the monosubstituted product. An $\\alpha$-halogenated ketone is generally more acidic than the starting, unsubstituted ketone because of the electron-withdrawing inductive effect of the halogen atom. Thus, the monohalogenated products are themselves rapidly turned into enolate ions and further halogenated.\n\nIf excess base and halogen are used, a methyl ketone is triply halogenated and then cleaved by base in the haloform reaction. The products are a carboxylic acid plus a so-called haloform (chloroform, $\\mathrm{CHCl}_{3}$; bromoform, $\\mathrm{CHBr}_{3}$; or iodoform, $\\mathrm{CHI}_{3}$ ). Note that the second step of the reaction is a nucleophilic acyl substitution of ${ }^{-} \\mathrm{CX}_{3}$ by ${ }^{-} \\mathrm{OH}$. That is, a halogen-stabilized carbanion acts as a leaving group.\n\n\nPROBLEM Why do you suppose ketone halogenations in acidic media are referred to as being acid-catalyzed, 22-9 whereas halogenations in basic media are base-promoted? In other words, why is a full equivalent of base required for halogenation?"}
{"id": 1456, "contents": "Solution - 22.7 Alkylation of Enolate lons\nPerhaps the most useful reaction of enolate ions is their alkylation by treatment with an alkyl halide, thereby forming a new $\\mathrm{C}-\\mathrm{C}$ bond and joining two smaller pieces into one larger molecule. Alkylation occurs when the nucleophilic enolate ion reacts with the electrophilic alkyl halide in an $\\mathrm{S}_{\\mathrm{N}} 2$ reaction and displaces the leaving group by backside attack.\n\n\nAlkylation reactions are subject to the same constraints that affect all $S_{N} 2$ reactions (Section 11.3). Thus, the leaving group X in the alkylating agent $\\mathrm{R}-\\mathrm{X}$ can be chloride, bromide, or iodide. The alkyl group R should be primary or methyl, and preferably should be allylic or benzylic. Secondary halides react poorly, and tertiary halides don't react at all because a competing E2 elimination of HX occurs instead. Vinylic and aryl halides are also unreactive because a backside approach is sterically prevented.\n\n$$\n\\mathrm{R}-\\mathrm{X}\\left\\{\\begin{array}{l}\n-\\mathrm{X}: \\text { Tosylate }>-\\mathrm{I}>-\\mathrm{Br}>-\\mathrm{Cl} \\\\\n\\mathrm{R}-: \\text { Allylic } \\approx \\text { Benzylic }>\\mathrm{H}_{3} \\mathrm{C}->\\mathrm{RCH}_{2}-\n\\end{array}\\right.\n$$"}
{"id": 1457, "contents": "The Malonic Ester Synthesis - \nOne of the oldest and best known carbonyl alkylation reactions is the malonic ester synthesis, a method for preparing a carboxylic acid from an alkyl halide while lengthening the carbon chain by two atoms.\n\n\nDiethyl propanedioate, commonly called diethyl malonate, or malonic ester, is relatively acidic ( $\\mathrm{p} K_{\\mathrm{a}}=13$ ) because its $\\alpha$ hydrogens are flanked by two carbonyl groups. Thus, malonic ester is easily converted into its enolate ion by reaction with sodium ethoxide in ethanol. The enolate ion, in turn, is a good nucleophile that reacts rapidly with an alkyl halide to give an $\\alpha$-substituted malonic ester. Note in the following examples that the abbreviation \"Et\" is used for an ethyl group, $-\\mathrm{CH}_{2} \\mathrm{CH}_{3}$.\n\n\nThe product of a malonic ester alkylation has one acidic $\\alpha$ hydrogen remaining, so the alkylation process can be repeated to yield a dialkylated malonic ester.\n\n\nOn heating with aqueous hydrochloric acid, the alkylated (or dialkylated) malonic ester undergoes hydrolysis of its two ester groups followed by decarboxylation (loss of $\\mathrm{CO}_{2}$ ) to yield a substituted monocarboxylic acid.\n\n\n$$\n\\begin{array}{lc}\n\\text { An alkylated } & \\text { A carboxylic } \\\\\n\\text { malonic ester } & \\text { acid }\n\\end{array}\n$$\n\nDecarboxylation is not a general reaction of carboxylic acids. Rather, it is unique to compounds that have a second carbonyl group two atoms away from the $-\\mathrm{CO}_{2} \\mathrm{H}$. That is, only substituted malonic acids and $\\beta$-keto acids undergo loss of $\\mathrm{CO}_{2}$ on heating. The decarboxylation reaction occurs by a cyclic mechanism and involves initial formation of an enol, thereby accounting for the need to have a second carbonyl group appropriately positioned."}
{"id": 1458, "contents": "The Malonic Ester Synthesis - \nAs noted previously, the overall effect of malonic ester synthesis is to convert an alkyl halide into a carboxylic acid while lengthening the carbon chain by two atoms ( $\\mathrm{RX} \\rightarrow \\mathrm{RCH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ ).\n\n\nA malonic ester synthesis can also be used to prepare cycloalkanecarboxylic acids. For example, when 1,4 -dibromobutane is treated with diethyl malonate in the presence of two equivalents of sodium ethoxide base, the second alkylation step occurs intramolecularly to yield a cyclic product. Hydrolysis and decarboxylation then give cyclopentanecarboxylic acid. Three-, four-, five-, and six-membered rings can be prepared in this way, but yields decrease for larger ring sizes.\n\n\n1,4-Dibromobutane"}
{"id": 1459, "contents": "Using Malonic Ester Synthesis to Prepare a Carboxylic Acid - \nHow would you prepare heptanoic acid using a malonic ester synthesis?"}
{"id": 1460, "contents": "Strategy - \nA malonic ester synthesis converts an alkyl halide into a carboxylic acid having two more carbons. Thus, a seven-carbon acid chain must be derived from the five-carbon alkyl halide 1-bromopentane.\n\nSolution\n\n$$\n\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{Br}+\\mathrm{CH}_{2}\\left(\\mathrm{CO}_{2} \\mathrm{Et}\\right)_{2} \\xrightarrow[\\text { 2. } \\mathrm{H}_{3} \\mathrm{O}^{+} \\text {, heat }]{\\text { 1. }{ }^{+}{ }^{-} \\text {OEt }} \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{COH}\n$$\n\nPROBLEM How could you use a malonic ester synthesis to prepare the following compounds? Show all steps."}
{"id": 1461, "contents": "22-10 (a) - \n(b)\n\n\n(c)\n\n\nPROBLEM Monoalkylated and dialkylated acetic acids can be prepared by malonic ester synthesis, but 22-11 trialkylated acetic acids $\\left(\\mathrm{R}_{3} \\mathrm{C}-\\mathrm{CO}_{2} \\mathrm{H}\\right)$ can't be prepared. Explain.\n\nPROBLEM How could you use a malonic ester synthesis to prepare the following compound?\n22-12"}
{"id": 1462, "contents": "The Acetoacetic Ester Synthesis - \nJust as the malonic ester synthesis converts an alkyl halide into a carboxylic acid, the acetoacetic ester synthesis converts an alkyl halide into a methyl ketone having three more carbons.\n\n\nEthyl 3-oxobutanoate, commonly called ethyl acetoacetate, or acetoacetic ester, is much like malonic ester in that its $\\alpha$ hydrogens are flanked by two carbonyl groups. It is therefore readily converted into its enolate ion, which can be alkylated by reaction with an alkyl halide. A second alkylation can also be carried out if desired, since acetoacetic ester has two acidic $\\alpha$ hydrogens.\n\n\nOn heating with aqueous HCl , the alkylated (or dialkylated) acetoacetic ester is hydrolyzed to a $\\beta$-keto acid, which then undergoes decarboxylation to yield a ketone product. The decarboxylation occurs in the same way as in the malonic ester synthesis and involves a ketone enol as the initial product.\n\n\nThe three-step sequence of (1) enolate ion formation, (2) alkylation, and (3) hydrolysis/decarboxylation is applicable to all $\\beta$-keto esters with acidic $\\alpha$ hydrogens, not just to acetoacetic ester itself. For example, cyclic $\\beta$-keto esters, such as ethyl 2-oxocyclohexanecarboxylate, can be alkylated and decarboxylated to give 2 -substituted cyclohexanones."}
{"id": 1463, "contents": "Using Acetoacetic Ester Synthesis to Prepare a Ketone - \nHow would you prepare 2-pentanone by an acetoacetic ester synthesis?"}
{"id": 1464, "contents": "Strategy - \nAn acetoacetic ester synthesis yields a methyl ketone by adding three carbons to an alkyl halide.\n\n\nThus, the acetoacetic ester synthesis of 2-pentanone must involve reaction of bromoethane.\nSolution\n\n\nPROBLEM What alkyl halides would you use to prepare the following ketones by an acetoacetic ester\n22-13 synthesis?\n(a)\n\n(b)\n\n\nPROBLEM Which of the following compounds can't be prepared by an acetoacetic ester synthesis? Explain.\n22-14\n(a) Phenylacetone\n(b) Acetophenone\n(c) 3,3-Dimethyl-2-butanone\n\nPROBLEM How would you prepare the following compound using an acetoacetic ester synthesis?\n22-15\n\n\nDirect Alkylation of Ketones, Esters, and Nitriles\nBoth the malonic ester synthesis and the acetoacetic ester synthesis are easy to carry out because they involve relatively acidic dicarbonyl compounds. As a result, sodium ethoxide in ethanol can be used to prepare the\nnecessary enolate ions. Alternatively, however, it's also possible in many cases to directly alkylate the $\\alpha$ position of monocarbonyl compounds. A strong, sterically hindered base such as LDA is needed so that complete conversion to the enolate ion takes place rather than a nucleophilic addition, and a nonprotic solvent must be used.\n\nKetones, esters, and nitriles can all be alkylated using LDA or related dialkylamide bases in tetrahydrofuran (THF). Aldehydes, however, rarely give high yields of pure products because their enolate ions undergo carbonyl condensation reactions instead of alkylation. (We'll study this condensation reaction in the next chapter.) Some specific examples of alkylation reactions are shown.\n\nLactone\n\n\n\n> Ethyl 2-methylpropanoate\n\nEthyl 2,2-dimethylpropanoate (87\\%)\nKetone\n\n\n2,2-Dimethylcyclohexanone (6\\%)"}
{"id": 1465, "contents": "Nitrile - \nPhenylacetonitrile\n2-Phenylpropanenitrile (71\\%)"}
{"id": 1466, "contents": "Nitrile - \nPhenylacetonitrile\n\n\n2-Phenylpropanenitrile (71\\%)\n\nNote in the ketone example that alkylation of 2-methylcyclohexanone leads to a mixture of products because both possible enolate ions are formed. In general, the major product in such cases occurs by alkylation at the less hindered, more accessible position. Thus, alkylation of 2-methylcyclohexanone occurs primarily at C6 (secondary) rather than C 2 (tertiary)."}
{"id": 1467, "contents": "Using an Alkylation Reaction to Prepare a Substituted Ester - \nHow might you use an alkylation reaction to prepare ethyl 1-methylcyclohexanecarboxylate?\n\n\nEthyl 1-methylcyclohexanecarboxylate"}
{"id": 1468, "contents": "Strategy - \nAn alkylation reaction is used to introduce a methyl or primary alkyl group onto the $\\alpha$ position of a ketone, ester, or nitrile by $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of an enolate ion with an alkyl halide. Thus, we need to look at the target molecule and identify any methyl or primary alkyl groups attached to an $\\alpha$ carbon. In the present instance, the target has an $\\alpha$ methyl group, which might be introduced by alkylation of an ester enolate ion with iodomethane."}
{"id": 1469, "contents": "Solution - \nEthyl cyclohexanecarboxylate\n\nEthyl 1-methylcyclo-\nhexanecarboxylate\n\nPROBLEM Show how you might prepare the following compounds using an alkylation reaction as the key step:\n22-16 (a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)"}
{"id": 1470, "contents": "Biological Alkylations - \nAlkylations are rare but not unknown in biological chemistry. One example occurs during biosynthesis of the antibiotic indolmycin from indolylpyruvate when a base abstracts an acidic hydrogen from an $\\alpha$ position and the resultant enolate ion carries out an $\\mathrm{S}_{\\mathrm{N}} 2$ alkylation reaction on the methyl group of $S$-adenosylmethionine (SAM; Section 11.6). Although it's convenient to speak of \"enolate ion\" intermediates in biological pathways, it's unlikely that they exist for long in an aqueous cellular environment. Rather, proton removal and alkylation\nprobably occur at essentially the same time (FIGURE 22.7).\n\n\n\n\nIndolylpyruvate\n\n\nIndolmycin (an antibiotic)\nFIGURE 22.7 The biosynthesis of indolmycin from indolylpyruvate occurs through a pathway that includes an alkylation reaction of a short-lived enolate ion intermediate."}
{"id": 1471, "contents": "Barbiturates - \nThe use of herbal remedies to treat illness and disease goes back thousands of years, but the medical use of chemicals prepared in the laboratory has a much shorter history. Barbiturates, a large class of drugs with a wide variety of uses, constitute one of the earliest successes of medicinal chemistry. The synthesis and medical use of barbiturates goes back to 1904 when Bayer, a German chemical company, first marketed a compound called barbital, trade named Veronal, as a treatment for insomnia. Since that time, more than 2500 different barbiturate analogs have been synthesized by drug companies, more than 50 have been used medicinally, and about a dozen are still in use as anesthetics, anticonvulsants, sedatives, and anxiolytics.\n\n\nFIGURE 22.8 Different barbiturates come in a multitude of colors, giving rise to similarly colorful street names when the drugs are abused. (credit: \"Medications\" by freestocks.org/Flickr, Public Domain)\n\nThe synthesis of barbiturates is relatively simple and relies on reactions that are now familiar: enolate alkylations and nucleophilic acyl substitutions. Starting with diethyl malonate, alkylation of the corresponding enolate ion with simple alkyl halides provides a wealth of different disubstituted malonic esters. Reaction with urea, $\\left(\\mathrm{H}_{2} \\mathrm{~N}\\right)_{2} \\mathrm{C}=\\mathrm{O}$, then gives the product barbiturates by a twofold nucleophilic acyl substitution reaction of the ester groups with the $-\\mathrm{NH}_{2}$ groups of urea (FIGURE 22.9). Amobarbital (Amytal), pentobarbital (Nembutal), and secobarbital (Seconal) are typical examples."}
{"id": 1472, "contents": "Diethyl malonate - \n1. $\\mathrm{Na}^{+-}{ }^{-} \\mathrm{OEt}$\n2. $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{Br}$\n3. $\\mathrm{Na}^{+}-\\mathrm{OEt}$\n4. $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CHCH}_{2} \\mathrm{CH}_{2} \\mathrm{Br}$\n\n\n\n\nAmobarbital\n\n\n\n\n\n\n\nSecobarbital\n\nFIGURE 22.9 The synthesis of barbiturates relies on malonic ester alkylations and nucleophilic acyl substitution reactions. More than 2500 different barbiturates have been synthesized over the past 100 years.\n\n\nBarbital (Veronal), the first barbiturate\n\nIn addition to their prescribed medical uses, many barbiturates have also found widespread illegal use as street drugs. Each barbiturate often comes as a tablet of regulated size, shape, and color. Although still used today, most barbiturates have been replaced by safer, more potent alternatives with markedly different structures."}
{"id": 1473, "contents": "Key Terms - \n```\n- acetoacetic ester synthesis\n- enolate ion\n- \\alpha-substitution reaction\n- malonic ester synthesis\n- decarboxylation\n- tautomer\n- enol\n```"}
{"id": 1474, "contents": "Summary - \nThe $\\alpha$-substitution reaction of a carbonyl compound through either an enol or enolate ion intermediate is one of the four fundamental reaction types in carbonyl-group chemistry.\n\n\nAn enol\nCarbonyl compounds are in an equilibrium with their enols, a process called keto-enol tautomerism. Although enol tautomers are normally present to only a small extent at equilibrium and can't usually be isolated in pure form, they nevertheless contain a highly nucleophilic double bond and react with electrophiles in an $\\boldsymbol{\\sigma}$-substitution reaction. An example is the $\\alpha$ halogenation of ketones on treatment with $\\mathrm{Cl}_{2}, \\mathrm{Br}_{2}$, or $\\mathrm{I}_{2}$ in acid solution. Alpha bromination of carboxylic acids can be similarly accomplished by the Hell-Volhard-Zelinskii (HVZ) reaction, in which an acid is treated with $\\mathrm{Br}_{2}$ and $\\mathrm{PBr}_{3}$. The $\\alpha$-halogenated products can then undergo base-induced E2 elimination to yield $\\alpha, \\beta$-unsaturated carbonyl compounds.\n\nAlpha hydrogen atoms of carbonyl compounds are weakly acidic and can be removed by strong bases, such as lithium diisopropylamide (LDA), to yield nucleophilic enolate ions. The most useful reaction of enolate ions is their $\\mathrm{S}_{\\mathrm{N}} 2$ alkylation with alkyl halides. The Malonic ester synthesis converts an alkyl halide into a carboxylic acid with the addition of two carbon atoms ( $\\mathrm{RX} \\rightarrow \\mathrm{RCH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ ). Similarly, the acetoacetic ester synthesis converts an alkyl halide into a methyl ketone with the addition of three carbon atoms (RX $\\rightarrow$ $\\mathrm{RCH}_{2} \\mathrm{COCH}_{3}$ ). In addition, many carbonyl compounds, including ketones, esters, and nitriles, can be directly alkylated by treatment with LDA and an alkyl halide."}
{"id": 1475, "contents": "Summary of Reactions - \n1. Aldehyde/ketone halogenation (Section 22.3)\n\n2. Hell-Volhard-Zelinskii bromination of acids (Section 22.4)\n\n\n3. Dehydrobromination of $\\alpha$-bromo ketones (Section 22.3)\n\n4. Haloform reaction (Section 22.6)\n\n5. Alkylation of enolate ions (Section 22.7)\na. Malonic ester synthesis\n\nb. Acetoacetic ester synthesis\n\nc. Direct alkylation of ketones\n\nd. Direct alkylation of esters\n\ne. Direct alkylation of nitriles"}
{"id": 1476, "contents": "Visualizing Chemistry - \nPROBLEM Show the steps in preparing each of the following substances using either a malonic ester synthesis 22-17 or an acetoacetic ester synthesis:\n(a)\n\n(b)\n\n\nPROBLEM Unlike most $\\beta$-diketones, the following $\\beta$-diketone has no detectable enol content and is about as 22-18 acidic as acetone. Explain.\n\n\nPROBLEM For a given $\\alpha$ hydrogen atom to be acidic, the $\\mathrm{C}-\\mathrm{H}$ bond must be parallel to the $p$ orbitals of the\n22-19 $\\mathrm{C}=\\mathrm{O}$ double bond (that is, perpendicular to the plane of the adjacent carbonyl group). Identify the most acidic hydrogen atom in the conformation shown for the following structure. Is it axial or equatorial?"}
{"id": 1477, "contents": "Mechanism Problems - \nPROBLEM Draw the corresponding keto or enol tautomer for each of the following molecules.\n\n22-20 (a)\n\n(b)\n\n\n\nPROBLEM Predict the product(s) and show the mechanism for each of the following reactions:\n\n22-21 (a)\n\n(b)\n\n\nPROBLEM Predict the product(s) and show the mechanism for each of the following reactions:\n\n22-22 (a)\n\n(b)\n\n\nPROBLEM The following two optically active $\\beta$-keto acids were decarboxylated under the conditions typically\n22-23 used for the acetoacetic ester synthesis. Will the ketone products be optically active? Explain.\n(a)\n\n(b)\n\n\nPROBLEM In the Hell-Volhard-Zelinskii reaction, only a catalytic amount of $\\mathrm{PBr}_{3}$ is necessary because of the\n22-24 following equilibrium. Propose a mechanism for formation of the equilibrium.\n\n\nPROBLEM When a ketone is treated with a halogen under acidic conditions, the $\\alpha$-monohalogenated product\n22-25 can be obtained in high yield. Under basic conditions however it is difficult to isolate the monohalogenated product. Explain.\n\nPROBLEM Nonconjugated $\\beta, \\gamma$-unsaturated ketones, such as 3 -cyclohexenone, are in an acid-catalyzed\n22-26 equilibrium with their conjugated $\\alpha, \\beta$-unsaturated isomers. Propose a mechanism for the isomerization.\n\n\nPROBLEM 2-substituted 2-cyclopentenones can be interconverted with 5-substituted 2-cyclopentenones 22-27 under basic conditions. Propose a mechanism for this isomerization.\n\n\n\nPROBLEM Using curved arrows, propose a mechanism for the following reaction, one of the steps in the\n22-28 metabolism of the amino acid alanine.\n\n\nPROBLEM Using curved arrows, propose a mechanism for the following reaction, one of the steps in the\n22-29 biosynthesis of the amino acid tyrosine.\n\n\nPROBLEM One of the later steps in glucose biosynthesis is the isomerization of fructose 6-phosphate to 22-30 glucose 6-phosphate. Propose a mechanism, using acid or base catalysis as needed."}
{"id": 1478, "contents": "Mechanism Problems - \nPROBLEM One of the later steps in glucose biosynthesis is the isomerization of fructose 6-phosphate to 22-30 glucose 6-phosphate. Propose a mechanism, using acid or base catalysis as needed.\n\n\nFructose 6-phosphate\n\nGlucose\n6-phosphate\n\nPROBLEM The Favorskii reaction involves treatment of an $\\alpha$-bromo ketone with base to yield a ring-contracted\n22-31 product. For example, reaction of 2 -bromocyclohexanone with aqueous NaOH yields cyclopentanecarboxylic acid. Propose a mechanism.\n\n\nPROBLEM Treatment of a cyclic ketone with diazomethane is a method for accomplishing a ring-expansion\n22-32 reaction. For example, treatment of cyclohexanone with diazomethane yields cycloheptanone. Propose a mechanism.\n\n\nPROBLEM The final step in an attempted synthesis of laurene, a hydrocarbon isolated from the marine alga\n22-33 Laurencia glandulifera, involved the Wittig reaction shown. The product obtained, however, was not laurene but an isomer. Propose a mechanism to account for the unexpected results.\n\n\n\n\nLaurene\n(Not formed)\nPROBLEM Amino acids can be prepared by reaction of alkyl halides with diethyl acetamidomalonate, followed\n22-34 by heating the initial alkylation product with aqueous HCl . Show how you would prepare alanine, $\\mathrm{CH}_{3} \\mathrm{CH}\\left(\\mathrm{NH}_{2}\\right) \\mathrm{CO}_{2} \\mathrm{H}$, one of the twenty amino acids found in proteins, and propose a mechanism for acid-catalyzed conversion of the initial alkylation product to the amino acid.\n\n\nDiethyl acetamidomalonate"}
{"id": 1479, "contents": "Mechanism Problems - \nDiethyl acetamidomalonate\n\nPROBLEM Amino acids can also be prepared by a two-step sequence that involves Hell-Volhard-Zelinskii\n22-35 reaction of a carboxylic acid followed by treatment with ammonia. Show how you would prepare leucine, $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CHCH}_{2} \\mathrm{CH}\\left(\\mathrm{NH}_{2}\\right) \\mathrm{CO}_{2} \\mathrm{H}$, and identify the mechanism of the second step.\n\nPROBLEM Heating carvone with aqueous sulfuric acid converts it into carvacrol. Propose a mechanism for the 22-36 isomerization."}
{"id": 1480, "contents": "Acidity of Carbonyl Compounds - \nPROBLEM Identify all the acidic hydrogens ( $\\mathrm{p} K_{\\mathrm{a}}<25$ ) in the following molecules:\n22-37 (a)\n\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n\nPROBLEM Rank the following compounds in order of increasing acidity:\n22-38\n(a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$\n(b) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$\n(c) $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2}\\right)_{2} \\mathrm{NH}$\n(d) $\\mathrm{CH}_{3} \\mathrm{COCH}_{3}$\n\n(f) $\\mathrm{CCl}_{3} \\mathrm{CO}_{2} \\mathrm{H}$\n\nPROBLEM Write resonance structures for the following anions:\n22-39 (a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n\nPROBLEM Base treatment of the following $\\alpha, \\beta$-unsaturated carbonyl compound yields an anion by removal of\n$\\mathbf{2 2 - 4 0} \\mathrm{H}^{+}$from the $\\gamma$ carbon. Why are hydrogens on the $\\gamma$ carbon atom acidic?\n\n\nPROBLEM Treatment of 1-phenyl-2-propenone with a strong base such as LDA does not yield an anion, even\n22-41 though it contains a hydrogen on the carbon atom next to the carbonyl group. Explain.\n\n\n1-Phenyl-2-propenone"}
{"id": 1481, "contents": "$\\boldsymbol{\\sigma}$-Substitution Reactions - \nPROBLEM Predict the product(s) of the following reactions:\n\n22-42 (a)\n\n(c)\n\n(b)\n\n(d)\n\n\nPROBLEM Which, if any, of the following compounds can be prepared by a malonic ester synthesis? Show the 22-43 alkyl halide you would use in each case.\n(a) Ethyl pentanoate\n(b) Ethyl 3-methylbutanoate\n(c) Ethyl 2-methylbutanoate\n(d) Ethyl 2,2-dimethylpropanoate\n\nPROBLEM Which, if any, of the following compounds can be prepared by an acetoacetic ester synthesis? 22-44 Explain.\n(a) ${ }_{B}$\n\n(b)\n\n(c)\n\n\nPROBLEM How would you prepare the following ketones using an acetoacetic ester synthesis?\n22-45 (a)\n\n\n(b)\n\n\nPROBLEM How would you prepare the following compounds using either an acetoacetic ester synthesis or a\n22-46 malonic ester synthesis?\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM Which of the following substances would undergo the haloform reaction?\n22-47\n(a) $\\mathrm{CH}_{3} \\mathrm{COCH}_{3}$\n(b) Acetophenone\n(c) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CHO}$\n(d) $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$\n(e) $\\mathrm{CH}_{3} \\mathrm{C} \\equiv \\mathrm{N}$\n\nPROBLEM How might you convert geraniol into either ethyl geranylacetate or geranylacetone?\n22-48\n\n\nPROBLEM Aprobarbital, a barbiturate once used in treating insomnia, is synthesized in three steps from 22-49 diethyl malonate. Show how you would synthesize the necessary dialkylated intermediate, and then propose a mechanism for the reaction of this intermediate with urea to give aprobarbital.\n\n\nAprobarbital"}
{"id": 1482, "contents": "General Problems - \nPROBLEM One way to determine the number of acidic hydrogens in a molecule is to treat the compound with\n22-50 NaOD in $\\mathrm{D}_{2} \\mathrm{O}$, isolate the product, and determine its molecular weight by mass spectrometry. For example, if cyclohexanone is treated with NaOD in $\\mathrm{D}_{2} \\mathrm{O}$, the product has $\\mathrm{MW}=102$. Explain how this method works.\n\nPROBLEM When optically active ( $R$ )-2-methylcyclohexanone is treated with either aqueous base or acid, 22-51 racemization occurs. Explain.\n\nPROBLEM Would you expect optically active ( $S$ )-3-methylcyclohexanone to be racemized on acid or base 22-52 treatment in the same way as 2-methylcyclohexanone (Problem 22-51)? Explain.\n\nPROBLEM When an optically active carboxylic acid such as ( $R$ )-2-phenylpropanoic acid is brominated under 22-53 Hell-Volhard-Zelinskii conditions, is the product optically active or racemic? Explain.\n\nPROBLEM Fill in the reagents a-c that are missing from the following scheme:\n22-54\n\n\nPROBLEM Although 2-substituted 2-cyclopentenones are in a base-catalyzed equilibrium with their\n22-55 5-substituted 2-cyclopentenone isomers, the analogous isomerization is not observed for 2-substituted 2-cyclohexenones. Explain.\n\n\nPROBLEM How would you synthesize the following compounds from cyclohexanone? More than one step may 22-56 be required.\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n\nPROBLEM The two isomers cis- and trans-4-tert-butyl-2-methylcyclohexanone are interconverted by base 22-57 treatment. Which isomer do you think is more stable, and why?\n\nPROBLEM The following synthetic routes are incorrect. What is wrong with each?\n22-58 (a)\n\n(b)\n\n(c)"}
{"id": 1483, "contents": "General Problems - \nPROBLEM The following synthetic routes are incorrect. What is wrong with each?\n22-58 (a)\n\n(b)\n\n(c)\n\n\nPROBLEM Attempted Grignard reaction of cyclohexanone with tert-butylmagnesium bromide yields only\n22-59 about $1 \\%$ of the expected addition product along with $99 \\%$ unreacted cyclohexanone. If $\\mathrm{D}_{3} \\mathrm{O}^{+}$is added to the reaction mixture after a suitable period, however, the \"unreacted\" cyclohexanone is found to have one deuterium atom incorporated into it. Explain.\n\n\nPROBLEM Ketones react slowly with benzeneselenenyl chloride in the presence of HCl to yield $\\alpha$-phenylseleno\n22-60 ketones. Propose a mechanism for this acid-catalyzed $\\alpha$-substitution reaction.\n\n\nPROBLEM South American Incas chewed the leaves of the coca bush, Erythroxylon coca, to combat fatigue.\n22-61 Chemical studies of Erythroxylon coca by Friedrich W\u00f6hler in 1862 resulted in the discovery of cocaine, $\\mathrm{C}_{17} \\mathrm{H}_{21} \\mathrm{NO}_{4}$, as the active component. Basic hydrolysis of cocaine leads to methanol, benzoic acid, and another compound called ecgonine, $\\mathrm{C}_{9} \\mathrm{H}_{15} \\mathrm{NO}_{3}$. Oxidation of ecgonine with $\\mathrm{CrO}_{3}$ yields a keto acid that readily loses $\\mathrm{CO}_{2}$ on heating, giving tropinone."}
{"id": 1484, "contents": "Tropinone - \n(a) What is a likely structure for the keto acid?\n(b) What is a likely structure for ecgonine, neglecting stereochemistry?\n(c) What is a likely structure for cocaine, neglecting stereochemistry?\n\nPROBLEM The key step in a reported laboratory synthesis of sativene, a hydrocarbon isolated from the mold\n22-62 Helminthosporium sativum, involves the following base treatment of a keto tosylate. What kind of reaction is occurring? How would you complete the synthesis?\n\n\nPROBLEM Sodium pentothal is a short-acting barbiturate derivative used as a general anesthetic and known 22-63 in popular culture as a truth serum. It is synthesized like other barbiturates (see the Chemistry Matters at the end of this chapter), using thiourea, $\\left(\\mathrm{H}_{2} \\mathrm{~N}\\right)_{2} \\mathrm{C}=\\mathrm{S}$, in place of urea. How would you synthesize sodium pentothal?"}
{"id": 1485, "contents": "CHAPTER 23 <br> Carbonyl Condensation Reactions - \nFIGURE 23.1 Many of the molecules needed by all growing organisms are biosynthesized using carbonyl condensation reactions. (credit: modification of work \"Newborn Sea Lion\" by Steven Bedard/Flickr, CC BY 2.0)"}
{"id": 1486, "contents": "CHAPTER CONTENTS - \n23.1 Carbonyl Condensations: The Aldol Reaction\n23.2 Carbonyl Condensations versus Alpha Substitutions\n23.3 Dehydration of Aldol Products: Synthesis of Enones\n23.4 Using Aldol Reactions in Synthesis\n23.5 Mixed Aldol Reactions\n23.6 Intramolecular Aldol Reactions\n23.7 The Claisen Condensation Reaction\n23.8 Mixed Claisen Condensations\n23.9 Intramolecular Claisen Condensations: The Dieckmann Cyclization\n23.10 Conjugate Carbonyl Additions: The Michael Reaction\n23.11 Carbonyl Condensations with Enamines: The Stork Enamine Reaction\n23.12 The Robinson Annulation Reaction\n23.13 Some Biological Carbonyl Condensation Reactions\n\nWHY THIS CHAPTER? We'll see later in this chapter and again in Chapter 29 that carbonyl condensation reactions occur in a large number of metabolic pathways. In fact, almost all classes of biomolecules-carbohydrates, lipids, proteins, nucleic acids, and many others-are biosynthesized through pathways that involve carbonyl condensation reactions. As with the $\\alpha$-substitution reaction discussed in the previous chapter, the great value of carbonyl condensations is that they are one of the few general methods for forming carbon-carbon bonds, thereby making it possible to build larger molecules from smaller precursors. In this chapter, we'll see how and why these reactions occur.\n\nWe've now studied three of the four general kinds of carbonyl-group reactions and have seen two general\nkinds of behavior. In nucleophilic addition and nucleophilic acyl substitution reactions, a carbonyl compound behaves as an electrophile when an electron-rich reagent adds to it. In $\\alpha$-substitution reactions, however, a carbonyl compound behaves as a nucleophile when it is converted into its enol or enolate ion. In the carbonyl condensation reactions that we'll study in this chapter, the carbonyl compound behaves both as an electrophile and as a nucleophile.\n\n\nElectrophilic carbonyl group reacts with nucleophiles.\n\n\nNucleophilic enolate ion reacts with electrophiles."}
{"id": 1487, "contents": "CHAPTER CONTENTS - 23.1 Carbonyl Condensations: The Aldol Reaction\nCarbonyl condensation reactions take place between two carbonyl partners and involve a combination of nucleophilic addition and $\\alpha$-substitution steps. One partner is converted into an enolate-ion nucleophile and adds to the electrophilic carbonyl group of the second partner. In so doing, the nucleophilic partner undergoes an $\\alpha$-substitution reaction and the electrophilic partner undergoes a nucleophilic addition. The general mechanism of the process is shown in FIGURE 23.2."}
{"id": 1488, "contents": "FIGURE 23.2 MECHANISM - \nThe general mechanism of a carbonyl condensation reaction. One partner becomes a nucleophilic donor and adds to the second partner as an electrophilic acceptor. After protonation, the final product is a $\\beta$-hydroxy carbonyl compound.\n\n(1) A carbonyl compound with an $\\alpha$ hydrogen atom is converted by base into its enolate ion.\n\n\n(2) The enolate ion acts as a nucleophilic donor and adds to the electrophilic carbonyl group of a second carbonyl (2 \\| compound.\n(3) Protonation of the tetrahedral alkoxide ion intermediate gives the neutral condensation product and regenerates the base catalyst.\n\n(3) $\\downarrow$ New C-C bond\n\n\nA $\\boldsymbol{\\beta}$-hydroxy\ncarbonyl compound\n\nAldehydes and ketones with an $\\alpha$ hydrogen atom undergo a base-catalyzed carbonyl condensation reaction called the aldol reaction. For example, treatment of acetaldehyde with a base such as sodium ethoxide or sodium hydroxide in a protic solvent leads to rapid and reversible formation of 3-hydroxybutanal, known commonly as aldol (aldehyde + alcohol), hence the general name of the reaction.\n\n\n3-Hydroxybutanal\n(aldol-a $\\beta$-hydroxy carbonyl compound)\nThe exact position of the aldol equilibrium depends both on reaction conditions and on substrate structure. The equilibrium generally favors the condensation product in the case of aldehydes with no $\\alpha$ substituent\n$\\left(\\mathrm{RCH}_{2} \\mathrm{CHO}\\right)$ but favors the reactant for disubstituted aldehydes $\\left(\\mathrm{R}_{2} \\mathrm{CHCHO}\\right)$ and for most ketones. Steric factors are probably responsible for these trends, since increased substitution near the reaction site increases steric congestion in the aldol product.\n\nAldehydes"}
{"id": 1489, "contents": "Predicting the Product of an Aldol Reaction - \nWhat is the structure of the aldol product from propanal?"}
{"id": 1490, "contents": "Strategy - \nAn aldol reaction combines two molecules of reactant by forming a bond between the $\\alpha$ carbon of one partner and the carbonyl carbon of the second partner. The product is a $\\beta$-hydroxy aldehyde or ketone, meaning that the two oxygen atoms in the product have a 1,3 relationship."}
{"id": 1491, "contents": "Solution - \nPROBLEM Predict the aldol reaction product of the following compounds:\n23-1 (a)\n\n\n(c)\n\n\nPROBLEM Using curved arrows to indicate the electron flow in each step, show how the base-catalyzed\n23-2 reverse-aldol reaction of 4-hydroxy-4-methyl-2-pentanone takes place to yield 2 equivalents of acetone."}
{"id": 1492, "contents": "Solution - 23.2 Carbonyl Condensations versus Alpha Substitutions\nTwo of the four general carbonyl-group reactions-carbonyl condensations and $\\alpha$ substitutions-take place under basic conditions and involve enolate-ion intermediates. Because the experimental conditions for the two reactions are similar, how can we predict which will occur in a given case? When we generate an enolate ion\nwith the intention of carrying out an $\\alpha$ alkylation, how can we be sure that a carbonyl condensation reaction won't occur instead?\n\nThere is no simple answer to this question, but the exact experimental conditions usually have much to do with the result. Alpha-substitution reactions require a full equivalent of strong base and are normally carried out so that the carbonyl compound is rapidly and completely converted into its enolate ion at a low temperature. An electrophile is then added rapidly to ensure that the reactive enolate ion is quenched quickly. In a ketone alkylation reaction, for instance, we might use 1 equivalent of lithium diisopropylamide (LDA) in tetrahydrofuran solution at $-78^{\\circ} \\mathrm{C}$. Rapid and complete generation of the ketone enolate ion would occur, and no unreacted ketone would remain, meaning that no condensation reaction could take place. We would then immediately add an alkyl halide to complete the alkylation reaction.\n\n\nOn the other hand, carbonyl condensation reactions require only a catalytic amount of a relatively weak base, rather than a full equivalent, so that a small amount of enolate ion is generated in the presence of unreacted carbonyl compound. Once a condensation occurs, the basic catalyst is regenerated. To carry out an aldol reaction on propanal, for instance, we might dissolve the aldehyde in methanol, add 0.05 equivalent of sodium methoxide, and then warm the mixture to give the aldol product."}
{"id": 1493, "contents": "Solution - 23.3 Dehydration of Aldol Products: Synthesis of Enones\nWhen heated under either acidic or basic conditions, the $\\beta$-hydroxy aldehydes or ketones formed in aldol reactions can be dehydrated to yield $\\alpha, \\beta$-unsaturated products, or conjugated enones (ene + ketone). In fact, it's this loss of water that gives the carbonyl condensation reaction its name, because water condenses out when an enone product forms.\n\n\nA $\\boldsymbol{\\beta}$-hydroxy ketone or aldehyde\n\n\nA conjugated enone\n\nMost alcohols are resistant to dehydration by base (Section 17.6) because hydroxide ion is a poor leaving group, but aldol products dehydrate easily because of their carbonyl group. Under basic conditions, an acidic $\\alpha$ hydrogen is removed, yielding an enolate ion that expels the nearby ${ }^{-} \\mathrm{OH}$ leaving group in an E1cB reaction (Section 11.10). Under acidic conditions, an enol is formed, the -OH group is protonated, and water is expelled in an E1 or E2 reaction.\n\n\nThe reaction conditions needed for aldol dehydration are often only a bit more vigorous (slightly higher temperature, for instance) than the conditions needed for the aldol formation itself. As a result, conjugated enones are usually obtained directly from aldol reactions without isolating the intermediate $\\beta$-hydroxy carbonyl compounds.\n\nConjugated enones are more stable than nonconjugated enones for the same reason that conjugated dienes are more stable than nonconjugated dienes (Section 14.1). Interaction between the $\\pi$ electrons of the $\\mathrm{C}=\\mathrm{C}$ bond and the $\\pi$ electrons of the $\\mathrm{C}=\\mathrm{O}$ group leads to a molecular-orbital description for a conjugated enone with an interaction of the $\\pi$ electrons over all four atomic centers (FIGURE 23.3).\n\n\n\nPropenal\n\n\n\n1,3-Butadiene\n\nFIGURE 23.3 The $\\boldsymbol{\\pi}$ bonding molecular orbitals of a conjugated enone (propenal) and a conjugated diene (1,3-butadiene) are similar in shape and are spread over the entire $\\pi$ system."}
{"id": 1494, "contents": "Solution - 23.3 Dehydration of Aldol Products: Synthesis of Enones\nThe real value of aldol dehydration is that removal of water from the reaction mixture can be used to drive the aldol equilibrium toward the product. Even though the initial aldol step itself may be unfavorable, as it usually is for ketones, the subsequent dehydration step nevertheless allows many aldol condensations to be carried out in good yield. Cyclohexanone, for example, gives cyclohexylidenecyclohexanone in $92 \\%$ yield, even though the initial equilibrium is unfavorable.\n\n\nCyclohexanone\nCyclohexylidenecyclohexanone\n(92\\%)"}
{"id": 1495, "contents": "Predicting the Product of an Aldol Reaction - \nWhat is the structure of the enone obtained from aldol condensation of acetaldehyde?"}
{"id": 1496, "contents": "Strategy - \nIn the aldol reaction, $\\mathrm{H}_{2} \\mathrm{O}$ is eliminated and a double bond is formed by removing two hydrogens from the acidic $\\alpha$ position of one partner and the carbonyl oxygen from the second partner. The product is thus an $\\alpha, \\beta$-unsaturated aldehyde or ketone."}
{"id": 1497, "contents": "Solution - \nPROBLEM What enone product would you expect from aldol condensation of each of the following 23-3 compounds?\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Aldol condensation of 3-methylcyclohexanone leads to a mixture of two enone products, not 23-4 counting double-bond isomers. Draw them."}
{"id": 1498, "contents": "Solution - 23.4 Using Aldol Reactions in Synthesis\nThe aldol reaction yields either a $\\beta$-hydroxy aldehyde/ketone or an $\\alpha, \\beta$-unsaturated aldehyde/ketone, depending on the experimental conditions. By learning how to work backward, it's possible to predict when the aldol reaction might be useful in synthesis. Whenever the target molecule contains either a $\\beta$-hydroxy aldehyde/ ketone or a conjugated enone functional group, it might come from an aldol reaction.\n\n\nWe can extend this kind of reasoning even further by imagining that subsequent transformations might be carried out on the aldol products. For example, a saturated ketone might be prepared by catalytic hydrogenation of the enone product. A good example can be found in the industrial preparation of 2-ethyl-1-hexanol, an alcohol used in the synthesis of plasticizers for polymers. Although 2-ethyl-1-hexanol bears little resemblance to an aldol product at first glance, it is in fact prepared commercially from butanal by an aldol reaction. Working backward, we can reason that 2 -ethyl-1-hexanol might come from 2 -ethylhexanal by a reduction. 2-Ethylhexanal, in turn, might be prepared by catalytic reduction of 2 -ethyl-2-hexenal, which is the aldol condensation product of butanal. The reactions that follow show the sequence in reverse order.\n\n\n\nButanal\n\n\n2-Ethyl-2-hexenal\n\nPROBLEM Which of the following compounds are aldol condensation products? What is the aldehyde or ketone 23-5 precursor of each?\n(a) 2-Hydroxy-2-methylpentanal\n(b) 5-Ethyl-4-methyl-4-hepten-3-one\n\nPROBLEM 1-Butanol is prepared commercially by a route that begins with an aldol reaction. Show the steps 23-6 that are likely to be involved.\n\nPROBLEM Show how you would synthesize the following compound using an aldol reaction:\n23-7"}
{"id": 1499, "contents": "Solution - 23.5 Mixed Aldol Reactions\nUntil now, we've considered only symmetrical aldol reactions, in which the two carbonyl components have been the same. What would happen, though, if an aldol reaction were carried out between two different carbonyl partners?\n\nIn general, a mixed aldol reaction between two similar aldehyde or ketone partners leads to a mixture of four possible products. For example, base treatment of a mixture of acetaldehyde and propanal gives a complex product mixture containing two \"symmetrical\" aldol products and two \"mixed\" aldol products. Clearly, such a reaction is of no practical value.\n\n\nOn the other hand, mixed aldol reactions can lead cleanly to a single product if either of two conditions is met:\n\n- If one of the carbonyl partners contains no $\\alpha$ hydrogens (and thus can't form an enolate ion to become a donor), but does contain an unhindered carbonyl group (and so is a good acceptor of nucleophiles), then a mixed aldol reaction is likely to be successful. This is the case, for instance, when either benzaldehyde or\nformaldehyde is used as one of the carbonyl partners.\nNeither benzaldehyde nor formaldehyde can form an enolate ion to add to another partner, yet both compounds have an unhindered carbonyl group. The ketone 2-methylcyclohexanone, for instance, gives the mixed aldol product on reaction with benzaldehyde.\n\n- If one of the carbonyl partners is much more acidic than the other, and is thus transformed into its enolate ion in preference to the other, then a mixed aldol reaction is likely to be successful. Ethyl acetoacetate, for instance, is completely converted into its enolate ion in preference to enolate ion formation from monocarbonyl partners. Thus, aldol condensations of monoketones with ethyl acetoacetate occur preferentially to give the mixed product.\n\n\nThe situation can be summarized by saying that a mixed aldol reaction leads to a mixture of products unless one of the partners either has no $\\alpha$ hydrogens but is a good electrophilic acceptor (such as benzaldehyde) or is an unusually acidic nucleophilic donor (such as ethyl acetoacetate)."}
{"id": 1500, "contents": "Solution - 23.5 Mixed Aldol Reactions\nPROBLEM Which of the following compounds can probably be prepared by a mixed aldol reaction? Show the 23-8 reactants you would use in each case.\n(a)\n\n(b)\n\n(c)"}
{"id": 1501, "contents": "Solution - 23.6 Intramolecular Aldol Reactions\nThe aldol reactions we've seen thus far have all been intermolecular, meaning that they have taken place between two different molecules. When certain dicarbonyl compounds are treated with base, however, an intramolecular aldol reaction can occur, leading to the formation of a cyclic product. For example, base treatment of a 1,4-diketone such as 2,5-hexanedione yields a cyclopentenone product, and base treatment of a 1,5-diketone such as 2,6-heptanedione yields a cyclohexenone.\n\n\n$$\n\\begin{aligned}\n& \\text { 2,5-Hexanedione } \\\\\n& \\text { (a 1,4-diketone) }\n\\end{aligned}\n$$"}
{"id": 1502, "contents": "2,6-Heptanedione (a 1,5-diketone) - \n3-Methyl-2-cyclohexenone\n\nThe mechanism of intramolecular aldol reactions is similar to that of intermolecular reactions. The only difference is that both the nucleophilic carbonyl anion donor and the electrophilic carbonyl acceptor are now in the same molecule. One complication, however, is that intramolecular aldol reactions might lead to a mixture of products, depending on which enolate ion is formed. For example, 2,5-hexanedione might yield either the five-membered-ring product 3-methyl-2-cyclopentenone or the three-membered-ring product (2-methylcyclopropenyl) ethanone (FIGURE 23.4). In practice, though, only the cyclopentenone is formed.\n\n\nFIGURE 23.4 Intramolecular aldol reaction of 2,5-hexanedione yields 3-methyl-2-cyclopentenone rather than the alternative cyclopropene.\n\nThe selectivity observed in the intramolecular aldol reaction of 2,5-hexanedione is due to the fact that all steps in the mechanism are reversible, so an equilibrium is reached. Thus, the relatively strain-free cyclopentenone product is considerably more stable than the highly strained cyclopropene alternative. For similar reasons, intramolecular aldol reactions of 1,5-diketones lead only to cyclohexenone products rather than to the more strained acylcyclobutenes.\n\nPROBLEM Treatment of a 1,3-diketone such as 2,4-pentanedione with base does not give an aldol\n23-9 condensation product. Explain.\nPROBLEM What product would you expect to obtain from base treatment of 1,6-cyclodecanedione? 23-10\n\n\n1,6-Cyclodecanedione"}
{"id": 1503, "contents": "2,6-Heptanedione (a 1,5-diketone) - 23.7 The Claisen Condensation Reaction\nEsters, like aldehydes and ketones, are weakly acidic. When an ester with an $\\alpha$ hydrogen is treated with 1 equivalent of a base such as sodium ethoxide, a reversible carbonyl condensation reaction occurs to yield a $\\beta$-keto ester. For instance, ethyl acetate yields ethyl acetoacetate on base treatment. This reaction between two ester molecules is known as the Claisen condensation reaction. (We'll use ethyl esters, abbreviated \"Et,\" for consistency, but other esters will also work.)\n\n\nThe mechanism of the Claisen condensation is similar to that of the aldol condensation and involves the nucleophilic addition of an ester enolate ion to the carbonyl group of a second ester molecule (FIGURE 23.5). The only difference between the aldol condensation of an aldehyde or ketone and the Claisen condensation of an ester involves the fate of the initially formed tetrahedral intermediate. The tetrahedral intermediate in the aldol reaction is protonated to give an alcohol product-exactly the behavior previously seen for aldehydes and ketones (Section 19.4). The tetrahedral intermediate in the Claisen reaction, however, expels an alkoxide leaving group to yield an acyl substitution product-as previously seen for esters (Section 21.6)."}
{"id": 1504, "contents": "FIGURE 23.5 MECHANISM - \nMechanism of the Claisen condensation reaction.\n\n(1) Base abstracts an acidic alpha hydrogen atom from an ester molecule, yielding an ester enolate ion.\n(2) The enolate ion is added in a nucleophilic addition reaction to a second ester molecule, giving a tetrahedral alkoxide intermediate.\n\n(2)\n\n(3) The tetrahedral intermediate expels ethoxide ion to yield a new carbonyl compound, ethyl acetoacetate.\n4) But ethoxide ion is a strong enough base to deprotonate ethyl acetoacetate, shifting the equilibrium and driving the overall\nreaction to completion.\n\n\n(5) Protonation of the enolate ion by addition of aqueous acid in a separate step yields the final $\\beta$-keto ester product.\n\n\n\nIf the starting ester has more than one acidic $\\alpha$ hydrogen, the product $\\beta$-keto ester has a highly acidic, doubly activated hydrogen atom that can be abstracted by base. This deprotonation of the product requires the use of a full equivalent of base rather than a catalytic amount. Furthermore, the deprotonation serves to drive the equilibrium completely to the product side so that high yields are usually obtained in Claisen condensations."}
{"id": 1505, "contents": "Predicting the Product of a Claisen Condensation Reaction - \nWhat product would you obtain from Claisen condensation of ethyl propanoate?"}
{"id": 1506, "contents": "Strategy - \nThe Claisen condensation of an ester results in loss of one molecule of alcohol and formation of a product in which an acyl group of one reactant bonds to the $\\alpha$ carbon of the second reactant. The product is a $\\beta$-keto ester.\n\nSolution\n\n\nPROBLEM Show the products you would expect to obtain by Claisen condensation of the following esters:\n23-11\n(a) $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CHCH}_{2} \\mathrm{CO}_{2} \\mathrm{Et}$\n(b) Ethyl phenylacetate\n(c) Ethyl cyclohexylacetate\n\nPROBLEM As shown in Figure 23.5, the Claisen reaction is reversible. That is, a $\\beta$-keto ester can be cleaved 23-12 by base into two fragments. Using curved arrows to indicate electron flow, show the mechanism by which this cleavage occurs."}
{"id": 1507, "contents": "Strategy - 23.8 Mixed Claisen Condensations\nThe mixed Claisen condensation of two different esters is similar to the mixed aldol condensation of two different aldehydes or ketones (Section 23.5). Mixed Claisen reactions are successful only when one of the two ester components has no $\\alpha$ hydrogens and thus can't form an enolate ion. For example, ethyl benzoate and ethyl formate can't form enolate ions and thus can't serve as donors. They can, however, act as the electrophilic acceptor components in reactions with other ester anions to give mixed $\\beta$-keto ester products.\n\n\nMixed Claisen-like reactions can also be carried out between an ester and a ketone, resulting in the synthesis of a $\\beta$-diketone. The reaction works best when the ester component has no $\\alpha$ hydrogens and thus can't act as the nucleophilic donor. For example, ethyl formate gives high yields in mixed Claisen condensations with ketones."}
{"id": 1508, "contents": "2,2-Dimethylcyclohexanone (donor) - \n$+$\n\n\nEthyl formate (acceptor)\n\n\nA $\\boldsymbol{b}$-keto aldehyde (91\\%)"}
{"id": 1509, "contents": "Predicting the Product of a Mixed Claisen Condensation Reaction - \nDiethyl oxalate, $\\left(\\mathrm{CO}_{2} \\mathrm{Et}\\right)_{2}$, often gives high yields in mixed Claisen reactions. What product would you expect to obtain from a mixed Claisen reaction of ethyl acetate with diethyl oxalate?"}
{"id": 1510, "contents": "Strategy - \nA mixed Claisen reaction is effective when only one of the two partners has an acidic $\\alpha$ hydrogen atom. In the present case, ethyl acetate can be converted into its enolate ion, but diethyl oxalate cannot. Thus, ethyl acetate acts as the donor and diethyl oxalate as the acceptor.\n\nSolution\n\n\nPROBLEM What product would you expect from the following mixed Claisen-like reaction?\n23-13"}
{"id": 1511, "contents": "Strategy - 23.9 Intramolecular Claisen Condensations: The Dieckmann Cyclization\nIntramolecular Claisen condensations can be carried out with diesters, just as intramolecular aldol condensations can be carried out with diketones (Section 23.6). Called the Dieckmann cyclization, this reaction works best on 1,6-diesters and 1,7-diesters. Intramolecular Claisen cyclization of a 1,6-diester gives a five-membered cyclic $\\beta$-keto ester, and cyclization of a 1,7-diester gives a six-membered cyclic $\\beta$-keto ester.\n\n\nThe mechanism of the Dieckmann cyclization, shown in FIGURE 23.6, is the same as that of the Claisen condensation. One of the two ester groups is converted into an enolate ion, which carries out a nucleophilic acyl substitution on the second ester group at the other end of the molecule. A cyclic $\\beta$-keto ester product results.\n\nFIGURE 23.6 MECHANISM\nMechanism of the Dieckmann cyclization of a 1,7-diester to yield a cyclic $\\beta$-keto ester product.\n\n(1) Base abstracts an acidic $\\alpha$ proton from\nthe carbon atom next to one of the ester groups, yielding an enolate ion.\n(1) $\\downarrow \\mathrm{Na}^{+- \\text {OEt }}$\n\n\nO |\n\n(3) $\\downarrow$\n\n(4) Deprotonation of the acidic $\\beta$-keto ester gives an enolate ion...\n-\n\n(5) $\\downarrow \\mathrm{H}_{3} \\mathrm{O}^{+}$"}
{"id": 1512, "contents": "Strategy - 23.9 Intramolecular Claisen Condensations: The Dieckmann Cyclization\n(4) Deprotonation of the acidic $\\beta$-keto ester gives an enolate ion...\n-\n\n(5) $\\downarrow \\mathrm{H}_{3} \\mathrm{O}^{+}$\n\n\nThe cyclic $\\beta$-keto ester produced in a Dieckmann cyclization can be further alkylated and decarboxylated by a series of reactions analogous to those used in the acetoacetic ester synthesis (Section 22.7). Alkylation and\nsubsequent decarboxylation of ethyl 2-oxocyclohexanecarboxylate, for instance, yields a 2-alkylcyclohexanone. The overall sequence of (1) Dieckmann cyclization, (2) $\\beta$-keto ester alkylation, and (3) decarboxylation is a powerful method for preparing 2 -substituted cyclopentanones and cyclohexanones.\n\n\nPROBLEM What product would you expect from the following reaction?\n\n\nPROBLEM Dieckmann cyclization of diethyl 3-methylheptanedioate gives a mixture of two $\\beta$-keto ester\n23-15 products. What are their structures, and why is a mixture formed?"}
{"id": 1513, "contents": "Strategy - 23.10 Conjugate Carbonyl Additions: The Michael Reaction\nWe saw in Section 19.13 that certain nucleophiles, such as amines, react with $\\alpha, \\beta$-unsaturated aldehydes and ketones to give a conjugate addition product, rather than a direct addition product.\n\n\nExactly the same kind of conjugate addition can occur when a nucleophilic enolate ion reacts with an $\\alpha, \\beta$-unsaturated carbonyl compound-a process known as the Michael reaction after Arthur Michael at Tufts College and Harvard University.\n\nThe best Michael reactions are those that take place when a particularly stable enolate ion, such as that derived from a $\\beta$-keto ester or other 1,3-dicarbonyl compound, adds to an unhindered $\\alpha, \\beta$-unsaturated ketone. For example, ethyl acetoacetate reacts with 3-buten-2-one in the presence of sodium ethoxide to yield the conjugate addition product.\n\n\nEthyl\n3-Buten-2-one"}
{"id": 1514, "contents": "acetoacetate - \nMichael reactions take place by addition of a nucleophilic enolate ion donor to the $\\beta$ carbon of an $\\alpha, \\beta$-unsaturated carbonyl acceptor, according to the mechanism shown in FIGURE 23.7."}
{"id": 1515, "contents": "FIGURE 23.7 MECHANISM - \nMechanism of the Michael reaction between a $\\boldsymbol{\\beta}$-keto ester and an\n$\\boldsymbol{\\alpha}, \\boldsymbol{\\beta}$-unsaturated ketone. The reaction is a conjugate addition of an enolate ion to\nthe unsaturated carbonyl compound.\n\nThe base catalyst removes an acidic alpha proton from the starting $\\beta$-keto ester to generate a stabilized enolate ion nucleophile.\n\n2 The nucleophile adds to the $\\alpha, \\beta$-unsaturated ketone electrophile in a Michael reaction to generate a new enolate as product.\n(3) The enolate product abstracts an acidic proton, either from solvent or from starting keto ester, to yield the final addition product.\n\n(1) $\\| \\mathrm{Na}^{+}{ }^{+}$-OEt\n\n\n(3) $\\downarrow \\mathrm{EtOH}$\n\n\nThe Michael reaction occurs with a variety of $\\alpha, \\beta$-unsaturated carbonyl compounds, not just conjugated ketones. Unsaturated aldehydes, esters, thioesters, nitriles, amides, and nitro compounds can all act as the electrophilic acceptor component in Michael reactions (TABLE 23.1). Similarly, a variety of different donors can be used, including $\\beta$-diketones, $\\beta$-keto esters, malonic esters, $\\beta$-keto nitriles, and nitro compounds.\n\nTABLE 23.1 Some Michael Acceptors and Michael Donors\nMichael acceptors Michael donors\n\n| <smiles>C=[C]C=O</smiles> | Propenal | <smiles>O=[14C]C=[18O]</smiles> | $\\beta$-Diketone |\n| :---: | :---: | :---: | :---: |\n| <smiles>C=CC(C)=O</smiles> | 3-Buten-2-one | <smiles>CCOC(=O)CC=O</smiles> | $\\beta$-Keto ester |\n| <smiles>C=C[C-]O[O-]</smiles> | Ethyl propenoate | <smiles></smiles> | Diethyl malonate |"}
{"id": 1516, "contents": "FIGURE 23.7 MECHANISM - \nTABLE 23.1 Some Michael Acceptors and Michael Donors\n\n| Michael accep |  | Michael donors |  |\n| :---: | :---: | :---: | :---: |\n| <smiles>C=C(N)O</smiles> | Propenamide | $\\stackrel{\\mathrm{O}}{\\mathrm{RCCH}}{ }_{2} \\mathrm{C} \\equiv \\mathrm{~N}$ | $\\beta$-Keto nitrile |\n| $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHC} \\equiv \\mathrm{N}$ | Propenenitrile | $\\mathrm{RCH}_{2} \\mathrm{NO}_{2}$ | Nitro compound |\n| <smiles>C=CN+[O-]</smiles> | Nitroethylene |  |  |"}
{"id": 1517, "contents": "Using the Michael Reaction - \nHow might you obtain the following compound using a Michael reaction?"}
{"id": 1518, "contents": "Strategy - \nA Michael reaction involves the conjugate addition of a stable enolate ion donor to an $\\alpha, \\beta$-unsaturated carbonyl acceptor, yielding a 1,5-dicarbonyl product. Usually, the stable enolate ion is derived from a $\\beta$-diketone, $\\beta$-keto ester, malonic ester, or similar compound. The $C-C$ bond formed in the conjugate addition step is the one between the $\\alpha$ carbon of the acidic donor and the $\\beta$ carbon of the unsaturated acceptor."}
{"id": 1519, "contents": "Solution - \nPROBLEM What product would you obtain from a base-catalyzed Michael reaction of 2,4-pentanedione with 23-16 each of the following $\\alpha, \\beta$-unsaturated acceptors?\n(a) 2-Cyclohexenone\n(b) Propenenitrile\n(c) Ethyl 2-butenoate\n\nPROBLEM What product would you obtain from a base-catalyzed Michael reaction of 3-buten-2-one with each 23-17 of the following nucleophilic donors?\n(a)\n\n(b)\n\n\nPROBLEM How would you prepare the following compound using a Michael reaction (gray = H, black = C, red 23-18 = O, blue = N)?"}
{"id": 1520, "contents": "Solution - 23.11 Carbonyl Condensations with Enamines: The Stork Enamine Reaction\nIn addition to enolate ions, other kinds of carbon nucleophiles also add to $\\alpha, \\beta$-unsaturated acceptors in Michaellike reactions. Among the most important and useful of such nucleophiles, particularly in biological chemistry, are enamines, which are readily prepared by reaction between a ketone and a secondary amine (Section 19.8). For example:\n\n\nAs the following resonance structures indicate, enamines are electronically similar to enolate ions. Overlap of the nitrogen lone-pair orbital with the double-bond $p$ orbitals leads to an increase in electron density on the $\\alpha$ carbon atom, making that carbon nucleophilic. An electrostatic potential map of $N, N$-dimethylaminoethylene shows this shift of electron density (red) toward the $\\alpha$ position.\n\n\nEnamines behave in much the same way as enolate ions and enter into many of the same kinds of reactions. In the Stork enamine reaction, named for its discoverer, Gilbert Stork of Columbia University, for example, an enamine adds to an $\\alpha, \\beta$-unsaturated carbonyl acceptor in a Michael-like process. The initial product is then hydrolyzed by aqueous acid to yield a 1,5 -dicarbonyl compound. The overall reaction is thus a threestep sequence of (1) enamine formation from a ketone, (2) Michael addition to an $\\alpha, \\beta$-unsaturated carbonyl compound, and (3) enamine hydrolysis back to a ketone.\n\nThe net effect of the Stork enamine reaction is the Michael addition of a ketone to an $\\alpha, \\beta$-unsaturated carbonyl compound. For example, cyclohexanone reacts with the cyclic amine pyrrolidine to yield an enamine; further reaction with an enone such as 3-buten-2-one yields a Michael adduct; and aqueous hydrolysis completes the sequence to give a 1,5-diketone (FIGURE 23.8)."}
{"id": 1521, "contents": "Solution - 23.11 Carbonyl Condensations with Enamines: The Stork Enamine Reaction\nFIGURE 23.8 The Stork enamine reaction between cyclohexanone and 3-buten-2-one. (1) Cyclohexanone is first converted into an enamine, (2) the enamine adds to the $\\alpha, \\beta$-unsaturated ketone in a Michael reaction, and (3) the conjugate addition product is hydrolyzed to yield a 1,5-diketone.\n\nThe enamine-Michael reaction has two advantages over the enolate-ion-Michael reaction that makes it particularly useful in biological pathways. First, an enamine is neutral, easily prepared, and easily handled, while an enolate ion is charged, sometimes difficult to prepare, and must be handled carefully. Second, an enamine from a monoketone can be used in the Michael addition, whereas only enolate ions from $\\beta$-dicarbonyl compounds can be used."}
{"id": 1522, "contents": "Using the Stork Enamine Reaction - \nHow might you use an enamine reaction to prepare the following compound?"}
{"id": 1523, "contents": "Strategy - \nThe overall result of an enamine reaction is the Michael addition of a ketone as donor to an $\\alpha, \\beta$-unsaturated carbonyl compound as acceptor, yielding a 1,5-dicarbonyl product. The $\\mathrm{C}-\\mathrm{C}$ bond formed in the Michael addition step is the one between the $\\alpha$ carbon of the ketone donor and the $\\beta$ carbon of the unsaturated acceptor.\n\nSolution\n\n\nPROBLEM What products would result after hydrolysis from reaction of the enamine prepared from\n23-19 cyclopentanone and pyrrolidine with the following $\\alpha, \\beta$-unsaturated acceptors?\n(a) $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCO}_{2} \\mathrm{Et}$\n(b) $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCHO}$\n(c) $\\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{CHCOCH}_{3}$\n\nPROBLEM Show how you might use an enamine reaction to prepare each of the following compounds:\n23-20\n(a)\n\n(b)"}
{"id": 1524, "contents": "Strategy - 23.12 The Robinson Annulation Reaction\nCarbonyl condensation reactions are perhaps the most versatile methods available for synthesizing complex molecules. By putting a few fundamental reactions together in the proper sequence, some remarkably useful transformations can be carried out. One such example is the Robinson annulation reaction for the synthesis of polycyclic molecules. The word annulation derives from the Latin annulus, meaning \"ring,\" so an annulation reaction is one that builds a new ring onto a molecule.\n\nNamed after Robert Robinson, a British chemist and 1947 Nobel Prize winner, the Robinson annulation is a two-step process that combines a Michael reaction with an intramolecular aldol reaction. It takes place between a nucleophilic donor, such as a $\\beta$-keto ester; an enamine, or a $\\beta$-diketone; and an $\\alpha, \\beta$-unsaturated ketone acceptor, such as 3-buten-2-one. The product is a substituted 2 -cyclohexenone.\n\n\nThe first step of the Robinson annulation is simply a Michael reaction. An enamine or an enolate ion from a $\\beta$-keto ester or $\\beta$-diketone effects a conjugate addition to an $\\alpha, \\beta$-unsaturated ketone, yielding a 1,5 -diketone. But as we saw in Section 23.6, 1,5-diketones undergo intramolecular aldol condensation to yield cyclohexenones when treated with base. Thus, the final product contains a six-membered ring, and an annulation has been accomplished. One example of this occurs during a synthesis of the steroid hormone estrone (FIGURE 23.9)."}
{"id": 1525, "contents": "Strategy - 23.12 The Robinson Annulation Reaction\nFIGURE 23.9 Synthesis of the steroid hormone estrone using a Robinson annulation reaction. The nucleophilic donor is a $\\beta$-diketone.\nIn this example, the $\\beta$-diketone 2 -methyl-1,3-cyclopentanedione is used to generate the enolate ion required for Michael reaction and an aryl-substituted $\\alpha, \\beta$-unsaturated ketone is used as the acceptor. Base-catalyzed Michael reaction between the two partners yields an intermediate triketone, which then cyclizes in an intramolecular aldol condensation to give a Robinson annulation product. Several further transformations are required to complete the synthesis of estrone.\n\nPROBLEM What product would you expect from a Robinson annulation reaction of\n23-21 2-methyl-1,3-cyclopentanedione with 3-buten-2-one?"}
{"id": 1526, "contents": "2-Methyl-1,3-cyclo- 3-Buten-2-one pentanedione - \nPROBLEM How would you prepare the following compound using a Robinson annulation reaction between\n23-22 a $\\beta$-diketone and an $\\alpha, \\beta$-unsaturated ketone? Draw the structures of both reactants and the intermediate Michael addition product."}
{"id": 1527, "contents": "Biological Aldol Reactions - \nAldol reactions occur in many biological pathways but are particularly common in carbohydrate metabolism, where enzymes called aldolases catalyze the addition of a ketone enolate ion to an aldehyde. Aldolases occur in all organisms and are of two types. Type I aldolases occur primarily in animals and higher plants; type II aldolases occur primarily in fungi and bacteria. Both types catalyze the same kind of reaction, but type I aldolases operate through an enamine while type II aldolases require a metal ion (usually $\\mathrm{Zn}^{2+}$ ) as a Lewis acid and operate through an enolate ion.\n\nAn example of an aldolase-catalyzed reaction occurs in glucose biosynthesis when dihydroxyacetone phosphate reacts with glyceraldehyde 3 -phosphate to give fructose 1,6-bisphosphate. In animals and higher plants, dihydroxyacetone phosphate is first converted into an enamine by reaction with the $-\\mathrm{NH}_{2}$ group on a lysine amino acid in the enzyme. The enamine then adds to glyceraldehyde 3-phosphate, and the iminium ion that results is hydrolyzed. In bacteria and fungi, the aldol reaction occurs directly, with the ketone carbonyl group of glyceraldehyde 3-phosphate complexed to a $\\mathrm{Zn}^{2+}$ ion to make it a better acceptor."}
{"id": 1528, "contents": "Type II aldolase - \nGlyceraldehyde 3-phosphate"}
{"id": 1529, "contents": "Fructose - \n1,6-bisphosphate\n\nNote that the aldolase-catalyzed reactions are mixed aldol reactions, which take place between two different partners, as opposed to the symmetrical aldol reactions between identical partners usually carried out in the laboratory. Mixed aldol reactions often give mixtures of products in the laboratory but are successful in living systems because of the selectivity of the enzyme catalysts."}
{"id": 1530, "contents": "Biological Claisen Condensations - \nClaisen condensations, like aldol reactions, also occur in a large number of biological pathways. In fatty-acid biosynthesis, for instance, an enolate ion generated by decarboxylation (Section 22.7) of malonyl ACP adds to the carbonyl group of another acyl group bonded through a thioester linkage to a synthase enzyme. The tetrahedral intermediate that results then expels the synthase, giving acetoacetyl ACP. (The abbreviation ACP stands for acyl carrier protein, which forms thioester bonds to acyl groups.)\n\n\nMixed Claisen condensations also occur frequently in living organisms, particularly in the pathway for fattyacid biosynthesis that we'll discuss in Section 29.4. Butyryl synthase, for instance, reacts with malonyl ACP in a mixed Claisen condensation to give 3-ketohexanoyl ACP."}
{"id": 1531, "contents": "A Prologue to Metabolism - \nBiochemistry is carbonyl chemistry. Almost all metabolic pathways used by living organisms involve one or more of the four fundamental carbonyl-group reactions we've seen in the chapters on Aldehydes and Ketones: Nucleophilic Addition Reactions through Carbonyl Condensation Reactions. The digestion and metabolic breakdown of all the major classes of food molecules-fats, carbohydrates, and proteins-take place by nucleophilic addition reactions, nucleophilic acyl substitutions, $\\alpha$ substitutions, and carbonyl condensations. Similarly, hormones and other crucial biological molecules are built up from smaller precursors by these same carbonyl-group reactions.\n\n\nFIGURE 23.10 You are what you eat. Food molecules are metabolized by pathways that involve the four major carbonyl-group reactions. (credit: \"Spring\" by Giuseppe Arcimboldo, Louvre Museum/Wikimedia Commons, Public Domain)\n\nTake glycolysis, for example, the metabolic pathway by which organisms convert glucose to pyruvate as the first step in extracting energy from carbohydrates.\n\n\nGlycolysis is a ten-step process that begins with isomerization of glucose from its cyclic hemiacetal form to its open-chain aldehyde form-the reverse of a nucleophilic addition reaction. The aldehyde then undergoes tautomerization to yield an enol, which undergoes yet another tautomerization to give the ketone fructose.\n\n\nFructose, a $\\beta$-hydroxy ketone, is then cleaved by a retro-aldol reaction into two three-carbon molecules-one ketone and one aldehyde. Further carbonyl-group reactions then occur until pyruvate is formed.\n\n\nThese few examples are only an introduction; we'll look at several of the major metabolic pathways in more detail in Chapter 29. The bottom line, however, is that you haven't seen the end of carbonyl-group chemistry if you have any interest in biological or medical fields. A solid grasp of carbonyl-group reactions is crucial to an understanding of biochemistry."}
{"id": 1532, "contents": "Key Terms - \n- aldol reaction\n- carbonyl condensation reaction\n- Claisen condensation reaction\n- Dieckmann cyclization\n- Michael reaction\n- Robinson annulation reaction\n- Stork enamine reaction"}
{"id": 1533, "contents": "Summary - \nIn this chapter, we've discussed the fourth and last of the common carbonyl-group reactions-the carbonyl condensation. A carbonyl condensation reaction takes place between two carbonyl partners and involves both nucleophilic addition and $\\alpha$-substitution processes. One carbonyl partner is converted by base into a nucleophilic enolate ion, which then adds to the electrophilic carbonyl group of the second partner. The first partner thus undergoes an $\\alpha$ substitution, while the second undergoes a nucleophilic addition.\n\n\n$$\n\\begin{array}{cc}\n\\text { Nucleophilic } & \\text { Electrophilic } \\\\\n\\text { donor } & \\text { acceptor }\n\\end{array}\n$$\n\nThe aldol reaction is a carbonyl condensation that occurs between two aldehyde or ketone molecules. Aldol reactions are reversible, leading first to $\\beta$-hydroxy aldehydes/ketones and then to $\\alpha, \\beta$-unsaturated products after dehydration. Mixed aldol condensations between two different aldehydes or ketones generally give a mixture of all four possible products. A mixed reaction can be successful, however, if one of the two partners is an unusually good donor (ethyl acetoacetate, for instance) or if it can act only as an acceptor (formaldehyde and benzaldehyde, for instance). Intramolecular aldol condensations of 1,4- and 1,5-diketones are also successful and provide a good way to make five- and six-membered rings.\n\nThe Claisen condensation reaction is a carbonyl condensation that occurs between two ester components and gives a $\\beta$-keto ester product. Mixed Claisen condensations between two different esters are successful only when one of the two partners has no acidic $\\alpha$ hydrogens (ethyl benzoate and ethyl formate, for instance) and thus can function only as the acceptor partner. Intramolecular Claisen condensations, called Dieckmann cyclization reactions, yield five- and six-membered cyclic $\\beta$-keto esters starting from 1,6-and 1,7-diesters."}
{"id": 1534, "contents": "Summary - \nThe conjugate addition of a carbon nucleophile to an $\\alpha, \\beta$-unsaturated acceptor is known as the Michael reaction. The best Michael reactions take place between relatively acidic donors ( $\\beta$-keto esters or $\\beta$-diketones) and unhindered $\\alpha, \\beta$-unsaturated acceptors. Enamines, prepared by reaction of a ketone with a disubstituted amine, are also good Michael donors in the Stork enamine reaction.\n\nCarbonyl condensation reactions are widely used in synthesis. One example of their versatility is the Robinson annulation reaction, which leads to the formation of a substituted cyclohexenone. Treatment of a $\\beta$-diketone or $\\beta$-keto ester with an $\\alpha, \\beta$-unsaturated ketone leads first to a Michael addition, which is followed by intramolecular aldol cyclization. Condensation reactions are also used widely in nature for the biosynthesis of such molecules as fats and steroids."}
{"id": 1535, "contents": "Summary of Reactions - \n1. Aldol reaction (Section 23.1)\n\n2. Mixed aldol reaction (Section 23.5)\n\n\n3. Intramolecular aldol reaction (Section 23.6)\n\n4. Dehydration of aldol products (Section 23.3)\n\n5. Claisen condensation reaction (Section 23.7)\n\n6. Mixed Claisen condensation reaction (Section 23.8)\n\n7. Intramolecular Claisen condensation (Dieckmann cyclization; Section 23.9)\n\n\n8. Michael reaction (Section 23.10)\n\n9. Carbonyl condensations with enamines (Stork enamine reaction; Section 23.11)"}
{"id": 1536, "contents": "Additional Problems - \nVisualizing Chemistry\nPROBLEM What ketones or aldehydes might the following enones have been prepared from by aldol reaction?\n23-23 (a)\n\n(b)\n\n\nPROBLEM The following structure represents an intermediate formed by addition of an ester enolate ion to a\n23-24 second ester molecule. Identify the reactant, the leaving group, and the product.\n\n\nPROBLEM The following molecule was formed by an intramolecular aldol reaction. What dicarbonyl precursor 23-25 was used for its preparation?\n\n\nPROBLEM The following molecule was formed by a Robinson annulation reaction. What reactants were used? 23-26"}
{"id": 1537, "contents": "Mechanism Problems - \nPROBLEM Predict the addition product for each of the following reactions and write the mechanism.\n\n23-27 (a)\n\n(b)\n\n\nPROBLEM Based on your answers to Problem 23-27, predict the dehydration product for both reactions and 23-28 write the mechanism.\n\nPROBLEM Predict the product(s) and provide the mechanisms for the following reactions.\n23-29 (a)\n\n(b)\n\n\nPROBLEM Predict the product(s) and provide the mechanisms for the following reactions.\n23-30 (a)\n\n(b)\n\n\nPROBLEM Predict the product(s) and provide the mechanisms for the following reactions.\n23-31 (a)\n\n(b)\n\n\nPROBLEM Knoevenagel condensation is a reaction involving an active methylene compound (a $\\mathrm{CH}_{2}$ flanked\n23-32 by two electron-withdrawing groups) and an aldehyde and ketone. Propose a mechanism for the reaction below.\n\n\nPROBLEM Azlactones are important starting materials used in the synthesis of dehydro $\\alpha$-aminoacids. They\n23-33 react with aldehydes to form an intermediate that is hydrolyzed under acidic conditions to give the final amino acid product. Provide the structure of the intermediate and propose a mechanism for its formation."}
{"id": 1538, "contents": "Azlactone - \nDeydro- $\\alpha$-amino acid\nPROBLEM Leucine, one of the twenty amino acids found in proteins, is metabolized by a pathway that includes 23-34 the following step. Propose a mechanism.\n\n\nPROBLEM Isoleucine, another of the twenty amino acids found in proteins, is metabolized by a pathway that 23-35 includes the following step. Propose a mechanism.\n\n\nPROBLEM The first step in the citric acid cycle of food metabolism is reaction of oxaloacetate with acetyl CoA 23-36 to give citrate. Propose a mechanism, using acid or base catalysis as needed.\n\n\nPROBLEM The amino acid leucine is biosynthesized from $\\alpha$-ketoisovalerate by the following sequence of steps.\n23-37 Show the mechanism of each.\n\n\nPROBLEM The Knoevenagel reaction was introduced in Problem 23-2. Show the mechanism for the 23-38 Knoevenagel reaction of diethyl malonate with benzaldehyde.\n\n\nBenzaldehyde\nCinnamic acid (91\\%)\nPROBLEM The Darzens reaction involves a two-step, base-catalyzed condensation of ethyl chloroacetate with\n23-39 a ketone to yield an epoxy ester. The first step is a carbonyl condensation reaction, and the second step is an $\\mathrm{S}_{\\mathrm{N}} 2$ reaction. Write both steps, and show their mechanisms.\n\n\nPROBLEM The following reaction involves a hydrolysis followed by an intramolecular nucleophilic acyl 23-40 substitution reaction. Write both steps, and show their mechanisms.\n\n\nPROBLEM The following reaction involves an intramolecular Michael reaction followed by an intramolecular 23-41 aldol reaction. Write both steps, and show their mechanisms.\n\n\nPROBLEM The following reaction involves a conjugate addition reaction followed by an intramolecular Claisen\n23-42 condensation. Write both steps, and show their mechanisms.\n\n\nPROBLEM The following reaction involves an intramolecular aldol reaction followed by a retro aldol-like 23-43 reaction. Write both steps, and show their mechanisms.\n\n\nPROBLEM Propose a mechanism for the following base-catalyzed isomerization:\n23-44"}
{"id": 1539, "contents": "Azlactone - \nPROBLEM Propose a mechanism for the following base-catalyzed isomerization:\n23-44\n\n\nPROBLEM The Mannich reaction of a ketone, an amine, and an aldehyde is one of the few three-component\n23-45 reactions in organic chemistry. Cyclohexanone, for example, reacts with dimethylamine and acetaldehyde to yield an amino ketone. The reaction takes place in two steps, both of which are typical carbonyl-group reactions.\n\n(a) The first step is reaction between the aldehyde and the amine to yield an intermediate iminium ion $\\left(\\mathrm{R}_{2} \\mathrm{C}=\\mathrm{NR}_{2}{ }^{+}\\right)$plus water. Propose a mechanism, and show the structure of the intermediate iminium ion.\n(b) The second step is reaction between the iminium ion intermediate and the ketone to yield the final product. Propose a mechanism.\n\nPROBLEM Cocaine has been prepared by a sequence beginning with a Mannich reaction (Problem 23-45)\n23-46 between dimethyl acetonedicarboxylate, an amine, and a dialdehyde. Show the structures of the amine and dialdehyde.\n\n\nPROBLEM Propose a mechanism to account for the following reaction of an enamine with an alkyl halide: 23-47"}
{"id": 1540, "contents": "Aldol Reactions - \nPROBLEM Which of the following compounds would you expect to undergo aldol self-condensation? Show the 23-48 product of each successful reaction.\n(a) Trimethylacetaldehyde\n(b) Cyclobutanone\n(c) Benzophenone (diphenyl ketone)\n(d) 3-Pentanone\n(e) Decanal\n(f) 3-Phenyl-2-propenal\n\nPROBLEM How might you synthesize each of the following compounds using an aldol reaction? Show the 23-49 structure of the starting aldehyde(s) or ketone(s) you would use in each case.\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM What product would you expect to obtain from aldol cyclization of hexanedial, 23-50 $\\mathrm{OHCCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CHO}$ ?\n\nPROBLEM Intramolecular aldol cyclization of 2,5-heptanedione with aqueous NaOH yields a mixture of two 23-51 enone products in the approximate ratio 9:1. Write their structures, and show how each is formed.\n\nPROBLEM The major product formed by intramolecular aldol cyclization of 2,5-heptanedione (Problem 23-51)\n23-52 has two singlet absorptions in the ${ }^{1} \\mathrm{H}$ NMR spectrum, at $1.65 \\delta$ and $1.90 \\delta$, and has no absorptions in the range 3 to $10 \\delta$. What is its structure?\n\nPROBLEM Treatment of the minor product formed in the intramolecular aldol cyclization of 2,5-heptanedione 23-53 (Problems 23-51 and 23-52) with aqueous NaOH converts it into the major product. Propose a mechanism to account for this base-catalyzed isomerization.\n\nPROBLEM How can you account for the fact that 2,2,6-trimethylcyclohexanone yields no detectable aldol 23-54 product even though it has an acidic $\\alpha$ hydrogen?"}
{"id": 1541, "contents": "Aldol Reactions - \nPROBLEM How can you account for the fact that 2,2,6-trimethylcyclohexanone yields no detectable aldol 23-54 product even though it has an acidic $\\alpha$ hydrogen?\n\nPROBLEM The aldol reaction is catalyzed by acid as well as base. What is the reactive nucleophile in the acid-23-55 catalyzed aldol reaction? Propose a mechanism.\n\nPROBLEM Cinnamaldehyde, the aromatic constituent of cinnamon oil, can be synthesized by a mixed aldol 23-56 condensation. Show the starting materials you would use, and write the reaction."}
{"id": 1542, "contents": "Cinnamaldehyde - \nPROBLEM The following bicyclic ketone does not undergo aldol self-condensation even though it has two $\\alpha$ 23-57 hydrogen atoms. Explain."}
{"id": 1543, "contents": "Claisen Condensations - \nPROBLEM Give the structures of the possible Claisen condensation products from the following reactions. Tell 23-58 which, if any, you would expect to predominate in each case.\n(a) $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{Et}+\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{Et}$\n(b) $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CO}_{2} \\mathrm{Et}+\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{Et}$\n(c) $\\mathrm{EtOCO}_{2} \\mathrm{Et}+$ cyclohexanone\n(d) $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CHO}+\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{Et}$\n\nPROBLEM In the mixed Claisen reaction of cyclopentanone with ethyl formate, a much higher yield of the\n23-59 desired product is obtained by first mixing the two carbonyl components and then adding base, rather than by first mixing base with cyclopentanone and then adding ethyl formate. Explain.\n\nPROBLEM Ethyl dimethylacetoacetate reacts instantly at room temperature when treated with ethoxide ion\n23-60 to yield two products, ethyl acetate and ethyl 2-methylpropanoate. Propose a mechanism for this cleavage reaction.\n\n\nPROBLEM In contrast to the rapid reaction shown in Problem 23-60, ethyl acetoacetate requires a temperature\n23-61 over $150^{\\circ} \\mathrm{C}$ to undergo the same kind of cleavage reaction. How can you explain the difference in reactivity?"}
{"id": 1544, "contents": "Michael and Enamine Reactions - \nPROBLEM How might the following compounds be prepared using Michael reactions? Show the nucleophilic 23-62 donor and the electrophilic acceptor in each case.\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n\nPROBLEM The so-called Wieland-Miescher ketone is a valuable starting material used in the synthesis of 23-63 steroid hormones. How might you prepare it from 1,3-cyclohexanedione?\n\n\nWieland-Miescher ketone\n\nPROBLEM The Stork enamine reaction and the intramolecular aldol reaction can be carried out in sequence 23-64 to allow the synthesis of cyclohexenones. For example, reaction of the pyrrolidine enamine of cyclohexanone with 3-buten-2-one, followed by enamine hydrolysis and base treatment, yields the product indicated. Write each step, and show the mechanism of each.\n\n\n\nPROBLEM How could you prepare the following cyclohexenones by combining a Stork enamine reaction with\n23-65 an intramolecular aldol condensation? (See Problem 23-64.)\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM The following reaction involves two successive intramolecular Michael reactions. Write both steps, 23-66 and show their mechanisms."}
{"id": 1545, "contents": "General Problems - \nPROBLEM What condensation products would you expect to obtain by treatment of the following substances 23-67 with sodium ethoxide in ethanol?\n(a) Ethyl butanoate\n(b) Cycloheptanone\n(c) 3,7-Nonanedione\n(d) 3-Phenylpropanal\n\nPROBLEM The following reactions are unlikely to provide the indicated product in high yield. What is wrong 23-68 with each?\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Fill in the missing reagents a-h in the following scheme:\n23-69\n\n\nPROBLEM How would you prepare the following compounds from cyclohexanone?\n\n23-70 (a)\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM The compound known as Hagemann's ester is prepared by treatment of a mixture of formaldehyde 23-71 and ethyl acetoacetate with base, followed by acid-catalyzed decarboxylation.\n\n\nHagemann's ester\n(a) The first step is an aldol-like condensation between ethyl acetoacetate and formaldehyde to yield an $\\alpha, \\beta$-unsaturated product. Write the reaction, and show the structure of the product.\n(b) The second step is a Michael reaction between ethyl acetoacetate and the unsaturated product of the first step. Show the structure of the product.\n\nPROBLEM The third and fourth steps in the synthesis of Hagemann's ester from ethyl acetoacetate and\n23-72 formaldehyde (Problem 23-71) are an intramolecular aldol cyclization to yield a substituted cyclohexenone, and a decarboxylation reaction. Write both reactions, and show the products of each step.\n\nPROBLEM When 2-methylcyclohexanone is converted into an enamine, only one product is formed despite\n23-73 the fact that the starting ketone is unsymmetrical. Build molecular models of the two possible products and explain the fact that the sole product is the one with the double bond opposite the methyl-substituted carbon."}
{"id": 1546, "contents": "Amines and Heterocycles - \nFIGURE 24.1 The characteristic and unmistakable odor of fish is due to a mixture of simple alkylamines. (credit: modification of work \"Pike Place\" by Sharat Ganapati/Flickr, CC BY 2.0)"}
{"id": 1547, "contents": "CHAPTER CONTENTS - 24.1 Naming Amines\n24.2 Structure and Properties of Amines\n24.3 Basicity of Amines\n24.4 Basicity of Arylamines\n24.5 Biological Amines and the Henderson-Hasselbalch Equation\n24.6 Synthesis of Amines\n24.7 Reactions of Amines\n24.8 Reactions of Arylamines\n24.9 Heterocyclic Amines\n24.10 Spectroscopy of Amines\n\nWHY THIS CHAPTER? By the end of this chapter, we will have seen all the common functional groups. Of those groups, carbonyl compounds and amines are the most abundant and have the richest chemistry. In addition to proteins and nucleic acids, the majority of pharmaceutical agents contain amine functional groups, and many of the common coenzymes necessary for biological catalysis are amines.\n\nAmines are organic derivatives of ammonia in the same way that alcohols and ethers are organic derivatives of water. Like ammonia, amines contain a nitrogen atom with a lone pair of electrons, making amines both basic and nucleophilic. We'll soon see, in fact, that most of the chemistry of amines depends on the presence of this lone pair of electrons.\n\nAmines occur widely in all living organisms. Trimethylamine, for instance, occurs in animal tissues and is partially responsible for the distinctive odor of fish; nicotine is found in tobacco; and cocaine is a stimulant found in the leaves of the South American coca bush. In addition, amino acids are the building blocks from\nwhich all proteins are made, and cyclic amine bases are constituents of nucleic acids.\n\n\nTrimethylamine\n\n\nNicotine\n\n\nCocaine"}
{"id": 1548, "contents": "CHAPTER CONTENTS - 24.1 Naming Amines\nAmines can be either alkyl-substituted (alkylamines) or aryl-substituted (arylamines). Although much of the chemistry of the two classes is similar, there are also substantial differences. Amines are classified as primary ( $\\mathbf{R N H}_{\\mathbf{2}}$ ), secondary ( $\\left.\\mathbf{R}_{\\mathbf{2}} \\mathbf{N H}\\right)$, or tertiary $\\left(\\mathbf{R}_{\\mathbf{3}} \\mathbf{N}\\right)$, depending on the number of organic substituents attached to nitrogen. Thus, methylamine $\\left(\\mathrm{CH}_{3} \\mathrm{NH}_{2}\\right)$ is a primary amine, dimethylamine $\\left[\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NH}\\right]$ is a secondary amine, and trimethylamine $\\left[\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{~N}\\right]$ is a tertiary amine. Note that this usage of the terms primary, secondary, and tertiary differs from our previous usage. When we speak of a tertiary alcohol or alkyl halide, we refer to the degree of substitution at the alkyl carbon atom, but when we speak of a tertiary amine, we refer to the degree of substitution at the nitrogen atom.\n\ntert-Butyl alcohol (a tertiary alcohol)\n\n\nTrimethylamine (a tertiary amine)\n\ntert-Butylamine\n(a primary amine)\n\nCompounds containing a nitrogen atom with four attached groups also exist, but the nitrogen atom must carry a formal positive charge. Such compounds are called quaternary ammonium salts."}
{"id": 1549, "contents": "A quaternary ammonium salt - \nPrimary amines are named in the IUPAC system in several ways. For simple amines, the suffix -amine is added to the name of the alkyl substituent. You might also recall from the chapter on Benzene and Aromaticity that the aromatic phenylamine, $\\mathrm{H}_{2} \\mathrm{~N}-\\mathrm{C}_{6} \\mathrm{H}_{5}$, has the common name aniline.\n\ntert-Butylamine\n\n\nCyclohexylamine\n\n\nAlternatively, the suffix -amine can be used in place of the final -e in the name of the parent compound.\n\n\n4,4-Dimethylcyclohexanamine\n$\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}$\n\n1,4-Butanediamine\n\nAmines with more than one functional group are named by considering the $-\\mathrm{NH}_{2}$ as an amino substituent on the parent molecule.\n\n\nSymmetrical secondary and tertiary amines are named by adding the prefix di- or tri- to the alkyl group.\n\n\nDiphenylamine\n\n\nTriethylamine\n\nUnsymmetrically substituted secondary and tertiary amines are referred to as $N$-substituted primary amines. The largest alkyl group takes the parent name, and the other alkyl groups are considered $N$-substituents on the parent ( $N$ because they're attached to nitrogen).\n\n\n$\\mathbf{N}, \\mathbf{N}$-Dimethylpropylamine\n$\\boldsymbol{N}$-Ethyl- $\\boldsymbol{N}$-methylcyclohexylamine\nHeterocyclic amines-compounds in which the nitrogen atom occurs as part of a ring-are also common, and each different heterocyclic ring system has its own parent name. The heterocyclic nitrogen atom is always numbered as position 1.\n\n\nPyridine\n\n\nPyrrole\n\n\nQuinoline\n\n\nImidazole"}
{"id": 1550, "contents": "Indole - \nPyrimidine\n\n\nPyrrolidine\n\n\nPiperidine\n\nPROBLEM Name the following compounds:\n24-1 (a) $\\mathrm{CH}_{3} \\mathrm{NHCH}_{2} \\mathrm{CH}_{3}$\n(b)\n\n(c)\n\n(d)\n\n\n\nPROBLEM Draw structures corresponding to the following IUPAC names:\n24-2 (a) Triisopropylamine (b) Triallylamine (c) $N$-Methylaniline\n(d) $N$-Ethyl- $N$-methylcyclopentylamine (e) $N$-Isopropylcyclohexylamine (f) $N$-Ethylpyrrole\n\nPROBLEM Draw structures for the following heterocyclic amines:\n24-3 (a) 5-Methoxyindole\n(b) 1,3-Dimethylpyrrole\n(c) 4-( $N, N$-Dimethylamino)pyridine\n(d) 5-aminopyrimidine"}
{"id": 1551, "contents": "Indole - 24.2 Structure and Properties of Amines\nThe bonding in alkylamines is similar to the bonding in ammonia. The nitrogen atom is $s p^{3}$-hybridized, with its three substituents occupying three corners of a regular tetrahedron and the lone pair of electrons occupying the fourth corner. As you might expect, the $\\mathrm{C}-\\mathrm{N}-\\mathrm{C}$ bond angles are close to the $109^{\\circ}$ tetrahedral value. For trimethylamine, the $\\mathrm{C}-\\mathrm{N}-\\mathrm{C}$ bond angle is $108^{\\circ}$ and the $\\mathrm{C}-\\mathrm{N}$ bond length is 147 pm .\n\n\n\nTrimethylamine\nOne consequence of tetrahedral geometry is that an amine with three different substituents on nitrogen is chiral, as we saw in Section 5.10. Unlike chiral carbon compounds, however, chiral amines can't usually be resolved because the two enantiomeric forms rapidly interconvert by a pyramidal inversion, much as an alkyl halide inverts in an $\\mathrm{S}_{\\mathrm{N}} 2$ reaction. Pyramidal inversion occurs by a momentary rehybridization of the nitrogen atom to planar, $s p^{2}$ geometry, followed by rehybridization of the planar intermediate to tetrahedral, $s p^{3}$ geometry (FIGURE 24.2). The barrier to inversion is about $25 \\mathrm{~kJ} / \\mathrm{mol}$ ( $6 \\mathrm{kcal} / \\mathrm{mol}$ ), an amount only twice as large as the barrier to rotation about a $\\mathrm{C}-\\mathrm{C}$ single bond."}
{"id": 1552, "contents": "Indole - 24.2 Structure and Properties of Amines\nFIGURE 24.2 Pyramidal inversion rapidly interconverts the two mirror-image (enantiomeric) forms of an amine.\nAlkylamines have a variety of applications in the chemical industry as starting materials for the preparation of insecticides and pharmaceuticals. Labetalol, for instance, a so-called $\\beta$-blocker used for the treatment of high blood pressure, is prepared by $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of an epoxide with a primary amine. The substance marketed for drug use is a mixture of all four possible stereoisomers, but the biological activity results primarily from the $(R, R)$ isomer.\n\n\nLabetalol\nLike alcohols, amines with fewer than five carbon atoms are generally water-soluble. Also like alcohols, primary and secondary amines form hydrogen bonds and are highly associated. As a result, amines have higher boiling points than alkanes of similar molecular weight. Diethylamine ( $\\mathrm{MW}=73 \\mathrm{amu}$ ) boils at $56.3^{\\circ} \\mathrm{C}$, for instance, while pentane ( $\\mathrm{MW}=72 \\mathrm{amu}$ ) boils at $36.1^{\\circ} \\mathrm{C}$.\n\n\nFIGURE 24.3 Hydrogen bonding in amines.\nOne other characteristic of amines is their odor. Low-molecular-weight amines such as trimethylamine have a distinctive fishlike aroma, while diamines such as cadaverine (1,5-pentanediamine) and putrescine (1,4-butanediamine) have the appalling odors you might expect from their common names. Both of these diamines arise from the decomposition of decaying tissue."}
{"id": 1553, "contents": "Indole - 24.3 Basicity of Amines\nThe chemistry of amines is dominated by the lone pair of electrons on nitrogen, which makes amines both basic and nucleophilic. They react with acids to form acid-base salts, and they react with electrophiles in many of the polar reactions seen in past chapters. Note in the following electrostatic potential map of trimethylamine how the negative (red) region corresponds to the lone pair of electrons on nitrogen.\n\n\nAmines are much stronger bases than alcohols and ethers, their oxygen-containing analogs. When an amine is dissolved in water, an equilibrium is established in which water acts as an acid and transfers a proton to the amine. Just as the acid strength of a carboxylic acid can be measured by defining an acidity constant $K_{\\mathrm{a}}$ (Section 2.8), the base strength of an amine can be measured by defining an analogous basicity constant $K_{b}$. The larger the value of $K_{\\mathrm{b}}$ and the smaller the value of $\\mathrm{p} K_{\\mathrm{b}}$, the more favorable the proton-transfer equilibrium and the stronger the base.\n\nFor the reaction\n\n$$\n\\begin{aligned}\n& \\mathrm{RNH}_{2}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftarrows \\mathrm{RNH}_{3}^{+}+\\mathrm{OH}^{-} \\\\\n& K_{\\mathrm{b}}=\\frac{\\left[\\mathrm{RNH}_{3}{ }^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]}{\\left[\\mathrm{RNH}_{2}\\right]} \\\\\n& \\mathrm{p} K_{\\mathrm{b}}=-\\log K_{\\mathrm{b}}\n\\end{aligned}\n$$\n\nIn practice, $K_{\\mathrm{b}}$ values are not often used. Instead, the most convenient way to measure the basicity of an amine $\\left(\\mathrm{RNH}_{2}\\right)$ is to look at the acidity of the corresponding ammonium ion $\\left(\\mathrm{RNH}_{3}{ }^{+}\\right)$.\n\nFor the reaction"}
{"id": 1554, "contents": "Indole - 24.3 Basicity of Amines\nFor the reaction\n\n$$\n\\begin{aligned}\n& \\mathrm{RNH}_{3}{ }^{+}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftarrows \\mathrm{RNH}_{2}+\\mathrm{H}_{3} \\mathrm{O}^{+} \\\\\n& K_{\\mathrm{a}}=\\frac{\\left[\\mathrm{RNH}_{2}\\right]\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]}{\\left[\\mathrm{RNH}_{3}{ }^{+}\\right]}\n\\end{aligned}\n$$\n\nso\n\n$$\n\\begin{aligned}\nK_{\\mathrm{a}} & \\cdot K_{\\mathrm{b}}=\\left[\\frac{\\left[\\mathrm{RNH}_{2}\\right]\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]}{\\left[\\mathrm{RNH}_{3}{ }^{+}\\right]}\\right]\\left[\\frac{\\left[\\mathrm{RNH}_{3}{ }^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]}{\\left[\\mathrm{RNH}_{2}\\right]}\\right] \\\\\n& =\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]=K_{\\mathrm{w}}=1.00 \\times 10^{-14}\n\\end{aligned}\n$$\n\nThus\n\n$$\nK_{\\mathrm{a}}=\\frac{K_{\\mathrm{w}}}{K_{\\mathrm{b}}} \\quad \\text { and } \\quad K_{\\mathrm{b}}=\\frac{K_{\\mathrm{w}}}{K_{\\mathrm{a}}}\n$$\n\nand\n\n$$\n\\mathrm{p} K_{\\mathrm{a}}+\\mathrm{p} K_{\\mathrm{b}}=14\n$$"}
{"id": 1555, "contents": "Indole - 24.3 Basicity of Amines\nand\n\n$$\n\\mathrm{p} K_{\\mathrm{a}}+\\mathrm{p} K_{\\mathrm{b}}=14\n$$\n\nThese equations say that the $K_{\\mathrm{b}}$ of an amine multiplied by the $K_{\\mathrm{a}}$ of the corresponding ammonium ion is equal to $K_{\\mathrm{W}}$, the ion-product constant for water ( $1.00 \\times 10^{-14}$ ). Thus, if we know $K_{\\mathrm{a}}$ for an ammonium ion, we also know $K_{\\mathrm{b}}$ for the corresponding amine base because $K_{\\mathrm{b}}=K_{\\mathrm{w}} / K_{\\mathrm{a}}$. The more acidic the ammonium ion, the less tightly the proton is held and the weaker the corresponding base. That is, a weaker base has an ammonium ion with a smaller $\\mathrm{p} K_{\\mathrm{a}}$ and a stronger base has an ammonium ion with a larger $\\mathrm{p} K_{\\mathrm{a}}$."}
{"id": 1556, "contents": "Weaker base Smaller $\\mathrm{p} K_{\\mathrm{a}}$ for ammonium ion <br> Stronger base Larger $\\mathrm{p} K_{\\mathrm{a}}$ for ammonium ion - \nTABLE 24.1 lists $\\mathrm{p} K_{\\mathrm{a}}$ values of the ammonium ions from a variety of amines and indicates that there is a substantial range of amine basicities. Most simple alkylamines are similar in their base strength, with $\\mathrm{p} K_{\\mathrm{a}}$ 's for their ammonium ions in the narrow range 10 to 11. Arylamines, however, are considerably less basic than alkylamines, as are the heterocyclic amines pyridine and pyrrole.\n\nTABLE 24.1 Basicity of Some Common Amines\n\n| Name | Structure | $\\mathrm{p} K_{\\mathrm{a}}$ of ammonium ion |\n| :--- | :--- | :--- |\n| Ammonia | $\\mathrm{NH}_{3}$ | 9.26 |\n| Primary alkylamine |  |  |\n| Methylamine |  | $\\mathrm{CH}_{3} \\mathrm{NH}_{2}$ |"}
{"id": 1557, "contents": "Secondary alkylamine - \nTABLE 24.1 Basicity of Some Common Amines\n\n| Name | Structure | $\\mathrm{p} K_{\\mathrm{a}}$ of ammonium ion |\n| :---: | :---: | :---: |\n| Diethylamine | $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2}\\right)_{2} \\mathrm{NH}$ | 10.98 |\n| Pyrrolidine | <smiles>C1CCNC1</smiles> | 11.27 |\n| Tertiary alkylamine |  |  |\n| Triethylamine | $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2}\\right)_{3} \\mathrm{~N}$ | 10.76 |\n| Arylamine |  |  |\n| Aniline | <smiles>Nc1ccccc1</smiles> | 4.63 |\n| Heterocyclic amine |  |  |\n| Pyridine | <smiles>c1ccncc1</smiles> | 5.25 |\n| Pyrimidine | <smiles>c1cncnc1</smiles> | 1.3 |\n| Pyrrole | <smiles>c1cc[nH]c1</smiles> | 0.4 |\n| Imidazole | <smiles>c1c[nH]cn1</smiles> | 6.95 |\n\nIn contrast with amines, amides $\\left(\\mathrm{RCONH}_{2}\\right)$ are nonbasic. Amides aren't protonated by aqueous acids, and they are poor nucleophiles. The main reason for this difference in basicity between amines and amides is that an amide is stabilized by delocalization of the nitrogen lone-pair electrons through orbital overlap with the carbonyl group. In resonance terms, amides are more stable and less reactive than amines because they are hybrids of two resonance forms. This amide resonance stabilization is lost when the nitrogen atom is protonated, so protonation is disfavored. The following electrostatic potential maps clearly show a reduced electron density on the amide nitrogen."}
{"id": 1558, "contents": "Secondary alkylamine - \nTo purify amines, it's often possible to take advantage of their basicity. For example, if a mixture of a basic amine and a neutral compound such as a ketone or alcohol is dissolved in an organic solvent and aqueous acid is added, the basic amine dissolves in the water layer as its protonated salt, while the neutral compound remains in the organic solvent layer. Separation of the water layer and neutralization of the ammonium ion by addition of NaOH then provides the pure amine (FIGURE 24.4).\n\n\nFIGURE 24.4 Separation and purification of an amine component from a mixture by extraction of its ammonium salt into water.\nIn addition to their behavior as bases, primary and secondary amines can also act as very weak acids because an $\\mathrm{N}-\\mathrm{H}$ proton can be removed by a sufficiently strong base. We've seen, for example, how diisopropylamine ( $\\mathrm{p} K_{\\mathrm{a}} \\approx 36$ ) reacts with butyllithium to yield lithium diisopropylamide (LDA; Section 22.5). Dialkylamine anions like LDA are very strong bases that are often used in laboratory organic chemistry for the generation of enolate ions from carbonyl compounds (Section 22.7). They are not, however, encountered in biological chemistry.\n\n\nPROBLEM Which compound in each of the following pairs is more basic?\n24-4 (a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{NH}_{2}$ or $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CONH}_{2}$\n(b) NaOH or $\\mathrm{CH}_{3} \\mathrm{NH}_{2}$\n(c) $\\mathrm{CH}_{3} \\mathrm{NHCH}_{3}$ or pyridine"}
{"id": 1559, "contents": "Secondary alkylamine - \nPROBLEM The benzylammonium ion $\\left(\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CH}_{2} \\mathrm{NH}_{3}{ }^{+}\\right)$has $\\mathrm{p} K_{\\mathrm{a}}=9.33$, and the propylammonium ion has $\\mathrm{p} K_{\\mathrm{a}}=$\n24-5 10.71. Which is the stronger base, benzylamine or propylamine? What are the $\\mathrm{p} K_{\\mathrm{b}}$ 's of benzylamine and propylamine?"}
{"id": 1560, "contents": "Secondary alkylamine - 24.4 Basicity of Arylamines\nAs noted previously, arylamines are generally less basic than alkylamines. Anilinium ion has $\\mathrm{p} K_{\\mathrm{a}}=4.63$, for instance, whereas methylammonium ion has $\\mathrm{p} K_{\\mathrm{a}}=10.64$. Arylamines are less basic than alkylamines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic ring's $\\pi$ electron system and are less available for bonding to $\\mathrm{H}^{+}$. In resonance terms, arylamines are stabilized relative to alkylamines\nbecause of their five resonance forms.\n\n\nMuch of the resonance stabilization is lost on protonation, however, so the energy difference between protonated and nonprotonated forms is higher for arylamines than it is for alkylamines. As a result, arylamines are less basic. FIGURE 24.5 illustrates this difference.\n\n\n\nAniline (delocalized electrons)\n\n\nAnilinium ion (localized charge)\n\nFIGURE 24.5 Arylamines have a larger positive $\\Delta G^{\\circ}$ for protonation and are therefore less basic than alkylamines, primarily because of resonance stabilization of the ground state. Electrostatic potential maps show that lone-pair electron density is delocalized in the amine but the charge is localized in the corresponding ammonium ion.\n\nSubstituted arylamines can be either more basic or less basic than aniline, depending on the substituent. Electron-donating substituents, such as $-\\mathrm{CH}_{3},-\\mathrm{NH}_{2}$, and $-\\mathrm{OCH}_{3}$, which increase the reactivity of an aromatic ring toward electrophilic substitution (Section 16.4), also increase the basicity of the corresponding arylamine. Electron-withdrawing substituents, such as $-\\mathrm{Cl},-\\mathrm{NO}_{2}$, and -CN , which decrease ring reactivity toward electrophilic substitution, also decrease arylamine basicity. TABLE 24.2 considers only $p$-substituted anilines, but similar trends are observed for ortho and meta derivatives.\n\nTABLE 24.2 Base Strength of Some p-Substituted Anilines"}
{"id": 1561, "contents": "Secondary alkylamine - 24.4 Basicity of Arylamines\nTABLE 24.2 Base Strength of Some p-Substituted Anilines\n\n\n|  | Substit | $\\mathrm{p} K_{\\mathrm{a}}$ |  |\n| :---: | :---: | :---: | :---: |\n| Stronger base | $-\\mathrm{NH}_{2}$ | 6.15 | Activating groups |\n|  | $-\\mathrm{OCH}_{3}$ | 5.34 |  |\n|  | $-\\mathrm{CH}_{3}$ | 5.08 |  |\n|  | -H | 4.63 |  |\n| Weaker base | $-\\mathrm{Cl}$ | 3.98 | Deactivating groups |\n|  | $-\\mathrm{Br}$ | 3.86 |  |\n|  | -CN | 1.74 |  |\n|  | $-\\mathrm{NO}_{2}$ | 1.00 |  |\n\nPROBLEM Without looking at Table 24.2, rank the following compounds in order of ascending basicity.\n24-6 (a) $p$-Nitroaniline, $p$-aminobenzaldehyde, $p$-bromoaniline\n(b) $p$-Chloroaniline, $p$-aminoacetophenone, $p$-methylaniline\n(c) $p$-(Trifluoromethyl)aniline, $p$-methylaniline, $p$-(fluoromethyl)aniline"}
{"id": 1562, "contents": "Secondary alkylamine - 24.5 Biological Amines and the Henderson-Hasselbalch Equation\nWe saw in Section 20.3 that the extent of dissociation of a carboxylic acid HA in an aqueous solution buffered to a given pH can be calculated with the Henderson-Hasselbalch equation. Furthermore, we concluded that at the physiological pH of 7.3 inside living cells, carboxylic acids are almost entirely dissociated into their carboxylate anions, $\\mathrm{RCO}_{2}{ }^{-}$."}
{"id": 1563, "contents": "Henderson-Hasselbalch equation: - \n$$\n\\mathrm{pH}=\\mathrm{p} K \\mathrm{a}+\\log \\frac{\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]} \\text { so } \\log \\frac{\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]}=\\mathrm{pH}-\\mathrm{p} K \\mathrm{a}\n$$\n\nWhat about amine bases? In what form do they exist at physiological pH ? As the amine $\\left(\\mathrm{A}^{-}=\\mathrm{RNH}_{2}\\right)$, or as the ammonium ion $\\left(\\mathrm{HA}=\\mathrm{RNH}_{3}{ }^{+}\\right)$? Let's take a 0.0010 M solution of methylamine at $\\mathrm{pH}=7.3$, for example. According to TABLE 24.1, the $\\mathrm{p} K_{\\mathrm{a}}$ of methylammonium ion is 10.64 , so from the Henderson-Hasselbalch equation, we have\n\n$$\n\\begin{aligned}\n& \\log \\frac{\\left[\\mathrm{RNH}_{2}\\right]}{\\left[\\mathrm{RNH}_{3}+\\right]}=\\mathrm{pH}-\\mathrm{p} K_{\\mathrm{a}}=7.3-10.64=-3.34 \\\\\n& \\frac{\\left[\\mathrm{RNH}_{2}\\right]}{\\left[\\mathrm{RNH}_{3}{ }^{+}\\right]}=\\operatorname{antilog}(-3.34)=4.6 \\times 10^{-4} \\text { so }\\left[\\mathrm{RNH}_{2}\\right]=\\left(4.6 \\times 10^{-4}\\right)\\left[\\mathrm{RNH}_{3}^{+}\\right]\n\\end{aligned}\n$$\n\nIn addition, we know that\n\n$$\n\\left[\\mathrm{RNH}_{2}\\right]+\\left[\\mathrm{RNH}_{3}{ }^{+}\\right]=0.0010 \\mathrm{M}\n$$"}
{"id": 1564, "contents": "Henderson-Hasselbalch equation: - \nIn addition, we know that\n\n$$\n\\left[\\mathrm{RNH}_{2}\\right]+\\left[\\mathrm{RNH}_{3}{ }^{+}\\right]=0.0010 \\mathrm{M}\n$$\n\nSolving the two simultaneous equations gives $\\left[\\mathrm{RNH}_{3}{ }^{+}\\right]=0.0010 \\mathrm{M}$ and $\\left[\\mathrm{RNH}_{2}\\right]=5 \\times 10^{-7} \\mathrm{M}$. In other words, at a physiological pH of 7.3 , essentially $100 \\%$ of the methylamine in a 0.0010 M solution exists in its protonated form as methylammonium ion. The same is true of other amine bases, so we always write cellular amines in their protonated form and amino acids in their ammonium carboxylate form to reflect their structures at\nphysiological pH.\n\n\nPROBLEM Calculate the percentages of neutral and protonated forms present in a solution of 0.0010 Molar 24-7 pyrimidine at $\\mathrm{pH}=7.3$. The $\\mathrm{p} K_{\\mathrm{a}}$ of pyrimidinium ion is 1.3."}
{"id": 1565, "contents": "Reduction of Nitriles, Amides, and Nitro Compounds - \nWe've already seen in Section 20.7 and Section 21.7 how amines can be prepared by reduction of nitriles and amides with $\\mathrm{LiAlH}_{4}$. The two-step sequence of $\\mathrm{S}_{\\mathrm{N}} 2$ displacement with $\\mathrm{CN}^{-}$followed by reduction thus converts an alkyl halide into a primary alkylamine having an additional carbon atom. Amide reduction converts carboxylic acids and their derivatives into amines with the same number of carbon atoms.\n\n\n\nArylamines are usually prepared by nitration of an aromatic starting material, followed by reduction of the nitro group (Section 16.2). The reduction step can be carried out in many different ways, depending on the circumstances. Catalytic hydrogenation over platinum works well but is often incompatible with the presence elsewhere in the molecule of other reducible groups, such as $\\mathrm{C}=\\mathrm{C}$ bonds or carbonyl groups. Iron, zinc, tin, and tin(II) chloride $\\left(\\mathrm{SnCl}_{2}\\right)$ are also effective when used in acidic aqueous solution. Tin(II) chloride is particularly mild and is often used when other reducible functional groups are present.\n\np-tert-Butylnitrobenzene\n\nm-Nitrobenzaldehyde\np-tert-Butylaniline (100\\%)\n\nm-Aminobenzaldehyde (90\\%)\n\nPROBLEM Propose structures for either a nitrile or an amide that might be a precursor of each of the following 24-8 amines\n(a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}$\n(b) $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2}\\right)_{2} \\mathrm{NH}$\n(c) Benzylamine, $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CH}_{2} \\mathrm{NH}_{2}$\n(d) N -Ethylaniline"}
{"id": 1566, "contents": "$\\mathrm{S}_{\\mathrm{N}} 2$ Reactions of Alkyl Halides - \nAmmonia and other amines are good nucleophiles in $S_{N} 2$ reactions. As a result, the simplest method of alkylamine synthesis is by $\\mathrm{S}_{\\mathrm{N}} 2$ alkylation of ammonia or an alkylamine with an alkyl halide. If ammonia is used, a primary amine results; if a primary amine is used, a secondary amine results; and so on. Even tertiary amines react rapidly with alkyl halides to yield quaternary ammonium salts, $\\mathrm{R}_{4} \\mathrm{~N}^{+} \\mathrm{X}^{-}$.\n\n\nUnfortunately, these reactions don't stop cleanly after a single alkylation has occurred. Because ammonia and primary amines have similar reactivity, the initially formed monoalkylated substance often undergoes further reaction to yield a mixture of products. Even secondary and tertiary amines undergo further alkylation, although to a lesser extent. For example, treatment of 1-bromooctane with a twofold excess of ammonia leads to a mixture containing only $45 \\%$ octylamine. A nearly equal amount of dioctylamine is produced by double alkylation, along with smaller amounts of trioctylamine and tetraoctylammonium bromide.\n\n\nA better method for preparing primary amines is to use azide ion, $\\mathrm{N}_{3}{ }^{-}$, rather than ammonia, as the nucleophile for $\\mathrm{S}_{\\mathrm{N}} 2$ reaction with a primary or secondary alkyl halide. The product is an alkyl azide, which is not nucleophilic, so overalkylation can't occur. Subsequent reduction of the alkyl azide with $\\mathrm{LiAlH}_{4}$ then leads to the desired primary amine. Although this method works well, low-molecular-weight alkyl azides are explosive and must be handled carefully."}
{"id": 1567, "contents": "$\\mathrm{S}_{\\mathrm{N}} 2$ Reactions of Alkyl Halides - \nAnother alternative for preparing a primary amine from an alkyl halide is the Gabriel amine synthesis, which uses a phthalimide alkylation. An imide $(\\mathbf{O}=\\mathbf{C}-\\mathbf{N}-\\mathbf{C}=\\mathbf{O})$ is similar to a $\\beta$-keto ester in that the acidic $\\mathrm{N}-\\mathrm{H}$ hydrogen is flanked by two carbonyl groups. Thus, imides are deprotonated by such bases as KOH, and the resultant anions are readily alkylated in a reaction similar to acetoacetic ester synthesis (Section 22.7). Basic hydrolysis of the $N$-alkylated imide then yields a primary amine product. The imide hydrolysis step is analogous to the hydrolysis of an amide (Section 21.7).\n\n\nPROBLEM Write the mechanism of the last step in Gabriel amine synthesis, the base-promoted hydrolysis of a 24-9 phthalimide to yield an amine plus phthalate ion.\n\nPROBLEM Show two methods for the synthesis of dopamine, a neurotransmitter involved in regulation of the\n24-10 central nervous system. Use any alkyl halide needed."}
{"id": 1568, "contents": "Reductive Amination of Aldehydes and Ketones - \nAmines can be synthesized in a single step by treatment of an aldehyde or ketone with ammonia or an amine in the presence of a reducing agent, a process called reductive amination. For example, amphetamine, a central nervous system stimulant, is prepared commercially by reductive amination of phenyl-2-propanone with ammonia using hydrogen gas over a nickel catalyst as the reducing agent. In the laboratory, either $\\mathrm{NaBH}_{4}$ or the related $\\mathrm{NaBH}(\\mathrm{OAc})_{3}$ is commonly used ( $\\mathrm{OAc}=$ acetate ).\n\n\nReductive amination takes place by the pathway shown in FIGURE 24.6. An imine intermediate is first formed by a nucleophilic addition reaction (Section 19.8), and the $\\mathrm{C}=\\mathrm{N}$ bond of the imine is then reduced to the amine, much as the $\\mathrm{C}=\\mathrm{O}$ bond of a ketone can be reduced to an alcohol."}
{"id": 1569, "contents": "FIGURE 24.6 MECHANISM - \nMechanism for reductive amination of a ketone to yield an amine. Details of the imine-forming step are shown in FIGURE 19.7.\n(1) Ammonia adds to the ketone carbonyl group in a nucleophilic addition reaction to yield an intermediate carbinolamine.\n\n2 The carbinolamine loses water to give an imine.\n(3) The imine is reduced by $\\mathrm{NaBH}_{4}$ or $\\mathrm{H}_{2} /$ Ni to yield the amine product.\n\n(1)\n\n(2.\n\n(3) $\\mathrm{NaBH}_{4}$ or $\\mathrm{H}_{2} / \\mathrm{Ni}$\n\n\nAmmonia, primary amines, and secondary amines can all be used in the reductive amination reaction, yielding primary, secondary, and tertiary amines, respectively.\n\n\nReductive aminations also occur in various biological pathways. In the biosynthesis of the amino acid proline, for instance, glutamate 5 -semialdehyde undergoes internal imine formation to give 1-pyrrolinium 5 -carboxylate, which is then reduced by nucleophilic addition of hydride ion to the $\\mathrm{C}=\\mathrm{N}$ bond. Reduced nicotinamide adenine dinucleotide, NADH, acts as the biological reducing agent."}
{"id": 1570, "contents": "Using a Reductive Amination Reaction - \nHow might you prepare $N$-methyl-2-phenylethylamine using a reductive amination reaction?"}
{"id": 1571, "contents": "Strategy - \nLook at the target molecule, and identify the groups attached to nitrogen. One of the groups must be derived from the aldehyde or ketone component, and the other must be derived from the amine component. In the case of N -methyl-2-phenylethylamine, two combinations can lead to the product: phenylacetaldehyde plus methylamine or formaldehyde plus 2-phenylethylamine. It's usually better to choose the combination with the simpler amine component-methylamine in this case-and to use an excess of that amine as reactant.\n\nSolution\n\n\nPROBLEM How might the following amines be prepared using reductive amination reactions? Show all 24-11 precursors if more than one is possible.\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM How could you prepare the following amine using a reductive amination reaction?\n24-12"}
{"id": 1572, "contents": "Hofmann and Curtius Rearrangements - \nCarboxylic acid derivatives can be converted into primary amines with loss of one carbon atom by both the Hofmann rearrangement and the Curtius rearrangement. Although the Hofmann rearrangement involves a primary amide and the Curtius rearrangement involves an acyl azide, both proceed through similar mechanisms.\nHofmann\nrearrangement\n\n\nAn amide"}
{"id": 1573, "contents": "An acyl azide - \nHofmann rearrangement occurs when a primary amide, $\\mathrm{RCONH}_{2}$, is treated with $\\mathrm{Br}_{2}$ and base (FIGURE 24.7). The overall mechanism is lengthy, but most of the individual steps have been encountered before. Thus, the bromination of an amide in steps 1 and 2 is analogous to the base-promoted bromination of a ketone enolate ion (Section 22.6), and the rearrangement of the bromoamide anion in step 4 is analogous to a carbocation rearrangement (Section 7.11). Nucleophilic addition of water to the isocyanate carbonyl group in step 5 is a typical carbonyl-group process (Section 19.4), as is the final decarboxylation step 6 (Section 22.7)."}
{"id": 1574, "contents": "FIGURE 24.7 MECHANISM - \nMechanism of the Hofmann rearrangement of an amide to an amine. Each step is analogous to a reaction studied previously.\n\nAmide\n\n\u00a9 || proton, yielding an amide anion.\n(2) The anion reacts with bromine in an $\\alpha$-substitution reaction to give an N -bromoamide.\n\n\nBromoamide\n\n(3) $\\downarrow$\n\n(4)\n\n(5)\n\n\nCarbamic acid\n(6) The carbamic acid\nspontaneously loses $\\mathrm{CO}_{2}$ to give an amine."}
{"id": 1575, "contents": "(6) - \nalkylamines. For example, the appetite-suppressant drug phentermine is prepared commercially by Hofmann rearrangement of a primary amide. Commonly known by the name Fen-Phen, the combination of phentermine with another appetite-suppressant, fenfluramine, is suspected of causing heart damage.\n\n\n2,2-Dimethyl-3-phenyl-\nPhentermine propanamide\nThe Curtius rearrangement, like the Hofmann rearrangement, involves migration of an -R group from the $\\mathrm{C}=\\mathrm{O}$ carbon atom to the neighboring nitrogen with simultaneous loss of a leaving group. The reaction takes place on heating an acyl azide that is itself prepared by nucleophilic acyl substitution of an acid chloride.\n\n\nAlso like the Hofmann rearrangement, the Curtius rearrangement is often used commercially. The antidepressant drug tranylcypromine, for instance, is made by Curtius rearrangement of 2-phenylcyclopropanecarbonyl chloride.\n\ntrans-2-Phenylcyclo-\nTranylcypromine propanecarbonyl chloride"}
{"id": 1576, "contents": "Using the Hofmann and Curtius Rearrangements - \nHow would you prepare o-methylbenzylamine from a carboxylic acid, using both Hofmann and Curtius rearrangements?"}
{"id": 1577, "contents": "Strategy - \nBoth Hofmann and Curtius rearrangements convert a carboxylic acid derivative-either an amide (Hofmann) or an acid chloride (Curtius)-into a primary amine with loss of one carbon, RCOY $\\rightarrow \\mathrm{RNH}_{2}$. Both reactions begin with the same carboxylic acid, which can be identified by replacing the $-\\mathrm{NH}_{2}$ group of the amine product by a $-\\mathrm{CO}_{2} \\mathrm{H}$ group. In the present instance, $o$-methylphenylacetic acid is needed."}
{"id": 1578, "contents": "Solution - \nPROBLEM How would you prepare the following amines, using both Hofmann and Curtius rearrangements on\n24-13 a carboxylic acid derivative?\n(a)\n\n(b)"}
{"id": 1579, "contents": "Alkylation and Acylation - \nWe've already studied the two most general reactions of amines-alkylation and acylation. As we saw earlier in this chapter, primary, secondary, and tertiary amines can be alkylated by reaction with a primary alkyl halide. Alkylations of primary and secondary amines are difficult to control and often give mixtures of products, but tertiary amines are cleanly alkylated to give quaternary ammonium salts. Primary and secondary (but not tertiary) amines can also be acylated by nucleophilic acyl substitution reaction with an acid chloride or an acid anhydride to yield an amide (Section 21.4 and Section 21.5). Note that overacylation of the nitrogen does not occur because the amide product is much less nucleophilic and less reactive than the starting amine."}
{"id": 1580, "contents": "Hofmann Elimination Reaction - \nLike alcohols, amines can be converted into alkenes by an elimination reaction. But because an amide ion, $\\mathrm{NH}_{2}{ }^{-}$, is such a poor leaving group, it must first be converted into a better leaving group. In the Hofmann elimination reaction, an amine is completely methylated by reaction with an excess amount of iodomethane to produce the corresponding quaternary ammonium salt. This salt then undergoes elimination to give an alkene on heating with a base, typically silver oxide, $\\mathrm{Ag}_{2} \\mathrm{O}$. For example, 1-methylpentylamine is converted into 1-hexene.\n\n\n1-Methylpentylamine\n(1-Methylpentyl)trimethyl-\n1-Hexene (60\\%) ammonium iodide\nSilver oxide acts by exchanging iodide ion for hydroxide ion in the quaternary salt, thus providing the base necessary for elimination. The actual elimination step is an E2 reaction (Section 11.8) in which hydroxide ion removes a proton while the positively charged nitrogen atom leaves.\n\n\nUnlike what happens in other E2 reactions, the major product of Hofmann elimination is the less highly substituted alkene rather than the more highly substituted one, as shown by the reaction of (1-methylbutyl)trimethylammonium hydroxide to give 1-pentene rather than the alternative 2 -pentene. The reason for this non-Zaitsev result is probably steric. Because of the large size of the trialkylamine leaving group, the base must abstract a hydrogen from the more accessible, least hindered position.\n\n\nThe Hofmann elimination reaction is not often currently used in the laboratory, but analogous biological eliminations occur frequently, although usually with protonated ammonium ions rather than quaternary ammonium salts. In the biosynthesis of nucleic acids, for instance, a substance called adenylosuccinate undergoes an elimination of a positively charged nitrogen to give fumarate plus adenosine monophosphate."}
{"id": 1581, "contents": "Predicting the Product of a Hofmann Elimination - \nWhat product would you expect from Hofmann elimination of the following amine?"}
{"id": 1582, "contents": "Strategy - \nThe Hofmann elimination is an E2 reaction that converts an amine into an alkene and occurs with non-Zaitsev regiochemistry to form the less highly substituted double bond. To predict the product, look at the reactant and identify the positions from which elimination might occur (the positions two carbons away from nitrogen). Then, carry out an elimination using the most accessible hydrogen. In the present instance, there are three possible positions from which elimination might occur-one primary, one secondary, and one tertiary. The primary position is the most accessible and leads to the least highly substituted alkene, ethylene.\n\nSolution\n\n\nPROBLEM What products would you expect from Hofmann elimination of the following amines? If more than 24-14 one product is formed, indicate which is major.\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM What product would you expect from Hofmann elimination of a heterocyclic amine such as 24-15 piperidine? Write all the steps."}
{"id": 1583, "contents": "Electrophilic Aromatic Substitutions - \nAn amino group is strongly activating and ortho- and para-directing in electrophilic aromatic substitution reactions (Section 16.4). This high reactivity of amino-substituted benzenes can be a drawback at times because it's often difficult to prevent polysubstitution. Reaction of aniline with $\\mathrm{Br}_{2}$, for instance, takes place rapidly and yields the 2,4,6-tribrominated product. The amino group is so strongly activating that it's not possible to stop at the monobromo stage."}
{"id": 1584, "contents": "Aniline $\\quad$ 2,4,6-Tribromoaniline <br> (100\\%) - \nAnother drawback to the use of amino-substituted benzenes in electrophilic aromatic substitution reactions is that Friedel-Crafts reactions are not successful (Section 16.3). The amino group forms an acid-base complex with the $\\mathrm{AlCl}_{3}$ catalyst, which prevents further reaction. Both drawbacks can be overcome, however, by carrying out electrophilic aromatic substitution reactions on the corresponding amide rather than on the free amine.\n\nAs we saw in Section 21.5, treatment of an amine with acetic anhydride yields the corresponding acetyl amide, or acetamide. Although still activating and ortho-, para-directing, amido substituents ( $\\mathrm{R}-\\mathrm{N}-\\mathrm{C}=\\mathrm{O}$ ) are less strongly activating and less basic than amino groups because their nitrogen lone-pair electrons are delocalized by the neighboring carbonyl group. As a result, bromination of an $N$-arylamide occurs cleanly to give a monobromo product, and hydrolysis of the amide with aqueous base then gives the free amine. For example, p-toluidine (4-methylaniline) can be acetylated, brominated, and hydrolyzed to yield 2-bromo-4-methylaniline. None of the 2,6-dibrominated product is obtained.\n\n\nFriedel-Crafts alkylations and acylations of $N$-arylamides also proceed normally. For example, benzoylation of acetanilide ( $N$-acetylaniline) under Friedel-Crafts conditions gives 4-aminobenzophenone in $80 \\%$ yield after hydrolysis.\n\n\nModerating the reactivity of an amino-substituted benzene by forming an amide is a useful trick that allows many kinds of electrophilic aromatic substitutions to be carried out that would otherwise be impossible. One example is the preparation of the sulfa drugs, such as sulfanilamide."}
{"id": 1585, "contents": "Aniline $\\quad$ 2,4,6-Tribromoaniline <br> (100\\%) - \nSulfa drugs were among the first pharmaceutical agents to be used clinically against bacterial infection. Although they have largely been replaced today by safer and more powerful antibiotics, sulfa drugs are credited with saving the lives of thousands of wounded during World War II and are still prescribed for urinary tract infections. They are prepared by chlorosulfonation of acetanilide, followed by reaction of\n$p$-( $N$-acetylamino)benzenesulfonyl chloride with ammonia or some other amine to give a sulfonamide. Hydrolysis of the amide then yields the sulfa drug. Note that hydrolysis of the amide can be carried out in the presence of the sulfonamide group because sulfonamides hydrolyze very slowly.\n\n\nPROBLEM Propose a synthesis of the drug sulfathiazole from benzene and any necessary amine.\n24-16"}
{"id": 1586, "contents": "Sulfathiazole - \nPROBLEM Propose syntheses of the following compounds from benzene:\n24-17 (a) $N, N$-Dimethylaniline\n(b) p-Chloroaniline\n(c) m-Chloroaniline\n(d) 2,4-Dimethylaniline"}
{"id": 1587, "contents": "Diazonium Salts: The Sandmeyer Reaction - \nPrimary arylamines react with nitrous acid, $\\mathrm{HNO}_{2}$, to yield stable arenediazonium salts, $\\mathrm{Ar}-\\stackrel{+}{\\mathrm{N}} \\equiv \\mathrm{N} \\mathrm{X}^{-}$, a process called a diazotization reaction. Alkylamines also react with nitrous acid, but the corresponding alkanediazonium products are so reactive they can't be isolated. Instead, they lose nitrogen instantly to yield carbocations. The analogous loss of $\\mathrm{N}_{2}$ from an arenediazonium ion to yield an aryl cation is disfavored by the instability of the cation.\n\n\nArenediazonium salts are useful because the diazonio group ( $-\\mathrm{N} \\equiv \\mathrm{N}$ ) can be replaced by a nucleophile in a substitution reaction.\n\n\nMany different nucleophiles-halide, hydride, cyanide, and hydroxide among others-react with arenediazonium salts, yielding many different kinds of substituted benzenes. The overall sequence of (1) nitration, (2) reduction, (3) diazotization, and (4) nucleophilic substitution is perhaps the single most versatile method of aromatic substitution.\n\nAryl chlorides and bromides are prepared by reaction of an arenediazonium salt with the corresponding copper(I) halide, CuX , a process called the Sandmeyer reaction. Aryl iodides can be prepared by direct reaction\nwith NaI without using a copper(I) salt. Yields generally fall between $60 \\%$ and $80 \\%$.\n\n\n\nAniline\nIodobenzene (67\\%)\nSimilar treatment of an arenediazonium salt with CuCN yields the nitrile ArCN, which can then be further converted into other functional groups such as carboxyl. For example, Sandmeyer reaction of $o$-methylbenzenediazonium bisulfate with CuCN yields o-methylbenzonitrile, which can be hydrolyzed to give $o$-methylbenzoic acid. This product can't be prepared from $o$-xylene by the usual side-chain oxidation route because both methyl groups would be oxidized."}
{"id": 1588, "contents": "Diazonium Salts: The Sandmeyer Reaction - \nThe diazonio group can also be replaced by -OH to yield a phenol and by -H to yield an arene. A phenol is prepared by reaction of the arenediazonium salt with copper(I) oxide in an aqueous solution of copper(II) nitrate, a reaction that is especially useful because few other general methods exist for introducing an -OH group onto an aromatic ring.\n\n$\\underset{\\text { ( } \\boldsymbol{p} \\text {-Toluidine) }}{\\boldsymbol{p} \\text {-Methylaniline }}$\np-Cresol\n(93\\%)\nReduction of a diazonium salt to give an arene occurs on treatment with hypophosphorous acid, $\\mathrm{H}_{3} \\mathrm{PO}_{2}$. This reaction is used primarily when there is a need for temporarily introducing an amino substituent onto a ring to take advantage of its directing effect. Suppose, for instance, that you needed to make 3,5-dibromotoluene. This product can't be made by direct bromination of toluene because reaction would occur at positions 2 and 4. Starting with $p$-methylaniline ( $p$-toluidine), however, dibromination occurs ortho to the strongly directing amino substituent, and diazotization followed by treatment with $\\mathrm{H}_{3} \\mathrm{PO}_{2}$ to remove the amino group yields the desired product.\n\n\n\nMechanistically, these diazonio replacement reactions occur through radical rather than polar pathways. In the presence of a copper(I) compound, for instance, it's thought that the arenediazonium ion is first converted to an aryl radical plus copper(II), followed by subsequent reaction to give product plus regenerated copper(I) catalyst."}
{"id": 1589, "contents": "Using Diazonium Replacement Reactions - \nHow would you prepare $m$-hydroxyacetophenone from benzene, using a diazonium replacement reaction in your scheme?\n\nm-Hydroxyacetophenone"}
{"id": 1590, "contents": "Strategy - \nAs always, organic syntheses are planned by working retrosynthetically from the final product, one step at a time. First, identify the functional groups in the product and recall how those groups can be synthesized. $m$-Hydroxyacetophenone has an -OH group and $\\mathrm{a}-\\mathrm{COCH}_{3}$ group in a meta relationship on a benzene ring. A hydroxyl group is generally introduced onto an aromatic ring by a four-step sequence of nitration, reduction, diazotization, and diazonio replacement. An acetyl group is introduced by a Friedel-Crafts acylation reaction.\n\nNext, ask yourself what an immediate precursor of the target might be. Since an acetyl group is a meta director while a hydroxyl group is an ortho and para director, acetophenone might be a precursor of $m$-hydroxyacetophenone. Benzene, in turn, is a precursor of acetophenone."}
{"id": 1591, "contents": "Solution - \nPROBLEM How would you prepare the following compounds from benzene, using a diazonium replacement 24-18 reaction in your scheme?\n(a) $p$-Bromobenzoic acid\n(b) $m$-Bromobenzoic acid\n(c) m-Bromochlorobenzene\n(d) $p$-Methylbenzoic acid\n(e) 1,2,4-Tribromobenzene"}
{"id": 1592, "contents": "Diazonium Coupling Reactions - \nArenediazonium salts undergo a coupling reaction with activated aromatic rings such as phenols and arylamines to yield brightly colored azo compounds, $\\mathbf{A r}-\\mathbf{N}=\\mathbf{N}-\\mathbf{A r}^{\\prime}$.\n\n\nAn azo compound\n\n$$\n\\text { where } \\mathrm{Y}=-\\mathrm{OH} \\text { or }-\\mathrm{NR}_{2}\n$$\n\nDiazonium coupling reactions are typical electrophilic aromatic substitutions in which the positively charged diazonium ion is the electrophile that reacts with the electron-rich ring of a phenol or arylamine. Reaction usually occurs at the para position.\n\n\nAzo-coupled products are widely used as dyes for textiles because their extended conjugated $\\pi$ electron system causes them to absorb in the visible region of the electromagnetic spectrum (Section 14.9). p-(Dimethylamino)azobenzene, for instance, is a bright yellow compound that was at one time used as a coloring agent in margarine.\n\n\nPROBLEM Propose a synthesis of $p$-(dimethylamino)azobenzene with benzene as your only organic starting 24-19 material."}
{"id": 1593, "contents": "Diazonium Coupling Reactions - 24.9 Heterocyclic Amines\nAs noted in Section 15.5 in connection with a discussion of aromaticity, a cyclic organic compound that contains atoms of two or more elements in its ring is a called a heterocycle. Heterocyclic amines are particularly common, and many have important biological properties. Pyridoxal phosphate, a coenzyme; sildenafil (Viagra), a well-known pharmaceutical; and heme, the oxygen carrier in blood, are a few examples.\n\n\nPyridoxal phosphate (a coenzyme)\n\n\nSildenafil\n(Viagra)\n\n\nHeme\n\nMost heterocycles have the same chemistry as their open-chain counterparts. Lactones and acyclic esters behave similarly, lactams and acyclic amides behave similarly, and cyclic and acyclic ethers behave similarly. In certain cases, however, particularly when the ring is unsaturated, heterocycles have unique and interesting properties."}
{"id": 1594, "contents": "Pyrrole and Imidazole - \nPyrrole, the simplest five-membered unsaturated heterocyclic amine, is obtained commercially by treatment of furan with ammonia over an alumina catalyst at $400^{\\circ} \\mathrm{C}$. Furan, the oxygen-containing analog of pyrrole, is obtained by acid-catalyzed dehydration of the five-carbon sugars found in oat hulls and corncobs.\n\n\nAlthough pyrrole appears to be both an amine and a conjugated diene, its chemical properties are not consistent with either of these structural features. Unlike most other amines, pyrrole is not basic-the $\\mathrm{p} K_{\\mathrm{a}}$ of the pyrrolinium ion is 0.4 ; unlike most other conjugated dienes, pyrrole undergoes electrophilic substitution reactions rather than additions. The reason for both of these properties, as noted in Section 15.5, is that pyrrole has six $\\pi$ electrons and is aromatic. Each of the four carbons contributes one $\\pi$ electron, and the $s p^{2}$-hybridized nitrogen contributes two more from its lone pair.\n\n\nPyrrole"}
{"id": 1595, "contents": "Six $\\pi$ electrons - \nBecause the nitrogen lone pair is a part of the aromatic sextet, protonation on nitrogen would destroy the aromaticity of the ring. The nitrogen atom in pyrrole is therefore less electron-rich, less basic, and less nucleophilic than the nitrogen in an aliphatic amine. By the same token, the carbon atoms of pyrrole are more electron-rich and more nucleophilic than typical double-bond carbons. The pyrrole ring is therefore reactive toward electrophiles in the same way as enamines (Section 23.11). Electrostatic potential maps show how the pyrrole nitrogen is electron-poor (less red) compared with the nitrogen in its saturated counterpart pyrrolidine, while the pyrrole carbon atoms are electron-rich (more red) compared with the carbons in 1,3-cyclopentadiene.\n\n\nThe chemistry of pyrrole is similar to that of activated benzene rings. In general, however, the heterocycles are more reactive toward electrophiles than benzene rings, and low temperatures are often necessary to control the reactions. Halogenation, nitration, sulfonation, and Friedel-Crafts acylation can all be accomplished. For example:\n\n\nElectrophilic substitutions normally occur at C2, the position next to the nitrogen, because reaction at this position leads to a more stable intermediate cation having three resonance forms, whereas reaction at C3 gives a less stable cation with only two resonance forms (FIGURE 24.8).\n\n\n3-Nitropyrrole\n(Not formed)\nFIGURE 24.8 Electrophilic nitration of pyrrole. The intermediate produced by reaction at C 2 is more stable than that produced by reaction at C3.\n\nOther common five-membered heterocyclic amines include imidazole and thiazole. Imidazole, a constituent of the amino acid histidine, has two nitrogens, only one of which is basic. Thiazole, the five-membered ring system on which the structure of thiamin (vitamin $B_{1}$ ) is based, also contains a basic nitrogen that is alkylated in thiamin to form a quaternary ammonium ion.\n\n\nImidazole\n\n\nThiazole\n\n\nHistidine\n\n\nThiamin (vitamin $\\mathrm{B}_{1}$ )"}
{"id": 1596, "contents": "Six $\\pi$ electrons - \nImidazole\n\n\nThiazole\n\n\nHistidine\n\n\nThiamin (vitamin $\\mathrm{B}_{1}$ )\n\nPROBLEM Draw an orbital picture of thiazole. Assume that both the nitrogen and sulfur atoms are 24-20 $s p^{2}$-hybridized, and show the orbitals that the lone pairs occupy.\n\nPROBLEM What is the percent protonation of the imidazole nitrogen atom in histidine at a physiological pH of 24-21 7.3 (Section 24.5)?"}
{"id": 1597, "contents": "Pyridine and Pyrimidine - \nPyridine is the nitrogen-containing heterocyclic analog of benzene. Like benzene, pyridine is a flat, aromatic molecule, with bond angles of $120^{\\circ}$ and $\\mathrm{C}-\\mathrm{C}$ bond lengths of 139 pm , intermediate between typical single and double bonds. The five carbon atoms and the $s p^{2}$-hybridized nitrogen atom each contribute one $\\pi$ electron to the aromatic sextet, and the lone-pair electrons occupy an $s p^{2}$ orbital in the plane of the ring (Section 15.5).\n\nAs shown previously in TABLE 24.1, pyridine ( $\\mathrm{p} K_{\\mathrm{a}}=5.25$ ) is a stronger base than pyrrole but a weaker base than the alkylamines. The diminished basicity of pyridine compared with that of alkylamines is due to the fact that the lone-pair electrons on the pyridine nitrogen are in an $s p^{2}$ orbital, while those on an alkylamine nitrogen are in an $s p^{3}$ orbital. Because $s$ orbitals have their maximum electron density at the nucleus but $p$ orbitals have a node at the nucleus, electrons in an orbital with more $s$ character are held more closely to the positively charged nucleus and are less available for bonding. As a result, the $s p^{2}$-hybridized nitrogen atom ( $33 \\% s$ character) in\npyridine is less basic than the $s p^{3}$-hybridized nitrogen in an alkylamine ( $25 \\% s$ character).\n\n\nPyridine\nUnlike benzene, pyridine undergoes electrophilic aromatic substitution reactions with difficulty. Halogenation can be carried out under drastic conditions, but nitration occurs in very low yield, and Friedel-Crafts reactions are not successful. Reactions usually give the 3 -substituted product."}
{"id": 1598, "contents": "Pyridine and Pyrimidine - \nThe low reactivity of pyridine toward electrophilic aromatic substitution is caused by a combination of factors. One is that acid-base complexation between the basic ring's nitrogen atom and the incoming electrophile places a positive charge on the ring, thereby deactivating it. Equally important is that the electron density of the ring is decreased by the electron-withdrawing inductive effect of the electronegative nitrogen atom. Thus, pyridine has a substantial dipole moment ( $\\mu=2.26 \\mathrm{D}$ ), with the ring carbons acting as the positive end of the dipole. Reaction of an electrophile with the positively polarized carbon atoms is therefore difficult.\n\n\nIn addition to pyridine, the six-membered diamine pyrimidine is also found commonly in biological molecules, particularly as a constituent of nucleic acids. With a $\\mathrm{p} K_{\\mathrm{a}}$ of 1.3 , pyrimidine is substantially less basic than pyridine because of the inductive effect of the second nitrogen.\n\n\nPROBLEM Electrophilic aromatic substitution reactions of pyridine normally occur at C3. Draw the 24-22 carbocation intermediates resulting from reaction of an electrophile at C2, C3, and C4, and explain the observed result."}
{"id": 1599, "contents": "Polycyclic Heterocycles - \nAs we saw in Section 15.6, quinoline, isoquinoline, indole, and purine are common polycyclic heterocycles. The first three contain both a benzene ring and a heterocyclic aromatic ring, while purine contains two heterocyclic rings joined together. All four ring systems occur commonly in nature, and many compounds with these rings have pronounced physiological activity. The quinoline alkaloid quinine, for instance, is widely used as an antimalarial drug; tryptophan is a common amino acid; and the purine adenine is a constituent of nucleic acids.\n\n\nQuinine (antimalarial)\n\n\nTryptophan\n(amino acid)\n\n\nAdenine (DNA constituent)\n\nThe chemistry of these polycyclic heterocycles is just what you might expect from a knowledge of the simpler heterocycles pyridine and pyrrole. Quinoline and isoquinoline both have basic, pyridine-like nitrogen atoms, and both undergo electrophilic substitutions. As with pyridine, both quinoline and isoquinoline are less reactive toward electrophilic substitution than benzene because of the electronegative nitrogen atom that withdraws electrons from the ring. Reaction occurs on the benzene ring rather than on the nitrogen-containing pyridine ring, and a mixture of substitution products is obtained.\n\n\nIndole has a nonbasic, pyrrole-like nitrogen and undergoes electrophilic substitution more easily than benzene. Substitution occurs at C3 of the electron-rich pyrrole ring rather than on the benzene ring.\n\n\nPurine has three basic, pyridine-like nitrogens with lone-pair electrons in $s p^{2}$ orbitals in the plane of the ring. The remaining purine nitrogen is nonbasic and pyrrole-like, with its lone-pair electrons as part of the aromatic\n$\\pi$ electron system.\n\n\n\nPurine\nPROBLEM Which nitrogen atom in the hallucinogenic indole alkaloid $N, N$-dimethyltryptamine is more basic? 24-23 Explain."}
{"id": 1600, "contents": "N,N-Dimethyltryptamine - \nPROBLEM Indole reacts with electrophiles at C3 rather than at C2. Draw resonance forms of the intermediate 24-24 cations resulting from reaction at C2 and C3, and explain the observed results."}
{"id": 1601, "contents": "Infrared Spectroscopy - \nPrimary and secondary amines can be identified by a characteristic N-H stretching absorption in the 3300 to $3500 \\mathrm{~cm}^{-1}$ range of the IR spectrum. Alcohols also absorb in this range (Section 17.11), but amine absorption bands are generally sharper and less intense than hydroxyl bands. Primary amines show a pair of bands at about 3350 and $3450 \\mathrm{~cm}^{-1}$ from the symmetric and asymmetric stretching modes, and secondary amines show a single band at $3350 \\mathrm{~cm}^{-1}$. Tertiary amines have no absorption in this region because they have no $\\mathrm{N}-\\mathrm{H}$ bonds. FIGURE 24.9 recalls the IR spectrum of cyclohexylamine from Section 12.8.\n\n\nFIGURE 24.9 IR spectrum of cyclohexylamine."}
{"id": 1602, "contents": "Nuclear Magnetic Resonance Spectroscopy - \nAmines are difficult to identify solely by ${ }^{1} \\mathrm{H}$ NMR spectroscopy because $\\mathrm{N}-\\mathrm{H}$ hydrogens tend to appear as broad signals without clear-cut coupling to neighboring $\\mathrm{C}-\\mathrm{H}$ hydrogens. As with $\\mathrm{O}-\\mathrm{H}$ absorptions (Section 17.11), amine $\\mathrm{N}-\\mathrm{H}$ absorptions can appear over a wide range and are best identified by adding a small amount of $\\mathrm{D}_{2} \\mathrm{O}$ to the sample. Exchange of $\\mathrm{N}-\\mathrm{D}$ for $\\mathrm{N}-\\mathrm{H}$ occurs, and the $\\mathrm{N}-\\mathrm{H}$ signal disappears from the NMR spectrum.\n\n\nHydrogens on the carbon next to nitrogen are deshielded because of the electron-withdrawing effect of the nitrogen, and they therefore absorb further downfield than alkane hydrogens. $N$-Methyl groups are particularly distinctive because they absorb as a sharp three-proton singlet at 2.2 to $2.6 \\delta$. This $N$-methyl resonance at 2.42 $\\delta$ is easily seen in the ${ }^{1} \\mathrm{H}$ NMR spectrum of $N$-methylcyclohexylamine (FIGURE 24.10).\n\n\nFIGURE 24.10 Proton NMR spectrum of $\\boldsymbol{N}$-methylcyclohexylamine.\nCarbons next to amine nitrogens are slightly deshielded in the ${ }^{13} \\mathrm{C}$ NMR spectrum and absorb about 20 ppm downfield from where they would absorb in an alkane of similar structure. In $N$-methylcyclohexylamine, for example, the ring carbon to which nitrogen is attached absorbs at a position 24 ppm downfield from any other ring carbon."}
{"id": 1603, "contents": "Nuclear Magnetic Resonance Spectroscopy - \nPROBLEM Compound $\\mathbf{A}, \\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}$, has an IR absorption at $1715 \\mathrm{~cm}^{-1}$ and gives compound $\\mathbf{B}, \\mathrm{C}_{6} \\mathrm{H}_{15} \\mathrm{~N}$, when 24-25 treated with ammonia and $\\mathrm{NaBH}_{4}$. The IR and ${ }^{1} \\mathrm{H}$ NMR spectra of $\\mathbf{B}$ are shown. What are the structures of $\\mathbf{A}$ and $\\mathbf{B}$ ?"}
{"id": 1604, "contents": "Mass Spectrometry - \nThe nitrogen rule of mass spectrometry says that a compound with an odd number of nitrogen atoms has an odd-numbered molecular weight. Thus, the presence of nitrogen in a molecule is detected simply by observing its mass spectrum. An odd-numbered molecular ion usually means that the unknown compound has one or three nitrogen atoms, and an even-numbered molecular ion usually means that a compound has either zero or two nitrogen atoms. The logic behind the rule derives from the fact that nitrogen is trivalent, thus requiring an odd number of hydrogen atoms. For example, morphine has the formula $\\mathrm{C}_{17} \\mathrm{H}_{19} \\mathrm{NO}_{3}$ and a molecular weight of 285 amu .\n\nAlkylamines undergo a characteristic $\\alpha$ cleavage in the mass spectrometer, similar to the cleavage observed for alcohols (Section 17.11). $\\mathrm{C}-\\mathrm{C}$ bond nearest the nitrogen atom is broken, yielding an alkyl radical and a resonance-stabilized, nitrogen-containing cation.\n\n\nAs an example, the mass spectrum of $N$-ethylpropylamine shown in FIGURE 24.11 has peaks at $m / z=58$ and $m / z=72$, corresponding to the two possible modes of $\\alpha$ cleavage.\n\n\n\nFIGURE 24.11 Mass spectrum of $N$-ethylpropylamine. The two possible modes of $\\alpha$ cleavage lead to the observed fragment ions at $m / z=$ 58 and $m / z=72$."}
{"id": 1605, "contents": "Green Chemistry II: Ionic Liquids - \nLiquids made of ions? Usually when we think of ionic compounds, we think of high-melting solids: sodium chloride, magnesium sulfate, lithium carbonate, and so forth. But yes, there are also ionic compounds that are liquid at room temperature, and they are gaining importance as reaction solvents, particularly for use in green chemistry processes (see the Chapter 11 Chemistry Matters). More than 1500 ionic liquids are known, and about 500 are available commercially.\n\nIonic liquids have been studied for nearly a century; the first to be discovered was ethylammonium nitrate, $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{NH}_{3}{ }^{+} \\mathrm{NO}_{3}{ }^{-}$, with a melting point of $12{ }^{\\circ} \\mathrm{C}$. More generally, however, the ionic liquids in use today are salts in which the cation is unsymmetrical and in which one or both of the ions are bulky so that the charges are dispersed over a large volume. Both factors minimize the crystal lattice energy and disfavor formation of the solid. Typical cations are quaternary ammonium ions from heterocyclic amines, either 1,3-dialkylimidazolium ions, $N$-alkylpyridinium ions, or ring-substituted $N$-alkylpyridinium ions.\n\n\n$$\n\\left[\\begin{array}{rl}\n\\mathrm{R}= & -\\mathrm{CH}_{3},-\\mathrm{CH}_{2} \\mathrm{CH}_{3},-\\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}, \\\\\n& -\\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}\n\\end{array}\\right]\n$$\n\n1,3-Dialkylimidazolium ions"}
{"id": 1606, "contents": "Green Chemistry II: Ionic Liquids - \n1,3-Dialkylimidazolium ions\n\n$\\left[\\begin{array}{c}\\mathrm{R}=- \\\\ -\\mathrm{CH}_{2} \\mathrm{CH}_{3},-\\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}, \\\\ \\end{array}\\right]$\n$\\boldsymbol{N}$-Alkylpyridinium ions\n\nAnions are just as varied as the cations. Hexafluorophosphate, tetrafluoroborate, alkyl sulfates, trifluoromethanesulfonates (triflates), and halides are some anion possibilities.\n\n\nHexafluorophosphate\n\n\nTetrafluoroborate\n\n\nMethyl\nsulfate\n\n\n$$\n\\mathrm{Cl}^{-}, \\mathrm{Br}^{-}, \\mathrm{I}^{-}\n$$\n\nTrifluoromethane- Halide sulfonate\n\nIonic liquids have several important features that make them attractive for use, both as solvents in green chemistry and as specialty chemicals in such applications as paint additives and refrigerants:\n\n- They dissolve both polar and nonpolar organic compounds, giving high solute concentrations and thereby minimizing the amount of solvent needed.\n- They can be optimized for specific reactions by varying cation and anion structures.\n- They are nonflammable.\n- They are thermally stable.\n- They have negligible vapor pressures and do not evaporate.\n- They are generally recoverable and can be reused many times.\n\nAs an example of their use in organic chemistry, the analgesic drug pravadoline has been synthesized in two steps using 1-butyl-3-methylimidazolium hexafluorophosphate, abbreviated bmim, as the solvent for both steps. The first step is a base-induced $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of 2-methylindole with a primary alkyl halide, and the second is a Friedel-Crafts acylation. Both steps take place with $95 \\%$ yield, and the ionic solvent is recovered simply by washing the reaction mixture, first with toluene and then with water."}
{"id": 1607, "contents": "Green Chemistry II: Ionic Liquids - \nThe first commercial process using an ionic liquid catalyst was introduced by PetroChina in 2008, when they opened a plant producing 65,000 tons per year of alkylate gasoline from isobutane. The aluminum-based ionic liquid catalyst replaced the sulfuric acid and hydrofluoric acid catalysts that had previously been used."}
{"id": 1608, "contents": "Key Terms - \n- alkylamine\n- amine\n- arylamine\n- azo compound ( $\\mathrm{R}-\\mathrm{N}=\\mathrm{N}-\\mathrm{R}^{\\prime}$ )\n- Curtius rearrangement\n- Gabriel amine synthesis\n- heterocyclic amine\n- Hofmann elimination reaction\n- Hofmann rearrangement\n- imide $(\\mathrm{O}=\\mathrm{C}-\\mathrm{N}-\\mathrm{C}=\\mathrm{O})$\n- primary amine $\\left(\\mathrm{RNH}_{2}\\right)$\n- quaternary ammonium salt\n- reductive amination\n- Sandmeyer reaction\n- secondary amine ( $\\mathrm{R}_{2} \\mathrm{NH}$ )\n- tertiary amine ( $\\mathrm{R}_{3} \\mathrm{~N}$ )"}
{"id": 1609, "contents": "Summary - \nWe've now seen all the common functional groups that occur in organic and biological chemistry. Of those groups, amines are among the most abundant and have among the richest chemistry. In addition to proteins and nucleic acids, the majority of pharmaceutical agents contain amine functional groups and many of the common coenzymes necessary for biological reactions are amines.\nAmines are organic derivatives of ammonia. They are named in the IUPAC system either by adding the suffix -amine to the name of the alkyl substituent or by considering the amino group as a substituent on a more complex parent molecule.\nThe chemistry of amines is dominated by the lone-pair electrons on nitrogen, which makes amines both basic and nucleophilic. The basicity of arylamines is generally lower than that of alkylamines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic $\\pi$ system. Electron-withdrawing substituents on the aromatic ring further weaken the basicity of a substituted aniline, while electron-donating substituents increase basicity. Alkylamines are sufficiently basic that they exist almost entirely in their protonated form at the physiological pH of 7.3.\n\nHeterocyclic amines are compounds that contain one or more nitrogen atoms as part of a ring. Saturated\nheterocyclic amines usually have the same chemistry as their open-chain analogs, but unsaturated heterocycles such as pyrrole, imidazole, pyridine, and pyrimidine are aromatic. All four are unusually stable, and all undergo aromatic substitution on reaction with electrophiles. Pyrrole is nonbasic because its nitrogen lone-pair electrons are part of the aromatic $\\pi$ system. Fused-ring heterocycles such as quinoline, isoquinoline, indole, and purine are also commonly found in biological molecules."}
{"id": 1610, "contents": "Summary - \nArylamines are prepared by nitration of an aromatic ring followed by reduction. Alkylamines are prepared by $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of ammonia or an amine with an alkyl halide or by the Gabriel amine synthesis. Amines can also be prepared by a number of reductive methods, including $\\mathrm{LiAlH}_{4}$ reduction of amides, nitriles, and azides. Also important is the reductive amination reaction in which a ketone or an aldehyde is treated with an amine in the presence of a reducing agent such as $\\mathrm{NaBH}_{4}$. In addition, amines result from Hofmann and Curtius rearrangements of carboxylic acid derivatives. Both methods involve migration of the -R group bonded to the carbonyl carbon and yield a product that has one less carbon atom than the starting material.\n\nMany of the reactions of amines are familiar from past chapters. Thus, amines react with alkyl halides in $\\mathrm{S}_{\\mathrm{N}} 2$ reactions and with acid chlorides in nucleophilic acyl substitution reactions. Amines also undergo E2 elimination to yield alkenes if they are first quaternized by treatment with iodomethane and then heated with silver oxide, a process called the Hofmann elimination.\n\nArylamines are converted by diazotization with nitrous acid into arenediazonium salts, $\\mathrm{ArN}_{2}{ }^{+} \\mathrm{X}^{-}$. The diazonio group can then be replaced by many other substituents by the Sandmeyer reaction to give a wide variety of substituted aromatic compounds. Aryl chlorides, bromides, iodides, and nitriles can be prepared from arenediazonium salts, as can arenes and phenols. In addition to their reactivity toward substitution reactions, diazonium salts undergo coupling with phenols and arylamines to give brightly colored azo compounds."}
{"id": 1611, "contents": "Summary of Reactions - \n1. Synthesis of amines (Section 24.6)\na. Reduction of nitriles\n\n$$\n\\mathrm{RCH}_{2} \\mathrm{X} \\xrightarrow{\\mathrm{NaCN}} \\mathrm{RCH}_{2} \\mathrm{C} \\equiv \\mathrm{~N} \\xrightarrow[\\text { 2. } \\mathrm{H}_{2} \\mathrm{O}]{\\text { 1. } \\mathrm{LiAlH}_{4} \\text {, ether }} \\underset{\\mathrm{RCH}_{2}}{\\text { - }}\n$$\n\nb. Reduction of amides\n\nc. Reduction of nitrobenzenes\n\nd. $\\mathrm{S}_{\\mathrm{N}} 2$ Alkylation of alkyl halides\n\ne. Gabriel amine synthesis\n\nf. Reduction of azides\n\n$$\n\\mathrm{RCH}_{2}-\\mathrm{X} \\xrightarrow[\\text { ethanol }]{\\mathrm{Na}^{+}{ }^{-} \\mathrm{N}_{3}} \\quad \\mathrm{RCH}_{2}-\\mathrm{N}=\\stackrel{+}{\\mathrm{N}}=\\stackrel{-}{\\mathrm{N}} \\quad \\xrightarrow[2 . \\mathrm{H}_{2} \\mathrm{O}]{\\text { 1. } \\mathrm{LiAlH}_{4} \\text {, ether }} \\quad \\mathrm{R}-\\mathrm{CH}_{2}-\\mathrm{NH}_{2}\n$$\n\ng. Reductive amination of aldehydes/ketones\n\nh. Hofmann rearrangement of amides\n\ni. Curtius rearrangement of acyl azides\n\n2. Reactions of amines\na. Alkylation with alkyl halides; see reaction 1(d)\nb. Hofmann elimination (Section 24.7)\n\nc. Diazotization (Section 24.8)\n\n3. Reactions of arenediazonium salts (Section 24.8)\na. Nucleophilic substitutions"}
{"id": 1612, "contents": "Summary of Reactions - \nc. Diazotization (Section 24.8)\n\n3. Reactions of arenediazonium salts (Section 24.8)\na. Nucleophilic substitutions\n\n\n| HCl <br> CuCl | HBr <br> CuBr | NaI | KCN <br> CuCN | $\\mathrm{Cu}_{2} \\mathrm{O}, \\mathrm{H}_{2} \\mathrm{O}$ <br> $\\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}$ |\n| :--- | :--- | :--- | :--- | :--- | $\\mathrm{H}_{3} \\mathrm{PO}_{2}$\n\n\nb. Diazonium coupling"}
{"id": 1613, "contents": "Visualizing Chemistry - \nPROBLEM Name the following amines, and identify each as primary, secondary, or tertiary:\n\n24-26 (a)\n\n(b)\n\n(c)\n\n\nPROBLEM The following compound contains three nitrogen atoms. Rank them in order of increasing basicity.\n24-27\n\n\nPROBLEM Name the following amine, including $R, S$ stereochemistry, and draw the product of its reaction with\n24-28 excess iodomethane followed by heating with $\\mathrm{Ag}_{2} \\mathrm{O}$ (Hofmann elimination). Is the stereochemistry of the alkene product $Z$ or $E$ ? Explain.\n\n\nPROBLEM Which nitrogen atom in the following compound is most basic? Explain.\n24-29"}
{"id": 1614, "contents": "Mechanism Problems - \nPROBLEM Predict the product(s) and write the mechanism for each of the following reactions:\n24-30 (a)\n\n\n(b)\n\n\nPROBLEM Predict the product(s) and write the mechanism for each of the following reactions:\n24-31 (a)\n\n(b)\n\n\nPROBLEM Predict the product(s) and write the mechanism for each of the following reactions:\n24-32 (a)\n\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM Predict the product(s) and write the mechanism for each of the following reactions:\n\n24-33 (a)\n\n(b)\n\n\nPROBLEM The diazotization of aniline first involves the formation of $\\mathrm{NO}^{+}$(nitrosonium ion) by the dehydration\n24-34 of nitrous acid with sulfuric acid. The aniline nitrogen then acts as a nucleophile and eventually loses water. Propose a mechanism for the formation of the dizaonium salt of aniline using curved arrows to show all electron movement.\n\nPROBLEM Substituted pyrroles are often prepared by treatment of a 1,4-diketone with ammonia. Propose a 24-35 mechanism.\n\n\nPROBLEM 3,5-Dimethylisoxazole is prepared by reaction of 2,4-pentanedione with hydroxylamine. Propose a\n24-36 mechanism.\n\n\n3,5-Dimethylisoxazole\nPROBLEM One problem with reductive amination as a method of amine synthesis is that by-products are\n24-37 sometimes obtained. For example, reductive amination of benzaldehyde with methylamine leads to a mixture of $N$-methylbenzylamine and $N$-methyldibenzylamine. How do you suppose the tertiary amine by-product is formed? Propose a mechanism."}
{"id": 1615, "contents": "Mechanism Problems - \nPROBLEM Chlorophyll, heme, vitamin $\\mathrm{B}_{12}$, and a host of other substances are biosynthesized from\n24-38 porphobilinogen (PBG), which is itself formed from condensation of two molecules of 5-aminolevulinate. The two 5-aminolevulinates are bound to lysine (Lys) amino acids in the enzyme, one in the enamine form and one in the imine form, and their condensation is thought to occur by the following steps. Use curved arrows to show the mechanism of each step.\n\n\nEnzyme-bound 5-aminolevulinate\n\n\nPROBLEM Choline, a component of the phospholipids in cell membranes, can be prepared by $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of\n24-39 trimethylamine with ethylene oxide. Show the structure of choline, and propose a mechanism for the reaction.\n\n\nPROBLEM The antitumor antibiotic mitomycin C functions by forming cross-links in DNA chains. 24-40\n\n\nMitomycin C\n\n(a) The first step is loss of methoxide and formation of an iminium ion intermediate that is deprotonated to give an enamine. Show the mechanism.\n(b) The second step is reaction of the enamine with DNA to open the three-membered, nitrogencontaining (aziridine) ring. Show the mechanism.\n(c) The third step is loss of carbamate $\\left(\\mathrm{NH}_{2} \\mathrm{CO}_{2}{ }^{-}\\right)$and formation of an unsaturated iminium ion, followed by a conjugate addition of another part of the DNA chain. Show the mechanism.\nPROBLEM $\\alpha$-Amino acids can be prepared by the Strecker synthesis, a two-step process in which an aldehyde\n24-41 is treated with ammonium cyanide followed by hydrolysis of the amino nitrile intermediate with aqueous acid. Propose a mechanism for the reaction.\n\n\nAn $\\alpha$-amino acid\nPROBLEM One of the reactions used in determining the sequence of nucleotides in a strand of DNA is reaction\n24-42 with hydrazine. Propose a mechanism for the following reaction, which occurs by an initial conjugate addition followed by internal amide formation."}
{"id": 1616, "contents": "Mechanism Problems - \nPROBLEM When an $\\alpha$-hydroxy amide is treated with $\\mathrm{Br}_{2}$ in aqueous NaOH under Hofmann rearrangement\n24-43 conditions, loss of $\\mathrm{CO}_{2}$ occurs and a chain-shortened aldehyde is formed. Propose a mechanism.\n\n\nPROBLEM The following transformation involves a conjugate nucleophilic addition reaction (Section 19.13)\n24-44 followed by an intramolecular nucleophilic acyl substitution reaction (Section 21.2). Show the mechanism.\n\n\nPROBLEM Propose a mechanism for the following reaction:\n24-45\n\n\nPROBLEM One step in the biosynthesis of morphine is the reaction of dopamine with\n24-46 $p$-hydroxyphenylacetaldehyde to give ( $S$ )-norcoclaurine. Assuming that the reaction is acidcatalyzed, propose a mechanism.\n\n\nNaming Amines\nPROBLEM Name the following compounds:\n24-47 (a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f) $\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CN}$\n\nPROBLEM Draw structures corresponding to the following IUPAC names:\n24-48\n(a) $N, N$-Dimethylaniline\n(b) (Cyclohexylmethyl)amine\n(c) $N$-Methylcyclohexylamine\n(d) (2-Methylcyclohexyl)amine (e) 3-(N,N-Dimethylamino)propanoic acid\n\nPROBLEM Classify each of the amine nitrogen atoms in the following substances as primary, secondary, or 24-49 tertiary:\n(a)\n\n(b)\n\n(c)\n\nLysergic acid diethylamide"}
{"id": 1617, "contents": "Amine Basicity - \nPROBLEM Although pyrrole is a much weaker base than most other amines, it is a much stronger acid ( $\\mathrm{p} K_{\\mathrm{a}} \\approx$\n24-50 15 for the pyrrole versus 35 for diethylamine). The $\\mathrm{N}-\\mathrm{H}$ hydrogen is readily abstracted by base to yield the pyrrole anion, $\\mathrm{C}_{4} \\mathrm{H}_{4} \\mathrm{~N}^{-}$. Explain.\n\nPROBLEM Histamine, whose release in the body triggers nasal secretions and constricted airways, has three\n24-51 nitrogen atoms. List them in order of increasing basicity and explain your ordering.\n\n\nHistamine\n\nPROBLEM Account for the fact that $p$-nitroaniline ( $\\mathrm{p} K_{\\mathrm{a}}=1.0$ ) is less basic than $m$-nitroaniline ( $\\mathrm{p} K_{\\mathrm{a}}=2.5$ ) by\n24-52 a factor of 30 . Draw resonance structures to support your argument. (The $\\mathrm{p} K_{\\mathrm{a}}$ values refer to the corresponding ammonium ions.)"}
{"id": 1618, "contents": "Synthesis of Amines - \nPROBLEM How would you prepare the following substances from 1-butanol?\n24-53 (a) Butylamine (b) Dibutylamine (c) Propylamine (d) Pentylamine\n(e) $N, N$-Dimethylbutylamine (f) Propene\n\nPROBLEM How would you prepare the following substances from pentanoic acid?\n24-54 (a) Pentanamide (b) Butylamine (c) Pentylamine (d) 2-Bromopentanoic acid\n(e) Hexanenitrile (f) Hexylamine\nPROBLEM How would you prepare aniline from the following starting materials?\n24-55 (a) Benzene (b) Benzamide (c) Toluene\nPROBLEM How would you prepare benzylamine, $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CH}_{2} \\mathrm{NH}_{2}$, from benzene? More than one step is needed.\n24-56\nPROBLEM How might you prepare pentylamine from the following starting materials?\n24-57 (a) Pentanamide\n(b) Pentanenitrile\n(c) 1-Butene\n(d) Hexanamide\n(e) 1-Butanol\n(f) 5-Decene (g) Pentanoic acid\n\nPROBLEM How might a reductive amination be used to synthesize ephedrine, an amino alcohol that was\n24-58 widely used for the treatment of bronchial asthma?\n\n\nEphedrine"}
{"id": 1619, "contents": "Reactions of Amines - \nPROBLEM How would you convert aniline into each of the following products?\n24-59\n(a) Benzene\n(b) Benzamide\n(c) Toluene\n\nPROBLEM Write the structures of the major organic products you would expect from reaction of $m$-toluidine 24-60 ( $m$-methylaniline) with the following reagents:\n(a) $\\mathrm{Br}_{2}$ (1 equivalent)\n(b) $\\mathrm{CH}_{3} \\mathrm{I}$ (excess)\n(c) $\\mathrm{CH}_{3} \\mathrm{COCl}$ in pyridine\n(d) The product of (c), then $\\mathrm{HSO}_{3} \\mathrm{Cl}$\n\nPROBLEM Show the products from reaction of $p$-bromoaniline with the following reagents:\n24-61 (a) $\\mathrm{CH}_{3} \\mathrm{I}$ (excess)\n(b) HCl\n(c) $\\mathrm{HNO}_{2}, \\mathrm{H}_{2} \\mathrm{SO}_{4}$\n(d) $\\mathrm{CH}_{3} \\mathrm{COCl}$\n(e) $\\mathrm{CH}_{3} \\mathrm{MgBr}$\n(f) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{Cl}, \\mathrm{AlCl}_{3}$ (g) Product of (c) with $\\mathrm{CuCl}, \\mathrm{HCl}$ (h) Product of (d) with $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{Cl}, \\mathrm{AlCl}_{3}$\n\nPROBLEM What are the major products you would expect from Hofmann elimination of the following amines?\n\n24-62 (a)\n\n(b)\n\n(c)\n\n\nPROBLEM How would you prepare the following compounds from toluene? A diazonio replacement reaction 24-63 is needed in some instances.\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Predict the product(s) of the following reactions. If more than one product is formed, tell which is 24-64 major.\n(a)\n\n(b)\n\n(c)\n\n(d)"}
{"id": 1620, "contents": "Reactions of Amines - \n(b)\n\n(c)\n\n\nPROBLEM Predict the product(s) of the following reactions. If more than one product is formed, tell which is 24-64 major.\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n\nSpectroscopy\nPROBLEM Phenacetin, a substance formerly used in over-the-counter headache remedies, has the formula\n24-65 $\\mathrm{C}_{10} \\mathrm{H}_{13} \\mathrm{NO}_{2}$. Phenacetin is neutral and does not dissolve in either acid or base. When warmed with aqueous NaOH , phenacetin yields an amine, $\\mathrm{C}_{8} \\mathrm{H}_{11} \\mathrm{NO}$, whose ${ }^{1} \\mathrm{H}$ NMR spectrum is shown. When heated with HI , the amine is cleaved to an aminophenol, $\\mathrm{C}_{6} \\mathrm{H}_{7} \\mathrm{NO}$. What is the structure of\nphenacetin, and what are the structures of the amine and the aminophenol?\n\n\nPROBLEM Propose structures for amines with the following ${ }^{1} \\mathrm{H}$ NMR spectra:\n24-66 (a) $\\mathrm{C}_{3} \\mathrm{H}_{9} \\mathrm{NO}$\n\n(b) $\\mathrm{C}_{4} \\mathrm{H}_{11} \\mathrm{NO}_{2}$\n\n(c) $\\mathrm{C}_{8} \\mathrm{H}_{11} \\mathrm{~N}$\n\n\nPROBLEM Draw the structure of the amine that produced the ${ }^{1} \\mathrm{H}$ NMR spectrum shown in Problem 24-66(c).\n24-67 This compound has a single strong peak in its IR spectrum at $3280 \\mathrm{~cm}^{-1}$."}
{"id": 1621, "contents": "General Problems - \nPROBLEM Fill in the missing reagents a-e in the following scheme:\n24-68\n\n\nPROBLEM Oxazole is a five-membered aromatic heterocycle. Would you expect oxazole to be more basic or 24-69 less basic than pyrrole? Explain.\n\n\nOxazole\n\nPROBLEM Protonation of an amide using strong acid occurs on oxygen rather than on nitrogen. Suggest a 24-70 reason for this behavior, taking resonance into account.\n\n\nPROBLEM What is the structure of the compound with formula $\\mathrm{C}_{8} \\mathrm{H}_{11} \\mathrm{~N}$ that produced the following IR 24-71 spectrum?\n\n\nPROBLEM Fill in the missing reagents a-d in the following synthesis of racemic methamphetamine from 24-72 benzene.\n\n\nPROBLEM Cyclopentamine is an amphetamine-like central nervous system stimulant. Propose a synthesis of 24-73 cyclopentamine from materials of five carbons or less."}
{"id": 1622, "contents": "Cyclopentamine - \nPROBLEM Tetracaine is a substance used as a spinal anesthetic.\n24-74\n\n(a) How would you prepare tetracaine from the corresponding aniline derivative, $\\mathrm{ArNH}_{2}$ ?\n(b) How would you prepare tetracaine from $p$-nitrobenzoic acid?\n(c) How would you prepare tetracaine from benzene?\n\nPROBLEM Atropine, $\\mathrm{C}_{17} \\mathrm{H}_{23} \\mathrm{NO}_{3}$, is a poisonous alkaloid isolated from the leaves and roots of Atropa\n24-75 belladonna, the deadly nightshade. In small doses, atropine acts as a muscle relaxant; 0.5 ng (nanogram, $10^{-9} \\mathrm{~g}$ ) is sufficient to cause pupil dilation. On basic hydrolysis, atropine yields tropic acid, $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CH}\\left(\\mathrm{CH}_{2} \\mathrm{OH}\\right) \\mathrm{CO}_{2} \\mathrm{H}$, and tropine, $\\mathrm{C}_{8} \\mathrm{H}_{15} \\mathrm{NO}$. Tropine is an optically inactive alcohol that yields tropidene on dehydration with $\\mathrm{H}_{2} \\mathrm{SO}_{4}$. Propose a structure for atropine.\n\n\nPROBLEM Tropidene (Problem 24-75) can be converted by a series of steps into tropilidene 24-76 (1,3,5-cycloheptatriene). How would you accomplish this conversion?\n\nPROBLEM Propose a structure for the product with formula $\\mathrm{C}_{9} \\mathrm{H}_{17} \\mathrm{~N}$ that results when 24-77 2-(2-cyanoethyl)cyclohexanone is reduced catalytically."}
{"id": 1623, "contents": "Cyclopentamine - \nPROBLEM Coniine, $\\mathrm{C}_{8} \\mathrm{H}_{17} \\mathrm{~N}$, is the toxic principle of the poison hemlock drunk by Socrates. When subjected\n24-78 to Hofmann elimination, coniine yields 5 -( $N, N$-dimethylamino)-1-octene. If coniine is a secondary amine, what is its structure?\n\nPROBLEM How would you synthesize coniine (Problem 24-78) from acrylonitrile ( $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCN}$ ) and ethyl\n24-79 3-oxohexanoate $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{COCH}_{2} \\mathrm{CO}_{2} \\mathrm{Et}\\right)$ ? (See Problem 24-77.)\nPROBLEM Tyramine is an alkaloid found, among other places, in mistletoe and ripe cheese. How would you\n24-80 synthesize tyramine from benzene? From toluene?\n\n\nPROBLEM Reaction of anthranilic acid (o-aminobenzoic acid) with $\\mathrm{HNO}_{2}$ and $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ yields a diazonium salt\n24-81 that can be treated with base to yield a neutral diazonium carboxylate.\n(a) What is the structure of the neutral diazonium carboxylate?\n(b) Heating the diazonium carboxylate results in the formation of $\\mathrm{CO}_{2}, \\mathrm{~N}_{2}$, and an intermediate that reacts with 1,3-cyclopentadiene to yield the following product:\n\n\nWhat is the structure of the intermediate, and what kind of reaction does it undergo with cyclopentadiene?\n\nPROBLEM Cyclooctatetraene was first synthesized in 1911 by a route that involved the following 24-82 transformation:\n\n\nHow might you use the Hofmann elimination to accomplish this reaction? How would you finish the synthesis by converting cyclooctatriene into cyclooctatetraene?"}
{"id": 1624, "contents": "Cyclopentamine - \nHow might you use the Hofmann elimination to accomplish this reaction? How would you finish the synthesis by converting cyclooctatriene into cyclooctatetraene?\n\nPROBLEM Propose structures for compounds that show the following ${ }^{1} \\mathrm{H}$ NMR spectra.\n24-83 (a) $\\mathrm{C}_{9} \\mathrm{H}_{13} \\mathrm{~N}$\n\n(b) $\\mathrm{C}_{15} \\mathrm{H}_{17} \\mathrm{~N}$\n\n\nPROBLEM 4-Dimethylaminopyridine (DMAP) acts as a catalyst in acyl transfer reactions. DMAP's catalytic\n24-84 activity stems from its nucleophilic character at the pyridine nitrogen, not the dimethylamino group. Explain this behavior, taking resonance into account."}
{"id": 1625, "contents": "CHAPTER 25 <br> Biomolecules: Carbohydrates - \nFIGURE 25.1 Produced by honeybees from the nectar of flowers, honey is primarily a mixture of the two simple sugars fructose and glucose. (credit: modification of work \"Apis mellifera\" by Thomas Bresson/Flickr, CC BY 2.0)"}
{"id": 1626, "contents": "CHAPTER CONTENTS - \n25.1 Classification of Carbohydrates\n25.2 Representing Carbohydrate Stereochemistry: Fischer Projections\n25.3 D,L Sugars\n25.4 Configurations of the Aldoses\n25.5 Cyclic Structures of Monosaccharides: Anomers\n25.6 Reactions of Monosaccharides\n25.7 The Eight Essential Monosaccharides\n25.8 Disaccharides\n25.9 Polysaccharides and Their Synthesis\n25.10 Some Other Important Carbohydrates\n\nWHY THIS CHAPTER? We've now seen all the common functional groups and reaction types that occur in organic and biological chemistry. In this and the next four chapters, we'll focus on the major classes of biological molecules, beginning with a look at the structures and primary biological functions of carbohydrates. Then, in Chapter 29, we'll return to the subject to see how carbohydrates are both synthesized and degraded in organisms.\n\nCarbohydrates occur in every living organism. The sugar and starch in food, and the cellulose in wood, paper, and cotton are nearly pure carbohydrates. Modified carbohydrates form part of the coating around living cells, other carbohydrates are part of the nucleic acids that carry our genetic information, and still others are used as medicines.\n\nThe word carbohydrate derives historically from the fact that glucose, the first simple carbohydrate to be obtained in pure form, has the molecular formula $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$ and was originally thought to be a \"hydrate of\ncarbon, $\\mathrm{C}_{6}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}$.\" This view was soon abandoned, but the name persisted. Today, the term carbohydrate is used to refer loosely to the broad class of polyhydroxylated aldehydes and ketones commonly called sugars. Glucose, also known as dextrose in medical work, is the most familiar example."}
{"id": 1627, "contents": "CHAPTER CONTENTS - \nGlucose (dextrose),\na pentahydroxyhexanal\nCarbohydrates are synthesized by green plants during photosynthesis, a complex process in which sunlight provides the energy to convert $\\mathrm{CO}_{2}$ and $\\mathrm{H}_{2} \\mathrm{O}$ into glucose plus oxygen. Many molecules of glucose are then chemically linked for storage by the plant in the form of either cellulose or starch. It has been estimated that more than $50 \\%$ of the dry weight of the earth's biomass-all plants and animals-consists of glucose polymers. When eaten and metabolized, carbohydrates then provide animals with a source of readily available energy. Thus, carbohydrates act as the chemical intermediaries by which solar energy is stored and used to support life.\n\n$$\n6 \\mathrm{CO}_{2}+6 \\mathrm{H}_{2} \\mathrm{O} \\xrightarrow{\\text { Sunlight }} 6 \\mathrm{O}_{2}+\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6} \\longrightarrow \\text { Cellulose, starch }\n$$\n\nGlucose\nBecause humans and most other mammals lack the enzymes needed for digestion of cellulose, they require starch as their dietary source of carbohydrates. Grazing animals such as cows, however, have microorganisms in their first stomach that are able to digest cellulose. The energy stored in cellulose is thus moved up the biological food chain when these ruminant animals eat grass and are themselves used for food."}
{"id": 1628, "contents": "CHAPTER CONTENTS - 25.1 Classification of Carbohydrates\nCarbohydrates are generally classified as either simple or complex. Simple sugars, or monosaccharides, are carbohydrates like glucose and fructose that can't be converted into smaller sugars by hydrolysis. Complex carbohydrates are made of two or more simple sugars linked together by acetal bonds (Section 19.10). Sucrose (table sugar), for instance, consists of one glucose linked to one fructose. Similarly, cellulose is made up of several thousand glucose units linked together. Enzyme-catalyzed hydrolysis of a complex carbohydrate breaks it down into its constituent monosaccharides.\n\n\nSucrose (a disaccharide)\n\n\nMonosaccharides are further classified as either aldoses or ketoses. The -ose suffix designates a carbohydrate, and the prefixes aldo- and keto- identify the kind of carbonyl group in the molecule, whether aldehyde or ketone.\n\nThe number of carbon atoms in the monosaccharide is indicated by the appropriate numerical prefix tri-, tetr-, pent-, hex-, and so forth, in the name. Putting it all together, glucose is an aldohexose, a six-carbon aldehyde sugar; fructose is a ketohexose, a six-carbon keto sugar; ribose is an aldopentose, a five-carbon aldehyde sugar; and sedoheptulose is a ketoheptose, a seven-carbon keto sugar. Most of the common simple sugars are either pentoses or hexoses.\n\n\nPROBLEM Classify each of the following monosaccharides:\n25-1 (a)\n\nThreose\n(b)\n\nRibulose\n(c)\n\nTagatose\n(d)\n\n2-Deoxyribose"}
{"id": 1629, "contents": "CHAPTER CONTENTS - 25.2 Representing Carbohydrate Stereochemistry: Fischer Projections\nBecause most carbohydrates have numerous chirality centers, it was recognized long ago that a quick method for representing their stereochemistry was needed. In 1891, the German chemist Emil Fischer suggested a method based on the projection of a tetrahedral carbon atom onto a flat surface. These Fischer projections were soon adopted and are still a common means of representing stereochemistry at chirality centers in carbohydrate chemistry.\n\nA tetrahedral carbon atom is represented in a Fischer projection by two crossed lines. The horizontal lines represent bonds coming out of the page, and the vertical lines represent bonds going into the page.\n\nFor example, (R)-glyceraldehyde, the simplest monosaccharide, can be drawn as in FIGURE 25.2.\n\n\nFIGURE 25.2 A Fischer projection of (R)-glyceraldehyde.\nBecause a given chiral molecule can be drawn in many ways, it's sometimes necessary to compare two projections to see if they represent the same or different enantiomers. To test for identity, Fischer projections can be moved around on the paper, but only two kinds of motions are allowed; moving a Fischer projection in any other way inverts its meaning.\n\n- A Fischer projection can be rotated on the page by $180^{\\circ}$, but not by $90^{\\circ}$ or $270^{\\circ}$. Only a $180^{\\circ}$ rotation\nmaintains the Fischer convention by keeping the same substituent groups going into and coming out of the plane. In the following Fischer projection of ( $R$ )-glyceraldehyde, for instance, the -H and -OH groups come out of the plane both before and after a $180^{\\circ}$ rotation.\n\n\nA $90^{\\circ}$ rotation breaks the Fischer convention by exchanging the groups that go into the plane with those that come out. In the following Fischer projections of $(R)$-glyceraldehyde, the -H and -OH groups originally come out of the plane but go into the plane after a $90^{\\circ}$ rotation. As a result, the rotated projection represents (S)-glyceraldehyde.\n\n\n(R)-Glyceraldehyde\n\n- A Fischer projection can also have one group held steady while the other three rotate in either a clockwise or a counterclockwise direction. The effect is simply to rotate around a single bond, which does not change the stereochemistry."}
{"id": 1630, "contents": "CHAPTER CONTENTS - 25.2 Representing Carbohydrate Stereochemistry: Fischer Projections\n- A Fischer projection can also have one group held steady while the other three rotate in either a clockwise or a counterclockwise direction. The effect is simply to rotate around a single bond, which does not change the stereochemistry.\n\n\n(R)-Glyceraldehyde\n\n(R)-Glyceraldehyde\n\n\n\n(S)-Glyceraldehyde\n(S)-Glyceraidehyde\n\n$R, S$ stereochemical designations (Section 5.5) can be assigned to the chirality center in a Fischer projection by following three steps, as shown in Worked Example 25.1."}
{"id": 1631, "contents": "STEP 1 - \nRank the four substituents in the usual way (Section 5.5)."}
{"id": 1632, "contents": "STEP 2 - \nPlace the group of lowest ranking, usually H, at the top of the Fischer projection by using one of the allowed motions. This means that the lowest-ranked group is oriented back, away from the viewer, as required for assigning configuration."}
{"id": 1633, "contents": "STEP 3 - \nDetermine the direction of rotation $1 \\rightarrow 2 \\rightarrow 3$ of the remaining three groups, and assign $R$ or $S$ configuration.\nCarbohydrates with more than one chirality center are shown in Fischer projections by stacking the centers on top of one another, with the carbonyl carbon at or near the top. Glucose, for example, has four chirality centers stacked on top of one another in a Fischer projection. Such representations don't, however, give an accurate\npicture of the molecule's true three-dimensional conformation, which is curled around on itself like a bracelet.\n\n\nGlucose\n(carbonyl group on top)"}
{"id": 1634, "contents": "Assigning $R$ or $S$ Configuration to a Fischer Projection - \nAssign $R$ or $S$ configuration to the following Fischer projection of alanine:\n\n\nAlanine"}
{"id": 1635, "contents": "Strategy - \nFollow the steps in the text. (1) Rank the four substituents on the chiral carbon. (2) Manipulate the Fischer projection to place the group of lowest ranking at the top by carrying out one of the allowed motions. (3) Determine the direction $1 \\rightarrow 2 \\rightarrow 3$ of the remaining three groups."}
{"id": 1636, "contents": "Solution - \nThe rankings of the groups are (1) $-\\mathrm{NH}_{2},(\\mathbf{2})-\\mathrm{CO}_{2} \\mathrm{H},(\\mathbf{3})-\\mathrm{CH}_{3}$, and $(\\mathbf{4})-\\mathrm{H}$. To bring the group of lowest ranking $(-\\mathrm{H})$ to the top, we might want to hold the $-\\mathrm{CH}_{3}$ group steady while rotating the other three groups counterclockwise.\n\n\nGoing from first- to second- to third-highest ranking requires a counterclockwise turn, corresponding to $S$ stereochemistry.\n\n\nPROBLEM Convert each of the following Fischer projections into a tetrahedral representation, and assign $R$ or 25-2 $S$ stereochemistry:\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM Which of the following Fischer projections of glyceraldehyde represent the same enantiomer?\n25-3\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM Redraw the following molecule as a Fischer projection, and assign $R$ or $S$ configuration to the\n25-4 chirality center (green $=\\mathrm{Cl}$ ):\n\n\nPROBLEM Redraw the following aldotetrose as a Fischer projection, and assign $R$ or $S$ configuration to each\n25-5 chirality center:"}
{"id": 1637, "contents": "25.3 d, L Sugars - \nGlyceraldehyde, the simplest aldose, has only one chirality center and thus has two enantiomeric (nonidentical mirror-image) forms. Only the dextrorotatory enantiomer occurs naturally, however. That is, a sample of naturally occurring glyceraldehyde placed in a polarimeter rotates plane-polarized light in a clockwise direction, denoted (+). Since (+)-glyceraldehyde has been found to have an $R$ configuration at C2, it can be represented by a Fischer projection as shown previously in FIGURE 25.2. For historical reasons dating back long before the adoption of the $R, S$ system, $(R)-(+)$-glyceraldehyde is also referred to as D-glyceraldehyde (D for dextrorotatory). The other enantiomer, (S)-(-)-glyceraldehyde, is known as L-glyceraldehyde (L for levorotatory).\n\nBecause of the way that monosaccharides are biosynthesized in nature, glucose, fructose, and most other naturally occurring monosaccharides all have the same $R$ stereochemical configuration as D-glyceraldehyde at the chirality center farthest from the carbonyl group. In Fischer projections, therefore, most naturally occurring sugars have the hydroxyl group at the bottom chirality center pointing to the right (FIGURE 25.3). Such compounds are referred to as D sugars.\n\n\nD-Glyceraldehyde [(R)-(+)-glyceraldehyde]\n\n\nD-Ribose\n\n\nD-Glucose\n\n\nD-Fructose\n\nFIGURE 25.3 Some naturally occurring D sugars. The -OH group at the chirality center farthest from the carbonyl group has the same configuration as (R)-(+)-glyceraldehyde and points toward the right in Fischer projections.\n\nIn contrast with D sugars, L sugars have an $S$ configuration at the lowest chirality center, with the bottom -OH group pointing to the left in Fischer projections. Thus, an l sugar is the mirror image (enantiomer) of the corresponding D sugar and has the opposite configuration from the $D$ sugar at all chirality centers."}
{"id": 1638, "contents": "25.3 d, L Sugars - \nNote that the $D$ and L notations have no relation to the direction in which a given sugar rotates plane-polarized light. A D sugar can be either dextrorotatory or levorotatory. The prefix D indicates only that the - OH group at the lowest chirality center has $R$ stereochemistry and points to the right when the molecule is drawn in a standard Fischer projection. Note also that the D,L system of carbohydrate nomenclature describes the configuration at only one chirality center and says nothing about the configuration of other chirality centers that may be present.\n\nPROBLEM Assign $R$ or $S$ configuration to each chirality center in the following monosaccharides, and tell\n25-6 whether each is a D sugar or an L sugar:\n(a)\n\n(b)\n\n(c)\n\n\nPROBLEM (+)-Arabinose, an aldopentose that is widely distributed in plants, is systematically named\n25-7 ( $2 R, 3 S, 4 S$ )-2,3,4,5-tetrahydroxypentanal. Draw a Fischer projection of (+)-arabinose, and identify it as a D sugar or an L sugar."}
{"id": 1639, "contents": "25.3 d, L Sugars - 25.4 Configurations of the Aldoses\nAn aldotetrose is a four-carbon sugar with two chirality centers. Thus, there are $2^{2}=4$ possible stereoisomeric aldotetroses, or two D,L pairs of enantiomers named erythrose and threose.\n\nAldopentoses have three chirality centers and a total of $2^{3}=8$ possible stereoisomers, or four D,L pairs of enantiomers. These four pairs are called ribose, arabinose, xylose, and lyxose. All except lyxose occur widely in nature. D-Ribose is an important constituent of RNA (ribonucleic acid), L-arabinose is found in many plants, and\n\nD-xylose is found in wood.\nAldohexoses have four chirality centers and a total of $2^{4}=16$ possible stereoisomers, or eight D,L pairs of enantiomers. The names of the eight are allose, altrose, glucose, mannose, gulose, idose, galactose, and talose. Only D-glucose, from starch and cellulose, and D-galactose, from gums and fruit pectins, are widely distributed in nature. D-mannose and D-talose also occur naturally but in lesser abundance.\n\nFischer projections of the four-, five-, and six-carbon D aldoses are shown in FIGURE 25.4. Starting with D-glyceraldehyde, we can imagine constructing the two D aldotetroses by inserting a new chirality center just below the aldehyde carbon. Each of the two D aldotetroses then leads to two D aldopentoses (four total), and each of the four $D$ aldopentoses leads to two $D$ aldohexoses (eight total). In addition, each of the $D$ aldoses in FIGURE 25.4 has an l enantiomer, which is not shown."}
{"id": 1640, "contents": "25.3 d, L Sugars - 25.4 Configurations of the Aldoses\nFIGURE 25.4 Configurations of $\\mathbf{D}$ aldoses. The structures are arranged from left to right so that the -OH groups on C2 alternate right/left $(R / L)$ across each series. Similarly, the -OH groups at C 3 alternate two right/two left ( $2 \\mathrm{R} / 2 \\mathrm{~L}$ ), the -OH groups at C 4 alternate $4 R / 4 \\mathrm{~L}$, and the -OH groups at C 5 are to the right in all eight ( 8 R ). Each D aldose has a corresponding L enantiomer that is not shown.\n\nLouis Fieser of Harvard University long ago suggested a simple procedure for remembering the names and structures of the eight $D$ aldohexoses:\n\nSTEP 1\nSet up eight Fischer projections with the -CHO group on top and the $-\\mathrm{CH}_{2} \\mathrm{OH}$ group at the bottom.\nSTEP 2\nAt C5, place all eight -OH groups to the right (D series).\n\nSTEP 3\nAt C 4 , alternate four -OH groups to the right and four to the left.\nSTEP 4\nAt C3, alternate two -OH groups to the right, two to the left.\nSTEP 5\nAt C2, alternate - OH groups right, left, right, left.\nSTEP 6\nName the eight isomers using the mnemonic \"All altruists gladly make gum in gallon tanks.\"\nThe structures of the four $D$ aldopentoses can be generated in a similar way and named by the memonic suggested by a Cornell University undergraduate: \"RIBs ARe eXtra Lean.\""}
{"id": 1641, "contents": "Drawing a Fischer Projection - \nDraw a Fischer projection of L-fructose."}
{"id": 1642, "contents": "Strategy - \nBecause L-fructose is the enantiomer of D-fructose, simply look at the structure of D-fructose and reverse the configuration at each chirality center."}
{"id": 1643, "contents": "Drawing a Fischer Projection of a Molecular Model - \nDraw the following aldotetrose as a Fischer projection, and identify it as a D sugar or an L sugar:"}
{"id": 1644, "contents": "Strategy - \nThe Fischer projection of a monosaccharide is drawn vertically, with the carbonyl group at or near the top and the $-\\mathrm{CH}_{2} \\mathrm{OH}$ group at the bottom. The interior -H and -OH are drawn to the sides, pointing out of the page toward the viewer."}
{"id": 1645, "contents": "Solution - \nFischer projection\n\nPROBLEM Only the D sugars are shown in Figure 25.4. Draw Fischer projections for the following l sugars:\n25-8 (a) L-Xylose (b) L-Galactose (c) L-Allose\nPROBLEM How many aldoheptoses are there? How many are D sugars, and how many are L sugars?\n25-9\nPROBLEM The following model is that of an aldopentose. Draw a Fischer projection of the sugar, name it, and\n25-10 identify it as a D sugar or an L sugar."}
{"id": 1646, "contents": "Solution - 25.5 Cyclic Structures of Monosaccharides: Anomers\nWe said in Section 19.10 that aldehydes and ketones undergo a rapid and reversible nucleophilic addition reaction with alcohols to form hemiacetals."}
{"id": 1647, "contents": "An aldehyde - \nA hemiacetal\nIf the carbonyl and the hydroxyl group are in the same molecule, an intramolecular nucleophilic addition can take place, leading to the formation of a cyclic hemiacetal. Five- and six-membered cyclic hemiacetals are relatively strain-free and particularly stable, and many carbohydrates therefore exist in an equilibrium between open-chain and cyclic forms. Glucose, for instance, exists in aqueous solution primarily in the six-membered pyranose form resulting from intramolecular nucleophilic addition of the -OH group at C 5 to the C 1 carbonyl group (FIGURE 25.5). The name pyranose is derived from pyran, the name of the unsaturated six-membered cyclic ether.\n\nLike cyclohexane rings (Section 4.6), pyranose rings have a chairlike geometry with axial and equatorial substituents. By convention, the rings are usually drawn by placing the hemiacetal oxygen atom at the right rear,\nas shown in FIGURE 25.5. Note that an -OH group on the right in a Fischer projection is on the bottom face of the pyranose ring, and an -OH group on the left in a Fischer projection is on the top face of the ring. For D sugars, the terminal $-\\mathrm{CH}_{2} \\mathrm{OH}$ group is on the top of the ring, whereas for L sugars, the $-\\mathrm{CH}_{2} \\mathrm{OH}$ group is on the bottom."}
{"id": 1648, "contents": "An aldehyde - \nWhen an open-chain monosaccharide cyclizes to a pyranose form, a new chirality center is generated at the former carbonyl carbon and two diastereomers, called anomers, are produced. The hemiacetal carbon atom is referred to as the anomeric center. For example, glucose cyclizes reversibly in aqueous solution to a $37: 63$ mixture of two anomers (FIGURE 25.5). The compound with its newly generated - OH group at C 1 cis to the -OH at the lowest chirality center in a Fischer projection is called the $\\boldsymbol{\\alpha}$ anomer; its full name is $\\alpha$-D-glucopyranose. The compound with its newly generated -OH group trans to the -OH at the lowest chirality center is called the $\\boldsymbol{\\beta}$ anomer; its full name is $\\beta$-D-glucopyranose. Note that in $\\beta$-D-glucopyranose, all the substituents on the ring are equatorial. Thus, $\\beta$-D-glucopyranose is the least sterically crowded and most stable of the eight D aldohexoses.\n\n\n\n$\\alpha$-D-Glucopyranose\n(37.3\\%)\n\n$\\beta$-D-Glucopyranose\n(62.6\\%)\n\nFIGURE 25.5 Glucose in its cyclic pyranose forms. As explained in the text, two anomers are formed by cyclization of glucose. The molecule whose newly formed -OH group at C 1 is cis to the oxygen atom on the lowest chirality center (C5) in a Fischer projection is the $\\alpha$ anomer. The molecule whose newly formed -OH group is trans to the oxygen atom on the lowest chirality center in a Fischer projection is the $\\beta$ anomer."}
{"id": 1649, "contents": "An aldehyde - \nSome monosaccharides also exist in a five-membered cyclic hemiacetal form called a furanose. D-Fructose, for instance, exists in water solution as $68 \\% \\beta$-pyranose, $2.7 \\% \\alpha$-pyranose, $0.5 \\%$ open-chain, $22.4 \\% \\beta$-furanose, and $6.2 \\% \\alpha$-furanose. The pyranose form results from addition of the - OH at C 6 to the carbonyl group, while the furanose form results from addition of the -OH at C 5 to the carbonyl group (FIGURE 25.6).\n\n\nFIGURE 25.6 Pyranose and furanose forms of fructose in aqueous solution. The two pyranose anomers result from addition of the $\\mathrm{C} 6-\\mathrm{OH}$ group to the C2 carbonyl; the two furanose anomers result from addition of the $\\mathrm{C} 5-\\mathrm{OH}$ group to the C 2 carbonyl.\n\nBoth anomers of $D$-glucopyranose can be crystallized and purified. Pure $\\alpha$-D-glucopyranose has a melting point of $146{ }^{\\circ} \\mathrm{C}$ and a specific rotation $[\\alpha]_{D}=+112.2$; pure $\\beta$-D-glucopyranose has a melting point of 148 to $155{ }^{\\circ} \\mathrm{C}$ and a specific rotation $[\\alpha]_{D}=+18.7$. When a sample of either pure anomer is dissolved in water, however, its optical rotation slowly changes until it reaches a constant value of +52.6 . That is, the specific rotation of the $\\alpha$-anomer solution decreases from +112.2 to +52.6 , and the specific rotation of the $\\beta$-anomer solution increases from +18.7 to +52.6 . Called mutarotation, this change in optical rotation is due to the slow interconversion of the pure anomers to give the same $37: 63$ equilibrium mixture."}
{"id": 1650, "contents": "An aldehyde - \nMutarotation occurs by a reversible ring-opening of each anomer to the open-chain aldehyde, followed by reclosure. Although the equilibration is slow at neutral pH , it is catalyzed by both acid and base."}
{"id": 1651, "contents": "Drawing the Chair Conformation of an Aldohexose - \nD-Mannose differs from D-glucose in its stereochemistry at C2. Draw D-mannose in its chairlike pyranose form."}
{"id": 1652, "contents": "Strategy - \nFirst draw a Fischer projection of D-mannose. Then lay it on its side, and curl it around so that the - CHO group (C1) is on the right front and the $-\\mathrm{CH}_{2} \\mathrm{OH}$ group (C6) is toward the left rear. Now, connect the -OH at C 5 to the C1 carbonyl group to form the pyranose ring. In drawing the chair form, raise the leftmost carbon (C4) up and drop the rightmost carbon (C1) down."}
{"id": 1653, "contents": "Drawing the Chair Conformation of a Pyranose - \nDraw $\\beta$-L-glucopyranose in its more stable chair conformation."}
{"id": 1654, "contents": "Strategy - \nIt's probably easiest to begin by drawing the chair conformation of $\\beta$-D-glucopyranose. Then draw its mirrorimage L enantiomer by changing the stereochemistry at every position on the ring, and carry out a ring-flip to give the more stable chair conformation. Note that the $-\\mathrm{CH}_{2} \\mathrm{OH}$ group is on the bottom face of the ring in the L enantiomer, as is the anomeric -OH ."}
{"id": 1655, "contents": "Solution - \n$\\beta$-D-Glucopyranose\n$\\beta$-L-Glucopyranose\n\nPROBLEM Ribose exists largely in a furanose form, produced by addition of the $\\mathrm{C} 4-\\mathrm{OH}$ group to the C 1 25-11 aldehyde. Draw D-ribose in its furanose form.\n\nPROBLEM Figure 25.6 shows only the $\\beta$-pyranose and $\\beta$-furanose anomers of D -fructose. Draw the $\\alpha$-pyranose\n$\\mathbf{2 5 - 1 2}$ and $\\alpha$-furanose anomers.\nPROBLEM Draw $\\beta$-D-galactopyranose and $\\beta$-D-mannopyranose in their more stable chair conformations. Label\n25-13 each ring substituent as either axial or equatorial. Which would you expect to be more stable, galactose or mannose?\n\nPROBLEM Draw $\\beta$-L-galactopyranose in its more stable chair conformation, and label the substituents as\n25-14 either axial or equatorial.\nPROBLEM Identify the following monosaccharide, write its full name, and draw its open-chain form as a\n25-15 Fischer projection."}
{"id": 1656, "contents": "Solution - 25.6 Reactions of Monosaccharides\nBecause monosaccharides have only two kinds of functional groups, hydroxyls and carbonyls, most of the chemistry of monosaccharides is the familiar chemistry of these two groups. As we've seen, alcohols can be converted into esters and ethers and can be oxidized; carbonyl compounds can react with nucleophiles and can be reduced."}
{"id": 1657, "contents": "Ester and Ether Formation - \nMonosaccharides behave as simple alcohols in much of their chemistry. For example, carbohydrate - OH groups can be converted into esters and ethers, which are often easier to work with than the free sugars. Because of their many hydroxyl groups, monosaccharides are usually soluble in water but insoluble in organic solvents such as ether. They are also difficult to purify and have a tendency to form syrups rather than crystals when water is removed. Ester and ether derivatives, however, are soluble in organic solvents and are easily purified and crystallized.\n\nEsterification is normally carried out by treating a carbohydrate with an acid chloride or acid anhydride in the presence of a base (Section 21.4 and Section 21.5). All the -OH groups react, including the anomeric one. For example, $\\beta$-D-glucopyranose is converted into its pentaacetate by treatment with acetic anhydride in pyridine solution.\n\n$\\beta$-D-Glucopyranose\nPenta- $O$-acetyl- $\\beta$-D-glucopyranose\n(91\\%)\nCarbohydrates are converted into ethers by treatment with an alkyl halide in the presence of base-the Williamson ether synthesis (Section 18.2). Standard Williamson conditions using a strong base tend to degrade sensitive sugar molecules, but silver oxide works well as a mild base and gives high yields of ethers. For example, $\\alpha$-D-glucopyranose is converted into its pentamethyl ether in $85 \\%$ yield on reaction with iodomethane and $\\mathrm{Ag}_{2} \\mathrm{O}$.\n\n\nPROBLEM Draw the products you would obtain by reaction of $\\beta$-d-ribofuranose with: (a) $\\mathrm{CH}_{3} \\mathrm{I}, \\mathrm{Ag}_{2} \\mathrm{O}$ (b)\n25-16 $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O}$, pyridine"}
{"id": 1658, "contents": "Glycoside Formation - \nWe saw in Section 19.10 that treatment of a hemiacetal with an alcohol and an acid catalyst yields an acetal.\n\n\nA hemiacetal\nAn acetal\n\nIn the same way, treatment of a monosaccharide hemiacetal with an alcohol and an acid catalyst yields an acetal called a glycoside, in which the anomeric -OH has been replaced by an -OR group. For example, reaction of $\\beta$-D-glucopyranose with methanol gives a mixture of $\\alpha$ and $\\beta$ methyl D-glucopyranosides. (Note that a glycoside is the functional group name for any sugar, whereas a glucoside is formed specifically from glucose.)\n\n\nGlycosides are named by first citing the alkyl group and then replacing the -ose ending of the sugar with -oside. Like all acetals, glycosides are stable in neutral water. They aren't in equilibrium with an open-chain form, and they don't show mutarotation. They can, however, be hydrolyzed to give back the free monosaccharide plus alcohol on treatment with aqueous acid (Section 19.10).\n\nGlycosides are abundant in nature, and many biologically important molecules contain glycosidic linkages. For example, digitoxin, the active component of the digitalis preparations used for treatment of heart disease, is a glycoside consisting of a steroid alcohol linked to a trisaccharide. Note also that the three sugars are linked to one another by glycoside bonds.\n\n\nDigitoxin, a glycoside\nThe laboratory synthesis of glycosides can be difficult because of the numerous -OH groups on the sugar molecule. One method that is particularly suitable for preparing glucose $\\beta$-glycosides involves treatment of glucose pentaacetate with HBr , followed by addition of the appropriate alcohol in the presence of silver oxide. Called the Koenigs-Knorr reaction, the sequence involves formation of a pyranosyl bromide, followed by nucleophilic substitution. For example, methylarbutin, a glycoside found in pears, has been prepared by reaction of tetraacetyl- $\\alpha$-D-glucopyranosyl bromide with $p$-methoxyphenol."}
{"id": 1659, "contents": "Glycoside Formation - \nAlthough the Koenigs-Knorr reaction appears to involve a simple backside $\\mathrm{S}_{\\mathrm{N}} 2$ displacement of bromide ion by alkoxide ion, the situation is actually more complex. Both $\\alpha$ and $\\beta$ anomers of tetraacetyl-D-glucopyranosyl bromide give the same $\\beta$-glycoside product, implying that they react by a common pathway.\n\nThis result can be understood by assuming that tetraacetyl-D-glucopyranosyl bromide (either $\\alpha$ or $\\beta$ anomer) undergoes a spontaneous $\\mathrm{S}_{\\mathrm{N}} 1$-like loss of $\\mathrm{Br}^{-}$, followed by internal reaction with the ester group at C 2 to form an oxonium ion. Since the acetate at C 2 is on the bottom of the glucose ring, the $\\mathrm{C}-\\mathrm{O}$ bond also forms from the bottom. Backside $\\mathrm{S}_{\\mathrm{N}} 2$ displacement of the oxonium ion then occurs with the usual inversion of configuration, yielding a $\\beta$-glycoside and regenerating the acetate at C2 (FIGURE 25.7).\n\n\nFIGURE 25.7 Mechanism of the Koenigs-Knorr reaction, showing the neighboring-group effect of a nearby acetate.\nThe participation shown by the nearby acetate group in the Koenigs-Knorr reaction is referred to as a neighboring-group effect and is a common occurrence in organic chemistry. Neighboring-group effects are usually noticeable only because they affect the rate or stereochemistry of a reaction; the nearby group itself does not undergo any evident change during the reaction."}
{"id": 1660, "contents": "Biological Ester Formation: Phosphorylation - \nIn living organisms, carbohydrates occur not only in the free form but also linked through their anomeric center to other molecules such as lipids (glycolipids) or proteins (glycoproteins). Collectively called glycoconjugates, these sugar-linked molecules are components of cell walls that are crucial to the mechanism by which different cell types recognize one another.\n\nGlycoconjugate formation occurs by reaction of the lipid or protein with a glycosyl nucleoside diphosphate. This diphosphate is itself formed by initial reaction of a monosaccharide with adenosine triphosphate (ATP) to give a glycosyl monophosphate, followed by reaction with uridine triphosphate (UTP), to give a glycosyl uridine diphosphate. (We'll see the structures of nucleoside phosphates in Section 28.1.) The purpose of the phosphorylation is to activate the anomeric - OH group of the sugar and make it a better leaving group in a nucleophilic substitution reaction by a protein or lipid (FIGURE 25.8).\n\n\nA glycoprotein\nFIGURE 25.8 Glycoprotein formation occurs by initial phosphorylation of the starting carbohydrate with ATP to a glycosyl monophosphate, followed by reaction with UTP to form a glycosyl uridine 5'-diphosphate. Nucleophilic substitution by an $-\\mathrm{OH}\\left(\\mathrm{or}-\\mathrm{NH}_{2}\\right)$ group on a protein then gives the glycoprotein."}
{"id": 1661, "contents": "Reduction of Monosaccharides - \nTreatment of an aldose or ketose with $\\mathrm{NaBH}_{4}$ reduces it to a polyalcohol called an alditol. The reduction occurs by reaction of the open-chain form present in the aldehyde/ketone $\\rightleftarrows$ hemiacetal equilibrium. Although only a small amount of the open-chain form is present at any given time, that small amount is reduced, more is produced by opening of the pyranose form, that additional amount is reduced, and so on, until the entire sample has undergone reaction.\n\n\nD-Glucitol, the alditol produced by reduction of D-glucose, is itself a naturally occurring substance found in many fruits and berries. It is used under the name D-sorbitol as a sweetener and sugar substitute in many foods.\n\nPROBLEM Reduction of D-glucose leads to an optically active alditol (D-glucitol), whereas reduction of 25-17 D-galactose leads to an optically inactive alditol. Explain.\n\nPROBLEM Reduction of L-gulose with $\\mathrm{NaBH}_{4}$ leads to the same alditol (D-glucitol) as reduction of D-glucose.\n25-18 Explain."}
{"id": 1662, "contents": "Oxidation of Monosaccharides - \nLike other aldehydes, aldoses are easily oxidized to yield the corresponding carboxylic acids, called aldonic acids. A buffered solution of aqueous $\\mathrm{Br}_{2}$ is often used for this purpose.\n\n\nHistorically, the oxidation of an aldose with either $\\mathrm{Ag}^{+}$in aqueous ammonia (called Tollens' reagent) or $\\mathrm{Cu}^{2+}$ with aqueous sodium citrate (Benedict's reagent) formed the basis of simple tests for what are called reducing sugars. (Reducing because the aldose reduces the metal oxidizing agent.) Some simple diabetes self-test kits sold in drugstores still use Benedict's reagent to detect glucose in urine, though more modern methods have largely replaced it.\n\nAll aldoses are reducing sugars because they contain an aldehyde group, but some ketoses are reducing sugars as well. Fructose reduces Tollens' reagent, for example, even though it contains no aldehyde group. Reduction occurs because fructose is readily isomerized to a mixture of aldoses (glucose and mannose) in basic solution by a series of keto-enol tautomeric shifts (FIGURE 25.9). Glycosides, however, are nonreducing because the acetal group is not hydrolyzed to an aldehyde under basic conditions.\n\n\nFIGURE 25.9 Fructose, a ketose, is a reducing sugar because it undergoes two base-catalyzed keto-enol tautomerizations that result in conversion to a mixture of aldoses.\n\nIf warm, dilute $\\mathrm{HNO}_{3}$ (nitric acid) is used as the oxidizing agent, an aldose is oxidized to a dicarboxylic acid called an aldaric acid. Both the aldehyde carbonyl and the terminal $-\\mathrm{CH}_{2} \\mathrm{OH}$ group are oxidized in this reaction.\n\n\nFinally, if only the $-\\mathrm{CH}_{2} \\mathrm{OH}$ end of the aldose is oxidized without affecting the -CHO group, the product is a monocarboxylic acid called a uronic acid. The reaction can only be done enzymatically; no chemical reagent is known that can accomplish this selective oxidation in the laboratory."}
{"id": 1663, "contents": "Oxidation of Monosaccharides - \nPROBLEM D-Glucose yields an optically active aldaric acid on treatment with $\\mathrm{HNO}_{3}$, but D -allose yields an\n25-19 optically inactive aldaric acid. Explain.\nPROBLEM Which of the other six D aldohexoses yield optically active aldaric acids on oxidation, and which\n$\\mathbf{2 5 - 2 0}$ yield optically inactive (meso) aldaric acids? (See Problem 25-19.)"}
{"id": 1664, "contents": "Chain Lengthening: The Kiliani-Fischer Synthesis - \nMuch early activity in carbohydrate chemistry was devoted to unraveling the stereochemical relationships among monosaccharides. One of the most important methods used was the Kiliani-Fischer synthesis, which results in the lengthening of an aldose chain by one carbon atom. The C1 aldehyde group of the starting sugar becomes C2 of the chain-lengthened sugar, and a new C1 carbon is added. For example, an aldopentose is converted by Kiliani-Fischer synthesis into two aldohexoses.\n\nDiscovery of the chain-lengthening sequence was initiated by the observation of Heinrich Kiliani in 1886 that aldoses react with HCN to form cyanohydrins (Section 19.6). Emil Fischer immediately realized the importance of Kiliani's discovery and devised a method for converting the cyanohydrin nitrile group into an aldehyde.\n\nFischer's original method for conversion of the nitrile into an aldehyde involved hydrolysis to a carboxylic acid, ring closure to a cyclic ester (lactone), and subsequent reduction. A modern improvement involves reducing the nitrile over a palladium catalyst, yielding an imine intermediate that is hydrolyzed to an aldehyde. Note that the cyanohydrin is formed as a mixture of stereoisomers at the new chirality center, so two new aldoses, differing only in their stereochemistry at C2, result from Kiliani-Fischer synthesis. Chain extension of D-arabinose, for example, yields a mixture of D -glucose and D -mannose.\n\n\nPROBLEM What product(s) would you expect from Kiliani-Fischer reaction of D-ribose?\n25-21\nPROBLEM What aldopentose would give a mixture of L-gulose and L-idose on Kiliani-Fischer chain extension? 25-22"}
{"id": 1665, "contents": "Chain Shortening: The Wohl Degradation - \nJust as the Kiliani-Fischer synthesis lengthens an aldose chain by one carbon, the Wohl degradation shortens an aldose chain by one carbon. Wohl degradation is almost the exact opposite of the Kiliani-Fischer sequence. That is, the aldose aldehyde carbonyl group is first converted into a nitrile, and the resulting cyanohydrin loses HCN under basic conditions-the reverse of a nucleophilic addition reaction.\n\nConversion of the aldehyde into a nitrile is accomplished by treatment of an aldose with hydroxylamine to give an imine called an oxime (Section 19.8), followed by dehydration of the oxime with acetic anhydride. The Wohl degradation does not give particularly high yields of chain-shortened aldoses, but the reaction is general for all aldopentoses and aldohexoses. For example, D-galactose is converted by Wohl degradation into D-lyxose.\n\n\nPROBLEM Two of the four D aldopentoses yield D-threose on Wohl degradation. What are their structures? 25-23"}
{"id": 1666, "contents": "Chain Shortening: The Wohl Degradation - 25.7 The Eight Essential Monosaccharides\nHumans need to obtain eight monosaccharides for proper functioning. Although all eight can be biosynthesized from simpler precursors if necessary, it's more energetically efficient to obtain them from the diet. The eight are L-fucose (6-deoxy-L-galactose), D-galactose, D-glucose, D-mannose, $N$-acetyl-D-glucosamine, N -acetyl-D-galactosamine, D-xylose, and N -acetyl-D-neuraminic acid (FIGURE 25.10). All are used for the synthesis of the glycoconjugate components of cell membranes, and glucose is also the body's primary source of energy.\n\n\nHO\n\n\nL-Fucose\n(6-deoxy-L-galactose)\n\n\n\nD-Galactose\n\n\n\nN-Acetyl-Dglucosamine (2-acetamido-\n2-deoxy-D-glucose)\n\n\n\n\n$N$-Acetyl-D-\ngalactosamine\n(2-acetamido-\n2-deoxy-D-galactose)\n\n\n\nD-Glucose\n\nD-Mannose"}
{"id": 1667, "contents": "Chain Shortening: The Wohl Degradation - 25.7 The Eight Essential Monosaccharides\n$N$-Acetyl-D-\ngalactosamine\n(2-acetamido-\n2-deoxy-D-galactose)\n\n\n\nD-Glucose\n\nD-Mannose\n\n\n\nD-Xylose\n$N$-Acetyl-D-neuraminic acid\nFIGURE 25.10 Structures of the eight monosaccharides essential to humans.\nOf the eight essential monosaccharides, galactose, glucose, and mannose are simple aldohexoses, while xylose is an aldopentose. Fucose is a deoxy sugar, meaning that it has an oxygen atom \"missing.\" That is, an -OH group (the one at C6) is replaced by an $-\\mathrm{H} . N$-Acetylglucosamine and $N$-acetylgalactosamine are amide derivatives of amino sugars in which an -OH (the one at C 2 ) is replaced by an $-\\mathrm{NH}_{2}$ group. $N$-Acetylneuraminic acid is the parent compound of the sialic acids, a group of more than 30 compounds with different modifications, including various oxidations, acetylations, sulfations, and methylations. Note that neuraminic acid has nine carbons and is an aldol reaction product of N -acetylmannosamine with pyruvate $\\left(\\mathrm{CH}_{3} \\mathrm{COCO}_{2}{ }^{-}\\right)$. Neuraminic acid is crucial to the mechanism by which an influenza virus spreads.\n\nAll the essential monosaccharides arise from glucose, by the conversions summarized in FIGURE 25.11. We'll not look specifically at these conversions, but might note that Problems 25-30 through 25-32 and 25-35 at the end of the chapter lead you through several of the biosynthetic pathways.\n\n\nFIGURE 25.11 An overview of biosynthetic pathways for the eight essential monosaccharides.\nPROBLEM Show how neuraminic acid can arise by an aldol reaction of $N$-acetylmannosamine with pyruvate\n25-24 $\\left(\\mathrm{CH}_{3} \\mathrm{COCO}_{2}^{-}\\right)$.\n\n$N$-Acetylmannosamine"}
{"id": 1668, "contents": "Chain Shortening: The Wohl Degradation - 25.8 Disaccharides\nWe saw in Section 25.6 that reaction of a monosaccharide with an alcohol yields a glycoside in which the anomeric - OH group is replaced by an -OR substituent. If the alcohol is itself a sugar, the glycosidic product is a disaccharide."}
{"id": 1669, "contents": "Maltose and Cellobiose - \nDisaccharides contain a glycosidic acetal bond between the anomeric carbon of one sugar and an - OH group at any position on the other sugar. A glycosidic bond between C 1 of the first sugar and the -OH at C 4 of the second sugar is particularly common. Such a bond is called a $1 \\rightarrow 4$ link.\n\nThe glycosidic bond to an anomeric carbon can be either $\\alpha$ or $\\beta$. Maltose, the disaccharide obtained by enzymecatalyzed hydrolysis of starch, consists of two $\\alpha$-D-glucopyranose units joined by a $1 \\rightarrow 4-\\alpha$-glycoside bond. Cellobiose, the disaccharide obtained by partial hydrolysis of cellulose, consists of two $\\beta$-D-glucopyranose units joined by a $1 \\rightarrow 4-\\beta$-glycoside bond.\n\n\nMaltose, a 1 $\\rightarrow 4-\\alpha$-glycoside\n[4-O-( $\\alpha$-D-glucopyranosyl)- $\\alpha$-D-glucopyranose]\n\n\nCellobiose, a $1 \\rightarrow 4-\\beta$-glycoside [4-O-( $\\beta$-D-glucopyranosyl)- $\\beta$-D-glucopyranose]\n\n\nMaltose and cellobiose are both reducing sugars because the anomeric carbons on their right-hand glucopyranose units have hemiacetal groups and are in equilibrium with aldehyde forms. For a similar reason, both maltose and cellobiose exhibit mutarotation of $\\alpha$ and $\\beta$ anomers of the glucopyranose unit on the right.\n\n\nDespite the similarities of their structures, cellobiose and maltose have dramatically different biological properties. Cellobiose can't be digested by humans and can't be fermented by yeast. Maltose, however, is digested without difficulty and is fermented readily."}
{"id": 1670, "contents": "Maltose and Cellobiose - \nPROBLEM Show the product you would obtain from the reaction of cellobiose with the following reagents:\n25-25\n(a) $\\mathrm{NaBH}_{4}$\n(b) $\\mathrm{Br}_{2}, \\mathrm{H}_{2} \\mathrm{O}$\n(c) $\\mathrm{CH}_{3} \\mathrm{COCl}$, pyridine"}
{"id": 1671, "contents": "Lactose - \nLactose is a disaccharide that occurs naturally in both human and cow's milk. It is widely used in baking and in commercial milk formulas for infants. Like maltose and cellobiose, lactose is a reducing sugar. It exhibits mutarotation and is a $1 \\rightarrow 4$ - $\\beta$-linked glycoside. Unlike maltose and cellobiose, however, lactose contains two different monosaccharides-D-glucose and D-galactose-joined by a $\\beta$-glycosidic bond between C 1 of galactose and C4 of glucose."}
{"id": 1672, "contents": "Sucrose - \nSucrose, or ordinary table sugar, is probably the most abundant pure organic chemical in the world. Whether from sugar cane ( $20 \\%$ sucrose by weight) or sugar beets ( $15 \\%$ by weight), and whether raw or refined, all table sugar is sucrose.\n\nSucrose is a disaccharide that yields 1 equivalent of glucose and 1 equivalent of fructose on hydrolysis. This 1:1 mixture of glucose and fructose is often referred to as invert sugar because the sign of optical rotation changes, or inverts, during the hydrolysis of sucrose $\\left([\\alpha]_{\\mathrm{D}}=+66.5\\right.$ ) to a glucose/ fructose mixture ( $[\\alpha]_{\\mathrm{D}}=-22.0$ ). Some insects, such as honeybees, have enzymes called invertases that catalyze the sucrose hydrolysis. Honey, in fact, is primarily a mixture of glucose, fructose, and sucrose.\n\nUnlike most other disaccharides, sucrose is not a reducing sugar and does not undergo mutarotation. These observations imply that sucrose is not a hemiacetal and that glucose and fructose must both be glycosides. This can happen only if the two sugars are joined by a glycoside link between the anomeric carbons of both sugars-C1 of glucose and C2 of fructose.\n\n\n\nSucrose, a ( $1 \\rightarrow 2$ ) glycoside\n[ $\\alpha$-D-Glucopyranosyl-( $1 \\rightarrow 2$ ) $\\boldsymbol{\\beta}$ - D -fructofuranoside]"}
{"id": 1673, "contents": "Sucrose - 25.9 Polysaccharides and Their Synthesis\nPolysaccharides are complex carbohydrates in which tens, hundreds, or even thousands of simple sugars are linked together through glycoside bonds. Because they have only the one free anomeric -OH group at the end of a very long chain, polysaccharides aren't reducing sugars and don't show noticeable mutarotation. Cellulose and starch are the two most widely occurring polysaccharides."}
{"id": 1674, "contents": "Cellulose - \nCellulose consists of several thousand D-glucose units linked by $1 \\rightarrow 4-\\beta$-glycoside bonds like those in cellobiose. Different cellulose molecules then interact to form a large aggregate structure held together by hydrogen bonds.\n\n\nCellulose, a $\\boldsymbol{\\beta}$-(1 $\\rightarrow 4$ )-D-Glucopyranoside polymer\nNature uses cellulose primarily as a structural material to impart strength and rigidity to plants. Leaves, grasses, and cotton, for instance, are primarily cellulose. Cellulose also serves as raw material for the manufacture of cellulose acetate, known commercially as acetate rayon, and cellulose nitrate, known as guncotton, which is the major ingredient in smokeless powder, the explosive propellant used in artillery shells and in ammunition for firearms."}
{"id": 1675, "contents": "Starch and Glycogen - \nPotatoes, corn, and cereal grains contain large amounts of starch, a polymer of glucose in which the monosaccharide units are linked by $1 \\rightarrow 4-\\alpha$-glycoside bonds like those in maltose. Starch can be separated into two fractions: amylose and amylopectin. Amylose accounts for about $20 \\%$ by weight of starch and consists of several hundred glucose molecules linked together by $1 \\rightarrow 4-\\alpha$-glycoside bonds."}
{"id": 1676, "contents": "Amylose, an $\\alpha$-(1 $\\rightarrow 4)$-D-Glucopyranoside polymer - \nAmylopectin accounts for the remaining $80 \\%$ of starch and is more complex in structure than amylose. Unlike cellulose and amylose, which are linear polymers, amylopectin contains $1 \\rightarrow 6-\\alpha$-glycoside branches approximately every 25 glucose units.\n\n\nAmylopectin: $a-(1 \\rightarrow 4)$ links with $a$-( $1 \\rightarrow 6$ ) branches\nStarch is digested in the mouth and stomach by $\\alpha$-glycosidases, which catalyze the hydrolysis of glycoside bonds and release individual molecules of glucose. Like most enzymes, $\\alpha$-glycosidases are highly selective in their action. They hydrolyze only the $\\alpha$-glycoside links in starch and leave the $\\beta$-glycoside links in cellulose\nuntouched. Thus, humans can digest potatoes and grains but not grass and leaves.\nGlycogen is a polysaccharide that serves the same energy storage function in animals that starch serves in plants. Dietary carbohydrates not needed for immediate energy are converted by the body into glycogen for long-term storage. Like the amylopectin found in starch, glycogen contains a complex branching structure with both $1 \\rightarrow 4$ and $1 \\rightarrow 6$ links (FIGURE 25.12). Glycogen molecules are larger than those of amylopectin-up to 100,000 glucose units-and contain even more branches.\n\n\nFIGURE 25.12 A representation of the structure of glycogen. The hexagons represent glucose units linked by $1 \\rightarrow 4$ and $1 \\rightarrow 6$ glycoside bonds."}
{"id": 1677, "contents": "Polysaccharide Synthesis - \nWith numerous - OH groups of similar reactivity, polysaccharides are so structurally complex that their laboratory synthesis has long been a particularly difficult problem. Several methods have recently been devised, however, that have greatly simplified the problem. Among these approaches is the glycal assembly method.\n\nEasily prepared from the appropriate monosaccharide, a glycal is an unsaturated sugar with a $\\mathrm{C} 1-\\mathrm{C} 2$ double bond. To ready it for use in polysaccharide synthesis, the glycal is first protected at its primary - OH group by formation of a silyl ether (Section 17.8) and at its two adjacent secondary -OH groups by formation of a cyclic carbonate ester. Then, the protected glycal is epoxidized.\n\n\nTreatment of the protected glycal epoxide in the presence of $\\mathrm{ZnCl}_{2}$ as a Lewis acid with a second glycal having a free -OH group causes acid-catalyzed opening of the epoxide ring by $\\mathrm{S}_{\\mathrm{N}} 2$ backside attack (Section 18.6) and yields a disaccharide. The disaccharide is itself a glycal, so it can be epoxidized and coupled again to yield a trisaccharide, and so on. Using the appropriate sugars at each step, a great variety of polysaccharides can be prepared. After these sugars are linked, the silyl ethers and cyclic carbonate protecting groups are removed by hydrolysis.\n\n\n\nA disaccharide glycal\nAmong the numerous complex polysaccharides that have been synthesized in the laboratory are several tumorassociated carbohydrate antigens (TACAs) that are currently being tested as potential vaccines for breast,\nprostate, lung, colon, and ovarian cancers."}
{"id": 1678, "contents": "Polysaccharide Synthesis - 25.10 Some Other Important Carbohydrates\nIn addition to the common carbohydrates mentioned in previous sections, there are a variety of important carbohydrate-derived materials. Their structural resemblance to sugars is clear, but they aren't simple aldoses or ketoses.\n\nDeoxy sugars, as we saw in Section 25.7, have an oxygen atom \"missing.\" That is, an -OH group is replaced by an $-H$. The most common deoxy sugar is 2-deoxyribose, a monosaccharide found in DNA (deoxyribonucleic acid). Note that 2-deoxyribose exists in water solution as a complex equilibrium mixture of both furanose and pyranose forms.\n\n\nAmino sugars, such as D-glucosamine, have an -OH group replaced by an $-\\mathrm{NH}_{2}$. The $N$-acetyl amide derived from D-glucosamine is the monosaccharide unit from which chitin, the hard crust that protects insects and shellfish, is made. Still other amino sugars are found in antibiotics such as streptomycin and gentamicin."}
{"id": 1679, "contents": "Sweetness - \nSay the word sugar and most people immediately think of sweet-tasting candies, desserts, and such. In fact, most simple carbohydrates do taste sweet but the degree of sweetness varies greatly from one sugar to another. With sucrose (table sugar) as a reference point, fructose is nearly twice as sweet, but lactose is only about one-sixth as sweet. Comparisons are difficult, though, because perceived sweetness varies depending on the concentration of the solution being tasted and on personal opinion. Nevertheless, the ordering in TABLE 25.1 is generally accepted.\n\nThe desire of many people to cut their caloric intake has led to the development of synthetic sweeteners such as saccharin, aspartame, acesulfame, and sucralose. All are far sweeter than natural sugars, so the choice of one or another depends on personal taste, government regulations, and (for baked goods) heat stability. Saccharin, the oldest synthetic sweetener, has been used for more than a century, although it has a somewhat metallic aftertaste. Doubts about its safety and potential carcinogenicity were raised in the early 1970s, but it has now been cleared of suspicion for human consumption.\n\n\nFIGURE 25.13 The real thing comes from sugarcane fields like this one. (credit: modification of work \"Flowering sugarcane field\" by \"carrotmadman6\"/Flickr, CC BY 2.0)\n\nTABLE 25.1 Sweetness of Some Sugars and Sugar Substitutes\n\n| Name | Type | Sweetness |\n| :--- | :--- | :---: |\n| Lactose | Disaccharide | 0.16 |\n| Glucose | Monosaccharide | 0.75 |\n| Sucrose | Disaccharide | 1.00 |\n| Fructose | Monosaccharide | 1.75 |\n| Cyclamate | Synthetic | 40 |\n| Aspartame | Semisynthetic | 180 |\n| Acesulfame-K | Synthetic | 200 |\n| Saccharin | Synthetic | 350 |\n\nTABLE 25.1 Sweetness of Some Sugars and Sugar Substitutes"}
{"id": 1680, "contents": "Sweetness - \nTABLE 25.1 Sweetness of Some Sugars and Sugar Substitutes\n\n| Name | Type | Sweetness |\n| :--- | :--- | :---: |\n| Sucralose | Semisynthetic | 600 |\n| Alitame | Semisynthetic | 2,000 |\n| Neotame | Semisynthetic | 6,000 |\n| Advantame | Semisynthetic | 20,000 |\n\nAcesulfame potassium, one of the more recently approved sweeteners, is proving to be extremely popular in soft drinks because it has little aftertaste. Sucralose, another recent sweetener, is particularly useful in baked goods because of its stability at high temperatures. Alitame, marketed in some countries under the name Aclame, is not approved for sale in the United States. It is some 2000 times as sweet as sucrose and, like acesulfame-K, has no aftertaste. Of the eight synthetic sweeteners listed in TABLE 25.1, only sucralose has clear structural resemblance to a carbohydrate, although it differs dramatically in containing three chlorine atoms. Aspartame, Alitame, and Advantame are dipeptides.\n\n\nSaccharin\n\n\nAspartame\n\n\nAcesulfame potassium\n\n\nSucralose\n\n\nAlitame"}
{"id": 1681, "contents": "Key Terms - \n- aldaric acid\nfuranose\n- alditol\n- aldonic acid\n- aldose\n- amino sugar\n- $\\alpha$ anomer, $\\beta$ anomer\n- anomeric center\n- anomer\n- carbohydrate\n- complex carbohydrate\n- D sugar\n- deoxy sugar\n- disaccharide\n- Fischer projection\n- glycoside\n- ketose\n- Kiliani-Fischer synthesis\n- Koenigs-Knorr reaction\n- L sugar\n- monosaccharide\n- mutarotation\n- polysaccharide\n- pyranose\n- reducing sugar\n- simple sugar\n- uronic acid\n- Wohl degradation"}
{"id": 1682, "contents": "Summary - \nNow that we've now seen all the common functional groups and reaction types, our focus has changed to looking at the major classes of biological molecules. Carbohydrates are polyhydroxy aldehydes and ketones. They are classified according to the number of carbon atoms and the kind of carbonyl group they contain. Glucose, for example, is an aldohexose, a six-carbon aldehydo sugar. Monosaccharides are further classified as either d sugars or l sugars, depending on the stereochemistry of the chirality center farthest from the carbonyl group. Carbohydrate stereochemistry is frequently depicted using Fischer projections, which represent a chirality center as the intersection of two crossed lines.\n\nMonosaccharides normally exist as cyclic hemiacetals rather than as open-chain aldehydes or ketones. The hemiacetal linkage results from reaction of the carbonyl group with an -OH group three or four carbon atoms away. A five-membered cyclic hemiacetal is called a furanose, and a six-membered cyclic hemiacetal is called a pyranose. Cyclization leads to the formation of a new chirality center called the anomeric center and the production of two diastereomeric hemiacetals called alpha ( $\\boldsymbol{\\alpha}$ ) and beta ( $\\boldsymbol{\\beta}$ ) anomers.\n\nMuch of the chemistry of monosaccharides is the familiar chemistry of alcohols and aldehydes/ketones. Thus, the hydroxyl groups of carbohydrates form esters and ethers. The carbonyl group of a monosaccharide can be reduced with $\\mathrm{NaBH}_{4}$ to form an alditol, oxidized with aqueous $\\mathrm{Br}_{2}$ to form an aldonic acid, oxidized with $\\mathrm{HNO}_{3}$ to form an aldaric acid, oxidized enzymatically to form a uronic acid, or treated with an alcohol in the presence of acid to form a glycoside. Monosaccharides can also be chain-lengthened by the multistep Kiliani-Fischer synthesis and can be chain-shortened by Wohl degradation."}
{"id": 1683, "contents": "Summary - \nDisaccharides are complex carbohydrates in which simple sugars are linked by a glycoside bond between the anomeric center of one unit and a hydroxyl of the second unit. The sugars can be the same, as in maltose and cellobiose, or different, as in lactose and sucrose. The glycosidic bond can be either $\\alpha$ (maltose) or $\\beta$ (cellobiose, lactose) and can involve any hydroxyl of the second sugar. A $1 \\rightarrow 4$ link is most common (cellobiose, maltose), but others such as $1 \\rightarrow 2$ (sucrose) are also known. Polysaccharides, such as cellulose, starch, and glycogen, are used in nature as structural materials, as a means of long-term energy storage, and as cell-surface markers."}
{"id": 1684, "contents": "Visualizing Chemistry - \nPROBLEM Identify the following aldoses, and tell whether each is a D or L sugar:\n\n25-26 (a)\n\n(b)\n\n\nPROBLEM Draw Fischer projections of the following molecules, placing the carbonyl group at the top in the 25-27 usual way. Identify each as a D or L sugar.\n(a)\n\n(b)\n\n\nPROBLEM The following structure is that of an laldohexose in its pyranose form. Identify it, and tell whether\n25-28 it is an $\\alpha$ or $\\beta$ anomer.\n\n\nPROBLEM The following model is that of an aldohexose:\n25-29\n\n(a) Draw Fischer projections of the sugar, its enantiomer, and a diastereomer.\n(b) Is this a D sugar or an L sugar? Explain.\n(c) Draw the $\\beta$ anomer of the sugar in its furanose form."}
{"id": 1685, "contents": "Mechanism Problems - \nPROBLEM Galactose, one of the eight essential monosaccharides (Section 25.7), is biosynthesized from UDP-\n25-30 glucose by galactose 4-epimerase, where UDP $=$ uridylyl diphosphate (a ribonucleotide diphosphate; Section 28.1). The enzyme requires NAD ${ }^{+}$for activity (Section 17.7), but it is not a stoichiometric reactant, and NADH is not a final reaction product. Propose a mechanism.\n\n\nPROBLEM Mannose, one of the eight essential monosaccharides (Section 25.7), is biosynthesized as its\n25-31 6-phosphate derivative from fructose 6-phosphate. No enzyme cofactor is required. Propose a mechanism.\n\n\nPROBLEM Glucosamine, one of the eight essential monosaccharides (Section 25.7), is biosynthesized as its 25-32 6-phosphate derivative from fructose 6-phosphate by reaction with ammonia. Propose a mechanism.\n\n\nPROBLEM D-Glucose reacts with acetone in the presence of acid to yield the nonreducing 1,2:\n25-33 5,6-diisopropylidene-D-glucofuranose. Propose a mechanism.\n\n\n1,2:5,6-Diisopropylidene-\nD-glucofuranose\nPROBLEM One of the steps in the biological pathway for carbohydrate metabolism is the conversion of fructose\n25-34 1,6-bisphosphate into dihydroxyacetone phosphate and glyceraldehyde 3-phosphate. Propose a mechanism for the transformation.\n\n\nPROBLEM L-Fucose, one of the eight essential monosaccharides (Section 25.7), is biosynthesized from 25-35 GDP-D-mannose by the following three-step reaction sequence, where GDP = guanosine diphosphate (a ribonucleoside diphosphate; Section 28.1):"}
{"id": 1686, "contents": "Mechanism Problems - \n(a) Step 1 involves an oxidation to a ketone, a dehydration to an enone, and a conjugate reduction. The step requires $\\mathrm{NADP}^{+}$, but no NADPH is formed as a final reaction product. Propose a mechanism.\n(b) Step 2 accomplishes two epimerizations and utilizes acidic and basic sites in the enzyme but does not require a coenzyme. Propose a mechanism.\n(c) Step 3 requires NADPH as coenzyme. Show the mechanism."}
{"id": 1687, "contents": "Carbohydrate Structures - \nPROBLEM Classify each of the following sugars. (For example, glucose is an aldohexose.)\n25-36 (a)\n\n(b)\n\n(c)\n\n\nPROBLEM Write open-chain structures for the following:\n25-37 (a) A ketotetrose\n(b) A ketopentose\n(c) A deoxyaldohexose\n(d) A five-carbon amino sugar\n\nPROBLEM What is the stereochemical relationship of D-ribose to L-xylose? What generalizations can you make 25-38 about the following properties of the two sugars?\n(a) Melting point\n(b) Solubility in water\n(c) Specific rotation\n(d) Density\n\nPROBLEM Does ascorbic acid (vitamin C) have a D or L configuration?\n25-39\n\n\nAscorbic acid\n\nPROBLEM Draw the three-dimensional furanose form of ascorbic acid (Problem 25-39), and assign $R$ or $S$ 25-40 stereochemistry to each chirality center.\n\nPROBLEM Assign $R$ or $S$ configuration to each chirality center in the following molecules:\n\n25-41 (a)\n\n(b)\n\n(c)\n\n\nPROBLEM Draw Fischer projections of the following molecules:\n25-42 (a) The $S$ enantiomer of 2-bromobutane (b) The $R$ enantiomer of alanine, $\\mathrm{CH}_{3} \\mathrm{CH}\\left(\\mathrm{NH}_{2}\\right) \\mathrm{CO}_{2} \\mathrm{H}$\n(c) The $R$ enantiomer of 2-hydroxypropanoic acid\n(d) The $S$ enantiomer of 3-methylhexane\n\nPROBLEM Draw Fischer projections for the two D aldoheptoses whose stereochemistry at C3, C4, C5, and C6 is\n25-43 the same as that of D-glucose at C2, C3, C4, and C5.\nPROBLEM The following cyclic structure is that of allose. Is this a furanose or pyranose form? Is it an $\\alpha$ or a $\\beta$\n25-44 anomer? Is it a D or an L sugar?\n\n\nPROBLEM What is the complete name of the following sugar?\n25-45"}
{"id": 1688, "contents": "Carbohydrate Structures - \nPROBLEM What is the complete name of the following sugar?\n25-45\n\n\nPROBLEM Write the following sugars in their open-chain forms:\n25-46 (a)\n\n(b)\n\n(c)\n\n\nPROBLEM Draw D-ribulose in its five-membered cyclic $\\beta$-hemiacetal form.\n25-47\n\n\nRibulose\n\nPROBLEM Look up the structure of D-talose in Figure 25.4, and draw the $\\beta$ anomer in its pyranose form. 25-48 Identify the ring substituents as axial or equatorial."}
{"id": 1689, "contents": "Carbohydrate Reactions - \nPROBLEM Draw structures for the products you would expect to obtain from reaction of $\\beta$-D-talopyranose with 25-49 each of the following reagents:\n(a) $\\mathrm{NaBH}_{4}$ in $\\mathrm{H}_{2} \\mathrm{O}$\n(b) Warm dilute $\\mathrm{HNO}_{3}$\n(c) $\\mathrm{Br}_{2}, \\mathrm{H}_{2} \\mathrm{O}$\n(d) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}, \\mathrm{HCl}$\n(e) $\\mathrm{CH}_{3} \\mathrm{I}, \\mathrm{Ag}_{2} \\mathrm{O}$\n(f) $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O}$, pyridine\n\nPROBLEM How many D-2-ketohexoses are possible? Draw them. 25-50\n\nPROBLEM One of the D-2-ketohexoses is called sorbose. On treatment with $\\mathrm{NaBH}_{4}$, sorbose yields a mixture of 25-51 gulitol and iditol. What is the structure of sorbose?\n\nPROBLEM Another D-2-ketohexose, psicose, yields a mixture of allitol and altritol when reduced with $\\mathrm{NaBH}_{4}$. 25-52 What is the structure of psicose?\n\nPROBLEM L-Gulose can be prepared from D-glucose by a route that begins with oxidation to D-glucaric acid, 25-53 which cyclizes to form two six-membered-ring lactones. Separating the lactones and reducing them with sodium amalgam gives D -glucose and L-gulose. What are the structures of the two lactones, and which one is reduced to L -gulose?"}
{"id": 1690, "contents": "Carbohydrate Reactions - \nPROBLEM Gentiobiose, a rare disaccharide found in saffron and gentian, is a reducing sugar and forms only\n25-54 D-glucose on hydrolysis with aqueous acid. Reaction of gentiobiose with iodomethane and $\\mathrm{Ag}_{2} \\mathrm{O}$ yields an octamethyl derivative, which can be hydrolyzed with aqueous acid to give 1 equivalent of 2,3,4,6-tetra-O-methyl-D-glucopyranose and 1 equivalent of 2,3,4-tri-O-methyl-D-glucopyranose. If gentiobiose contains a $\\beta$-glycoside link, what is its structure?"}
{"id": 1691, "contents": "General Problems - \nPROBLEM All aldoses exhibit mutarotation. For example, $\\alpha$-D-galactopyranose has $[\\alpha]_{D}=+150.7$, and\n25-55 $\\beta$-D-galactopyranose has $[\\alpha]_{D}=+52.8$. If either anomer is dissolved in water and allowed to reach equilibrium, the specific rotation of the solution is +80.2 . What are the percentages of each anomer at equilibrium? Draw the pyranose forms of both anomers.\n\nPROBLEM What other D aldohexose gives the same alditol as D-talose? 25-56\n\nPROBLEM Which of the eight D aldohexoses give the same aldaric acids as their L enantiomers? 25-57\n\nPROBLEM Which of the other three D aldopentoses gives the same aldaric acid as D-lyxose? 25-58\n\nPROBLEM Draw the structure of L-galactose, and then answer the following questions:\n25-59 (a) Which other aldohexose gives the same aldaric acid as L-galactose on oxidation with warm $\\mathrm{HNO}_{3}$ ?\n(b) Is this other aldohexose a D sugar or an L sugar?\n(c) Draw this other aldohexose in its most stable pyranose conformation.\n\nPROBLEM Amygdalin, or laetrile, is a cyanogenic glycoside first isolated in 1830 from almond and apricot 25-60 seeds. If acidic hydrolysis of amygdalin liberates HCN, along with benzaldehyde cyanohydrin with gentiobiose (Problem 25-54), what is its structure?\n\nPROBLEM Trehalose is a nonreducing disaccharide that is hydrolyzed by aqueous acid to yield 2 equivalents of\n25-61 D-glucose. Methylation followed by hydrolysis yields 2 equivalents of 2,3,4,6-tetra- $O$-methylglucose. How many structures are possible for trehalose?"}
{"id": 1692, "contents": "General Problems - \nPROBLEM Trehalose (Problem 25-61) is cleaved by enzymes that hydrolyze $\\alpha$-glycosides but not by enzymes\n25-62 that hydrolyze $\\beta$-glycosides. What is the structure and systematic name of trehalose?\nPROBLEM Isotrehalose and neotrehalose are chemically similar to trehalose (Problems 25-61 and 25-62)\n25-63 except that neotrehalose is hydrolyzed only by $\\beta$-glycosidase enzymes, whereas isotrehalose is hydrolyzed by both $\\alpha$ - and $\\beta$-glycosidase enzymes. What are the structures of isotrehalose and neotrehalose?\n\nPROBLEM D-Mannose reacts with acetone to give a diisopropylidene derivative (Problem 25-33) that is still 25-64 reducing toward Tollens' reagent. Propose a likely structure for this derivative.\n\nPROBLEM Glucose and mannose can be interconverted (in low yield) by treatment with dilute aqueous NaOH .\n25-65 Propose a mechanism.\nPROBLEM Propose a mechanism to account for the fact that D-gluconic acid and D-mannonic acid are\n25-66 interconverted when either is heated in pyridine solvent.\nPROBLEM The cyclitols are a group of carbocyclic sugar derivatives having the general formulation 25-67 1,2,3,4,5,6-cyclohexanehexol. How many stereoisomeric cyclitols are possible? Draw them in their chair forms.\n\nPROBLEM Compound $\\mathbf{A}$ is a D aldopentose that can be oxidized to an optically inactive aldaric acid $\\mathbf{B}$. On\n25-68 Kiliani-Fischer chain extension, $\\mathbf{A}$ is converted into $\\mathbf{C}$ and $\\mathbf{D} ; \\mathbf{C}$ can be oxidized to an optically active aldaric acid $\\mathbf{E}$, but $\\mathbf{D}$ is oxidized to an optically inactive aldaric acid $\\mathbf{F}$. What are the structures of A-F?"}
{"id": 1693, "contents": "General Problems - \nPROBLEM Simple sugars undergo reaction with phenylhydrazine, $\\mathrm{PhNH}-\\mathrm{NH}_{2}$, to yield crystalline derivatives 25-69 called osazones. The reaction is a bit complex, however, as shown by the fact that glucose and fructose yield the same osazone.\n\n(a) Draw the structure of a third sugar that yields the same osazone as glucose and fructose.\n(b) Using glucose as the example, the first step in osazone formation is reaction of the sugar with phenylhydrazine to yield an imine called a phenylhydrazone. Draw the structure of the product.\n(c) The second and third steps in osazone formation are tautomerization of the phenylhydrazone to give an enol, followed by elimination of aniline to give a keto imine. Draw the structures of both the enol tautomer and the keto imine.\n(d) The final step is reaction of the keto imine with 2 equivalents of phenylhydrazine to yield the osazone plus ammonia. Propose a mechanism for this step.\n\nPROBLEM When heated to $100{ }^{\\circ} \\mathrm{C}$, D-idose undergoes a reversible loss of water and exists primarily as 25-70 1,6-anhydro-D-idopyranose.\n\n(a) Draw D-idose in its pyranose form, showing the more stable chair conformation of the ring.\n(b) Which is more stable, $\\alpha$-D-idopyranose or $\\beta$-D-idopyranose? Explain.\n(c) Draw 1,6-anhydro-D-idopyranose in its most stable conformation.\n(d) When heated to $100^{\\circ} \\mathrm{C}$ under the same conditions as those used for D-idose, D-glucose does not lose water and does not exist in a 1,6-anhydro form. Explain.\n\nPROBLEM Acetyl coenzyme A (acetyl CoA) is the key intermediate in food metabolism. What sugar is present 25-71 in acetyl CoA?"}
{"id": 1694, "contents": "CHAPTER 26 - \nBiomolecules: Amino Acids, Peptides, and Proteins\n\n\nFIGURE 26.1 The building blocks of life that we call proteins are aptly named after Proteus, the early Greek sea-god whose name means \"first\" or \"primordial.\" (credit: modification of work by Coyau/Wikimedia Commons, CC BY-SA 3.0)"}
{"id": 1695, "contents": "CHAPTER CONTENTS - \n26.1 Structures of Amino Acids\n26.2 Amino Acids and the Henderson-Hasselbalch Equation: Isoelectric Points\n26.3 Synthesis of Amino Acids\n26.4 Peptides and Proteins\n26.5 Amino Acid Analysis of Peptides\n26.6 Peptide Sequencing: The Edman Degradation\n26.7 Peptide Synthesis\n26.8 Automated Peptide Synthesis: The Merrifield Solid-Phase Method\n26.9 Protein Structure\n26.10 Enzymes and Coenzymes\n26.11 How Do Enzymes Work? Citrate Synthase\n\nWHY THIS CHAPTER? Continuing our look at the main classes of biomolecules, we'll focus in this chapter on amino acids, the fundamental building blocks from which up to 150,000 or so different proteins in our bodies are made. We'll then see how amino acids are incorporated into proteins and examine the structures of those proteins. Any understanding of biological chemistry would be impossible without this knowledge.\n\nProteins occur in every living organism, are of many different types, and have many different biological\nfunctions. The keratin of skin and fingernails, the fibroin of silk and spider webs, and the estimated 50,000 or so enzymes that catalyze the biological reactions in our bodies are all proteins. Regardless of their function, all proteins have a fundamentally similar structure and are made up of many amino acids linked together in a long chain.\n\nAmino acids, as their name implies, are difunctional. They contain both a basic amino group and an acidic carboxyl group."}
{"id": 1696, "contents": "Alanine, an amino acid - \nTheir value as building blocks to make proteins stems from the fact that amino acids can join together into long chains by forming amide bonds between the $-\\mathrm{NH}_{2}$ of one amino acid and the $-\\mathrm{CO}_{2} \\mathrm{H}$ of another. For classification purposes, chains with fewer than 50 amino acids are often called peptides, while the term protein is generally used for larger chains.\n\nMany"}
{"id": 1697, "contents": "Alanine, an amino acid - 26.1 Structures of Amino Acids\nWe saw in Section 20.3 and Section 24.5 that a carboxyl group is deprotonated and exists as the carboxylate anion at a physiological pH of 7.3 , while an amino group is protonated and exists as the ammonium cation. Thus, amino acids exist in aqueous solution primarily in the form of a dipolar ion, or zwitterion (from the German zwitter, meaning \"hybrid\").\n(uncharged)\n\n(zwitterion)\n\nAlanine\nAmino acid zwitterions are internal salts and therefore have many of the physical properties associated with salts. They have large dipole moments, are relatively soluble in water but insoluble in hydrocarbons, and are crystalline with relatively high melting points. In addition, amino acids are amphiprotic; they can react either as acids or as bases, depending on the circumstances. In aqueous acid solution, an amino acid zwitterion is a base that accepts a proton onto its $-\\mathrm{CO}_{2}{ }^{-}$group to yield a cation. In aqueous base solution, the zwitterion is an acid that loses a proton from its $-\\mathrm{NH}_{3}{ }^{+}$group to form an anion.\n\nIn acid solution\n\n\nIn base solution\n\n\nThe structures, abbreviations (both three- and one-letter), and $\\mathrm{p} K_{\\mathrm{a}}$ values of the 20 amino acids commonly found in proteins are shown in TABLE 26.1. All are $\\boldsymbol{\\alpha}$-amino acids, meaning that the amino group in each is a substituent on the $\\alpha$ carbon-the one next to the carbonyl group. Nineteen of the twenty amino acids are primary amines, $\\mathrm{RNH}_{2}$, and differ only in the nature of their side chain-the substituent attached to the $\\alpha$ carbon. Proline is a secondary amine whose nitrogen and $\\alpha$ carbon atoms are part of a five-membered pyrrolidine ring.\n\nTABLE 26.1 The 20 Common Amino Acids in Proteins"}
{"id": 1698, "contents": "Alanine, an amino acid - 26.1 Structures of Amino Acids\nTABLE 26.1 The 20 Common Amino Acids in Proteins\n\n| Name | Abbreviations |  | MW | Structure | $\\begin{aligned} & \\mathrm{p} K_{\\mathrm{a}} \\\\ & \\boldsymbol{\\alpha}-\\mathrm{CO}_{2} \\mathrm{H} \\end{aligned}$ | $\\begin{aligned} & \\mathrm{p} K_{\\mathrm{a}} \\\\ & \\boldsymbol{\\alpha}-\\mathrm{NH}_{3}+ \\end{aligned}$ | $\\mathrm{p} K_{\\mathrm{a}}$ <br> side <br> chain | pl |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| Neutral Amino Acids |  |  |  |  |  |  |  |  |\n| Alanine | Ala | A | 89 | <smiles></smiles> | 2.34 | 9.69 | - | 6.01 |\n| Asparagine | Asn | N | 132 | <smiles>NC(=O)CC@@HC(=O)[O-]</smiles> | 2.02 | 8.80 | - | 5.41 |\n| Cysteine | Cys | C | 121 | <smiles>[NH3+]C(CS)C(=O)[O-]</smiles> | 1.96 | 10.28 | 8.18 | 5.07 |\n| Glutamine | GIn | Q | 146 | <smiles>NC(=O)CCC@HC(=O)[O-]</smiles> | 2.17 | 9.13 | - | 5.65 |\n| Glycine | Gly | G | 75 | <smiles></smiles> | 2.34 | 9.60 | - | 5.97 |\n| Isoleucine | Ile | I | 131 | <smiles>CCC(C)C([NH3+])C(=O)[O-]</smiles> | 2.36 | 9.60 | - | 6.02 |"}
{"id": 1699, "contents": "Alanine, an amino acid - 26.1 Structures of Amino Acids\n| Name | Abb | tions | MW | Structure | $\\begin{aligned} & \\mathrm{p} K_{\\mathrm{a}} \\\\ & \\boldsymbol{\\alpha}-\\mathrm{CO}_{2} \\mathrm{H} \\end{aligned}$ | $\\begin{aligned} & \\mathrm{p} K_{\\mathrm{a}} \\\\ & \\boldsymbol{\\alpha}-\\mathrm{NH}_{3}+ \\end{aligned}$ | $\\mathrm{p} K_{\\mathrm{a}}$ <br> side <br> chain | p/ |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| Leucine | Leu | L | 131 | <smiles>CC(C)CC@@HC(=O)[O-]</smiles> | 2.36 | 9.60 | - | 5.98 |\n| Methionine | Met | M | 149 | <smiles>CSCCC@HC(=O)[O-]</smiles> | 2.28 | 9.21 | - | 5.74 |\n| Phenylalanine | Phe | F | 165 | <smiles>NC@@HC(=O)[O-]</smiles> | 1.83 | 9.13 | - | 5.48 |\n| Proline | Pro | P | 115 | <smiles>O=C([O-])C1CCC[NH2+]1</smiles> | 1.99 | 10.60 | - | 6.30 |\n| Serine | Ser | S | 105 | <smiles>NC@@HC(=O)[O-]</smiles> | 2.21 | 9.15 | - | 5.68 |\n| Threonine | Thr | T | 119 | <smiles></smiles> | 2.09 | 9.10 | - | 5.60 |"}
{"id": 1700, "contents": "Alanine, an amino acid - 26.1 Structures of Amino Acids\n| Threonine | Thr | T | 119 | <smiles></smiles> | 2.09 | 9.10 | - | 5.60 |\n| Tryptophan | Trp | W | 204 | <smiles>[NH3+]C(Cc1c[nH]c2ccccc12)C(=O)[O-]</smiles> | 2.83 | 9.39 | - | 5.89 |\n| Tyrosine | Tyr | Y | 181 | <smiles>NC@@Hcc1)C(=O)[O-]</smiles> | 2.20 | 9.11 | 10.07 | 5.66 |"}
{"id": 1701, "contents": "Alanine, an amino acid - 26.1 Structures of Amino Acids\n| Name | Abb | ions | MW | Structure | $\\begin{aligned} & \\mathrm{p} K_{\\mathrm{a}} \\\\ & \\alpha-\\mathrm{CO}_{2} \\mathrm{H} \\end{aligned}$ | $\\begin{aligned} & \\mathrm{p} K_{\\mathrm{a}} \\\\ & \\boldsymbol{\\alpha}-\\mathrm{NH}_{3}+ \\end{aligned}$ | $\\mathrm{p} K_{\\mathrm{a}}$ <br> side <br> chain | pl |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| Valine | Val | V | 117 | <smiles>CC(C)C@HC(=O)[O-]</smiles> | 2.32 | 9.62 | - | 5.96 |"}
{"id": 1702, "contents": "Acidic Amino Acids - \n| Aspartic acid | Asp | D | 133 | <smiles></smiles> | 1.88 | 9.60 | 3.65 | 2.77 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| Glutamic acid | Glu | E | 147 | <smiles>COC(=O)CCC@HC(=O)[O-]</smiles> | 2.19 | 9.67 | 4.25 | 3.22 |\n\nBasic Amino Acids\n\n\nIn addition to the 20 amino acids commonly found in proteins, 2 others-selenocysteine and pyrrolysine-are found in some organisms, and more than 700 nonprotein amino acids are also found in nature. $\\gamma$-Aminobutyric acid (GABA), for instance, is found in the brain and acts as a neurotransmitter; homocysteine is found in blood and is linked to coronary heart disease; and thyroxine is found in the thyroid gland, where it acts as a hormone.\n\n\nSelenocysteine\n\n\nPyrrolysine\n\n$\\gamma$-Aminobutyric\nacid\n\n\nHomocysteine\n\n\nThyroxine\n\nExcept for glycine, $\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$, the $\\alpha$ carbons of amino acids are chirality centers. Two enantiomers of each are therefore possible, but nature uses only one to build proteins. In Fischer projections, naturally occurring amino acids are represented by placing the $-\\mathrm{CO}_{2}{ }^{-}$group at the top and pointing the side chain downwards, as if drawing a carbohydrate (Section 25.2) and then placing the $-\\mathrm{NH}_{3}{ }^{+}$group on the left. Because of their stereochemical similarity to L sugars (Section 25.3), the naturally occurring $\\alpha$-amino acids are often referred to as L amino acids. The nonnaturally occurring enantiomers are called D amino acids.\n\n\nL-Serine\n(S)-Serine"}
{"id": 1703, "contents": "Acidic Amino Acids - \nL-Serine\n(S)-Serine\n\n\nL-Cysteine\n(R)-Cysteine\n\n\nL-Alanine\n(S)-Alanine\n\n\nD-Alanine\n( $R$ )-Alanine\n\nThe 20 common amino acids can be further classified as neutral, acidic, or basic, depending on the structure of their side chains. Fifteen of the twenty have neutral side chains, two (aspartic acid and glutamic acid) have an extra carboxylic acid function in their side chains, and three (lysine, arginine, and histidine) have basic amino groups in their side chains. Note that both cysteine (a thiol) and tyrosine (a phenol), although usually classified as neutral amino acids, nevertheless have weakly acidic side chains that can be deprotonated in a sufficiently basic solution.\n\nAt the physiological pH of 7.3 , the side-chain carboxyl groups of aspartic acid and glutamic acid are deprotonated and the basic side-chain nitrogens of lysine and arginine are protonated. Histidine, however, which contains a heterocyclic imidazole ring in its side chain, is not quite basic enough to be protonated at pH 7.3. Note that only the pyridine-like, doubly bonded nitrogen in histidine is basic. The pyrrole-like singly bonded nitrogen is nonbasic because its lone pair of electrons is part of the six- $\\pi$-electron aromatic imidazole ring (Section 24.9).\n\n\nHistidine\nHumans are able to biosynthesize only 11 of the 20 protein amino acids, called nonessential amino acids.\n\nThe other 9, called essential amino acids, are biosynthesized only in plants and microorganisms and must be obtained in our diet. The division between essential and nonessential amino acids is not clear-cut, however. Tyrosine, for instance, is sometimes considered nonessential because humans can produce it from phenylalanine, but phenylalanine itself is essential and must be obtained in the diet. Arginine can be synthesized by humans, but much of the arginine we need also comes from our diet."}
{"id": 1704, "contents": "Acidic Amino Acids - \nPROBLEM How many of the $\\alpha$-amino acids shown in Table 26.1 contain aromatic rings? How many contain 26-1 sulfur? How many contain alcohol groups? How many contain hydrocarbon side chains?\n\nPROBLEM Of the 19 l amino acids, 18 have the $S$ configuration at the $\\alpha$ carbon. Cysteine is the only lamino 26-2 acid that has an $R$ configuration. Explain.\n\nPROBLEM The amino acid threonine, ( $2 S, 3 R$ )-2-amino-3-hydroxybutanoic acid, has two chirality centers. 26-3\n(a) Draw threonine, using normal, wedged, and dashed lines to show dimensionality.\n(b) Draw a diastereomer of threonine, and label its chirality centers as $R$ or $S$."}
{"id": 1705, "contents": "Acidic Amino Acids - 26.2 Amino Acids and the Henderson-Hasselbalch Equation: Isoelectric Points\nAccording to the Henderson-Hasselbalch equation (Section 20.3 and Section 24.5), if we know both the pH of a solution and the $\\mathrm{p} K_{\\mathrm{a}}$ of an acid HA, we can calculate the ratio of [ $\\mathrm{A}^{-}$] to $[\\mathrm{HA}]$ in the solution. Furthermore, when $\\mathrm{pH}=\\mathrm{p} K_{\\mathrm{a}}$, the two forms $\\mathrm{A}^{-}$and HA are present in equal amounts because $\\log 1=0$.\n\n$$\n\\mathrm{pH}=\\mathrm{p} K_{\\mathrm{a}}+\\log \\frac{\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]} \\quad \\text { or } \\quad \\log \\frac{\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]}=\\mathrm{pH}-\\mathrm{p} K_{\\mathrm{a}}\n$$\n\nTo apply the Henderson-Hasselbalch equation to an amino acid, let's find out what species are present in a 1.00 M solution of alanine at $\\mathrm{pH}=9.00$. According to TABLE 26.1, protonated alanine $\\left[{ }^{+} \\mathrm{H}_{3} \\mathrm{NCH}\\left(\\mathrm{CH}_{3}\\right) \\mathrm{CO}_{2} \\mathrm{H}\\right]$ has $\\mathrm{p} K_{\\mathrm{a} 1}$ $=2.34$ and neutral zwitterionic alanine $\\left[{ }^{+} \\mathrm{H}_{3} \\mathrm{NCH}\\left(\\mathrm{CH}_{3}\\right) \\mathrm{CO}_{2}{ }^{-}\\right]$has $\\mathrm{p} K_{\\mathrm{a} 2}=9.69$ :\n\n\nBecause the pH of the solution is much closer to $\\mathrm{p} K_{\\mathrm{a} 2}$ than to $\\mathrm{p} K_{\\mathrm{a} 1}$, we need to use $\\mathrm{p} K_{\\mathrm{a} 2}$ for the calculation. From the Henderson-Hasselbalch equation, we have:"}
{"id": 1706, "contents": "Acidic Amino Acids - 26.2 Amino Acids and the Henderson-Hasselbalch Equation: Isoelectric Points\n$$\n\\log \\frac{\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]}=\\mathrm{pH}-\\mathrm{p} K_{\\mathrm{a}}=9.00-9.69=-0.69\n$$\n\nSo\n\n$$\n\\frac{\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]}=\\operatorname{antilog}(-0.69)=0.20 \\text { and }\\left[\\mathrm{A}^{-}\\right]=0.20[\\mathrm{HA}]\n$$\n\nIn addition, we know that\n\n$$\n\\left[\\mathrm{A}^{-}\\right]+[\\mathrm{HA}]=1.00 \\mathrm{M}\n$$\n\nSolving the two simultaneous equations gives $[\\mathrm{HA}]=0.83$ and $\\left[\\mathrm{A}^{-}\\right]=0.17$. In other words, at $\\mathrm{pH}=9.00,83 \\%$ of alanine molecules in a 1.00 M solution are neutral (zwitterionic) and $17 \\%$ are deprotonated. Similar calculations can be done at other pH values and the results plotted to give the titration curve shown in FIGURE 26.2."}
{"id": 1707, "contents": "Acidic Amino Acids - 26.2 Amino Acids and the Henderson-Hasselbalch Equation: Isoelectric Points\nEach leg of the titration curve is calculated separately. The first leg, from pH 1 to 6 , corresponds to the dissociation of protonated alanine, $\\mathrm{H}_{2} \\mathrm{~A}^{+}$. The second leg, from pH 6 to 11, corresponds to the dissociation of zwitterionic alanine, HA. It's as if we started with $\\mathrm{H}_{2} \\mathrm{~A}^{+}$at low pH and then titrated with NaOH . When 0.5\nequivalent of NaOH is added, the deprotonation of $\\mathrm{H}_{2} \\mathrm{~A}^{+}$is $50 \\%$ complete; when 1.0 equivalent of NaOH is added, the deprotonation of $\\mathrm{H}_{2} \\mathrm{~A}^{+}$is finished and HA predominates; when 1.5 equivalents of NaOH is added, the deprotonation of HA is $50 \\%$ complete; and when 2.0 equivalents of NaOH is added, the deprotonation of HA is finished.\n\n\nFIGURE 26.2 A titration curve for alanine, plotted using the Henderson-Hasselbalch equation. Each of the two legs is plotted separately. At $\\mathrm{pH}<1$, alanine is entirely protonated; at $\\mathrm{pH}=2.34$, alanine is a $50: 50 \\mathrm{mix}$ of protonated and neutral forms; at $\\mathrm{pH}=6.01$, alanine is entirely neutral; at $\\mathrm{pH}=9.69$, alanine is a $50: 50 \\mathrm{mix}$ of neutral and deprotonated forms; at $\\mathrm{pH}>11.5$, alanine is entirely deprotonated."}
{"id": 1708, "contents": "Acidic Amino Acids - 26.2 Amino Acids and the Henderson-Hasselbalch Equation: Isoelectric Points\nLook carefully at the titration curve in FIGURE 26.2. In acid solution, the amino acid is protonated and exists primarily as a cation. In basic solution, the amino acid is deprotonated and exists primarily as an anion. In between the two is an intermediate pH at which the amino acid is exactly balanced between anionic and cationic forms, existing primarily as the neutral, dipolar zwitterion. This pH is called the amino acid's isoelectric point (pI) and has a value of 6.01 for alanine.\n\n\nThe isoelectric point of an amino acid depends on its structure, with values for the 20 common amino acids given previously in TABLE 26.1. The 15 neutral amino acids have isoelectric points near neutrality, in the pH range 5.0 to 6.5 . The two acidic amino acids have isoelectric points at lower pH so that deprotonation of the side-chain $-\\mathrm{CO}_{2} \\mathrm{H}$ does not occur at their pI, and the three basic amino acids have isoelectric points at higher pH so that protonation of the side-chain amino group does not occur at their pI.\n\nMore specifically, the $\\mathrm{p} I$ of any amino acid is the average of the two acid-dissociation constants that involve the neutral zwitterion. For the 13 amino acids with a neutral side chain, $\\mathrm{p} I$ is the average of $\\mathrm{p} K_{\\mathrm{a} 1}$ and $\\mathrm{p} K_{\\mathrm{a} 2}$. For the four amino acids with either a strongly or weakly acidic side chain, $\\mathrm{p} I$ is the average of the two lowest $\\mathrm{p} K_{\\mathrm{a}}$ values. For the three amino acids with a basic side chain, $\\mathrm{p} I$ is the average of the two highest $\\mathrm{p} K_{\\mathrm{a}}$ values."}
{"id": 1709, "contents": "Acidic Amino Acids - 26.2 Amino Acids and the Henderson-Hasselbalch Equation: Isoelectric Points\nJust as individual amino acids have isoelectric points, proteins have an overall $\\mathrm{p} I$ due to the cumulative effect of all the acidic or basic amino acids they may contain. The enzyme lysozyme, for instance, has a preponderance of basic amino acids and thus has a high isoelectric point ( $\\mathrm{p} I=11.0$ ). Pepsin, however, has a preponderance of acidic amino acids and a low isoelectric point ( $\\mathrm{p} I \\sim 1.0$ ). Not surprisingly, the solubilities and properties of proteins with different prs are strongly affected by the pH of the medium. Solubility in water is usually lowest at the isoelectric point, where the protein has no net charge, and is higher both above and below the pI, where the protein is charged.\n\nWe can take advantage of the differences in isoelectric points to separate a mixture of proteins into its pure components. Using a technique known as electrophoresis, a mixture of proteins is placed near the center of a strip of paper or gel. The paper or gel is moistened with an aqueous buffer of a given pH , and electrodes are connected to the ends of the strip. When an electric potential is applied, the proteins with negative charges (those that are deprotonated because the pH of the buffer is above their isoelectric point) migrate slowly toward the positive electrode. At the same time, those amino acids with positive charges (those that are protonated because the pH of the buffer is below their isoelectric point) migrate toward the negative electrode.\n\nDifferent proteins migrate at different rates, depending on their isoelectric points and on the pH of the aqueous buffer, thereby effecting a separation of the mixture into its components. FIGURE 26.3 illustrates this process for a mixture containing basic, neutral, and acidic components.\n\n\nFIGURE 26.3 Separation of a protein mixture by electrophoresis. At $\\mathrm{pH}=6.00$, a neutral protein does not migrate, a basic protein is protonated and migrates toward the negative electrode, and an acidic protein is deprotonated and migrates toward the positive electrode."}
{"id": 1710, "contents": "Acidic Amino Acids - 26.2 Amino Acids and the Henderson-Hasselbalch Equation: Isoelectric Points\nPROBLEM Hemoglobin has $\\mathrm{p} I=6.8$. Does hemoglobin have a net negative charge or net positive charge at pH 26-4 $=5.3$ ? At $\\mathrm{pH}=7.3$ ?"}
{"id": 1711, "contents": "Acidic Amino Acids - 26.3 Synthesis of Amino Acids\n$\\alpha$-Amino acids can be synthesized in the laboratory using some of the reactions discussed in previous chapters. One of the oldest methods of $\\alpha$-amino acid synthesis begins with $\\alpha$ bromination of a carboxylic acid by treatment with $\\mathrm{Br}_{2}$ and $\\mathrm{PBr}_{3}$ (the Hell-Volhard-Zelinskii reaction; Section 22.4). $\\mathrm{S}_{\\mathrm{N}} 2$ substitution of the $\\alpha$-bromo acid with ammonia then yields an $\\alpha$-amino acid.\n\n\n4-Methylpentanoic acid\n\n\n2-Bromo-4-methylpentanoic acid\n\nPROBLEM Show how you could prepare the following $\\alpha$-amino acids from the appropriate carboxylic acids:\n26-5 (a) Phenylalanine (b) Valine"}
{"id": 1712, "contents": "The Amidomalonate Synthesis - \nA more general method for preparation of $\\alpha$-amino acids is the amidomalonate synthesis, a straightforward extension of the malonic ester synthesis (Section 22.7). The reaction begins with the conversion of diethyl acetamidomalonate into an enolate ion by treatment with base, followed by $\\mathrm{S}_{\\mathrm{N}} 2$ alkylation with a primary alkyl halide. Hydrolysis of both the amide protecting group and the esters occurs when the alkylated product is warmed with aqueous acid, and decarboxylation then takes place to yield an $\\alpha$-amino acid. For example, aspartic acid can be prepared from ethyl bromoacetate, $\\mathrm{BrCH}_{2} \\mathrm{CO}_{2} \\mathrm{Et}$ :\n\n\nPROBLEM What alkyl halides would you use to prepare the following $\\alpha$-amino acids by the amidomalonate 26-6 method?\n(a) Leucine\n(b) Histidine\n(c) Tryptophan\n(d) Methionine"}
{"id": 1713, "contents": "Reductive Amination of $\\boldsymbol{\\alpha}$-Keto Acids - \nYet another method for the synthesis of $\\alpha$-amino acids is by reductive amination of an $\\alpha$-keto acid with ammonia and a reducing agent. Alanine, for instance, is prepared by treatment of pyruvic acid with ammonia in the presence of $\\mathrm{NaBH}_{4}$. As described in Section 24.6, the reaction proceeds through formation of an intermediate imine which is then reduced."}
{"id": 1714, "contents": "Enantioselective Synthesis - \nThe synthesis of an $\\alpha$-amino acid from an achiral precursor by any of the methods just described yields a racemic mixture, with equal amounts of $S$ and $R$ enantiomers. To use an amino acid in the laboratory synthesis of a naturally occurring protein, however, the pure $S$ enantiomer must be obtained.\n\nTwo methods are used in practice to obtain enantiomerically pure amino acids. One way requires resolving the racemic mixture into its pure enantiomers (Section 5.8). A more direct approach, however, is to use an enantioselective synthesis to prepare only the desired $S$ enantiomer directly. As discussed in the Chapter 19 Chemistry Matters, the idea behind enantioselective synthesis is to find a chiral reaction catalyst that will temporarily hold a substrate molecule in an unsymmetrical, chiral environment. While in that chiral environment, the substrate may be more open to reaction on one side than on another, leading to an excess of one enantiomeric product.\n\nWilliam Knowles at the Monsanto Company discovered in 1968 that $\\alpha$-amino acids can be prepared enantioselectively by hydrogenation of a $Z$ enamido acid with a chiral hydrogenation catalyst. (S)Phenylalanine, for instance, is prepared at $98.7 \\%$ purity, contaminated by only $1.3 \\%$ of the $(R)$ enantiomer, when using a chiral rhodium catalyst. For this discovery, Knowles shared the 2001 Nobel Prize in Chemistry.\n\n\nA (Z) enamido acid\n(S)-Phenylalanine\n\nThe most effective catalysts for enantioselective amino acid synthesis are coordination complexes of rhodium(I) with 1,5-cyclooctadiene (COD) and a chiral diphosphine such as ( $R, R$ )-1,2-bis( $o$-anisylphenylphosphino)ethane, the so-called DiPAMP ligand. This complex owes its chirality to the presence of trisubstituted phosphorus atoms (Section 5.10).\n\n\nPROBLEM Show how you could prepare the following amino acid enantioselectively:\n26-7"}
{"id": 1715, "contents": "Enantioselective Synthesis - 26.4 Peptides and Proteins\nProteins and peptides are amino acid polymers in which the individual amino acids, called residues, are linked together by amide bonds, or peptide bonds. An amino group from one residue forms an amide bond with the carboxyl of a second residue, the amino group of the second forms an amide bond with the carboxyl of a third, and so on. For example, alanylserine is the dipeptide that results when an amide bond forms between the alanine carboxyl and the serine amino group.\n\n\nNote that two dipeptides can result from reaction between alanine and serine, depending on which carboxyl group reacts with which amino group. If the alanine amino group reacts with the serine carboxyl, serylalanine results.\n\n\nThe long, repetitive sequence of $-\\mathrm{N}-\\mathrm{CH}-\\mathrm{CO}-$ atoms that makes up a continuous chain is called the protein's backbone. By convention, peptides are written with the $\\mathbf{N}$-terminal amino acid (the one with the free $-\\mathrm{NH}_{3}{ }^{+}$ group) on the left and the $\\mathbf{C}$-terminal amino acid (the one with the free $-\\mathrm{CO}_{2}{ }^{-}$group) on the right. The name of a peptide is denoted by the abbreviations listed in TABLE 26.1 for each amino acid. Thus, alanylserine is abbreviated Ala-Ser or A-S, and serylalanine is abbreviated Ser-Ala or S-A. The one-letter abbreviations are more convenient, though less immediately recognizable, than the three-letter abbreviations.\n\nThe amide bond that links amino acids together in peptides is no different from any other amide bond (Section 24.3). An amide nitrogen is nonbasic because its unshared electron pair is delocalized by resonance with the carbonyl group. This overlap of the nitrogen $p$ orbital with the $p$ orbitals of the carbonyl group imparts a certain amount of double-bond character to the $\\mathrm{C}-\\mathrm{N}$ bond and restricts rotation around it. The amide bond is therefore planar, and the $\\mathrm{N}-\\mathrm{H}$ is oriented $180^{\\circ}$ to the $\\mathrm{C}=\\mathrm{O}$."}
{"id": 1716, "contents": "Enantioselective Synthesis - 26.4 Peptides and Proteins\nA second kind of covalent bonding in peptides occurs when a disulfide linkage, RS-SR, is formed between two cysteine residues. As we saw in Section 18.7, a disulfide is formed by mild oxidation of a thiol, RSH, and is cleaved by mild reduction.\n\n\nA disulfide bond between cysteine residues in different peptide chains links the otherwise separate chains together, whereas a disulfide bond between cysteine residues in the same chain forms a loop. Insulin, for instance, is composed of two chains that total 51 amino acids and are linked by two cysteine disulfide bridges.\n\n\nPROBLEM Six isomeric tripeptides contain valine, tyrosine, and glycine. Name them using both three- and 26-8 one-letter abbreviations.\n\nPROBLEM Draw the structure of M-P-V-G, and indicate its amide bonds. 26-9"}
{"id": 1717, "contents": "Enantioselective Synthesis - 26.5 Amino Acid Analysis of Peptides\nTo determine the structure of a protein or peptide, we need to answer three questions: What amino acids are present? How much of each is present? In what sequence do the amino acids occur in the peptide chain? The answers to the first two questions are provided by an automated instrument called an amino acid analyzer.\n\nAn amino acid analyzer is based on techniques worked out in the 1950s by William Stein and Stanford Moore at the Rockefeller University, who shared the 1972 Nobel Prize in Chemistry for their work. In preparation for analysis, the peptide is broken into its constituent amino acids by reducing all disulfide bonds, capping the - SH groups of cysteine residues by $\\mathrm{S}_{\\mathrm{N}} 2$ reaction with iodoacetic acid, and hydrolyzing the amide bonds by heating with aqueous 6 M HCl at $110^{\\circ} \\mathrm{C}$ for 24 hours. The resultant amino acid mixture is then separated into its components by a technique called chromatography, either high-pressure liquid chromatography (HPLC) or ion-exchange chromatography.\n\nIn both HPLC and ion-exchange chromatography, the mixture to be separated is dissolved in a solvent, called the mobile phase, and passed through a metal tube or glass column that contains an adsorbent material, called the stationary phase. Because different compounds adsorb to the stationary phase to different extents, they migrate through the chromatography column at different rates and are separated as they emerge (elute) from the end.\n\nIn the ion-exchange technique, separated amino acids eluting from the chromatography column mix with a solution of a substance called ninhydrin and undergo a rapid reaction that produces an intense purple color. The color is measured by a spectrometer, and a plot of elution time versus spectrometer absorbance is obtained."}
{"id": 1718, "contents": "Enantioselective Synthesis - 26.5 Amino Acid Analysis of Peptides\nBecause the time required for a given amino acid to elute from a standard column is reproducible, the identities of the amino acids in a peptide can be determined. The amount of each amino acid in the sample is determined by measuring the intensity of the purple color resulting from its reaction with ninhydrin. FIGURE 26.4 shows the results of amino acid analysis of a standard equimolar mixture of $17 \\alpha$-amino acids. Typically, amino acid analysis requires about 100 picomoles $(2-3 \\mu \\mathrm{~g})$ of sample for a protein containing about 200 residues.\n\n\nFIGURE 26.4 Amino acid analysis of an equimolar mixture of 17 amino acids.\nPROBLEM Show the structure of the product you would expect to obtain by $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of a cysteine residue 26-10 with iodoacetic acid.\n\nPROBLEM Show the structures of the products obtained on reaction of valine with ninhydrin.\n26-11"}
{"id": 1719, "contents": "Enantioselective Synthesis - 26.6 Peptide Sequencing: The Edman Degradation\nWith the identities and relative amounts of amino acids known, a peptide can then be sequenced to find out in what order the amino acids are linked. Much peptide sequencing is now done by mass spectrometry, using either electrospray ionization (ESI) or matrix-assisted laser desorption ionization (MALDI) linked to a time-offlight (TOF) mass analyzer, as described in Section 12.4. Also in common use is a chemical method of peptide sequencing called the Edman degradation.\n\nThe general idea of peptide sequencing by Edman degradation is to cleave one amino acid at a time from an end of the peptide chain. That terminal amino acid is then separated and identified, and the cleavage reactions are repeated on the chain-shortened peptide until the entire peptide sequence is known. Automated protein sequencers are available that allow as many as 50 repetitive sequencing cycles to be carried out before a buildup of unwanted by-products interferes with the results. So efficient are these instruments that sequence information can be obtained from as little as 1 to 5 picomoles of sample-less than $0.1 \\mu \\mathrm{~g}$.\n\nAs shown in FIGURE 26.5, Edman degradation involves treatment of a peptide with phenyl isothiocyanate (PITC), $\\mathrm{C}_{6} \\mathrm{H}_{5}-\\mathrm{N}=\\mathrm{C}=\\mathrm{S}$, followed by reaction with trifluoroacetic acid. The first step attaches the PITC to the $-\\mathrm{NH}_{2}$ group of the N -terminal amino acid, and the second step splits the N -terminal residue from the peptide chain, yielding an anilinothiazolinone (ATZ) derivative plus the chain-shortened peptide. Further acidcatalyzed rearrangement of the ATZ derivative with aqueous acid converts it into a phenylthiohydantoin (PTH), which is identified by comparison of its elution time with the known elution times of PTH derivatives of the 20 common amino acids. The chain-shortened peptide is then automatically resubmitted for another round of Edman degradation."}
{"id": 1720, "contents": "FIGURE 26.5 MECHANISM - \nMechanism of the Edman degradation for $\\mathbf{N}$-terminal analysis of peptides.\n(1) Nucleophilic addition of the peptide terminal amino group to phenyl isothiocyanate (PITC) gives an N -phenylthiourea derivative.\n\n(2) Acid-catalyzed cyclization of the phenylthiourea yields a tetrahedral intermediate...\n(2) $\\downarrow{ }_{\\downarrow} \\mathrm{CF}_{3} \\mathrm{CO}_{2} \\mathrm{H}$\n(3)... which expels the chainshortened peptide and forms an anilinothiazolinone (ATZ) derivative.\n\n(3) $\\downarrow$\n\n\nAnilinothiazolinone (ATZ)\nThe ATZ rearranges in the presence of aqueous acid to an isomeric $N$-phenylthiohydantoin (PTH) as the final product.\n\n$N$-Phenylthiohydantoin (PTH)\n\nComplete sequencing of large proteins by Edman degradation is impractical because of the buildup of unwanted by-products. To get around this problem, a large peptide chain is first cleaved by partial hydrolysis into a number of smaller fragments, the sequence of each fragment is determined, and the individual fragments are\nfitted together by matching their overlapping ends. In this way, protein chains with more than 400 amino acids have been sequenced.\n\nPartial hydrolysis of a peptide can be carried out either chemically with aqueous acid or enzymatically. Acid hydrolysis is unselective and gives a more-or-less random mixture of small fragments, but enzymatic hydrolysis is quite specific. The enzyme trypsin, for instance, catalyzes hydrolysis of peptides only at the carboxyl side of the basic amino acids arginine and lysine; chymotrypsin cleaves only at the carboxyl side of the aryl-substituted amino acids phenylalanine, tyrosine, and tryptophan.\n\n\nPROBLEM The octapeptide angiotensin II has the sequence Asp-Arg-Val-Tyr-Ile-His-Pro-Phe. What fragments 26-12 would result if angiotensin II were cleaved with trypsin? With chymotrypsin?"}
{"id": 1721, "contents": "FIGURE 26.5 MECHANISM - \nPROBLEM What is the N -terminal residue on a peptide that gives the following PTH derivative upon Edman 26-13 degradation?\n\n\nPROBLEM Draw the structure of the PTH derivative that would be formed by Edman degradation of 26-14 angiotensin II (Problem 26-12).\n\nPROBLEM Give the amino acid sequence of hexapeptides that produce the following sets of fragments upon\n26-15 partial acid hydrolysis:\n(a) Arg, Gly, Ile, Leu, Pro, Val gives Pro-Leu-Gly, Arg-Pro, Gly-Ile-Val\n(b) $\\mathrm{N}, \\mathrm{L}, \\mathrm{M}, \\mathrm{W}, \\mathrm{V}_{2}$ gives V-L, V-M-W, W-N-V"}
{"id": 1722, "contents": "FIGURE 26.5 MECHANISM - 26.7 Peptide Synthesis\nOnce the structure of a peptide is known, its synthesis can be undertaken-perhaps to obtain a larger amount for biological evaluation. A simple amide might be formed by treating an amine and a carboxylic acid with a carbodiimide (either DCC or EDC; Section 21.3), but peptide synthesis is a more difficult problem because many different amide bonds must be formed in a specific order, rather than at random.\n\nThe solution to the specificity problem is protection (Section 17.8). If we want to couple alanine with leucine to synthesize Ala-Leu, for instance, we could protect the $-\\mathrm{NH}_{2}$ group of alanine and the $-\\mathrm{CO}_{2} \\mathrm{H}$ group of leucine to shield them from reacting, then form the desired Ala-Leu amide bond by reaction with EDC or DCC, and then remove the protecting groups.\n\n\nA number of different amino- and carboxyl-protecting groups have been devised, but only a few are used in peptide synthesis. Carboxyl groups are often protected simply by converting them into methyl or benzyl esters. Both groups are easily introduced by standard methods of ester formation (Section 21.6) and are easily removed by mild hydrolysis with aqueous NaOH . Benzyl esters can also be cleaved by catalytic hydrogenolysis of the weak benzylic $\\mathrm{C}-\\mathrm{O}$ bond $\\left(\\mathrm{RCO}_{2}-\\mathrm{CH}_{2} \\mathrm{Ph}+\\mathrm{H}_{2} \\rightarrow \\mathrm{RCO}_{2} \\mathrm{H}+\\mathrm{PhCH}_{3}\\right)$."}
{"id": 1723, "contents": "FIGURE 26.5 MECHANISM - 26.7 Peptide Synthesis\nAmino groups are sometimes protected as their tert-butyloxycarbonyl amide (Boc) or, more commonly, as their fluorenylmethyloxycarbonyl amide (Fmoc). The Boc protecting group is introduced by reaction of the amino acid with di-tert-butyl dicarbonate in a nucleophilic acyl substitution reaction and is removed by brief treatment with a strong acid such as trifluoroacetic acid, $\\mathrm{CF}_{3} \\mathrm{CO}_{2} \\mathrm{H}$. The Fmoc protecting group is introduced by reaction with fluorenylmethyloxycarbonyl chloride and is removed by treatment with a $20 \\%$ solution of the amine piperidine in dimethylformamide as solvent.\n\n\nBoc-Ala\n\n\nFmoc-Ala\nThus, five steps are needed to synthesize a dipeptide such as Ala-Leu:"}
{"id": 1724, "contents": "FIGURE 26.5 MECHANISM - 26.7 Peptide Synthesis\nBoc-Ala\n\n\nFmoc-Ala\nThus, five steps are needed to synthesize a dipeptide such as Ala-Leu:\n\n|  | Ala | <smiles>[14CH3]O14C[18O]</smiles> | Leu | $+\\mathrm{CH}_{3} \\mathrm{OH}$ |\n| :---: | :---: | :---: | :---: | :---: |\n| (1) The amino group of alanine is protected as the Boc derivative, and <br> (2) the carboxyl group of leucine is protected as the methyl ester. |  | (1) $\\downarrow$ |  | (2) $\\downarrow \\begin{aligned} & \\mathrm{H}^{+} \\\\ & \\text {catalyst }\\end{aligned}$ |\n|  |  | Boc-Ala |  | Leu-OCH3 |\n| (3) The two protected amino acids are coupled using DCC. | (3) $\\downarrow$ CC |  |  |  |\n|  | Boc-Ala-Leu-OCH3 |  |  |  |\n| (4) The Boc protecting group is removed by acid treatment. | (4) $\\downarrow \\mathrm{CF}_{3} \\mathrm{CO}_{2} \\mathrm{H}$ |  |  |  |\n|  | Ala-Leu- $\\mathrm{OCH}_{3}$ |  |  |  |\n| (5) The methyl ester is removed by basic hydrolysis. | 5 |  | $\\downarrow{ }^{\\mathrm{NaOH}} \\mathrm{H}$ |  |\n|  |  | Ala- | -Leu |  |\n\nThese steps can be repeated to add one amino acid at a time to the growing chain or to link two peptide chains together. Many remarkable achievements in peptide synthesis have been reported, including a complete synthesis of human insulin. Insulin is composed of two chains totaling 51 amino acids linked by two disulfide bridges. The three-dimensional structure of insulin, shown previously, was determined by Dorothy Crowfoot Hodgkin, a British chemist who received the 1964 Nobel Prize in Chemistry for her work on this and other complex biological molecules."}
{"id": 1725, "contents": "FIGURE 26.5 MECHANISM - 26.7 Peptide Synthesis\nPROBLEM Show the mechanism for formation of a Boc derivative by reaction of an amino acid with 26-16 di-tert-butyl dicarbonate.\n\nPROBLEM Write all five steps required for the synthesis of Leu-Ala from alanine and leucine."}
{"id": 1726, "contents": "FIGURE 26.5 MECHANISM - 26.8 Automated Peptide Synthesis: The Merrifield Solid-Phase Method\nAs you might imagine, the synthesis of a large peptide chain by sequential addition of one amino acid at a time is a long and arduous process. An immense simplification is possible, however, using methods introduced by R. Bruce Merrifield, who received the 1984 Nobel Prize in Chemistry. In the Merrifield solid-phase method, peptide synthesis is carried out with the growing amino acid chain covalently bonded to small beads of a polymer resin, rather than in solution.\n\nIn the original procedure, polystyrene resin was used, prepared so that one of every hundred or so benzene rings contained a chloromethyl $\\left(-\\mathrm{CH}_{2} \\mathrm{Cl}\\right)$ group. A Boc-protected C-terminal amino acid was then attached to the resin through an ester bond formed by $\\mathrm{S}_{\\mathrm{N}} 2$ reaction.\n\n\nWith the first amino acid bonded to the resin, a repeating series of four steps is then carried out to build a peptide.\n(1) A Boc-protected amino acid is covalently linked to the polystyrene polymer by formation of an ester bond $\\left(S_{N} 2\\right.$ reaction).\n(2) The polymer-bonded amino acid is washed free of excess reagent and then treated with trifluoroacetic acid to remove the Boc group.\n(3) A second Boc-protected amino acid is coupled to the first by reaction with DCC. Excess reagents are removed by washing them from the insoluble polymer.\n(4) The cycle of deprotection, coupling, and washing is repeated as many times as desired to add amino acid units to the growing chain.\n(5) After the desired peptide has been made, treatment with anhydrous HF removes the final Boc group and cleaves the ester bond to the polymer, yielding the free peptide.\n\n\n\n\n\nThe steps in the solid-phase procedure have been improved substantially over the years, but the fundamental idea remains the same. A common procedure is use of the Wang resin, developed by Su-Sun Wang, with the Fmoc protecting group. Less common is using the PAM (phenylacetamidomethyl) resin with Boc as the N protecting group."}
{"id": 1727, "contents": "FIGURE 26.5 MECHANISM - 26.8 Automated Peptide Synthesis: The Merrifield Solid-Phase Method\nRobotic peptide synthesizers are now used to automatically repeat the coupling, washing, and deprotection steps with different amino acids. Each step occurs in high yield, and mechanical losses are minimized because the peptide intermediates are never removed from the insoluble polymer until the final step. Peptides with 20 amino acids can be routinely prepared in a few hours."}
{"id": 1728, "contents": "FIGURE 26.5 MECHANISM - 26.9 Protein Structure\nProteins are usually classified as either fibrous or globular, according to their three-dimensional shape. Fibrous proteins, such as the collagen in tendons and connective tissue and the myosin in muscle tissue, consist of polypeptide chains arranged side by side in long filaments. Because these proteins are tough and insoluble in water, they are used in nature for structural materials. Globular proteins, by contrast, are usually coiled into compact, roughly spherical shapes. These proteins are generally soluble in water and are mobile within cells. Most of the more than 3000 or so enzymes that have been characterized are globular proteins, including the 1300 different enzymes in the human body.\n\nProteins are so large that the word structure takes on a broader meaning than with simpler organic compounds. In fact, chemists speak of four different levels of structure when describing proteins.\n\n- The primary structure of a protein is simply the amino acid sequence.\n- The secondary structure of a protein describes how segments of the peptide backbone orient into a regular pattern.\n- The tertiary structure describes how the entire protein molecule coils into an overall three-dimensional shape.\n- The quaternary structure describes how different protein molecules come together to yield large aggregate structures.\n\nPrimary structure is determined, as we've seen, by sequencing the protein. Secondary, tertiary, and quaternary structures are determined either by NMR or by X-ray crystallography (Chapter 12 Chemistry Matters)."}
{"id": 1729, "contents": "FIGURE 26.5 MECHANISM - 26.9 Protein Structure\nPrimary structure is determined, as we've seen, by sequencing the protein. Secondary, tertiary, and quaternary structures are determined either by NMR or by X-ray crystallography (Chapter 12 Chemistry Matters).\n\nThe most common secondary structures are the $\\alpha$ helix and the $\\beta$-pleated sheet. An $\\boldsymbol{\\alpha}$ helix is a right-handed coil of the protein backbone, much like the coil of a spiral staircase (FIGURE 26.6a). Each turn of the helix contains 3.6 amino acid residues, with a distance between coils of 540 pm , or 5.4 A. The structure is stabilized by hydrogen bonds between amide $\\mathrm{N}-\\mathrm{H}$ groups and $\\mathrm{C}=\\mathrm{O}$ groups four residues away, with an $\\mathrm{N}-\\mathrm{H} \\cdots \\mathrm{O}$ distance of $2.8 \\AA$. The $\\alpha$ helix is an extremely common secondary structure, and nearly all globular proteins contain many helical segments. Myoglobin, a small globular protein containing 153 amino acid residues in a single chain, is one example (FIGURE 26.6b).\n\nA $\\boldsymbol{\\beta}$-pleated sheet differs from an $\\alpha$ helix in that the peptide chain is fully extended rather than coiled and the hydrogen bonds occur between residues in adjacent chains (FIGURE 26.7a). The neighboring chains can run either in the same direction (parallel) or in opposite directions (antiparallel), although the antiparallel arrangement is more common and somewhat more energetically favorable. Concanavalin A, for instance,\nconsists of two identical chains of 237 residues, with extensive regions of antiparallel $\\beta$ sheets (FIGURE 26.7b).\n\n\nFIGURE 26.6 (a) The $\\alpha$-helical secondary structure of proteins is stabilized by hydrogen bonds between the $\\mathrm{N}-\\mathrm{H}$ group of one residue and the $\\mathrm{C}=\\mathrm{O}$ group four residues away. (b) The structure of myoglobin, a globular protein with extensive helical regions that are shown as coiled ribbons in this representation.\n(a)\n\n(b)"}
{"id": 1730, "contents": "FIGURE 26.5 MECHANISM - 26.9 Protein Structure\n(b)\n\n\nFIGURE 26.7 (a) The $\\beta$-pleated sheet secondary structure of proteins is stabilized by hydrogen bonds between parallel or antiparallel chains. (b) The structure of concanavalin A, a protein with extensive regions of antiparallel $\\beta$ sheets, shown as flat ribbons.\n\nWhat about tertiary structure? Why does a protein adopt the shape it does? The forces that determine the tertiary structure of a protein are the same forces that act on all molecules, regardless of size, to provide maximum stability. Particularly important are the hydrophilic (water-loving; Section 2.12) interactions of the polar side chains on acidic or basic amino acids and the hydrophobic (water-fearing) interactions of nonpolar\nside chains. These acidic or basic amino acids with charged side chains tend to congregate on the exterior of the protein, where they can be solvated by water. Amino acids with neutral, nonpolar side chains tend to congregate on the hydrocarbon-like interior of a protein molecule, away from the aqueous medium.\n\nAlso important for stabilizing a protein's tertiary structure are the formation of disulfide bridges between cysteine residues, the formation of hydrogen bonds between nearby amino acid residues, and the presence of ionic attractions, called salt bridges, between positively and negatively charged sites on various amino acid side chains within the protein.\n\nBecause the tertiary structure of a globular protein is delicately maintained by weak intramolecular attractions, a modest change in temperature or pH is often enough to disrupt that structure and cause the protein to become denatured. Denaturation occurs under such mild conditions that the primary structure remains intact, but the tertiary structure unfolds from a specific globular shape to a randomly looped chain (FIGURE 26.8).\n\n\nFIGURE 26.8 A representation of protein denaturation. A globular protein loses its specific three-dimensional shape and becomes randomly looped."}
{"id": 1731, "contents": "FIGURE 26.5 MECHANISM - 26.9 Protein Structure\nFIGURE 26.8 A representation of protein denaturation. A globular protein loses its specific three-dimensional shape and becomes randomly looped.\n\nDenaturation is accompanied by changes in both physical and biological properties. Solubility is drastically decreased, as occurs when egg white is cooked and the albumins unfold and coagulate. Most enzymes lose all catalytic activity when denatured, since a precisely defined tertiary structure is required for their action. Although most denaturation is irreversible, some cases are known where spontaneous renaturation of an unfolded protein to its stable tertiary structure occurs, accompanied by a full recovery of biological activity."}
{"id": 1732, "contents": "FIGURE 26.5 MECHANISM - 26.10 Enzymes and Coenzymes\nAn enzyme is a substance-usually a large protein-that acts as a catalyst for a biological reaction. Like all catalysts, an enzyme doesn't affect the equilibrium constant of a reaction and can't bring about a chemical change that is otherwise unfavorable. An enzyme acts only to lower the activation energy for a reaction, thereby making the reaction take place more rapidly than it otherwise would. Sometimes, in fact, the rate acceleration brought about by enzymes is extraordinary. Millionfold rate increases are common, and the glycosidase enzymes that hydrolyze polysaccharides increase the reaction rate by a factor of more than $10^{17}$, changing the time required for the reaction from millions of years to milliseconds!\n\nUnlike many of the catalysts that chemists use in the laboratory, enzymes are usually specific in their action. Often, in fact, an enzyme will catalyze only a single reaction of a single compound, called the enzyme's substrate. For example, the enzyme amylase, found in the human digestive tract, catalyzes only the hydrolysis of starch to yield glucose; cellulose and other polysaccharides are untouched by amylase.\n\nDifferent enzymes have different specificities. Some, such as amylase, are specific for a single substrate, but others operate on a range of substrates. Papain, for instance, a globular protein of 212 amino acids isolated from papaya fruit, catalyzes the hydrolysis of many kinds of peptide bonds. In fact, it's this ability to hydrolyze peptide bonds that makes papain useful as a cleaner for contact lenses.\n\n\nEnzymes function through a pathway that involves initial formation of an enzyme-substrate complex (E $\\cdot \\mathrm{S}$ ), followed by a multistep chemical conversion of the enzyme-bound substrate into enzyme-bound product ( $\\mathrm{E} \\cdot \\mathrm{P}$ )\nand final release of product from the complex.\n\n$$\n\\mathrm{E}+\\mathrm{S} \\rightleftarrows \\mathrm{E} \\cdot \\mathrm{~S} \\rightleftarrows \\mathrm{E} \\cdot \\mathrm{P} \\rightleftarrows \\mathrm{E}+\\mathrm{P}\n$$"}
{"id": 1733, "contents": "FIGURE 26.5 MECHANISM - 26.10 Enzymes and Coenzymes\nThe overall rate constant for conversion of the $\\mathrm{E} \\cdot \\mathrm{S}$ complex to products $\\mathrm{E} \\cdot \\mathrm{P}$ is called the turnover number because it represents the number of substrate molecules a single enzyme molecule turns over into product per unit time. A value of about $10^{3}$ per second is typical, although carbonic anhydrase can reach a value of up to 600,000.\n\nThe extraordinary rate accelerations achieved by enzymes are due to a combination of several factors. One important factor is simple geometry: an enzyme will adjust its shape to hold the substrate, other reactants, and various catalytic sites on acidic or basic residues in the precise geometry needed for reaction. In addition, the wrapping of the enzyme around the substrate can create specialized microenvironments that protect the substrate from the aqueous medium and can dramatically change the behavior of acid-base catalytic residues in the active site. But perhaps most important is that the enzyme stabilizes and thus lowers the energy of the rate-limiting transition state for reaction. That is, it's not the ability of the enzyme to bind the substrate that matters but rather its ability to bind and stabilize the transition state. Often, in fact, the enzyme binds the transition structure as much as $10^{12}$ times more tightly than it binds the substrate or products. An energy diagram for an enzyme-catalyzed process might resemble that in FIGURE 26.9.\n\n\nFIGURE 26.9 Energy diagrams for uncatalyzed and enzyme-catalyzed processes. The enzyme makes available an alternative, lowerenergy pathway. Rate enhancement is due to the ability of the enzyme to bind to the transition state for product formation, thereby lowering its energy."}
{"id": 1734, "contents": "FIGURE 26.5 MECHANISM - 26.10 Enzymes and Coenzymes\nEnzymes are classified into six categories depending on the kind of reaction they catalyze, as shown in TABLE 26.2. Oxidoreductases catalyze oxidations and reductions; transferases catalyze the transfer of a group from one substrate to another; hydrolases catalyze hydrolysis reactions of esters, amides, and related substrates; lyases catalyze the elimination or addition of a small molecule such as $\\mathrm{H}_{2} \\mathrm{O}$ from or to a substrate; isomerases catalyze isomerizations; and ligases catalyze the bonding of two molecules, often coupled with the hydrolysis of ATP. The systematic name of an enzyme has two parts, ending with -ase. The first part identifies the enzyme's substrate, and the second part identifies its class. Hexose kinase, for example, is a transferase that catalyzes the transfer of a phosphate group from ATP to a hexose sugar.\n\nTABLE 26.2 Classification of Enzymes\n\n| Class | Some subclasses | Function |\n| :--- | :--- | :--- |\n| Oxidoreductases | Dehydrogenases <br> Oxidases <br> Reductases | Introduction of double bond <br> Oxidation <br> Reduction |\n| Transferases | Kinases <br> Transaminases | Transfer of phosphate group <br> Transfer of amino group |\n| Hydrolases | Lipases <br> Nucleases | Hydrolysis of ester <br> Hydrolysis of phosphate <br> Hydrolysis of amide |\n| Lyases | Decarboxylases | Loss of $\\mathrm{CO}_{2}$ |\n| Isomerases | Epimerases | Isomerization of chirality center |\n| Ligases | Carboxylases <br> Synthetases | Addition of CO <br> 2 |\n\nIn addition to their protein part, most enzymes also contain a small non-protein part called a cofactor. A cofactor can be either an inorganic ion, such as $\\mathrm{Zn}^{2+}$, or a small organic molecule, called a coenzyme. A coenzyme is not a catalyst but is a reactant that undergoes chemical change during the reaction and requires an additional step to return to its initial state."}
{"id": 1735, "contents": "FIGURE 26.5 MECHANISM - 26.10 Enzymes and Coenzymes\nMany coenzymes are derived from vitamins-substances that an organism requires in small amounts for growth but is unable to synthesize and must receive in its diet (Chapter 20 Chemistry Matters). Coenzyme A from pantothenate (vitamin $\\mathrm{B}_{3}$ ), $\\mathrm{NAD}^{+}$from niacin, FAD from riboflavin (vitamin $\\mathrm{B}_{2}$ ), tetrahydrofolate from folic acid, pyridoxal phosphate from pyridoxine (vitamin $B_{6}$ ), and thiamin diphosphate from thiamin (vitamin $B_{1}$ ) are examples. TABLE 26.3 shows the structures of some common coenzymes.\n\nPROBLEM To what classes do the following enzymes belong?\n26-18 (a) Pyruvate decarboxylase\n(b) Chymotrypsin (c) Alcohol dehydrogenase"}
{"id": 1736, "contents": "FIGURE 26.5 MECHANISM - 26.11 How Do Enzymes Work? Citrate Synthase\nAs we saw in the previous section, enzymes work by bringing substrate and other reactant molecules together, holding them in the orientation necessary for reaction, providing any necessary acidic or basic sites to catalyze specific steps, and stabilizing the transition state for reaction. As an example, let's look at citrate synthase, an enzyme that catalyzes the aldol-like addition of acetyl CoA to oxaloacetate to give citrate. This reaction is the first step in the citric acid cycle, in which acetyl groups produced by degradation of food molecules are metabolized to yield $\\mathrm{CO}_{2}$ and $\\mathrm{H}_{2} \\mathrm{O}$. We'll look at the details of the citric acid cycle in Section 29.7.\n\n\nCitrate synthase is a globular protein of 433 amino acids with a deep cleft lined by an array of functional groups that can bind to the substrate, oxaloacetate. On binding oxaloacetate, the original cleft closes and another opens up nearby to bind acetyl CoA. This second cleft is also lined by appropriate functional groups, including a histidine at position 274 and an aspartic acid at position 375 . The two reactants are now held by the enzyme in close proximity and with a suitable orientation for reaction. FIGURE 26.10 shows the structure of citrate synthase as determined by X-ray crystallography, along with a close-up of the active site.\n\nTABLE 26.3 Structures and Functions of Some Common Coenzymes"}
{"id": 1737, "contents": "Coenzyme A (acyl transfer) - \nNicotinamide adenine dinucleotide-NAD ${ }^{+}$(oxidation/reduction)\n(NADP ${ }^{+}$)\n\n\nFlavin adenine dinucleotide-FAD (oxidation/reduction)"}
{"id": 1738, "contents": "S-Adenosylmethionine (methyl transfer) - \nLipoic acid (acyl transfer)\n\n\nThiamin diphosphate (decarboxylation)\n\n\nPyridoxal phosphate (amino acid metabolism)\n\n\nBiotin (carboxylation)\n\n(a)\n\n(b)\n\n(c)\n\nOxaloacetate\n\nFIGURE 26.10 X-ray crystal structure of citrate synthase. Part (a) is a space-filling model and part (b) is a ribbon model, which emphasizes the $\\alpha$-helical segments of the protein chain and indicates that the enzyme is dimeric; that is, it consists of two identical chains held together by hydrogen bonds and other intermolecular attractions. Part (c) is a close-up of the active site in which oxaloacetate and an unreactive acetyl CoA mimic are bound.\n\nAs shown in FIGURE 26.11, the first step in the aldol reaction of acetyl CoA and oxaloacetate is generation of the enol of acetyl CoA. The side-chain carboxyl of an aspartate residue acts as base to abstract an acidic $\\alpha$ proton, while at the same time the side-chain imidazole ring of a histidine donates $\\mathrm{H}^{+}$to the carbonyl oxygen. The enol thus produced then performs a nucleophilic addition to the ketone carbonyl group of oxaloacetate. The first histidine acts as a base to remove the -OH hydrogen from the enol, while a second histidine residue simultaneously donates a proton to the oxaloacetate carbonyl group, giving citryl CoA. Water then hydrolyzes the thiol ester group in citryl CoA in a nucleophilic acyl substitution reaction, releasing citrate and coenzyme A as the final products."}
{"id": 1739, "contents": "FIGURE 26.11 MECHANISM - \nMechanism of the addition of acetyl CoA to oxaloacetate to give (S)-citryl CoA, catalyzed by citrate synthase.\n(1) The side-chain carboxylate group of an aspartic acid acts as a base and removes an acidic $\\alpha$ proton from acetyl CoA, while the $\\mathrm{N}-\\mathrm{H}$ group on the side chain of a histidine acts as an acid and donates a proton to the carbonyl oxygen, giving an enol.\n\n\nAcetyl CoA\n\n\nA histidine deprotonates the acetyl-CoA enol, which adds to the ketone carbonyl group of oxaloacetate in an aldol-like reaction. Simultaneously, an acid N-H proton of another histidine protonates the carbonyl oxygen, producing (S)-citryl CoA.\n\nOxaloacetate\n(2)\n\n(S)-Citryl CoA\n(3) $\\mathrm{H}_{2} \\mathrm{O}$\n\n\nCitrate"}
{"id": 1740, "contents": "The Protein Data Bank - \nEnzymes are so large, so structurally complex, and so numerous that the use of computer databases and molecular visualization programs has become an essential tool for studying biological chemistry. Of the various databases available online, the Kyoto Encyclopedia of Genes and Genomes (KEGG) database (http://www.genome.jp/kegg/pathway.html), maintained by the Kanehisa Laboratory of Kyoto University Bioinformatics Center, is useful for obtaining information on biosynthetic pathways of the sort we'll be describing in Chapter 29. For obtaining information on a specific enzyme, the BRENDA database (http://www.brenda-enzymes.org), maintained by the Institute of Biochemistry at the University of Cologne, Germany, is particularly valuable.\n\nPerhaps the most useful of all biological databases is the Protein Data Bank (PDB), operated by the Research Collaboratory for Structural Bioinformatics (RCSB). The PDB is a worldwide repository of X-ray and NMR structural data for biological macromolecules. In mid-2022, data for more than 192,888 structures were available, and more than 7000 new ones were being added yearly. To access the Protein Data Bank, go to www.rcsb.org and a home page like that shown in FIGURE 26.12 will appear. As with much that is available online, however, the PDB site is changing rapidly, so you may not see quite the same thing."}
{"id": 1741, "contents": "The Protein Data Bank - \nFIGURE 26.12 The Protein Data Bank home page. (credit: RCSB Protein Data Bank)\nTo learn how to use the PDB, begin by running the short tutorial listed under Learn on the left side column. After that introduction, start exploring. Let's say you want to view citrate synthase, the enzyme that catalyzes the addition of acetyl CoA to oxaloacetate to give citrate. Type \"citrate synthase\" (with quotation marks) into the small search box on the top line, click on \"Search,\" and a list of 2,400 or so structures will appear. Scroll down near the end of the list until you find the entry with a PDB code of 5CTS and the title \"Proposed Mechanism for the Condensation Reaction of Citrate Synthase: 1.9-Angstroms Structure of the Ternary Complex with Oxaloacetate and Carboxymethyl Coenzyme A.\" Alternatively, if you know the code of the enzyme you want, you can enter it directly into the search box. Click on the PDB code of entry 5CTS, and a new page containing information about the enzyme will open.\n\nIf you choose, you can download the structure file to your computer and open it with any of numerous molecular graphics programs to see an image like that in FIGURE 26.13. The biologically active molecule is a dimer of two identical subunits consisting primarily of $\\alpha$-helical regions displayed as coiled ribbons. For now, just select \"View in Jmol\" under the enzyme image on the right side of the screen to see some of the options for visualizing and further exploring the enzyme.\n\n\nFIGURE 26.13 An image of citrate synthase, downloaded from the Protein Data Bank. (credit: Protein Data Bank, 1AL6 B. PDB ID: 1AL6 B. Schwartz, K.W. Vogel, K.C. Usher, C. Narasimhan, H.M. Miziorko, S.J. Remington, D.G. Drueckhammer. Mechanisms of Enzyme-Catalyzed Deprotonation of Acetyl-Coenzyme A, CC BY 1.0)"}
{"id": 1742, "contents": "Key Terms - \n- $\\alpha$-amino acids\n- $\\alpha$ helix\n- backbone\n- $\\beta$-pleated sheet\n- C-terminal amino acid\n- coenzyme\n- cofactor\n- denatured\n- Edman degradation\n- enzyme\n- fibrous proteins\n- globular proteins\n- isoelectric point (pI)\n- N -terminal amino acid\n- peptides\n- primary structure\n- proteins\n- quaternary structure\n- residues\n- secondary structure\n- side chain\n- tertiary structure\n- turnover number\n- zwitterion"}
{"id": 1743, "contents": "Summary - \nProteins and peptides are large biomolecules made of $\\boldsymbol{\\alpha}$-amino acid residues linked together by amide, or peptide, bonds. Twenty amino acids are commonly found in proteins, and all except glycine have stereochemistry similar to that of L sugars. In neutral solution, amino acids exist as dipolar zwitterions.\n\nAmino acids can be synthesized in racemic form by several methods, including ammonolysis of an $\\alpha$-bromo acid, alkylation of diethyl acetamidomalonate, and reductive amination of an $\\alpha$-keto acid. Alternatively, an enantioselective synthesis of amino acids can be carried out using a chiral hydrogenation catalyst.\n\nDetermining the structure of a peptide or protein begins with amino acid analysis. The peptide is hydrolyzed to its constituent $\\alpha$-amino acids, which are separated and identified. Next, the peptide is sequenced. Edman degradation by treatment with phenyl isothiocyanate (PITC) cleaves one residue from the N terminus of the peptide and forms an easily identifiable phenylthiohydantoin (PTH) derivative of the $\\mathbf{N}$-terminal amino acid. An automated series of Edman degradations can sequence peptide chains up to 50 residues in length.\n\nPeptide synthesis involves the use of protecting groups. An N-protected amino acid with a free $-\\mathrm{CO}_{2} \\mathrm{H}$ group is coupled using DCC or EDC to an O-protected amino acid with a free $-\\mathrm{NH}_{2}$ group. Amide formation occurs, the protecting groups are removed, and the sequence is repeated. Amines are usually protected as their tert-butyloxycarbonyl (Boc) or fluorenylmethyloxycarbonyl (Fmoc) derivatives; acids are usually protected as esters. The synthesis is often carried out by the Merrifield solid-phase method, in which the peptide is bonded to insoluble polymer beads."}
{"id": 1744, "contents": "Summary - \nProteins have four levels of structure. Primary structure describes a protein's amino acid sequence; secondary structure describes how segments of the protein chain orient into regular patterns-either $\\boldsymbol{\\alpha}$ helix or $\\boldsymbol{\\beta}$-pleated sheet; tertiary structure describes how the entire protein molecule coils into an overall three-dimensional shape; and quaternary structure describes how individual protein molecules aggregate into larger structures.\n\nProteins are classified as either globular or fibrous. Fibrous proteins such as $\\alpha$-keratin are tough, rigid, and water-insoluble; globular proteins such as myoglobin are water-soluble and roughly spherical in shape. Many globular proteins are enzymes-substances that act as catalysts for biological reactions. Enzymes are grouped into six classes according to the kind of reaction they catalyze. In addition to their protein part, many enzymes contain cofactors, which can be either metal ions or small organic molecules called coenzymes."}
{"id": 1745, "contents": "Summary of Reactions - \n1. Amino acid synthesis (Section 26.3)\na. From $\\alpha$-bromo acids\n\nb. Diethyl acetamidomalonate synthesis\n\nc. Reductive amination of an $\\alpha$-keto acid\n\nd. Enantioselective synthesis\n\n$A(Z)$ enamido acid\n$\\xrightarrow[\\text { 2. } \\mathrm{NaOH}, \\mathrm{H}_{2} \\mathrm{O}]{\\text { 1. } \\mathrm{H}_{2},[\\mathrm{Rh}(\\text { DiPAMP })(\\mathrm{COD})]^{+} \\mathrm{BF}_{4}^{-}}$\n\n\nAn (S)-amino acid\n2. Peptide sequencing by Edman degradation (Section 26.6)\n\n3. Peptide synthesis (Section 26.7)\na. Amine protection\n\n\nBoc-protected amino acid\nb. Carboxyl protection"}
{"id": 1746, "contents": "Additional Problems - \nVisualizing Chemistry\nPROBLEM Identify the following amino acids:"}
{"id": 1747, "contents": "26-19 (a) - \n(b)\n\n(c)\n\n\nPROBLEM Give the sequence of the following tetrapeptide (yellow = $S$ ):\n26-20\n\n\nPROBLEM Isoleucine and threonine are the only two amino acids with two chirality centers. Assign $R$ or $S$\n26-21 configuration to the methyl-bearing carbon atom of isoleucine.\n\n\nPROBLEM Is the following structure a D amino acid or an L amino acid? Identify it.\n26-22\n\n\nPROBLEM Give the sequence of the following tetrapeptide:\n26-23"}
{"id": 1748, "contents": "Mechanism Problems - \nPROBLEM The reaction of ninhydrin with an $\\alpha$-amino acid occurs in several steps.\n26-24 (a) The first step is formation of an imine by reaction of the amino acid with ninhydrin. Show its structure and the mechanism of its formation.\n(b) The second step is a decarboxylation. Show the structure of the product and the mechanism of the decarboxylation reaction.\n(c) The third step is hydrolysis of an imine to yield an amine and an aldehyde. Show the structures of both products and the mechanism of the hydrolysis reaction.\n(d) The final step is formation of the purple anion. Show the mechanism of the reaction.\n\n\nNinhydrin\nPROBLEM The chloromethylated polystyrene resin originally used for Merrifield solid-phase peptide\n26-25 synthesis was prepared by treatment of polystyrene with chloromethyl methyl ether and a Lewis acid catalyst. Propose a mechanism for the reaction.\n\n\nPROBLEM An Fmoc protecting group can be removed from an amino acid by treatment with the amine base 26-26 piperidine. Propose a mechanism.\n\n\nFmoc-protected amino acid\n\nPROBLEM Proteins can be cleaved specifically at the amide bond on the carboxyl side of methionine residues 26-27 by reaction with cyanogen bromide, $\\mathrm{BrC} \\equiv \\mathrm{N}$."}
{"id": 1749, "contents": "Mechanism Problems - \nThe reaction occurs in several steps:\n(a) The first step is a nucleophilic substitution reaction of the sulfur on the methionine side chain with BrCN to give a cyanosulfonium ion, $\\left[\\mathrm{R}_{2} \\mathrm{SCN}\\right]^{+}$. Show the structure of the product, and propose a mechanism for the reaction.\n(b) The second step is an internal $\\mathrm{S}_{\\mathrm{N}} 2$ reaction, with the carbonyl oxygen of the methionine residue displacing the positively charged sulfur leaving group and forming a five-membered ring product. Show the structure of the product and the mechanism of its formation.\n(c) The third step is a hydrolysis reaction to split the peptide chain. The carboxyl group of the former methionine residue is now part of a lactone (cyclic ester) ring. Show the structure of the lactone product and the mechanism of its formation.\n(d) The final step is a hydrolysis of the lactone to give the product shown. Show the mechanism of the reaction.\n\nPROBLEM A clever recent method of peptide synthesis involves formation of an amide bond by reaction of an\n26-28 $\\alpha$-keto acid with an $N$-alkylhydroxylamine:\n\n\nAn $\\boldsymbol{\\alpha}$-keto acid $\\quad$ A hydroxylamine $\\quad$ An amide\nThe reaction is thought to occur by nucleophilic addition of the $N$-alkylhydroxylamine to the keto acid as if forming an oxime (Section 19.8), followed by decarboxylation and elimination of water. Show the mechanism."}
{"id": 1750, "contents": "Amino Acid Structures and Chirality - \nPROBLEM Except for cysteine, only $S$ amino acids occur in proteins. Several $R$ amino acids are also found in 26-29 nature, however. ( $R$ )-Serine is found in earthworms, and ( $R$ )-alanine is found in insect larvae. Draw Fischer projections of $(R)$-serine and $(R)$-alanine. Are these D or L amino acids?\n\nPROBLEM Cysteine is the only amino acid that has L stereochemistry but an $R$ configuration. Make up a 26-30 structure for another $L$ amino acid of your own creation that also has an $R$ configuration.\n\nPROBLEM Draw a Fischer projection of ( $S$ ) -proline. 26-31\n\nPROBLEM Show the structures of the following amino acids in their zwitterionic forms:\n26-32 (a) $\\operatorname{Trp}$\n(b) Ile\n(c) Cys\n(d) His\n\nPROBLEM Proline has $\\mathrm{p} K_{\\mathrm{a} 1}=1.99$ and $\\mathrm{p} K_{\\mathrm{a} 2}=10.60$. Use the Henderson-Hasselbalch equation to calculate the\n26-33 ratio of protonated and neutral forms at $\\mathrm{pH}=2.50$. Calculate the ratio of neutral and deprotonated forms at $\\mathrm{pH}=9.70$.\n\nPROBLEM Using both three- and one-letter codes for amino acids, write the structures of all possible peptides\n26-34 that contain the following amino acids:\n(a) Val, Ser, Leu\n(b) $\\mathrm{Ser}, \\mathrm{Leu}_{2}$, Pro\n\nPROBLEM Look at the side chains of the 20 amino acids in Table 26.1, and then think about what is not 26-35 present. None of the 20 contain either an aldehyde or a ketone carbonyl group, for instance. Is this just one of nature's oversights, or is there a likely chemical reason? What complications might an aldehyde or ketone carbonyl group cause?"}
{"id": 1751, "contents": "Amino Acid Synthesis and Reactions - \nPROBLEM Show how you could use the acetamidomalonate method to prepare the following amino acids:\n26-36 (a) Leucine (b) Tryptophan\nPROBLEM Show how you could prepare the following amino acids using a reductive amination:\n26-37 (a) Methionine (b) Isoleucine\nPROBLEM Show how you could prepare the following amino acids enantioselectively:\n26-38 (a) Pro (b) Val\nPROBLEM Serine can be synthesized by a simple variation of the amidomalonate method using formaldehyde 26-39 rather than an alkyl halide. How might this be done?\n\nPROBLEM Predict the product of the reaction of valine with the following reagents:\n26-40 (a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$, acid (b) Di-tert-butyl dicarbonate (c) $\\mathrm{KOH}, \\mathrm{H}_{2} \\mathrm{O}$\n(d) $\\mathrm{CH}_{3} \\mathrm{COCl}$, pyridine; then $\\mathrm{H}_{2} \\mathrm{O}$\n\nPROBLEM Draw resonance forms for the purple anion obtained by reaction of ninhydrin with an $\\alpha$-amino acid 26-41 (see Problem 26-24)."}
{"id": 1752, "contents": "Peptides and Enzymes - \nPROBLEM Write full structures for the following peptides:\n26-42 (a) C-H-E-M (b) P-E-P-T-I-D-E\n\nPROBLEM Propose two structures for a tripeptide that gives Leu, Ala, and Phe on hydrolysis but does not react 26-43 with phenyl isothiocyanate.\n\nPROBLEM Show the steps involved in a synthesis of Phe-Ala-Val using the Merrifield procedure. 26-44\n\nPROBLEM Draw the structure of the PTH derivative product you would obtain by Edman degradation of the 26-45 following peptides:\n(a) I-L-P-F (b) D-T-S-G-A\n\nPROBLEM Which amide bonds in the following polypeptide are cleaved by trypsin? By chymotrypsin?\n26-46\nPhe-Leu-Met-Lys-Tyr-Asp-Gly-Gly-Arg-Val-Ile-Pro-Tyr\nPROBLEM What kinds of reactions do the following classes of enzymes catalyze?\n26-47 (a) Hydrolases (b) Lyases (c) Transferases\n\nPROBLEM Which of the following amino acids are more likely to be found on the exterior of a globular protein, 26-48 and which on the interior? Explain.\n(a) Valine\n(b) Aspartic acid\n(c) Phenylalanine\n(d) Lysine\n\nPROBLEM Leuprolide is a synthetic nonapeptide used to treat both endometriosis in women and prostate 26-49 cancer in men.\n\n(a) Both C-terminal and N -terminal amino acids in leuprolide have been structurally modified. Identify the modifications.\n(b) One of the nine amino acids in leuprolide has D stereochemistry rather than the usual L. Which one?\n(c) Write the structure of leuprolide using both one- and three-letter abbreviations.\n(d) What charge would you expect leuprolide to have at neutral pH ?"}
{"id": 1753, "contents": "General Problems - \nPROBLEM The $\\alpha$-helical parts of myoglobin and other proteins stop whenever a proline residue is encountered\n26-50 in the chain. Why is proline never present in a protein $\\alpha$ helix?\nPROBLEM Arginine, the most basic of the 20 common amino acids, contains a guanidino functional group in\n26-51 its side chain. Explain, using resonance structures to show how the protonated guanidino group is stabilized.\n\n\nArginine\n\nPROBLEM Cytochrome $c$ is an enzyme found in the cells of all aerobic organisms. Elemental analysis of 26-52 cytochrome $c$ shows that it contains $0.43 \\%$ iron. What is the minimum molecular weight of this enzyme?\n\nPROBLEM Evidence for restricted rotation around amide $\\mathrm{O}=\\mathrm{C}-\\mathrm{N}$ bonds comes from NMR studies. At room 26-53 temperature, the ${ }^{1} \\mathrm{H}$ NMR spectrum of $\\mathrm{N}, \\mathrm{N}$-dimethylformamide shows three peaks: $2.9 \\mathrm{\\delta}$ (singlet, 3 H ), $3.0 \\mathrm{\\delta}$ (singlet, 3 H ), and $8.0 \\delta$ (singlet, 1 H ). As the temperature is raised, however, the two singlets at $2.9 \\delta$ and $3.0 \\delta$ slowly merge. At $180^{\\circ} \\mathrm{C}$, the ${ }^{1} \\mathrm{H}$ NMR spectrum shows only two peaks: $2.95 \\delta$ (singlet, 6 H ) and $8.0 \\delta$ (singlet, 1 H ). Explain this temperature-dependent behavior."}
{"id": 1754, "contents": "General Problems - \nPROBLEM Propose a structure for an octapeptide that shows the composition Asp, Gly $y_{2}$, Leu, Phe, $\\mathrm{Pro}_{2}$, Val 26-54 on amino acid analysis. Edman analysis shows a glycine N-terminal group, and leucine is the C-\nterminal group. Acidic hydrolysis gives the following fragments:\nVal-Pro-Leu, Gly, Gly-Asp-Phe-Pro, Phe-Pro-Val\nPROBLEM Look at the structure of human insulin, and indicate where in each chain the molecule is cleaved by 26-55 trypsin and chymotrypsin."}
{"id": 1755, "contents": "Insulin - \nPROBLEM What is the structure of a nonapeptide that gives the following fragments when cleaved?\n26-56\nTrypsin cleavage: Val-Val-Pro-Tyr-Leu-Arg, Ser-Ile-Arg\nChymotrypsin cleavage: Leu-Arg, Ser-Ile-Arg-Val-Val-Pro-Tyr\n\nPROBLEM Oxytocin, a nonapeptide hormone secreted by the pituitary gland, functions by stimulating uterine\n26-57 contraction and lactation during childbirth. Its sequence was determined from the following evidence:\n\n1. Oxytocin is a cyclic compound containing a disulfide bridge between two cysteine residues.\n2. When the disulfide bridge is reduced, oxytocin has the constitution Asn, Cys ${ }_{2}$, Gln, Gly, Ile, Leu, Pro, Tyr.\n3. Partial hydrolysis of reduced oxytocin yields seven fragments: Asp-Cys, Ile-Glu, Cys-Tyr, Leu-Gly, Tyr-Ile-Glu, Glu-Asp-Cys, and Cys-Pro-Leu.\n4. Gly is the C-terminal group.\n5. Both Glu and Asp are present as their side-chain amides (Gln and Asn) rather than as free sidechain acids.\n\nWhat is the amino acid sequence of reduced oxytocin? What is the structure of oxytocin itself?\n\nPROBLEM Aspartame, a nonnutritive sweetener marketed under such trade names as Equal, NutraSweet, and 26-58 Canderel, is the methyl ester of a simple dipeptide, Asp-Phe-OCH3.\n(a) Draw the structure of aspartame.\n(b) The isoelectric point of aspartame is 5.9. Draw the principal structure present in aqueous solution at this pH .\n(c) Draw the principal form of aspartame present at physiological $\\mathrm{pH}=7.3$."}
{"id": 1756, "contents": "Insulin - \nPROBLEM Refer to Figure 26.5 and propose a mechanism for the final step in Edman degradation-the acid-\n26-59 catalyzed rearrangement of the ATZ derivative to the PTH derivative.\nPROBLEM Amino acids are metabolized by a transamination reaction in which the $-\\mathrm{NH}_{2}$ group of the amino\n26-60 acid changes places with the keto group of an $\\alpha$-keto acid. The products are a new amino acid and a new $\\alpha$-keto acid. Show the product from transamination of isoleucine.\n\nPROBLEM The first step in the biological degradation of histidine is formation of a 26-61 4-methylidene-5-imidazolone (MIO) by cyclization of a segment of the peptide chain in the histidine ammonia lyase enzyme. Propose a mechanism.\n\n\n4-Methylidene-5-imidazolone (MIO)\nPROBLEM The first step in the biological degradation of lysine is reductive amination with $\\alpha$-ketoglutarate to\n26-62 give saccharopine. Nicotinamide adenine dinucleotide phosphate (NADPH), a relative of NADH, is the reducing agent. Show the mechanism.\n\n\nLysine\n$+$\n\n\n$\\alpha$-Ketoglutarate"}
{"id": 1757, "contents": "CHAPTER 27 - \nBiomolecules: Lipids\n\n\nFIGURE 27.1 Soap bubbles, so common yet so beautiful, are made from animal fat, a lipid. (credit: \"Reflections in soap bubbles\" by Image Picture Photography/Flickr, CC BY 2.0)"}
{"id": 1758, "contents": "CHAPTER CONTENTS - 27.1 Waxes, Fats, and Oils\n27.2 Soap\n27.3 Phospholipids\n27.4 Prostaglandins and Other Eicosanoids\n27.5 Terpenoids\n27.6 Steroids\n27.7 Biosynthesis of Steroids\n\nWHY THIS CHAPTER? We've now covered two of the four major classes of biomolecules-proteins and carbohydrates-and have two remaining. In this chapter, we'll cover lipids, the largest and most diverse class of biomolecules, looking both at their structure and function and at their metabolism.\n\nLipids are naturally occurring organic molecules that have limited solubility in water and can be isolated from organisms by extraction with nonpolar organic solvents. Fats, oils, waxes, many vitamins and hormones, and most nonprotein cell-membrane components are some examples. Note that this definition differs from the sort used for carbohydrates and proteins in that lipids are defined by a physical property (solubility) rather than by structure. Of the many kinds of lipids, we'll be concerned in this chapter with only a few: triacylglycerols, eicosanoids, terpenoids, and steroids.\n\nLipids are classified into two broad types: those like fats and waxes, which contain ester linkages and can be hydrolyzed, and those like cholesterol and other steroids, which don't have ester linkages and can't be hydrolyzed.\n\n\nAnimal fat-a triester ( $\\mathrm{R}, \\mathrm{R}^{\\prime}, \\mathrm{R}^{\\prime \\prime}=\\mathrm{C}_{11}-\\mathrm{C}_{19}$ chains)\n\n\nCholesterol"}
{"id": 1759, "contents": "CHAPTER CONTENTS - 27.1 Waxes, Fats, and Oils\nWaxes are mixtures of esters of long-chain carboxylic acids with long-chain alcohols. The carboxylic acid usually has an even number of carbons from 16 to 36 , while the alcohol has an even number of carbons from 24 to 36 . One of the major components of beeswax, for instance, is triacontyl hexadecanoate, the ester of the $\\mathrm{C}_{30}$ alcohol (1-triacontanol) and the $\\mathrm{C}_{16}$ acid (hexadecanoic acid). The waxy protective coatings on most fruits, berries, leaves, and animal furs have similar structures."}
{"id": 1760, "contents": "Triacontyl hexadecanoate (from beeswax) - \nAnimal fats and vegetable oils are the most widely occurring lipids. Although they appear different-animal fats like butter and lard are solids, whereas vegetable oils like corn and peanut oil are liquid-their structures are closely related. Chemically, fats and oils are triglycerides, or triacylglycerols-triesters of glycerol with three long-chain carboxylic acids called fatty acids. Animals use fats for long-term energy storage because they are far less highly oxidized than carbohydrates and provide about six times as much energy as an equal weight of stored, hydrated glycogen."}
{"id": 1761, "contents": "A triacylglycerol - \nHydrolysis of a fat or oil with aqueous NaOH yields glycerol and three fatty acids. The fatty acids are generally unbranched and contain an even number of carbon atoms between 12 and 20 . If double bonds are present, they have mostly, although not entirely, $Z$, or cis, geometry. The three fatty acids of a specific triacylglycerol molecule need not be the same, and the fat or oil from a given source is likely to be a complex mixture of many different triacylglycerols. TABLE 27.1 lists some of the commonly occurring fatty acids, and TABLE 27.2 lists the approximate composition of fats and oils from different sources.\n\nTABLE 27.1 Structures of Some Common Fatty Acids\n\n| Name | No. of carbons | Melting point $\\left({ }^{\\circ} \\mathrm{C}\\right)$ | Structure |\n| :--- | :---: | :--- | :--- |\n| Saturated |  |  |  |\n| Lauric | 12 | 43.2 | $\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{10} \\mathrm{CO}_{2} \\mathrm{H}$ |\n| Myristic | 14 | 53.9 | $\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{12} \\mathrm{CO}_{2} \\mathrm{H}$ |\n| Palmitic | 16 | 63.1 | $\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{14} \\mathrm{CO}_{2} \\mathrm{H}$ |\n| Stearic | 18 | 68.8 | $\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{16} \\mathrm{CO}_{2} \\mathrm{H}$ |\n| Arachidic | 20 | 76.5 | $\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{18} \\mathrm{CO}_{2} \\mathrm{H}$ |\n\nUnsaturated"}
{"id": 1762, "contents": "A triacylglycerol - \nUnsaturated\n\n| Palmitoleic | 16 | -0.1 | $(Z)-\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{5} \\mathrm{CH}=\\mathrm{CH}\\left(\\mathrm{CH}_{2}\\right)_{7} \\mathrm{CO}_{2} \\mathrm{H}$ |\n| :--- | :---: | :---: | :--- |\n| Oleic | 18 | 13.4 | $(Z)-\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{7} \\mathrm{CH}=\\mathrm{CH}\\left(\\mathrm{CH}_{2}\\right)_{7} \\mathrm{CO}_{2} \\mathrm{H}$ |\n| Linoleic | 18 | -12 | $(Z, Z)-\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{4}\\left(\\mathrm{CH}=\\mathrm{CHCH}_{2}\\right)_{2}\\left(\\mathrm{CH}_{2}\\right)_{6} \\mathrm{CO}_{2} \\mathrm{H}$ |\n| Linolenic | 18 | -11 | $($ all $Z)-\\mathrm{CH}_{3} \\mathrm{CH}_{2}\\left(\\mathrm{CH}=\\mathrm{CHCH}_{2}\\right)_{3}\\left(\\mathrm{CH}_{2}\\right)_{6} \\mathrm{CO}_{2} \\mathrm{H}$ |\n| Arachidonic | 20 | -49.5 | $($ all $Z)-\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{4}\\left(\\mathrm{CH}=\\mathrm{CHCH}_{2}\\right)_{4} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ |\n\nTABLE 27.2 Composition of Some Fats and Oils"}
{"id": 1763, "contents": "A triacylglycerol - \nTABLE 27.2 Composition of Some Fats and Oils\n\n|  | Saturated fatty acids (\\%) |  |  | Unsaturated fatty acids (\\%) |  |  |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Source | $\\mathrm{C}_{12}$ lauric | $\\mathrm{C}_{14}$ myristic | $\\mathrm{C}_{16}$ palmitic | $\\mathrm{C}_{18}$ stearic | $\\mathrm{C}_{18}$ oleic | $\\mathrm{C}_{18}$ linoleic |\n\nAnimal fat\n\n| Lard | - | 1 | 25 | 15 | 50 | 6 |\n| :--- | ---: | ---: | ---: | ---: | ---: | ---: |\n| Butter | 2 | 10 | 25 | 10 | 25 | 5 |\n| Human fat | 1 | 3 | 25 | 8 | 46 | 10 |\n| Whale blubber | - | 8 | 12 | 3 | 35 | 10 |\n\n\n| Vegetable oil |  |  |  |  |  |  |\n| :--- | ---: | ---: | ---: | ---: | ---: | ---: |\n| Coconut | 50 | 18 | 8 | 2 | 6 | 1 |\n| Corn | - | 1 | 10 | 4 | 35 | 45 |\n| Olive | - | 1 | 5 | 5 | 80 | 7 |\n| Peanut | - | - | 7 | 5 | 60 | 20 |"}
{"id": 1764, "contents": "A triacylglycerol - \nMore than 100 different fatty acids are known and about 40 occur widely. Palmitic acid ( $\\mathrm{C}_{16}$ ) and stearic acid $\\left(\\mathrm{C}_{18}\\right)$ are the most abundant saturated fatty acids; oleic and linoleic acids (both $\\mathrm{C}_{18}$ ) are the most abundant unsaturated ones. Oleic acid is monounsaturated because it has only one double bond, whereas linoleic, linolenic, and arachidonic acids are polyunsaturated fatty acids because they have more than one double bond. Linoleic and linolenic acids occur in cream and are essential in the human diet; infants grow poorly and develop skin lesions if fed a diet of nonfat milk for prolonged periods. Linolenic acid, in particular, is an example of an omega- $\\mathbf{3}$ fatty acid, which has been found to lower blood triglyceride levels and reduce the risk of heart attack. The name omega-3 means that there is a double bond three carbons in from the noncarboxyl end of the chain.\n\n\nStearic acid\n\n\nLinolenic acid, an omega-3 polyunsaturated fatty acid\nThe data in TABLE 27.1 show that unsaturated fatty acids generally have lower melting points than their saturated counterparts, a trend that is also true for triacylglycerols. Since vegetable oils generally have a higher proportion of unsaturated to saturated fatty acids than animal fats (TABLE 27.2), they have lower melting points. The difference is a consequence of structure. Saturated fats have a uniform shape that allows them to pack together efficiently in a crystal lattice. In unsaturated vegetable oils, however, the $\\mathrm{C}=\\mathrm{C}$ bonds introduce bends and kinks into the hydrocarbon chains, making crystal formation more difficult. The more double bonds there are, the harder it is for the molecules to crystallize and the lower the melting point of the oil."}
{"id": 1765, "contents": "A triacylglycerol - \nThe $\\mathrm{C}=\\mathrm{C}$ bonds in vegetable oils can be reduced by catalytic hydrogenation, typically carried out at high temperature using a nickel catalyst, to produce saturated solid or semisolid fats. Prior to 2015 , industrially produced margarine and shortening were produced by hydrogenating soybean, peanut, or cottonseed oil until the proper consistency is obtained. As of 2015, trans fats are no longer considered safe by the FDA and have since been phased out of industrially produced foods. Unfortunately, the hydrogenation reaction is accompanied by some cis-trans isomerization of the remaining double bonds, producing fats with about 10\\% to $15 \\%$ trans unsaturated fatty acids. Dietary intake of trans fatty acids increases cholesterol levels in the blood, thereby increasing the risk of heart problems. The conversion of linoleic acid into elaidic acid is an example.\n\n\nPROBLEM Carnauba wax, used in floor and furniture polishes, contains an ester of a $\\mathrm{C}_{32}$ straight-chain alcohol\n27-1 with a $\\mathrm{C}_{20}$ straight-chain carboxylic acid. Draw its structure.\nPROBLEM Draw structures of glyceryl tripalmitate and glyceryl trioleate. Which would you expect to have a\n27-2 higher melting point?"}
{"id": 1766, "contents": "A triacylglycerol - 27.2 Soap\nSoap has been in use for nearly 5000 years. As early as 2800 BC, the Babylonians boiled fats with ashes to create a soap-like substance. Ancient Egyptian medical papyri dating from 1550 BC reveals that Egyptians bathed regularly with soap made from a mixture of animal fats, vegetable oils, and alkaline salts. Chemically, soap is a mixture of the sodium or potassium salts of the long-chain fatty acids produced by hydrolysis (saponification) of animal fat with alkali. Wood ash was used as a source of alkali until the early 1800s, when the LeBlanc process for making $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ by heating sodium sulfate with limestone became available."}
{"id": 1767, "contents": "A fat ( $\\mathrm{R}=\\mathrm{C}_{11}-\\mathrm{C}_{19}$ Aliphatic chains) - \nGlycerol\n\nCrude soap curds contain glycerol and excess alkali as well as soap but can be purified by boiling with water and adding NaCl or KCl to precipitate the pure carboxylate salts. The smooth soap that precipitates is dried, perfumed, and pressed into bars for household use. Dyes are added to make colored soaps, antiseptics are added for medicated soaps, pumice is added for scouring soaps, and air is blown in for soaps that float. Regardless of these extra treatments and regardless of price, though, all soaps are basically the same.\n\nSoaps act as cleansers because the two ends of a soap molecule are so different. The carboxylate end of the longchain molecule is ionic and therefore hydrophilic (Section 2.12), or attracted to water. The long hydrocarbon portion of the molecule, however, is nonpolar and hydrophobic, avoiding water and therefore more soluble in oils. The net effect of these two opposing tendencies is that soaps are attracted to both oils and water and are therefore useful as cleansers.\n\nWhen soaps are dispersed in water, the long hydrocarbon tails cluster together on the inside of a tangled, hydrophobic ball, while the ionic heads on the surface of the cluster protrude into the water layer. These spherical clusters, called micelles, are shown schematically in FIGURE 27.2. Grease and oil droplets are solubilized in water when they are coated by the nonpolar, hydrophobic tails of soap molecules in the center of micelles. Once solubilized, the grease and dirt can be rinsed away.\n\n\nFIGURE 27.2 A soap micelle solubilizing a grease particle in water. An electrostatic potential map of a fatty acid carboxylate shows how the negative charge is located in the head group."}
{"id": 1768, "contents": "A fat ( $\\mathrm{R}=\\mathrm{C}_{11}-\\mathrm{C}_{19}$ Aliphatic chains) - \nFIGURE 27.2 A soap micelle solubilizing a grease particle in water. An electrostatic potential map of a fatty acid carboxylate shows how the negative charge is located in the head group.\n\nAs useful as they are, soaps also have drawbacks. In hard water, which contains metal ions, soluble sodium carboxylates are converted into insoluble magnesium and calcium salts, leaving the familiar ring of scum around bathtubs and a gray tinge on white clothes. Chemists have circumvented this problem by synthesizing a class of synthetic detergents based on salts of long-chain alkylbenzenesulfonic acids. The mechanism of synthetic detergents is the same as that of soaps: the alkylbenzene end of the molecule is attracted to grease, while the anionic sulfonate end is attracted to water. Unlike soaps, though, sulfonate detergents don't form insoluble metal salts in hard water and don't leave an unpleasant scum.\n\n\nA synthetic detergent\n( $R=$ various $C_{12}$ chains)\nPROBLEM Draw the structure of magnesium oleate, a component of bathtub scum.\n27-3\nPROBLEM Write the saponification reaction of glyceryl dioleate monopalmitate with aqueous NaOH .\n27-4"}
{"id": 1769, "contents": "A fat ( $\\mathrm{R}=\\mathrm{C}_{11}-\\mathrm{C}_{19}$ Aliphatic chains) - 27.3 Phospholipids\nJust as waxes, fats, and oils are esters of carboxylic acids, phospholipids are esters of phosphoric acid, $\\mathrm{H}_{3} \\mathrm{PO}_{4}$.\n\n\nA phosphoric acid monoester\n\n\nA phosphoric acid diester\n\n\nA phosphoric acid triester\n\nPhospholipids are of two general types: glycerophospholipids and sphingomyelins. Glycerophospholipids are based on phosphatidic acid, which contains a glycerol backbone linked by ester bonds to two fatty acids and one phosphoric acid. Although the fatty-acid residues can be any of the $\\mathrm{C}_{12}-\\mathrm{C}_{20}$ units typically present in fats, the acyl group at C 1 is usually saturated and the one at C 2 is usually unsaturated. The phosphate group at C 3 is also\nbonded to an amino alcohol such as choline $\\left[\\mathrm{HOCH}_{2} \\mathrm{CH}_{2} \\mathrm{~N}\\left(\\mathrm{CH}_{3}\\right)_{3}\\right]^{+}$, ethanolamine $\\left(\\mathrm{HOCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right)$, or serine $\\left[\\mathrm{HOCH}_{2} \\mathrm{CH}\\left(\\mathrm{NH}_{2}\\right) \\mathrm{CO}_{2} \\mathrm{H}\\right]$. The compounds are chiral and have an L , or $R$, configuration at C 2 .\n\n\nSphingomyelins are the second major group of phospholipids. These compounds have sphingosine or a related dihydroxyamine as their backbone and are particularly abundant in brain and nerve tissue, where they are a major constituent of the coating around nerve fibers.\n\n\nSphingosine\n\n\nA sphingomyelin"}
{"id": 1770, "contents": "A fat ( $\\mathrm{R}=\\mathrm{C}_{11}-\\mathrm{C}_{19}$ Aliphatic chains) - 27.3 Phospholipids\nSphingosine\n\n\nA sphingomyelin\n\nPhospholipids are found widely in both plant and animal tissues and make up approximately $50 \\%$ to $60 \\%$ of cell membranes. Because they are like soaps in having a long, nonpolar hydrocarbon tail bound to a polar ionic head, phospholipids in the cell membrane organize into a lipid bilayer about $5.0 \\mathrm{~nm}(50 \\AA$ ) thick. As shown in FIGURE 27.3, the nonpolar tails aggregate in the center of the bilayer in much the same way that soap tails aggregate in the center of a micelle. This bilayer serves as an effective barrier to the passage of water, ions, and other components into and out of cells.\n\n\n\nFIGURE 27.3 Aggregation of glycerophospholipids into the lipid bilayer that composes cell membranes."}
{"id": 1771, "contents": "A fat ( $\\mathrm{R}=\\mathrm{C}_{11}-\\mathrm{C}_{19}$ Aliphatic chains) - 27.4 Prostaglandins and Other Eicosanoids\nProstaglandins are a group of $\\mathrm{C}_{20}$ lipids that contain a five-membered ring with two long side chains. First isolated in the 1930s by Ulf von Euler at the Karolinska Institute in Sweden, much of the structural and chemical\nwork on prostaglandins was carried out by Sune Bergstr\u00f6m and Bengt Samuelsson. All three shared Nobel Prizes for their work. The name prostaglandin derives from the fact that the compounds were first isolated from sheep prostate glands, but they have subsequently been found to be present in small amounts in all body tissues and fluids.\n\nThe several dozen known prostaglandins have an extraordinarily wide range of biological effects. Among their many properties, they can lower blood pressure, affect blood platelet aggregation during clotting, lower gastric secretions, control inflammation, affect kidney function, affect reproductive systems, and stimulate uterine contractions during childbirth.\n\nProstaglandins, together with related compounds called thromboxanes and leukotrienes, make up a class of compounds called eicosanoids because they are derived biologically from 5,8,11,14-eicosatetraenoic acid, or arachidonic acid (FIGURE 27.4). Prostaglandins (PG) have a cyclopentane ring with two long side chains; thromboxanes (TX) have a six-membered, oxygen-containing ring; and leukotrienes (LT) are acyclic.\n\n\nArachidonic acid\n\n\nProstaglandin $\\mathrm{E}_{\\mathbf{1}}\\left(\\mathrm{PGE}_{1}\\right)$\n\n\nThromboxane $\\mathbf{B}_{\\mathbf{2}}\\left(\\right.$ TXB $\\left._{2}\\right)$\n\n\nProstaglandin $\\mathrm{I}_{2}\\left(\\mathrm{PGI}_{2}\\right)$ (prostacyclin)\n\n\nLeukotriene $\\mathrm{E}_{\\mathbf{4}}\\left(\\mathrm{LTE}_{4}\\right)$"}
{"id": 1772, "contents": "A fat ( $\\mathrm{R}=\\mathrm{C}_{11}-\\mathrm{C}_{19}$ Aliphatic chains) - 27.4 Prostaglandins and Other Eicosanoids\nLeukotriene $\\mathrm{E}_{\\mathbf{4}}\\left(\\mathrm{LTE}_{4}\\right)$\n\nFIGURE 27.4 Structures of some representative eicosanoids. All are derived biologically from arachidonic acid.\nEicosanoids are named based on their ring system (PG, TX, or LT), substitution pattern, and number of double bonds. The various substitution patterns on the ring are indicated by letter as in FIGURE 27.5, and the number of double bonds is indicated by a subscript. Thus, $\\mathrm{PGE}_{1}$ is a prostaglandin with the \"E\" substitution pattern and one double bond. The numbering of the atoms in the various eicosanoids is the same as in arachidonic acid, starting with the $-\\mathrm{CO}_{2} \\mathrm{H}$ carbon as C 1 , continuing around the ring, and ending with the $-\\mathrm{CH}_{3}$ carbon at the other end of the chain as C20.\n\n\nA prostaglandin (PG)\n\n\nA thromboxane (TX)\n\n\nA leukotriene (LT)\n\nFIGURE 27.5 The naming system for eicosanoids.\n\nEicosanoid biosynthesis begins with the conversion of arachidonic acid to $\\mathrm{PGH}_{2}$, catalyzed by the multifunctional PGH synthase (PGHS), also called cyclooxygenase (COX). There are two distinct enzymes, PGHS-1 and PGHS-2 (or COX-1 and COX-2), both of which accomplish the same reaction but appear to function independently. COX-1 carries out the normal physiological production of prostaglandins, and COX-2 produces additional prostaglandin in response to arthritis or other inflammatory conditions. Vioxx, Bextra, and several other drugs selectively inhibit the COX-2 enzyme but cause potentially serious heart and gastrointestinal problems in weakened patients. (See the Chapter 15 Chemistry Matters.)"}
{"id": 1773, "contents": "A fat ( $\\mathrm{R}=\\mathrm{C}_{11}-\\mathrm{C}_{19}$ Aliphatic chains) - 27.4 Prostaglandins and Other Eicosanoids\nPGHS accomplishes two transformations, an initial reaction of arachidonic acid with $\\mathrm{O}_{2}$ to yield $\\mathrm{PGG}_{2}$ and a subsequent reduction of the hydroperoxide group $(-\\mathrm{OOH})$ to the alcohol $\\mathrm{PGH}_{2}$. The sequence of steps involved in these transformations was shown in FIGURE 8.12.\n\nFurther processing of $\\mathrm{PGH}_{2}$ leads to other eicosanoids. $\\mathrm{PGE}_{2}$, for instance, arises by an isomerization of $\\mathrm{PGH}_{2}$ catalyzed by PGE synthase (PGES) (FIGURE 27.6).\n\n\nFIGURE 27.6 Mechanism of the conversion of PGH $_{2}$ into PGE $_{2}$.\nPROBLEM Assign $R$ or $S$ configuration to each chirality center in prostaglandin $\\mathrm{E}_{2}$ (Figure 27.6), the most 27-5 abundant and biologically potent of mammalian prostaglandins."}
{"id": 1774, "contents": "A fat ( $\\mathrm{R}=\\mathrm{C}_{11}-\\mathrm{C}_{19}$ Aliphatic chains) - 27.5 Terpenoids\nWe saw in the Chapter 8 Chemistry Matters that terpenoids are a vast and diverse group of lipids found in all living organisms. Despite their apparent structural differences, all terpenoids contain a multiple of five carbons and are derived biosynthetically from the five-carbon precursor isopentenyl diphosphate (FIGURE 27.7). Although formally a terpenoid contains oxygen, while a hydrocarbon is called a terpene, we'll use the term terpenoid to refer to both for simplicity.\n\n\nIsopentenyl diphosphate\n\n\nPatchouli alcohol (a sesquiterpenoid- $\\mathrm{C}_{15}$ )\n\n\nLanosterol\n(a triterpenoid- $\\mathrm{C}_{30}$ )\n\nCamphor (a monoterpenoid- $\\mathrm{C}_{10}$ )\n\n$\\beta$-Carotene\n(a tetraterpenoid- $\\mathrm{C}_{40}$ )\nFIGURE 27.7 Structures of some representative terpenoids.\nYou might recall from the chapter on Alkenes: Reactions and Synthesis that terpenoids are classified according to the number of five-carbon multiples they contain. Monoterpenoids contain 10 carbons and are derived from two isopentenyl diphosphates, sesquiterpenoids contain 15 carbons and are derived from three isopentenyl diphosphates, diterpenoids contain 20 carbons and are derived from four isopentenyl diphosphates, and so on, up to triterpenoids $\\left(\\mathrm{C}_{30}\\right)$ and tetraterpenoids $\\left(\\mathrm{C}_{40}\\right)$. Lanosterol, for instance, is a triterpenoid from which steroid hormones are made, and $\\beta$-carotene is a tetraterpenoid that serves as a dietary source of vitamin A (FIGURE 27.7)."}
{"id": 1775, "contents": "A fat ( $\\mathrm{R}=\\mathrm{C}_{11}-\\mathrm{C}_{19}$ Aliphatic chains) - 27.5 Terpenoids\nThe terpenoid precursor isopentenyl diphosphate, formerly called isopentenyl pyrophosphate and thus abbreviated IPP, is biosynthesized by two different pathways, depending on the organism and the structure of the final product. In animals and higher plants, sesquiterpenoids and triterpenoids arise primarily from the mevalonate pathway, whereas monoterpenoids, diterpenoids, and tetraterpenoids are biosynthesized by the 1-deoxyxylulose 5-phosphate ( $D X P$ ) pathway, also called the methylerithritol phosphate, or MEP, pathway. In bacteria, both pathways are used. We'll look only at the mevalonate pathway, which is more common and better understood at present.\n\n(R)-Mevalonate\n\n\n\n1-Deoxy-D-xylulose 5-phosphate\nThe Mevalonate Pathway to Isopentenyl Diphosphate\nAs shown in FIGURE 27.8, the mevalonate pathway begins with the conversion of acetate to acetyl CoA, followed by Claisen condensation to yield acetoacetyl CoA. A second carbonyl condensation reaction with a third molecule of acetyl CoA, this one an aldol-like process, then yields the six-carbon compound 3 -hydroxy-3-methylglutaryl CoA, which is reduced to give mevalonate. Phosphorylation, followed by loss of $\\mathrm{CO}_{2}$ and phosphate ion, completes the process."}
{"id": 1776, "contents": "Step 1 of FIGURE 27.8: Claisen Condensation - \nThe first step in mevalonate biosynthesis is a Claisen condensation to yield acetoacetyl CoA, a reaction catalyzed by acetoacetyl-CoA acetyltransferase. An acetyl group is first bound to the enzyme by a nucleophilic acyl substitution reaction with a cysteine -SH group. Formation of an enolate ion from a second molecule of acetyl CoA, followed by Claisen condensation, then yields the product."}
{"id": 1777, "contents": "Step 2 of FIGURE 27.8: Aldol Condensation - \nAcetoacetyl CoA next undergoes an aldol-like addition of an acetyl CoA enolate ion in a reaction catalyzed by 3-hydroxy-3-methylglutaryl-CoA synthase. The reaction occurs by initial binding of the substrate to a cysteine -SH group in the enzyme, followed by enolate-ion addition and subsequent hydrolysis to give (3S)-3-hydroxy-3-methylglutaryl CoA (HMG-CoA)."}
{"id": 1778, "contents": "FIGURE 27.8 MECHANISM - \nThe mevalonate pathway for the biosynthesis of isopentenyl diphosphate from three molecules of acetyl CoA. Individual steps are explained in the text.\n(1) Claisen condensation of two molecules of acetyl CoA gives acetoacetyl CoA.\n\n2 Aldol-like condensation of acetoacetyl CoA with a third molecule of acetyl CoA, followed by hydrolysis, gives\n(3S)-3-hydroxy-3-methylglutaryl CoA."}
{"id": 1779, "contents": "Acetyl CoA - \nAcetoacetyl CoA\n\n\n(3S)-3-Hydroxy-3-methylglutaryl CoA\n3 Reduction of the thioester group by 2 equivalents of NADPH gives (R)-mevalonate, a dihydroxy acid.\n\n\n\nPhosphorylation of the tertiary hydroxyl and diphosphorylation of the primary hydroxyl, followed by decarboxylation and simultaneous expulsion of phosphate, gives isopentenyl diphosphate, the precursor of terpenoids.\n\n\nIsopentenyl diphosphate\n\n\nStep 3 of FIGURE 27.8: Reduction\nReduction of HMG-CoA to give ( $R$ )-mevalonate is catalyzed by 3-hydroxy-3-methylglutaryl-CoA reductase and requires 2 equivalents of reduced nicotinamide adenine dinucleotide phosphate (NADPH), a close relative of NADH (Section 19.12). The reaction occurs in two steps and proceeds through an aldehyde intermediate. The first step is a nucleophilic acyl substitution reaction involving hydride transfer from NADPH to the thioester carbonyl group of HMG-CoA. Following expulsion of HSCoA as leaving group, the aldehyde intermediate undergoes a second hydride addition to give mevalonate.\n\n\nStep 4 of FIGURE 27.8: Phosphorylation and Decarboxylation\nThree additional reactions are needed to convert mevalonate to isopentenyl diphosphate. The first two are straightforward phosphorylations by ATP that occur through nucleophilic substitution reactions on the terminal phosphorus. Mevalonate is first converted to mevalonate 5-phosphate (phosphomevalonate) by reaction with ATP; mevalonate 5-phosphate then reacts with a second ATP to give mevalonate 5-diphosphate (diphosphomevalonate). The third reaction results in phosphorylation of the tertiary hydroxyl group, followed by decarboxylation and loss of phosphate ion."}
{"id": 1780, "contents": "Acetyl CoA - \nMevalonate 5-diphosphate\nIsopentenyl diphosphate\nThe final decarboxylation of mevalonate 5-diphosphate seems unusual because decarboxylations of acids do not typically occur except in $\\beta$-keto acids and malonic acids, in which the carboxylate group is two atoms away from an additional carbonyl group. As discussed in Section 22.7, the function of this second carbonyl group is to act as an electron acceptor and stabilize the charge resulting from loss of $\\mathrm{CO}_{2}$. In fact, though, the decarboxylation of a $\\beta$-keto acid and the decarboxylation of mevalonate 5 -diphosphate are closely related.\n\nCatalyzed by mevalonate-5-diphosphate decarboxylase, the substrate is first phosphorylated on the free - OH group by reaction with ATP to give a tertiary phosphate, which undergoes spontaneous $\\mathrm{S}_{\\mathrm{N}} 1$-like dissociation to give a tertiary carbocation. The positive charge then acts as an electron acceptor to facilitate decarboxylation in the same way a $\\beta$ carbonyl group does, giving isopentenyl diphosphate. (In the following structures, the diphosphate group is abbreviated OPP.)\n\n\nMevalonate 5-diphosphate\n\n\nPROBLEM The conversion of mevalonate 5-phosphate to isopentenyl diphosphate occurs with the following\n27-6 result. Which hydrogen, pro- $R$ or pro- $S$, ends up cis to the methyl group, and which ends up trans?\n\n\nMevalonate 5-diphosphate\nIsopentenyl diphosphate"}
{"id": 1781, "contents": "Conversion of Isopentenyl Diphosphate to Terpenoids - \nThe conversion of isopentenyl diphosphate (IPP) to terpenoids begins with its isomerization to dimethylallyl diphosphate, abbreviated DMAPP and formerly called dimethylallyl pyrophosphate. These two $\\mathrm{C}_{5}$ building blocks then combine to give the $\\mathrm{C}_{10}$ unit geranyl diphosphate (GPP). The corresponding alcohol, geraniol, is itself a fragrant terpenoid that occurs in rose oil.\n\nFurther combination of GPP with another IPP gives the $\\mathrm{C}_{15}$ unit farnesyl diphosphate (FPP), and so on, up to $\\mathrm{C}_{25}$.\n\nTerpenoids with more than 25 carbons-that is, triterpenoids $\\left(\\mathrm{C}_{30}\\right)$ and tetraterpenoids ( $\\mathrm{C}_{40}$ ) - are synthesized by dimerization of $\\mathrm{C}_{15}$ and $\\mathrm{C}_{20}$ units, respectively (FIGURE 27.9). Triterpenoids and steroids, in particular, arise from dimerization of farnesyl diphosphate to give squalene.\n\n\nGeranyl diphosphate (GPP)\n\n\n\nFarnesyl diphosphate (FPP)\n$\\downarrow$ Dimerization\n\n\nSqualene\nFIGURE 27.9 An overview of terpenoid biosynthesis from isopentenyl diphosphate.\nThe isomerization of isopentenyl diphosphate to dimethylallyl diphosphate is catalyzed by IPP isomerase and occurs through a carbocation pathway. Protonation of the IPP double bond by a hydrogen-bonded cysteine residue in the enzyme gives a tertiary carbocation intermediate, which is deprotonated by a glutamate residue as base to yield DMAPP. X-ray structural studies on the enzyme show that it holds the substrate in an unusually deep, well-protected pocket to shield the highly reactive carbocation from reaction with solvent or other external substances."}
{"id": 1782, "contents": "Conversion of Isopentenyl Diphosphate to Terpenoids - \nBoth the initial coupling of DMAPP with IPP to give geranyl diphosphate and the subsequent coupling of GPP with a second molecule of IPP to give farnesyl diphosphate are catalyzed by farnesyl diphosphate synthase. The process requires $\\mathrm{Mg}^{2+}$ ion, and the key step is a nucleophilic substitution reaction in which the double bond of IPP behaves as a nucleophile in displacing diphosphate ion leaving group ( $\\mathrm{PP}_{\\mathrm{i}}$ ) or DMAPP. Evidence suggests that the DMAPP develops a considerable cationic character and that spontaneous dissociation of the\nallylic diphosphate ion in an $\\mathrm{S}_{\\mathrm{N}} 1$-like pathway probably occurs (FIGURE 27.10).\n\n\n\n\nFarnesyl diphosphate (FPP)\nFIGURE 27.10 Mechanism of the coupling reaction of dimethylallyl diphosphate (DMAPP) and isopentenyl diphosphate (IPP), to give geranyl diphosphate (GPP).\n\nFurther conversion of geranyl diphosphate into monoterpenoids typically involves carbocation intermediates and multistep reaction pathways that are catalyzed by terpene cyclases. Monoterpene cyclases function by first isomerizing geranyl diphosphate to its allylic isomer linalyl diphosphate (LPP), a process that occurs by spontaneous $\\mathrm{S}_{\\mathrm{N}} 1$-like dissociation to an allylic carbocation, followed by recombination. The effect of this isomerization is to convert the C2-C3 double bond of GPP into a single bond, thereby making cyclization possible and allowing $E / Z$ isomerization of the double bond.\n\nFurther dissociation and cyclization by electrophilic addition of the cationic carbon to the terminal double bond then gives a cyclic cation, which might either rearrange, undergo a hydride shift, be captured by a nucleophile, or be deprotonated to give any of the several hundred known monoterpenoids. As just one example, limonene, a monoterpenoid found in many citrus oils, arises by the biosynthetic pathway shown in FIGURE 27.11."}
{"id": 1783, "contents": "Proposing a Terpenoid Biosynthesis Pathway - \nPropose a mechanistic pathway for the biosynthesis of $\\alpha$-terpineol from geranyl diphosphate.\n\n\nStrategy\n$\\alpha$-Terpineol, a monoterpenoid, must be derived biologically from geranyl diphosphate through its isomer linalyl diphosphate. Draw the precursor in a conformation that approximates the structure of the target molecule, and then carry out a cationic cyclization, using the appropriate double bond to displace the diphosphate leaving group. Since the target is an alcohol, the carbocation resulting from cyclization evidently reacts with water."}
{"id": 1784, "contents": "Solution - \nPROBLEM Propose mechanistic pathways for the biosynthetic formation of the following terpenoids:\n27-7\n\n\n$\\alpha$-Pinene\n(b)\n\n$\\gamma$-Bisabolene"}
{"id": 1785, "contents": "Solution - 27.6 Steroids\nIn addition to fats, phospholipids, eicosanoids, and terpenoids, the lipid extracts of plants and animals also contain steroids, molecules that are derived from the triterpenoid lanosterol (FIGURE 27.7) and whose structures are based on a tetracyclic ring system. The four rings are designated $A, B, C$, and $D$, beginning at the lower left. The three 6 -membered rings ( $\\mathrm{A}, \\mathrm{B}$, and C ) adopt chair conformations but are prevented by their rigid geometry from undergoing the usual cyclohexane ring-flips (Section 4.6).\n\n\n\nA steroid ( $\\mathrm{R}=$ various side chains)\n\nTwo cyclohexane rings can be joined in either a cis or a trans manner. With cis fusion to give cis-decalin, both groups at the ring-junction positions (the angular groups) are on the same side of the two rings. With trans fusion to give trans-decalin, the groups at the ring junctions are on opposite sides.\n\n\nAs shown in FIGURE 27.12, steroids can have either a cis or a trans fusion of the $A$ and $B$ rings, but the other ring fusions (B-C and C-D) are usually trans.\n\nAn A-B trans steroid\n\n\n\nAn A-B cis steroid\n\n\n\nFIGURE 27.12 Steroid conformations. The three 6-membered rings have chair conformations but are unable to undergo ring-flips. The $A$ and $B$ rings can be either cis-fused or trans-fused.\n\nSubstituent groups on the steroid ring system can be either axial or equatorial. As with simple cyclohexanes, equatorial substitution is generally more favorable than axial substitution for steric reasons. The hydroxyl group of cholesterol, for example, has the more stable equatorial orientation.\n\n\n\nCholesterol\nPROBLEM Draw the following molecules in chair conformations, and tell whether the ring substituents are 27-8 axial or equatorial:\n(a)\n\n(b)"}
{"id": 1786, "contents": "Solution - 27.6 Steroids\nCholesterol\nPROBLEM Draw the following molecules in chair conformations, and tell whether the ring substituents are 27-8 axial or equatorial:\n(a)\n\n(b)\n\n\nPROBLEM Lithocholic acid is an A-B cis steroid found in human bile. Draw lithocholic acid showing chair 27-9 conformations, as in Figure 27.12, and tell whether the hydroxyl group at C3 is axial or equatorial.\nLithocholic acid"}
{"id": 1787, "contents": "Steroid Hormones - \nIn humans, most steroids function as hormones, chemical messengers that are secreted by endocrine glands\nand carried through the bloodstream to target tissues. There are two main classes of steroid hormones: the sex hormones, which control maturation, tissue growth, and reproduction, and the adrenocortical hormones, which regulate a variety of metabolic processes."}
{"id": 1788, "contents": "Sex Hormones - \nTestosterone and androsterone are the two most important male sex hormones, or androgens. Androgens are responsible for the development of male secondary sex characteristics during puberty and for promoting tissue and muscle growth. Both are synthesized in the testes from cholesterol. Androstenedione is another minor hormone that has received particular attention because of its use by prominent athletes.\n\n\nTestosterone\n\n\nAndrosterone\n\n\nAndrostenedione\n(Androgens)\nEstrone and estradiol are the two most important female sex hormones, or estrogens. Synthesized in the ovaries from testosterone, estrogenic hormones are responsible for the development of female secondary sex characteristics and for regulation of the menstrual cycle. Note that both have a benzene-like aromatic A ring. In addition, another kind of sex hormone called a progestin is essential in preparing the uterus for implantation of a fertilized ovum during pregnancy. Progesterone is the most important progestin.\n\n\nEstrone\n\n\nEstradiol\n\n\nProgesterone\n(a progestin)\n(Estrogens)"}
{"id": 1789, "contents": "Adrenocortical Hormones - \nAdrenocortical steroids are secreted by the adrenal glands, small organs located near the upper end of each kidney. There are two types of adrenocortical steroids, called mineralocorticoids and glucocorticoids. Mineralocorticoids, such as aldosterone, control tissue swelling by regulating cellular salt balance between $\\mathrm{Na}^{+}$ and $\\mathrm{K}^{+}$. Glucocorticoids, such as hydrocortisone, are involved in the regulation of glucose metabolism and in the control of inflammation. Glucocorticoid ointments are widely used to bring down the swelling from exposure to poison oak or poison ivy.\n\n\nAldosterone (a mineralocorticoid)\n\n\nHydrocortisone (a glucocorticoid)"}
{"id": 1790, "contents": "Synthetic Steroids - \nIn addition to the hundreds of steroids isolated from plants and animals, many more have been synthesized in pharmaceutical laboratories in the search for new drugs. Among the best-known synthetic steroids are oral contraceptives and anabolic agents. Most birth control pills are a mixture of two compounds, a synthetic estrogen, such as ethynylestradiol, and a synthetic progestin, such as norethindrone. Anabolic steroids, such as methandrostenolone (Dianabol), are synthetic androgens that mimic the tissue-building effects of natural testosterone.\n\n\nEthynylestradiol (a synthetic estrogen)\n\n\nNorethindrone\n(a synthetic progestin)\n\n\nMethandrostenolone (Dianabol)"}
{"id": 1791, "contents": "Synthetic Steroids - 27.7 Biosynthesis of Steroids\nSteroids are heavily modified triterpenoids that are biosynthesized in living organisms from farnesyl diphosphate $\\left(\\mathrm{C}_{15}\\right)$. A reductive dimerization first converts farnesyl diphosphate to the acyclic hydrocarbon squalene ( $\\mathrm{C}_{30}$ ), which is converted into lanosterol (FIGURE 27.13). Further rearrangements and degradations then take place to yield various steroids. The conversion of squalene to lanosterol is among the most intensively studied of biosynthetic transformations. Starting from an achiral, open-chain polyene, the entire process requires only two enzymes and results in the formation of six carbon-carbon bonds, four rings, and seven chirality centers.\n\n\n2 Farnesyl diphosphate\nDimerization\n\n\nSqualene\n\n\n\nLanosterol\nFIGURE 27.13 An overview of steroid biosynthesis from farnesyl diphosphate.\nLanosterol biosynthesis begins with the selective epoxidation of squalene to give ( $3 S$ )-2,3-oxidosqualene, catalyzed by squalene epoxidase. Molecular $\\mathrm{O}_{2}$ provides the epoxide oxygen atom, and NADPH is required,\nalong with a flavin coenzyme. The proposed mechanism involves reaction of $\\mathrm{FADH}_{2}$ with $\\mathrm{O}_{2}$ to produce a flavin hydroperoxide intermediate ( ROOH ), which transfers an oxygen to squalene in a pathway initiated by nucleophilic attack of the squalene double bond on the terminal hydroperoxide oxygen (FIGURE 27.14). The flavin alcohol formed as a by-product loses $\\mathrm{H}_{2} \\mathrm{O}$ to give FAD, which is reduced back to FADH 2 by NADPH. As noted in Section 8.7, this biological epoxidation mechanism is closely analogous to the mechanism by which peroxyacids $\\left(\\mathrm{RCO}_{3} \\mathrm{H}\\right)$ react with alkenes to give epoxides in the laboratory."}
{"id": 1792, "contents": "Synthetic Steroids - 27.7 Biosynthesis of Steroids\nThe second part of lanosterol biosynthesis is catalyzed by oxidosqualene-lanosterol cyclase and occurs as shown in FIGURE 27.15. Squalene is folded by the enzyme into a conformation that aligns the various double bonds for a cascade of successive intramolecular electrophilic additions, followed by a series of hydride and methyl migrations. Except for the initial epoxide protonation/cyclization, the process is probably stepwise and appears to involve discrete carbocation intermediates that are stabilized by electrostatic interactions with electron-rich aromatic amino acids in the enzyme.\n\n\nSqualene\n\n(3S)-2,3-0xidosqualene\n\nFIGURE 27.14 Proposed mechanism of the oxidation of squalene by flavin hydroperoxide.\n\nSteps 1, 2 of FIGURE 27.15: Epoxide Opening and Initial Cyclizations\nCyclization begins in step 1 with protonation of the epoxide ring by an aspartic acid residue in the enzyme. Nucleophilic opening of the protonated epoxide by the nearby 5,10 double bond (steroid numbering; Section 27.6) then yields a tertiary carbocation at C10. Further addition of C10 to the 8,9 double bond in step 2 next gives a bicyclic tertiary cation at C8.\n\n(3S)-2,3-Oxidosqualene\nStep 3 of FIGURE 27.15: Third Cyclization\nThe third cationic cyclization is somewhat unusual because it occurs with non-Markovnikov regiochemistry and gives a secondary cation at C13 rather than the alternative tertiary cation at C14. There is growing evidence, however, that the tertiary carbocation may in fact be formed initially and that the secondary cation arises by subsequent rearrangement. The secondary cation is probably stabilized in the enzyme pocket by the proximity of an electron-rich aromatic ring.\n\nMechanism of the conversion of 2,3-oxidosqualene to lanosterol. Four cationic cyclizations are followed by four rearrangements and a final loss of $\\mathrm{H}^{+}$from C9. The steroid numbering system is used for referring to specific positions in the intermediates (Section 27.6). Individual steps are explained in the text."}
{"id": 1793, "contents": "Synthetic Steroids - 27.7 Biosynthesis of Steroids\nProtonation on oxygen opens the epoxide ring and gives a tertiary carbocation at C4. Intramolecular electrophilic addition of C4 to the 5,10 double bond then yields a tertiary monocyclic carbocation at C10.\n\nThe C10 carbocation adds to the 8,9 double bond, giving a C8 tertiary bicyclic carbocation.\n\nFurther intramolecular addition of the C8 carbocation to the 13,14 double bond occurs with non-Markovnikov regiochemistry and gives a tricyclic secondary carbocation at C13.\n\n4 The fourth and final cyclization occurs by addition of the C13 cation to the 17,20 double bond, giving the protosteryl cation with $17 \\beta$ stereochemistry.\n\nHydride migration from C17 to C20 occurs, establishing $R$ stereochemistry at C20.\n\n(3S)-2,3-Oxidosqualene\n\n\n2 $\\downarrow$\n\n(3)\n\n(4)\n\n\nProtosteryl cation\n(5)\n\nFIGURE 27.16"}
{"id": 1794, "contents": "Protosteryl cation - \n\u00a9\n\n\u00a9\nfrom C13 to C17, establishing the final $17 \\beta$ stereochemistry of the side chain.\n\n\nMethyl migration from C14 to C13 occurs.\n(0)\n\n\n8 A second methyl migration occurs, from C8 to C14.\n(8)\n\n\nLoss of a proton from C9 forms an 8,9 double bond and gives lanosterol.\n\u00a9\n\n\n\nStep 4 of FIGURE 27.15: Final Cyclization\nThe fourth and last cyclization occurs in step 4 by addition of the cationic center at C13 to the 17,20 double bond, giving what is known as the protosteryl cation. The side-chain alkyl group at C17 has $\\beta$ (up) stereochemistry, although this stereochemistry is lost in step 5 and then reset in step 6.\n\n\nSteps 5-9 of FIGURE 27.15: Carbocation Rearrangements\nOnce the tetracyclic carbon skeleton of lanosterol has been formed, a series of carbocation rearrangements occur (Section 7.11). The first rearrangement, hydride migration from C 17 to C 20 , occurs in step 5 and results in establishment of $R$ stereochemistry at C20 in the side chain. In step 6, a second hydride migration occurs from C13 to C17 on the $\\alpha$ (bottom) face of the ring and reestablishes the $17 \\beta$ orientation of the side chain. Finally, two methyl migrations, the first from C 14 to C 13 on the top $(\\beta)$ face and the second from C 8 to C 14 on the bottom $(\\alpha)$ face, place the positive charge at C8. A basic histidine residue in the enzyme then removes the neighboring $\\beta$ proton from C9 to give lanosterol.\n\n\nFrom lanosterol, the pathway for steroid biosynthesis continues on to yield cholesterol. Cholesterol then becomes a branch point, serving as the common precursor from which all other steroids are derived.\n\n\nPROBLEM Compare the structures of lanosterol and cholesterol, and catalog the changes needed for the 27-10 transformation shown."}
{"id": 1795, "contents": "Saturated Fats, Cholesterol, and Heart Disease - \nWe hear a lot these days about the relationship between saturated fats, cholesterol, and heart disease. What are the facts? It's well established that a diet rich in saturated animal fats often leads to an increase in blood serum cholesterol, particularly in sedentary, overweight individuals. Conversely, a diet lower in saturated fats and higher in polyunsaturated fats leads to a lower serum cholesterol level. Studies have shown that a serum cholesterol level greater than $240 \\mathrm{mg} / \\mathrm{dL}$ (a desirable value is $<200 \\mathrm{mg} / \\mathrm{dL}$ ) is correlated with an increased incidence of coronary artery disease, in which cholesterol deposits build up on the inner walls of coronary arteries, blocking the flow of blood to the heart muscles.\n\nA better indication of a person's risk of heart disease comes from a measurement of blood lipoprotein levels. Lipoproteins are complex molecules with both lipid and protein components that transport lipids through the body. They can be divided into three types according to density, as shown in TABLE 27.3. Very-low-density lipoproteins (VLDLs) act primarily as carriers of triglycerides from the intestines to peripheral tissues, whereas low-density lipoproteins (LDLs) and high-density lipoproteins (HDLs) act as carriers of cholesterol to and from the liver.\n\n\nFIGURE 27.17 It's hard to resist, but a high intake of saturated animal fat doesn't do much for your cholesterol level. (credit: modification of work \"food-chicken-meat-outdoors\" by www.pixel.la/Flickr, Public Domain)\n\nEvidence suggests that LDLs transport cholesterol as its fatty-acid ester to peripheral tissues, whereas HDLs remove cholesterol as its stearate ester from dying cells. If LDLs deliver more cholesterol than is needed, and if insufficient HDLs are present to remove it, the excess is deposited in arteries. Thus, a low level of low-density lipoproteins is good because it means that less cholesterol is being transported, and a high level of high-density lipoproteins is good because it means that more cholesterol is being removed. In addition, HDL contains an enzyme that has antioxidant properties, offering further protection against heart disease."}
{"id": 1796, "contents": "Saturated Fats, Cholesterol, and Heart Disease - \nAs a rule of thumb, a person's risk drops about $25 \\%$ for each increase of $5 \\mathrm{mg} / \\mathrm{dL}$ in HDL concentration. Normal values are $>40 \\mathrm{mg} / \\mathrm{dL}$ for men and $<50 \\mathrm{mg} / \\mathrm{dL}$ for women, perhaps explaining why premenopausal women appear to be somewhat less susceptible than men to heart disease.\n\nTABLE 27.3 Serum Lipoproteins\n\n| Name | Density (g/mL) | \\% Lipid | \\% Protein | Optimal (mg/dL) | Poor (mg/dL) |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| VLDL | $0.930-1.006$ | 90 | 10 | - | - |\n| LDL | $1.019-1.063$ | 75 | 25 | $<100$ | $>130$ |\n| HDL | $1.063-1.210$ | 67 | 33 | $>60$ | $<40$ |\n\nNot surprisingly, the most important factor in gaining high HDL levels is a generally healthful lifestyle. Obesity, smoking, and lack of exercise lead to low HDL levels, whereas regular exercise and a sensible diet lead to high HDL levels. Distance runners and other endurance athletes have HDL levels nearly $50 \\%$ higher than the general population. Failing that-some of us, but not everyone, wants to run 30 miles or bike 80 miles per week-diet is also important. Diets high in cold-water fish, like salmon and whitefish, raise HDL and lower blood cholesterol because these fish contain almost entirely polyunsaturated fat, including a large percentage of omega-3 fatty acids. Animal fat from red meat and cooking fats should be minimized because saturated fats and monounsaturated trans fats raise blood cholesterol."}
{"id": 1797, "contents": "Key Terms - \n- anabolic steroid\n- androgen\n- eicosanoid\n- estrogen\n- fatty acid\n- hormone\n- lipid\n- lipid bilayer\n- lipoprotein\n- micelle\n- mineralocorticoid\n- omega-3 fatty acid\n- phospholipid\n- polyunsaturated fatty acid\n- progestin\n- prostaglandin\n- saponification\n- steroid\n- terpenoid\n- triacylglycerol\n- wax"}
{"id": 1798, "contents": "Summary - \nLipids are the naturally occurring materials isolated from plants and animals by extraction with a nonpolar organic solvent. Animal fats and vegetable oils are the most widely occurring lipids. Both are triacylglycerols-triesters of glycerol with long-chain fatty acids. Animal fats are usually saturated, whereas vegetable oils usually have unsaturated fatty acid residues.\n\nPhospholipids are important constituents of cell membranes and are of two kinds. Glycerophospholipids, such as phosphatidylcholine and phosphatidylethanolamine, are closely related to fats in that they have a glycerol backbone esterified to two fatty acids (one saturated and one unsaturated) and to one phosphate ester. Sphingomyelins have the amino alcohol sphingosine for their backbone.\n\nEicosanoids and terpenoids are still other classes of lipids. Eicosanoids, of which prostaglandins are the most abundant kind, are derived biosynthetically from arachidonic acid, are found in all body tissues, and have a wide range of physiological activity. Terpenoids are often isolated from the essential oils of plants, have an immense diversity of structure, and are produced biosynthetically from the five-carbon precursor isopentenyl diphosphate (IPP). Isopentenyl diphosphate is itself biosynthesized from 3 equivalents of acetate in the mevalonate pathway.\n\nSteroids are plant and animal lipids with a characteristic tetracyclic carbon skeleton. Like the eicosanoids, steroids occur widely in body tissues and have a large variety of physiological activities. Steroids are closely related to terpenoids and arise biosynthetically from the triterpenoid lanosterol. Lanosterol, in turn, arises from cationic cyclization of the acyclic hydrocarbon squalene."}
{"id": 1799, "contents": "Visualizing Chemistry - \nPROBLEM The following model is that of cholic acid, a constituent of human bile. Locate the three hydroxyl\n27-11 groups, and identify each as axial or equatorial. Is cholic acid an A-B trans steroid or an A-B cis steroid?\n\n\nPROBLEM Propose a biosynthetic pathway for the sesquiterpenoid helminthogermacrene from farnesyl 27-12 diphosphate.\n\n\nPROBLEM Identify the following fatty acid, and tell whether it is more likely to be found in peanut oil or in red\n27-13 meat:"}
{"id": 1800, "contents": "Mechanism Problems - \nPROBLEM Propose a mechanistic pathway for the biosynthesis of caryophyllene, a substance found in clove 27-14 oil."}
{"id": 1801, "contents": "Caryophyllene - \nPROBLEM Suggest a mechanism by which $\\psi$-ionone is transformed into $\\beta$-ionone on treatment with acid. 27-15\n\n\nPROBLEM Isoborneol is converted into camphene on treatment with dilute sulfuric acid. Propose a 27-16 mechanism for the reaction, which involves a carbocation rearrangement.\n\n\nIsoborneol\n\n\nCamphene\n\nFats, Oils, and Related Lipids\nPROBLEM Fatty fish like salmon and albacore are rich in omega-3 fatty acids, which have a double bond three\n27-17 carbons in from the noncarboxyl end of the chain and have been shown to lower blood cholesterol levels. Draw the structure of $5,8,11,14,17$-eicosapentaenoic acid, a common example. $($ Eicosane $=$ $\\mathrm{C}_{20} \\mathrm{H}_{42}$.)\n\nPROBLEM Fats can be either optically active or optically inactive, depending on their structure. Draw the\n27-18 structure of an optically active fat that yields 2 equivalents of stearic acid and 1 equivalent of oleic acid on hydrolysis. Draw the structure of an optically inactive fat that yields the same products.\n\nPROBLEM Spermaceti, a fragrant substance from sperm whales, was widely used in cosmetics until it was\n27-19 banned in 1976 to protect whales from extinction. Chemically, spermaceti is cetyl palmitate, the ester of cetyl alcohol ( $n-\\mathrm{C}_{16} \\mathrm{H}_{33} \\mathrm{OH}$ ) with palmitic acid. Draw its structure."}
{"id": 1802, "contents": "Caryophyllene - \nPROBLEM Show the products you would expect to obtain from reaction of glyceryl trioleate with the following\n27-20 reagents:\n(a) Excess $\\mathrm{Br}_{2}$ in $\\mathrm{CH}_{2} \\mathrm{Cl}_{2}$\n(b) $\\mathrm{H}_{2} / \\mathrm{Pd}$\n(c) $\\mathrm{NaOH} / \\mathrm{H}_{2} \\mathrm{O}$\n(d) $\\mathrm{O}_{3}$, then $\\mathrm{Zn} / \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$\n(e) $\\mathrm{LiAlH}_{4}$, then $\\mathrm{H}_{3} \\mathrm{O}^{+}$\n$\\mathrm{CH}_{3} \\mathrm{MgBr}$, then $\\mathrm{H}_{3} \\mathrm{O}^{+}$\n\nPROBLEM How would you convert oleic acid into the following substances?\n27-21 (a) Methyl oleate\n(b) Methyl stearate (c) Nonanal\n(d) Nonanedioic acid\n(e) 9-Octadecynoic acid (stearolic acid) (f) 2-Bromostearic acid\n(g) 18-Pentatriacontanone, $\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{16} \\mathrm{CO}\\left(\\mathrm{CH}_{2}\\right)_{16} \\mathrm{CH}_{3}$\n\nPROBLEM Plasmalogens are a group of lipids found in nerve and muscle cells. How do plasmalogens differ 27-22 from fats?"}
{"id": 1803, "contents": "A plasmalogen - \nPROBLEM What products would you obtain from hydrolysis of a plasmalogen (Problem 27-22) with aqueous 27-23 NaOH ? With $\\mathrm{H}_{3} \\mathrm{O}^{+}$?\n\nPROBLEM Cardiolipins are a group of lipids found in heart muscles. What products would be formed if all ester 27-24\nbonds, including phosphates, were saponified by treatment with aqueous NaOH ?\n\n\nA cardiolipin\n\nPROBLEM Stearolic acid, $\\mathrm{C}_{18} \\mathrm{H}_{32} \\mathrm{O}_{2}$, yields stearic acid on catalytic hydrogenation and undergoes oxidative\n27-25 cleavage with ozone to yield nonanoic acid and nonanedioic acid. What is the structure of stearolic acid?\n\nPROBLEM How would you synthesize stearolic acid (Problem 27-25) from 1-decyne and\n27-26 1-chloro-7-iodoheptane?\nTerpenoids and Steroids\nPROBLEM Without proposing an entire biosynthetic pathway, draw the appropriate precursor, either geranyl\n27-27 diphosphate or farnesyl diphosphate, in a conformation that shows a likeness to each of the following terpenoids:\n\n\n\nGuaiol\n(b)\n\n\nSabinene\n\nPROBLEM Indicate by asterisks the chirality centers present in each of the terpenoids shown in Problem\n27-28 27-27. What is the maximum possible number of stereoisomers for each?\nPROBLEM Assume that the three terpenoids in Problem 27-27 are derived biosynthetically from isopentenyl\n27-29 diphosphate and dimethylallyl diphosphate, each of which was isotopically labeled at the diphosphate-bearing carbon atom (C1). At what positions would the terpenoids be isotopically labeled?\n\nPROBLEM Assume that acetyl CoA containing a ${ }^{14} \\mathrm{C}$ isotopic label in the carboxyl carbon atom is used as\n27-30 starting material for the biosynthesis of mevalonate, as shown in Figure 27.8. At what positions in mevalonate would the isotopic label appear?"}
{"id": 1804, "contents": "A plasmalogen - \nPROBLEM Assume that acetyl CoA containing a ${ }^{14} \\mathrm{C}$ isotopic label in the carboxyl carbon atom is used as\n27-31 starting material and that the mevalonate pathway is followed. Identify the positions in $\\alpha$-cadinol where the label would appear.\n\n\nPROBLEM Assume that acetyl CoA containing a ${ }^{14} \\mathrm{C}$ isotopic label in the carboxyl carbon atom is used as\n27-32 starting material and that the mevalonate pathway is followed. Identify the positions in squalene\nwhere the label would appear.\n\n\nSqualene\nPROBLEM Assume that acetyl CoA containing a ${ }^{14} \\mathrm{C}$ isotopic label in the carboxyl carbon atom is used as 27-33 starting material and that the mevalonate pathway is followed. Identify the positions in lanosterol where the label would appear.\n\n\nLanosterol"}
{"id": 1805, "contents": "General Problems - \nPROBLEM Flexibilene, a compound isolated from marine coral, is the first known terpenoid to contain a\n27-34 15-membered ring. What is the structure of the acyclic biosynthetic precursor of flexibilene? Show the mechanistic pathway for the biosynthesis.\n\n\nFlexibilene\n\nPROBLEM Draw the most stable chair conformation of dihydrocarvone.\n27-35\n\n\nPROBLEM Draw the most stable chair conformation of menthol, and label each substituent as axial or\n27-36 equatorial.\n\n\nMenthol (from peppermint oil)\n\nPROBLEM As a general rule, equatorial alcohols are esterified more readily than axial alcohols. What product\n27-37 would you expect to obtain from reaction of the following two compounds with 1 equivalent of acetic anhydride?\n(a)\n\n(b)\n\n\nPROBLEM Propose a mechanistic pathway for the biosynthesis of isoborneol. A carbocation rearrangement is\n27-38 needed at one point in the scheme."}
{"id": 1806, "contents": "Isoborneol - \nPROBLEM Digitoxigenin is a heart stimulant obtained from the purple foxglove Digitalis purpurea and used\n27-39 in the treatment of heart disease. Draw the three-dimensional conformation of digitoxigenin, and identify the two -OH groups as axial or equatorial.\n\n\nPROBLEM What product would you obtain by reduction of digitoxigenin (Problem 27-39) with $\\mathrm{LiAlH}_{4}$ ? By\n27-40 oxidation with the Dess-Martin periodinane?\nPROBLEM Vaccenic acid, $\\mathrm{C}_{18} \\mathrm{H}_{34} \\mathrm{O}_{2}$, is a rare fatty acid that gives heptanal and 11-oxoundecanoic acid\n27-41 $\\left[\\mathrm{OHC}\\left(\\mathrm{CH}_{2}\\right)_{9} \\mathrm{CO}_{2} \\mathrm{H}\\right]$ on ozonolysis followed by zinc treatment. When allowed to react with $\\mathrm{CH}_{2} \\mathrm{I}_{2} / \\mathrm{Zn}(\\mathrm{Cu})$, vaccenic acid is converted into lactobacillic acid. What are the structures of vaccenic and lactobacillic acids?\n\nPROBLEM Eleostearic acid, $\\mathrm{C}_{18} \\mathrm{H}_{30} \\mathrm{O}_{2}$, is a rare fatty acid found in the tung oil used for finishing furniture. On\n27-42 ozonolysis followed by treatment with zinc, eleostearic acid furnishes one part pentanal, two parts glyoxal ( $\\mathrm{OHC}-\\mathrm{CHO}$ ), and one part 9 -oxononanoic acid $\\left[\\mathrm{OHC}\\left(\\mathrm{CH}_{2}\\right)_{7} \\mathrm{CO}_{2} \\mathrm{H}\\right]$. What is the structure of eleostearic acid?\n\nPROBLEM Diterpenoids are derived biosynthetically from geranylgeranyl diphosphate (GGPP), which is itself\n27-43 biosynthesized by reaction of farnesyl diphosphate with isopentenyl diphosphate. Show the structure of GGPP, and propose a mechanism for its biosynthesis from FPP and IPP."}
{"id": 1807, "contents": "Isoborneol - \nPROBLEM Diethylstilbestrol (DES) has estrogenic activity even though it is structurally unrelated to steroids.\n27-44 Once used as an additive in animal feed, DES has been implicated as a causative agent in several types of cancer. Show how DES can be drawn so that it is sterically similar to estradiol.\n\n\nPROBLEM Cembrene, $\\mathrm{C}_{20} \\mathrm{H}_{32}$, a diterpenoid hydrocarbon isolated from pine resin, has a UV absorption at 245 27-45 nm, but dihydrocembrene $\\left(\\mathrm{C}_{20} \\mathrm{H}_{34}\\right)$, the product of hydrogenation with 1 equivalent of $\\mathrm{H}_{2}$, has no UV absorption. On exhaustive hydrogenation, 4 equivalents of $\\mathrm{H}_{2}$ react, and octahydrocembrene, $\\mathrm{C}_{20} \\mathrm{H}_{40}$, is produced. On ozonolysis of cembrene, followed by treatment of the ozonide with zinc, four carbonyl-containing products are obtained:\n\n\nPropose a structure for cembrene that is consistent with its formation from geranylgeranyl diphosphate.\n\nPROBLEM $\\alpha$-Fenchone is a pleasant-smelling terpenoid isolated from oil of lavender. Propose a pathway for\n27-46 the formation of $\\alpha$-fenchone from geranyl diphosphate. A carbocation rearrangement is required.\n\n\nPROBLEM Fatty acids are synthesized by a multistep route that starts with acetate. The first step is a reaction\n27-47 between protein-bound acetyl and malonyl units to give a protein-bound 3 -ketobutyryl unit. Show the mechanism, and tell what kind of reaction is occurring.\n\n\n3-Ketobutyryl-protein\nPROBLEM Propose a mechanism for the biosynthesis of the sesquiterpenoid trichodiene from farnesyl 27-48 diphosphate. The process involves cyclization to give an intermediate secondary carbocation, followed by several carbocation rearrangements.\n\n\nFarnesyl diphosphate (FPP)\n\n\nTrichodiene"}
{"id": 1808, "contents": "CHAPTER 28 - \nBiomolecules: Nucleic Acids\n\n\nFIGURE 28.1 If these Afghan hounds look similar, that's because they're identical-all cloned from somatic cells of the same donor. (credit: modification of work \"Figure 1\" in \"Birth of clones of the world's first cloned dog\" by Kim, M.J., Oh, H.J., Kim, G.A. et al./Sci Rep 7, 15235 (2017), CC BY 4.0)"}
{"id": 1809, "contents": "CHAPTER CONTENTS - \n28.1 Nucleotides and Nucleic Acids\n28.2 Base Pairing in DNA\n28.3 Replication of DNA\n28.4 Transcription of DNA\n28.5 Translation of RNA: Protein Biosynthesis\n28.6 DNA Sequencing\n28.7 DNA Synthesis\n28.8 The Polymerase Chain Reaction\n\nWHY THIS CHAPTER? Nucleic acids are the last of the four major classes of biomolecules we'll consider. So much has been written and spoken about DNA in the media that the basics of DNA replication and transcription are probably known to you. Thus, we'll move fairly quickly through the fundamentals and then look more closely at the chemical details of DNA sequencing, synthesis, and metabolism. This field is moving rapidly, and there's a lot you may not be familiar with.\n\nThe nucleic acids, deoxyribonucleic acid (DNA) and ribonucleic acid (RNA), are the chemical carriers of a cell's genetic information. Coded in a cell's DNA is the information that determines the nature of the cell, controls its growth and division, and directs biosynthesis of the enzymes and other proteins required for cellular functions.\n\nIn addition to nucleic acids themselves, nucleic acid derivatives such as ATP are involved as phosphorylating agents in many biochemical pathways, and several important coenzymes, including NAD ${ }^{+}$, FAD, and coenzyme A, have nucleic acid components. See TABLE 26.3 for their structures."}
{"id": 1810, "contents": "CHAPTER CONTENTS - 28.1 Nucleotides and Nucleic Acids\nJust as proteins are biopolymers made of amino acids, the nucleic acids are biopolymers made of nucleotides, joined together to form a long chain. Each nucleotide is composed of a nucleoside bonded to a phosphate group, and each nucleoside is composed of an aldopentose sugar linked through its anomeric carbon to the nitrogen atom of a heterocyclic purine or pyrimidine base.\n\n\nThe sugar component in RNA is ribose, and the sugar in DNA is $2^{\\prime}$-deoxyribose. (In naming and numbering nucleotides, numbers with a prime superscript refer to positions on the sugar and numbers without a prime superscript refer to positions on the heterocyclic base. Thus, the prefix $2^{\\prime}$-deoxy indicates that oxygen is missing from C2' of ribose.) DNA contains four different amine bases: two substituted purines (A, adenine, and G, guanine) and two substituted pyrimidines ( $\\mathbf{C}$, cytosine, and $\\mathbf{T}$, thymine). Adenine, guanine, and cytosine also occur in RNA, but thymine is replaced in RNA by a closely related pyrimidine base called $\\mathbf{U}$, uracil.\n\n\nRibose\n\n\n2-Deoxyribose\n\n\nPurine\n\n\n\nThymine (T) DNA\n\n\nUracil (U) RNA\n\nThe structures of the four deoxyribonucleotides and the four ribonucleotides are shown in FIGURE 28.2. Although similar chemically, DNA and RNA differ dramatically in size. Molecules of DNA are enormous, containing as many as 245 million nucleotides and having molecular weights as high as 75 billion. Molecules of RNA, by contrast, are much smaller, containing as few as 21 nucleotides and having molecular weights as low as 7000 .\n\n\n2'-Deoxyadenosine 5'-phosphate\nDeoxyribonucleotides\n\nC\n\n\n2'-Deoxycytidine 5'-phosphate\n\n\n2'-Deoxyguanosine 5'-phosphate\n\nT\n\n\nThymidine 5'-phosphate\n\n\nAdenosine 5'-phosphate\nRibonucleotides\n\n\nCytidine 5'-phosphate"}
{"id": 1811, "contents": "CHAPTER CONTENTS - 28.1 Nucleotides and Nucleic Acids\n2'-Deoxyguanosine 5'-phosphate\n\nT\n\n\nThymidine 5'-phosphate\n\n\nAdenosine 5'-phosphate\nRibonucleotides\n\n\nCytidine 5'-phosphate\n\n\nGuanosine 5'-phosphate\n\nU\n\n\nUridine $5^{\\prime}$-phosphate\n\nFIGURE 28.2 Structures of the four deoxyribonucleotides and the four ribonucleotides.\nNucleotides are linked together in DNA and RNA by phosphodiester bonds [ $\\mathrm{RO}-\\left(\\mathrm{PO}_{2}{ }^{-}\\right)-\\mathrm{OR}$ '] between phosphate, the $5^{\\prime}$-hydroxyl group on one nucleoside, and the $3^{\\prime}$-hydroxyl group on another nucleoside. One end of the nucleic acid polymer has a free hydroxyl at C3' (the $\\mathbf{3}^{\\prime}$ end), and the other end has a phosphate at $\\mathbf{C 5}^{\\prime}$ (the $\\mathbf{5}^{\\prime}$ end). The sequence of nucleotides in a chain is described by starting at the $5^{\\prime}$ end and identifying the bases in order of occurrence, using the abbreviations G, C, A, T (or U in RNA). Thus, a typical DNA sequence might be written as TAGGCT.\n\n\nPROBLEM Draw the full structure of the DNA dinucleotide AG. 28-1\n\nPROBLEM Draw the full structure of the RNA dinucleotide UA.\n28-2"}
{"id": 1812, "contents": "CHAPTER CONTENTS - 28.2 Base Pairing in DNA\nSamples of DNA isolated from different tissues of the same species have the same proportions of heterocyclic bases, but samples from different species often have greatly differing proportions of bases. Human DNA, for example, contains about $30 \\%$ each of adenine and thymine and about $20 \\%$ each of guanine and cytosine. The bacterium Clostridium perfringens, however, contains about $37 \\%$ each of adenine and thymine and only $13 \\%$ each of guanine and cytosine. Note that in both examples the bases occur in pairs. Adenine and thymine are present in equal amounts, as are cytosine and guanine. Why?\n\nIn 1953, Rosalind Franklin, Maurice Wilkins, James Watson, and Francis Crick published scientific reports describing the secondary structure of DNA. According to their model, DNA under physiological conditions consists of two polynucleotide strands, running in opposite directions and coiled around each other in a double helix like the handrails on a spiral staircase. The two strands are complementary rather than identical and are held together by hydrogen bonds between specific pairs of bases, A with $T$ and $C$ with $G$. That is, whenever an A base occurs in one strand, a T base occurs opposite it in the other strand; when a C base occurs in one, a G occurs in the other (FIGURE 28.3). This complementary base-pairing thus explains why A and $T$ are always found in equal amounts, as are $G$ and $C$.\n\nA\nT\n\n\nG\n\nFIGURE 28.3 Hydrogen-bonding between base pairs in the DNA double helix. Electrostatic potential maps show that the faces of the bases are relatively neutral, while the edges have positive and negative regions. Pairing $G$ with $C$ and $A$ with $T$ brings together oppositely charged regions."}
{"id": 1813, "contents": "CHAPTER CONTENTS - 28.2 Base Pairing in DNA\nA full turn of a DNA double helix is shown in FIGURE 28.4. The helix is $20 \\AA$ wide, there are 10 base pairs per turn, and each turn is 34 \u00c5 in length. Notice in FIGURE 28.4 that the two strands of the double helix coil in such a way that two kinds of \"grooves\" result, a major groove $12 \\AA$ wide and a minor groove $6 \\AA$ wide. The major groove is slightly deeper than the minor groove, and both are lined by flat heterocyclic bases. As a result, a variety of other polycyclic aromatic molecules are able to slip sideways, or intercalate, between the stacked bases. Many cancer-causing and cancer-preventing agents function by interacting with DNA in this way.\n\n\nFIGURE 28.4 A turn of the DNA double helix in both space-filling and wire-frame formats. The sugar-phosphate backbone runs along the outside of the helix, and the amine bases hydrogen bond to one another on the inside. Both major and minor grooves are visible.\n\nAn organism's genetic information is stored as a sequence of deoxyribonucleotides strung together in the DNA chain. For the information to be preserved and passed on to future generations, a mechanism must exist for\ncopying DNA. For the information to be used, a mechanism must exist for decoding the DNA message and implementing the instructions it contains.\n\nWhat Crick called the \"central dogma of molecular genetics\" says that the function of DNA is to store information and pass it on to RNA. The function of RNA, in turn, is to read, decode, and use the information received from DNA to make proteins. This view is greatly oversimplified but is nevertheless a good place to start. Three fundamental processes take place.\n\n- Replication is the process by which identical copies of DNA are made so that information can be preserved and handed down to offspring.\n- Transcription is the process by which genetic messages are read and carried out of the cell nucleus to ribosomes, where protein synthesis occurs.\n- Translation is the process by which the genetic messages are decoded and used to synthesize proteins."}
{"id": 1814, "contents": "Predicting the Complementary Base Sequence in Double-Stranded DNA - \nWhat sequence of bases on one strand of DNA is complementary to the sequence TATGCAT on another strand?"}
{"id": 1815, "contents": "Strategy - \nRemember that A and G form complementary pairs with T and C , respectively, and then go through the sequence replacing A by T, G by C, T by A, and C by G. Remember also that the $5^{\\prime}$ end is on the left and the $3^{\\prime}$ end is on the right in the original strand."}
{"id": 1816, "contents": "Solution - \nOriginal: (5') TATGCAT (3')\nComplement: (3') ATACGTA (5') or (5') ATGCATA (3')\n\nPROBLEM What sequence of bases on one strand of DNA is complementary to the following sequence on 28-3 another strand?\n(5') GGCTAATCCGT (3')"}
{"id": 1817, "contents": "Solution - 28.3 Replication of DNA\nDNA replication is an enzyme-catalyzed process that begins with a partial unwinding of the double helix at various points along the chain, brought about by enzymes called helicases. Hydrogen bonds are broken, the two strands separate to form a \"bubble,\" and bases are exposed. New nucleotides then line up on each strand in a complementary manner, A to T and G to C , and two new strands begin to grow from the ends of the bubble, called the replication forks. Each new strand is complementary to its old template strand, so two identical DNA double helices are produced (FIGURE 28.5). Because each of the new DNA molecules contains one old strand and one new strand, the process is described as semiconservative replication.\n\n\nFIGURE 28.5 A representation of semiconservative DNA replication. The original double-stranded DNA partially unwinds, bases are exposed, nucleotides line up on each strand in a complementary manner, and two new strands begin to grow. Both strands are synthesized in the same $5^{\\prime} \\rightarrow 3^{\\prime}$ direction, one continuously and one in fragments.\n\nAddition of nucleotides to the growing chain takes place in the $5^{\\prime} \\rightarrow 3^{\\prime}$ direction and is catalyzed by the DNA polymerase enzyme. The key step is the addition of a nucleoside $5^{\\prime}$-triphosphate to the free $3^{\\prime}$-hydroxyl group of the growing chain with the loss of a diphosphate leaving group."}
{"id": 1818, "contents": "Solution - 28.3 Replication of DNA\nBecause both new DNA strands are synthesized in the $5^{\\prime} \\rightarrow 3^{\\prime}$ direction, they can't be made in exactly the same way. One new strand must have its $3^{\\prime}$ end nearer a replication fork, while the other new strand has its $5^{\\prime}$ end nearer the replication fork. What happens is that the complement of the original $5^{\\prime} \\rightarrow 3^{\\prime}$ strand is synthesized continuously in a single piece to give a newly synthesized copy called the leading strand, while the complement of the original $3^{\\prime} \\rightarrow 5^{\\prime}$ strand is synthesized discontinuously in small pieces called Okazaki fragments (named for Tsuneko Okazaki, in recognition of her discovery of them) that are subsequently linked by DNA ligases to form the lagging strand.\n\nThe magnitude of the replication process is staggering. The nucleus of every human cell contains 2 copies of 22 chromosomes plus an additional 2 sex chromosomes, for a total of 46 . Each chromosome consists of one very large DNA molecule, and the sum of the DNA in each of the two sets of chromosomes is estimated to be 3.0 billion base pairs, or 6.0 billion nucleotides. Despite the size of these enormous molecules, their base sequence\nis faithfully copied during replication. The entire copying process takes only a few hours and, after proofreading and repair, an error gets through only about once per 10 to 100 billion bases. In fact, only about 60 of these random mutations are passed on from parent to child per human generation."}
{"id": 1819, "contents": "Solution - 28.4 Transcription of DNA\nAs noted previously, RNA is structurally similar to DNA but contains ribose rather than deoxyribose and uracil rather than thymine. RNA has three major types, each of which serves a specific purpose. In addition, there are a number of small RNAs that appear to control a wide variety of important cellular functions. All RNA molecules are much smaller than DNA, and all remain single-stranded rather than double-stranded.\n\n- Messenger RNA (mRNA) carries genetic messages from DNA to ribosomes, small granular particles in the cytoplasm of a cell where protein synthesis takes place.\n- Ribosomal RNA (rRNA) complexed with protein provides the physical makeup of the ribosomes.\n- Transfer RNA (tRNA) transports amino acids to the ribosomes, where they are joined together to make proteins.\n- Small RNAs, also called functional RNAs, have a variety of functions within the cell, including silencing transcription and catalyzing chemical modifications of other RNA molecules.\n\nThe genetic information in DNA is contained in segments called genes, each of which consists of a specific nucleotide sequence that encodes a specific protein. The conversion of that information from DNA into proteins begins in the nucleus of cells with the synthesis of mRNA by transcription of DNA. In bacteria, the process begins when RNA polymerase recognizes and binds to a promoter sequence on DNA, typically consisting of around 40 base pairs located upstream ( 5 ') of the transcription start site. Within the promoter are two hexameric consensus sequences, one located 10 base pairs upstream of the start and the second located 35 base pairs upstream.\n\nFollowing formation of the polymerase-promoter complex, several turns of the DNA double helix untwist, forming a bubble and exposing 14 or so base pairs of the two strands. Appropriate ribonucleotides then line up by hydrogen-bonding to their complementary bases on DNA, bond formation occurs in the $5^{\\prime} \\rightarrow 3^{\\prime}$ direction, the RNA polymerase moves along the DNA chain, and the growing RNA molecule unwinds from DNA (FIGURE 28.6). At any one time, about 12 base pairs of the growing RNA remain hydrogen-bonded to the DNA template."}
{"id": 1820, "contents": "Solution - 28.4 Transcription of DNA\nFIGURE 28.6 Biosynthesis of RNA using a DNA segment as a template.\nUnlike what happens in DNA replication, where both strands are copied, only one of the two DNA strands is transcribed into mRNA. The DNA strand that contains the gene is often called the sense strand, or coding strand, and the DNA strand that gets transcribed to give RNA is called the antisense strand, or noncoding strand. Because the sense strand and the antisense strand in DNA are complementary, and because the DNA antisense strand and the newly formed RNA strand are also complementary, the RNA molecule produced during transcription is a copy of the DNA sense strand. That is, the complement of the complement is the same as the original. The only difference is that the RNA molecule has a U everywhere that the DNA sense strand has a T.\n\nAnother part of the picture in vertebrates and flowering plants is that genes are often not continuous segments of the DNA chain. Instead, a gene will begin in one small section of DNA called an exon, then be interrupted\nby a noncoding section called an intron, and then take up again farther down the chain in another exon. The final mRNA molecule results only after the noncoded sections are cut out of the transcribed mRNA and the remaining pieces are joined together by spliceosome enzymes. The gene for triose phosphate isomerase in maize, for instance, contains eight noncoding introns accounting for approximately $70 \\%$ of the DNA base pairs and nine coding exons accounting for only $30 \\%$ of the base pairs.\n\nPROBLEM Show how uracil can form strong hydrogen bonds to adenine.\n28-4\nPROBLEM What RNA base sequence is complementary to the following DNA base sequence?\n28-5\n(5') GATTACCGTA (3')\n\nPROBLEM From what DNA base sequence was the following RNA sequence transcribed?\n28-6\n(5') UUCGCAGAGU (3')"}
{"id": 1821, "contents": "Solution - 28.5 Translation of RNA: Protein Biosynthesis\nThe primary cellular function of mRNA is to direct the biosynthesis of the thousands of diverse peptides and proteins required by an organism-as many as 150,000 in a human. The mechanics of protein biosynthesis take place on ribosomes, small granular particles in the cytoplasm of a cell that consist of about $60 \\%$ ribosomal RNA and $40 \\%$ protein.\n\nThe specific ribonucleotide sequence in mRNA forms a message that determines the order in which amino acid residues are to be joined. Each \"word,\" or codon, along the mRNA chain consists of a sequence of three ribonucleotides that is specific for a given amino acid. For example, the series UUC on mRNA is a codon directing incorporation of the amino acid phenylalanine into the growing protein. Of the $4^{3}=64$ possible triplets of the four bases in RNA, 61 code for specific amino acids and 3 code for chain termination. TABLE 28.1 shows the meaning of each codon.\n\n|  |  | Third base ( 3 ' end) |  |  |  |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| First base (5' end) | Second base | U | C | A | G |\n| U | U | Phe | Phe | Leu | Leu |\n|  | C | Ser | Ser | Ser | Ser |\n|  | A | Tyr | Tyr | Stop | Stop |\n|  | G | Cys | Cys | Stop | Trp |\n| C | U | Leu | Leu | Leu | Leu |\n|  | C | Pro | Pro | Pro | Pro |\n|  | A | His | His | Gln | Gln |\n|  | G | Arg | Arg | Arg | Arg |\n| A | U | Ile | Ile | Ile | Met |\n|  | C | Thr | Thr | Thr | Thr |"}
{"id": 1822, "contents": "Solution - 28.5 Translation of RNA: Protein Biosynthesis\n| TABLE 28.1 Codon Assignments of Base Triplets |  |  |  |  |  |  |  |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| Third base (3' end) |  |  |  |  |  |  |  |\n| First base (5' end) | Second base | U | C | A | G |  |  |\n|  | A | Asn | Asn | Lys | Lys |  |  |\n|  | G | Ser | Ser | Arg | Arg |  |  |\n| G | U | Val | Val | Val | Val |  |  |\n|  | C | Ala | Ala | Ala | Ala |  |  |\n|  | A | Asp | Asp | Glu | Glu |  |  |"}
{"id": 1823, "contents": "Solution - 28.5 Translation of RNA: Protein Biosynthesis\nThe message embedded in mRNA is read by transfer RNA (tRNA) in a process called translation. There are 61 different tRNAs, one for each of the 61 codons that specify an amino acid. A typical tRNA is single-stranded and has roughly the shape of a cloverleaf, as shown in FIGURE 28.7. It consists of about 70 to 100 ribonucleotides and is bonded to a specific amino acid by an ester linkage through the $3^{\\prime}$ hydroxyl on ribose at the $3^{\\prime}$ end of the tRNA. Each tRNA also contains on its middle leaf a segment called an anticodon, a sequence of three ribonucleotides complementary to the codon sequence. For example, the codon sequence UUC present on mRNA is read by a phenylalanine-bearing tRNA having the complementary anticodon base sequence GAA. [Remember that nucleotide sequences are written in the $5^{\\prime} \\rightarrow 3^{\\prime}$ direction, so the sequence in an anticodon must be reversed. That is, the complement to $\\left(5^{\\prime}\\right)$-UUC- $\\left(3^{\\prime}\\right)$ is ( $\\left.3^{\\prime}\\right)$-AAG- $\\left(5^{\\prime}\\right)$, which is written as $\\left(5^{\\prime}\\right)$-GAA- $\\left(3^{\\prime}\\right)$.]\n\n\nFIGURE 28.7 Structure of a tRNA molecule. The tRNA is a roughly cloverleaf-shaped molecule containing an anticodon triplet on one \"leaf\" and an amino acid unit attached covalently at its $\\mathbf{3 '}^{\\prime}$ end. The example shown is a yeast tRNA that codes for phenylalanine. The\nnucleotides not specifically identified are chemically modified analogs of the four common nucleotides.\nAs each successive codon on mRNA is read, different tRNAs bring the correct amino acids into position for enzyme-mediated transfer to the growing peptide. When synthesis of the proper protein is completed, a \"stop\" codon signals the end, and the protein is released from the ribosome. This process is illustrated in FIGURE 28.8."}
{"id": 1824, "contents": "Solution - 28.5 Translation of RNA: Protein Biosynthesis\nFIGURE 28.8 A representation of protein biosynthesis. The codon base sequences on mRNA are read by tRNAs containing complementary anticodon base sequences. Transfer RNAs assemble the proper amino acids into position for incorporation into the growing peptide."}
{"id": 1825, "contents": "Predicting the Amino Acid Sequence Transcribed from DNA - \nWhat amino acid sequence is coded by the following segment of a DNA coding strand (sense strand)?\n(5') CTA-ACT-AGC-GGG-TCG-CCG (3')"}
{"id": 1826, "contents": "Strategy - \nThe mRNA produced during translation is a copy of the DNA coding strand, with each T replaced by U . Thus, the mRNA has the sequence\n(5') CUA-ACU-AGC-GGG-UCG-CCG (3')\nEach set of three bases forms a codon, whose meaning can be found in TABLE 28.1."}
{"id": 1827, "contents": "Solution - \nLeu-Thr-Ser-Gly-Ser-Pro.\n\nPROBLEM List codon sequences for the following amino acids:\n28-7 (a) Ala\n(b) Phe\n(c) Leu\n(d) Tyr\n\nPROBLEM List anticodon sequences on the tRNAs carrying the following amino acids.\n28-8\n(a) Ala\n(b) Phe\n(c) Leu\n(d) Tyr\n\nPROBLEM What amino acid sequence is coded by the following mRNA base sequence?"}
{"id": 1828, "contents": "CUU-AUG-GCU-UGG-CCC-UAA - \nPROBLEM What is the base sequence in the original DNA strand on which the mRNA sequence in Problem 28-10 28-9 was made?"}
{"id": 1829, "contents": "CUU-AUG-GCU-UGG-CCC-UAA - 28.6 DNA Sequencing\nA scientific revolution is now under way in molecular biology, as scientists learn how to manipulate and harness the genetic machinery of organisms. None of the extraordinary advances of the past several decades would have been possible, however, were it not for the discovery in 1977 of methods for sequencing immense DNA chains.\n\nThe first step in DNA sequencing is to cleave the enormous chain at known points to produce smaller, more manageable pieces, a task accomplished by the use of restriction endonucleases, or restriction enzymes. Each different restriction enzyme, of which more than 4000 are known and approximately 600 are commercially available, cleaves a DNA molecule at a point in the chain where a specific base sequence occurs. For example, the restriction enzyme AluI cleaves between $G$ and $C$ in the four-base sequence AG-CT. Note that the sequence is a palindrome, meaning that the sequence ( $5^{\\prime}$ )-AGCT-( $3^{\\prime}$ ) is the same as its complement ( $3^{\\prime}$ )-TCGA-( $5^{\\prime}$ ) when both are read in the same $5^{\\prime} \\rightarrow 3^{\\prime}$ direction. The same is true for other restriction endonucleases.\n\nIf the original DNA molecule is cut with another restriction enzyme having a different specificity for cleavage, still other segments are produced whose sequences partially overlap those produced by the first enzyme. Sequencing all the segments, followed by identification of the overlapping regions, allows for complete DNA sequencing.\n\nA dozen or so different methods of DNA sequencing are now available, and many others are under development. The Sanger dideoxy method is among the most frequently used and was the method responsible for first sequencing the entire human genome of 3.0 billion base pairs. In commercial sequencing instruments, the dideoxy method begins with a mixture of the following:"}
{"id": 1830, "contents": "CUU-AUG-GCU-UGG-CCC-UAA - 28.6 DNA Sequencing\n- The restriction fragment to be sequenced\n- A small piece of DNA called a primer, whose sequence is complementary to that on the $3^{\\prime}$ end of the restriction fragment\n- The four 2'-deoxyribonucleoside triphosphates (dNTPs)\n- Very small amounts of the four $2^{\\prime}, 3^{\\prime}$-dideoxyribonucleoside triphosphates (ddNTPs), each of which is labeled with a fluorescent dye of a different color (A $2^{\\prime}, 3^{\\prime}$-dideoxyribonucleoside triphosphate is one in which both $2^{\\prime}$ and $3^{\\prime}-\\mathrm{OH}$ groups are missing from ribose.)\n\n\nA 2'-deoxyribonucleoside triphosphate (dNTP)\n\n\nA 2', 3'-dideoxyribonucleoside triphosphate (ddNTP)\n\nDNA polymerase is added to the mixture, and a strand of DNA complementary to the restriction fragment begins to grow from the end of the primer. Most of the time, only normal deoxyribonucleotides are incorporated into the growing chain because of their much higher concentration in the mixture, but every so often, a dideoxyribonucleotide is incorporated. When that happens, DNA synthesis stops because the chain end no longer has a $3^{\\prime}$-hydroxyl group for adding further nucleotides.\n\nWhen reaction is complete, the product consists of a mixture of DNA fragments of all possible lengths, each terminated by one of the four dye-labeled dideoxyribonucleotides. This product mixture is then separated according to the size of the pieces by gel electrophoresis (Section 26.2), and the identity of the terminal dideoxyribonucleotide in each piece-and thus the sequence of the restriction fragment-is determined by\nnoting the color with which the attached dye fluoresces. FIGURE 28.9 shows a typical result.\n\n\nFIGURE 28.9 The sequence of a restriction fragment determined by the Sanger dideoxy method can be read simply by noting the colors of the dye attached to each of the various terminal nucleotides."}
{"id": 1831, "contents": "CUU-AUG-GCU-UGG-CCC-UAA - 28.6 DNA Sequencing\nFIGURE 28.9 The sequence of a restriction fragment determined by the Sanger dideoxy method can be read simply by noting the colors of the dye attached to each of the various terminal nucleotides.\n\nSo efficient is the automated dideoxy method that sequences up to 1100 nucleotides in length, with a throughput of up to 19,000 bases per hour, can be sequenced with $98 \\%$ accuracy. After a decade of work and a cost of about $\\$ 500$ million, preliminary sequence information for the entire human genome of 3.0 billion base pairs was announced early in 2001 and complete information was released in 2003 . More recently, the genome sequencing of individuals, including that of James Watson, one of the discoverers of the double helix, has been accomplished. The sequencing price per genome is dropping rapidly and is currently less than $\\$ 1,000$, meaning that the routine sequencing of individuals is within reach.\n\nRemarkably, our genome appears to contain only about 21,000 genes, less than one-fourth the previously predicted number and only about twice the number found in the common roundworm. It's also interesting to note that the number of genes in a human $(21,000)$ is much smaller than the number of kinds of proteins (perhaps 500,000). This discrepancy arises because most proteins are modified in various ways after translation (posttranslational modifications), so a single gene can ultimately give many different proteins."}
{"id": 1832, "contents": "CUU-AUG-GCU-UGG-CCC-UAA - 28.7 DNA Synthesis\nThe ongoing revolution in molecular biology has brought with it an increased demand for the efficient chemical synthesis of short DNA segments, called oligonucleotides, or simply oligos. The problems of DNA synthesis are similar to those of peptide synthesis (Section 26.7) but are more difficult because of the complexity of the nucleotide monomers. Each nucleotide has multiple reactive sites that must be selectively protected and deprotected at specific times, and coupling of the four nucleotides must be carried out in the proper sequence. Automated DNA synthesizers are available, however, that allow the fast and reliable synthesis of DNA segments up to 200 nucleotides in length.\n\nDNA synthesizers operate on a principle similar to that of the Merrifield solid-phase peptide synthesizer (Section 26.8). In essence, a protected nucleotide is covalently bonded to a solid support, and one nucleotide at a time is added to the growing chain by the use of a coupling reagent. After the final nucleotide has been added, all the protecting groups are removed and the synthetic DNA is cleaved from the solid support. Five steps are needed:"}
{"id": 1833, "contents": "STEP 1 - \nThe first step in DNA synthesis is to attach a protected deoxynucleoside to a silica $\\left(\\mathrm{SiO}_{2}\\right)$ support by an ester linkage to the $3^{\\prime}-\\mathrm{OH}$ group of the deoxynucleoside. Both the $5^{\\prime}-\\mathrm{OH}$ group on the sugar and free $-\\mathrm{NH}_{2}$ groups on the heterocyclic bases must be protected. Adenine and cytosine bases are protected by benzoyl groups, guanine is protected by an isobutyryl group, and thymine requires no protection. The deoxyribose $5^{\\prime}-\\mathrm{OH}$ is protected as its p-dimethoxytrityl (DMT) ether.\n\n\n\n\nadenine\n\n\nN-protected\nguanine\n\n\nN-protected cytosine\n\n\nThymine\n\nSTEP 2\nThe second step is removal of the DMT protecting group by treatment with dichloroacetic acid in $\\mathrm{CH}_{2} \\mathrm{Cl}_{2}$. The reaction occurs by an $\\mathrm{S}_{\\mathrm{N}} 1$ mechanism and proceeds rapidly because of the stability of the tertiary, benzylic dimethoxytrityl cation.\n\n\nSTEP 3\nThe third step is the coupling of the polymer-bonded deoxynucleoside with a protected deoxynucleoside containing a phosphoramidite group $\\left[\\mathrm{R}_{2} \\mathrm{NP}(\\mathrm{OR})_{2}\\right]$ at its $3^{\\prime}$ position. The coupling reaction takes place in the polar aprotic solvent acetonitrile, requires catalysis by the heterocyclic amine tetrazole, and yields a phosphite, $\\mathrm{P}(\\mathrm{OR})_{3}$, as product. Note that one of the phosphorus oxygen atoms is protected by a $\\beta$-cyanoethyl group, $-\\mathrm{OCH}_{2} \\mathrm{CH}_{2} \\mathrm{C} \\equiv \\mathrm{N}$. The coupling step takes place with better than $99 \\%$ yield."}
{"id": 1834, "contents": "STEP 4 - \nWith the coupling accomplished, the phosphite product is oxidized to a phosphate by treatment with iodine in aqueous tetrahydrofuran in the presence of 2,6-dimethylpyridine. The cycle of (1) deprotection, (2) coupling, and (3) oxidation is then repeated until an oligonucleotide chain of the desired sequence has been constructed.\n\n\nSTEP 5\nThe final step is removal of all protecting groups and cleavage of the ester bond holding the DNA to the silica. All these reactions are done at the same time by treatment with aqueous $\\mathrm{NH}_{3}$. Purification by electrophoresis then yields the synthetic DNA.\n\n\nPROBLEM p-Dimethoxytrityl (DMT) ethers are easily cleaved by mild acid treatment. Show the mechanism of 28-11 the cleavage reaction.\n\nPROBLEM Propose a mechanism to account for cleavage of the $\\beta$-cyanoethyl protecting group from the 28-12 phosphate groups on treatment with aqueous ammonia. (Acrylonitrile, $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCN}$, is a byproduct.) What kind of reaction is occurring?"}
{"id": 1835, "contents": "STEP 4 - 28.8 The Polymerase Chain Reaction\nIt often happens that only a tiny amount of DNA can be obtained directly, as might occur at a crime scene, so methods for obtaining larger amounts are sometimes needed to carry out sequencing and characterization. The invention of the polymerase chain reaction (PCR) by Kary Mullis in 1986 has been described as being to genes what Gutenberg's invention of the printing press was to the written word. Just as the printing press produces multiple copies of a book, PCR produces multiple copies of a given DNA sequence. Starting from less than 1 picogram ( pg ) of DNA with a chain length of 10,000 nucleotides ( $1 \\mathrm{pg}=10^{-12} \\mathrm{~g}$; about $10^{5}$ molecules), PCR makes it possible to obtain several micrograms ( $1 \\mu \\mathrm{~g}=10^{-6} \\mathrm{~g}$; about $10^{11}$ molecules) in just a few hours.\n\nThe key to the polymerase chain reaction is Taq DNA polymerase, a heat-stable enzyme isolated from the thermophilic bacterium Thermus aquaticus found in a hot spring in Yellowstone National Park. Taq polymerase is able to take a single strand of DNA having a short, primer segment of complementary chain at one end and then finish constructing the entire complementary strand. The overall process takes three steps, as shown in FIGURE 28.10. More recently, improved heat-stable DNA polymerases have become available, including Vent polymerase and Pfu polymerase, both isolated from bacteria growing near geothermal vents in the ocean floor. The error rate of both enzymes is substantially less than that of Taq.\n\n\nFIGURE 28.10 The polymerase chain reaction. Details are explained in the text.\nSTEP 1\nThe double-stranded DNA to be amplified is heated in the presence of Taq polymerase, $\\mathrm{Mg}^{2+}$ ion, the four deoxynucleotide triphosphate monomers (dNTPs), and a large excess of two short oligonucleotide primers of about 20 bases each. Each primer is complementary to the sequence at the end of one of the target DNA segments. At a temperature of $95^{\\circ} \\mathrm{C}$, double-stranded DNA denatures, spontaneously breaking apart into two single strands."}
{"id": 1836, "contents": "STEP 4 - 28.8 The Polymerase Chain Reaction\nSTEP 2\nThe temperature is lowered to between 37 and $50^{\\circ} \\mathrm{C}$, allowing the primers, because of their relatively high concentration, to anneal by hydrogen-bonding to their complementary sequence at the end of each target strand.\n\nSTEP 3\nThe temperature is then raised to $72^{\\circ} \\mathrm{C}$, and Taq polymerase catalyzes the addition of further nucleotides to the two primed DNA strands. When replication of each strand is complete, two copies of the original DNA now exist. Repeating the denature-anneal-synthesize cycle a second time yields four DNA copies, repeating a third time yields eight copies, and so on, in an exponential series.\n\nPCR has been automated, and 30 or so cycles can be carried out in an hour, resulting in a theoretical amplification factor of $2^{30}\\left(\\sim 10^{9}\\right)$. In practice, however, the efficiency of each cycle is less than $100 \\%$, and an experimental amplification of about $10^{6}$ to $10^{8}$ is routinely achieved for 30 cycles."}
{"id": 1837, "contents": "DNA Fingerprinting - \nThe invention of DNA sequencing has affected society in many ways, few more dramatic than those stemming from the development of DNA fingerprinting. DNA fingerprinting arose from the discovery in 1984 that human genes contain short, repeating sequences of noncoding DNA, called short tandem repeat (STR) loci. Furthermore, the STR loci are slightly different for everyone except identical twins. By sequencing these loci, a pattern unique to each person can be obtained.\n\nPerhaps the most common and well-publicized use of DNA fingerprinting is that carried out by crime laboratories to link suspects to biological evidence-blood, hair follicles, skin, or semen-found at a crime scene. Many thousands of court cases have been decided based on DNA evidence.\n\nFor use in criminal cases, forensic laboratories in the United States have agreed on 13 core STR loci that are most accurate for the identification of an individual. Based on these 13 loci, a Combined DNA Index System (CODIS) has been established to serve as a registry of convicted offenders. When a DNA sample is obtained from a crime scene, the sample is subjected to cleavage with restriction endonucleases to cut out fragments containing the STR loci, the fragments are amplified using the polymerase chain reaction, and the sequences of the fragments are determined.\n\nIf the profile of sequences from a known individual and the profile from DNA obtained at a crime scene match, the probability is approximately 82 billion to 1 that the DNA is from the same individual. In paternity cases, where the DNA of male parent and offspring are related but not fully identical, the identity of the male parent can be established with a probability of around 100,000 to 1 . Even after several generations, paternity can still be inferred from DNA analysis of the $Y$ chromosome of direct male-line descendants. The most well-known such case is that of Thomas Jefferson, one of the nation's founders, who likely impregnated enslaved person Sally Hemings. Although Jefferson himself has no male-line descendants, DNA analysis of the male-line descendants of Jefferson's paternal uncle contained the same Y chromosome as a male-line descendant of Eston Hemings, the youngest son of Sally Hemings. Thus, a mixing of the two genomes is clear, although the male individual responsible for that mixing can't be identified with $100 \\%$ certainty."}
{"id": 1838, "contents": "DNA Fingerprinting - \nAmong its many other applications, DNA fingerprinting is widely used for the diagnosis of genetic disorders, both prenatally and in newborns. Cystic fibrosis, hemophilia, Huntington's disease, Tay-Sachs disease, sickle cell anemia, and thalassemia are among the many diseases that can be detected, enabling early treatment of an affected child. Furthermore, by studying the DNA fingerprints of relatives with a history of a particular disorder, it's possible to identify DNA patterns associated with the disease and perhaps obtain clues for an eventual cure. In addition, the U.S. Department of Defense requires blood and saliva samples from all military personnel. The samples are stored, and DNA is extracted if the need for identification of a casualty arises."}
{"id": 1839, "contents": "Key Terms - \n- anticodon\n- antisense strand\n- codon\n- deoxyribonucleic acid (DNA)\n- double helix\n- 3' end\n- 5' end\n- messenger RNA (mRNA)\n- nucleoside\n- nucleotides\n- polymerase chain reaction (PCR)\n- replication\n- ribonucleic acid (RNA)\n- ribosomal RNA (rRNA)\n- Sanger dideoxy method\n- sense strand\n- small RNAs\n- transcription\n- transfer RNA (tRNA)\n- translation"}
{"id": 1840, "contents": "Summary - \nDNA (deoxyribonucleic acid) and RNA (ribonucleic acid) are biological polymers that act as chemical carriers of an organism's genetic information. Enzyme-catalyzed hydrolysis of nucleic acids yields nucleotides, the monomer units from which RNA and DNA are constructed. Further enzyme-catalyzed hydrolysis of the nucleotides yields nucleosides plus phosphate. Nucleosides, in turn, consist of a purine or pyrimidine base linked to the C1 of an aldopentose sugar-ribose in RNA and 2-deoxyribose in DNA. The nucleotides are joined by phosphate links between the $5^{\\prime}$ phosphate of one nucleotide and the $3^{\\prime}$ hydroxyl on the sugar of another nucleotide.\n\nMolecules of DNA consist of two complementary polynucleotide strands held together by hydrogen bonds between heterocyclic bases on the different strands and coiled into a double helix. Adenine and thymine form hydrogen bonds to each other, as do cytosine and guanine.\n\nThree processes take place in deciphering the genetic information of DNA:\n\n- Replication of DNA is the process by which identical DNA copies are made. The DNA double helix unwinds, complementary deoxyribonucleotides line up in order, and two new DNA molecules are produced.\n- Transcription is the process by which RNA is produced to carry genetic information from the nucleus to the ribosomes. A short segment of the DNA double helix unwinds, and complementary ribonucleotides line up to produce messenger RNA (mRNA).\n- Translation is the process by which mRNA directs protein synthesis. Each mRNA is divided into codons, ribonucleotide triplets that are recognized by small amino acid-carrying molecules of transfer RNA (tRNA), which deliver the appropriate amino acids needed for protein synthesis.\n\nSequencing of DNA is carried out by the Sanger dideoxy method, and small DNA segments can be synthesized in the laboratory by automated instruments. Small amounts of DNA can be amplified by factors of $10^{6}$ using the polymerase chain reaction (PCR)."}
{"id": 1841, "contents": "Additional Problems - \nVisualizing Chemistry\nPROBLEM Identify the following bases, and tell whether each is found in DNA, RNA, or both:\n28-13 (a)\n\n(b)\n\n(c)\n\n\nPROBLEM Identify the following nucleotide, and tell how it is used:"}
{"id": 1842, "contents": "28-14 - \nPROBLEM Amine bases in nucleic acids can react with alkylating agents in typical $\\mathrm{S}_{\\mathrm{N}} 2$ reactions. Look at the\n28-15 following electrostatic potential maps, and tell which is the better nucleophile, guanine or adenine. The reactive positions in each are indicated.\n\n\n9-Methylguanine\n\n\n9-Methyladenine"}
{"id": 1843, "contents": "Mechanism Problems - \nPROBLEM The final step in DNA synthesis is deprotection by treatment with aqueous ammonia. Show the\n28-16 mechanisms by which deprotection occurs at the points indicated in the following structure:\n\n\nPROBLEM The final step in the metabolic degradation of uracil is the oxidation of malonic semialdehyde to\n28-17 give malonyl CoA. Propose a mechanism.\n\n\nPROBLEM One of the steps in the biosynthesis of a nucleotide called inosine monophosphate is the formation 28-18 of aminoimidazole ribonucleotide from formylglycinamidine ribonucleotide. Propose a mechanism.\n\n\nPROBLEM One of the steps in the metabolic degradation of guanine is hydrolysis to give xanthine. Propose a 28-19 mechanism.\n\n\nPROBLEM One of the steps in the biosynthesis of uridine monophosphate is the reaction of aspartate with\n28-20 carbamoyl phosphate to give carbamoyl aspartate followed by cyclization to form dihydroorotate. Propose mechanisms for both steps."}
{"id": 1844, "contents": "General Problems - \nPROBLEM Human brain natriuretic peptide (BNP) is a small peptide of 32 amino acids used in the treatment\n28-21 of congestive heart failure. How many nitrogen bases are present in the DNA that codes for BNP?\nPROBLEM Human and horse insulin both have two polypeptide chains, with one chain containing 21 amino\n28-22 acids and the other containing 30 amino acids. They differ in primary structure at two places. At position 9 in one chain, human insulin has Ser and horse insulin has Gly; at position 30 in the other chain, human insulin has Thr and horse insulin has Ala. How must the DNA for the two insulins differ?\n\nPROBLEM The DNA of sea urchins contains about $32 \\%$ A. What percentages of the other three bases would you\n28-23 expect in sea urchin DNA? Explain.\nPROBLEM The codon UAA stops protein synthesis. Why does the sequence UAA in the following stretch of 28-24 mRNA not cause any problems?\n\nPROBLEM Which of the following base sequences would most likely be recognized by a restriction 28-25 endonuclease? Explain.\n(a) GAATTC\n(b) GATTACA\n(c) CTCGAG\n\nPROBLEM For what amino acids do the following ribonucleotide triplets code?\n28-26\n(a) AAU\n(b) GAG\n(c) UCC\n(d) CAU\n\nPROBLEM From what DNA sequences were each of the mRNA codons in Problem 28-26 transcribed? 28-27\n\nPROBLEM What anticodon sequences of tRNAs are coded for by the codons in Problem 28-26? 28-28\n\nPROBLEM Draw the complete structure of the ribonucleotide codon UAC. For what amino acid does this 28-29 sequence code?\n\nPROBLEM Draw the complete structure of the deoxyribonucleotide sequence from which the mRNA codon in 28-30 Problem 28-29 was transcribed.\n\nPROBLEM Give an mRNA sequence that will code for the synthesis of metenkephalin. 28-31\nTyr-Gly-Gly-Phe-Met\n\nPROBLEM Give an mRNA sequence that will code for the synthesis of angiotensin II. 28-32"}
{"id": 1845, "contents": "General Problems - \nPROBLEM Give an mRNA sequence that will code for the synthesis of angiotensin II. 28-32\n\n> Asp-Arg-Val-Tyr-Ile-His-Pro-Phe\n\nPROBLEM What amino acid sequence is coded for by the following DNA coding strand (sense strand)?\n28-33\n(5') CTT-CGA-CCA-GAC-AGC-TTT (3')\n\nPROBLEM What amino acid sequence is coded for by the following mRNA base sequence?\n28-34\n(5') CUA-GAC-CGU-UCC-AAG-UGA (3')\nPROBLEM If the DNA coding sequence -CAA-CCG-GAT- were miscopied during replication and became -CGA-28-35 CCG-GAT-, what effect would there be on the sequence of the protein produced?\n\nPROBLEM Show the steps involved in a laboratory synthesis of the DNA fragment with the sequence CTAG. 28-36\n\nPROBLEM Draw the structure of cyclic adenosine monophosphate (cAMP), a messenger involved in the 28-37 regulation of glucose production in the body. Cyclic AMP has a phosphate ring connecting the $3^{\\prime}$ and 5 '-hydroxyl groups on adenosine.\n\nPROBLEM Valganciclovir, marketed as Valcyte, is an antiviral agent used for the treatment of cytomegalovirus.\n28-38 Called a prodrug, valganciclovir is inactive by itself but is rapidly converted in the intestine by hydrolysis of its ester bond to produce an active drug, called ganciclovir, along with an amino acid.\n\n\nValganciclovir\n(a) What amino acid is produced by hydrolysis of the ester bond in valganciclovir?\n(b) What is the structure of ganciclovir?\n(c) What atoms present in the nucleotide deoxyguanine are missing from ganciclovir?\n(d) What role do the atoms missing from deoxyguanine play in DNA replication?\n(e) How might valganciclovir interfere with DNA synthesis?"}
{"id": 1846, "contents": "CHAPTER 29 - \nThe Organic Chemistry of Metabolic Pathways\n\n\nFIGURE 29.1 Acyl CoA dehydrogenase is an enzyme that catalyzes the introduction of a $\\mathbf{C =} \\mathbf{C}$ double bond into fatty acids during their metabolism. (credit: modification of image from the RCSB PDF (rcsb.org) of PDB ID 2WBI. Muniz, J.R.C., Guo, K., Savitsky, P., Roos, A., Yue, W., Pilka, E., Vondelft, F., Edwards, A.M., Bountra, C., Arrowsmith, C.H., Weigelt, J., Oppermann, U. Crystal structure of human Acyl-CoA dehydrogenase 11)"}
{"id": 1847, "contents": "CHAPTER CONTENTS - \n29.1 An Overview of Metabolism and Biochemical Energy\n29.2 Catabolism of Triacylglycerols: The Fate of Glycerol\n29.3 Catabolism of Triacylglycerols: $\\beta$-Oxidation\n29.4 Biosynthesis of Fatty Acids\n29.5 Catabolism of Carbohydrates: Glycolysis\n29.6 Conversion of Pyruvate to Acetyl CoA\n29.7 The Citric Acid Cycle\n29.8 Carbohydrate Biosynthesis: Gluconeogenesis\n29.9 Catabolism of Proteins: Deamination\n29.10 Some Conclusions about Biological Chemistry\n\nWHY THIS CHAPTER? In this chapter, we'll look at some of the pathways by which organisms carry out their chemistry, focusing primarily on how they metabolize fats and carbohydrates. The treatment will be far from complete, but it should give you an idea of the kinds of processes that occur.\n\nAnyone who wants to understand or contribute to the revolution now taking place in the biological sciences must first understand life processes at the molecular level. This understanding, in turn, must be based on a detailed knowledge of the chemical reactions and pathways used by living organisms. Just knowing what occurs is not enough; it's also necessary to understand how and why organisms use the chemistry they do.\n\nBiochemical reactions are not mysterious. Even though the biological reactions that take place in living organisms often appear complicated, they follow the same rules of reactivity as laboratory reactions and they operate by the same mechanisms.\n\nSome of the molecules we'll be encountering are substantially larger and more complex than those we've been dealing with thus far. But don't be intimidated; keep your focus on the parts of the molecules where changes occur, and ignore the parts where nothing changes. The reactions themselves are exactly the same additions, eliminations, substitutions, carbonyl condensations, and so forth, that we've been dealing with all along. By the end of this chapter, it should be clear that the chemistry of living organisms is organic chemistry."}
{"id": 1848, "contents": "CHAPTER CONTENTS - 29.1 An Overview of Metabolism and Biochemical Energy\nThe many reactions that occur in the cells of living organisms are collectively called metabolism. The pathways that break down larger molecules into smaller ones are called catabolism, and the pathways that synthesize larger biomolecules from smaller ones are known as anabolism. Catabolic reaction pathways are usually exergonic and release energy, while anabolic pathways are often endergonic and absorb energy. Catabolism can be divided into the four stages shown in FIGURE 29.2.\n\n\nFIGURE 29.2 An overview of catabolic pathways for the degradation of food and the production of biochemical energy. The ultimate products of food catabolism are $\\mathrm{CO}_{2}$ and $\\mathrm{H}_{2} \\mathrm{O}$, with the energy released in the citric acid cycle used to drive the endergonic synthesis of adenosine triphosphate (ATP) from adenosine diphosphate (ADP) plus hydrogen phosphate ion, $\\mathrm{HOPO}_{3}{ }^{2-}$.\n\nIn the first catabolic stage, commonly called digestion, food is broken down in the mouth, stomach, and small intestine by hydrolysis of ester, acetal (glycoside), and amide (peptide) bonds to yield fatty acids, simple sugars,\nand amino acids. These smaller molecules are then absorbed and further degraded in the second stage of catabolism to yield acetyl groups attached by a thioester bond to the large carrier molecule, coenzyme A. The resultant compound, acetyl coenzyme A (acetyl CoA), is a key substance in the metabolism of food molecules and in many other biological pathways. As noted in Section 21.8, the acetyl group in acetyl CoA is linked to the sulfur atom of phosphopantetheine, which is itself linked to adenosine 3',5'-bisphosphate."}
{"id": 1849, "contents": "Acetyl CoA-a thioester - \nAcetyl groups are oxidized inside cellular mitochondria in the third stage of catabolism, the citric acid cycle, to yield $\\mathrm{CO}_{2}$. (We'll see the details of the process in Section 29.7.) Like most oxidations, this stage releases a large amount of energy, which is used in the fourth stage, the electron-transport chain, to accomplish the endergonic phosphorylation of adenosine diphosphate (ADP) with hydrogen phosphate ion ( $\\mathrm{HOPO}_{3}{ }^{2-}$, abbreviated $\\mathrm{P}_{\\mathrm{i}}$ ) to give adenosine triphosphate (ATP).\n\nAs the final result of food catabolism, ATP has been called the \"energy currency\" of the cell. Catabolic reactions \"buy\" ATP with the energy they release to synthesize it from ADP and hydrogen phosphate ion. Anabolic reactions then spend the ATP by transferring a phosphate group to another molecule, thereby regenerating ADP. Energy production and use in living organisms thus revolves around the ATP $\\rightleftarrows$ ADP interconversion.\n\n\nbreaking a $\\mathrm{C}-\\mathrm{O}$ bond and forming a carboxylic ester, ROCOR' (Section 21.5), phosphoric acid anhydrides react with alcohols by breaking a $\\mathrm{P}-\\mathrm{O}$ bond and forming a phosphate ester, $\\mathrm{ROPO}_{3}{ }^{2-}$. The reaction is, in effect, a nucleophilic acyl substitution at phosphorus. Note that phosphorylation reactions with ATP generally require the presence of a divalent metal cation in the enzyme, usually $\\mathrm{Mg}^{2+}$, to form a Lewis acid-base complex with the phosphate oxygen atoms and to neutralize negative charge.\n\n\nHow does the body use ATP? Recall from Section 6.7 that the free-energy change $\\Delta G$ must be negative and energy must be released for a reaction to be energetically favorable and occur spontaneously. If $\\Delta G$ is positive, the reaction is energetically unfavorable and the process can't occur spontaneously."}
{"id": 1850, "contents": "Acetyl CoA-a thioester - \nFor an energetically unfavorable reaction to occur, it must be \"coupled\" to an energetically favorable reaction so that the overall free-energy change for the two reactions together is favorable. To understand what it means for reactions to be coupled, imagine that reaction 1 does not occur to any reasonable extent because it has a small equilibrium constant and is energetically unfavorable; that is, the reaction has $\\Delta G>0$.\n\n$$\n(1) \\mathbf{A}+m \\longleftrightarrow \\mathbf{B}+n \\quad \\Delta G>0\n$$\n\nwhere $\\mathbf{A}$ and $\\mathbf{B}$ are the biochemically \"important\" substances while $m$ and $n$ are enzyme cofactors, $\\mathrm{H}_{2} \\mathrm{O}$, or other small molecules.\n\nImagine also that product $n$ can react with substance $o$ to yield $p$ and $q$ in a second, highly favorable reaction that has a large equilibrium constant and $\\Delta G \\ll 0$.\n(2) $n+o \\rightleftarrows p+q \\quad \\Delta G \\ll 0$\n\nTaking the two reactions together, they share, or are coupled through, the common intermediate $n$, which is a product in the first reaction and a reactant in the second. When even a tiny amount of $n$ is formed in reaction 1 , it undergoes essentially complete conversion in reaction 2 , thereby removing it from the first equilibrium and forcing reaction 1 to continually replenish $n$ until reactant $\\mathbf{A}$ is gone. That is, the two reactions added together have a favorable $\\Delta G<0$, and we say that the favorable reaction 2 \"drives\" the unfavorable reaction 1 . Because the two reactions are coupled through $n$, the transformation of $\\mathbf{A}$ to $\\mathbf{B}$ becomes favorable."}
{"id": 1851, "contents": "Acetyl CoA-a thioester - \n| (1) $\\mathbf{A}+m \\longleftrightarrow \\mathbf{B}+\\not h$ | $\\Delta G>0$ |\n| :--- | :--- |\n| (2) $\\not \\propto+o \\rightleftarrows p+q$ | $\\Delta G \\ll 0$ |\n| Net: $\\mathbf{A}+m+o \\rightleftarrows \\mathbf{B}+p+q$ | $\\Delta G<0$ |\n\nFor an example of two reactions that are coupled, look at the phosphorylation reaction of glucose to yield glucose 6-phosphate plus water, an important step in the breakdown of dietary carbohydrates.\n\n\nGlucose\nGlucose 6-phosphate\nThe reaction of glucose with $\\mathrm{HOPO}_{3}{ }^{2-}$ does not occur spontaneously because it is energetically unfavorable, with $\\Delta G^{\\circ{ }^{\\prime}}=13.8 \\mathrm{~kJ} / \\mathrm{mol}$. (The standard free-energy change for a biological reaction is denoted $\\Delta G^{\\circ}$ and refers to a process in which reactants and products have a concentration of 1.0 M in a solution with $\\mathrm{pH}=7$.) At the same time, however, the reaction of water with ATP to yield ADP plus $\\mathrm{HOPO}_{3}{ }^{2-}$ is strongly favorable, with $\\Delta G^{\\circ}$\n$=-30.5 \\mathrm{~kJ} / \\mathrm{mol}$. When the two reactions are coupled, glucose reacts with ATP to yield glucose 6-phosphate plus ADP in a reaction that is favorable by about $16.7 \\mathrm{~kJ} / \\mathrm{mol}(4.0 \\mathrm{kcal} / \\mathrm{mol})$. That is, ATP drives the phosphorylation reaction of glucose.\n\n\nIt's this ability to drive otherwise unfavorable phosphorylation reactions that makes ATP so useful. The resultant phosphates are much more reactive as leaving groups in nucleophilic substitutions and eliminations than the alcohols they're derived from and are therefore more chemically useful."}
{"id": 1852, "contents": "Acetyl CoA-a thioester - \nPROBLEM One of the steps in fat metabolism is the reaction of glycerol (1,2,3-propanetriol) with ATP to yield 29-1 glycerol 1-phosphate. Write the reaction, and draw the structure of glycerol 1-phosphate."}
{"id": 1853, "contents": "Acetyl CoA-a thioester - 29.2 Catabolism of Triacylglycerols: The Fate of Glycerol\nThe metabolic breakdown of triacylglycerols begins with their hydrolysis in the stomach and small intestine to yield glycerol plus fatty acids. The reaction is catalyzed by a lipase, whose mechanism is shown in FIGURE 29.3. The active site of the enzyme contains a catalytic triad of aspartic acid, histidine, and serine residues, which act cooperatively to provide the necessary acid and base catalysis for the individual steps. Hydrolysis is accomplished by two sequential nucleophilic acyl substitution reactions, one that covalently binds an acyl group to the side chain -OH of a serine residue on the enzyme and a second that frees the fatty acid from the enzyme."}
{"id": 1854, "contents": "FIGURE 29.3 MECHANISM - \nMechanism for the action of lipase. The active site of the enzyme contains a catalytic triad of aspartic acid, histidine, and serine, which react cooperatively to carry out two nucleophilic acyl substitution reactions. Individual steps are explained in the text.\n(1) The enzyme active site contains an aspartic acid, a histidine, and a serine. First, histidine acts as a base to deprotonate the -OH group of serine, with the negatively charged carboxylate of aspartic acid stabilizing the nearby histidine cation that results. Serine then adds to the carbonyl group of the triacylglycerol, yielding a tetrahedral intermediate.\n\nThis intermediate expels a diacylglycerol as leaving group in a nucleophilic acyl substitution reaction, giving an acyl enzyme. The diacylglycerol is protonated by the histidine cation.\n\n3 Histidine deprotonates a water molecule, which adds to the acyl group. A tetrahedral intermediate is again formed, and the histidine cation is again stabilized by the nearby carboxylate.\n4) The tetrahedral intermediate expels the serine as leaving group in a second nucleophilic acyl substitution reaction, yielding a free fatty acid. The serine accepts a proton from histidine, and the enzyme has now returned to its starting structure.\n\n\nTetrahedral intermediate\n(2)\n\n(3) $\\downarrow \\mathrm{H}_{2} \\mathrm{O}$\n\n\nTetrahedral intermediate"}
{"id": 1855, "contents": "(4) - \nEnzyme\nFatty acid\n\nSteps 1-2 of FIGURE 29.3: Acyl Enzyme Formation\nThe first nucleophilic acyl substitution step-reaction of the triacylglycerol with the active-site serine to give an acyl enzyme-begins with deprotonation of the serine alcohol by histidine to form the more strongly nucleophilic alkoxide ion. This proton transfer is facilitated by a nearby side-chain carboxylate anion of aspartic acid, which makes the histidine more basic and stabilizes the resultant histidine cation by electrostatic interactions. The deprotonated serine adds to a carbonyl group of a triacylglycerol to give a tetrahedral intermediate, which expels a diacylglycerol as a leaving group and produces an acyl enzyme. This step is catalyzed by a proton transfer from histidine to make the leaving group a neutral alcohol.\n\n\nSteps 3-4 of FIGURE 29.3: Hydrolysis\nThe second nucleophilic acyl substitution step hydrolyzes the acyl enzyme and gives the free fatty acid by a mechanism analogous to that of the first two steps. Water is deprotonated by histidine to give hydroxide ion, which adds to the enzyme-bound acyl group. The tetrahedral intermediate then expels the neutral serine residue as the leaving group, freeing the fatty acid and returning the enzyme to its active form.\n\n\nThe fatty acids released on triacylglycerol hydrolysis are transported to mitochondria and degraded to acetyl CoA, while the glycerol is carried to the liver for further metabolism. In the liver, glycerol is first phosphorylated by reaction with ATP and then oxidized by $\\mathrm{NAD}^{+}$. The resulting dihydroxyacetone phosphate (DHAP) enters the carbohydrate glycolysis pathway, which we'll discuss in Section 29.5.\n\n\nYou might note that C2 of glycerol is a prochiral center (Section 5.11) with two identical \"arms.\" As is typical for enzyme-catalyzed reactions, the phosphorylation of glycerol is selective. Only the pro- $R$ arm undergoes reaction, although this can't be predicted in advance."}
{"id": 1856, "contents": "(4) - \nNote also that the phosphorylation product is named $s n$-glycerol 3-phosphate, where the $s n$ - prefix means \"stereospecific numbering.\" In this convention, the molecule is drawn as a Fischer projection with the -OH group at C2 pointing to the left and the glycerol carbon atoms numbered from the top."}
{"id": 1857, "contents": "(4) - 29.3 Catabolism of Triacylglycerols: $\\boldsymbol{\\beta}$-Oxidation\nThe fatty acids that result from triacylglycerol hydrolysis are converted into thioesters with coenzyme A and then catabolized by a repetitive four-step sequence of reactions called the $\\boldsymbol{\\beta}$-oxidation pathway, shown in FIGURE 29.4. Each passage along the pathway results in the cleavage of an acetyl group from the end of the fatty-acid chain, until the entire molecule is ultimately degraded. As each acetyl group is produced, it enters the citric acid cycle and is further degraded to $\\mathrm{CO}_{2}$, as we'll see in Section 29.7."}
{"id": 1858, "contents": "FIGURE 29.4 MECHANISM - \nThe four steps of the $\\boldsymbol{\\beta}$-oxidation pathway, resulting in the cleavage of an acetyl group from the end of the fatty-acid chain. The key chain-shortening step is a retro-Claisen reaction of a $\\beta$-keto thioester. Individual steps are explained in the text.\n\n\nFatty acyl CoA\n(1) A conjugated double bond is introduced by removal of hydrogens from C2 and C3 by the coenzyme flavin adenine\ndinucleotide (FAD).\n\n$\\alpha, \\beta$-Unsaturated acyl CoA\n(2) $\\downarrow \\mathrm{H}_{2} \\mathrm{O}$ $\\beta$-hydroxyacyl CoA.\n\n$\\beta$-Hydroxyacyl CoA\n(3) The alcohol is oxidized by $\\mathrm{NAD}^{+}$to give a $\\beta$-keto thioester.\n\n(4) Nucleophilic addition of coenzyme A to the keto group occurs, followed by a retro-Claisen condensation reaction. The products are acetyl CoA and a chain-shortened fatty acyl CoA.\n\n\nAcetyl CoA\n\nStep 1 of FIGURE 29.4: Introduction of a Double Bond\nThe $\\beta$-oxidation pathway begins when two hydrogen atoms are removed from C 2 and C 3 of the fatty acyl CoA by one of a family of acyl-CoA dehydrogenases to yield an $\\alpha, \\beta$-unsaturated acyl CoA. This kind of oxidation-the introduction of a conjugated double bond into a carbonyl compound-occurs frequently in biochemical pathways and usually involves the coenzyme flavin adenine dinucleotide (FAD). Reduced $\\mathrm{FADH}_{2}$ is the byproduct."}
{"id": 1859, "contents": "FIGURE 29.4 MECHANISM - \nThe mechanisms of FAD-catalyzed reactions are often difficult to establish because flavin coenzymes can follow both two-electron (polar) and one-electron (radical) pathways. As a result, extensive studies of the family of acyl-CoA dehydrogenases have not yet provided a clear picture of how these enzymes function. What is known is that: (1) The first step is abstraction of the pro- $R$ hydrogen from the acidic $\\alpha$ position of the acyl CoA to give a thioester enolate ion. Hydrogen-bonding between the acyl carbonyl group and the ribitol hydroxyls of FAD increases the acidity of the acyl group. (2) The pro- $R$ hydrogen at the $\\beta$ position is transferred to FAD. (3) The $\\alpha, \\beta$-unsaturated acyl CoA that results has a trans double bond.\n\n\nOne suggested mechanism has the reaction taking place by a conjugate nucleophilic addition of hydride, analogous to what occurs during alcohol oxidations with $\\mathrm{NAD}^{+}$. Electrons on the enolate ion might expel a $\\beta$ hydride ion, which could add to the doubly bonded N5 nitrogen on FAD. Protonation of the intermediate at N1 would give the product.\n\n\nStep 2 of FIGURE 29.4: Conjugate Addition of Water\nThe $\\alpha, \\beta$-unsaturated acyl CoA produced in step 1 reacts with water by a conjugate addition pathway (Section 19.13) to yield a $\\beta$-hydroxyacyl CoA in a process catalyzed by enoyl CoA hydratase. Water as nucleophile adds to the $\\beta$ carbon of the double bond, yielding an intermediate thioester enolate ion that is protonated on the $\\alpha$ position."}
{"id": 1860, "contents": "FIGURE 29.4 MECHANISM - \n(3S)-Hydroxyacyl CoA\nStep 3 of FIGURE 29.4: Alcohol Oxidation\nThe $\\beta$-hydroxyacyl CoA from step 2 is oxidized to a $\\beta$-ketoacyl CoA in a reaction catalyzed by one of a family of L-3-hydroxyacyl-CoA dehydrogenases, which differ in substrate specificity according to the chain length of the acyl group. As in the oxidation of $s n$-glycerol 3-phosphate to dihydroxyacetone phosphate mentioned at the end of Section 29.2, this alcohol oxidation requires $\\mathrm{NAD}^{+}$as a coenzyme and yields reduced $\\mathrm{NADH} / \\mathrm{H}^{+}$as byproduct. Deprotonation of the hydroxyl group is carried out by a histidine residue at the active site."}
{"id": 1861, "contents": "Step 4 of FIGURE 29.4: Chain Cleavage - \nAcetyl CoA is split off from the chain in the final step of $\\beta$-oxidation, leaving an acyl CoA that is two carbon atoms shorter than the original. The reaction is catalyzed by $\\beta$-ketoacyl-CoA thiolase and is mechanistically the reverse of a Claisen condensation reaction (Section 23.7). In the forward direction, a Claisen condensation joins two esters together to form a $\\beta$-keto ester product. In the reverse direction, a retro-Claisen reaction splits apart a $\\beta$-keto ester (or $\\beta$-keto thioester in this case) to form two esters (or two thioesters).\n\n\nThe retro-Claisen reaction occurs by nucleophilic addition of a cysteine - SH group on the enzyme to the keto group of the $\\beta$-ketoacyl CoA to yield an alkoxide ion intermediate. Cleavage of the $\\mathrm{C} 2-\\mathrm{C} 3$ bond then follows, with the expulsion of an acetyl CoA enolate ion that is immediately protonated. The enzyme-bound acyl group then undergoes nucleophilic acyl substitution by reaction with a molecule of coenzyme A, and the chain-shortened acyl CoA that results re-enters the $\\beta$-oxidation pathway for further degradation.\n\n\nLook at the catabolism of myristic acid shown in FIGURE 29.5 to see the overall results of the $\\beta$-oxidation pathway. The first passage converts the 14 -carbon myristoyl CoA into the 12 -carbon lauroyl CoA plus acetyl CoA, the second passage converts lauroyl CoA into the 10-carbon caproyl CoA plus acetyl CoA, the third passage converts caproyl CoA into the 8 -carbon capryloyl CoA, and so on. Note that the final passage produces two molecules of acetyl CoA because the precursor has four carbons.\n\n\nMyristoyl CoA\n\n$$\n\\begin{aligned}\n& \\beta \\text {-Oxidation } \\\\\n& \\text { (passage 1) }\n\\end{aligned}\n$$"}
{"id": 1862, "contents": "Step 4 of FIGURE 29.4: Chain Cleavage - \nMyristoyl CoA\n\n$$\n\\begin{aligned}\n& \\beta \\text {-Oxidation } \\\\\n& \\text { (passage 1) }\n\\end{aligned}\n$$\n\n\n\nLauroyl CoA\n$\\downarrow \\begin{aligned} & \\beta \\text {-Oxidation } \\\\ & \\text { (passage 2) }\\end{aligned}$\n\n\nCaproyl CoA\n\n$$\n\\begin{aligned}\n& \\beta \\text {-Oxidation } \\\\\n& \\text { (passage 3) }\n\\end{aligned}\n$$"}
{"id": 1863, "contents": "Capryloyl CoA - \nFIGURE 29.5 Catabolism of the 14 -carbon myristic acid by the $\\boldsymbol{\\beta}$-oxidation pathway yields seven molecules of acetyl CoA after six passages.\n\nMost fatty acids have an even number of carbon atoms, so none are left over after $\\beta$-oxidation. Those fatty acids with an odd number of carbon atoms yield the three-carbon propionyl CoA in the final $\\beta$-oxidation. Propionyl CoA is then converted to succinate by a multistep radical pathway, and succinate enters the citric acid cycle (Section 29.7). Note that the three-carbon propionyl group should technically be called propanoyl, but biochemists generally use the nonsystematic name.\n\nPROBLEM Write the equations for the remaining passages of the $\\beta$-oxidation pathway following those shown\n29-2 in Figure 29.5.\nPROBLEM How many molecules of acetyl CoA are produced by catabolism of the following fatty acids, and how 29-3 many passages of the $\\beta$-oxidation pathway are needed?\n(a) Palmitic acid, $\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{14} \\mathrm{CO}_{2} \\mathrm{H}$\n(b) Arachidic acid, $\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{18} \\mathrm{CO}_{2} \\mathrm{H}$"}
{"id": 1864, "contents": "Capryloyl CoA - 29.4 Biosynthesis of Fatty Acids\nOne of the most striking features of the common fatty acids is that they have an even number of carbon atoms (TABLE 27.1). This occurs because all fatty acids are derived biosynthetically from acetyl CoA by sequential addition of two-carbon units to a growing chain. The acetyl CoA, in turn, arises primarily from the metabolic breakdown of carbohydrates in the glycolysis pathway, which we'll see in Section 29.5. Thus, dietary carbohydrates consumed in excess of immediate energy needs are turned into fats for storage.\n\nAs a general rule in biological chemistry, the anabolic pathway by which a substance is made is not the reverse of the catabolic pathway by which the same substance is degraded. The two paths must differ in some respects for both to be energetically favorable. Thus, the $\\beta$-oxidation pathway for converting fatty acids into acetyl CoA and the biosynthesis of fatty acids from acetyl CoA are related but are not exact opposites. Differences include the identity of the acyl-group carrier, the stereochemistry of the $\\beta$-hydroxyacyl reaction intermediate, and the identity of the redox coenzyme. FAD is used to introduce a double bond in $\\beta$-oxidation, while NADPH is used to reduce the double bond in fatty-acid biosynthesis.\n\nIn bacteria, each step in fatty-acid synthesis is catalyzed by a separate enzyme. In vertebrates, however, fattyacid synthesis is catalyzed by an immense, multienzyme complex called a synthase that contains two identical subunits of 2505 amino acids each and catalyzes all steps in the pathway. In fact, for an 18-carbon fatty acid, the synthase catalyzes 42 separate steps! An overview of fatty-acid biosynthesis is shown in FIGURE 29.6."}
{"id": 1865, "contents": "Steps 1-2 of FIGURE 29.6: Acyl Transfers - \nThe starting material for fatty-acid biosynthesis is the thioester acetyl CoA, the final product of carbohydrate breakdown, as we'll see in Section 29.6. The pathway begins with several priming reactions, which transport acetyl CoA and convert it into more reactive species. The first priming reaction is a nucleophilic acyl substitution reaction that converts acetyl CoA into acetyl ACP (acyl carrier protein).\n\nNotice that the mechanism of the nucleophilic acyl substitution in step 1 can be given in an abbreviated form that saves space by not explicitly showing the tetrahedral reaction intermediate. Instead, electron movement is shown as a heart-shaped path around the carbonyl oxygen to imply the two steps of the full mechanism. Biochemists commonly use this kind of abbreviated format, and we'll also use it on occasion through the rest of this chapter."}
{"id": 1866, "contents": "FIGURE 29.6 MECHANISM - \nThe pathway for fatty-acid biosynthesis from the two-carbon precursor, acetyl CoA. Individual steps are explained in the text.\n\nAn acetyl group is transferred from CoA to ACP (acyl carrier protein).\n(2) The acetyl group is\ntransferred again, from\nACP to a synthase enzyme.\n\n\nAcetyl ACP"}
{"id": 1867, "contents": "(7) - \n(5) Claisen-like condensation of malonyl ACP with acetyl synthase occurs, followed by decarboxylation to yield acetoacetyl ACP, a $\\beta$-keto thioester.\n(6) Reduction of the ketone by NADPH yields the corresponding $\\beta$-hydroxy thioester.\n\nDehydration of $\\beta$-hydroxybutyryl ACP gives crotonyl ACP, an $a, b$-unsaturated thioester.\n\n8 Reduction of the double bond yields the saturated, chain-elongated butyryl ACP."}
{"id": 1868, "contents": "Acetyl synthase - \n\u00a9\n\nAcetyl CoA\n\n(3) Acetyl CoA is carboxylated to give malonyl CoA.\n(4) The malonyl group is transferred from CoA to ACP.\n\n\nAcetoacetyl ACP\n\n\n$\\boldsymbol{\\beta}$-Hydroxybutyryl ACP\n\n\nCrotonyl ACP\n\n\n\nButyryl ACP\n\n\nTetrahedral intermediate\n\n\nIn bacteria, ACP is a small protein of 77 residues that transports an acyl group from one enzyme to another. In vertebrates, however, ACP appears to be a long arm on a multienzyme synthase complex, whose apparent function is to shepherd an acyl group from site to site within the complex. As in acetyl CoA, the acyl group in acetyl ACP is linked by a thioester bond to the sulfur atom of phosphopantetheine. The phosphopantetheine is in turn linked to ACP through the side-chain -OH group of a serine residue in the enzyme."}
{"id": 1869, "contents": "Acetyl ACP - \nStep 2, another priming reaction, involves a further exchange of thioester linkages by another nucleophilic acyl substitution and results in covalent bonding of the acetyl group to a cysteine residue in the synthase complex that catalyzes the upcoming condensation step.\n\nSteps 3-4 of FIGURE 29.6: Carboxylation and Acyl Transfer\nStep 3 is a loading reaction in which acetyl CoA is carboxylated by reaction with $\\mathrm{HCO}_{3}{ }^{-}$and ATP to yield malonyl CoA plus ADP. This step requires the coenzyme biotin, which is bonded to the lysine residue of acetyl CoA carboxylase and acts as a carrier of $\\mathrm{CO}_{2}$. Biotin first reacts with bicarbonate ion to give $N$-carboxybiotin, which then reacts with the enolate ion of acetyl CoA and transfers the $\\mathrm{CO}_{2}$ group. Thus, biotin acts as a carrier of $\\mathrm{CO}_{2}$, binding it in one step and releasing it in another.\n\nThe mechanism of the $\\mathrm{CO}_{2}$ transfer reaction with acetyl CoA to give malonyl CoA is thought to involve $\\mathrm{CO}_{2}$ as the reactive species. One proposal is that the loss of $\\mathrm{CO}_{2}$ is favored by hydrogen-bond formation between the N -carboxybiotin carbonyl group and a nearby acidic site in the enzyme. Simultaneous deprotonation of acetyl CoA by a basic site in the enzyme gives a thioester enolate ion that can react with $\\mathrm{CO}_{2}$ as it forms (FIGURE 29.7)."}
{"id": 1870, "contents": "Mechanism of step 3 in FIGURE 29.6, the biotin-dependent carboxylation of acetyl CoA to yield malonyl - \nCoA.\n(1) A basic site in the enzyme deprotonates acetyl CoA.\n\n(3) The enolate ion adds in an\nbond of carbon dioxide, yielding malonyl CoA.\n\n\nMalonyl CoA\n\nFollowing the formation of malonyl CoA, another nucleophilic acyl substitution reaction occurs in step 4 to form the more reactive malonyl ACP, thereby binding the malonyl group to an ACP arm of the multienzyme synthase. At this point, both acetyl and malonyl groups are bound to the enzyme, and the stage is set for their condensation."}
{"id": 1871, "contents": "Step 5 of FIGURE 29.6: Condensation - \nThe key carbon-carbon bond-forming reaction that builds the fatty-acid chain occurs in step 5 . This step is simply a Claisen condensation between acetyl synthase as the electrophilic acceptor and malonyl ACP as the nucleophilic donor. The mechanism of the condensation is thought to involve decarboxylation of malonyl ACP to give an enolate ion, followed by immediate nucleophilic addition of the enolate ion to the carbonyl group of acetyl synthase. Breakdown of the tetrahedral intermediate then gives the four-carbon condensation product acetoacetyl ACP and frees the synthase binding site for attachment of the chain-elongated acyl group at the end of the sequence.\n\n\nSteps 6-8 of FIGURE 29.6: Reduction and Dehydration\nThe ketone carbonyl group in acetoacetyl ACP is next reduced to the alcohol $\\beta$-hydroxybutyryl ACP by $\\beta$-keto thioester reductase and NADPH, a reducing coenzyme closely related to NADH. $R$ stereochemistry results at the newly formed chirality center in the $\\beta$-hydroxy thioester product. (Note that the systematic name of a butyryl group is butanoyl.)\n\n\nSubsequent dehydration of $\\beta$-hydroxybutyryl ACP by an E1cB reaction in step 7 yields trans-crotonyl ACP, and the carbon-carbon double bond of crotonyl ACP is reduced by NADPH in step 8 to yield butyryl ACP. The doublebond reduction occurs by conjugate nucleophilic addition of a hydride ion from NADPH to the $\\beta$ carbon of trans-crotonyl ACP. In vertebrates, this reduction occurs by an overall syn addition, but other organisms carry out similar chemistry with different stereochemistry."}
{"id": 1872, "contents": "Step 5 of FIGURE 29.6: Condensation - \nThe net effect of the eight steps in the fatty-acid biosynthesis pathway is to take two 2 -carbon acetyl groups and combine them into a 4-carbon butyryl group. Further condensation of the butyryl group with another malonyl ACP yields a 6-carbon unit, and still further repetitions add two carbon atoms at a time until the 16-carbon palmitoyl ACP is produced.\n\n\n\nPalmitoyl ACP\n\nFurther chain elongation of palmitic acid occurs by reactions similar to those just described, but CoA rather than ACP acts as the carrier group, and separate enzymes are needed for each step rather than a multienzyme synthase complex.\n\nPROBLEM Write a mechanism for the dehydration reaction of $\\beta$-hydroxybutyryl ACP to yield crotonyl ACP in 29-4 step 7 of fatty-acid synthesis.\n\nPROBLEM Evidence for the role of acetate in fatty-acid biosynthesis comes from isotope-labeling experiments.\n29-5 If acetate labeled with ${ }^{13} \\mathrm{C}$ in the methyl group $\\left({ }^{13} \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right)$ were incorporated into fatty acids, at what positions in the fatty-acid chain would you expect the ${ }^{13} \\mathrm{C}$ label to appear?\n\nPROBLEM Does the reduction of acetoacetyl ACP in step 6 occur on the Re face or the Si face of the molecule? 29-6"}
{"id": 1873, "contents": "Step 5 of FIGURE 29.6: Condensation - 29.5 Catabolism of Carbohydrates: Glycolysis\nGlucose is the body's primary short-term energy source. Its catabolism begins with glycolysis, a series of ten enzyme-catalyzed reactions that break down glucose into 2 equivalents of pyruvate, $\\mathrm{CH}_{3} \\mathrm{COCO}_{2}{ }^{-}$. The steps of glycolysis, also called the Embden-Meyerhoff pathway after its discoverers, are summarized in FIGURE 29.8.\n\nThe ten-step glycolysis pathway for catabolizing glucose to two molecules of pyruvate. Individual steps are described in the text.\n\n(1) Glucose is phosphorylated by reaction with ATP to yield glucose 6-phosphate.\n\n$\\alpha$-Glucose 6-phosphate\n\n\nGlucose 6-phosphate is isomerized to fructose 6-phosphate by ring opening followed by a keto-enol tautomerization.\n(2)"}
{"id": 1874, "contents": "$\\alpha$-Fructose 6-phosphate - \nFructose 6-phosphate is phosphorylated by reaction with ATP to yield fructose\n\n\n1,6-bisphosphate.\n\n$\\beta$-Fructose 1,6-bisphosphate\n\n(4) Fructose 1,6-bisphosphate undergoes ring opening and is cleaved by a retro-aldol reaction into glyceraldehyde 3-phosphate and dihydroxyacetone phosphate (DHAP).\n(5) DHAP then isomerizes to glyceraldehyde 3-phosphate."}
{"id": 1875, "contents": "Glyceraldehyde 3-phosphate - \n6) Glyceraldehyde 3-phosphate is oxidized to a carboxylic acid and then phosphorylated to yield 1,3-bisphosphoglycerate.\n$6 \\curvearrowleft \\mathrm{NAD}^{+}, \\mathrm{P}_{\\mathrm{i}}$\n\n1,3-Bisphosphoglycerate ${ }^{2-} \\mathrm{O}_{3} \\mathrm{POCH}_{2} \\mathrm{CHCO}_{2} \\mathrm{PO}_{3}{ }^{2-}$\n(7) A phosphate is transferred from the carboxyl group to ADP, resulting in synthesis of an ATP and yielding 3-phosphoglycerate.\n\n\n\n3-Phosphoglycerate\n8 Isomerization of 3-phosphoglycerate gives 2-phosphoglycerate.\n\n2-Phosphoglycerate\n9 Dehydration occurs to yield phosphoenolpyruvate (PEP).\n\n\nPhosphoenolpyruvate\n(10) A phosphate is transferred from PEP to ADP, yielding pyruvate and ATP.\n\n\n\nPyruvate\n\n\n\n\n\n\nSteps 1-2 of FIGURE 29.8: Phosphorylation and Isomerization\nGlucose, produced by the digestion of dietary carbohydrates, is phosphorylated at the C6 hydroxyl group by reaction with ATP in a process catalyzed by hexokinase. As noted in Section 29.1, the reaction requires $\\mathrm{Mg}^{2+}$ as a cofactor to complex with the negatively charged phosphate oxygens. The glucose 6-phosphate that results is then isomerized by glucose 6-phosphate isomerase to give fructose 6-phosphate. This isomerization takes place by initial opening of the glucose hemiacetal ring to its open-chain form, followed by keto-enol tautomerization to a cis enediol, $\\mathrm{HO}-\\mathrm{C}=\\mathrm{C}-\\mathrm{OH}$. But because glucose and fructose share a common enediol, further tautomerization to a different keto form produces open-chain fructose, and cyclization completes the process (FIGURE 29.10).\n\n\n$\\alpha$-Fructose\n6-phosphate"}
{"id": 1876, "contents": "Glyceraldehyde 3-phosphate - \n$\\alpha$-Fructose\n6-phosphate\n\nFIGURE 29.10 Mechanism of step 2 in glycolysis, the isomerization of glucose 6-phosphate to fructose 6-phosphate.\nStep 3 of FIGURE 29.8: Phosphorylation\nFructose 6-phosphate is converted in step 3 to fructose 1,6-bisphosphate (FBP) by a phosphofructokinasecatalyzed reaction with ATP (recall that the prefix bis- means two). The mechanism is similar to that in step 1 , with $\\mathrm{Mg}^{2+}$ ion again required as cofactor. Interestingly, the product of step 2 is the $\\alpha$ anomer of fructose 6-phosphate, but it is the $\\beta$ anomer that is phosphorylated in step 3 , implying that the two anomers equilibrate rapidly through the open-chain form. The result is a molecule ready to be split into the two three-carbon intermediates that will ultimately become two molecules of pyruvate.\n\n\nStep 4 of FIGURE 29.8: Cleavage\nFructose 1,6-bisphosphate is cleaved in step 4 into two 3-carbon pieces, dihydroxyacetone phosphate (DHAP) and glyceraldehyde 3-phosphate (GAP). The bond between C3 and C4 of fructose 1,6-bisphosphate breaks, and a $\\mathrm{C}=\\mathrm{O}$ group is formed at C 4 . Mechanistically, the cleavage is the reverse of an aldol reaction (Section 23.1) and is catalyzed by an aldolase. A forward aldol reaction joins two aldehydes or ketones to give a $\\beta$-hydroxy carbonyl compound, while a retro-aldol reaction, as in this case, cleaves a $\\beta$-hydroxy carbonyl compound into two aldehydes or ketones."}
{"id": 1877, "contents": "Glyceraldehyde 3-phosphate - \nTwo classes of aldolases are used by organisms for catalysis of the retro-aldol reaction. In fungi, algae, and some bacteria, the retro-aldol reaction is catalyzed by class II aldolases, which function by coordination of the fructose carbonyl group with $\\mathrm{Zn}^{2+}$ as Lewis acid. In plants and animals, the reaction is catalyzed by class I aldolases and does not take place on the free ketone. Instead, fructose 1,6 -bisphosphate undergoes reaction with the side-chain $-\\mathrm{NH}_{2}$ group of a lysine residue on the aldolase to yield a protonated enzyme-bound imine (Section 19.8), which is often called a Schiff base in biochemistry.\n\nBecause of its positive charge, the iminium ion is a better electron acceptor than a ketone carbonyl group. Retroaldol reaction ensues, giving glyceraldehyde 3-phosphate and an enamine, which is protonated to give another iminium ion that is hydrolyzed to yield dihydroxyacetone phosphate (FIGURE 29.11).\n\n\nFIGURE 29.11 Mechanism of step 4 in FIGURE 29.8, the cleavage of fructose 1,6-bisphosphate to yield glyceraldehyde 3-phosphate and dihydroxyacetone phosphate. The reaction occurs through an iminium ion formed by reaction with a lysine residue in the enzyme.\n\nStep 5 of FIGURE 29.8: Isomerization\nDihydroxyacetone phosphate is isomerized in step 5 by triose phosphate isomerase to form a second equivalent of glyceraldehyde 3-phosphate. As in the conversion of glucose 6-phosphate to fructose 6-phosphate in step 2, the isomerization takes place by keto-enol tautomerization through a common enediol intermediate. A base deprotonates C 1 and then reprotonates C2 using the same hydrogen. The net result of steps 4 and 5 together is the production of two glyceraldehyde 3-phosphate molecules, both of which pass down the rest of the pathway. Thus, each of the remaining five steps of glycolysis takes place twice for every glucose molecule entering at step 1."}
{"id": 1878, "contents": "Steps 6-7 of FIGURE 29.9: Oxidation, Phosphorylation, and Dephosphorylation - \nGlyceraldehyde 3-phosphate is oxidized and phosphorylated in step 6 to give 1,3-bisphosphoglycerate (FIGURE 29.12). The reaction is catalyzed by glyceraldehyde 3-phosphate dehydrogenase and begins by nucleophilic addition of the -SH group of a cysteine residue in the enzyme to the aldehyde carbonyl group to yield a hemithioacetal (Section 19.10), the sulfur analog of a hemiacetal. Oxidation of the hemithioacetal -OH group by $\\mathrm{NAD}^{+}$then yields a thioester, which reacts with phosphate ion in a nucleophilic acyl substitution step to yield 1,3-bisphosphoglycerate, a mixed anhydride derived from a carboxylic acid and a phosphoric acid.\n\n\nFIGURE 29.12 Mechanism of step 6 in FIGURE 29.9, the oxidation and phosphorylation of glyceraldehyde 3-phosphate to give 1,3-bisphosphoglycerate. The process occurs through initial formation of a hemiacetal that is oxidized to a thioester and converted into an acyl phosphate.\nLike all anhydrides (Section 21.5), the mixed carboxylic-phosphoric anhydride is a reactive substrate in nucleophilic acyl (or phosphoryl) substitution reactions. Reaction of 1,3-bisphosphoglycerate with ADP occurs in step 7 by substitution on phosphorus, resulting in transfer of a phosphate group to ADP and giving ATP plus 3-phosphoglycerate. The process is catalyzed by phosphoglycerate kinase and requires $\\mathrm{Mg}^{2+}$ as cofactor. Together, steps 6 and 7 accomplish the oxidation of an aldehyde to a carboxylic acid."}
{"id": 1879, "contents": "Steps 6-7 of FIGURE 29.9: Oxidation, Phosphorylation, and Dephosphorylation - \nStep 8 of FIGURE 29.9: Isomerization\n3-Phosphoglycerate isomerizes to 2-phosphoglycerate in a step catalyzed by phosphoglycerate mutase. In\nplants, 3-phosphoglycerate transfers its phosphoryl group from its C3 oxygen to a histidine residue on the enzyme in one step and then accepts the same phosphoryl group back onto the C2 oxygen in a second step. In animals and yeast, however, the enzyme contains a phosphorylated histidine, which transfers its phosphoryl group to the C2 oxygen of 3-phosphoglycerate and forms 2,3-bisphosphoglycerate as intermediate. The same histidine then accepts a phosphoryl group from the C3 oxygen to yield the isomerized product plus the regenerated enzyme. As explained in Section 29.4, we'll occasionally use an abbreviated mechanism for nucleophilic acyl substitution reactions to save space.\n\n\nSteps 9-10 of FIGURE 29.9: Dehydration and Dephosphorylation\nLike most $\\beta$-hydroxy carbonyl compounds, 2-phosphoglycerate undergoes a ready dehydration in step 9 by an E 1 cB mechanism (Section 23.3). The process is catalyzed by enolase, and the product is phosphoenolpyruvate, abbreviated PEP. Two $\\mathrm{Mg}^{2+}$ ions are associated with the 2-phosphoglycerate to neutralize the negative charges.\n\n\nTransfer of the phosphoryl group to ADP in step 10 then generates ATP and gives enolpyruvate, which tautomerizes to pyruvate. The reaction is catalyzed by pyruvate kinase and requires that a molecule of fructose 1,6-bisphosphate also be present, as well as 2 equivalents of $\\mathrm{Mg}^{2+}$. One $\\mathrm{Mg}^{2+}$ ion coordinates to ADP, and the other increases the acidity of a water molecule necessary for protonation of the enolate ion.\n\n\nThe overall result of glycolysis can be summarized by the following equation:\n\n\nGlucose\n\nPROBLEM Identify the two steps of glycolysis in which ATP is produced. 29-7"}
{"id": 1880, "contents": "Steps 6-7 of FIGURE 29.9: Oxidation, Phosphorylation, and Dephosphorylation - \nThe overall result of glycolysis can be summarized by the following equation:\n\n\nGlucose\n\nPROBLEM Identify the two steps of glycolysis in which ATP is produced. 29-7\n\nPROBLEM Look at the entire glycolysis pathway, and make a list of the kinds of organic reactions that take 29-8 place-nucleophilic acyl substitutions, aldol reactions, E1cB reactions, and so forth."}
{"id": 1881, "contents": "Steps 6-7 of FIGURE 29.9: Oxidation, Phosphorylation, and Dephosphorylation - 29.6 Conversion of Pyruvate to Acetyl CoA\nPyruvate, produced by catabolism of glucose (and by degradation of several amino acids), can undergo several further transformations depending on the conditions and on the organism. In the absence of oxygen, pyruvate can either be reduced by NADH to yield lactate $\\left[\\mathrm{CH}_{3} \\mathrm{CH}(\\mathrm{OH}) \\mathrm{CO}_{2}{ }^{-}\\right.$] or, in yeast, fermented to give ethanol. Under typical aerobic conditions in mammals, however, pyruvate is converted by a process called oxidative decarboxylation to give acetyl CoA plus $\\mathrm{CO}_{2}$. (Oxidative because the oxidation state of the carbonyl carbon rises from that of a ketone to that of a thioester.)\n\nThe conversion occurs through a multistep sequence of reactions catalyzed by a complex of enzymes and cofactors called the pyruvate dehydrogenase complex. The process occurs in three stages, each catalyzed by one of the enzymes in the complex, as outlined in FIGURE 29.13. Acetyl CoA, the ultimate product, then acts as fuel for the final stage of catabolism, the citric acid cycle.\n\nStep 1 of FIGURE 29.13: Addition of Thiamin Diphosphate\nThe conversion of pyruvate to acetyl CoA begins by reaction of pyruvate with thiamin diphosphate, a derivative of vitamin $B_{1}$. Formerly called thiamin pyrophosphate, thiamin diphosphate is usually abbreviated as TPP. The spelling thiamine is also correct and frequently used."}
{"id": 1882, "contents": "Steps 6-7 of FIGURE 29.9: Oxidation, Phosphorylation, and Dephosphorylation - 29.6 Conversion of Pyruvate to Acetyl CoA\nThe key structural element in thiamin diphosphate is the thiazolium ring-a five-membered, unsaturated heterocycle containing a sulfur atom and a positively charged nitrogen atom. The thiazolium ring is weakly acidic, with a $\\mathrm{p} K_{\\mathrm{a}}$ of approximately 18 for the ring hydrogen between N and S . Bases can therefore deprotonate thiamin diphosphate, leading to formation of an ylide much like the phosphonium ylides used in Wittig reactions (Section 19.11). As in the Wittig reaction, the TPP ylide is a nucleophile and adds to the ketone carbonyl group of pyruvate to yield an alcohol addition product.\n\n\n\n$$\n\\text { Pyruvate } \\begin{gathered}\n\\text { Thiamin } \\\\\n\\text { diphosphate } \\\\\n\\text { ylide }\n\\end{gathered}\n$$"}
{"id": 1883, "contents": "FIGURE 29.13 MECHANISM - \nMechanism for the conversion of pyruvate to acetyl CoA through a multistep sequence of reactions that requires three different enzymes and four different coenzymes. The individual steps are explained in the text.\n(1) Nucleophilic addition of thiamin diphosphate (TPP) ylide to pyruvate gives an alcohol addition product.\n\nDecarboxylation occurs in a step analogous to the loss of $\\mathrm{CO}_{2}$ from a $\\beta$-keto acid, yielding the enamine hydroxyethylthiamin diphosphate (HETPP).\n(3) The enamine double bond attacks a sulfur atom of lipoamide and carries out an $\\mathrm{S}_{\\mathrm{N}}{ }^{2-}$ like displacement of the second sulfur to yield a hemithioacetal.\n(4) Elimination of thiamin diphosphate ylide from the hemithioacetal intermediate yields acetyl dihydrolipoamide . . .\n5. . . which reacts with coenzyme A in a nucleophilic acyl substitution reaction to exchange one thioester for another and give acetyl CoA plus dihydrolipoamide.\n\n\nAcetyl CoA\n$+$\n\n\nDihydrolipoamide\n\nStep 2 of FIGURE 29.13: Decarboxylation\nThe TPP addition product, which contains an iminium ion $\\beta$ to a carboxylate anion, undergoes decarboxylation in much the same way that a $\\beta$-keto acid decarboxylates in the acetoacetic ester synthesis (Section 22.7). The $\\mathrm{C}=\\mathrm{N}^{+}$bond of the pyruvate addition product acts like the $\\mathrm{C}=\\mathrm{O}$ bond of a $\\beta$-keto acid to accept electrons as $\\mathrm{CO}_{2}$ leaves, giving hydroxyethylthiamin diphosphate (HETPP)."}
{"id": 1884, "contents": "FIGURE 29.13 MECHANISM - \nStep 3 of FIGURE 29.13: Reaction with Lipoamide\nHydroxyethylthiamin diphosphate is an enamine $\\left(\\mathrm{R}_{2} \\mathrm{~N}-\\mathrm{C}=\\mathrm{C}\\right)$, which, like all enamines, is nucleophilic (Section 23.11). It therefore reacts with the enzyme-bound disulfide lipoamide by nucleophilic attack on a sulfur atom, displacing the second sulfur in an $\\mathrm{S}_{\\mathrm{N}} 2$-like process.\n\n\n\nStep 4 of FIGURE 29.13: Elimination of Thiamin Diphosphate\nThe product of the HETPP reaction with lipoamide is a hemithioacetal, which eliminates thiamin diphosphate ylide. This elimination is the reverse of the ketone addition in step 1 and generates acetyl dihydrolipoamide.\n\n\nStep 5 of FIGURE 29.13: Acyl Transfer\nAcetyl dihydrolipoamide, a thioester, undergoes a nucleophilic acyl substitution reaction with coenzyme A to yield acetyl CoA plus dihydrolipoamide. The dihydrolipoamide is then oxidized back to lipoamide by FAD (Section 29.3), and the $\\mathrm{FADH}_{2}$ that results is in turn oxidized back to FAD by $\\mathrm{NAD}^{+}$, completing the catalytic cycle.\n\n\nPROBLEM Which carbon atoms in glucose end up as $-\\mathrm{CH}_{3}$ carbons in acetyl CoA? Which carbons end up as 29-9 $\\mathrm{CO}_{2}$ ?"}
{"id": 1885, "contents": "FIGURE 29.13 MECHANISM - 29.7 The Citric Acid Cycle\nThe initial stages of catabolism result in the conversion of both fats and carbohydrates into acetyl groups that are bonded through a thioester link to coenzyme A. Acetyl CoA then enters the next stage of catabolism-the citric acid cycle, also called the tricarboxylic acid (TCA) cycle, or Krebs cycle, after Hans Krebs, who unraveled its complexities in 1937. The overall result of the cycle is the conversion of an acetyl group into two molecules of $\\mathrm{CO}_{2}$ plus reduced coenzymes by the eight-step reaction sequence shown in FIGURE 29.14.\n\nAs its name implies, the citric acid cycle is a closed loop of reactions in which the product of the final step (oxaloacetate) is a reactant in the first step. The intermediates are constantly regenerated, flowing continuously through the cycle, which operates as long as the oxidizing coenzymes $\\mathrm{NAD}^{+}$and FAD are available. To meet this condition, the reduced coenzymes NADH and $\\mathrm{FADH}_{2}$ must be reoxidized via the electron-transport chain, which in turn relies on oxygen as the ultimate electron acceptor. Thus, the cycle is dependent on the availability of oxygen and on the operation of the electron-transport chain.\n\nFIGURE 29.14 MECHANISM\n\nThe citric acid cycle is an eight-step series of reactions that results in the conversion of an acetyl group into two molecules of $\\mathbf{C O}_{2}$ plus reduced coenzymes. Individual steps are explained in the text.\n\n\n\nOxaloacetate\n8 Oxidation of (S)-malate gives oxaloacetate, completing the cycle.\n\n(S)-Malate\n(L-malate)\n(7) Fumarate undergoes conjugate addition of water to its double bond to give (S)-malate.\n\n\nFumarate\n(6) Succinate is dehydrogenated by FAD to give fumarate.\n\n\n\n\nCitrate\n2\n\nCitrate is isomerized by dehydration and rehydration to give isocitrate.\n\n\nIsocitrate\n\n$\\alpha$-Ketoglutarate\n\n\n\nSuccinate"}
{"id": 1886, "contents": "FIGURE 29.13 MECHANISM - 29.7 The Citric Acid Cycle\nCitrate\n2\n\nCitrate is isomerized by dehydration and rehydration to give isocitrate.\n\n\nIsocitrate\n\n$\\alpha$-Ketoglutarate\n\n\n\nSuccinate\n\n\nSuccinyl CoA\nSuccinyl CoA is converted to\nsuccinate in a reaction coupled to the phosphorylation of GDP to give GTP.\n\nStep 1 of FIGURE 29.14: Addition to Oxaloacetate\nAcetyl CoA enters the citric acid cycle in step 1 by nucleophilic addition to the oxaloacetate carbonyl group, to give (S)-citryl CoA. This addition is an aldol reaction and is catalyzed by citrate synthase, as discussed in Section 26.11. (S)-Citryl CoA is then hydrolyzed to citrate by a typical nucleophilic acyl substitution reaction with water, catalyzed by the same citrate synthase enzyme.\n\nNote that the hydroxyl-bearing carbon of citrate is a prochirality center and contains two identical arms. Because the initial aldol reaction of acetyl CoA to oxaloacetate occurs specifically from the Si face of the\nketone carbonyl group, the pro- $S$ arm of citrate is derived from acetyl CoA and the pro- $R$ arm is derived from oxaloacetate."}
{"id": 1887, "contents": "FIGURE 29.13 MECHANISM - 29.7 The Citric Acid Cycle\nCitrate\nStep 2 of FIGURE 29.14: Isomerization\nCitrate, a prochiral tertiary alcohol, is next converted into its isomer, ( $2 R, 3 S$ )-isocitrate, a chiral secondary alcohol. The isomerization occurs in two steps, both of which are catalyzed by the same aconitase enzyme. The initial step is an E1cB dehydration of a $\\beta$-hydroxy acid to give cis-aconitate, the same sort of reaction that occurs in step 9 of glycolysis (FIGURE 29.8). The second step is a conjugate nucleophilic addition of water to the $\\mathrm{C}=\\mathrm{C}$ bond (Section 19.13). The dehydration of citrate takes place specifically on the pro- $R$ arm-the one derived from oxaloacetate-rather than on the pro- $S$ arm derived from acetyl CoA.\n\n\nStep 3 of FIGURE 29.14: Oxidation and Decarboxylation\n( $2 R, 3 S$ )-Isocitrate, a secondary alcohol, is oxidized by $\\mathrm{NAD}^{+}$in step 3 to give the ketone oxalosuccinate, which loses $\\mathrm{CO}_{2}$ to give $\\alpha$-ketoglutarate. Catalyzed by isocitrate dehydrogenase, the decarboxylation is a typical reaction of a $\\beta$-keto acid, just like that in the acetoacetic ester synthesis (Section 22.7). The enzyme requires a divalent cation as a cofactor to polarize the ketone carbonyl group and make it a better electron acceptor."}
{"id": 1888, "contents": "FIGURE 29.13 MECHANISM - 29.7 The Citric Acid Cycle\nStep 4 of FIGURE 29.14: Oxidative Decarboxylation\nThe transformation of $\\alpha$-ketoglutarate to succinyl CoA in step 4 is a multistep process just like the transformation of pyruvate to acetyl CoA that we saw in FIGURE 29.13. In both cases, an $\\alpha$-keto acid loses $\\mathrm{CO}_{2}$ and is oxidized to a thioester in a series of steps catalyzed by a multienzyme dehydrogenase complex. As in the conversion of pyruvate to acetyl CoA, the reaction involves an initial nucleophilic addition reaction of thiamin diphosphate ylide to $\\alpha$-ketoglutarate, followed by decarboxylation. Reaction with lipoamide, elimination of TPP ylide, and finally a transesterification of the dihydrolipoamide thioester with coenzyme A yields succinyl CoA.\n\n\nStep 5 of FIGURE 29.14: Acyl CoA Cleavage\nSuccinyl CoA is converted to succinate in step 5. This reaction is catalyzed by succinyl CoA synthetase and is coupled with phosphorylation of guanosine diphosphate (GDP) to give guanosine triphosphate (GTP). The overall transformation is similar to that of steps 6 through 8 in glycolysis (FIGURE 29.8), in which a thioester is converted into an acyl phosphate and a phosphate group is then transferred to ADP. The overall result is a \"hydrolysis\" of the thioester group without involvement of water.\n\n\nStep 6 of FIGURE 29.14: Dehydrogenation\nSuccinate is dehydrogenated in step 6 by the FAD-dependent succinate dehydrogenase to give fumarate. This process is analogous to what occurs during the $\\beta$-oxidation pathway of fatty-acid catabolism (Section 29.3). The reaction is stereospecific, removing the pro- $S$ hydrogen from one carbon and the pro- $R$ hydrogen from the other.\n\n\nSuccinate\n\n\nFumarate"}
{"id": 1889, "contents": "FIGURE 29.13 MECHANISM - 29.7 The Citric Acid Cycle\nSuccinate\n\n\nFumarate\n\nSteps 7-8 of FIGURE 29.14: Hydration and Oxidation\nThe final two steps in the citric acid cycle are the conjugate nucleophilic addition of water to fumarate to yield $(S)$-malate and the oxidation of $(S)$-malate by $\\mathrm{NAD}^{+}$to give oxaloacetate. The addition is catalyzed by fumarase and is mechanistically similar to the addition of water to cis-aconitate in step 2 . This reaction occurs through an enolate-ion intermediate, which is protonated on the side opposite the OH , leading to a net anti addition.\n\n\nThe final step is the oxidation of $(S)$-malate by $\\mathrm{NAD}^{+}$to give oxaloacetate, a reaction catalyzed by malate dehydrogenase. The citric acid cycle has now returned to its starting point, ready to revolve again. The overall result of the cycle is\n\n\nPROBLEM Which of the substances in the citric acid cycle are tricarboxylic acids, thus giving the cycle its 29-10 alternative name?\n\nPROBLEM Write mechanisms for step 2 of the citric acid cycle, the dehydration of citrate and the addition of\n29-11 water to aconitate.\nPROBLEM Is the pro- $R$ or pro- $S$ hydrogen removed from citrate during the dehydration in step 2 of the citric 29-12 acid cycle? Does the elimination reaction occur with syn or anti geometry?"}
{"id": 1890, "contents": "FIGURE 29.13 MECHANISM - 29.8 Carbohydrate Biosynthesis: Gluconeogenesis\nGlucose is the body's primary fuel when food is plentiful, but in times of fasting or prolonged exercise, glucose stores can become depleted. Most tissues then begin metabolizing fats as their source of acetyl CoA, but the brain is different. The brain relies almost entirely on glucose for fuel and is dependent on receiving a continuous supply in the blood. When the supply of glucose fails, even for a brief time, irreversible damage can occur. Thus, a pathway for synthesizing glucose from simple precursors is crucial.\n\nHigher organisms are not able to synthesize glucose from acetyl CoA but must instead use one of the threecarbon precursors lactate, glycerol, or alanine, all of which are readily converted into pyruvate.\n\n\n\nPyruvate then becomes the starting point for gluconeogenesis, the 11-step biosynthetic pathway by which organisms make glucose (FIGURE 29.15). The gluconeogenesis pathway is not the reverse of the glycolysis pathway by which glucose is degraded. As with the catabolic and anabolic pathways for fatty acids (Section 29.3 and Section 29.4), the catabolic and anabolic pathways for carbohydrates differ in some details so that both are energetically favorable."}
{"id": 1891, "contents": "FIGURE 29.15 MECHANISM - \nThe gluconeogenesis pathway for the biosynthesis of glucose from pyruvate. Individual steps are explained in the text.\n\nPyruvate\n\n(1) Pyruvate undergoes a biotin-dependent carboxylation on the methyl group to give oxaloacetate...\n\n\nOxaloacetate\n\n2. . . which is decarboxylated and then phosphorylated by GTP to give phosphoenolpyruvate.\n\n\nPhosphoenolpyruvate\n(3) Conjugate nucleophilic addition of water\nto the double bond of phosphoenolpyruvate 3 (\ngives 2 -phosphoglycerate $\\ldots$.\n\n\n2-Phosphoglycerate\n\n4. . . which is isomerized by transfer of the phosphoryl group to give 3-phosphoglycerate.\n\n3-Phosphoglycerate\n\n\n\n5 Phosphorylation of the carboxyl group by reaction with ATP yields 1,3-bisphosphoglycerate.\n\n\n1,3-Bisphosphoglycerate\n\n(6) Reduction of the acyl phosphate gives glyceraldehyde 3-phosphate, which\n(7) undergoes keto-enol tautomerization to yield dihydroxyacetone phosphate.\n\n\n\nDihydroxyacetone phosphate\n\n$\\stackrel{7}{4}$\n\n\n\nGlyceraldehyde 3-phosphate\n\n$\\downarrow$\n\n\nStep 1 of FIGURE 29.15: Carboxylation\nGluconeogenesis begins with the carboxylation of pyruvate to yield oxaloacetate. The reaction is catalyzed by pyruvate carboxylase and requires ATP, bicarbonate ion, and the coenzyme biotin, which acts as a carrier to transport $\\mathrm{CO}_{2}$ to the enzyme active site. The mechanism is analogous to that of step 3 in fatty-acid biosynthesis (FIGURE 29.6), in which acetyl CoA is carboxylated to yield malonyl CoA."}
{"id": 1892, "contents": "FIGURE 29.15 MECHANISM - \nStep 2 of FIGURE 29.15: Decarboxylation and Phosphorylation\nDecarboxylation of oxaloacetate, a $\\beta$-keto acid, occurs by the typical retro-aldol mechanism like that in step 3 in the citric acid cycle (FIGURE 29.14), and phosphorylation of the resultant pyruvate enolate ion by GTP occurs concurrently to give phosphoenolpyruvate. This reaction is catalyzed by phosphoenolpyruvate carboxykinase.\n\n\nSteps 3-4 of FIGURE 29.15: Hydration and Isomerization\nConjugate nucleophilic addition of water to the double bond of phosphoenolpyruvate gives 2-phosphoglycerate by a process similar to that of step 7 in the citric acid cycle. Phosphorylation of C3 and dephosphorylation of C2 then yields 3-phosphoglycerate. Mechanistically, these steps are the reverse of steps 9 and 8 in glycolysis (FIGURE 29.8), which have equilibrium constants near 1 so that substantial amounts of reactant and product are both present.\n\n\nSteps 5-7 of FIGURE 29.15: Phosphorylation, Reduction, and Tautomerization\nReaction of 3-phosphoglycerate with ATP generates the corresponding acyl phosphate, 1,3-bisphosphoglycerate, which binds to the glyceraldehyde 3-phosphate dehydrogenase by a thioester bond to a cysteine residue. Reduction of the thioester by $\\mathrm{NADH} / \\mathrm{H}^{+}$yields the corresponding aldehyde, and keto-enol tautomerization of the aldehyde gives dihydroxyacetone phosphate. All three steps comprise a mechanistic reversal of the corresponding steps 7,6 , and 5 of glycolysis and have equilibrium constants near 1."}
{"id": 1893, "contents": "FIGURE 29.15 MECHANISM - \nStep 8 of FIGURE 29.16: Aldol Reaction\nDihydroxyacetone phosphate and glyceraldehyde 3-phosphate, the two 3-carbon units produced in step 7, join by an aldol reaction to give fructose 1,6-bisphosphate, the reverse of step 4 in glycolysis (FIGURE 29.11). As in glycolysis, the reaction is catalyzed in plants and animals by a class I aldolase and takes place on an iminium ion formed by reaction of dihydroxyacetone phosphate with a side-chain lysine $-\\mathrm{NH}_{2}$ group on the enzyme. Loss of a proton from the neighboring carbon then generates an enamine, an aldol-like reaction ensues, and the product is hydrolyzed.\n\n\nSteps 9-10 of FIGURE 29.16: Hydrolysis and Isomerization\nHydrolysis of the phosphate group at C1 of fructose 1,6-bisphosphate gives fructose 6-phosphate. Although the result of the reaction is the exact opposite of step 3 in glycolysis, the mechanism is not. In glycolysis, phosphorylation is accomplished by reaction of fructose with ATP, with formation of ADP as by-product. The reverse of that process, however-the reaction of fructose 1,6-bisphosphate with ADP to give fructose 6-phosphate and ATP-is energetically unfavorable because ATP is too high in energy. Thus, an alternative pathway is used in which the C1 phosphate group is removed by a direct hydrolysis reaction, catalyzed by fructose 1,6-bisphosphatase.\n\nFollowing hydrolysis, keto-enol tautomerization of the carbonyl group from C 2 to C 1 gives glucose 6-phosphate. The isomerization is the reverse of step 2 in glycolysis."}
{"id": 1894, "contents": "FIGURE 29.15 MECHANISM - \nFollowing hydrolysis, keto-enol tautomerization of the carbonyl group from C 2 to C 1 gives glucose 6-phosphate. The isomerization is the reverse of step 2 in glycolysis.\n\n\nStep 11 of FIGURE 29.16: Hydrolysis\nThe final step in gluconeogenesis is the conversion of glucose 6-phosphate to glucose by a second phosphatasecatalyzed hydrolysis reaction. As just discussed for the hydrolysis of fructose 1,6-bisphosphate in step 9, and for the same energetic reasons, the mechanism of the glucose 6-phosphate hydrolysis is not the exact opposite\nof the corresponding step 1 in glycolysis.\nInterestingly, however, the mechanisms of the two phosphate hydrolysis reactions in steps 9 and 11 are not the same. In step 9, water is the nucleophile, but in the glucose 6-phosphate reaction of step 11, a histidine residue on the enzyme attacks phosphorus, giving a phosphoryl enzyme intermediate that subsequently reacts with water.\n\n\nThe overall result of gluconeogenesis is summarized by the following equation:\n\n\nPROBLEM Write a mechanism for step 6 of gluconeogenesis, the reduction of 3-phosphoglyceryl phosphate 29-13 with $\\mathrm{NADH} / \\mathrm{H}^{+}$to yield glyceraldehyde 3-phosphate."}
{"id": 1895, "contents": "FIGURE 29.15 MECHANISM - 29.9 Catabolism of Proteins: Deamination\nThe catabolism of proteins is much more complex than that of fats and carbohydrates because each of the 20 $\\alpha$-amino acids is degraded through its own unique pathway. The general idea, however, is that (1) the $\\alpha$ amino group is first removed as ammonia by a deamination process, (2) the ammonia is converted into urea, and (3) the remaining amino acid carbon skeleton (usually an $\\alpha$-keto acid) is converted into a compound that enters the citric acid cycle."}
{"id": 1896, "contents": "Transamination - \nDeamination is usually accomplished by a transamination reaction in which the $-\\mathrm{NH}_{2}$ group of the amino acid is exchanged with the keto group of $\\alpha$-ketoglutarate, forming a new $\\alpha$-keto acid plus glutamate. The overall process occurs in two parts, is catalyzed by aminotransferases, and involves participation of the coenzyme pyridoxal phosphate, abbreviated PLP, a derivative of pyridoxine (vitamin $\\mathrm{B}_{6}$ ). The aminotransferases differ in their specificity for amino acids, but the mechanism remains the same.\n\n\n\nThe mechanism of the first part of transamination is shown in FIGURE 29.17. The process begins with reaction between the $\\alpha$-amino acid and pyridoxal phosphate, which is covalently bonded to the aminotransferase by an imine linkage between the side-chain $-\\mathrm{NH}_{2}$ group of a lysine residue in the enzyme and the PLP aldehyde group. Deprotonation/reprotonation of the PLP-amino acid imine effects tautomerization of the imine $\\mathrm{C}=\\mathrm{N}$ bond, and hydrolysis of the tautomerized imine gives an $\\alpha$-keto acid plus pyridoxamine phosphate (PMP).\n\nFIGURE 29.17 MECHANISM\n\nMechanism for the enzyme-catalyzed, PLP-dependent transamination of an $\\alpha$-amino acid to give an $\\boldsymbol{\\alpha}$-keto acid. Individual steps are explained in the text.\n(1) An amino acid reacts with the enzyme-bound PLP imine by nucleophilic addition of its $-\\mathrm{NH}_{2}$ group to the $\\mathrm{C}=\\mathrm{N}$ bond of the imine, giving a PLP-amino acid imine and releasing the enzyme\n\n\nPLP-amino acid imine (Schiff base)\n\n(2) Deprotonation of the acidic $\\alpha$ carbon of the amino acid gives an intermediate $\\alpha$-keto acid imine . . ."}
{"id": 1897, "contents": "Transamination - \nPLP-amino acid imine (Schiff base)\n\n(2) Deprotonation of the acidic $\\alpha$ carbon of the amino acid gives an intermediate $\\alpha$-keto acid imine . . .\n\n(3).. that is reprotonated on the PLP carbon. The net result of this deprotonation/reprotonation sequence is tautomerization of the imine $C=N$ bond.\n$\\alpha$-Keto acid imine tautomer\n(4) Hydrolysis of the $\\alpha$-keto acid imine by nucleophilic addition of water to the $\\mathrm{C}=\\mathrm{N}$ bond gives the transamination products pyridoxamine\n\u0640\nphosphate (PMP) and $\\alpha$-keto acid.\n\nPyridoxamine phosphate\n(PMP)\n\n\nStep 1 of FIGURE 29.17: Transimination\nThe first step in transamination is transimination-the reaction of the PLP-enzyme imine with an $\\alpha$-amino acid\nto give a PLP-amino acid imine plus expelled enzyme as the leaving group. The reaction occurs by nucleophilic addition of the amino acid $-\\mathrm{NH}_{2}$ group to the $\\mathrm{C}=\\mathrm{N}$ bond of the PLP imine, much as an amine adds to the $\\mathrm{C}=\\mathrm{O}$ bond of a ketone or aldehyde in a nucleophilic addition reaction (Section 19.8). The protonated diamine intermediate undergoes a proton transfer and expels the lysine amino group in the enzyme to complete the step."}
{"id": 1898, "contents": "Transamination - \nSteps 2-4 of FIGURE 29.17: Tautomerization and Hydrolysis\nFollowing formation of the PLP-amino acid imine in step 1, a tautomerization of the $\\mathrm{C}=\\mathrm{N}$ bond occurs in step 2. The basic lysine residue in the enzyme that was expelled as a leaving group during transimination deprotonates the acidic $\\alpha$ position of the amino acid, with the protonated pyridine ring of PLP acting as the electron acceptor. Reprotonation occurs on the carbon atom next to the ring, generating a tautomeric product that is the imine of an $\\alpha$-keto acid with pyridoxamine phosphate, abbreviated PMP.\n\nHydrolysis of this PMP- $\\alpha$-keto acid imine then completes the first part of the transamination reaction. This is the mechanistic reverse of imine formation and occurs by nucleophilic addition of water to the imine, followed by proton transfer and expulsion of PMP as leaving group."}
{"id": 1899, "contents": "Regeneration of PLP from PMP - \nWith PLP plus the $\\alpha$-amino acid now converted into PMP plus an $\\alpha$-keto acid, PMP must be transformed back into PLP to complete the catalytic cycle. The conversion occurs by another transamination reaction, this one between PMP and an $\\alpha$-keto acid, usually $\\alpha$-ketoglutarate. The products are PLP plus glutamate, and the mechanism is the exact reverse of that shown in FIGURE 29.17. That is, PMP and $\\alpha$-ketoglutarate give an imine; the PMP- $\\alpha$-ketoglutarate imine undergoes tautomerization of the $\\mathrm{C}=\\mathrm{N}$ bond to give a PLP-glutamate imine; and the PLP-glutamate imine reacts with a lysine residue on the enzyme in a transimination process to yield PLP-enzyme imine plus glutamate.\n\n\nPROBLEM Write all the steps in the transamination reaction of PMP with $\\alpha$-ketoglutarate plus a lysine residue\n29-14 in the enzyme to give the PLP-enzyme imine plus glutamate.\nPROBLEM What $\\alpha$-keto acid is formed on transamination of leucine?\n29-15\nPROBLEM From what amino acid is the following $\\alpha$-keto acid derived?\n29-16"}
{"id": 1900, "contents": "Regeneration of PLP from PMP - 29.10 Some Conclusions about Biological Chemistry\nAs promised in the chapter introduction, the past few sections have been a fast-paced tour of a large number of reactions. Following it all undoubtedly required a lot of work and a lot of page-turning to look at earlier sections.\n\nAfter examining the various metabolic pathways, perhaps the main conclusion about biological chemistry is the remarkable similarity between the mechanisms of biological reactions and the mechanisms of laboratory reactions. If you were to look at the steps of vitamin $B_{12}$ biosynthesis, you would see the same kinds of reactions we've been seeing throughout the text-nucleophilic substitutions, eliminations, aldol reactions, nucleophilic acyl substitutions, and so forth. There are, of course, some complexities, but the fundamental mechanisms of organic chemistry remain the same, whether in the laboratory with smaller molecules or in organisms with larger molecules.\n\n\nSuccinyl CoA\n$+\\quad \\longrightarrow$\n\n\nGlycine\n\n\nVitamin $B_{12}$-cyanocobalamin\n\nSo what is there to be learned from studying metabolism? One good answer is given in the following Chemistry Matters, which relates the story of how knowledge of a biosynthetic pathway led to the design of new drugs that have saved millions of lives."}
{"id": 1901, "contents": "Statin Drugs - \nCoronary heart disease-the buildup of cholesterol-containing plaques on the walls of heart arteries-is the leading cause of death for people older than 20 in industrialized countries. It's estimated that up to one-third of women and one-half of men will develop the disease at some point in their lives.\n\nThe onset of coronary heart disease is directly correlated with blood cholesterol levels (see the Chapter 27 Chemistry Matters), and the first step in disease prevention is to lower those levels. It turns out that only about $25 \\%$ of your blood cholesterol comes from what you eat; the remaining $75 \\%$-about 1 gram each day-is biosynthesized in your liver from dietary fats and carbohydrates. Thus, any effective plan for lowering your cholesterol level means limiting the amount that your body makes, which is where a detailed chemical knowledge of cholesterol biosynthesis comes in.\n\n\nFIGURE 29.18 The buildup of cholesterol deposits inside arteries can cause coronary heart disease, a leading cause of death for adults in industrialized nations.\n\nWe saw in Section 27.5 and Section 27.7 that all steroids, including cholesterol, are biosynthesized from the triterpenoid lanosterol, which in turn comes from acetyl CoA through isopentenyl diphosphate. If you knew all the mechanisms for all the chemical steps in cholesterol biosynthesis, you might be able to devise a drug that would block one of those steps, thereby short-circuiting the biosynthetic process and controlling the amount of cholesterol produced.\n\nBut we do know those mechanisms! Look back at the pathway for the biosynthesis of isopentenyl diphosphate from acetyl CoA, shown in FIGURE 27.8. It turns out that the rate-limiting step in the pathway is the reduction of 3-hydroxy-3-methylglutaryl CoA (abbreviated HMG-CoA) to mevalonate, brought about by the enzyme HMGCoA reductase. If that enzyme could be stopped from functioning, cholesterol biosynthesis would also be stopped."}
{"id": 1902, "contents": "Statin Drugs - \nTo find a drug that blocks HMG-CoA reductase, chemists did two simultaneous experiments on a large number of potential drug candidates isolated from soil microbes. In one experiment, the drug candidate and mevalonate were added to liver extract; in the second experiment, only the drug candidate was added without mevalonate. If cholesterol was produced only in the presence of added mevalonate but not in the absence of mevalonate, the drug candidate must have blocked the enzyme for mevalonate synthesis.\n\nThe drugs that block HMG-CoA reductase, and thus control cholesterol synthesis in the body, are called statins. In just the 10-year period following their introduction in 1994, the death rate from coronary heart disease decreased by $33 \\%$ in the United States.\n\nLike many drugs, statins don't come without risks and have some serious side effects that people considering their use should be aware of, but approaching 30 years later, they remain among the most widely prescribed drugs in the world, with estimated annual sales of $\\$ 14$ billion. Atorvastatin (Lipitor), simvastatin (Zocor), rosuvastatin (Crestor), pravastatin (Pravachol), and lovastatin (Mevacor) are some examples. An X-ray crystal structure of the active site in the HMG-CoA reductase enzyme is shown in the accompanying graphic, along with a molecule of atorvastatin (purple) that is tightly bound in the active site and stops the enzyme from functioning. A good understanding of organic chemistry certainly paid off in this instance.\n\n\n\nAtorvastatin (Lipitor)"}
{"id": 1903, "contents": "Key Terms - \n- anabolism\n- gluconeogenesis\n- $\\beta$-oxidation pathway\n- glycolysis\n- catabolism\n- metabolism\n- citric acid cycle\n- Schiff base\n- deamination\n- transamination"}
{"id": 1904, "contents": "Summary - \nMetabolism is the sum of all chemical reactions in the body. Reactions that break down large molecules into\nsmaller fragments are called catabolism, and those that build up large molecules from small pieces are called anabolism. Although the details of specific biochemical pathways are sometimes complex, all the reactions that occur follow the normal rules of organic chemical reactivity.\n\nThe catabolism of fats begins with digestion, in which ester bonds are hydrolyzed to give glycerol and fatty acids. The fatty acids are degraded in the four-step $\\boldsymbol{\\beta}$-oxidation pathway by removal of two carbons at a time, yielding acetyl CoA. Catabolism of carbohydrates begins with the hydrolysis of glycoside bonds to give glucose, which is degraded in the ten-step glycolysis pathway. Pyruvate, the initial product of glycolysis, is then converted into acetyl CoA. Acetyl CoA next enters the eight-step citric acid cycle, where it is further degraded into $\\mathrm{CO}_{2}$. The cycle is a closed loop of reactions in which the product of the final step (oxaloacetate) is a reactant in the first step.\n\nCatabolism of proteins is more complex than that of fats or carbohydrates because each of the 20 different amino acids is degraded by its own unique pathway. In general, though, the amino nitrogen atoms are removed and the substances that remain are converted into compounds that enter the citric acid cycle. Most amino acids lose their nitrogen atom by transamination, a reaction in which the $-\\mathrm{NH}_{2}$ group of the amino acid trades places with the keto group of an $\\alpha$-keto acid such as $\\alpha$-ketoglutarate. The products are a new $\\alpha$-keto acid and glutamate.\n\nThe energy released in catabolic pathways is used in the electron-transport chain to make molecules of adenosine triphosphate, ATP. ATP, the final result of food catabolism, couples to and drives many otherwise unfavorable reactions."}
{"id": 1905, "contents": "Summary - \nBiomolecules are synthesized as well as degraded, but the pathways for anabolism and catabolism are not the exact reverse of one another. Fatty acids are biosynthesized from acetate by an 8 -step pathway, and carbohydrates are made from pyruvate by the 11-step gluconeogenesis pathway."}
{"id": 1906, "contents": "Visualizing Chemistry - \nPROBLEM Identify the amino acid that is a catabolic precursor of each of the following $\\alpha$-keto acids:\n29-17 (a)\n\n(b)\n\n\nPROBLEM Identify the following intermediate in the citric acid cycle, and tell whether it has $R$ or $S$\n29-18 stereochemistry:\n\n\nPROBLEM The following compound is an intermediate in the biosynthesis of one of the 20 common $\\alpha$-amino\n29-19 acids. Which one is it likely to be, and what kind of chemical change must take place to complete the biosynthesis?\n\n\nPROBLEM The following compound is an intermediate in the pentose phosphate pathway, an alternative route 29-20 for glucose metabolism. Identify the sugar it is derived from."}
{"id": 1907, "contents": "Mechanism Problems - \nPROBLEM In the pentose phosphate pathway for degrading sugars, ribulose 5-phosphate is converted to\n29-21 ribose 5-phosphate. Propose a mechanism for the isomerization.\n\n\nRibulose 5-phosphate\n\n\nRibose 5-phosphate\n\nPROBLEM Another step in the pentose phosphate pathway for degrading sugars (see Problem 29-21) is the\n29-22 conversion of ribose 5-phosphate to glyceraldehyde 3-phosphate. What kind of organic process is occurring? Propose a mechanism for the conversion.\n\n\nPROBLEM One of the steps in the pentose phosphate pathway for glucose catabolism is the reaction of 29-23 sedoheptulose 7-phosphate with glyceraldehyde 3-phosphate in the presence of a transaldolase to yield erythrose 4-phosphate and fructose 6-phosphate.\n\n(a) The first part of the reaction is the formation of a protonated Schiff base of sedoheptulose 7-phosphate with a lysine residue in the enzyme followed by a retro-aldol cleavage to give an enamine plus erythrose 4-phosphate. Show the structure of the enamine and the mechanism by which it is formed.\n(b) The second part of the reaction is a nucleophilic addition of the enamine to glyceraldehyde 3 -phosphate followed by hydrolysis of the Schiff base to give fructose 6-phosphate. Show the mechanism.\n\nPROBLEM One of the steps in the pentose phosphate pathway for glucose catabolism is the reaction of xylulose\n29-24 5-phosphate with ribose 5-phosphate in the presence of a transketolase to give glyceraldehyde 3 -phosphate and sedoheptulose 7-phosphate."}
{"id": 1908, "contents": "Mechanism Problems - \n(a) The first part of the reaction is nucleophilic addition of thiamin diphosphate (TPP) ylide to xylulose 5-phosphate, followed by a retro-aldol cleavage to give glyceraldehyde 3-phosphate and a TPP-containing enamine. Show the structure of the enamine and the mechanism by which it is formed.\n(b) The second part of the reaction is addition of the enamine to ribose 5-phosphate followed by loss of TPP ylide to give sedoheptulose 7-phosphate. Show the mechanism.\n\nPROBLEM The amino acid tyrosine is biologically degraded by a series of steps that include the following\n29-25 transformations:\n\n\n\nThe double-bond isomerization of maleoylacetoacetate to fumaroylacetoacetate is catalyzed by practically any nucleophile, : $\\mathrm{Nu}^{-}$. Propose a mechanism.\n\nPROBLEM Propose a mechanism for the conversion of fumaroylacetoacetate to fumarate plus acetoacetate 29-26 (see Problem 29-25).\n\nPROBLEM Propose a mechanism for the conversion of acetoacetate to acetyl CoA (see Problem 29-25). 29-27\n\nPROBLEM Design your own degradative pathway. You know the rules (organic mechanisms), and you've seen 29-28 the kinds of reactions that occur in the biological degradation of fats and carbohydrates into acetyl CoA. If you were Mother Nature, what series of steps would you use to degrade the amino acid serine into acetyl CoA?\n\n\nPROBLEM The amino acid serine is biosynthesized by a route that involves reaction of 29-29 3-phosphohydroxypyruvate with glutamate to give 3-phosphoserine. Propose a mechanism.\n\n\n3-Phosphohydroxypyruvate"}
{"id": 1909, "contents": "Mechanism Problems - \n3-Phosphohydroxypyruvate\n\n\n\n3-Phosphoserine\nPROBLEM The amino acid leucine is biosynthesized from $\\alpha$-ketoisocaproate, which is itself prepared from\n29-30 $\\alpha$-ketoisovalerate by a multistep route that involves (1) reaction with acetyl CoA, (2) hydrolysis, (3) dehydration, (4) hydration, (5) oxidation, and (6) decarboxylation. Show the steps in the transformation, and propose a mechanism for each.\n\n\nPROBLEM The amino acid cysteine, $\\mathrm{C}_{3} \\mathrm{H}_{7} \\mathrm{NO}_{2} \\mathrm{~S}$, is biosynthesized from a substance called cystathionine by a 29-31 multistep pathway.\n\n(a) The first step is a transamination. What is the product?\n(b) The second step is an E1cB reaction. Show the products and the mechanism of the reaction.\n(c) The final step is a double-bond reduction. What organic cofactor is required for this reaction, and what is the product represented by the question mark in the equation?"}
{"id": 1910, "contents": "Enzymes and Coenzymes - \nPROBLEM What chemical events occur during the digestion of food?\n29-32\n\nPROBLEM What is the difference between digestion and metabolism? 29-33\n\nPROBLEM What is the difference between anabolism and catabolism? 29-34\n\nPROBLEM Draw the structure of adenosine $5^{\\prime}$-monophosphate (AMP), an intermediate in some biochemical 29-35 pathways.\n\nPROBLEM Cyclic adenosine monophosphate (cyclic AMP), a modulator of hormone action, is related to AMP 29-36 (Problem 29-35) but has its phosphate group linked to two hydroxyl groups at C3' and C5' of the sugar. Draw the structure of cyclic AMP.\n\nPROBLEM What general kind of reaction does ATP carry out? 29-37\n\nPROBLEM What general kind of reaction does $\\mathrm{NAD}^{+}$carry out? 29-38\n\nPROBLEM What general kind of reaction does FAD carry out? 29-39\n\nPROBLEM What enzyme cofactor is associated with each of the following kinds of reactions?\n29-40 (a) Transamination\n(b) Carboxylation of a ketone\n(c) Decarboxylation of an $\\alpha$-keto acid\n\nPROBLEM Lactate, a product of glucose catabolism in oxygen-starved muscles, can be converted into pyruvate 29-41 by oxidation. What coenzyme do you think is needed? Write the equation in the normal biochemical format using a curved arrow.\n\n\nLactate"}
{"id": 1911, "contents": "Metabolism - \nPROBLEM Write the equation for the final step in the $\\beta$-oxidation pathway of any fatty acid with an even 29-42 number of carbon atoms.\n\nPROBLEM Show the products of each of the following reactions:\n29-43\n(a)\n\n(b)\nProduct of $(\\mathbf{a})+\\mathrm{H}_{2} \\mathrm{O} \\xrightarrow{\\begin{array}{l}\\text { Enoyl-CoA } \\\\ \\text { hydratase }\\end{array}}$ ?\n(c)\nProduct of (b)\n\n\nPROBLEM Why aren't the glycolysis and gluconeogenesis pathways the exact reverse of each other? 29-44\n\nPROBLEM How many moles of acetyl CoA are produced by catabolism of the following substances?\n29-45 (a) 1.0 mol of glucose\n(b) 1.0 mol of palmitic acid\n(c) 1.0 mol of maltose\n\nPROBLEM How many grams of acetyl CoA (MW $=809.6 \\mathrm{amu})$ are produced by catabolism of the following 29-46 substances? Which substance is the most efficient precursor of acetyl CoA on a weight basis?\n(a) 100.0 g of glucose\n(b) 100.0 g of palmitic acid\n(c) 100.0 g of maltose\n\nPROBLEM What is the structure of the $\\alpha$-keto acid formed by transamination of each of the following amino 29-47 acids?\n(a) Threonine\n(b) Phenylalanine\n(c) Asparagine\n\nPROBLEM The glycolysis pathway shown in Figure 29.8 has a number of intermediates that contain phosphate 29-48 groups. Why can 3-phosphoglyceryl phosphate and phosphoenolpyruvate transfer a phosphate group to ADP while glucose 6-phosphate cannot?\n\nPROBLEM Write a mechanism for the conversion of $\\alpha$-ketoglutarate to succinyl CoA in step 4 of the citric acid 29-49 cycle (Figure 29.14)."}
{"id": 1912, "contents": "Metabolism - \nPROBLEM Write a mechanism for the conversion of $\\alpha$-ketoglutarate to succinyl CoA in step 4 of the citric acid 29-49 cycle (Figure 29.14).\n\nPROBLEM In step 2 of the citric acid cycle (Figure 29.14), cis-aconitate reacts with water to give ( $2 R, 3 S$ )-29-50 isocitrate. Does -OH add from the Re face of the double bond or from the Si face? What about -H? Does the addition of water occur with syn or anti geometry?"}
{"id": 1913, "contents": "General Problems - \nPROBLEM In glycerol metabolism, the oxidation of sn-glycerol 3-phosphate to give dihydroxyacetone\n29-51 phosphate is catalyzed by sn-glycerol-3-phosphate dehydrogenase, with NAD ${ }^{+}$as cofactor. The reaction is stereospecific, occurring exclusively on the Re face of the nicotinamide ring.\n\n$N A D^{+}$\n\n\nNADH\n\nWhich hydrogen in the NADH product comes from sn-glycerol 3-phosphate? Does it have pro- $R$ or pro- $S$ stereochemistry?\n\nPROBLEM The primary fate of acetyl CoA under normal metabolic conditions is degradation in the citric 29-52 acid cycle to yield $\\mathrm{CO}_{2}$. When the body is stressed by prolonged starvation, however, acetyl CoA is converted into compounds called ketone bodies, which can be used by the brain as a temporary fuel. Fill in the missing information indicated by the four question marks in the following biochemical pathway for the synthesis of ketone bodies from acetyl CoA:\n\n\n\n\nAcetone\n\n\n\n3-Hydroxybutyrate\n\nKetone bodies\nPROBLEM The initial reaction in Problem 29-52, conversion of two molecules of acetyl CoA to one molecule of 29-53 acetoacetyl CoA, is a Claisen reaction. Assuming that there is a base present, show the mechanism\nof the reaction.\nPROBLEM In step 6 of fatty-acid biosynthesis (Figure 29.6), acetoacetyl ACP is reduced stereospecifically by 29-54 NADPH to yield an alcohol. Does hydride ion add to the Si face or the Re face of acetoacetyl ACP?\n\n\nAcetoacetyl ACP\n\n$\\beta$-Hydroxybutyryl ACP\n\nPROBLEM In step 7 of fatty-acid biosynthesis (Figure 29.6), dehydration of a $\\beta$-hydroxy thioester occurs to give 29-55 trans-crotonyl ACP. Is the dehydration a syn elimination or an anti elimination?"}
{"id": 1914, "contents": "General Problems - \nPROBLEM In step 8 of fatty-acid biosynthesis (Figure 29.6), reduction of trans-crotonyl ACP gives butyryl ACP.\n29-56 A hydride from NADPH adds to C3 of the crotonyl group from the Re face, and protonation on C2 occurs on the Si face. Is the reduction a syn addition or an anti addition?"}
{"id": 1915, "contents": "CHAPTER 30 - \nOrbitals and Organic Chemistry: Pericyclic Reactions\n\n\nFIGURE 30.1 It may look easy, but these sunbathers are working hard to carry out a pericyclic reaction. (credit: modification of work \"Tanning\" by Meraj Chhaya/Flickr, CC BY 2.0)."}
{"id": 1916, "contents": "CHAPTER CONTENTS - \n30.1 Molecular Orbitals of Conjugated Pi Systems\n30.2 Electrocyclic Reactions\n30.3 Stereochemistry of Thermal Electrocyclic Reactions\n30.4 Photochemical Electrocyclic Reactions\n30.5 Cycloaddition Reactions\n30.6 Stereochemistry of Cycloadditions\n30.7 Sigmatropic Rearrangements\n30.8 Some Examples of Sigmatropic Rearrangements\n30.9 A Summary of Rules for Pericyclic Reactions\n\nWHY THIS CHAPTER? Broad outlines of both polar and radical reactions have been in place for more than a century, but our understanding of pericyclic reactions has emerged more recently. Prior to the mid-1960s, in fact, they were even occasionally referred to as \"no-mechanism reactions.\" They occur largely in the laboratory rather than in biological processes, but a knowledge of them is necessary, both for completeness in studying organic chemistry and in understanding biological pathways where they do occur.\n\nMost organic reactions take place by polar mechanisms, in which a nucleophile donates two electrons to an electrophile in forming a new bond. Other reactions take place by radical mechanisms, in which each of two reactants donates one electron in forming a new bond. Both kinds of reactions occur frequently in the laboratory and in living organisms. Less common, however, is the third major class of organic reactions-pericyclic reactions.\n\nA pericyclic reaction is one that occurs by a concerted process through a cyclic transition state. A concerted reaction is one in which all bonding changes occur simultaneously; no intermediates are involved. Rather than try to expand this definition now, we'll begin by briefly reviewing some of the ideas of molecular orbital theory introduced in the chapters on Structure and Bonding and Conjugated Compounds and Ultraviolet Spectroscopy and then looking individually at the three main classes of pericyclic reactions: electrocyclic reactions, cycloadditions, and sigmatropic rearrangements."}
{"id": 1917, "contents": "CHAPTER CONTENTS - 30.1 Molecular Orbitals of Conjugated Pi Systems\nA conjugated polyene, as we saw in Section 14.1, is one with alternating double and single bonds. According to molecular orbital (MO) theory, the $p$ orbitals on the $s p^{2}$-hybridized carbons of a conjugated polyene interact to form a set of $\\pi$ molecular orbitals whose energies depend on the number of nodes they have between nuclei. Molecular orbitals with fewer nodes are lower in energy than isolated $p$ atomic orbitals and are bonding MOs; molecular orbitals with more nodes are higher in energy than isolated $p$ orbitals and are antibonding MOs. Pi molecular orbitals of ethylene and 1,3-butadiene are shown in FIGURE 30.2.\n\n\nA similar sort of molecular orbital description can be derived for any conjugated $\\pi$ electron system. 1,3,5-Hexatriene, for example, has three double bonds and six $\\pi$ MOs, as shown in FIGURE 30.3. In the ground state, only the three bonding orbitals, $\\psi_{1}, \\psi_{2}$, and $\\psi_{3}$, are filled. On irradiation with ultraviolet light, however, an electron is promoted from the highest-energy filled orbital $\\left(\\psi_{3}\\right)$ to the lowest-energy unfilled orbital ( $\\psi_{4} *$ ) to give an excited state (Section 14.7), in which $\\psi_{3}$ and $\\psi_{4}{ }^{*}$ are each half-filled. (An asterisk denotes an antibonding orbital.)\n\nWhat do molecular orbitals and their nodes have to do with pericyclic reactions? The answer is, everything. According to a series of rules formulated in the mid-1960s by R. B. Woodward and Roald Hoffmann at Harvard University, a pericyclic reaction can take place only if the symmetries of the reactant MOs are the same as the symmetries of the product MOs. In other words, the lobes of reactant MOs must be of the correct algebraic sign for bonding to occur in the transition state leading to product."}
{"id": 1918, "contents": "CHAPTER CONTENTS - 30.1 Molecular Orbitals of Conjugated Pi Systems\nIf the symmetries of reactant and product orbitals match up, or correlate, the reaction is said to be symmetryallowed. If the symmetries of reactant and product orbitals don't correlate, the reaction is symmetrydisallowed. Symmetry-allowed reactions often occur under relatively mild conditions, but symmetrydisallowed reactions can't occur by concerted paths. They either take place by nonconcerted, higher-energy\npathways, or they don't take place at all.\n\n\nGround state Excited state\nFIGURE 30.3 The six $\\boldsymbol{\\pi}$ molecular orbitals of 1,3,5-hexatriene. In the ground state, the three bonding MOs, $\\psi_{1}, \\psi_{2}$, and $\\psi_{3}$, are filled. In the excited state, $\\psi_{3}$ and $\\psi_{4}{ }^{*}$ are both half-filled.\n\nThe Woodward-Hoffmann rules for pericyclic reactions require an analysis of all reactant and product molecular orbitals, but Kenichi Fukui at Kyoto Imperial University in Japan introduced a simplified version. Hoffman and Fukui shared a Nobel Prize for their work, and Woodward would have joined them (for his second Nobel Prize) had he not passed away before the award was made. According to Fukui's simplified version, we need to consider only two molecular orbitals, called the frontier orbitals. These frontier orbitals are the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO). In ground-state $1,3,5$-hexatriene, for example, $\\psi_{3}$ is the HOMO and $\\psi_{4}{ }^{*}$ is the LUMO (FIGURE 30.3). In excitedstate 1,3,5-hexatriene, however, $\\psi_{4}{ }^{*}$ is the HOMO and $\\psi_{5}{ }^{*}$ is the LUMO.\n\nPROBLEM Look at Figure 30.2, and tell which molecular orbital is the HOMO and which is the LUMO for both 30-1 ground and excited states of ethylene and 1,3-butadiene."}
{"id": 1919, "contents": "CHAPTER CONTENTS - 30.2 Electrocyclic Reactions\nThe best way to understand how orbital symmetry affects pericyclic reactions is to look at some examples. Let's look first at a group of polyene rearrangements called electrocyclic reactions. An electrocyclic reaction is a pericyclic process that involves the cyclization of a conjugated acyclic polyene. One $\\pi$ bond is broken, the other $\\pi$ bonds change position, a new $\\sigma$ bond is formed, and a cyclic compound results. For example, a conjugated triene can be converted into a cyclohexadiene, and a conjugated diene can be converted into a cyclobutene."}
{"id": 1920, "contents": "A conjugated triene - \nA cyclohexadiene"}
{"id": 1921, "contents": "A cyclobutene - \nPericyclic reactions are reversible, and the position of the equilibrium depends on the specific case. In general, the triene $\\rightleftarrows$ cyclohexadiene equilibrium favors the cyclic product, whereas the diene $\\rightleftarrows$ cyclobutene equilibrium favors the less strained open-chain product.\n\nThe most striking feature of electrocyclic reactions is their stereochemistry. For example, ( $2 E, 4 Z, 6 E$ )-2,4,6-octatriene yields only cis-5,6-dimethyl-1,3-cyclohexadiene when heated, and ( $2 E, 4 Z, 6 Z$ )-2,4,6-octatriene yields only trans- 5,6 -dimethyl-1,3-cyclohexadiene. Remarkably, however, the stereochemical results change completely when the reactions are carried out under what are called photochemical, rather than thermal, conditions. Irradiation of ( $2 E, 4 Z, 6 E$ )-2,4,6-octatriene with ultraviolet light (denoted hv) yields trans-5,6-dimethyl-1,3-cyclohexadiene (FIGURE 30.4)."}
{"id": 1922, "contents": "A cyclobutene - \n$\\begin{array}{cc}(2 E, 4 Z, 6 Z)-2,4,6- & \\text { trans-5,6-Dimethyl-1,3- } \\\\ \\text { Octatriene } & \\text { cyclohexadiene }\\end{array}$\nFIGURE 30.4 Electrocyclic interconversions of 2,4,6-octatriene isomers and 5,6-dimethyl-1,3-cyclohexadiene isomers.\nA similar result is obtained for the thermal electrocyclic ring-opening of 3,4-dimethylcyclobutene. The trans isomer yields only $(2 E, 4 E)-2,4$-hexadiene when heated, and the cis isomer yields only ( $2 E, 4 Z$ )-2,4-hexadiene. On UV irradiation, however, the results are opposite. Cyclization of the $2 E, 4 E$ isomer under photochemical conditions yields cis product (FIGURE 30.5).\n\n\nTo account for these results, we need to look at the two outermost lobes of the polyene MOs-the lobes that interact when cyclization occurs. There are two possibilities: lobes of like sign can be either on the same side of the molecule or on opposite sides.\n\n\nLike lobes on same side\n\n\nLike lobes on opposite side\n\nFor a bond to form, the outermost $\\pi$ lobes must rotate so that favorable bonding interaction is achieved-a positive lobe with a positive lobe or a negative lobe with a negative lobe. If two lobes of like sign are on the same side of the molecule, the two orbitals must rotate in opposite directions-one clockwise and one counterclockwise. This kind of motion is referred to as disrotatory.\n\n\nConversely, if lobes of like sign are on opposite sides of the molecule, both orbitals must rotate in the same direction, either both clockwise or both counterclockwise. This kind of motion is called conrotatory."}
{"id": 1923, "contents": "A cyclobutene - 30.3 Stereochemistry of Thermal Electrocyclic Reactions\nHow can we predict whether conrotatory or disrotatory motion will occur in a given case? According to frontier orbital theory, the stereochemistry of an electrocyclic reaction is determined by the symmetry of the polyene HOMO. The electrons in the HOMO are the highest-energy, most loosely held electrons and are therefore most easily moved during reaction. For thermal reactions, the ground-state electron configuration is used to identify the HOMO; for photochemical reactions, the excited-state electron configuration is used.\n\nLet's look again at the thermal ring-closure of conjugated trienes. According to FIGURE 30.3, the HOMO of a conjugated triene in its ground state has lobes of like sign on the same side of the molecule, a symmetry that predicts disrotatory ring-closure. This disrotatory cyclization is precisely what is observed in the thermal cyclization of $2,4,6$-octatriene. The $2 E, 4 Z, 6 E$ isomer yields cis product; the $2 E, 4 Z, 6 Z$ isomer yields trans product (FIGURE 30.6).\n\n\nIn the same way, the ground-state HOMO of conjugated dienes (FIGURE 30.2) has a symmetry that predicts conrotatory ring-closure. In practice, however, the conjugated diene reaction can be observed only in the reverse direction (cyclobutene $\\rightarrow$ diene) because of the position of the equilibrium. We therefore find that the 3,4-dimethylcyclobutene ring opens in a conrotatory fashion. cis-3,4-Dimethylcyclobutene yields ( $2 E, 4 Z$ )-2,4-hexadiene, and trans-3,4-dimethylcyclobutene yields ( $2 E, 4 E$ )-2,4-hexadiene by conrotatory opening (FIGURE 30.7).\n\n\nFIGURE 30.7 Thermal ring-openings of cis- and trans-dimethylcyclobutene occur by conrotatory paths."}
{"id": 1924, "contents": "A cyclobutene - 30.3 Stereochemistry of Thermal Electrocyclic Reactions\nFIGURE 30.7 Thermal ring-openings of cis- and trans-dimethylcyclobutene occur by conrotatory paths.\n\nNote that a conjugated diene and a conjugated triene react with opposite stereochemistry. The diene opens and closes by a conrotatory path, whereas the triene opens and closes by a disrotatory path. This is due to the different symmetries of the diene and triene HOMOs.\n\n\nDiene \u043d\u043e\u043c\u043e\n\n\nTriene HOMO\n\nIt turns out that there is an alternating relationship between the number of electron pairs (double bonds) undergoing bond reorganization and the stereochemistry of ring-opening or -closure. Polyenes with an even number of electron pairs undergo thermal electrocyclic reactions in a conrotatory sense, whereas polyenes with an odd number of electron pairs undergo the same reactions in a disrotatory sense.\n\nPROBLEM Draw the products you would expect from conrotatory and disrotatory cyclizations of 30-2 ( $2 Z, 4 Z, 6 Z$ )-2,4,6-octatriene. Which of the two paths would you expect the thermal reaction to follow?\n\nPROBLEM trans-3,4-Dimethylcyclobutene can open by two conrotatory paths to give either 30-3 ( $2 E, 4 E$ )-2,4-hexadiene or ( $2 Z, 4 Z$ )-2,4-hexadiene. Explain why both products are symmetryallowed, and then account for the fact that only the $2 E, 4 E$ isomer is obtained in practice."}
{"id": 1925, "contents": "A cyclobutene - 30.4 Photochemical Electrocyclic Reactions\nWe noted previously that photochemical electrocyclic reactions take a different stereochemical course than their thermal counterparts, and we can now explain this difference. Ultraviolet irradiation of a polyene causes an excitation of one electron from the ground-state HOMO to the ground-state LUMO, thus changing their symmetries. But because electronic excitation changes the symmetries of HOMO and LUMO, it also changes the reaction stereochemistry. ( $2 E, 4 E$ )-2,4-Hexadiene, for instance, undergoes photochemical cyclization by a disrotatory path, whereas the thermal reaction is conrotatory. Similarly, ( $2 E, 4 Z, 6 E$ )-2,4,6-octatriene undergoes photochemical cyclization by a conrotatory path, whereas the thermal reaction is disrotatory (FIGURE 30.8).\n\n\nFIGURE 30.8 Photochemical cyclizations of conjugated dienes and trienes. The two processes occur with different stereochemistry because of their different orbital symmetries.\n\nThermal and photochemical electrocyclic reactions always take place with opposite stereochemistry because the symmetries of the frontier orbitals are always different. TABLE 30.1 gives some simple rules that make it possible to predict the stereochemistry of electrocyclic reactions.\n\n| TABLE 30.1 Stereochemical Rules for Electrocyclic Reactions |  |  |  |\n| :--- | :--- | :--- | :---: |\n| Electron pairs (double bonds) | Thermal reaction | Photochemical reaction |  |\n| Even number | Conrotatory | Disrotatory |  |\n| Odd number | Disrotatory | Conrotatory |  |\n\nPROBLEM What product would you expect to obtain from the photochemical cyclization of 30-4 (2E,4Z,6E)-2,4,6-octatriene? Of ( $2 E, 4 Z, 6 Z$ )-2,4,6-octatriene?"}
{"id": 1926, "contents": "A cyclobutene - 30.5 Cycloaddition Reactions\nA cycloaddition reaction is one in which two unsaturated molecules add to one another to yield a cyclic product. As with electrocyclic reactions, cycloadditions are governed by the orbital symmetry of the reactants. Symmetry-allowed processes often take place readily, but symmetry-disallowed processes take place with difficulty, if at all, and then only by nonconcerted pathways. Let's look at two examples to see how they differ.\n\nAs we saw in Section 14.4, the Diels-Alder cycloaddition reaction is a pericyclic process that takes place between a diene (four $\\pi$ electrons) and a dienophile (two $\\pi$ electrons) to yield a cyclohexene product. Many thousands of Diels-Alder reactions are known. They often take place easily at room temperature or slightly above, and they are stereospecific with respect to substituents. For example, room-temperature reaction between 1,3-butadiene and diethyl maleate (cis) exclusively yields the cis-disubstituted cyclohexene product. A similar reaction between 1,3-butadiene and diethyl fumarate (trans) exclusively yields the trans-disubstituted product.\n\n\nIn contrast to the [4 +2$]$ - $\\pi$-electron Diels-Alder reaction, the $[2+2]$ - $\\pi$-electron cycloaddition between two alkenes does not occur thermally. The [2 + 2] cycloaddition takes place only on irradiation, yielding cyclobutane products.\n\n\nA cyclobutane\n\nFor a successful cycloaddition, the terminal $\\pi$ lobes of the two reactants must have the correct symmetry for bonding to occur. This can happen in either of two ways, called suprafacial and antarafacial. Suprafacial cycloadditions take place when a bonding interaction occurs between lobes on the same face of one reactant and lobes on the same face of the other reactant. Antarafacial cycloadditions take place when a bonding interaction occurs between lobes on the same face of one reactant and lobes on opposite faces of the other reactant (FIGURE 30.9).\n\n(b) Antarafacial\nLike lobes\non opposite\nfaces\nLike lobes on same face"}
{"id": 1927, "contents": "A cyclobutene - 30.5 Cycloaddition Reactions\n(b) Antarafacial\nLike lobes\non opposite\nfaces\nLike lobes on same face\n\n\nFIGURE 30.9 (a) Suprafacial cycloaddition occurs when there is bonding between lobes on the same face of one reactant and lobes on the same face of the other reactant. (b) Antarafacial cycloaddition occurs when there is bonding between lobes on the same face of one reactant and lobes on opposite faces of the other, which requires a twist in one $\\pi$ system.\n\nNote that both suprafacial and antarafacial cycloadditions are symmetry-allowed. Geometric constraints often make antarafacial reactions difficult, however, because there must be a twisting of the $\\pi$ orbital system in one of the reactants. Thus, suprafacial cycloadditions are much more common for small $\\pi$ systems."}
{"id": 1928, "contents": "A cyclobutene - 30.6 Stereochemistry of Cycloadditions\nHow can we predict whether a given cycloaddition reaction will occur with suprafacial or with antarafacial geometry? According to frontier orbital theory, a cycloaddition reaction takes place when a bonding interaction occurs between the HOMO of one reactant and the LUMO of the other. An intuitive explanation of this rule is to imagine that one reactant donates electrons to the other. As with electrocyclic reactions, it's the electrons in the HOMO of the first reactant that are least tightly held and most likely to be donated. Of course when the second reactant accepts those electrons, they must go into a vacant, unoccupied orbital-the LUMO.\n\nFor a [4 + 2] cycloaddition (Diels-Alder reaction), let's arbitrarily select the diene LUMO and the alkene HOMO. The symmetries of the two ground-state orbitals are such that bonding of the terminal lobes can occur with suprafacial geometry (FIGURE 30.10), so the Diels-Alder reaction takes place readily under thermal conditions. Note that, as with electrocyclic reactions, we need be concerned only with the terminal lobes. For purposes of prediction, interactions among the interior lobes need not be considered."}
{"id": 1929, "contents": "A cyclobutene - 30.6 Stereochemistry of Cycloadditions\nFIGURE 30.10 Interaction of diene LUMO and alkene HOMO in a suprafacial [4 + 2] cycloaddition reaction (Diels-Alder reaction).\nIn contrast with the thermal [4 + 2] Diels-Alder reaction, the [2 + 2] cycloaddition of two alkenes to yield a cyclobutane can only occur photochemically. The explanation for this follows from orbital-symmetry arguments. Looking at the ground-state HOMO of one alkene and the LUMO of the second alkene, it's apparent that a thermal $[2+2]$ cycloaddition must take place by an antarafacial pathway (FIGURE 30.11a). Geometric constraints make the antarafacial transition state difficult, however, and so concerted thermal [2 + 2] cycloadditions are not observed.\nIn contrast with the thermal process, photochemical [2 + 2] cycloadditions are observed. Irradiation of an alkene with UV light excites an electron from $\\psi_{1}$, the ground-state HOMO, to $\\psi_{2}{ }^{*}$, which becomes the excited-state HOMO. Interaction between the excited-state HOMO of one alkene and the LUMO of the second alkene allows a photochemical [2 + 2] cycloaddition reaction to occur by a suprafacial pathway (FIGURE 30.11b).\n(a) Thermal reaction\n\n\nStrained, no reaction\n(b) Photochemical reaction\n\nAlkene 2: Ground-state LUMO\n\nAlkene 1:\nExcited-state HOMO\n\n\nA cyclobutane\n\nFIGURE 30.11 (a) Interaction of a ground-state HOMO and a ground-state LUMO in a potential [2 + 2] cycloaddition does not occur thermally because the antarafacial geometry is too strained. (b) Interaction of an excited-state HOMO and a ground-state LUMO in a photochemical [2+2] cycloaddition reaction is less strained, however, and occurs with suprafacial geometry."}
{"id": 1930, "contents": "A cyclobutene - 30.6 Stereochemistry of Cycloadditions\nThe photochemical [2+2] cycloaddition reaction occurs smoothly, particularly with $\\alpha, \\beta$-unsaturated carbonyl compounds, and represents one of the best methods known for synthesizing cyclobutane rings. For example:\n\n\n2-Cyclohexenone\n2-Methylpropene\n(40\\%)\nThermal and photochemical cycloaddition reactions always take place with opposite stereochemistry. As with electrocyclic reactions, we can categorize cycloadditions according to the total number of electron pairs (double bonds) involved in the rearrangement. Thus, a thermal [ $4+2]$ Diels-Alder reaction between a diene and a dienophile involves an odd number (three) of electron pairs and takes place by a suprafacial pathway. A thermal $[2+2]$ reaction between two alkenes involves an even number (two) of electron pairs and must take place by an antarafacial pathway. For photochemical cyclizations, these selectivities are reversed. These general rules are given in TABLE 30.2.\n\nTABLE 30.2 Stereochemical Rules for Cycloaddition Reactions\n\n| Electron pairs (double bonds) | Thermal reaction | Photochemical reaction |\n| :--- | :--- | :--- |\n| Even number | Antarafacial | Suprafacial |\n| Odd number | Suprafacial | Antarafacial |\n\nPROBLEM What stereochemistry would you expect for the product of the Diels-Alder reaction between 30-5 ( $2 E, 4 E$ )-2,4-hexadiene and ethylene? What stereochemistry would you expect if $(2 E, 4 Z)-2,4$-hexadiene were used instead?\n\nPROBLEM 1,3-Cyclopentadiene reacts with cycloheptatrienone to give the product shown. Tell what kind of 30-6 reaction is involved, and explain the observed result. Is the reaction suprafacial or antarafacial?"}
{"id": 1931, "contents": "A cyclobutene - 30.7 Sigmatropic Rearrangements\nA sigmatropic rearrangement, the third general kind of pericyclic reaction, is a process in which a $\\sigma$-bonded substituent atom or group migrates across a $\\pi$ electron system from one position to another. A $\\sigma$ bond is broken in the reactant, the $\\pi$ bonds move, and a new $\\sigma$ bond is formed in the product. The $\\sigma$-bonded group can be either at the end or in the middle of the $\\pi$ system, as the following $[1,5]$ and $[3,3]$ rearrangements illustrate:"}
{"id": 1932, "contents": "A [3,3] sigmatropic rearrangement - \nThe notations [1,5] and [3,3] describe the kind of rearrangement that is occurring. The numbers refer to the two groups connected by the $\\sigma$ bond in the reactant and designate the positions in those groups to which migration occurs. For example, in the [1,5] sigmatropic rearrangement of a 1,3-diene, the two groups connected by the $\\sigma$ bond are a hydrogen atom and a pentadienyl group. Migration occurs to position 1 of the H group (the only possibility) and to position 5 of the pentadienyl group. In the [3,3] rearrangement of an allylic vinylic ether, the two groups connected by the $\\sigma$ bond are an allylic group and the vinylic ether. Migration occurs to position 3 of the allylic group and also to position 3 of the vinylic ether.\n\nLike electrocyclic reactions and cycloadditions, sigmatropic rearrangements are controlled by orbital symmetries. There are two possible modes of reaction: migration of a group across the same face of the $\\pi$ system is suprafacial, and migration of a group from one face of the $\\pi$ system to the other face is antarafacial (FIGURE 30.12).\n\n\nFIGURE 30.12 Suprafacial and antarafacial sigmatropic rearrangements.\nBoth suprafacial and antarafacial sigmatropic rearrangements are symmetry-allowed, but suprafacial rearrangements are often easier for geometric reasons. The rules for sigmatropic rearrangements are identical to those for cycloaddition reactions (TABLE 30.3).\n\nTABLE 30.3 Stereochemical Rules for Sigmatropic Rearrangements\n\n| Electron pairs (double bonds) | Thermal reaction | Photochemical reaction |\n| :--- | :--- | :--- |\n| Even number | Antarafacial | Suprafacial |\n| Odd number | Suprafacial | Antarafacial |\n\nPROBLEM Classify the following sigmatropic reaction by order $[x, y]$, and tell whether it will proceed with 30-7 suprafacial or antarafacial stereochemistry:"}
{"id": 1933, "contents": "A [3,3] sigmatropic rearrangement - 30.8 Some Examples of Sigmatropic Rearrangements\nBecause a $[1,5]$ sigmatropic rearrangement involves three electron pairs (two $\\pi$ bonds and one $\\sigma$ bond), the orbital-symmetry rules in TABLE 30.3 predict a suprafacial reaction. In fact, the [1,5] suprafacial shift of a hydrogen atom across two double bonds of a $\\pi$ system is a commonly observed sigmatropic rearrangements. For example, 5-methyl-1,3-cyclopentadiene rapidly rearranges at room temperature to yield a mixture of 1-methyl-, 2-methyl-, and 5-methyl-isomers.\n\n\nAs another example, heating 5,5,5-trideuterio-(3Z)-1,3-pentadiene causes scrambling of deuterium between\npositions 1 and 5.\n\n\nBoth these $[1,5]$ hydrogen shifts occur by a symmetry-allowed suprafacial pathway, as illustrated in FIGURE 30.13. In contrast with these thermal [1,5] sigmatropic hydrogen shifts, however, thermal [1,3] hydrogen shifts are unknown. If they were to occur, they would have to proceed by a strained antarafacial reaction pathway.\n\n\nFIGURE 30.13 An orbital view of a suprafacial [1,5] hydrogen shift.\nTwo other important sigmatropic reactions are the Claisen rearrangement of either an allyl aryl ether $\\left(\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}_{2}-\\mathrm{O}-\\mathrm{Ar}\\right)$ or an allyl vinyl $\\left(\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}_{2}-\\mathrm{O}-\\mathrm{CH}=\\mathrm{CH}_{2}\\right)$ ether, and the Cope rearrangement of a 1,5 hexadiene to an isomeric 1,5-diene. These two rearrangements, along with the Diels-Alder reaction, are among the most generally useful pericyclic reactions for organic synthesis. Thousands of examples of all three are known."}
{"id": 1934, "contents": "Claisen rearrangement - \nAn allylic\nvinylic ether\n\nAn unsaturated ketone\n\nCope rearrangement\n\n\nLike the Diels-Alder reaction discussed in Section 14.4 and Section 14.5, the Claisen rearrangement takes place in a single step through a pericyclic mechanism in which a reorganization of bonding electrons occurs in a six-membered, cyclic transition state. The 6-allyl-2,4-cyclohexadienone intermediate then isomerizes to o-allylphenol (FIGURE 30.14).\n\n\nEvidence for this mechanism comes from the observation that the rearrangement takes place with transposition of the allyl group. That is, allyl phenyl ether containing a ${ }^{14} \\mathrm{C}$ label on the allyl ether carbon atom yields $o$-allylphenol in which the label is on the terminal vinylic carbon (green in FIGURE 30.14).\n\nPROBLEM Draw a curved-arrow mechanism for the Cope arrangement just shown.\n\n\nThe Cope rearrangement that converts a 1,5-diene to an isomeric 1,5-diene is somewhat limited but a modification called the Oxy Cope Rearrangement is wider in scope. As shown in the following example, a 1-5 diene with an -OH next to the double bond can be converted into an oxy-anion by reaction with a strong base such as potassium hydride (KH). A Cope rearrangement then occurs, and reaction with aqueous acid gives an enol that tautomerizes to an aldehyde.\n\n\nAlthough biological examples of pericyclic reactions are relatively rare, a much-studied example occurs in bacteria during biosynthesis of the essential amino acid phenylalanine. Phenylalanine arises from the precursor chorismate through a Claisen rearrangement to prephenate, followed by decarboxylation to phenylpyruvate and reductive amination (FIGURE 30.16). You might note that the reductive amination of phenylpyruvate is the exact reverse of the transamination process shown in FIGURE 29.17, by which amino acids are deaminated. In addition, the reductive amination of ketones is a standard method for preparing amines in the laboratory, as we saw in Section 24.6."}
{"id": 1935, "contents": "Claisen rearrangement - \nPhenylalanine\nFIGURE 30.16 Pathway for the bacterial biosynthesis of phenylalanine from chorismate, involving a Claisen rearrangement.\nPROBLEM Propose a mechanism to account for the fact that heating 1-deuterioindene scrambles the isotope 30-9 label to all three positions on the five-membered ring.\n\n\n1-Deuterioindene\nPROBLEM What product would you expect from Claisen rearrangement of 2-butenyl phenyl ether? 30-10\n\n\n2-Butenyl phenyl ether\nPROBLEM When a 2,6-disubstituted allyl phenyl ether is heated in an attempted Claisen rearrangement,\n30-11 migration occurs to give the $p$-allyl product as the result of two sequential pericyclic reactions. Explain."}
{"id": 1936, "contents": "Claisen rearrangement - 30.9 A Summary of Rules for Pericyclic Reactions\nHow can you keep straight all the rules about pericyclic reactions? The summary in TABLE 30.1, TABLE 30.2, and TABLE 30.3 can be distilled into a mnemonic phrase that provides an easy way to predict the stereochemical outcome of any pericyclic reaction:\n\nThe Electrons Circle Around (TECA)\nThermal reactions with an Even number of electron\npairs are Conrotatory or Antarafacial.\nA change either from thermal to photochemical or from an even to an odd number of electron pairs changes the outcome from conrotatory/antarafacial to disrotatory/suprafacial. A change from both thermal and even to photochemical and odd causes no change because two negatives make a positive.\n\nThese selection rules are summarized in TABLE 30.4; knowing them gives you the ability to predict the stereochemistry of literally thousands of pericyclic reactions.\n\n| TABLE 30.4 Stereochemical Rules for Pericyclic Reactions |  |  |\n| :--- | :--- | :--- |\n| Electronic state | Electron pairs | Stereochemistry |\n| Ground state (thermal) | Even number <br> Odd number | Antara-con <br> Supra-dis |\n| Excited state (photochemical) | Even number <br> Odd number | Supra-dis <br> Antara-con |\n\nPROBLEM Predict the stereochemistry of the following pericyclic reactions:\n30-12 (a) The thermal cyclization of a conjugated tetraene\n(b) The photochemical cyclization of a conjugated tetraene\n(c) A photochemical $[4+4]$ cycloaddition\n(d) A thermal $[2+6]$ cycloaddition\n(e) A photochemical $[3,5]$ sigmatropic rearrangement"}
{"id": 1937, "contents": "Vitamin D, the Sunshine Vitamin - \nVitamin D, discovered in 1918, is a general name for two related compounds, cholecalciferol (vitamin $\\mathrm{D}_{3}$ ) and ergocalciferol (vitamin $D_{2}$ ). Both are derived from steroids (Section 27.6) and differ only in the nature of the hydrocarbon side chain attached to the five-membered ring. Cholecalciferol comes primarily from dairy products and fish; ergocalciferol comes from some vegetables.\n\n\nFIGURE 30.17 Here are the sunbathers, again, with their chemical work complete. (credit: \"Tanning\" by Meraj Chhaya/Flickr, CC BY 2.0)\nThe function of vitamin $D$ in the body is to control the calcification of bones by increasing intestinal absorption of calcium. When sufficient vitamin $D$ is present, approximately $30 \\%$ of ingested calcium is absorbed, but in the absence of vitamin D, calcium absorption falls to about $10 \\%$. A deficiency of vitamin $D$ thus leads to poor bone growth and to the diseases rickets in children and osteoporosis in adults.\n\nActually, neither vitamin $D_{2}$ nor $D_{3}$ is present in foods. Rather, foods contain the precursor molecules 7-dehydrocholesterol and ergosterol. In the presence of sunlight, both precursors are converted in the outer, epidermal layer of skin to the active vitamins, hence the nickname for vitamin $D$, the \"sunshine vitamin.\"\n\n\nPericyclic reactions are unusual in living organisms, and the photochemical synthesis of vitamin $D$ is one of only a few well-studied examples. The reaction takes place in two steps, an electrocyclic ring-opening of a cyclohexadiene to yield an open-chain hexatriene, followed by a sigmatropic [1,7] H shift to yield an isomeric hexatriene. Only the initial electrocyclic ring-opening requires irradiation by so-called UVB light of 295 to 300 nm wavelength. The subsequent sigmatropic [1,7] H shift occurs spontaneously by a thermal isomerization."}
{"id": 1938, "contents": "Vitamin D, the Sunshine Vitamin - \nFollowing synthesis under the skin, further metabolic processing of cholecalciferol and ergocalciferol in the liver and kidney introduces two additional -OH groups to give the active forms of the vitamin, calcitriol and ergocalcitriol."}
{"id": 1939, "contents": "Key Terms - \n```\n- antarafacial \u2022 lowest unoccupied molecular orbital (LUMO)\n\u2022 conrotatory \u2022 pericyclic reaction\n- cycloaddition reaction\n- photochemical reaction\n- disrotatory\n- sigmatropic rearrangement\n- electrocyclic reaction\n- suprafacial\n- frontier orbitals\n- symmetry-allowed\n- highest occupied molecular orbital (HOMO)\n```\n\n- conrotatory\n- cycloaddition reaction\n- disrotatory\n- electrocyclic reaction\n- frontier orbitals\n- highest occupied molecular orbital (HOMO)\n- lowest unoccupied molecular orbital (LUMO)\n- pericyclic reaction\n- photochemical reaction\n- sigmatropic rearrangement\n- suprafacial\n- symmetry-allowed\n- symmetry-disallowed"}
{"id": 1940, "contents": "Summary - \nA pericyclic reaction takes place in a single step through a cyclic transition state without intermediates. There are three major classes of pericyclic processes: electrocyclic reactions, cycloaddition reactions, and sigmatropic rearrangements. The stereochemistry of these reactions is controlled by the symmetry of the orbitals involved in bond reorganization.\n\nElectrocyclic reactions involve the cyclization of conjugated acyclic polyenes. For example, 1,3,5-hexatriene cyclizes to 1,3 -cyclohexadiene on heating. Electrocyclic reactions can occur by either conrotatory or disrotatory pathways, depending on the symmetry of the terminal lobes of the $\\pi$ system. Conrotatory cyclization requires that both lobes rotate in the same direction, whereas disrotatory cyclization requires that the lobes rotate in opposite directions. The reaction course in a specific case can be found by looking at the symmetry of the highest occupied molecular orbital (HOMO).\n\nCycloaddition reactions are those in which two unsaturated molecules add together to yield a cyclic product. For example, Diels-Alder reaction between a diene (four $\\pi$ electrons) and a dienophile (two $\\pi$ electrons) yields a cyclohexene. Cycloadditions can take place either by suprafacial or antarafacial pathways. Suprafacial cycloaddition involves interaction between lobes on the same face of one component and on the same face of the second component. Antarafacial cycloaddition involves interaction between lobes on the same face of one component and on opposite faces of the other component. The reaction course in a specific case can be found by looking at the symmetry of the HOMO of one component and the lowest unoccupied molecular orbital (LUMO) of the other."}
{"id": 1941, "contents": "Summary - \nSigmatropic rearrangements involve the migration of a $\\sigma$-bonded group across a $\\pi$ electron system. For example, Claisen rearrangement of an allylic vinylic ether yields an unsaturated carbonyl compound, and Cope rearrangement of a 1,5-hexadiene yields an isomeric 1,5-hexadiene. Sigmatropic rearrangements can occur with either suprafacial or antarafacial stereochemistry; the selection rules for a given case are the same as those for cycloaddition reactions.\n\nThe stereochemistry of any pericyclic reaction can be predicted by counting the total number of electron pairs (bonds) involved in bond reorganization and then applying the mnemonic \"The Electrons Circle Around.\" That is, thermal (ground-state) reactions involving an even number of electron pairs occur with either conrotatory or antarafacial stereochemistry. Exactly the opposite rules apply to photochemical (excited-state) reactions."}
{"id": 1942, "contents": "Visualizing Chemistry - \nPROBLEM Predict the product obtained when the following substance is heated:"}
{"id": 1943, "contents": "30-13 - \nPROBLEM The ${ }^{13} \\mathrm{C}$ NMR spectrum of homotropilidene taken at room temperature shows only three peaks. 30-14 Explain."}
{"id": 1944, "contents": "Mechanism Problems - \nPROBLEM The following rearrangement of $N$-allyl- $N, N$-dimethylanilinium ion has been observed. Propose a\n30-15 mechanism.\n\n$N$-Allyl- $N, N$-dimethylanilinium ion\no-Allyl-N,N-dimethylanilinium ion\nPROBLEM Plastic photochromic sunglasses are based on the following reversible rearrangement of a dye 30-16 inside the lenses that occurs when the lenses are exposed to sunlight. The original dye absorbs UV light but not visible light and is thus colorless, while the rearrangement product absorbs visible light and is thus darkened.\n\n(a) Show the mechanism of the rearrangement.\n(b) Why does the rearrangement product absorb at a longer wavelength (visible light) than the original dye (UV)?\n\nPROBLEM The sex hormone estrone has been synthesized by a route that involves the following step. Identify\n30-17 the pericyclic reactions involved, and propose a mechanism.\n\n\nPROBLEM Coronafacic acid, a bacterial toxin, was synthesized using a key step that involves three sequential 30-18 pericyclic reactions. Identify them, and propose a mechanism for the overall transformation. How\nwould you complete the synthesis?\n\n\nPROBLEM The following thermal rearrangement involves two pericyclic reactions in sequence. Identify them, 30-19 and propose a mechanism to account for the observed result."}
{"id": 1945, "contents": "Electrocyclic Reactions - \nPROBLEM Do the following electrocyclic reactions take place in a conrotatory or disrotatory manner? Under\n$\\mathbf{3 0 - 2 0}$ what conditions, thermal or photochemical, would you carry out each reaction?\n(a)\n\n(b)\n\n\nPROBLEM The following thermal isomerization occurs under relatively mild conditions. Identify the pericyclic 30-21 reactions involved, and show how the rearrangement occurs.\n\n\nPROBLEM Would you expect the following reaction to proceed in a conrotatory or disrotatory manner? Show\n$\\mathbf{3 0 - 2 2}$ the stereochemistry of the cyclobutene product, and explain your answer.\n\n\nPROBLEM Heating ( $1 Z, 3 Z, 5 Z$ )-1,3,5-cyclononatriene to $100^{\\circ} \\mathrm{C}$ causes cyclization and formation of a bicyclic\n$\\mathbf{3 0 - 2 3}$ product. Is the reaction conrotatory or disrotatory? What is the stereochemical relationship of the two hydrogens at the ring junctions, cis or trans?\n\n(1Z,3Z,5Z)-1,3,5-Cyclononatriene\nPROBLEM ( $2 E, 4 Z, 6 Z, 8 E$ )-2,4,6,8-Decatetraene has been cyclized to give 7,8-dimethyl-1,3,5-cyclooctatriene.\n$\\mathbf{3 0 - 2 4}$ Predict the manner of ring-closure-conrotatory or disrotatory-for both thermal and photochemical reactions, and predict the stereochemistry of the product in each case.\n\nPROBLEM Answer Problem 30-24 for the thermal and photochemical cyclizations of 30-25 (2E,4Z,6Z,8Z)-2,4,6,8-decatetraene.\n\nPROBLEM The cyclohexadecaoctaene shown isomerizes to two different isomers, depending on reaction $\\mathbf{3 0 - 2 6}$ conditions. Explain the observed results, and indicate whether each reaction is conrotatory or disrotatory."}
{"id": 1946, "contents": "Cycloaddition Reactions - \nPROBLEM Which of the following reactions is more likely to occur? Explain.\n30-27\n\n\nPROBLEM The following reaction takes place in two steps, one of which is a cycloaddition while the other is a\n$\\mathbf{3 0 - 2 8}$ reverse cycloaddition. Identify the two pericyclic reactions, and show how they occur.\n\n\nPROBLEM Two sequential pericyclic reactions are involved in the following furan synthesis. Identify them, and 30-29 propose a mechanism for the transformation."}
{"id": 1947, "contents": "Sigmatropic Rearrangements - \nPROBLEM Predict the product of the following pericyclic reaction. Is this [5,5] shift a suprafacial or an\n30-30 antarafacial process?\n\n\nPROBLEM Propose a pericyclic mechanism to account for the following transformation:\n30-31\n\n\nPROBLEM Vinyl-substituted cyclopropanes undergo thermal rearrangement to yield cyclopentenes. Propose 30-32 a mechanism for the reaction, and identify the pericyclic process involved.\n\n\nVinylcyclopropane Cyclopentene\nPROBLEM The following synthesis of dienones occurs readily. Propose a mechanism to account for the results,\n30-33 and identify the kind of pericyclic reaction involved.\n\n\nPROBLEM Karahanaenone, a terpenoid isolated from oil of hops, has been synthesized by the thermal reaction\n30-34 shown. Identify the kind of pericyclic reaction, and explain how karahanaenone is formed."}
{"id": 1948, "contents": "General Problems - \nPROBLEM What stereochemistry-antarafacial or suprafacial-would you expect to observe in the following\n30-35 reactions?\n(a) A photochemical $[1,5]$ sigmatropic rearrangement\n(b) A thermal $[4+6]$ cycloaddition\n(c) A thermal $[1,7]$ sigmatropic rearrangement\n(d) A photochemical [2 + 6] cycloaddition\n\nPROBLEM Bicyclohexadiene, also known as Dewar benzene, is extremely stable despite the fact that its 30-36\nrearrangement to benzene is energetically favored. Explain why the rearrangement is so slow.\n\n\nPROBLEM Ring-opening of the trans-cyclobutene isomer shown takes place at much lower temperature than\n30-37 a similar ring-opening of the cis-cyclobutene isomer. Explain the temperature effect, and identify the stereochemistry of each reaction as either conrotatory or disrotatory.\n\n\nPROBLEM Photolysis of the cis-cyclobutene isomer in Problem 30-37 yields cis-cyclododecaen-7-yne, but\n30-38 photolysis of the trans isomer yields trans-cyclododecaen-7-yne. Explain these results, and identify the type and stereochemistry of the pericyclic reaction.\n\n\n\nPROBLEM The ${ }^{1} \\mathrm{H}$ NMR spectrum of bullvalene at $100^{\\circ} \\mathrm{C}$ consists only of a single peak at $4.22 \\delta$. Explain.\n30-39\n\n\nPROBLEM The following rearrangement was devised and carried out to prove the stereochemistry of $[1,5]$ 30-40 sigmatropic hydrogen shifts. Explain how the observed result confirms the predictions of orbital symmetry.\n\n\nPROBLEM The following reaction is an example of a $[2,3]$ sigmatropic rearrangement. Would you expect the\n30-41 reaction to be suprafacial or antarafacial? Explain."}
{"id": 1949, "contents": "General Problems - \nPROBLEM The following reaction is an example of a $[2,3]$ sigmatropic rearrangement. Would you expect the\n30-41 reaction to be suprafacial or antarafacial? Explain.\n\n\nPROBLEM When the compound having a cyclobutene fused to a five-membered ring is heated,\n30-42 ( $1 Z, 3 Z$ )-1,3-cycloheptadiene is formed. When the related compound having a cyclobutene fused to an eight-membered ring is heated, however, ( $1 E, 3 Z$ )-1,3-cyclodecadiene is formed. Explain these results, and suggest a reason why opening of the eight-membered ring occurs at a lower\ntemperature.\n\n\nPROBLEM In light of your answer to Problem 30-42, explain why a mixture of products occurs in the following 30-43 reaction:\n\n\nPROBLEM In nature, the enzyme chorismate mutase catalyzes a Claisen rearrangement of chorismate that\n30-44 involves both the terminal double bond and the double bond with the highlighted carbon. What is the structure of prephenate, the biological precursor to the amino acids phenylalanine and tyrosine?\n\n\nChorismate\nPROBLEM Predict the product(s) if the starting materials underwent a Claisen rearrangement. Draw arrows to\n30-45 illustrate the rearrangement of electrons.\n(a)\n\n(b)\n\n(c)"}
{"id": 1950, "contents": "CHAPTER 31 <br> Synthetic Polymers - \nFIGURE 31.1 If you ride a bike, wear your helmet! Most bike helmets are made of two different polymers, a hard polycarbonate shell and an inner layer of polystyrene. (credit: modification of work \"View from the top\" by Gene Bisbee/Flickr, CC BY 2.0)"}
{"id": 1951, "contents": "CHAPTER CONTENTS - 31.1 Chain-Growth Polymers\n31.2 Stereochemistry of Polymerization: Ziegler-Natta Catalysts\n31.3 Copolymers\n31.4 Step-Growth Polymers\n31.5 Olefin Metathesis Polymerization\n31.6 Intramolecular Olefin Metathesis\n31.7 Polymer Structure and Physical Properties\n\nWHY THIS CHAPTER? Our treatment of polymers has thus far been dispersed over several chapters, but it's also important to take a comprehensive view. In the present chapter, we'll look further at how polymers are made, and we'll see how polymer structure correlates with physical properties. No course in organic chemistry would be complete without a look at polymers.\n\nPolymers are a fundamental part of the modern world, used in everything from coffee cups to cars to clothing. In medicine, too, their importance is growing, with uses as diverse as cardiac pacemakers, artificial heart valves, and biodegradable sutures.\n\nWe've seen on several occasions in previous chapters that a polymer, whether synthetic or biological, is a large molecule built up by repetitive bonding of many smaller units, or monomers. Polyethylene, for instance, is a synthetic polymer made from ethylene (Section 8.10), nylon is a synthetic polyamide made from a diacid and a diamine (Section 21.9), and proteins are biological polyamides made from amino acids. Note that polymers are\noften drawn by indicating their repeating unit in parentheses. The repeating unit in polystyrene, for example, comes from the monomer styrene.\n\n\nStyrene\nPolystyrene"}
{"id": 1952, "contents": "CHAPTER CONTENTS - 31.1 Chain-Growth Polymers\nSynthetic polymers are classified by their method of synthesis as either chain-growth or step-growth. These categories are somewhat imprecise but nevertheless provide a useful distinction. Chain-growth polymers are produced by chain-reaction polymerization in which an initiator adds to the carbon-carbon double bond of an unsaturated substrate (a vinyl monomer) to yield a reactive intermediate. This intermediate reacts with a second molecule of monomer to yield a new intermediate, which reacts with a third monomer unit, and so on.\n\nThe initiator can be a radical, an acid, or a base. Historically, as we saw in Section 8.10, radical polymerization was the most common method because it can be carried out with practically any vinyl monomer.\n\n\nAcid-catalyzed (cationic) polymerization, by contrast, is effective only with vinyl monomers that contain an electron-donating group (EDG) capable of stabilizing the chain-carrying carbocation intermediate.\n\nwhere $E D G=$ an electron-donating group\nIsobutylene (2-methylpropene) is a good example of a monomer that polymerizes rapidly under cationic conditions. The reaction is carried out commercially at $-80^{\\circ} \\mathrm{C}$, using $\\mathrm{BF}_{3}$ and a small amount of water to generate $\\mathrm{BF}_{3} \\mathrm{OH}^{-} \\mathrm{H}^{+}$catalyst. The product is used in the manufacture of truck and bicycle inner tubes.\n\n\nVinyl monomers with electron-withdrawing groups (EWG) can be polymerized by basic (anionic) catalysts. The chain-carrying step is a conjugate nucleophilic addition of an anion to the unsaturated monomer (Section 19.13)."}
{"id": 1953, "contents": "CHAPTER CONTENTS - 31.1 Chain-Growth Polymers\nwhere EWG $=$ an electron-withdrawing group\nAcrylonitrile $\\left(\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCN}\\right)$, methyl methacrylate $\\left[\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{C}\\left(\\mathrm{CH}_{3}\\right) \\mathrm{CO}_{2} \\mathrm{CH}_{3}\\right]$, and styrene $\\left(\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHC}_{6} \\mathrm{H}_{5}\\right)$ can all be polymerized anionically. The polystyrene used in foam coffee cups, for example, is prepared by anionic\npolymerization of styrene using butyllithium as catalyst.\n\n\nAn interesting example of anionic polymerization accounts for the remarkable properties of \"super glue,\" one drop of which can support up to 2000 lb . Super glue is simply a solution of pure methyl $\\alpha$-cyanoacrylate, which has two electron-withdrawing groups that make anionic addition particularly easy. Trace amounts of water or bases on the surface of an object are sufficient to initiate polymerization of the cyanoacrylate and bind articles together. Skin is a good source of the necessary basic initiators, and many people have found their fingers stuck together after inadvertently touching super glue. So good is super glue at binding tissues that related cyanoacrylate esters such as Dermabond are often used in place of sutures to close wounds.\n\n\nMethyla-cyanoacrylate"}
{"id": 1954, "contents": "Dermabond <br> (2-ethylhexyl $\\alpha$-cyanoacrylate) - \nPROBLEM Order the following monomers with respect to their expected reactivity toward cationic\n31-1 polymerization, and explain your answer:\n\n$$\n\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}_{3}, \\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCl}, \\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}-\\mathrm{C}_{6} \\mathrm{H}_{5}, \\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCO}_{2} \\mathrm{CH}_{3}\n$$\n\nPROBLEM Order the following monomers with respect to their expected reactivity toward anionic 31-2 polymerization, and explain your answer:\n$\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}_{3}, \\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHC} \\equiv \\mathrm{N}, \\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHC}_{6} \\mathrm{H}_{5}$\nPROBLEM Polystyrene is produced commercially by reaction of styrene with butyllithium as an anionic 31-3 initiator. Using resonance structures, explain how the chain-carrying intermediate is stabilized."}
{"id": 1955, "contents": "Dermabond <br> (2-ethylhexyl $\\alpha$-cyanoacrylate) - 31.2 Stereochemistry of Polymerization: Ziegler-Natta Catalysts\nAlthough we didn't point it out when discussing chain-growth polymers in Section 8.10, the polymerization of a substituted vinyl monomer can lead to a polymer with numerous chirality centers in its chain. Propylene, for example, might polymerize with any of the three stereochemical outcomes shown in FIGURE 31.2. A polymer with all methyl groups on the same side of the zigzag backbone is called isotactic, one in which the methyl groups alternate regularly on opposite sides of the backbone is called syndiotactic, and one with its methyl groups randomly oriented is called atactic.\n\nIsotactic (same side)\n\n\nAtactic (random)\nFIGURE 31.2 Isotactic, syndiotactic, and atactic forms of polypropylene.\n\nThe three different stereochemical forms of polypropylene all have somewhat different properties, and all can be made by using the right polymerization catalyst. Propylene polymerization using radical initiators does not work well, but polymerization using Ziegler-Natta catalysts allows preparation of isotactic, syndiotactic, and atactic polypropylene.\n\nZiegler-Natta catalysts-there are many different formulations-are organometallic transition-metal complexes prepared by treatment of an alkylaluminum with a titanium compound. Triethylaluminum with titanium tetrachloride is a typical preparation.\n\n$$\n\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2}\\right)_{3} \\mathrm{Al}+\\mathrm{TiCl}_{4} \\rightarrow \\text { A Ziegler-Natta catalyst }\n$$\n\nFollowing their introduction in 1953, Ziegler-Natta catalysts revolutionized the field of polymer chemistry because of two advantages: first, the resultant polymers are linear, with practically no chain branching, and second, they are stereochemically controllable. Isotactic, syndiotactic, and atactic forms can all be produced, depending on the catalyst system used.\n\nThe active form of a Ziegler-Natta catalyst is an alkyltitanium intermediate with a vacant coordination site on the metal. Coordination of alkene monomer to the titanium occurs, and the coordinated alkene then inserts into the carbon-titanium bond to extend the alkyl chain. A new coordination site opens up during the insertion step, so the process repeats indefinitely."}
{"id": 1956, "contents": "Dermabond <br> (2-ethylhexyl $\\alpha$-cyanoacrylate) - 31.2 Stereochemistry of Polymerization: Ziegler-Natta Catalysts\nThe linear polyethylene produced by the Ziegler-Natta process, called high-density polyethylene, is a highly crystalline polymer with 4000 to 7000 ethylene units per chain and molecular weights in the range 100,000 to 200,000 amu. High-density polyethylene has greater strength and heat resistance than the branched product of radical-induced polymerization, called low-density polyethylene, and is used to produce plastic squeeze bottles and molded housewares.\n\nPolyethylenes of even higher molecular weights are produced for specialty applications. So-called high-molecular-weight (HMW) polyethylene contains 10,000 to 18,000 monomer units per chain and molecular weights in the range of 300,000-500,000. It is used for underground pipes and large containers. Ultrahigh-molecular-weight (UHMW) polyethylene contains more than 100,000 monomer units per chain and has molecular weights ranging from $3,000,000$ to $6,000,000 \\mathrm{amu}$. It is used in bearings, conveyor belts, and bulletproof vests, among other applications requiring exceptional wear resistance.\n\nPROBLEM Vinylidene chloride, $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CCl}_{2}$, does not polymerize in isotactic, syndiotactic, and atactic forms.\n31-4 Explain.\nPROBLEM Polymers such as polypropylene contain a large number of chirality centers. Would you therefore\n31-5 expect samples of isotactic, syndiotactic, or atactic polypropylene to rotate plane-polarized light? Explain."}
{"id": 1957, "contents": "Dermabond <br> (2-ethylhexyl $\\alpha$-cyanoacrylate) - 31.3 Copolymers\nUp to this point we've discussed only homopolymers-polymers that are made up of identical repeating units. In practice, however, copolymers are more important commercially. Copolymers are obtained when two or more different monomers are allowed to polymerize together. For example, copolymerization of vinyl chloride with vinylidene chloride (1,1-dichloroethylene) in a 1 : 4 ratio leads to the polymer Saran.\n\n\nCopolymerization of monomer mixtures often leads to materials with properties quite different from those of either corresponding homopolymer, giving the polymer chemist a vast flexibility for devising new materials. TABLE 31.1 lists some common copolymers and their commercial applications.\n\nTABLE 31.1 Some Common Copolymers and Their Uses\n\n| Monomers | Structures | Trade name | Uses |\n| :---: | :---: | :---: | :---: |\n| Vinyl chloride <br> Vinylidene chloride |  | Saran | Fibers, food packaging |\n| Styrene <br> 1,3-Butadiene |  | SBR (styrene-butadiene rubber) | Tires, rubber articles |\n| Hexafluoropropene Vinylidene fluoride |  | Viton | Gaskets, seals |\n| Acrylonitrile 1,3-Butadiene | <smiles>C=CC=C</smiles> | Nitrile rubber | Adhesives, hoses |\n\nTABLE 31.1 Some Common Copolymers and Their Uses\n\n| Monomers | Structures | Trade name | Uses |\n| :---: | :---: | :---: | :---: |\n| Isobutylene Isoprene |  | Butyl rubber | Inner tubes |\n| Acrylonitrile 1,3-Butadiene Styrene |  | ABS (monomer initials) | Pipes, high-impact applications |\n\nSeveral different types of copolymers can be defined, depending on the distribution of monomer units in the chain. If monomer A is copolymerized with monomer B, for instance, the resultant product might have a random distribution of the two units throughout the chain, or it might have an alternating distribution.\n\n$$\n(A-A-A-B-A-B-B-A-B-A-A-A-B-B-B+\n$$"}
{"id": 1958, "contents": "Random copolymer - \n$$\n(A-B-A-B-A-B-A-B-A-B-A-B-A-B-A+\n$$"}
{"id": 1959, "contents": "Alternating copolymer - \nThe exact distribution of monomer units depends on the initial proportions of the two reactant monomers and their relative reactivities. In practice, neither perfectly random nor perfectly alternating copolymers are usually found. Most copolymers have many random imperfections.\n\nTwo other forms of copolymers that can be prepared under certain conditions are called block copolymers and graft copolymers. Block copolymers are those in which different blocks of identical monomer units alternate with each other; graft copolymers are those in which homopolymer branches of one monomer unit are \"grafted\" onto a homopolymer chain of another monomer unit.\n\n\nBlock copolymers are prepared by initiating the polymerization of one monomer as if growing a homopolymer chain and then adding an excess of the second monomer to the still-active reaction mix. Graft copolymers are made by gamma irradiation of a completed homopolymer chain in the presence of the second monomer. The high-energy irradiation knocks hydrogen atoms off the homopolymer chain at random points, thus generating new radical sites that can initiate polymerization of the added monomer.\n\nPROBLEM Draw the structure of an alternating segment of butyl rubber, a copolymer of isoprene 31-6 (2-methyl-1,3-butadiene) and isobutylene (2-methylpropene) prepared using a cationic initiator.\n\nPROBLEM Irradiation of poly(1,3-butadiene), followed by addition of styrene, yields a graft copolymer that 31-7 is used to make rubber soles for shoes. Draw the structure of a representative segment of this styrene-butadiene graft copolymer."}
{"id": 1960, "contents": "Alternating copolymer - 31.4 Step-Growth Polymers\nStep-growth polymers are produced by reactions in which each bond in the polymer is formed stepwise, independently of the other bonds. Like the polyamides (nylons) and polyesters that we saw in Section 21.9, most step-growth polymers are produced by reaction between two difunctional reactants. The polymer Nylon 66, for instance, is made by reaction between the six-carbon adipic acid and the six-carbon hexamethylenediamine (1,6-hexanediamine). Alternatively, a single reactant with two different functional groups can polymerize. Nylon 6 is made by polymerization of the six-carbon caprolactam. The reaction is initiated by adding a small amount of water, which hydrolyzes some caprolactam to 6-aminohexanoic acid. Nucleophilic addition of the amino group to caprolactam then propagates the polymerization.\n\n\nNylon 66"}
{"id": 1961, "contents": "Polycarbonates - \nPolycarbonates are like polyesters, but their carbonyl group is linked to two $-O R$ groups, $\\left[\\mathrm{O}=\\mathrm{C}(\\mathrm{OR})_{2}\\right]$. Lexan, for instance, is a polycarbonate prepared from diphenyl carbonate and a diphenol called bisphenol A. Lexan has unusually high impact strength, making it valuable for use in machinery housings, telephones, bicycle safety helmets, and bulletproof glass."}
{"id": 1962, "contents": "Polyurethanes - \nA urethane is a carbonyl-containing functional group in which the carbonyl carbon is bonded to both an -OR group and an $-\\mathrm{NR}_{2}$ group. As such, a urethane is halfway between a carbonate and a urea.\n\n\nA carbonate\n\n\nA urethane\n\n\nA urea\n\nA urethane is typically prepared by nucleophilic addition reaction of an alcohol with an isocyanate $(\\mathrm{R}-\\mathrm{N}=\\mathrm{C}=\\mathrm{O})$, so a polyurethane is prepared by reaction of a diol with a diisocyanate. The diol is usually a low-molecular-weight polymer (MW $\\approx 1000 \\mathrm{amu}$ ) with hydroxyl end-groups; the diisocyanate is often toluene-2,4-diisocyanate.\n\n\nToluene-2,4-diisocyanate\nA polyurethane\nSeveral different kinds of polyurethanes can be produced, depending on the nature of the polymeric alcohol used. One major use of polyurethane is in the stretchable spandex fibers used for bathing suits and athletic gear. These polyurethanes have a fairly low degree of cross-linking so that the resultant polymer is soft and elastic. A second major use of polyurethanes is in the foams used for insulation. Foaming occurs when a small amount of water is added during polymerization, giving a carbamic acid intermediate that spontaneously loses bubbles of $\\mathrm{CO}_{2}$.\n\n\nPolyurethane foams are generally made using a polyalcohol rather than a diol as the monomer so that the polymer has a high amount of three-dimensional cross-linking. The result is a rigid but very light foam suitable for use as thermal insulation in building construction and portable ice chests.\n\nPROBLEM Poly(ethylene terephthalate), or PET, is a polyester used to make soft-drink bottles. It is prepared\n31-8 by reaction of ethylene glycol with 1,4-benzenedicarboxylic acid (terephthalic acid). Draw the structure of PET.\n\nPROBLEM Show the mechanism of the nucleophilic addition reaction of an alcohol with an isocyanate to yield 31-9 a urethane."}
{"id": 1963, "contents": "Polyurethanes - 31.5 Olefin Metathesis Polymerization\nPerhaps the most important advance in polymer synthesis in recent years has been the development of olefin metathesis polymerization by Yves Chauvin of the Institut Fran\u00e7ais du P\u00e9trole, Robert H. Grubbs of CalTech, and Richard R.Schrock of MIT who shared the 2005 Nobel Prize in Chemistry for their work. At its simplest, an olefin metathesis reaction is two olefins (alkenes) exchanging substituents on their double bonds."}
{"id": 1964, "contents": "An olefin metathesis reaction - \nOlefin metathesis catalysts, such as the Grubbs catalyst now in common use, contain a carbon-metal double\nbond (usually to ruthenium, Ru ) and have the general structure $\\mathrm{M}=\\mathrm{CH}-\\mathrm{R}$. They function by reacting reversibly with an alkene to form a four-membered, metal-containing intermediate called a metallacycle, which immediately opens to give a different catalyst and a different alkene. The mechanism is shown in FIGURE 31.3."}
{"id": 1965, "contents": "Initiation - \nFIGURE 31.3 Mechanism of the olefin metathesis reaction. The process is initiated by a two-step sequence that involves (1) reaction of the catalyst and olefin 1 to give a four-membered metallacycle intermediate, followed by (2) ring-opening to give a different form of catalyst that contains part of olefin 1. (3) Reaction of this new catalyst with olefin 2 gives another metallacycle intermediate, (4) which opens to give metathesis product and another form of catalyst. $(5,6)$ The repeating ring-forming and ring-opening steps then continue.\n\nThere are several methods for implementing the olefin metathesis reaction to prepare polymers. One method, called ring-opening metathesis polymerization, or ROMP, involves the use of a moderately strained cycloalkene, such as cyclopentene. The strain of the ring favors ring-opening, thereby driving formation of the open-chain product. The polymer that results has double bonds spaced regularly along the chain, allowing for either hydrogenation or further functionalization if desired."}
{"id": 1966, "contents": "Ring-opening metathesis polymerization (ROMP) - \nA second method of using olefin metathesis to prepare polymers is by acyclic diene metathesis, or ADMET. As the name suggests, ADMET involves olefin metathesis of an open-chain substrate with two double bonds at the ends of a long chain, such as 1,8-nonadiene. As the reaction proceeds, the gaseous ethylene by-product\nescapes, thereby driving the equilibrium toward polymer product. So efficient is this reaction that polymers with molecular weights as high as $80,000 \\mathrm{amu}$ have been prepared.\n\nAcyclic diene metathesis (ADMET)\n\n\nThe ROMP and ADMET procedures are particularly valuable because the metathesis reaction is compatible with the presence in the olefin monomer of many different functional groups. In addition, the double bonds in the polymers provide still more flexibility for further manipulations. Among the commercial polymers produced by olefin metathesis are Vestenamer, used in the manufacture of tires and other molded rubber objects, and Norsorex, used in the automobile industry as a sealing material.\n\n\nVestenamer\n\n\nNorsorex\n\nPROBLEM Look at the structures of Vestenamer and Norsorex and show how they might be made by olefin 31-10 metathesis polymerization."}
{"id": 1967, "contents": "Ring-opening metathesis polymerization (ROMP) - 31.6 Intramolecular Olefin Metathesis\nThe olefin metathesis reaction has been used primarily as an intermolecular process for polymer synthesis, but it is also finding increased usage in complex organic synthesis as an intramolecular process in what is called ring-closing metathesis (RCM). The optimum catalyst is an $\\mathrm{Ru}(\\mathrm{II})$ complex, and cyclic alkenes with up to 30 atoms in ring size are easily produced."}
{"id": 1968, "contents": "Ring-opening metathesis polymerization (ROMP) - 31.7 Polymer Structure and Physical Properties\nPolymers aren't really that different from other organic molecules. They're much larger, of course, but their chemistry is similar to that of analogous small molecules. Thus, the alkane chains of polyethylene undergo radical-initiated halogenation, the aromatic rings of polystyrene undergo typical electrophilic aromatic substitution reactions, and the amide linkages of nylon are hydrolyzed by aqueous base.\n\nThe major difference between small and large organic molecules is in their physical properties. For instance, their large size means that polymers experience substantially greater van der Waals forces than do small molecules (Section 2.12). But because van der Waals forces operate only at close distances, they are strongest in polymers like high-density polyethylene, in which chains can pack together closely in a regular way. Many\npolymers, in fact, have regions that are essentially crystalline. These regions, called crystallites, consist of highly ordered portions in which the zigzag polymer chains are held together by van der Waals forces (FIGURE 31.4).\n\n\nFIGURE 31.4 Crystallites in linear polyethylene. The long polymer chains are arranged in parallel lines in the crystallite regions.\nAs you might expect, polymer crystallinity is strongly affected by the steric requirements of substituent groups on the chains. Linear polyethylene is highly crystalline, but poly(methyl methacrylate) is noncrystalline because the chains can't pack closely together in a regular way. Polymers with a high degree of crystallinity are generally hard and durable. When heated, the crystalline regions melt at the melt transition temperature, $\\boldsymbol{T}_{\\mathbf{m}}$, to give an amorphous material.\n\nNoncrystalline, amorphous polymers like poly(methyl methacrylate), sold under the trade name Plexiglas, have little or no long-range ordering among chains but can nevertheless be very hard at room temperature. When heated, the hard amorphous polymer becomes soft and flexible at a point called the glass transition temperature, $\\boldsymbol{T}_{\\mathbf{g}}$. Much of the art in polymer synthesis lies in finding methods for controlling the degree of crystallinity and the glass transition temperature, thereby imparting useful properties to the polymer."}
{"id": 1969, "contents": "Ring-opening metathesis polymerization (ROMP) - 31.7 Polymer Structure and Physical Properties\nIn general, polymers can be divided into four major categories, depending on their physical behavior: thermoplastics, fibers, elastomers, and thermosetting resins. Thermoplastics are the polymers most people think of when the word plastic is mentioned. These polymers have a high glass transition temperature and are therefore hard at room temperature but become soft and viscous when heated. As a result, they can be molded into toys, beads, telephone housings, or any of a thousand other items. Because thermoplastics have little or no cross-linking, the individual chains can slip past one another in the melt. Some thermoplastic polymers, such as poly(methyl methacrylate) and polystyrene, are amorphous and noncrystalline; others, such as polyethylene and nylon, are partially crystalline. Among the better-known thermoplastics is poly(ethylene terephthalate), or PET, used for making plastic soft-drink bottles.\n\n\nPoly(ethylene terephthalate)\n\nPlasticizers-small organic molecules that act as lubricants between chains-are usually added to thermoplastics to keep them from becoming brittle at room temperature. An example is poly(vinyl chloride), which is brittle when pure but becomes supple and pliable when a plasticizer is added. In fact, most drip bags used in hospitals to deliver intravenous saline solutions are made of poly(vinyl chloride), although replacements such as polypropylene are appearing.\n\nDialkyl phthalates such as di(2-ethylhexyl) phthalate (generally called dioctyl phthalate) are commonly used as plasticizers, although questions about their safety have been raised. The U.S. Food and Drug Administration\n(FDA) has advised the use of alternative materials in compromised patients and infants but has found no evidence of toxicity for healthy individuals. In addition, children's toys that contain phthalates have been banned in the United States.\n\n\nFibers are thin threads produced by extruding a molten polymer through small holes in a die, or spinneret. The fibers are then cooled and drawn out, which orients the crystallite regions along the axis of the fiber and adds considerable tensile strength FIGURE 31.5. Nylon, Dacron, and polyethylene all have the semicrystalline structure necessary for being drawn into oriented fibers."}
{"id": 1970, "contents": "Ring-opening metathesis polymerization (ROMP) - 31.7 Polymer Structure and Physical Properties\nElastomers are amorphous polymers that have the ability to stretch out and spring back to their original shapes. These polymers must have low glass transition temperatures and a small amount of cross-linking to prevent the chains from slipping over one another. In addition, the chains must have an irregular shape to prevent crystallite formation. When stretched, the randomly coiled chains straighten out and orient along the direction of the pull. Van der Waals forces are too weak and too few to maintain this orientation, however, and the elastomer therefore reverts to its random coiled state when the stretching force is released (FIGURE 31.6).\n\n\nFIGURE 31.6 Unstretched and stretched forms of an elastomer.\nNatural rubber (Section 14.6) is the most common example of an elastomer. Rubber has the long chains and occasional cross-links needed for elasticity, but its irregular geometry prevents close packing of the chains into crystallites. Gutta-percha, by contrast, is highly crystalline and is not an elastomer (FIGURE 31.7).\n\n(b)\n\n\nFIGURE 31.7 (a) Natural rubber is elastic and noncrystalline because of its cis double-bond geometry, but (b) gutta-percha is nonelastic and crystalline because its geometry allows for better packing together of chains.\n\nThermosetting resins are polymers that become highly cross-linked and solidify into a hard, insoluble mass when heated. Bakelite, a thermosetting resin first produced in 1907, has been in commercial use longer than any other synthetic polymer. It is widely used for molded parts, adhesives, coatings, and even high-temperature applications such as missile nose cones.\n\nChemically, Bakelite is a phenolic resin, produced by reaction of phenol and formaldehyde. On heating, water is eliminated, many cross-links form, and the polymer sets into a rocklike mass. The cross-linking in Bakelite and other thermosetting resins is three-dimensional and is so extensive that we can't really speak of polymer \"chains.\" A piece of Bakelite is essentially one large molecule."}
{"id": 1971, "contents": "Ring-opening metathesis polymerization (ROMP) - 31.7 Polymer Structure and Physical Properties\nPROBLEM What product would you expect to obtain from catalytic hydrogenation of natural rubber? Would\n31-11 the product be syndiotactic, atactic, or isotactic?\nPROBLEM Propose a mechanism to account for the formation of Bakelite from acid-catalyzed polymerization\n31-12 of phenol and formaldehyde."}
{"id": 1972, "contents": "Degradable Polymers - \nThe high chemical stability of many polymers is both a blessing and a curse. Heat resistance, wear resistance, and long life are valuable characteristics of clothing fibers, bicycle helmets, underground pipes, food wrappers, and many other items. Yet when these items outlive their usefulness, disposal becomes a problem.\n\n\nFIGURE 31.8 What happens to the plastics that end up here? (credit: modification of work \"King of the trash hill\" by Alan Levine/Flickr, CC BY 2.0)\n\nRecycling of unwanted polymers is the best solution, and six types of plastics in common use are frequently stamped with identifying codes assigned by the Society of the Plastics Industry TABLE 31.2. After being sorted by type, the items to be recycled are shredded into small chips, washed, dried, and melted for reuse. Soft-drink bottles, for instance, are made from recycled poly(ethylene terephthalate), trash bags are made from recycled low-density polyethylene, and garden furniture is made from recycled polypropylene and mixed plastics.\n\n| TABLE 31.2 Recyclable Plastics |  | Recycling code |\n| :--- | :--- | :--- |\n| Polymer | Use |  |\n| Poly(ethylene terephthalate) | 1-PET | Soft-drink bottles |\n| High-density polyethylene | 2-HDPE | Bottles |\n| Poly(vinyl chloride) | $3-\\mathrm{V}$ | Floor mats |\n| Low-density polyethylene | 4-LDPE | Grocery bags |\n| Polypropylene | $5-\\mathrm{PP}$ | Furniture |\n| Polystyrene | 6-PS | Molded articles |\n| Mixed plastics | 7 | Benches, plastic lumber |"}
{"id": 1973, "contents": "Degradable Polymers - \nFrequently, however, plastics are simply thrown away rather than recycled, and much work has therefore been carried out on developing biodegradable polymers, which can be broken down by soil microorganisms. Among the most common biodegradable polymers are polyglycolic acid (PGA), polylactic acid (PLA), and polyhydroxybutyrate (PHB). All are polyesters and are therefore susceptible to hydrolysis of their ester links. Copolymers of PGA with PLA have found a particularly wide range of uses. A 90/10 copolymer of polyglycolic acid with polylactic acid is used to make absorbable sutures that are degraded and absorbed by the body within 90 days after surgery.\n\n\nGlycolic acid\n\n\n\nPoly(glycolic acid)\n\n\nLactic acid\n\n\n\nPoly(lactic acid)\n\n\n3-Hydroxybutyric acid\n\n\nPoly(hydroxybutyrate)\n\nBiodegradation is important, but even better than the breakdown of a small group of specialized polymers would be the development of a generalized method of breakdown that could be used on any polymer, including even hydrocarbon polymers such as polyethylene and polypropylene. Such a method has recently been reported by chemists at the University of Delaware, using hydrogen at $200{ }^{\\circ} \\mathrm{C}$ and 2 atm pressure in the presence of an organozirconium catalyst to turn polyethylene into simple hydrocarbons. The method uses a large amount of hydrogen at present, but it is an impressive start."}
{"id": 1974, "contents": "Key Terms - \n- atactic polymer\n- block copolymer\n- chain-growth polymer\n- copolymer\n- crystallite\n- elastomer\n- fiber\n- glass transition temperature ( $T_{\\mathrm{g}}$ )\n- graft copolymer\n- homopolymer\n- isotactic polymer\n- melt transition temperature ( $T_{\\mathrm{m}}$ )\n- monomer\n- olefin metathesis polymerization\n- plasticizer\n- polycarbonate\n- polymer\n- polyurethane\n- syndiotactic polymer\n- thermoplastic\n- thermosetting resin\n- urethane\n- Ziegler-Natta catalyst"}
{"id": 1975, "contents": "Summary - \nSynthetic polymers can be classified as either chain-growth or step-growth. Chain-growth polymers are prepared by chain-reaction polymerization of vinyl monomers in the presence of a radical, an anion, or a cation initiator. Radical polymerization is sometimes used, but alkenes such as 2 -methylpropene that have electron-donating substituents on the double bond polymerize easily by a cationic route through carbocation intermediates. Similarly, monomers such as methyl $\\alpha$-cyanoacrylate that have electron-withdrawing substituents on the double bond polymerize by an anionic, conjugate addition pathway.\n\nCopolymerization of two monomers gives a product with properties different from those of either homopolymer. Graft copolymers and block copolymers are two examples.\n\nAlkene polymerization can be carried out in a controlled manner using a Ziegler-Natta catalyst. Ziegler-Natta polymerization minimizes the amount of chain branching in the polymer and leads to stereoregular chains-either isotactic (substituents on the same side of the chain) or syndiotactic (substituents on alternate sides of the chain), rather than atactic (substituents randomly disposed).\n\nStep-growth polymers, the second major class of polymers, are prepared by reactions between difunctional molecules, with individual bonds in the polymer formed independently of one another. Polycarbonates are formed from a diester and a diol, and polyurethanes are formed from a diisocyanate and a diol.\n\nThe chemistry of synthetic polymers is similar to the chemistry of small molecules with the same functional groups, but the physical properties of polymers are greatly affected by size. Polymers can be classified by physical property into four groups: thermoplastics, fibers, elastomers, and thermosetting resins. The properties of each group can be accounted for by the structure, the degree of crystallinity, and the amount of cross-linking they contain."}
{"id": 1976, "contents": "Visualizing Chemistry - \nPROBLEM Identify the structural class to which the following polymer belongs, and show the structure of the 31-13 monomer units used to make it:\n\n\nPROBLEM Show the structures of the polymers that could be made from the following monomers (green $=\\mathrm{Cl}$ ):\n\n31-14 (a)\n\n(b)\n\n\nMechanism Problems\nPROBLEM Poly(ethylene glycol), or Carbowax, is made by anionic polymerization of ethylene oxide using\n31-15 NaOH as catalyst. Propose a mechanism.\n$\\left.+\\mathrm{O}-\\mathrm{CH}_{2} \\mathrm{CH}_{2}\\right)_{n} \\quad$ Poly(ethylene glycol)\nPROBLEM The polyurethane foam used for home insulation uses methanediphenyldiisocyanate (MDI) as\n31-16 monomer. The MDI is prepared by acid-catalyzed reaction of aniline with formaldehyde, followed by treatment with phosgene, $\\mathrm{COCl}_{2}$. Propose mechanisms for both steps.\n\n\nMDI\nPROBLEM Write the structure of a representative segment of polyurethane prepared by reaction of ethylene\n31-17 glycol with MDI (Problem 31-16).\nPROBLEM The polymeric resin used for Merrifield solid-phase peptide synthesis (Section 26.8) is prepared\n31-18 by treating polystyrene with $N$-(hydroxymethyl)phthalimide and trifluoromethanesulfonic acid, followed by reaction with hydrazine. Propose a mechanism for both steps.\n\n\nPolystyrene\n\n$\\xrightarrow{\\mathrm{CF}_{3} \\mathrm{SO}_{3} \\mathrm{H}}$"}
{"id": 1977, "contents": "Visualizing Chemistry - \nPolystyrene\n\n$\\xrightarrow{\\mathrm{CF}_{3} \\mathrm{SO}_{3} \\mathrm{H}}$\n\n\nPROBLEM Polydicyclopentadiene (PDCPD), marketed as Telene or Metton, is a highly cross-linked\n31-19 thermosetting resin used for molding such impact-resistant parts as cabs for large trucks and earth-moving equipment. PDCPD is prepared by ring-opening metathesis polymerization of dicyclopentadiene, which is itself prepared from 1,3-cyclopentadiene. The polymerization occurs by initial metathesis of the more highly strained double bond in the bicyclo[2.2.1]heptane part of the molecule (Section 4.9) to give a linear polymer, followed by cross-linking of different chains in a second metathesis of the remaining cyclopentene double bond.\n\n(a) Show the mechanism of the formation of dicyclopentadiene from cyclopentadiene.\n(b) Draw the structure of a representative sample of the initially formed linear polymer containing three monomer units.\n(c) Draw the structure of a representative sample of PDCPD that shows how cross-linking of the linear chains takes place."}
{"id": 1978, "contents": "General Problems - \nPROBLEM Identify the monomer units from which each of the following polymers is made, and tell whether 31-20 each is a chain-growth or a step-growth polymer:\n(a)\n$\\left.+\\mathrm{CH}_{2}-\\mathrm{O}\\right)_{n}$\n(b) $\\left.+\\mathrm{CF}_{2}-\\mathrm{CFCl}\\right)_{n}$\n(c)\n\n(d)\n\n(e)\n\n\nPROBLEM Draw a three-dimensional representation of segments of the following polymers:\n31-21\n(a) Syndiotactic polyacrylonitrile\n(b) Atactic poly(methyl methacrylate)\n(c) Isotactic poly(vinyl chloride)\n\nPROBLEM Draw the structure of Kodel, a polyester prepared by heating dimethyl 1,4-benzenedicarboxylate 31-22 with 1,4-bis(hydroxymethyl)cyclohexane.\n\n\n1,4-Bis(hydroxymethyl)cyclohexane\n\nPROBLEM Show the structure of the polymer that results from heating the following diepoxide and diamine:\n31-23\n\n\nPROBLEM Nomex, a polyamide used in such applications as fire-retardant clothing, is prepared by reaction of\n31-24 1,3-benzenediamine with 1,3-benzenedicarbonyl chloride. Show the structure of Nomex.\nPROBLEM Nylon 10,10 is an extremely tough, strong polymer used to make reinforcing rods for concrete.\n31-25 Draw a segment of nylon 10,10 , and show its monomer units.\nPROBLEM 1,3-Cyclopentadiene undergoes thermal polymerization to yield a polymer that has no double\n31-26 bonds in the chain. Upon strong heating, the polymer breaks down to regenerate cyclopentadiene. Propose a structure for the polymer.\n\nPROBLEM When styrene, $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CH}=\\mathrm{CH}_{2}$, is copolymerized in the presence of a few percent $p$-divinylbenzene,\n31-27 a hard, insoluble, cross-linked polymer is obtained. Show how this cross-linking of polystyrene chains occurs."}
{"id": 1979, "contents": "General Problems - \nPROBLEM Nitroethylene, $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHNO}_{2}$, is a sensitive compound that must be prepared with great care.\n31-28 Attempted purification of nitroethylene by distillation often results in low recovery of product and a white coating on the inner walls of the distillation apparatus. Explain.\n\nPROBLEM Poly(vinyl butyral) is used as the plastic laminate in the preparation of automobile windshield safety\n31-29 glass. How would you synthesize this polymer?"}
{"id": 1980, "contents": "Poly(vinyl butyral) - \nPROBLEM What is the structure of the polymer produced by anionic polymerization of $\\beta$-propiolactone using\n31-30 NaOH as catalyst?"}
{"id": 1981, "contents": "$\\beta$-Propiolactone - \nPROBLEM Glyptal is a highly cross-linked thermosetting resin produced by heating glycerol and phthalic\n31-31 anhydride (1,2-benzenedicarboxylic acid anhydride). Show the structure of a representative segment of glyptal.\n\nPROBLEM Melmac, a thermosetting resin often used to make plastic dishes, is prepared by heating melamine\n31-32 with formaldehyde. Look at the structure of Bakelite shown in Section 31.7, and then propose a structure for Melmac.\n\n\nMelamine\n\nPROBLEM Epoxy adhesives are cross-linked resins prepared in two steps. The first step involves $\\mathrm{S}_{\\mathrm{N}} 2$ reaction 31-33\nof the disodium salt of bisphenol A with epichlorohydrin to form a low-molecular-weight prepolymer. This prepolymer is then \"cured\" into a cross-linked resin by treatment with a triamine such as $\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NHCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}$.\n\n\nBisphenol A\n\n\nEpichlorohydrin\n(a) What is the structure of the prepolymer?\n(b) How does addition of the triamine to the prepolymer result in cross-linking?\n\nPROBLEM The smoking salons of the Hindenburg and other hydrogen-filled dirigibles of the 1930s were\n31-34 insulated with urea-formaldehyde polymer foams. The structure of this polymer is highly crosslinked, like that of Bakelite (Section 31.7). Propose a structure.\n\n\nPROBLEM 2-Ethyl-1-hexanol, used in the synthesis of di(2-ethylhexyl) phthalate plasticizer, is made 31-35 commercially from butanal. Show the likely synthesis route."}
{"id": 1982, "contents": "Nomenclature of Polyfunctional Organic Compounds - \nWith more than 40 million organic compounds now known and thousands more being created daily, naming them all is a real problem. Part of the problem is due to the sheer complexity of organic structures, but part is also due to the fact that chemical names have more than one purpose. For the Chemical Abstracts Service (CAS), which catalogs and indexes the worldwide chemical literature, each compound must have only one correct name. It would be chaos if half the entries for $\\mathrm{CH}_{3} \\mathrm{Br}$ were indexed under \" M \" for methyl bromide and half under \"B\" for bromomethane. Furthermore, a CAS name must be strictly systematic so that it can be assigned and interpreted by computers; common names are not allowed.\n\nPeople, however, have different requirements than computers. For people-which is to say students and professional chemists in their spoken and written communications-it's best that a chemical name be pronounceable and as easy as possible to assign and interpret. Furthermore, it's convenient if names follow historical precedents, even if that means a particularly well-known compound might have more than one name. People can readily understand that bromomethane and methyl bromide both refer to $\\mathrm{CH}_{3} \\mathrm{Br}$.\n\nAs noted in the text, chemists overwhelmingly use the nomenclature system devised and maintained by the International Union of Pure and Applied Chemistry, or IUPAC. Rules for naming monofunctional compounds were given throughout the text as each new functional group was introduced, and a list of where these rules can be found is given in TABLE A1.\n\n| TABLE A1 Nomenclature Rules for Functional <br> Groups |  |\n| :--- | :---: |\n| Functional group | Text section |\n| Acid anhydrides | $21-1$ |\n| Acid halides | $21-1$ |\n| Acyl phosphates | $21-1$ |\n| Alcohols | $17-1$ |\n| Aldehydes | $7-1$ |\n| Alkanes | $10-1$ |\n| Alkenes | $9-1$ |\n| Alkyl halides | $21-1$ |\n| Amides | $24-1$ |\n| Amines | $15-1$ |\n\nTABLE A1 Nomenclature Rules for Functional Groups"}
{"id": 1983, "contents": "Nomenclature of Polyfunctional Organic Compounds - \nTABLE A1 Nomenclature Rules for Functional Groups\n\n| Functional group | Text section |\n| :--- | :---: |\n| Carboxylic acids | $20-1$ |\n| Cycloalkanes | $4-1$ |\n| Esters | $21-1$ |\n| Ethers | $18-1$ |\n| Ketones | $20-1$ |\n| Nitriles | $17-1$ |\n| Phenols | $18-7$ |\n| Sulfides | $18-7$ |\n| Thiols | $21-1$ |\n| Thioesters |  |\n\nNaming a monofunctional compound is reasonably straightforward, but even experienced chemists often encounter problems when faced with naming a complex polyfunctional compound. Take the following compound, for instance. It has three functional groups, ester, ketone, and $\\mathrm{C}=\\mathrm{C}$, but how should it be named? As an ester with an -oate ending, a ketone with an -one ending, or an alkene with an -ene ending? It's actually named methyl 3-(2-oxo-6-cyclohexenyl)propanoate.\n\n\nThe name of a polyfunctional organic molecule has four parts-suffix, parent, prefixes, and locants-which must be identified and expressed in the proper order and format. Let's look at each of the four."}
{"id": 1984, "contents": "Name Part 1. The Suffix: Functional-Group Precedence - \nAlthough a polyfunctional organic molecule might contain several different functional groups, we must choose just one suffix for nomenclature purposes. It's not correct to use two suffixes. Thus, keto ester $\\mathbf{1}$ must be named either as a ketone with an -one suffix or as an ester with an -oate suffix, but it can't be named as an -onoate. Similarly, amino alcohol 2 must be named either as an alcohol (-ol) or as an amine (-amine), but it can't be named as an-olamine or -aminol.\n\n2.\n\n\nThe only exception to the rule requiring a single suffix is when naming compounds that have double or triple bonds. Thus, the unsaturated acid $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ is 3-butenoic acid, and the acetylenic alcohol $\\mathrm{HC} \\equiv \\mathrm{CCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}$ is 5-pentyn-1-ol.\n\nHow do we choose which suffix to use? Functional groups are divided into two classes, principal groups and subordinate groups, as shown in TABLE A2. Principal groups can be cited either as prefixes or as suffixes, while subordinate groups are cited only as prefixes. Within the principal groups, an order of priority has been\nestablished: the proper suffix for a given compound is determined by choosing the principal group of highest priority. For example, TABLE A2 indicates that keto ester $\\mathbf{1}$ should be named as an ester rather than as a ketone because an ester functional group is higher in priority than a ketone. Similarly, amino alcohol 2 should be named as an alcohol rather than as an amine. Thus, the name for $\\mathbf{1}$ is methyl 4-oxopentanoate and the name for $\\mathbf{2}$ is 5-amino-2-pentanol. Further examples are shown:\n\nTABLE A2 Classification of Functional Groups ${ }^{\\text {a }}$"}
{"id": 1985, "contents": "Name Part 1. The Suffix: Functional-Group Precedence - \nTABLE A2 Classification of Functional Groups ${ }^{\\text {a }}$\n\n| Functional group | Name as suffix | Name as prefix |\n| :---: | :---: | :---: |\n| Principal groups |  |  |\n| Carboxylic acids | -oic acid | carboxy |\n|  | -carboxylic acid |  |\n| Acid anhydrides | -oic anhydride | - |\n|  | -carboxylic anhydride |  |\n| Esters | -oate | alkoxycarbonyl |\n|  | -carboxylate |  |\n| Thioesters | -thioate | alkylthiocarbonyl |\n|  | -carbothioate |  |\n| Acid halides | -oyl halide | halocarbonyl |\n|  | -carbonyl halide |  |\n| Amides | -amide | carbamoyl |\n|  | -carboxamide |  |\n| Nitriles | -nitrile | cyano |\n|  | -carbonitrile |  |\n| Aldehydes | -al | oxo |\n|  | -carbaldehyde |  |\n| Ketones | -one | oxo |\n| Alcohols | -ol | hydroxy |\n| Phenols | -ol | hydroxy |\n| Thiols | -thiol | mercapto |\n\n$\\overline{{ }^{2} \\text { Principal groups are listed in order of decreasing priority; subordinate groups have no priority order. }}$\n\nTABLE A2 Classification of Functional Groups ${ }^{\\text {a }}$\n\n| Functional group | Name as suffix | Name as prefix |\n| :--- | :--- | :--- |\n| Amines | -amine | amino |\n| Imines | -imine | alkoxy |\n| Ethers | ether | alkylthio |\n| Sulfides | sulfide | - |\n| Disulfides | disulfide | - |\n| Alkenes | -ene | - |\n| Alkynes | -yne | - |\n| Alkanes | -ane |  |\n\nSubordinate groups"}
{"id": 1986, "contents": "Name Part 1. The Suffix: Functional-Group Precedence - \nSubordinate groups\n\n| Azides | - | azido |\n| :--- | :--- | :--- |\n| Halides | - | halo |\n| Nitro compounds | - | nitro |\n\n${ }^{\\text {a Principal groups are listed in order of decreasing priority; subordinate groups have no priority order. }}$\n\n\n1. Methyl 4-oxopentanoate (an ester with a ketone group)\n\n2. Methyl 5-methyl-6-oxohexanoate (an ester with an aldehyde group)\n\n3. 5-Amino-2-pentanol\n(an alcohol with an amine group)\n\n4. 5-Carbamoyl-4-hydroxypentanoic acid (a carboxylic acid with amide and alcohol groups)\n\n5. 3-Oxocyclohexanecarbaldehyde\n(an aldehyde with a ketone group)"}
{"id": 1987, "contents": "Name Part 2. The Parent: Selecting the Main Chain or Ring - \nThe parent, or base, name of a polyfunctional organic compound is usually easy to identify. If the principal group of highest priority is part of an open chain, the parent name is that of the longest chain containing the largest number of principal groups. For example, compounds $\\mathbf{6}$ and $\\mathbf{7}$ are isomeric aldehydo amides, which must be named as amides rather than as aldehydes according to TABLE A2. The longest chain in compound\n\n6 has six carbons, and the substance is named 5-methyl-6-oxohexanamide. Compound $\\mathbf{7}$ also has a chain of six carbons, but the longest chain that contains both principal functional groups has only four carbons. Thus, compound 7 is named 4-oxo-3-propylbutanamide.\n\n6. 5-Methyl-6-oxohexanamide\n\n7. 4-0xo-3-propylbutanamide"}
{"id": 1988, "contents": "Name Part 2. The Parent: Selecting the Main Chain or Ring - \n6. 5-Methyl-6-oxohexanamide\n\n7. 4-0xo-3-propylbutanamide\n\nIf the highest-priority principal group is attached to a ring, the parent name is that of the ring system. Compounds $\\mathbf{8}$ and $\\mathbf{9}$, for instance, are isomeric keto nitriles and must both be named as nitriles according to TABLE A2. Substance $\\mathbf{8}$ is named as a benzonitrile because the -CN functional group is a substituent on the aromatic ring, but substance $\\mathbf{9}$ is named as an acetonitrile because the -CN functional group is on an open chain. Thus, their names are 2 -acetyl-(4-bromomethyl)benzonitrile (8) and (2-acetyl-4-bromophenyl)acetonitrile (9). As further examples, compounds $\\mathbf{1 0}$ and $\\mathbf{1 1}$ are both keto acids and must be named as acids, but the parent name in $\\mathbf{1 0}$ is that of a ring system (cyclohexanecarboxylic acid) and the parent name in $\\mathbf{1 1}$ is that of an open chain (propanoic acid). Thus, their names are trans-2-(3-oxopropyl)cyclohexanecarboxylic acid (10) and 3-(2-oxocyclohexyl)propanoic acid (11).\n\n8. 2-Acetyl-(4-bromomethyl)benzonitrile\n\n10. trans-2-(3-oxopropyl)cyclohexanecarboxylic acid\n\n9. (2-Acetyl-4-bromophenyl)acetonitrile\n\n11. 3-(2-0xocyclohexyl)propanoic acid"}
{"id": 1989, "contents": "Name Parts 3 and 4. The Prefixes and Locants - \nWith the parent name and the suffix established, the next step is to identify and give numbers, or locants, to all substituents on the parent chain or ring. The substituents include all alkyl groups and all functional groups other than the one cited in the suffix. For example, compound $\\mathbf{1 2}$ contains three different functional groups (carboxyl, keto, and double bond). Because the carboxyl group is highest in priority and the longest chain containing the functional groups has seven carbons, compound $\\mathbf{1 2}$ is a heptenoic acid. In addition, the parent chain has a keto (oxo) substituent and three methyl groups. Numbering from the end nearer the highest-priority functional group gives the name ( $E$ )-2,5,5-trimethyl-4-oxo-2-heptenoic acid. Look back at some of the other compounds we've named to see other examples of how prefixes and locants are assigned.\n\n12. (E)-2,5,5-Trimethyl-4-oxo-2-heptenoic acid"}
{"id": 1990, "contents": "Writing the Name - \nWith the name parts established, the entire name can be written out. Several additional rules apply:\n\n1. Order of prefixes. When the substituents have been identified, the parent chain has been numbered, and the proper multipliers such as di- and tri- have been assigned, the name is written with the substituents listed in alphabetical, rather than numerical, order. Multipliers such as di- and tri- are not used for\nalphabetization, but the italicized prefixes iso- and sec- are used.\n\n2. 5-Amino-3-methyl-2-pentanol\n3. Use of hyphens; single- and multiple-word names. The general rule is to determine whether the parent is itself an element or compound. If it is, then the name is written as a single word; if it isn't, then the name is written as multiple words. Methylbenzene is written as one word, for instance, because the parent-benzene-is a compound. Diethyl ether, however, is written as two words because the parent-ether-is a class name rather than a compound name. Some further examples follow:\n\n$$\n\\mathrm{H}_{3} \\mathrm{C}-\\mathrm{Mg}-\\mathrm{CH}_{3}\n$$\n\n14. Dimethylmagnesium (one word, because magnesium is an element)\n\n15. 4-(Dimethylamino)pyridine (one word, because pyridine is a compound)\n\n16. Isopropyl 3-hydroxypropanoate (two words, because \"propanoate\" is not a compound)\n\n17. Methyl cyclopentanecarbothioate (two words, because \"cyclopentanecarbothioate\" is not a compound)\n18. Parentheses. Parentheses are used to denote complex substituents when ambiguity would otherwise arise. For example, chloromethylbenzene has two substituents on a benzene ring, but (chloromethyl)benzene has only one complex substituent. Note that the expression in parentheses is not set off by hyphens from the rest of the name.\n\n19. $p$-Chloromethylbenzene\n\n20. (Chloromethyl)benzene"}
{"id": 1991, "contents": "Writing the Name - \n19. $p$-Chloromethylbenzene\n\n20. (Chloromethyl)benzene\n\n21. 2-(1-Methylpropyl)pentanedioic acid"}
{"id": 1992, "contents": "Additional Reading - \nFurther explanations of the rules of organic nomenclature can be found online at ACD Labs (https://www.acdlabs.com/iupac/nomenclature/) (accessed May 2023) and in the following references:\n\n1. \"A Guide to IUPAC Nomenclature of Organic Compounds,\" CRC Press, Boca Raton, FL, 1993.\n2. \"Nomenclature of Organic Chemistry, Sections A, B, C, D, E, F, and H,\" International Union of Pure and Applied Chemistry, Pergamon Press, Oxford, 1979."}
{"id": 1993, "contents": "Acidity Constants for Some Organic Compounds - \n| Compound | $\\mathrm{p} K_{\\mathrm{a}}$ |\n| :---: | :---: |\n| $\\mathrm{CH}_{3} \\mathrm{SO}_{3} \\mathrm{H}$ | -1.8 |\n| $\\mathrm{CH}\\left(\\mathrm{NO}_{2}\\right)_{3}$ | 0.1 |\n| <smiles>O=N+c1cc(N+[O-])c(O)c(N+[O-])c1</smiles> | 0.3 |\n| $\\mathrm{CCl}_{3} \\mathrm{CO}_{2} \\mathrm{H}$ | 0.5 |\n| $\\mathrm{CF}_{3} \\mathrm{CO}_{2} \\mathrm{H}$ | 0.5 |\n| $\\mathrm{CBr}_{3} \\mathrm{CO}_{2} \\mathrm{H}$ | 0.7 |\n| $\\mathrm{HO}_{2} \\mathrm{CC} \\equiv \\mathrm{CCO}_{2} \\mathrm{H}$ | 1.2; 2.5 |\n| $\\mathrm{HO}_{2} \\mathrm{CCO}_{2} \\mathrm{H}$ | 1.2;3.7 |\n| $\\mathrm{CHCl}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ | 1.3 |\n| $\\mathrm{CH}_{2}\\left(\\mathrm{NO}_{2}\\right) \\mathrm{CO}_{2} \\mathrm{H}$ | 1.3 |\n| $\\mathrm{HC} \\equiv \\mathrm{CCO}_{2} \\mathrm{H}$ | 1.9 |\n| (z) $\\mathrm{HO}_{2} \\mathrm{CCH}=\\mathrm{CHCO}_{2} \\mathrm{H}$ | 1.9; 6.3 |\n| <smiles>O=C(O)c1ccccc1N+[O-]</smiles> | 2.4 |\n| $\\mathrm{CH}_{3} \\mathrm{COCO}_{2} \\mathrm{H}$ | 2.4 |\n| $\\mathrm{NCCH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ | 2.5 |"}
{"id": 1994, "contents": "Acidity Constants for Some Organic Compounds - \n| $\\mathrm{NCCH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ | 2.5 |\n| $\\mathrm{CH}_{3} \\mathrm{C} \\equiv \\mathrm{CCO}_{2} \\mathrm{H}$ | 2.6 |\n| $\\mathrm{CH}_{2} \\mathrm{FCO}_{2} \\mathrm{H}$ | 2.7 |\n| $\\mathrm{CH}_{2} \\mathrm{ClCO}_{2} \\mathrm{H}$ | 2.8 |"}
{"id": 1995, "contents": "Acidity Constants for Some Organic Compounds - \n| Compound | $\\mathrm{p} K_{\\mathrm{a}}$ |\n| :---: | :---: |\n| $\\mathrm{HO}_{2} \\mathrm{CCH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ | 2.8; 5.6 |\n| $\\mathrm{CH}_{2} \\mathrm{BrCO}_{2} \\mathrm{H}$ | 2.9 |\n| <smiles>O=C(O)c1ccccc1Cl</smiles> | 3.0 |\n| <smiles>O=C(O)c1ccccc1O</smiles> | 3.0 |\n| $\\mathrm{CH}_{2} \\mathrm{ICO}_{2} \\mathrm{H}$ | 3.2 |\n| $\\mathrm{CHOCO}_{2} \\mathrm{H}$ | 3.2 |\n| <smiles>O=C(O)c1ccc(N+[O-])cc1</smiles> | 3.4 |\n| <smiles>O=C(O)c1ccc(N+[O-])c(N+[O-])c1</smiles> | 3.5 |\n| $\\mathrm{HSCH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ | 3.5; 10.2 |\n| $\\mathrm{CH}_{2}\\left(\\mathrm{NO}_{2}\\right)_{2}$ | 3.6 |\n| $\\mathrm{CH}_{3} \\mathrm{OCH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ | 3.6 |\n| $\\mathrm{CH}_{3} \\mathrm{COCH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ | 3.6 |\n| $\\mathrm{HOCH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ | 3.7 |\n| $\\mathrm{HCO}_{2} \\mathrm{H}$ | 3.7 |\n| <smiles>O=C(O)c1cccc(Cl)c1</smiles> | 3.8 |\n| <smiles>O=C(O)c1ccc(Cl)cc1</smiles> | 4.0 |"}
{"id": 1996, "contents": "Acidity Constants for Some Organic Compounds - \n| <smiles>O=C(O)c1cccc(Cl)c1</smiles> | 3.8 |\n| <smiles>O=C(O)c1ccc(Cl)cc1</smiles> | 4.0 |\n| $\\mathrm{CH}_{2} \\mathrm{BrCH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ | 4.0 |"}
{"id": 1997, "contents": "Acidity Constants for Some Organic Compounds - \nTABLE B1"}
{"id": 1998, "contents": "Acidity Constants for Some Organic Compounds - \n| Compound | $\\mathrm{p} K_{\\mathrm{a}}$ |\n| :---: | :---: |\n| <smiles>O=N+c1ccc(O)c(N+[O-])c1</smiles> | 4.1 |\n| <smiles>O=C(O)c1ccccc1</smiles> | 4.2 |\n| $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCO}_{2} \\mathrm{H}$ | 4.2 |\n| $\\mathrm{HO}_{2} \\mathrm{CCH}_{2} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ | 4.2; 5.7 |\n| $\\mathrm{HO}_{2} \\mathrm{CCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ | 4.3; 5.4 |\n| <smiles>Oc1c(Cl)c(Cl)c(Cl)c(Cl)c1Cl</smiles> | 4.5 |\n| $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{C}\\left(\\mathrm{CH}_{3}\\right) \\mathrm{CO}_{2} \\mathrm{H}$ | 4.7 |\n| $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$ | 4.8 |\n| $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ | 4.8 |\n| $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CCO}_{2} \\mathrm{H}$ | 5.0 |\n| $\\mathrm{CH}_{3} \\mathrm{COCH}_{2} \\mathrm{NO}_{2}$ | 5.1 |\n| <smiles>O=C1CCCC(=O)C1</smiles> | 5.3 |\n| $\\mathrm{O}_{2} \\mathrm{NCH}_{2} \\mathrm{CO}_{2} \\mathrm{CH}_{3}$ | 5.8 |"}
{"id": 1999, "contents": "Acidity Constants for Some Organic Compounds - \n| $\\mathrm{O}_{2} \\mathrm{NCH}_{2} \\mathrm{CO}_{2} \\mathrm{CH}_{3}$ | 5.8 |\n| <smiles>O=CC1CCCC1=O</smiles> | 5.8 |\n| <smiles>Oc1c(Cl)cc(Cl)cc1Cl</smiles> | 6.2 |"}
{"id": 2000, "contents": "Acidity Constants for Some Organic Compounds - \nCompound\n\nTABLE B1"}
{"id": 2001, "contents": "Acidity Constants for Some Organic Compounds - \n| Compound | $\\mathrm{p} K_{\\mathrm{a}}$ |\n| :---: | :---: |\n| <smiles>Oc1ccccc1</smiles> | 9.9 |\n| $\\mathrm{CH}_{3} \\mathrm{COCH}_{2} \\mathrm{SOCH}_{3}$ | 10.0 |\n| <smiles>Cc1ccccc1O</smiles> | 10.3 |\n| $\\mathrm{CH}_{3} \\mathrm{NO}_{2}$ | 10.3 |\n| $\\mathrm{CH}_{3} \\mathrm{SH}$ | 10.3 |\n| $\\mathrm{CH}_{3} \\mathrm{COCH}_{2} \\mathrm{CO}_{2} \\mathrm{CH}_{3}$ | 10.6 |\n| $\\mathrm{CH}_{3} \\mathrm{COCHO}$ | 11.0 |\n| $\\mathrm{CH}_{2}(\\mathrm{CN})_{2}$ | 11.2 |\n| $\\mathrm{CCl}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$ | 12.2 |\n| Glucose | 12.3 |\n| $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{C}=\\mathrm{NOH}$ | 12.4 |\n| $\\mathrm{CH}_{2}\\left(\\mathrm{CO}_{2} \\mathrm{CH}_{3}\\right)_{2}$ | 12.9 |\n| $\\mathrm{CHCl}_{2} \\mathrm{CH}_{2} \\mathrm{OH}$ | 12.9 |\n| $\\mathrm{CH}_{2}(\\mathrm{OH})_{2}$ | 13.3 |\n| $\\mathrm{HOCH}_{2} \\mathrm{CH}(\\mathrm{OH}) \\mathrm{CH}_{2} \\mathrm{OH}$ | 14.1 |\n| $\\mathrm{CH}_{2} \\mathrm{ClCH}_{2} \\mathrm{OH}$ | 14.3 |\n| <smiles>C1=CCC=C1</smiles> | 15.0 |\n| <smiles>OCc1ccccc1</smiles> | 15.4 |"}
{"id": 2002, "contents": "Acidity Constants for Some Organic Compounds - \n| <smiles>C1=CCC=C1</smiles> | 15.0 |\n| <smiles>OCc1ccccc1</smiles> | 15.4 |\n| $\\mathrm{CH}_{3} \\mathrm{OH}$ | 15.5 |\n| $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}_{2} \\mathrm{OH}$ | 15.5 |\n| $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$ | 16.0 |"}
{"id": 2003, "contents": "Acidity Constants for Some Organic Compounds - \n| Compound | $\\mathrm{p} K_{\\mathrm{a}}$ |\n| :---: | :---: |\n| $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}$ | 16.1 |\n| $\\mathrm{CH}_{3} \\mathrm{COCH}_{2} \\mathrm{Br}$ | 16.1 |\n| <smiles>O=C1CCCCC1</smiles> | 16.7 |\n| $\\mathrm{CH}_{3} \\mathrm{CHO}$ | 17 |\n| $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CHCHO}$ | 17 |\n| $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CHOH}$ | 17.1 |\n| $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{COH}$ | 18.0 |\n| $\\mathrm{CH}_{3} \\mathrm{COCH}_{3}$ | 19.3 |\n| <smiles>c1ccc2c(c1)Cc1ccccc1-2</smiles> | 23 |\n| $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$ | 25 |\n| $\\mathrm{HC} \\equiv \\mathrm{CH}$ | 25 |\n| $\\mathrm{CH}_{3} \\mathrm{CN}$ | 25 |\n| $\\mathrm{CH}_{3} \\mathrm{SO}_{2} \\mathrm{CH}_{3}$ | 28 |\n| $\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{3} \\mathrm{CH}$ | 32 |\n| $\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{2} \\mathrm{CH}_{2}$ | 34 |\n| $\\mathrm{CH}_{3} \\mathrm{SOCH}_{3}$ | 35 |\n| $\\mathrm{NH}_{3}$ | 36 |\n| $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{NH}_{2}$ | 36 |"}
{"id": 2004, "contents": "Acidity Constants for Some Organic Compounds - \n| $\\mathrm{NH}_{3}$ | 36 |\n| $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{NH}_{2}$ | 36 |\n| $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2}\\right)_{2} \\mathrm{NH}$ | 40 |\n| <smiles>Cc1ccccc1</smiles> | 41 |\n| <smiles>c1ccccc1</smiles> | 43 |"}
{"id": 2005, "contents": "Acidity Constants for Some Organic Compounds - \nTABLE B1\n\n| Compound | $\\mathrm{p} K_{\\mathrm{a}}$ |\n| :--- | :--- |\n| $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}_{2}$ | 44 |\n| $\\mathrm{CH}_{4}$ | $\\sim 60$ |\n\nAn acidity list covering more than 5000 organic compounds has been published: E.P. Serjeant and B. Dempsey (eds.), \"Ionization Constants of Organic Acids in Aqueous Solution,\" IUPAC Chemical Data Series No. 23, Pergamon Press, Oxford, 1979."}
{"id": 2006, "contents": "Glossary - \nAbsolute configuration (Section 5.5): The exact threedimensional structure of a chiral molecule. Absolute configurations are specified verbally by the Cahn-Ingold-Prelog $R, S$ convention.\nAbsorbance (A) (Section 14.7): In optical spectroscopy, the logarithm of the intensity of the incident light divided by the intensity of the light transmitted through a sample; $A=\\log I_{0} / I$.\nAbsorption spectrum (Section 12.5): A plot of wavelength of incident light versus amount of light absorbed. Organic molecules show absorption spectra in both the infrared and the ultraviolet regions of the electromagnetic spectrum.\nAcetals, $\\mathbf{R}_{\\mathbf{2}} \\mathbf{C}\\left(\\mathbf{O R}^{\\prime}\\right)_{\\mathbf{2}}$ (Section 19.10): A type of functional group consisting of two -OR groups bonded to the same carbon, $\\mathrm{R}_{2} \\mathrm{C}\\left(\\mathrm{OR}^{\\prime}\\right)_{2}$. Acetals are often used as protecting groups for ketones and aldehydes.\nAcetoacetic ester synthesis (Section 22.7): The synthesis of a methyl ketone by alkylation of an alkyl halide with ethyl acetoacetate, followed by hydrolysis and decarboxylation.\nAcetyl group (Section 19.1): The $\\mathrm{CH}_{3} \\mathrm{CO}$ - group.\nAcetylide anion (Section 9.7): The anion formed by removal of a proton from a terminal alkyne, $\\mathrm{R}-\\mathrm{C} \\equiv \\mathrm{C} \\mathrm{C}^{-}$.\nAchiral (Section 5.2): Having a lack of handedness. A molecule is achiral if it has a plane of symmetry and is thus superimposable on its mirror image.\nAcid anhydrides (Chapter 21 Introduction): A type of functional group with two acyl groups bonded to a common oxygen atom, $\\mathrm{RCO}_{2} \\mathrm{COR}^{\\prime}$."}
{"id": 2007, "contents": "Glossary - \nAcid anhydrides (Chapter 21 Introduction): A type of functional group with two acyl groups bonded to a common oxygen atom, $\\mathrm{RCO}_{2} \\mathrm{COR}^{\\prime}$.\nAcid halides (Chapter 21 Introduction): A type of functional group with an acyl group bonded to a halogen atom, RCOX.\nAcidity constant, $\\boldsymbol{K}_{\\mathbf{a}}$ (Section 2.8): A measure of acid strength. For any acid HA, the acidity constant is given by the expression $K_{\\mathrm{a}}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]}$.\nActivating groups (Section 16.4): Electron-donating groups such as hydroxyl $(-\\mathrm{OH})$ or amino $\\left(-\\mathrm{NH}_{2}\\right)$ that increase the reactivity of an aromatic ring toward electrophilic aromatic substitution."}
{"id": 2008, "contents": "Glossary - \nActivation energy (Section 6.9): The difference in energy between ground state and transition state in a reaction. The amount of activation energy determines the rate at which the reaction proceeds. Most organic reactions have activation energies of $40-100 \\mathrm{~kJ} / \\mathrm{mol}$.\nActive site (Section 6.11, Section 26.11): The pocket in an enzyme where a substrate is bound and undergoes reaction.\nAcyclic diene metathesis (ADMET) (Section 31.5): A method of polymer synthesis that uses the olefin metathesis reaction of\nan open-chain diene.\nAcyl group (Section 16.3, Section 19.1): A -COR group.\nAcyl phosphates (Chapter 21 Introduction): A type of functional group with an acyl group bonded to a phosphate, $\\mathrm{RCO}_{2} \\mathrm{PO}_{3}{ }^{2-}$.\nAcylation (Section 16.3, Section 24.7): The introduction of an acyl group, -COR, onto a molecule. For example, acylation of an alcohol yields an ester, acylation of an amine yields an amide, and acylation of an aromatic ring yields an alkyl aryl ketone.\nAcylium ion (Section 16.3): A resonance-stabilized carbocation in which the positive charge is located at a carbonyl-group carbon, $\\mathrm{R}-\\stackrel{+}{\\mathrm{C}}=\\mathrm{O} \\leftrightarrow \\mathrm{R}-\\mathrm{C} \\equiv \\mathrm{O}^{+}$. Acylium ions are intermediates in Friedel-Crafts acylation reactions.\nAdams' catalyst (Section 6.11): The $\\mathrm{PtO}_{2}$ catalyst used for alkene hydrogenations.\n1,2-Addition (Section 14.2, Section 19.13): Addition of a reactant to the two ends of a double bond."}
{"id": 2009, "contents": "Glossary - \n1,2-Addition (Section 14.2, Section 19.13): Addition of a reactant to the two ends of a double bond.\n1,4-Addition (Section 14.2, Section 19.13): Addition of a reactant to the ends of a conjugated $\\pi$ system. Conjugated dienes yield 1,4 adducts when treated with electrophiles such as HCl . Conjugated enones yield 1,4 adducts when treated with nucleophiles such as amines.\nAddition reactions (Section 6.1): Occur when two reactants add together to form a single product with no atoms left over.\nAdrenocortical hormones (Section 27.6): Steroid hormones secreted by the adrenal glands. There are two types of these hormones: mineralocorticoids and glucocorticoids.\nAlcohols (Section 3.1, Chapter 17 Introduction): A class of compounds with an -OH group bonded to a saturated, $s p^{3}$-hybridized carbon, ROH.\nAldaric acid (Section 25.6): The dicarboxylic acid resulting from oxidation of an aldose.\nAldehydes (RCHO) (Section 3.1, Chapter 19 Introduction): A class of compounds containing the -CHO functional group.\nAlditol (Section 25.6): The polyalcohol resulting from reduction of the carbonyl group of a sugar.\nAldol reaction (Section 23.1): The carbonyl condensation reaction of an aldehyde or ketone to give a $\\beta$-hydroxy carbonyl compound.\nAldonic acids (Section 25.6): Monocarboxylic acids resulting from oxidation of the - CHO group of an aldose.\nAldoses (Section 25.1): A type of carbohydrate with an aldehyde functional group.\nAlicyclic (Section 4.1): A nonaromatic cyclic hydrocarbon such as a cycloalkane or cycloalkene.\nAliphatic (Section 3.2): A nonaromatic hydrocarbon such as a simple alkane, alkene, or alkyne."}
{"id": 2010, "contents": "Glossary - \nAlkaloids (Chapter 2 Chemistry Matters): Naturally occurring organic bases, such as morphine.\nAlkanes (Section 3.2): A class of compounds of carbon and hydrogen that contains only single bonds.\nAlkene (Section 3.1, Chapter 7 Introduction): A hydrocarbon that contains a carbon-carbon double bond, $\\mathrm{R}_{2} \\mathrm{C}=\\mathrm{CR}_{2}$.\nAlkoxide ion, $\\mathbf{R O}^{-}$(Section 17.2): The anion formed by deprotonation of an alcohol.\nAlkoxymercuration (Section 18.2): A method for synthesizing ethers by mercuric-ion catalyzed addition of an alcohol to an alkene followed by demercuration on treatment with $\\mathrm{NaBH}_{4}$.\nAlkyl group (Section 3.3): The partial structure that remains when a hydrogen atom is removed from an alkane.\nAlkyl halide (Section 3.1, Chapter 10 Introduction): A compound with a halogen atom bonded to a saturated, $s p^{3}$-hybridized carbon atom.\nAlkylamines (Section 24.1): Amino-substituted alkanes $\\mathrm{RNH}_{2}$, $\\mathrm{R}_{2} \\mathrm{NH}$, or $\\mathrm{R}_{3} \\mathrm{~N}$.\nAlkylation (Section 9.8, Section 16.3, Section 18.2, Section 22.7): Introduction of an alkyl group onto a molecule. For example, aromatic rings can be alkylated to yield arenes, and enolate anions can be alkylated to yield $\\alpha$-substituted carbonyl compounds.\nAlkyne (Section 3.1, Chapter 9 Introduction): A hydrocarbon that contains a carbon-carbon triple bond, $\\mathrm{RC} \\equiv \\mathrm{CR}$.\nAllyl group (Section 7.3): $\\mathrm{A}_{2} \\mathrm{C}=\\mathrm{CHCH}_{2}$ - substituent."}
{"id": 2011, "contents": "Glossary - \nAllyl group (Section 7.3): $\\mathrm{A}_{2} \\mathrm{C}=\\mathrm{CHCH}_{2}$ - substituent.\nAllylic (Section 10.3): The position next to a double bond. For example, $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}_{2} \\mathrm{Br}$ is an allylic bromide.\n$\\boldsymbol{\\alpha}$-Amino acids (Section 26.1): A type of difunctional compound with an amino group on the carbon atom next to a carboxyl group, $\\mathrm{RCH}\\left(\\mathrm{NH}_{2}\\right) \\mathrm{CO}_{2} \\mathrm{H}$.\n$\\boldsymbol{\\alpha}$ Anomer (Section 25.5): The cyclic hemiacetal form of a sugar that has the hemiacetal -OH group cis to the -OH at the lowest chirality center in a Fischer projection.\n$\\boldsymbol{\\alpha}$ Helix (Section 26.9): The coiled secondary structure of a protein.\n$\\boldsymbol{\\alpha}$ Position (Chapter 22 Introduction): The position next to a carbonyl group.\n$\\boldsymbol{\\alpha}$-Substitution reaction (Chapter 22 Introduction): The substitution of the $\\alpha$ hydrogen atom of a carbonyl compound by reaction with an electrophile.\nAmides (Section 3.1, Chapter 21 Introduction): A class of compounds containing the $-\\mathrm{CONR}_{2}$ functional group.\nAmidomalonate synthesis (Section 26.3): A method for preparing $\\alpha$-amino acids by alkylation of diethyl amidomalonate with an alkyl halide followed by deprotection and decarboxylation.\nAmines (Section 3.1, Chapter 24 Introduction): A class of compounds containing one or more organic substituents bonded to a nitrogen atom, $\\mathrm{RNH}_{2}, \\mathrm{R}_{2} \\mathrm{NH}$, or $\\mathrm{R}_{3} \\mathrm{~N}$."}
{"id": 2012, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nAmino sugar (Section 25.7): A sugar with one of its -OH groups replaced by $-\\mathrm{NH}_{2}$."}
{"id": 2013, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nAmphiprotic (Section 26.1): Capable of acting either as an acid or as a base. Amino acids are amphiprotic.\nAmplitude (Section 12.5): The height of a wave measured from the midpoint to the maximum. The intensity of radiant energy is proportional to the square of the wave's amplitude.\nAnabolic steroids (Section 27.6): Synthetic androgens that mimic the tissue-building effects of natural testosterone.\nAnabolism (Section 29.1): The group of metabolic pathways that build up larger molecules from smaller ones.\nAndrogen (Section 27.6): A male steroid sex hormone.\nAngle strain (Section 4.3): The strain introduced into a molecule when a bond angle is deformed from its ideal value. Angle strain is particularly important in small-ring cycloalkanes, where it results from compression of bond angles to less than their ideal tetrahedral values.\nAnnulation (Section 23.12): The building of a new ring onto an existing molecule.\nAnomeric center (Section 25.5): The hemiacetal carbon atom in the cyclic pyranose or furanose form of a sugar.\nAnomers (Section 25.5): Cyclic stereoisomers of sugars that differ only in their configuration at the hemiacetal (anomeric) carbon.\nAntarafacial (Section 30.5): A pericyclic reaction that takes place on opposite faces of the two ends of a $\\pi$ electron system.\nAnti conformation (Section 3.7): The geometric arrangement around a carbon-carbon single bond in which the two largest substituents are $180^{\\circ}$ apart as viewed in a Newman projection.\nAnti periplanar (Section 11.8): Describing the stereochemical relationship in which two bonds on adjacent carbons lie in the same plane at an angle of $180^{\\circ}$.\nAnti stereochemistry (Section 8.2): The opposite of syn. An anti addition reaction is one in which the two ends of the double bond are attacked from different sides. An anti elimination reaction is one in which the two groups leave from opposite sides of the molecule."}
{"id": 2014, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nAntiaromatic (Section 15.3): Referring to a planar, conjugated molecule with $4 n \\pi$ electrons. Delocalization of the $\\pi$ electrons leads to an increase in energy.\nAntibonding MO (Section 1.11): A molecular orbital that is higher in energy than the atomic orbitals from which it is formed.\nAnticodon (Section 28.5): A sequence of three bases on tRNA that reads the codons on mRNA and brings the correct amino acids into position for protein synthesis.\nAntisense strand (Section 28.4): The template, noncoding strand of double-helical DNA that does not contain the gene.\nArene (Section 15.1): An alkyl-substituted benzene.\nArenediazonium salt (Section 24.8): An aromatic compound $\\mathrm{Ar}-\\stackrel{+}{\\mathrm{N}} \\equiv \\mathrm{N} \\quad \\mathrm{X}^{-}$; used in the Sandmeyer reaction.\nAromaticity (Chapter 15 Introduction): The special characteristics of cyclic conjugated molecules, including unusual stability and a tendency to undergo substitution reactions rather than addition reactions on treatment with\nelectrophiles. Aromatic molecules are planar, cyclic, conjugated species with $4 n+2 \\pi$ electrons.\nArylamines (Section 24.1): Amino-substituted aromatic compounds, $\\mathrm{ArNH}_{2}$.\nAtactic (Section 31.2): A chain-growth polymer in which the stereochemistry of the substituents is oriented randomly along the backbone.\nAtomic mass (Section 1.1): The weighted average mass of an element's naturally occurring isotopes.\nAtomic number, $\\boldsymbol{Z}$ (Section 1.1): The number of protons in the nucleus of an atom.\nATZ Derivative (Section 26.6): An anilinothiazolinone, formed from an amino acid during Edman degradation of a peptide.\nAufbau principle (Section 1.3): The rules for determining the electron configuration of an atom.\nAxial bonds (Section 4.6): Bonds or positions in chair cyclohexane that lie along the ring axis, perpendicular to the rough plane of the ring."}
{"id": 2015, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nAxial bonds (Section 4.6): Bonds or positions in chair cyclohexane that lie along the ring axis, perpendicular to the rough plane of the ring.\nAzide synthesis (Section 24.6): A method for preparing amines by $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of an alkyl halide with azide ion, followed by reduction.\nAzo compounds (Section 24.8): A class of compounds with the general structure $\\mathrm{R}-\\mathrm{N}=\\mathrm{N}-\\mathrm{R}^{\\prime}$.\nBackbone (Section 26.4): The continuous chain of atoms running the length of a protein or other polymer.\nBase peak (Section 12.1): The most intense peak in a mass spectrum.\nBasicity constant, $\\boldsymbol{K}_{\\mathbf{b}}$ (Section 24.3): A measure of base strength in water. For any base B , the basicity constant is given"}
{"id": 2016, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \n$$\n\\mathrm{B}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftarrows \\mathrm{BH}^{+}+\\mathrm{OH}^{-}\n$$"}
{"id": 2017, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nby the expression $K_{\\mathrm{b}}=\\frac{\\left[\\mathrm{BH}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]}{[\\mathrm{B}]}$\nBent bonds (Section 4.4): The bonds in small rings such as cyclopropane that bend away from the internuclear line and overlap at a slight angle, rather than head-on. Bent bonds are highly strained and highly reactive.\nBenzoyl (Section 19.1): The $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CO}$ - group.\nBenzyl (Section 15.1): The $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CH}_{2}$ - group.\nBenzylic (Section 11.5): The position next to an aromatic ring.\nBenzyne (Section 16.7): An unstable compound having a triple bond in a benzene ring.\n$\\boldsymbol{\\beta}$ Anomer (Section 25.5): The cyclic hemiacetal form of a sugar that has the hemiacetal -OH group trans to the -OH at the lowest chirality center in a Fischer projection.\n$\\boldsymbol{\\beta}$ Diketone (Section 22.5): A 1,3-diketone.\n$\\boldsymbol{\\beta}$-Keto ester (Section 22.5): A 3-oxoester.\n$\\boldsymbol{\\beta}$ Lactam (Chapter 21 Chemistry Matters): A four-membered lactam, or cyclic amide. Penicillin and cephalosporin antibiotics contain $\\beta$-lactam rings.\n$\\boldsymbol{\\beta}$-Oxidation pathway (Section 29.3): The metabolic pathway for degrading fatty acids.\n$\\boldsymbol{\\beta}$-Pleated sheet (Section 26.9): A type of secondary structure of a protein.\nBetaine (Section 19.11): A neutral dipolar molecule with\nnonadjacent positive and negative charges. For example, the adduct of a Wittig reagent with a carbonyl compound is a betaine.\nBicycloalkane (Section 4.9): A cycloalkane that contains two rings."}
{"id": 2018, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nBicycloalkane (Section 4.9): A cycloalkane that contains two rings.\nBimolecular reaction (Section 11.2): A reaction whose ratelimiting step occurs between two reactants.\nBlock copolymers (Section 31.3): Polymers in which different blocks of identical monomer units alternate with one another.\nBoat cyclohexane (Section 4.5): A conformation of cyclohexane that bears a slight resemblance to a boat. Boat cyclohexane has no angle strain but has a large number of eclipsing interactions that make it less stable than chair cyclohexane.\nBoc derivative (Section 26.7): A butyloxycarbonyl N-protected amino acid.\nBond angle (Section 1.6): The angle formed between two adjacent bonds.\nBond dissociation energy, $\\boldsymbol{D}$ (Section 6.8): The amount of energy needed to break a bond and produce two radical fragments.\nBond length (Section 1.5): The equilibrium distance between the nuclei of two atoms that are bonded to each other.\nBond strength (Section 1.5): An alternative name for bond dissociation energy.\nBonding MO (Section 1.11): A molecular orbital that is lower in energy than the atomic orbitals from which it is formed.\nBranched-chain alkanes (Section 3.2): Alkanes that contain a branching connection of carbons as opposed to straight-chain alkanes.\nBridgehead (Section 4.9): An atom that is shared by more than one ring in a polycyclic molecule.\nBromohydrin (Section 8.3): A 1,2-bromoalcohol; obtained by addition of HOBr to an alkene.\nBromonium ion (Section 8.2): A species with a divalent, positively charged bromine, $\\mathrm{R}_{2} \\mathrm{Br}^{+}$.\nBr\u00f8nsted-Lowry acid (Section 2.7): A substance that donates a hydrogen ion (proton; $\\mathrm{H}^{+}$) to a base."}
{"id": 2019, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nBr\u00f8nsted-Lowry acid (Section 2.7): A substance that donates a hydrogen ion (proton; $\\mathrm{H}^{+}$) to a base.\nBr\u00f8nsted-Lowry base (Section 2.7): A substance that accepts $\\mathrm{H}^{+}$from an acid.\nC-terminal amino acid (Section 26.4): The amino acid with a free $-\\mathrm{CO}_{2} \\mathrm{H}$ group at the end of a protein chain.\nCahn-Ingold-Prelog sequence rules (Section 5.5, Section 7.5): A series of rules for assigning relative rankings to substituent groups on a chirality center or a double-bond carbon atom.\nCannizzaro reaction (Section 19.12): The disproportionation reaction of an aldehyde on treatment with base to yield an alcohol and a carboxylic acid.\nCarbanion (Section 10.6, Section 19.7): A carbon anion, or substance that contains a trivalent, negatively charged carbon atom $\\left(\\mathrm{R}_{3} \\mathrm{C}:^{-}\\right)$. Alkyl carbanions are $s p^{3}$-hybridized and have eight electrons in the outer shell of the negatively charged carbon."}
{"id": 2020, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nCarbene, $\\mathbf{R}_{\\mathbf{2}} \\mathbf{C}$ (Section 8.9): A neutral substance that contains a divalent carbon atom having only six electrons in its outer shell ( $\\mathrm{R}_{2} \\mathrm{C}:$ ).\nCarbinolamine (Section 19.8): A molecule that contains the $\\mathrm{R}_{2} \\mathrm{C}(\\mathrm{OH}) \\mathrm{NH}_{2}$ functional group. Carbinolamines are produced as intermediates during the nucleophilic addition of amines to carbonyl compounds.\nCarbocation (Section 6.4, Section 7.9): A carbon cation, or substance that contains a trivalent, positively charged carbon atom having six electrons in its outer shell $\\left(\\mathrm{R}_{3} \\mathrm{C}^{+}\\right)$.\nCarbohydrates (Chapter 25 Introduction): Polyhydroxy aldehydes or ketones. Carbohydrates can be either simple sugars, such as glucose, or complex sugars, such as cellulose.\n\nCarbonyl condensation reactions (Section 23.1): A type of reaction that joins two carbonyl compounds together by a combination of $\\alpha$-substitution and nucleophilic addition reactions."}
{"id": 2021, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nCarbonyl group (Section 3.1, Preview of Carbonyl Chemistry): The $\\mathrm{C}=\\mathrm{O}$ functional group.\nCarboxyl group (Section 20.1): The $-\\mathrm{CO}_{2} \\mathrm{H}$ functional group.\nCarboxylation (Section 20.5): The addition of $\\mathrm{CO}_{2}$ to a molecule.\nCarboxylic acids, $\\quad \\mathbf{R C O}_{\\mathbf{2}} \\mathbf{H}$ (Section 3.1, Chapter 20 Introduction): Compounds containing the $-\\mathrm{CO}_{2} \\mathrm{H}$ functional group.\nCarboxylic acid derivative (Chapter 21 Introduction): A compound in which an acyl group is bonded to an electronegative atom or substituent that can act as a leaving group in a substitution reaction. Esters, amides, and acid halides are examples.\nCatabolism (Section 29.1): The group of metabolic pathways that break down larger molecules into smaller ones.\nCatalyst (Section 6.11): A substance that increases the rate of a chemical transformation by providing an alternative mechanism but is not itself changed in the reaction.\nCation radical (Section 12.1): A reactive species, typically formed in a mass spectrometer by loss of an electron from a neutral molecule and having both a positive charge and an odd number of electrons.\nChain-growth polymers (Section 8.10, Section 21.9, Section 31.1): Polymers whose bonds are produced by chain reaction mechanisms. Polyethylene and other alkene polymers are examples.\nChain reaction (Section 6.6): A reaction that, once initiated, sustains itself in an endlessly repeating cycle of propagation steps. The radical chlorination of alkanes is an example of a chain reaction that is initiated by irradiation with light and then continues in a series of propagation steps.\nChair conformation (Section 4.5): A three-dimensional conformation of cyclohexane that resembles the rough shape of a chair. The chair form of cyclohexane is the lowest-energy conformation of the molecule.\nChemical shift (Section 13.3): The position on the NMR chart"}
{"id": 2022, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nChemical shift (Section 13.3): The position on the NMR chart\nwhere a nucleus absorbs. By convention, the chemical shift of tetramethylsilane (TMS) is set at zero, and all other absorptions usually occur downfield (to the left on the chart). Chemical shifts are expressed in delta units ( $\\delta$ ), where $1 \\delta$ equals 1 ppm of the spectrometer operating frequency.\nChiral (Section 5.2): Having handedness. Chiral molecules are those that do not have a plane of symmetry and are therefore not superimposable on their mirror image. A chiral molecule thus exists in two forms, one right-handed and one left-handed. The most common cause of chirality in a molecule is the presence of a carbon atom that is bonded to four different substituents."}
{"id": 2023, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nChiral environment (Section 5.12): The chiral surroundings or conditions in which a molecule resides.\nChirality center (Section 5.2): An atom (usually carbon) that is bonded to four different groups.\nChlorohydrin (Section 8.3): A 1,2-chloroalcohol; obtained by addition of HOCl to an alkene.\nChromatography (Section 26.5): A technique for separating a mixture of compounds into pure components. Different compounds adsorb to a stationary support phase and are then carried along it at different rates by a mobile phase.\nCis-trans isomers (Section 4.2, Section 7.4): Stereoisomers that differ in their stereochemistry about a ring or double bond. Citric acid cycle (Section 29.7): The metabolic pathway by which acetyl CoA is degraded to $\\mathrm{CO}_{2}$.\nClaisen condensation reaction (Section 23.7): The carbonyl condensation reaction of two ester molecules to give a $\\beta$-keto ester product.\nClaisen rearrangement (Section 30.8): The pericyclic conversion of an allyl phenyl ether to an $o$-allylphenol or an allyl vinyl ether to a $\\gamma, \\delta$-unsaturated ketone by heating.\nCoding strand (Section 28.4): The sense strand of doublehelical DNA that contains the gene.\nCodon (Section 28.5): A three-base sequence on a messenger RNA chain that encodes the genetic information necessary to cause a specific amino acid to be incorporated into a protein. Codons on mRNA are read by complementary anticodons on tRNA.\nCoenzyme (Section 26.10): A small organic molecule that acts as a cofactor in a biological reaction.\nCofactor (Section 26.10): A small nonprotein part of an enzyme that is necessary for biological activity.\nCombinatorial chemistry (Chapter 16 Chemistry Matters): A procedure in which anywhere from a few dozen to several hundred thousand substances are prepared simultaneously.\nComplex carbohydrates (Section 25.1): Carbohydrates that are made of two or more simple sugars linked together by glycoside bonds."}
{"id": 2024, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nComplex carbohydrates (Section 25.1): Carbohydrates that are made of two or more simple sugars linked together by glycoside bonds.\nConcerted reaction (Chapter 30 Introduction): A reaction that takes place in a single step without intermediates. For example, the Diels-Alder cycloaddition reaction is a concerted process.\nCondensed structures (Section 1.12): A shorthand way of\nwriting structures in which carbon-hydrogen and carbon-carbon bonds are understood rather than shown explicitly. Propane, for example, has the condensed structure $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$.\nConfiguration (Section 5.5): The three-dimensional arrangement of atoms bonded to a chirality center.\nConformations (Section 3.6): The three-dimensional shape of a molecule at any given instant, assuming that rotation around single bonds is frozen.\nConformational analysis (Section 4.8): A means of assessing the energy of a substituted cycloalkane by totaling the steric interactions present in the molecule.\nConformers (Section 3.6): Conformational isomers.\nConjugate acid (Section 2.7): The product that results from protonation of a Br\u00f8nsted-Lowry base.\nConjugate addition (Section 19.13): Addition of a nucleophile to the $\\beta$ carbon atom of an $\\alpha, \\beta$-unsaturated carbonyl compound.\nConjugate base (Section 2.7): The product that results from deprotonation of a Br\u00f8nsted-Lowry acid.\nConjugation (Chapter 14 Introduction): A series of overlapping $p$ orbitals, usually in alternating single and multiple bonds. For example, 1,3-butadiene is a conjugated diene, 3 -buten-2-one is a conjugated enone, and benzene is a cyclic conjugated triene.\nConrotatory (Section 30.2): A term used to indicate that $p$ orbitals must rotate in the same direction during electrocyclic ring-opening or ring-closure."}
{"id": 2025, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nConrotatory (Section 30.2): A term used to indicate that $p$ orbitals must rotate in the same direction during electrocyclic ring-opening or ring-closure.\nConstitutional isomers (Section 3.2, Section 3.6, Section 5.9): Isomers that have their atoms connected in a different order. For example, butane and 2-methylpropane are constitutional isomers.\nCope rearrangement (Section 30.8): The sigmatropic rearrangement of a 1,5-hexadiene.\nCopolymers (Section 31.3): Polymers obtained when two or more different monomers are allowed to polymerize together.\nCoupled reactions (Section 29.1): Two reactions that share a common intermediate so that the energy released in the favorable step allows the unfavorable step to occur.\nCoupling constant, $\\boldsymbol{J}$ (Section 13.6): The magnitude (expressed in hertz) of the interaction between nuclei whose spins are coupled.\nCovalent bond (Section 1.4, Section 1.5): A bond formed by sharing electrons between atoms.\nCracking (Chapter 3 Chemistry Matters): A process used in petroleum refining in which large alkanes are thermally cracked into smaller fragments.\nCrown ethers (Section 18.6): Large-ring polyethers; used as phase-transfer catalysts.\nCrystallites (Section 31.7): Highly ordered crystal-like regions within a long polymer chain.\nCurtius rearrangement (Section 24.6): The conversion of an acid chloride into an amine by reaction with azide ion, followed by heating with water."}
{"id": 2026, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nCyanohydrins (Section 19.6): A class of compounds with an -OH group and a - CN group bonded to the same carbon atom; formed by addition of HCN to an aldehyde or ketone.\nCycloaddition reaction (Section 14.4, Section 30.5): A pericyclic reaction in which two reactants add together in a single step to yield a cyclic product. The Diels-Alder reaction between a diene and a dienophile to give a cyclohexene is an example.\nCycloalkane (Section 4.1): An alkane that contains a ring of carbons.\nD Sugars (Section 25.3): Sugars whose hydroxyl group at the chirality center farthest from the carbonyl group has the same configuration as D -glyceraldehyde and points to the right when drawn in Fischer projection.\n$\\boldsymbol{d}, \\boldsymbol{1}$ form (Section 5.8): The racemic mixture of a chiral compound.\nDeactivating groups (Section 16.4): Electron-withdrawing substituents that decrease the reactivity of an aromatic ring toward electrophilic aromatic substitution.\nDeamination (Section 29.9): The removal of an amino group from a molecule, as occurs with amino acids during metabolic degradation.\nDebyes (D) (Section 2.2): Units for measuring dipole moments; $1 \\mathrm{D}=3.336 \\times 10^{-30}$ coulomb meter ( $\\mathrm{C} \\cdot \\mathrm{m}$ ).\nDecarboxylation (Section 22.7): The loss of carbon dioxide from a molecule. $\\beta$-Keto acids decarboxylate readily on heating.\nDegenerate orbitals (Section 15.2): Two or more orbitals that have the same energy level.\nDegree of unsaturation (Section 7.2): The number of rings and/or multiple bonds in a molecule.\nDehydration (Section 8.1, Section 12.3, Section 17.6): The loss of water from an alcohol to yield an alkene."}
{"id": 2027, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nDehydration (Section 8.1, Section 12.3, Section 17.6): The loss of water from an alcohol to yield an alkene.\nDehydrohalogenation (Section 8.1, Section 9.2): The loss of HX from an alkyl halide. Alkyl halides undergo dehydrohalogenation to yield alkenes on treatment with strong base.\nDelocalization (Section 10.4, Section 15.2): A spreading out of electron density over a conjugated $\\pi$ electron system. For example, allylic cations and allylic anions are delocalized because their charges are spread out over the entire $\\pi$ electron system. Aromatic compounds have $4 n+2 \\pi$ electrons delocalized over their ring.\nDelta (\u05e1) scale (Section 13.3): An arbitrary scale used to calibrate NMR charts. One delta unit ( $\\delta$ ) is equal to 1 part per million (ppm) of the spectrometer operating frequency.\nDenatured (Section 26.9): The physical changes that occur in a protein when secondary and tertiary structures are disrupted.\nDeoxy sugar (Section 25.7): A sugar with one of its -OH groups replaced by an $-H$.\nDeoxyribonucleic acid (DNA) (Chapter 28 Introduction): The biopolymer consisting of deoxyribonucleotide units linked together through phosphate-sugar bonds. Found in the nucleus of cells, DNA contains an organism's genetic\ninformation.\nDEPT-NMR (Section 13.12): An NMR method for distinguishing among signals due to $\\mathrm{CH}_{3}, \\mathrm{CH}_{2}, \\mathrm{CH}$, and quaternary carbons. That is, the number of hydrogens attached to each carbon can be determined.\nDeshielding (Section 13.2): An effect observed in NMR that causes a nucleus to absorb toward the left (downfield) side of the chart. Deshielding is caused by a withdrawal of electron density from the nucleus."}
{"id": 2028, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nDess-Martin periodinane (Section 17.7): An iodine-based reagent commonly used for the laboratory oxidation of a primary alcohol to an aldehyde or a secondary alcohol to a ketone."}
{"id": 2029, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nDeuterium isotope effect (Section 11.8): A tool used in mechanistic investigations to establish whether a $\\mathrm{C}-\\mathrm{H}$ bond is broken in the rate-limiting step of a reaction.\nDextrorotatory (Section 5.3): A word used to describe an optically active substance that rotates the plane of polarization of plane-polarized light in a right-handed (clockwise) direction.\nDiastereomers (Section 5.6): Non-mirror-image stereoisomers; diastereomers have the same configuration at one or more chirality centers but differ at other chirality centers.\nDiastereotopic (Section 13.7): Hydrogens in a molecule whose replacement by some other group leads to different diastereomers.\n1,3-Diaxial interaction (Section 4.7): The strain energy caused by a steric interaction between axial groups three carbon atoms apart in chair cyclohexane.\nDiazonium salts (Section 24.8): A type of compound with the general structure $\\mathrm{RN}_{2}^{+} \\mathrm{X}^{-}$.\nDiazotization (Section 24.8): The conversion of a primary amine, $\\mathrm{RNH}_{2}$, into a diazonium ion, $\\mathrm{RN}_{2}{ }^{+}$, by treatment with nitrous acid.\nDieckmann cyclization reaction (Section 23.9): An intramolecular Claisen condensation reaction of a diester to give a cyclic $\\beta$-keto ester.\nDiels-Alder reaction (Section 14.4, Section 30.5): The cycloaddition reaction of a diene with a dienophile to yield a cyclohexene.\nDienophile (Section 14.5): A compound containing a double bond that can take part in the Diels-Alder cycloaddition reaction. The most reactive dienophiles are those that have electron-withdrawing groups on the double bond.\nDigestion (Section 29.1): The first stage of catabolism, in which food is broken down by hydrolysis of ester, glycoside (acetal), and peptide (amide) bonds to yield fatty acids, simple sugars, and amino acids."}
{"id": 2030, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nDihedral angle (Section 3.6): The angle between two bonds on adjacent carbons as viewed along the $\\mathrm{C}-\\mathrm{C}$ bond.\nDipole moment, $\\boldsymbol{\\mu}$ (Section 2.2): A measure of the net polarity of a molecule. A dipole moment arises when the centers of mass of positive and negative charges within a molecule do not\ncoincide\nDipole-dipole forces (Section 2.12): Noncovalent electrostatic interactions between dipolar molecules.\nDisaccharide (Section 25.8): A carbohydrate formed by linking two simple sugars through an acetal bond.\nDispersion forces (Section 2.12): Noncovalent interactions between molecules that arise because of constantly changing electron distributions within the molecules.\nDisrotatory (Section 30.2): A term used to indicate that $p$ orbitals rotate in opposite directions during electrocyclic ringopening or ring-closing reactions.\nDisulfides (RSSR') (Section 18.7): A class of compounds of the general structure RSSR'.\nDeoxyribonucleic acid (DNA) (Chapter 28 Introduction): Chemical carriers of a cell's genetic information.\nDouble bond (Section 1.8): A covalent bond formed by sharing two electron pairs between atoms.\nDouble helix (Section 28.2): The structure of DNA in which two polynucleotide strands coil around each other.\nDoublet (Section 13.6): A two-line NMR absorption caused by spin-spin splitting when the spin of the nucleus under observation couples with the spin of a neighboring magnetic nucleus.\nDownfield (Section 13.3): Referring to the left-hand portion of the NMR chart.\n$\\boldsymbol{E}$ geometry (Section 7.5): A term used to describe the stereochemistry of a carbon-carbon double bond. The two groups on each carbon are ranked according to the Cahn-Ingold-Prelog sequence rules, and the two carbons are compared. If the higher-ranked groups on each carbon are on opposite sides of the double bond, the bond has E geometry."}
{"id": 2031, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nE1 reaction (Section 11.10): A unimolecular elimination reaction in which the substrate spontaneously dissociates to give a carbocation intermediate, which loses a proton in a separate step.\nE1cB reaction (Section 11.10): A unimolecular elimination reaction in which a proton is first removed to give a carbanion intermediate, which then expels the leaving group in a separate step.\nE2 reaction (Section 11.8): A bimolecular elimination reaction in which $\\mathrm{C}-\\mathrm{H}$ and $\\mathrm{C}-\\mathrm{X}$ bond cleavages are simultaneous.\nEclipsed conformation (Section 3.6): The geometric arrangement around a carbon-carbon single bond in which the bonds to substituents on one carbon are parallel to the bonds to substituents on the neighboring carbon as viewed in a Newman projection.\nEclipsing strain (Section 3.6): The strain energy in a molecule caused by electron repulsions between eclipsed bonds. Eclipsing strain is also called torsional strain.\nEdman degradation (Section 26.6): A method for $N$-terminal sequencing of peptide chains by treatment with $N$-phenylisothiocyanate.\nEicosanoid (Section 27.4): A lipid derived biologically from"}
{"id": 2032, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \n5,8,11,14-eicosatetraenoic acid, or arachidonic acid. Prostaglandins, thromboxanes, and leukotrienes are examples. Elastomer (Section 31.7): An amorphous polymer that has the ability to stretch out and spring back to its original shape.\nElectrocyclic reaction (Section 30.2): A unimolecular pericyclic reaction in which a ring is formed or broken by a concerted reorganization of electrons through a cyclic transition state. For example, the cyclization of 1,3,5-hexatriene to yield 1,3-cyclohexadiene is an electrocyclic reaction.\nElectromagnetic spectrum (Section 12.5): The range of electromagnetic energy, including infrared, ultraviolet, and visible radiation.\nElectron configuration (Section 1.3): A list of the orbitals occupied by electrons in an atom.\nElectron-dot structure (Section 1.4): A representation of a molecule showing valence electrons as dots.\nElectron shells (Section 1.2): A group of an atom's electrons with the same principal quantum number.\nElectron-transport chain (Section 29.1): The final stage of catabolism in which ATP is produced.\nElectronegativity (EN) (Section 2.1): The ability of an atom to attract electrons in a covalent bond. Electronegativity increases across the periodic table from left to right and from bottom to top.\nElectrophile (Section 6.3, Section 6.4): An \"electron-lover,\" or substance that accepts an electron pair from a nucleophile in a polar bond-forming reaction.\nElectrophilic addition reactions (Section 7.7): Addition of an electrophile to a carbon-carbon double bond to yield a saturated product.\nElectrophilic aromatic substitution reaction (Chapter 16 Introduction): A reaction in which an electrophile ( $\\mathrm{E}^{+}$) reacts with an aromatic ring and substitutes for one of the ring hydrogens."}
{"id": 2033, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nElectrophilic aromatic substitution reaction (Chapter 16 Introduction): A reaction in which an electrophile ( $\\mathrm{E}^{+}$) reacts with an aromatic ring and substitutes for one of the ring hydrogens.\nElectrophoresis (Section 26.2, Section 28.6): A technique used for separating charged organic molecules, particularly proteins and DNA fragments. The mixture to be separated is placed on a buffered gel or paper, and an electric potential is applied across the ends of the apparatus. Negatively charged molecules migrate toward the positive electrode, and positively charged molecules migrate toward the negative electrode.\nElectrostatic potential maps (Section 2.1): Molecular representations that use color to indicate the charge distribution in molecules as derived from quantummechanical calculations.\nElimination reactions (Section 6.1): What occurs when a single reactant splits into two products.\nElution (Section 26.5): The passage of a substance from a chromatography column.\nEmbden-Meyerhof pathway (Section 29.5): An alternative name for glycolysis.\nEnamines (Section 19.8): Compounds with the $\\mathrm{R}_{2} \\mathrm{~N}-\\mathrm{CR}=\\mathrm{CR}_{2}$\nfunctional group.\nEnantiomers (Section 5.1): Stereoisomers of a chiral substance that have a mirror-image relationship. Enantiomers have opposite configurations at all chirality centers.\nEnantioselective synthesis (Chapter 19 Chemistry Matters): A reaction method that yields only a single enantiomer of a chiral product starting from an achiral reactant.\nEnantiotopic (Section 13.7): Hydrogens in a molecule whose replacement by some other group leads to different enantiomers.\n$\\mathbf{3}^{\\prime}$ End (Section 28.1): The end of a nucleic acid chain with a free hydroxyl group at C3'.\n$5^{\\prime}$ End (Section 28.1): The end of a nucleic acid chain with a free hydroxyl group at C5'.\nEndergonic (Section 6.7): A reaction that has a positive freeenergy change and is therefore nonspontaneous. In an energy diagram, the product of an endergonic reaction has a higher energy level than the reactants."}
{"id": 2034, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nEndo (Section 14.5): A term indicating the stereochemistry of a substituent in a bridged bicycloalkane. An endo substituent is syn to the larger of the two bridges.\nEndothermic (Section 6.7): A reaction that absorbs heat and therefore has a positive enthalpy change.\nEnergy diagram (Section 6.9): A representation of the course of a reaction, in which free energy is plotted as a function of reaction progress. Reactants, transition states, intermediates, and products are represented, and their appropriate energy levels are indicated.\nEnol (Section 9.4, Section 22.1): A vinylic alcohol that is in equilibrium with a carbonyl compound, $\\mathrm{C}=\\mathrm{C}-\\mathrm{OH}$.\nEnolate ion (Section 22.1): The anion of an enol, $\\mathrm{C}=\\mathrm{C}-\\mathrm{O}^{-}$.\nEnthalpy change ( $\\boldsymbol{\\Delta H} \\boldsymbol{H}$ ) (Section 6.7): The heat of reaction. The enthalpy change that occurs during a reaction is a measure of the difference in total bond energy between reactants and products.\nEntropy change ( $\\boldsymbol{\\Delta} \\boldsymbol{S}$ ) (Section 6.7): The change in amount of molecular randomness. The entropy change that occurs during a reaction is a measure of the difference in randomness between reactants and products.\nEnzyme (Section 6.11, Section 26.10): A biological catalyst. Enzymes are large proteins that catalyze specific biochemical reactions.\nEpimers (Section 5.6): Diastereomers that differ in configuration at only one chirality center but are the same at all others.\nEpoxide (Section 8.7): A three-membered-ring ether functional group.\nEquatorial bonds (Section 4.6): Bonds or positions in chair cyclohexane that lie along the rough equator of the ring.\nESI (Section 12.4): Electrospray ionization; a \"soft\" ionization method used for mass spectrometry of biological samples of very high molecular weight.\nEssential amino acid (Section 26.1): One of nine amino acids"}
{"id": 2035, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nEssential amino acid (Section 26.1): One of nine amino acids\nthat are biosynthesized only in plants and microorganisms and must be obtained by humans in the diet.\nEssential monosaccharide (Section 25.7): One of eight simple sugars that is best obtained in the diet rather than by biosynthesis.\nEssential oil (Chapter 8 Chemistry Matters): The volatile oil obtained by steam distillation of a plant extract."}
{"id": 2036, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nEsters (Section 3.1, Chapter 21 Introduction): A class of compounds containing the $-\\mathrm{CO}_{2} \\mathrm{R}$ functional group.\n\nEstrogens (Section 27.6): Female steroid sex hormones.\nEthers (Section 3.1, Chapter 18 Introduction): A class of compounds that has two organic substituents bonded to the same oxygen atom, ROR'\n\nExergonic (Section 6.7): A reaction that has a negative freeenergy change and is therefore spontaneous. On an energy diagram, the product of an exergonic reaction has a lower energy level than that of the reactants.\nExo (Section 14.5): A term indicating the stereochemistry of a substituent in a bridged bicycloalkane. An exo substituent is anti to the larger of the two bridges.\nExon (Section 28.4): A section of DNA that contains genetic information."}
{"id": 2037, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nExothermic (Section 6.7): A reaction that releases heat and therefore has a negative enthalpy change.\nFats (Section 27.1): Solid triacylglycerols derived from an animal source.\nFatty acids (Section 27.1): A long, straight-chain carboxylic acid found in fats and oils.\nFiber (Section 31.7): A thin thread produced by extruding a molten polymer through small holes in a die.\nFibrous proteins (Section 26.9): A type of protein that consists of polypeptide chains arranged side by side in long threads. Such proteins are tough, insoluble in water, and used in nature for structural materials such as hair, hooves, and fingernails.\nFingerprint region (Section 12.7): The complex region of the infrared spectrum from $1500-400 \\mathrm{~cm}^{-1}$.\nFirst-order reaction (Section 11.4): Designates a reaction whose rate-limiting step is unimolecular and whose kinetics therefore depend on the concentration of only one reactant.\nFischer esterification reaction (Section 21.3): The acidcatalyzed nucleophilic acyl substitution reaction of a carboxylic acid with an alcohol to yield an ester.\nFischer projections (Section 25.2): A means of depicting the absolute configuration of a chiral molecule on a flat page. A Fischer projection uses a cross to represent the chirality center. The horizontal arms of the cross represent bonds coming out of the plane of the page, and the vertical arms of the cross represent bonds going back into the plane of the page.\nFmoc derivative (Section 26.7): A fluorenylmethyloxycarbonyl N-protected amino acid.\nFormal charges (Section 2.3): The difference in the number of electrons owned by an atom in a molecule and by the same atom in its elemental state."}
{"id": 2038, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nFormyl (Section 19.1): A -CHO group.\nFrequency, $\\boldsymbol{\\nu}$ (Section 12.5): The number of electromagnetic wave cycles that travel past a fixed point in a given unit of time. Frequencies are expressed in units of cycles per second, or hertz.\nFriedel-Crafts reaction (Section 16.3): An electrophilic aromatic substitution reaction to alkylate or acylate an aromatic ring.\nFrontier orbitals (Section 30.1): The highest occupied (HOMO) and lowest unoccupied (LUMO) molecular orbitals.\nFT-NMR (Section 13.10): Fourier-transform NMR; a rapid technique for recording NMR spectra in which all magnetic nuclei absorb at the same time.\nFunctional (Section 3.1): An atom or group of atoms that is part of a larger molecule and has a characteristic chemical reactivity.\nFunctional RNAs (Section 28.4): An alternative name for small RNAs."}
{"id": 2039, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nFuranose (Section 25.5): The five-membered-ring form of a simple sugar.\nGabriel amine synthesis (Section 24.6): A method for preparing an amine by $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of an alkyl halide with potassium phthalimide, followed by hydrolysis.\nGauche conformation (Section 3.7): The conformation of butane in which the two methyl groups lie $60^{\\circ}$ apart as viewed in a Newman projection. This conformation has $3.8 \\mathrm{~kJ} / \\mathrm{mol}$ steric strain.\nGeminal (Section 19.5): Referring to two groups attached to the same carbon atom. For example, the hydrate formed by nucleophilic addition of water to an aldehyde or ketone is a geminal diol.\nGibbs free-energy change ( $\\boldsymbol{\\Delta} \\boldsymbol{G}$ ) (Section 6.7): The free-energy change that occurs during a reaction, given by the equation $\\Delta G=\\Delta H-T \\Delta S$. A reaction with a negative free-energy change is spontaneous, and a reaction with a positive freeenergy change is nonspontaneous.\nGilman reagent ( $\\mathbf{L i R}_{\\mathbf{2}} \\mathbf{C u}$ ) (Section 10.7): A diorganocopper reagent.\nGlass transition temperature, $\\boldsymbol{T}_{\\mathbf{g}}$ (Section 31.7): The temperature at which a hard, amorphous polymer becomes soft and flexible.\nGlobular proteins (Section 26.9): A type of protein that is coiled into a compact, nearly spherical shape. Globular proteins, which are generally water-soluble and mobile within the cell, are the structural class to which enzymes belong.\nGluconeogenesis (Section 29.8): The anabolic pathway by which organisms make glucose from simple three-carbon precursors.\nGlycal (Section 25.9): An unsaturated sugar with a C1-C2 double bond.\nGlycal assembly method (Section 25.9): A method for linking monosaccharides together to synthesize polysaccharides."}
{"id": 2040, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nGlycal assembly method (Section 25.9): A method for linking monosaccharides together to synthesize polysaccharides.\nGlycerophospholipids (Section 27.3): Lipids that contain a\nglycerol backbone linked to two fatty acids and a phosphoric acid.\nGlycoconjugate (Section 25.6): A molecule in which a carbohydrate is linked through its anomeric center to another biological molecule such as a lipid or protein.\nGlycol (Section 8.7): A diol, such as ethylene glycol, $\\mathrm{HOCH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}$.\nGlycolipid (Section 25.6): A biological molecule in which a carbohydrate is linked through a glycoside bond to a lipid.\nGlycolysis (Section 29.5): A series of ten enzyme-catalyzed reactions that break down glucose into 2 equivalents of pyruvate, $\\mathrm{CH}_{3} \\mathrm{COCO}_{2}{ }^{-}$.\nGlycoprotein (Section 25.6): A biological molecule in which a carbohydrate is linked through a glycoside bond to a protein.\nGlycoside (Section 25.6): A cyclic acetal formed by reaction of a sugar with another alcohol.\nGraft copolymers (Section 31.3): Copolymers in which homopolymer branches of one monomer unit are \"grafted\" onto a homopolymer chain of another monomer unit.\nGreen chemistry (Chapter 11 Chemistry Matters, Chapter 24 Chemistry Matters): The design and implementation of chemical products and processes that reduce waste and minimize or eliminate the generation of hazardous substances.\nGrignard reagent (RMgX) (Section 10.6): An organomagnesium halide.\nGround-state electron configuration (Section 1.3): The most stable, lowest-energy electron configuration of a molecule or atom.\nHaloform reaction (Section 22.6): The reaction of a methyl ketone with halogen and base to yield a haloform $\\left(\\mathrm{CHX}_{3}\\right)$ and a carboxylic acid."}
{"id": 2041, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nHaloform reaction (Section 22.6): The reaction of a methyl ketone with halogen and base to yield a haloform $\\left(\\mathrm{CHX}_{3}\\right)$ and a carboxylic acid.\nHalogenation (Section 8.2, Section 16.1): The reaction of halogen with an alkene to yield a 1,2-dihalide addition product or with an aromatic compound to yield a substitution product.\nHalohydrin (Section 8.3): A 1,2-haloalcohol, such as that obtained on addition of HOBr to an alkene.\nHalonium ion (Section 8.2): A species containing a positively charged, divalent halogen. Three-membered-ring bromonium ions are intermediates in the electrophilic addition of $\\mathrm{Br}_{2}$ to alkenes.\nHammond postulate (Section 7.10): A postulate stating that we can get a picture of what a given transition state looks like by looking at the structure of the nearest stable species. Exergonic reactions have transition states that resemble reactant; endergonic reactions have transition states that resemble product.\nHeat of combustion (Section 4.3): The amount of heat released when a compound burns completely in oxygen.\nHeat of hydrogenation (Section 7.6): The amount of heat released when a carbon-carbon double bond is hydrogenated.\nHeat of reaction (Section 6.7): An alternative name for the enthalpy change in a reaction, $\\Delta H$.\nHell-Volhard-Zelinskii (HVZ) reaction (Section 22.4): The\nreaction of a carboxylic acid with $\\mathrm{Br}_{2}$ and phosphorus to give an $\\alpha$-bromo carboxylic acid.\nHemiacetal (Section 19.10): A functional group having one -OR and one - OH group bonded to the same carbon.\nHenderson-Hasselbalch equation (Section 20.3, Section 24.5, Section 26.2): An equation for determining the extent of dissociation of a weak acid at various pH values.\nHertz, Hz (Section 12.5): A unit of measure of electromagnetic frequency, the number of waves that pass by a fixed point per second."}
{"id": 2042, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nHertz, Hz (Section 12.5): A unit of measure of electromagnetic frequency, the number of waves that pass by a fixed point per second.\nHeterocycle (Section 15.5, Section 24.9): A cyclic molecule whose ring contains more than one kind of atom. For example, pyridine is a heterocycle that contains five carbon atoms and one nitrogen atom in its ring.\nHeterolytic bond breakage (Section 6.2): The kind of bondbreaking that occurs in polar reactions when one fragment leaves with both of the bonding electrons: $\\mathrm{A}: \\mathrm{B} \\rightarrow \\mathrm{A}^{+}+\\mathrm{B}:^{-}$.\nHofmann elimination reaction (Section 24.7): The elimination reaction of an amine to yield an alkene by reaction with iodomethane followed by heating with $\\mathrm{Ag}_{2} \\mathrm{O}$.\nHofmann rearrangement (Section 24.6): The conversion of an amide into an amine by reaction with $\\mathrm{Br}_{2}$ and base.\nHighest occupied molecular orbital (HOMO) (Section 14.7, Section 30.1): The symmetries of the HOMO and LUMO are important in pericyclic reactions.\nHomolytic bond breakage (Section 6.2): The kind of bondbreaking that occurs in radical reactions when each fragment leaves with one bonding electron: $\\mathrm{A}: \\mathrm{B} \\rightarrow \\mathrm{A}^{+}+\\mathrm{B}:^{-}$.\nHomopolymers (Section 31.3): A polymer made up of identical repeating units.\nHomotopic (Section 13.7): Hydrogens in a molecule that give the identical structure on replacement by X and thus show identical NMR absorptions.\nHormones (Section 27.6): Chemical messengers that are secreted by an endocrine gland and carried through the bloodstream to a target tissue.\nHPLC (Section 26.5): High-pressure liquid chromatography; a variant of column chromatography using high pressure to force solvent through very small absorbent particles."}
{"id": 2043, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nHPLC (Section 26.5): High-pressure liquid chromatography; a variant of column chromatography using high pressure to force solvent through very small absorbent particles.\nH\u00fcckel $4 \\boldsymbol{n}+2$ rule (Section 15.3): A rule stating that monocyclic conjugated molecules having $4 n+2 \\pi$ electrons ( $n$ = an integer) are aromatic.\nHund's rule (Section 1.3): If two or more empty orbitals of equal energy are available, one electron occupies each, with their spins parallel, until all are half-full.\nHybrid orbital (Section 1.6): An orbital derived from a combination of atomic orbitals. Hybrid orbitals, such as the $s p^{3}, s p^{2}$, and $s p$ hybrids of carbon, are strongly directed and form stronger bonds than atomic orbitals do.\nHydration (Section 8.4): Addition of water to a molecule, such as occurs when alkenes are treated with aqueous sulfuric acid to give alcohols.\nHydride shift (Section 7.11): The shift of a hydrogen atom and\nits electron pair to a nearby cationic center.\nHydroboration (Section 8.5): Addition of borane $\\left(\\mathrm{BH}_{3}\\right)$ or an alkylborane to an alkene. The resultant trialkylborane products can be oxidized to yield alcohols."}
{"id": 2044, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nHydrocarbons (Section 3.2): A class of compounds that contain only carbon and hydrogen.\n\nHydrogen bond (Section 2.12): A weak attraction between a hydrogen atom bonded to an electronegative atom and an electron lone pair on another electronegative atom.\n\nHydrogenated (Section 6.11): Addition of hydrogen to a double or triple bond to yield a saturated product.\n\nHydrogenolysis (Section 26.7): Cleavage of a bond by reaction with hydrogen. Benzylic ethers and esters, for instance, are cleaved by hydrogenolysis\n\nHydrophilic (Section 2.12): Water-loving; attracted to water.\nHydrophobic (Section 2.12): Water-fearing; repelled by water.\nHydroquinones (Section 17.10): 1,4-dihydroxybenzene.\nHydroxylation (Section 8.7): Addition of two -OH groups to a double bond."}
{"id": 2045, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nHyperconjugation (Section 7.6, Section 7.9): An electronic interaction that results from overlap of a vacant $p$ orbital on one atom with a neighboring $\\mathrm{C}-\\mathrm{H} \\sigma$ bond. Hyperconjugation is important in stabilizing carbocations and substituted alkenes.\nImide (Section 24.6): A compound with the -CONHCOfunctional group.\nImines (Section 19.8): A class of compounds with the $\\mathrm{R}_{2} \\mathrm{C}=\\mathrm{NR}$ functional group.\nInductive effect (Section 2.1, Section 7.9, Section 16.4): The electron-attracting or electron-withdrawing effect transmitted through $\\sigma$ bonds. Electronegative elements have an electronwithdrawing inductive effect.\nInfrared (IR) spectroscopy (Section 12.6): A kind of optical spectroscopy that uses infrared energy. IR spectroscopy is particularly useful in organic chemistry for determining the kinds of functional groups present in molecules.\nInitiator (Section 8.10, Section 31.1): A substance that is used to initiate a radical chain reaction or polymerization. For example, radical chlorination of alkanes is initiated when light energy breaks the weak $\\mathrm{Cl}-\\mathrm{Cl}$ bond to form Cl - radicals.\nIntegrating (Section 13.5): A technique for measuring the area under an NMR peak to determine the relative number of each kind of proton in a molecule.\nIntermediate (Section 6.10): A species that is formed during the course of a multistep reaction but is not the final product. Intermediates are more stable than transition states but may or may not be stable enough to isolate.\nIntramolecular, intermolecular (Section 23.6): A reaction that occurs within the same molecule is intramolecular; a reaction that occurs between two molecules is intermolecular.\nIntron (Section 28.4): A section of DNA that does not contain genetic information.\nIon pairs (Section 11.4): A loose association between two ions in solution. Ion pairs are implicated as intermediates in $\\mathrm{S}_{\\mathrm{N}} 1$\nreactions to account for the partial retention of stereochemistry that is often observed."}
{"id": 2046, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nreactions to account for the partial retention of stereochemistry that is often observed.\nIonic bond (Section 1.4): The electrostatic attraction between ions of unlike charge.\nIsoelectric point (pI) (Section 26.2): The pH at which the number of positive charges and the number of negative charges on a protein or an amino acid are equal."}
{"id": 2047, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nIsomers (Section 3.2, Section 5.9): Compounds that have the same molecular formula but different structures.\n\nIsoprene rule (Chapter 8 Chemistry Matters): An observation to the effect that terpenoids appear to be made up of isoprene (2-methyl-1,3-butadiene) units connected head-to-tail.\n\nIsotactic (Section 31.2): A chain-growth polymer in which the stereochemistry of the substituents is oriented regularly along the backbone.\n\nIsotopes (Section 1.1): Atoms of the same element that have different mass numbers.\n\nIUPAC system of nomenclature (Section 3.4): Rules for naming compounds, devised by the International Union of Pure and Applied Chemistry.\nKekul\u00e9 structure (Section 1.4): An alternative name for a linebond structure, which represents a molecule by showing covalent bonds as lines between atoms."}
{"id": 2048, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nKetals (Section 19.10): An alternative name for acetals derived from a ketone rather than an aldehyde and consisting of two -OR groups bonded to the same carbon, $\\mathrm{R}_{2} \\mathrm{C}\\left(\\mathrm{OR}^{\\prime}\\right)_{2}$. Ketals are often used as protecting groups for ketones.\nKeto-enol tautomerism (Section 9.4, Section 22.1): The equilibration between a carbonyl form and vinylic alcohol form of a molecule.\nKetones ( $\\mathbf{R}_{\\mathbf{2}} \\mathbf{C O}$ ) (Section 3.1, Chapter 19 Introduction): A class of compounds with two organic substituents bonded to a carbonyl group, $\\mathrm{R}_{2} \\mathrm{C}=\\mathrm{O}$.\nKetoses (Section 25.1): Carbohydrates with a ketone functional group.\nKiliani-Fischer synthesis (Section 25.6): A method for lengthening the chain of an aldose sugar.\nKinetic control (Section 14.3): A reaction that follows the lowest activation energy pathway is said to be kinetically controlled. The product is the most rapidly formed but is not necessarily the most stable.\nKinetics (Section 11.2): Referring to reaction rates. Kinetic measurements are useful for helping to determine reaction mechanisms.\nKoenigs-Knorr reaction (Section 25.6): A method for the synthesis of glycosides by reaction of an alcohol with a pyranosyl bromide.\nKrebs cycle (Section 29.7): An alternative name for the citric acid cycle, by which acetyl CoA is degraded to $\\mathrm{CO}_{2}$.\nL Sugar (Section 25.3): A sugar whose hydroxyl group at the chirality center farthest from the carbonyl group points to the left when drawn in Fischer projection.\n\nLactams (Section 21.7): Cyclic amides."}
{"id": 2049, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nLactones (Section 21.6): Cyclic esters.\nLagging strand (Section 28.3): The complement of the original $3^{\\prime} \\rightarrow 5^{\\prime}$ DNA strand that is synthesized discontinuously in small pieces that are subsequently linked by DNA ligases.\nLD50 (Chapter 1 Chemistry Matters): The amount of a substance per kilogram body weight that is lethal to $50 \\%$ of test animals.\nLDA (Section 22.5): Lithium diisopropylamide, $\\operatorname{LiN}\\left(i-\\mathrm{C}_{3} \\mathrm{H}_{7}\\right)_{2}$, a strong base commonly used to convert carbonyl compounds into their enolate ions.\nLeading strand (Section 28.3): The complement of the original $5^{\\prime} \\rightarrow 3^{\\prime}$ DNA strand that is synthesized continuously in a single piece.\nLeaving group (Section 11.3): The group that is replaced in a substitution reaction.\nLevorotatory (Section 5.3): An optically active substance that rotates the plane of polarization of plane-polarized light in a left-handed (counterclockwise) direction.\nLewis acid (Section 2.11): A substance with a vacant lowenergy orbital that can accept an electron pair from a base. All electrophiles are Lewis acids.\nLewis base (Section 2.11): A substance that donates an electron lone pair to an acid. All nucleophiles are Lewis bases.\nLewis structures (Section 1.4): Representations of molecules showing valence electrons as dots.\nLindlar catalyst (Section 9.5): A hydrogenation catalyst used to convert alkynes to cis alkenes.\nLine-bond structure (Section 1.4): An alternative name for a Kekul\u00e9 structure, which represents a molecule by showing covalent bonds as lines between atoms.\n$\\mathbf{1 \\rightarrow 4}$ Link (Section 25.8): A glycoside link between the $\\mathrm{C} 1-\\mathrm{OH}$ group of one sugar and the $\\mathrm{C} 4-\\mathrm{OH}$ group of another sugar."}
{"id": 2050, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nLipids (Chapter 27 Introduction): Naturally occurring substances isolated from cells and tissues by extraction with a nonpolar solvent. Lipids belong to many different structural classes, including fats, terpenoids, prostaglandins, and steroids.\nLipid bilayer (Section 27.3): The ordered lipid structure that forms a cell membrane.\nLipoprotein (Chapter 27 Chemistry Matters): A complex molecule with both lipid and protein parts that transports lipids through the body.\nLocant (Section 3.4): A number in a chemical name that locates the positions of the functional groups and substituents in the molecule.\nLone-pair electrons (Section 1.4): Nonbonding valence-shell electron pairs. Lone-pair electrons are used by nucleophiles in their reactions with electrophiles.\nLowest unoccupied molecular orbital (LUMO) (Section 14.7, Section 30.1): The symmetries of the LUMO and the HOMO are important in determining the stereochemistry of pericyclic reactions.\nMagnetic resonance imaging, MRI (Chapter 13 Chemistry"}
{"id": 2051, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nMatters): A medical diagnostic technique based on nuclear magnetic resonance.\nMagnetogyric ratio (Section 13.1): A ratio of the isotope's magnetic moment to its angular momentum.\nMALDI (Section 12.4): Matrix-assisted laser desorption ionization; a soft ionization method used for mass spectrometry of biological samples of very high molecular weight.\nMalonic ester synthesis (Section 22.7): The synthesis of a carboxylic acid by alkylation of an alkyl halide with diethyl malonate, followed by hydrolysis and decarboxylation.\nMarkovnikov's rule (Section 7.8): A guide for determining the regiochemistry (orientation) of electrophilic addition reactions. In the addition of HX to an alkene, the hydrogen atom bonds to the alkene carbon that has fewer alkyl substituents.\nMass number (A) (Section 1.1): The total of protons plus neutrons in an atom.\nMass spectrometry (MS) (Section 12.1): A technique for measuring the mass, and therefore the molecular weight (MW), of ions.\nMcLafferty rearrangement (Section 12.3, Section 19.4): A mass-spectral fragmentation pathway for carbonyl compounds.\nMechanism (Section 6.2): A complete description of how a reaction occurs. A mechanism accounts for all starting materials and all products and describes the details of each individual step in the overall reaction process.\nMeisenheimer complex (Section 16.6): An intermediate formed by addition of a nucleophile to a halo-substituted aromatic ring.\nMelt transition temperature, $\\boldsymbol{T}_{\\mathbf{m}}$ (Section 31.7): The temperature at which crystalline regions of a polymer melt to give an amorphous material.\nMercapto group (Section 18.7): An alternative name for the thiol group, -SH.\nMeso compounds (Section 5.7): Compounds that contain chirality centers but are nevertheless achiral because they contain a symmetry plane.\nMessenger RNA (mRNA) (Section 28.4): A kind of RNA formed by transcription of DNA and used to carry genetic messages from DNA to ribosomes."}
{"id": 2052, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nMessenger RNA (mRNA) (Section 28.4): A kind of RNA formed by transcription of DNA and used to carry genetic messages from DNA to ribosomes.\nMeta (m) (Section 15.1): A naming prefix used for 1,3-disubstituted benzenes.\nMetabolism (Section 29.1): A collective name for the many reactions that go on in the cells of living organisms.\nMetallacycle (Section 31.5): A cyclic compound that contains a metal atom in its ring.\nMethylene group (Section 7.3): $\\mathrm{A}-\\mathrm{CH}_{2}$ - or $=\\mathrm{CH}_{2}$ group.\nMicelles (Section 27.2): Spherical clusters of soaplike molecules that aggregate in aqueous solution. The ionic heads of the molecules lie on the outside, where they are solvated by water, and the organic tails bunch together on the inside of the micelle."}
{"id": 2053, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nMichael reaction (Section 23.10): The conjugate addition reaction of an enolate ion to an unsaturated carbonyl compound.\nMolar absorptivity ( $\\boldsymbol{\\varepsilon}$ ) (Section 14.7): A quantitative measure of the amount of UV light absorbed by a sample.\nMolecular ion (Section 12.1): The cation produced in a mass spectrometer by loss of an electron from the parent molecule. The mass of the molecular ion corresponds to the molecular weight of the sample.\nMolecular mechanics (Chapter 4 Chemistry Matters): A computer-based method for calculating the minimum-energy conformation of a molecule.\nMolecular orbital (MO) theory (Section 1.11, Section 14.1): A description of covalent bond formation as resulting from a mathematical combination of atomic orbitals (wave functions) to form molecular orbitals.\nMolecule (Section 1.4): A neutral collection of atoms held together by covalent bonds.\nMolozonide (Section 8.8): The initial addition product of ozone with an alkene.\nMonomers (Section 8.10, Section 21.9; Chapter 31 Introduction): The simple starting units from which polymers are made.\nMonosaccharides (Section 25.1): Simple sugars.\nMonoterpenoids (Chapter 8 Chemistry Matters, Section 27.5): Ten-carbon lipids.\nMultiplet (Section 13.6): A pattern of peaks in an NMR spectrum that arises by spin-spin splitting of a single absorption because of coupling between neighboring magnetic nuclei.\nMutarotation (Section 25.5): The change in optical rotation observed when a pure anomer of a sugar is dissolved in water. Mutarotation is caused by the reversible opening and closing of the acetal linkage, which yields an equilibrium mixture of anomers.\nn +1 rule (Section 13.6): A hydrogen with $n$ other hydrogens on neighboring carbons shows $n+1$ peaks in its ${ }^{1} H$ NMR spectrum.\nN -terminal amino acid (Section 26.4): The amino acid with a free $-\\mathrm{NH}_{2}$ group at the end of a protein chain."}
{"id": 2054, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nN -terminal amino acid (Section 26.4): The amino acid with a free $-\\mathrm{NH}_{2}$ group at the end of a protein chain.\nNatural gas (Chapter 3 Chemistry Matters): A naturally occurring hydrocarbon mixture consisting chiefly of methane, along with smaller amounts of ethane, propane, and butane.\nNatural product (Chapter 7 Chemistry Matters): A catchall term generally taken to mean a secondary metabolite found in bacteria, plants, and other living organisms.\nNew molecular entity, NME (Chapter 6 Chemistry Matters): A new biologically active chemical substance approved for sale as a drug by the U.S. Food and Drug Administration.\nNewman projection (Section 3.6): A means of indicating stereochemical relationships between substituent groups on neighboring carbons. The carbon-carbon bond is viewed endon, and the carbons are indicated by a circle. Bonds radiating\nfrom the center of the circle are attached to the front carbon, and bonds radiating from the edge of the circle are attached to the rear carbon.\nNitration (Section 16.2): The substitution of a nitro group onto an aromatic ring.\nNitriles (Section 3.1, Section 20.1): A class of compounds containing the $\\mathrm{C} \\equiv \\mathrm{N}$ functional group.\nNitrogen rule (Section 24.10): A compound with an odd number of nitrogen atoms has an odd-numbered molecular weight.\nNode (Section 1.2): A surface of zero electron density within an orbital. For example, a $p$ orbital has a nodal plane passing through the center of the nucleus, perpendicular to the axis of the orbital."}
{"id": 2055, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nNonbonding electrons (Section 1.4): Valence electrons that are not used in forming covalent bonds.\nNoncoding strand (Section 28.4): An alternative name for the antisense strand of DNA."}
{"id": 2056, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nNoncovalent interactions (Section 2.12): One of a variety of nonbonding interactions between molecules, such as dipole-dipole forces, dispersion forces, and hydrogen bonds.\nNonessential amino acid (Section 26.1): One of the eleven amino acids that are biosynthesized by humans.\nNormal alkanes (Section 3.2): Straight-chain alkanes, as opposed to branched alkanes. Normal alkanes are denoted by the suffix $n$, as in $n-\\mathrm{C}_{4} \\mathrm{H}_{10}$ ( $n$-butane).\nNSAID (Chapter 15 Chemistry Matters): A nonsteroidal antiinflammatory drug, such as aspirin or ibuprofen.\nNuclear magnetic resonance (NMR) spectroscopy (Chapter 13 Introduction): A spectroscopic technique that provides information about the carbon-hydrogen framework of a molecule. NMR works by detecting the energy absorptions accompanying the transitions between nuclear spin states that occur when a molecule is placed in a strong magnetic field and irradiated with radiofrequency waves.\nNucleic acid (Section 28.1): Deoxyribonucleic acid (DNA) and ribonucleic acid (RNA); biological polymers made of nucleotides joined together to form long chains.\nNucleophile (Section 6.3, Section 6.4): An electron-rich species that donates an electron pair to an electrophile in a polar bondforming reaction. Nucleophiles are also Lewis bases.\nNucleophilic acyl substitution reaction (Section 21.2): A reaction in which a nucleophile attacks a carbonyl compound and substitutes for a leaving group bonded to the carbonyl carbon.\nNucleophilic addition reaction (Section 19.4): A reaction in which a nucleophile adds to the electrophilic carbonyl group of a ketone or aldehyde to give an alcohol.\nNucleophilic aromatic substitution reactions (Section 16.6): The substitution reactions of an aryl halide by a nucleophile.\nNucleophilic substitution reactions (Section 11.1): Reactions in which one nucleophile replaces another attached to a saturated carbon atom."}
{"id": 2057, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nNucleophilicity (Section 11.3): The ability of a substance to act as a nucleophile in an $\\mathrm{S}_{\\mathrm{N}} 2$ reaction.\nNucleoside (Section 28.1): A nucleic acid constituent consisting of a sugar residue bonded to a heterocyclic purine or pyrimidine base.\nNucleotides (Section 28.1): Nucleic acid constituents consisting of a sugar residue bonded both to a heterocyclic purine or pyrimidine base and to a phosphoric acid. Nucleotides are the monomer units from which DNA and RNA are constructed.\nNylons (Section 21.9): Synthetic polyamide step-growth polymers.\nOkazaki fragments (Section 28.3): Short segments of a DNA lagging strand that is biosynthesized discontinuously and then linked by DNA ligases.\nOlefin (Chapter 7 Introduction): An alternative name for an alkene.\nOlefin metathesis polymerization (Section 31.5): A method of polymer synthesis based on using an olefin metathesis reaction.\nOlefin metathesis reaction (Section 31.5): A reaction in which two olefins (alkenes) exchange substituents on their double bonds.\nOligonucleotides (Section 28.7): Short segments of DNA.\nOptical isomers (Section 5.4): An alternative name for enantiomers. Optical isomers are isomers that have a mirrorimage relationship.\nOptically active (Section 5.3): A property of some organic molecules wherein the plane of polarization is rotated through an angle when a beam of plane-polarized light is passed through a solution of the molecules.\nOrbital (Section 1.2): A wave function, which describes the volume of space around a nucleus in which an electron is most likely to be found.\nOrganic chemistry (Chapter 1 Introduction): The study of carbon compounds.\nOrganohalides (Chapter 10 Introduction): Compounds that contain one or more halogen atoms bonded to carbon.\nOrganometallic compound (Section 10.6): A compound that contains a carbon-metal bond. Grignard reagents, RMgX, are examples."}
{"id": 2058, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nOrganometallic compound (Section 10.6): A compound that contains a carbon-metal bond. Grignard reagents, RMgX, are examples.\nOrganophosphate (Section 1.10): A compound that contains a phosphorus atom bonded to four oxygens, with one of the oxygens also bonded to carbon.\nOrtho (o) (Section 15.1): A naming prefix used for 1,2-disubstituted benzenes.\nOxidation (Section 8.7, Section 10.8): A reaction that causes a decrease in electron ownership by carbon, either by bond formation between carbon and a more electronegative atom (usually oxygen, nitrogen, or a halogen) or by bond-breaking between carbon and a less electronegative atom (usually hydrogen).\nOximes (Section 19.8): Compounds with the $\\mathrm{R}_{2} \\mathrm{C}=\\mathrm{NOH}$\nfunctional group.\nOxirane (Section 8.7): An alternative name for an epoxide.\nOxymercuration (Section 8.4): A method for double-bond hydration by reaction of an alkene with aqueous mercuric acetate followed by treatment with $\\mathrm{NaBH}_{4}$.\nOzonide (Section 8.8): The product initially formed by addition of ozone to a carbon-carbon double bond. Ozonides are usually treated with a reducing agent, such as zinc in acetic acid, to produce carbonyl compounds.\nPara (p) (Section 15.1): A naming prefix used for 1,4-disubstituted benzenes.\nParaffins (Section 3.5): A common name for alkanes.\nParent peak (Section 12.1): The peak in a mass spectrum corresponding to the molecular ion. The mass of the parent peak therefore represents the molecular weight of the compound.\nPauli exclusion principle (Section 1.3): No more than two electrons can occupy the same orbital, and those two must have spins of opposite sign.\nPeptide bond (Section 26.4): An amide bond in a peptide chain.\nPeptides (Chapter 26 Introduction): A type of short amino acid polymer in which the individual amino acid residues are linked by amide bonds."}
{"id": 2059, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nPeptide bond (Section 26.4): An amide bond in a peptide chain.\nPeptides (Chapter 26 Introduction): A type of short amino acid polymer in which the individual amino acid residues are linked by amide bonds.\nPericyclic reaction (Chapter 30 Introduction): A reaction that occurs in a single step by a reorganization of bonding electrons in a cyclic transition state.\nPeriplanar (Section 11.8): A conformation in which bonds to neighboring atoms have a parallel arrangement. In an eclipsed conformation, the neighboring bonds are syn periplanar; in a staggered conformation, the bonds are anti periplanar.\nPeroxides (Section 18.1): Molecules containing an oxygen-oxygen bond functional group, ROOR' or ROOH.\nPeroxyacid (Section 8.7): A compound with the $-\\mathrm{CO}_{3} \\mathrm{H}$ functional group. Peroxyacids react with alkenes to give epoxides.\nPhenols (Chapter 17 Introduction): A class of compounds with an -OH group directly bonded to an aromatic ring, ArOH.\nPhenoxide ion, $\\mathrm{ArO}^{-}$(Section 17.2): The anion of a phenol.\nPhenyl (Section 15.1): The name for the $-\\mathrm{C}_{6} \\mathrm{H}_{5}$ unit when the benzene ring is considered as a substituent. A phenyl group is abbreviated as -Ph.\nPhosphine (Section 5.10): A trivalent phosphorus compound, $\\mathrm{R}_{3} \\mathrm{P}$.\nPhosphite (Section 28.7): A compound with the structure $\\mathrm{P}(\\mathrm{OR})_{3}$.\nPhospholipids (Section 27.3): Lipids that contain a phosphate residue. For example, glycerophospholipids contain a glycerol backbone linked to two fatty acids and a phosphoric acid.\nPhosphoramidite (Section 28.7): A compound with the structure $\\mathrm{R}_{2} \\mathrm{NP}(\\mathrm{OR})_{2}$."}
{"id": 2060, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nPhosphoramidite (Section 28.7): A compound with the structure $\\mathrm{R}_{2} \\mathrm{NP}(\\mathrm{OR})_{2}$.\nPhosphoric acid anhydride (Section 29.1): A substance that contains $\\mathrm{PO}_{2} \\mathrm{PO}$ link, analogous to the $\\mathrm{CO}_{2} \\mathrm{CO}$ link in carboxylic acid anhydrides."}
{"id": 2061, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nPhotochemical reactions (Section 30.3): A reaction carried out by irradiating the reactants with light.\nPhysiological pH (Section 20.3): The pH of 7.3 that exists inside cells.\nPi ( $\\boldsymbol{\\pi}$ ) bond (Section 1.8): The covalent bond formed by sideways overlap of atomic orbitals. For example, carbon-carbon double bonds contain a $\\pi$ bond formed by sideways overlap of two $p$ orbitals.\nPITC (Section 26.6): Phenylisothiocyanate; used in the Edman degradation.\n$\\mathbf{p} K_{\\mathbf{a}}$ (Section 2.8): The negative common logarithm of the $K_{\\mathrm{a}}$; used to express acid strength.\nPlane of symmetry (Section 5.2): A plane that bisects a molecule such that one half of the molecule is the mirror image of the other half. Molecules containing a plane of symmetry are achiral.\nPlane-polarized light (Section 5.3): Light that has its electromagnetic waves oscillating in a single plane rather than in random planes. The plane of polarization is rotated when the light is passed through a solution of a chiral substance.\nPlasticizers (Section 31.7): Small organic molecules added to polymers to act as a lubricant between polymer chains.\nPolar aprotic solvents (Section 11.3): Polar solvents that can't function as hydrogen ion donors. Polar aprotic solvents such as dimethyl sulfoxide (DMSO) and dimethylformamide (DMF) are particularly useful in $\\mathrm{S}_{\\mathrm{N}} 2$ reactions because of their ability to solvate cations.\nPolar covalent bond (Section 2.1): A covalent bond in which the electron distribution between atoms is unsymmetrical.\nPolar reactions (Section 6.2, Section 6.3): Reactions in which bonds are made when a nucleophile donates two electrons to an electrophile and in which bonds are broken when one fragment leaves with both electrons from the bond."}
{"id": 2062, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nPolarity (Section 2.1): The unsymmetrical distribution of electrons in a molecule that results when one atom attracts electrons more strongly than another.\nPolarizability (Section 6.3): The measure of the change in a molecule's electron distribution in response to changing electrostatic interactions with solvents or ionic reagents.\nPolycarbonates (Section 31.4): Polyesters in which the carbonyl groups are linked to two -OR groups, $\\left[\\mathrm{O}=\\mathrm{C}(\\mathrm{OR})_{2}\\right]$.\nPolycyclic (Section 4.9): Containing more than one ring.\nPolycyclic aromatic compound (Section 15.6): A compound with two or more benzene-like aromatic rings fused together.\nPolymer (Section 8.10, Section 21.9; Chapter 31 Introduction): A large molecule made up of repeating smaller units. For example, polyethylene is a synthetic polymer made from repeating ethylene units, and DNA is a biopolymer made of repeating deoxyribonucleotide units.\nPolymerase chain reaction (PCR) (Section 28.8): A method for amplifying small amounts of DNA to produce larger amounts.\nPolysaccharides (Section 25.9): A type of carbohydrate that is made of many simple sugars linked together by glycoside\n(acetal) bonds.\nPolyunsaturated fatty acids (Section 27.1): Fatty acids that contain more than one double bond.\nPolyurethane (Section 31.4): A step-growth polymer prepared by reaction between a diol and a diisocyanate.\nPosttranslational modification (Section 28.6): A chemical modification of a protein that occurs after translation from DNA.\nPrimary, secondary, tertiary, and quaternary (Section 3.3): Terms used to describe the substitution pattern at a specific site. A primary site has one organic substituent attached to it, a secondary site has two organic substituents, a tertiary site has three, and a quaternary site has four."}
{"id": 2063, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \n|  | Primary | Secondary | Tertiary | Quaternary |\n| :--- | :--- | :--- | :--- | :--- |\n| Carbon | $\\mathrm{RCH}_{3}$ | $\\mathrm{R}_{2} \\mathrm{CH}_{2}$ | $\\mathrm{R}_{3} \\mathrm{CH}$ | $\\mathrm{R}_{4} \\mathrm{C}$ |\n| Carbocation | $\\mathrm{RCH}_{2}{ }^{+}$ | $\\mathrm{R}_{2} \\mathrm{CH}^{+}$ | $\\mathrm{R}_{3} \\mathrm{C}^{+}$ |  |\n| Hydrogen | $\\mathrm{RCH}_{3}$ | $\\mathrm{R}_{2} \\mathrm{CH}_{2}$ | $\\mathrm{R}_{3} \\mathrm{CH}$ |  |\n| Alcohol | $\\mathrm{RCH}_{2} \\mathrm{OH}$ | $\\mathrm{R}_{2} \\mathrm{CHOH}$ | $\\mathrm{R}_{3} \\mathrm{COH}$ |  |\n| Amine | $\\mathrm{RNH}_{2}$ | $\\mathrm{R}_{2} \\mathrm{NH}$ | $\\mathrm{R}_{3} \\mathrm{~N}$ |  |"}
{"id": 2064, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nPrimary structure (Section 26.9): The amino acid sequence in a protein.\npro- $\\boldsymbol{R}$ (Section 5.11): One of two identical atoms or groups of atoms in a compound whose replacement leads to an $R$ chirality center.\npro-S (Section 5.11): One of two identical atoms or groups of atoms in a compound whose replacement leads to an $S$ chirality center.\nProchiral (Section 5.11): A molecule that can be converted from achiral to chiral in a single chemical step.\nProchirality center (Section 5.11): An atom in a compound that can be converted into a chirality center by changing one of its attached substituents.\nPromoter sequence (Section 28.4): A short sequence on DNA located upstream of the transcription start site and recognized by RNA polymerase.\nPropagation step (Section 8.10): A step in a radical chain reaction that carries on the chain. The propagation steps must yield both product and a reactive intermediate.\nProstaglandins (Section 27.4): Lipids derived from arachidonic acid. Prostaglandins are present in nearly all body tissues and fluids, where they serve many important hormonal functions.\nProtecting group (Section 17.8, Section 19.10, Section 26.7): A group that is introduced to protect a sensitive functional group toward reaction elsewhere in the molecule. After serving its protective function, the group is removed.\nProteins (Chapter 26 Introduction): Large peptides containing"}
{"id": 2065, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \n50 or more amino acid residues. Proteins serve both as structural materials and as enzymes that control an organism's chemistry.\nProtein Data Bank (Chapter 26 Chemistry Matters): A worldwide online repository of X-ray and NMR structural data for biological macromolecules. To access the Protein Data Bank, go to https://www.rcsb.org (https://www.rcsb.org).\nProtic solvents (Section 11.3): Solvents such as water or alcohol that can act as a proton donor.\nPyramidal inversion (Section 24.2): The rapid stereochemical inversion of a trivalent nitrogen compound.\nPyranose (Section 25.5): The six-membered, cyclic hemiacetal form of a simple sugar.\nQuadrupole mass analyzer (Section 12.1): A type of mass spectrometer that uses four cylindrical rods to create an oscillating electrostatic field. Ion trajectories are determined by their $m / z$ ratios. At a given field, only one $m / z$ value will make it through the quadrupole region-the others will crash into the quadrupole rods or the walls of the instrument and never reach the detector.\nQuartet (Section 13.6): A set of four peaks in an NMR spectrum, caused by spin-spin splitting of a signal by three adjacent nuclear spins.\nQuaternary: See Primary.\nQuaternary ammonium salt (Section 24.1): An ionic compound containing a positively charged nitrogen atom with four attached groups, $\\mathrm{R}_{4} \\mathrm{~N}^{+} \\mathrm{X}^{-}$.\nQuaternary structure (Section 26.9): The highest level of protein structure, involving an ordered aggregation of individual proteins into a larger cluster.\nQuinone (Section 17.10): A 2,5-cyclohexadiene-1,4-dione.\n$\\boldsymbol{R}$ configuration (Section 5.5): The configuration at a chirality center as specified using the Cahn-Ingold-Prelog sequence rules.\n$\\mathbf{R}$ (Section 3.3): A generalized abbreviation for an organic partial structure."}
{"id": 2066, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \n$\\mathbf{R}$ (Section 3.3): A generalized abbreviation for an organic partial structure.\nRacemate (Section 5.8): A mixture consisting of equal parts (+) and (-) enantiomers of a chiral substance; also called a racemic mixture.\nRadical (Section 2.6, Section 6.2): A species that has an odd number of electrons, such as the chlorine radical, $\\mathrm{Cl} \\cdot$.\nRadical reactions (Section 6.2, Section 6.3): Reactions in which bonds are made by donation of one electron from each of two reactants and in which bonds are broken when each fragment leaves with one electron.\nRate constant (Section 11.2): The constant $k$ in a rate equation. Rate equation (Section 11.2): An equation that expresses the dependence of a reaction's rate on the concentration of reactants.\nRate-limiting step (Section 11.4): The slowest step in a multistep reaction sequence; also called the rate-determining step. The rate-limiting step acts as a kind of bottleneck in multistep reactions."}
{"id": 2067, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nRe face (Section 5.11): One of two faces of a planar, $s p^{2}$-hybridized atom.\nRearrangement reactions (Section 6.1): What occurs when a single reactant undergoes a reorganization of bonds and atoms to yield an isomeric product.\nReducing sugars (Section 25.6): Sugars that reduce silver ion in the Tollens test or cupric ion in the Fehling or Benedict tests.\nReduction (Section 8.6, Section 10.8): A reaction that causes an increase of electron ownership by carbon, either by bondbreaking between carbon and a more electronegative atom or by bond formation between carbon and a less electronegative atom.\nReductive amination (Section 24.6, Section 26.3): A method for preparing an amine by reaction of an aldehyde or ketone with ammonia and a reducing agent.\nRefining (Chapter 3 Chemistry Matters): The process by which petroleum is converted into gasoline and other useful products. Regiochemistry (Section 7.8): A term describing the orientation of a reaction that occurs on an unsymmetrical substrate.\nRegiospecific (Section 7.8): A term describing a reaction that occurs with a specific regiochemistry to give a single product rather than a mixture of products.\nReplication (Section 28.3): The process by which doublestranded DNA uncoils and is replicated to produce two new copies.\nReplication forks (Section 28.3): The point of unraveling in a DNA chain where replication occurs.\nResidues (Section 26.4): Amino acids in a protein chain.\nResolution (Section 5.8): The process by which a racemate is separated into its two pure enantiomers.\nResonance effect (Section 16.4): The donation or withdrawal of electrons through orbital overlap with neighboring $\\pi$ bonds. For example, an oxygen or nitrogen substituent donates electrons to an aromatic ring by overlap of the O or N orbital with the aromatic ring $p$ orbitals.\nResonance forms (Section 2.4): Individual structural forms of a resonance hybrid."}
{"id": 2068, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nResonance forms (Section 2.4): Individual structural forms of a resonance hybrid.\nResonance hybrid (Section 2.4): A molecule, such as benzene, that can't be represented adequately by a single Kekul\u00e9 structure but must instead be considered as an average of two or more resonance forms. The resonance forms themselves differ only in the positions of their electrons, not their nuclei.\nRestriction endonucleases (Section 28.6): Enzymes that are able to cleave a DNA molecule at points in the chain where a specific base sequence occurs.\nRetrosynthetic (Section 9.9, Section 16.10): Planning an organic synthesis by working backward from the final product to the starting material.\nRibonucleic acid (RNA) (Chapter 28 Introduction): The biopolymer found in cells that serves to transcribe the genetic information found in DNA and uses that information to direct the synthesis of proteins."}
{"id": 2069, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nRibosomal RNA (rRNA) (Section 28.4): A kind of RNA used in the physical makeup of ribosomes.\nRing-current (Section 15.7): The circulation of $\\pi$ electrons induced in aromatic rings by an external magnetic field. This effect accounts for the downfield shift of aromatic ring protons in the ${ }^{1} \\mathrm{H}$ NMR spectrum.\nRing-flip (Section 4.6): A molecular motion that interconverts two chair conformations of cyclohexane. The effect of a ringflip is to convert an axial substituent into an equatorial substituent.\nRing-opening metathesis polymerization (ROMP) (Section 31.5): A method of polymer synthesis that uses an olefin metathesis reaction of a cycloalkene."}
{"id": 2070, "contents": "Amino acid (Section 26.1): See $\\boldsymbol{\\alpha}$-Amino acid. - \nRNA (Section 28.1): Ribonucleic acid.\nRobinson annulation reaction (Section 23.12): A method for synthesis of cyclohexenones by sequential Michael reaction and intramolecular aldol reaction.\n$\\boldsymbol{S}$ configuration (Section 5.5): The configuration at a chirality center as specified using the Cahn-Ingold-Prelog sequence rules.\n$\\boldsymbol{s}$-Cis conformation (Section 14.5): The conformation of a conjugated diene that is cis-like around the single bond.\nSaccharide (Section 25.1): A sugar.\nSalt bridge (Section 26.9): An ionic attraction between two oppositely charged groups in a protein chain.\nSandmeyer reaction (Section 24.8): The nucleophilic substitution reaction of an arenediazonium salt with a cuprous halide to yield an aryl halide.\nSanger dideoxy method (Section 28.6): A commonly used method of DNA sequencing.\nSaponification (Section 21.6, Section 27.2): An old term for the base-induced hydrolysis of an ester to yield a carboxylic acid salt.\nSaturated (Section 3.2): A molecule that has only single bonds and thus can't undergo addition reactions. Alkanes are saturated, but alkenes are unsaturated.\nSawhorse representations (Section 3.6): A manner of representing stereochemistry that uses a stick drawing and gives a perspective view of the conformation around a single bond.\nSchiff bases (Section 19.8, Section 29.5): An alternative name for an imine, $\\mathrm{R}_{2} \\mathrm{C}=\\mathrm{NR}^{\\prime}$, used primarily in biochemistry.\nSecond-order reaction (Section 11.2): A reaction whose ratelimiting step is bimolecular and whose kinetics are therefore dependent on the concentration of two reactants."}
{"id": 2071, "contents": "Secondary: See Primary. - \nSecondary metabolite (Chapter 7 Chemistry Matters): A small naturally occurring molecule that is not essential to the growth and development of the producing organism and is not classified by structure.\nSecondary structure (Section 26.9): The level of protein substructure that involves organization of chain sections into ordered arrangements such as $\\beta$-pleated sheets or $\\alpha$ helices.\n\nSemiconservative replication (Section 28.3): The process by which DNA molecules are made containing one strand of old DNA and one strand of new DNA.\nSense strand (Section 28.4): The coding strand of doublehelical DNA that contains the gene.\nSequence rules (Section 5.5, Section 7.5): A series of rules for assigning relative rankings to substituent groups on a doublebond carbon atom or on a chirality center."}
{"id": 2072, "contents": "Secondary: See Primary. - \nSesquiterpenoids (Section 27.5): 15-carbon lipids.\nSharpless epoxidation (Chapter 19 Chemistry Matters): A method for enantioselective synthesis of a chiral epoxide by treatment of an allylic alcohol with tert-butyl hydroperoxide, $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{C}-\\mathrm{OOH}$, in the presence of titanium tetraisopropoxide and diethyl tartrate.\nShielding (Section 13.2): An effect observed in NMR that causes a nucleus to absorb toward the right (upfield) side of the chart. Shielding is caused by donation of electron density to the nucleus.\nSi face (Section 5.11): One of two faces of a planar, $s p^{2}$-hybridized atom.\nSialic acid (Section 25.7): One of a group of more than 300 carbohydrates based on acetylneuramic acid.\nSide chain (Section 26.1): The substituent attached to the $\\alpha$ carbon of an amino acid.\nSigma ( $\\boldsymbol{\\sigma}$ ) bond (Section 1.5): A covalent bond formed by headon overlap of atomic orbitals.\nSigmatropic reaction (Section 30.8): A pericyclic reaction that involves the migration of a group from one end of a $\\pi$ electron system to the other.\nSilyl ether (Section 17.8): A substance with the structure $\\mathrm{R}_{3} \\mathrm{Si}-\\mathrm{O}-\\mathrm{R}$. The silyl ether acts as a protecting group for alcohols."}
{"id": 2073, "contents": "Secondary: See Primary. - \nSimmons-Smith reaction (Section 8.9): The reaction of an alkene with $\\mathrm{CH}_{2} \\mathrm{I}_{2}$ and $\\mathrm{Zn}-\\mathrm{Cu}$ to yield a cyclopropane.\nSimple sugars (Section 25.1): Carbohydrates that cannot be broken down into smaller sugars by hydrolysis.\nSingle bond (Section 1.8): A covalent bond formed by sharing one electron pair between atoms.\nSkeletal structures (Section 1.12): A shorthand way of writing structures in which carbon atoms are assumed to be at each intersection of two lines (bonds) and at the end of each line.\nSmall RNAs (Section 28.4): A type of RNA that has a variety of functions within the cell, including silencing transcription and catalyzing chemical modifications of other RNA molecules.\n$\\mathbf{S}_{\\mathbf{N}} 1$ reaction (Section 11.4): A unimolecular nucleophilic substitution reaction.\n$\\mathrm{S}_{\\mathrm{N}} 2$ reaction (Section 11.2): A bimolecular nucleophilic substitution reaction.\nSolid-phase synthesis (Section 26.8): A technique of synthesis whereby the starting material is covalently bound to a solid polymer bead and reactions are carried out on the bound substrate. After the desired transformations have been effected, the product is cleaved from the polymer."}
{"id": 2074, "contents": "Secondary: See Primary. - \nSolvation (Section 11.3): The clustering of solvent molecules around a solute particle to stabilize it.\n$\\boldsymbol{s p}$ hybrid orbitals (Section 1.9): Hybrid orbitals derived from the combination of an $s$ and a $p$ atomic orbital. The two $s p$ orbitals that result from hybridization are oriented at an angle of $180^{\\circ}$ to each other.\n$\\boldsymbol{s p}^{2}$ hybrid orbitals (Section 1.8): Hybrid orbitals derived by combination of an $s$ atomic orbital with two $p$ atomic orbitals. The three $s p^{2}$ hybrid orbitals that result lie in a plane at angles of $120^{\\circ}$ to each other.\n$\\boldsymbol{s p}^{\\mathbf{3}}$ hybrid orbitals (Section 1.6): Hybrid orbitals derived by combination of an $s$ atomic orbital with three $p$ atomic orbitals. The four $s p^{3}$ hybrid orbitals that result are directed toward the corners of a regular tetrahedron at angles of $109^{\\circ}$ to each other.\nSpecific rotation, [ $\\boldsymbol{\\alpha}] \\mathbf{D}$ (Section 5.3): The optical rotation of a chiral compound under standard conditions.\nSphingomyelins (Section 27.3): Phospholipids that have sphingosine as the backbone rather than glycerol.\nSpin-spin splitting (Section 13.6): The splitting of an NMR signal into a multiplet because of an interaction between nearby magnetic nuclei whose spins are coupled. The magnitude of spin-spin splitting is given by the coupling constant, J.\nStaggered conformation (Section 3.6): The three-dimensional arrangement of atoms around a carbon-carbon single bond in which the bonds on one carbon bisect the bond angles on the second carbon as viewed end-on.\nStatin (Chapter 29 Chemistry Matters): A drug that controls cholesterol biosynthesis in the body by blocking the HMG-CoA reductase enzyme.\nStep-growth polymers (Section 21.9, Section 31.4): Polymers in which each bond is formed independently of the others. Polyesters and polyamides (nylons) are examples."}
{"id": 2075, "contents": "Secondary: See Primary. - \nStep-growth polymers (Section 21.9, Section 31.4): Polymers in which each bond is formed independently of the others. Polyesters and polyamides (nylons) are examples.\nStereocenter (Section 5.2): An alternative name for a chirality center.\nStereochemistry (Section 3.5; Chapters 3, 4, 5): The branch of chemistry concerned with the three-dimensional arrangement of atoms in molecules.\nStereogenic center (Section 5.2): An alternative name for a chirality center.\nStereoisomers (Section 4.2): Isomers that have their atoms connected in the same order but have different threedimensional arrangements. The term stereoisomer includes both enantiomers and diastereomers.\nStereospecific (Section 8.9, Section 14.5): A term indicating that only a single stereoisomer is produced in a given reaction rather than a mixture.\nSteric strain (Section 3.7, Section 4.3, Section 4.7): The strain imposed on a molecule when two groups are too close together and try to occupy the same space. Steric strain is responsible both for the greater stability of trans versus cis alkenes and for the greater stability of equatorially substituted versus axially substituted cyclohexanes."}
{"id": 2076, "contents": "Secondary: See Primary. - \nSteroids (Section 27.6): Lipids whose structure is based on a tetracyclic carbon skeleton with three 6 -membered and one 5 -membered ring. Steroids occur in both plants and animals and have a variety of important hormonal functions.\nStork enamine reaction (Section 23.11): The conjugate addition of an enamine to an $\\alpha, \\beta$-unsaturated carbonyl compound, followed by hydrolysis to yield a 1,5-dicarbonyl product.\nSTR loci (Chapter 28 Chemistry Matters): Short tandem repeat sequences of noncoding DNA that are unique to every individual and allow DNA fingerprinting.\nStraight-chain alkanes (Section 3.2): Alkanes whose carbon atoms are connected without branching.\nSubstitution reactions (Section 6.1): What occurs when two reactants exchange parts to give two new products. $\\mathrm{S}_{\\mathrm{N}} 1$ and $\\mathrm{S}_{\\mathrm{N}} 2$ reactions are examples.\nSulfides (Section 3.1, Section 18.7): A class of compounds that has two organic substituents bonded to the same sulfur atom, RSR'.\nSulfonation (Section 16.2): The substitution of a sulfonic acid group $\\left(-\\mathrm{SO}_{3} \\mathrm{H}\\right)$ onto an aromatic ring.\nSulfone (Section 18.7): A compound of the general structure $\\mathrm{RSO}_{2} \\mathrm{R}^{\\prime}$.\nSulfonium ions (Section 18.7): A species containing a positively charged, trivalent sulfur atom, $\\mathrm{R}_{3} \\mathrm{~S}^{+}$.\nSulfoxide (Section 18.7): A compound of the general structure RSOR'.\nSuprafacial (Section 30.5): A word used to describe the geometry of pericyclic reactions. Suprafacial reactions take place on the same side of the two ends of a $\\pi$ electron system.\nSuzuki-Miyaura reaction (Section 10.7): The palladiumcatalyzed coupling reaction of an aromatic or vinylic halide with an aromatic or vinylic boronic acid."}
{"id": 2077, "contents": "Secondary: See Primary. - \nSuzuki-Miyaura reaction (Section 10.7): The palladiumcatalyzed coupling reaction of an aromatic or vinylic halide with an aromatic or vinylic boronic acid.\nSymmetry-allowed, symmetry-disallowed (Section 30.1): A symmetry-allowed reaction is a pericyclic process that has a favorable orbital symmetry for reaction through a concerted pathway. A symmetry-disallowed reaction is one that does not have favorable orbital symmetry for reaction through a concerted pathway.\nSymmetry plane (Section 5.2): A plane that bisects a molecule such that one half of the molecule is the mirror image of the other half. Molecules containing a plane of symmetry are achiral.\nSyn periplanar (Section 11.8): Describing a stereochemical relationship in which two bonds on adjacent carbons lie in the same plane and are eclipsed.\nSyn stereochemistry (Section 8.5): The opposite of anti. A syn addition reaction is one in which the two ends of the double bond react from the same side. A syn elimination is one in which the two groups leave from the same side of the molecule. Syndiotactic (Section 31.2): A chain-growth polymer in which the stereochemistry of the substituents alternates regularly on opposite sides of the backbone."}
{"id": 2078, "contents": "Secondary: See Primary. - \nTautomers (Section 9.4, Section 22.1): Isomers that interconvert spontaneously, usually with the change in position of a hydrogen.\n\nTerpenoids (Chapter 8 Chemistry Matters, Section 27.5): Lipids that are formally derived by head-to-tail polymerization of isoprene units."}
{"id": 2079, "contents": "Tertiary: See Primary - \nTertiary structure (Section 26.9): The level of protein structure that involves the manner in which the entire protein chain is folded into a specific three-dimensional arrangement.\n\nThermodynamic control (Section 14.3): An equilibrium reaction that yields the lowest-energy, most stable product is said to be thermodynamically controlled.\n\nThermoplastics (Section 31.7): Polymers that have a high $T_{\\mathrm{g}}$ and are hard at room temperature but become soft and viscous when heated.\n\nThermosetting resins (Section 31.7): Polymers that become highly cross-linked and solidify into a hard, insoluble mass when heated.\n\nThioesters (Chapter 21 Introduction): A class of compounds with the RCOSR' functional group.\nThiols (Section 3.1, Chapter 18 Introduction): A class of compounds containing the -SH functional group.\nThiolate ion (Section 18.7): The anion of a thiol, $\\mathrm{RS}^{-}$.\nTMS (Section 13.3): Tetramethylsilane; used as an NMR calibration standard."}
{"id": 2080, "contents": "Tertiary: See Primary - \nTOF (Section 12.4): Time-of-flight mass spectrometry; a sensitive method of mass detection accurate to about 3 ppm .\nTollens' reagent (Section 25.6): A solution of $\\mathrm{Ag}_{2} \\mathrm{O}$ in aqueous ammonia; used to oxidize aldehydes to carboxylic acids.\nTorsional strain (Section 3.6, Section 4.3): The strain in a molecule caused by electron repulsion between eclipsed bonds. Torsional strain is also called eclipsing strain.\nTosylate (Section 11.1): A p-toluenesulfonate ester; useful as a leaving group in nucleophilic substitution reactions.\nTransamination (Section 29.9): The exchange of an amino group and a keto group between reactants.\nTranscription (Section 28.4): The process by which the genetic information encoded in DNA is read and used to synthesize RNA in the nucleus of the cell. A small portion of doublestranded DNA uncoils, and complementary ribonucleotides line up in the correct sequence for RNA synthesis.\nTransfer RNA (tRNA) (Section 28.4): A kind of RNA that transports amino acids to the ribosomes, where they are joined together to make proteins.\nTransimination (Section 29.9): The exchange of an amino group and an imine group between reactants.\n\nTransition state (Section 6.9): An activated complex between reactants, representing the highest energy point on a reaction curve. Transition states are unstable complexes that can't be isolated.\n\nTranslation (Section 28.5): The process by which the genetic information transcribed from DNA onto mRNA is read by tRNA\nand used to direct protein synthesis.\nTree diagram (Section 13.8): A diagram used in NMR to sort out the complicated splitting patterns that can arise from multiple couplings.\nTriacylglycerols (Section 27.1): Lipids, such as those found in animal fat and vegetable oil, that are a triester of glycerol with long-chain fatty acids.\n\nTricarboxylic acid cycle (Section 29.7): An alternative name for the citric acid cycle by which acetyl CoA is degraded to $\\mathrm{CO}_{2}$."}
{"id": 2081, "contents": "Tertiary: See Primary - \nTricarboxylic acid cycle (Section 29.7): An alternative name for the citric acid cycle by which acetyl CoA is degraded to $\\mathrm{CO}_{2}$.\n\nTriple bonds (Section 1.8): A type of covalent bond formed by sharing three electron pairs between atoms.\n\nTriplet (Section 13.6): A symmetrical three-line splitting pattern observed in the ${ }^{1} \\mathrm{H}$ NMR spectrum when a proton has two equivalent neighbor protons.\n\nTurnover number (Section 26.10): The number of substrate molecules acted on by an enzyme molecule per unit time.\nTwist-boat conformation (Section 4.5): A conformation of cyclohexane that is somewhat more stable than a pure boat conformation.\n\nUltraviolet (UV) spectroscopy (Section 14.7): An optical spectroscopy employing ultraviolet irradiation. UV spectroscopy provides structural information about the extent of $\\pi$ electron conjugation in organic molecules.\nUnimolecular reaction (Section 11.4): A reaction that occurs by spontaneous transformation of the starting material without the intervention of other reactants. For example, the dissociation of a tertiary alkyl halide in the $\\mathrm{S}_{\\mathrm{N}} 1$ reaction is a unimolecular process.\nUnsaturated (Section 7.2): A molecule that has one or more multiple bonds.\nUpfield (Section 13.3): The right-hand portion of the NMR chart.\n\nUrethane (Section 31.4): A functional group in which a carbonyl group is bonded to both an -OR and an $-\\mathrm{NR}_{2}$.\nUronic acid (Section 25.6): A monocarboxylic acid formed by oxidizing the $-\\mathrm{CH}_{2} \\mathrm{OH}$ end of an aldose without affecting the - CHO end"}
{"id": 2082, "contents": "Tertiary: See Primary - \nValence bond theory (Section 1.5): A bonding theory that describes a covalent bond as resulting from the overlap of two atomic orbitals.\nValence shell (Section 1.4): The outermost electron shell of an atom.\nvan der Waals forces (Section 2.12): Intermolecular forces that are responsible for holding molecules together in the liquid and solid states.\nVegetable oils (Section 27.1): Liquid triacylglycerols derived from a plant source.\nVicinal (Section 9.2): A term used to refer to a 1,2-disubstitution pattern. For example, 1,2-dibromoethane is a vicinal dibromide.\nVinyl group (Section 7.3): $\\mathrm{A}_{2} \\mathrm{C}=\\mathrm{CH}$ - substituent.\nVinyl monomer (Section 8.10, Section 31.1): A substituted alkene monomer used to make a chain-growth polymer."}
{"id": 2083, "contents": "Tertiary: See Primary - \nVinylic (Section 9.3): A term that refers to a substituent at a double-bond carbon atom. For example, chloroethylene is a vinylic chloride, and enols are vinylic alcohols.\nVitamin (Section 26.10): A small organic molecule that must be obtained in the diet and is required in trace amounts for proper growth and function.\nVulcanization (Section 14.6): A technique for cross-linking and hardening a diene polymer by heating with a few percent by weight of sulfur.\nWalden inversion (Section 11.1): The inversion of configuration at a chirality center that accompanies an $\\mathrm{S}_{\\mathrm{N}} 2$ reaction.\nWave equation (Section 1.2): A mathematical expression that defines the behavior of an electron in an atom.\nWave function (Section 1.2): A solution to the wave equation for defining the behavior of an electron in an atom. The square of the wave function defines the shape of an orbital.\nWavelength, $\\lambda$ (Section 12.5): The length of a wave from peak to peak. The wavelength of electromagnetic radiation is inversely proportional to frequency and inversely proportional to energy.\nWavenumber, $\\widetilde{\\boldsymbol{v}}$ (Section 12.6): The reciprocal of the wavelength in centimeters.\nWaxes (Section 27.1): A mixture of esters of long-chain carboxylic acids with long-chain alcohols.\nWilliamson ether synthesis (Section 18.2): A method for synthesizing ethers by $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of an alkyl halide with an alkoxide ion.\nWittig reaction (Section 19.11): The reaction of a phosphorus\nylide with a ketone or aldehyde to yield an alkene.\nWohl degradation (Section 25.6): A method for shortening the chain of an aldose sugar by one carbon.\nWolff-Kishner reaction (Section 19.9): The conversion of an aldehyde or ketone into an alkane by reaction with hydrazine and base."}
{"id": 2084, "contents": "Tertiary: See Primary - \nWolff-Kishner reaction (Section 19.9): The conversion of an aldehyde or ketone into an alkane by reaction with hydrazine and base.\nX-ray crystallography (Chapter 12 Chemistry Matters): A technique that uses $X$ rays to determine the structure of molecules.\nYlide (Section 19.11): A neutral species with adjacent + and charges, such as the phosphoranes used in Wittig reactions.\n$\\boldsymbol{Z}$ geometry (Section 7.5): A term used to describe the stereochemistry of a carbon-carbon double bond. The two groups on each carbon are ranked according to the Cahn-Ingold-Prelog sequence rules, and the two carbons are compared. If the higher ranked groups on each carbon are on the same side of the double bond, the bond has $Z$ geometry.\nZaitsev's rule (Section 11.7): A rule stating that E2 elimination reactions normally yield the more highly substituted alkene as major product.\nZiegler-Natta catalysts (Section 31.2): Catalysts of an alkylaluminum and a titanium compound used for preparing alkene polymers.\nZwitterion (Section 26.1): A neutral dipolar molecule in which the positive and negative charges are not adjacent. For example, amino acids exist as zwitterions, $\\mathrm{H}_{3} \\stackrel{+}{\\mathrm{N}}-\\mathrm{CHR}-\\mathrm{CO}_{2}{ }^{-}$"}
{"id": 2085, "contents": "APPENDIX D - \nPeriodic Table"}
{"id": 2086, "contents": "Chapter 1 - \nPROBLEM (a) $1 s^{2} 2 s^{2} 2 p^{4}$ (b) $1 s^{2} 2 s^{2} 2 p^{3}$ (c) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{6} 3 p^{4}$\n1-1\nPROBLEM (a) 2 (b) 2 (c) 6\n1-2\nPROBLEM\n\n\nPROBLEM\n1-4\n\n\nPROBLEM\n(a) $\\mathrm{CCl}_{4}$\n(b) $\\mathrm{AlH}_{3}$\n(c) $\\mathrm{CH}_{2} \\mathrm{Cl}_{2}$\n(d) $\\mathrm{SiF}_{4}$\n(e) $\\mathrm{CH}_{3} \\mathrm{NH}_{2}$\n\n1-5\nPROBLEM (a)\n\n(c)\n\n(d)\n\n\nPROBLEM $\\mathrm{C}_{2} \\mathrm{H}_{7}$ has too many hydrogens for a compound with two carbons.\n1-7\nPROBLEM\n1-8\n\n\n\nAll bond angles are near $109^{\\circ}$.\nPROBLEM\n1-9\n\n\nPROBLEM The $\\mathrm{CH}_{3}$ carbon is $s p^{3}$; the double-bond carbons are $s p^{2}$; the $\\mathrm{C}=\\mathrm{C}-\\mathrm{C}$ and $\\mathrm{C}=\\mathrm{C}-\\mathrm{H}$ bond angles are\n1-10 approximately $120^{\\circ}$; other bond angles are near $109^{\\circ}$.\n\n\nPROBLEM All carbons are $s p^{2}$, and all bond angles are near $120^{\\circ}$.\n1-11\n\n\nPROBLEM All carbons except $\\mathrm{CH}_{3}$ are $s p^{2}$.\n1-12\n\n\nPROBLEM The $\\mathrm{CH}_{3}$ carbon is $s p^{3}$; the triple-bond carbons are $s p$; the $\\mathrm{C} \\equiv \\mathrm{C}-\\mathrm{C}$ and $\\mathrm{H}-\\mathrm{C} \\equiv \\mathrm{C}$ bond angles are\n1-13 approximately $180^{\\circ}$."}
{"id": 2087, "contents": "Chapter 1 - \nPROBLEM (a) O has 2 lone pairs and is $s p^{3}$-hybridized. (b) N has 1 lone pair and is $s p^{3}$-hybridized.\n1-14 (c) P has 1 lone pair and is $s p^{3}$-hybridized. (d) S has 2 lone pairs and is $s p^{3}$-hybridized.\n\nPROBLEM (a)\n1-15\n\n(b)\n\n\nEstrone $-\\mathrm{C}_{18} \\mathrm{H}_{22} \\mathrm{O}_{2}$\n\nPROBLEM (a) There are numerous possibilities, such as: (b) There are numerous possibilities, such as:\n1-16\n\n\n\n(c) There are numerous possibilities, such as:\n\n\n\n\n\n\n\n(d) There are numerous possibilities, such as:\n\n\n\n\n\nPROBLEM\n1-17"}
{"id": 2088, "contents": "Chapter 2 - \nPROBLEM (a) H (b) Br (c) Cl (d) C\n2-1\nPROBLEM\n(c)\n$\\mathrm{H}_{2}{ }_{\\mathrm{S}}^{\\mathrm{N}}-\\mathrm{S}_{\\mathrm{H}}{ }^{\\delta+}$\n(d) $\\mathrm{H}_{3} \\mathrm{C}-\\mathrm{SH}$\n\nCarbon and sulfur have identical electronegativities.\n\nPROBLEM $\\mathrm{H}_{3} \\mathrm{C}-\\mathrm{OH}<\\mathrm{H}_{3} \\mathrm{C}-\\mathrm{MgBr}<\\mathrm{H}_{3} \\mathrm{C}-\\mathrm{Li}=\\mathrm{H}_{3} \\mathrm{C}-\\mathrm{F}<\\mathrm{H}_{3} \\mathrm{C}-\\mathrm{K}$\n2-3\nPROBLEM The nitrogen is electron-rich, and the carbon is electron-poor.\n2-4\n\n\nPROBLEM The two $\\mathrm{C}-\\mathrm{O}$ dipoles cancel because of the symmetry of the molecule:\n2-5\n\n\nPROBLEM (a)\n\n(b)\n\n2-6\n\n\n(c)\n\n(d)\n\n\nPROBLEM (a) For carbon: $\\mathrm{FC}=4-8 / 2-0=0$ For the middle nitrogen: $\\mathrm{FC}=5-8 / 2-0=+1$ For the end 2-7 nitrogen: $\\mathrm{FC}=5-4 / 2-4=-1$\n(b) For nitrogen: $\\mathrm{FC}=5-8 / 2-0=+1$ For oxygen: $\\mathrm{FC}=6-2 / 2-6=-1$\n(c) For nitrogen: $\\mathrm{FC}=5-8 / 2-0=+1$ For the triply bonded carbon: $\\mathrm{FC}=4-6 / 2-2=-1$\n\nPROBLEM\n2-8\n\n\nPROBLEM The structures in (a) are resonance forms.\n2-9\nPROBLEM (a)\n2-10\n\n(b)\n\n(c)\n\n(d)\n\n\n\nPROBLEM\n2-11\n\n\nPROBLEM Phenylalanine is stronger.\n2-12\nPROBLEM Water is a stronger acid."}
{"id": 2089, "contents": "Chapter 2 - \n(b)\n\n(c)\n\n(d)\n\n\n\nPROBLEM\n2-11\n\n\nPROBLEM Phenylalanine is stronger.\n2-12\nPROBLEM Water is a stronger acid.\n\n2-13\nPROBLEM Neither reaction will take place.\n2-14\nPROBLEM Reaction will take place.\n2-15\nPROBLEM $K_{\\mathrm{a}}=4.9 \\times 10^{-10}$\n2-16\nPROBLEM (a)\n2-17\n\n(b)\n\n\n\n\n\nPROBLEM (a)\n2-18\n\n(b)\n\n\n\nPROBLEM Vitamin C is water-soluble (hydrophilic); vitamin A is fat-soluble (hydrophilic)."}
{"id": 2090, "contents": "Chapter 3 - \nPROBLEM (a) Sulfide, carboxylic acid, amine (b) Aromatic ring, carboxylic acid\n3-1 (c) Ether, alcohol, aromatic ring, amide, $\\mathrm{C}=\\mathrm{C}$ bond\nPROBLEM (a) $\\mathrm{CH}_{3} \\mathrm{OH}$\n3-2\n(b)\n\n(c)\n\n(d) $\\mathrm{CH}_{3} \\mathrm{NH}_{2}$\n(e)\n\n(f)"}
{"id": 2091, "contents": "PROBLEM - \n3-3\n\n\nPROBLEM\n3-4\n$\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$\n\n\n\n\nPROBLEM (a)\n3-5\n\n\n\n(b)\n\n$\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{C} \\equiv \\mathrm{N}$\n(c) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{SSCH}_{2} \\mathrm{CH}_{3}$\n$\\mathrm{CH}_{3} \\mathrm{SSCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$\n\n\nPROBLEM\n(a) Two (b) Four\n(c) Four\n\n3-6\nPROBLEM\n3-7\n\n\n\n\n\n\n\n\n\nPROBLEM (a)\n\n(b)\n\n(c)\n\n\nPROBLEM Primary carbons have primary hydrogens, secondary carbons have secondary hydrogens, and 3-9 tertiary carbons have tertiary hydrogens.\n\nPROBLEM (a)\n3-10\n\n(b)\n\n(c)\n\n\nPROBLEM (a) Pentane, 2-methylbutane, 2,2-dimethylpropane (b) 2,3-Dimethylpentane\n3-11 (c) 2,4-Dimethylpentane (d) 2,2,5-Trimethylhexane\nPROBLEM (a)\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM Pentyl, 1-methylbutyl, 1-ethylpropyl, 2-methylbutyl, 3-methylbutyl, 1,1-dimethylpropyl,\n3-13 1,2-dimethylpropyl, 2,2-dimethylpropyl\nPROBLEM\n3-14\n\n\n3,3,4,5-Tetramethylheptane\n\nPROBLEM\n3-15\n\n\n\nPROBLEM (a)\n3-16\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM\n3-17\n\n\nPROBLEM\n3-18"}
{"id": 2092, "contents": "PROBLEM - \nPROBLEM\n3-15\n\n\n\nPROBLEM (a)\n3-16\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM\n3-17\n\n\nPROBLEM\n3-18\n\n\nChapter 4\nPROBLEM (a) 1,4-Dimethylcyclohexane (b) 1-Methyl-3-propylcyclopentane (c) 3-Cyclobutylpentane\n4-1 (d) 1-Bromo-4-ethylcyclodecane (e) 1-Isopropyl-2-methylcyclohexane\n(f) 4-Bromo-1-tert-butyl-2-methylcycloheptane\n\nPROBLEM (a)\n4-2\n\n(b)\n\n\n(c)\n\n(d)\n\n\nPROBLEM 3-Ethyl-1,1-dimethylcyclopentane\n4-3\nPROBLEM (a) trans-1-Chloro-4-methylcyclohexane (b) cis-1-Ethyl-3-methylcycloheptane\n4-4\nPROBLEM (a)\n4-5\n\n(b)\n\n(c)\n\n\nPROBLEM The two hydroxyl groups are cis. The two side chains are trans.\n4-6\nPROBLEM (a) cis-1,2-Dimethylcyclopentane (b) cis-1-Bromo-3-methylcyclobutane\n4-7\nPROBLEM Six interactions; 21\\% of strain\n4-8\nPROBLEM The cis isomer is less stable because the methyl groups nearly eclipse each other.\n4-9\nPROBLEM Ten eclipsing interactions; $40 \\mathrm{~kJ} / \\mathrm{mol} ; 35 \\%$ is relieved.\n4-10\nPROBLEM Conformation (a) is more stable because the methyl groups are farther apart.\n4-11\nPROBLEM\n4-12\n\n\n\nPROBLEM\n4-13"}
{"id": 2093, "contents": "PROBLEM - \nPROBLEM\n4-13\n\n\n\nPROBLEM Before the ring-flip, red and blue are equatorial and green is axial. After the ring-flip, red and blue\n4-14 are axial and green is equatorial.\nPROBLEM $4.2 \\mathrm{~kJ} / \\mathrm{mol}$\n4-15\nPROBLEM Cyano group points straight up.\n4-16\nPROBLEM Equatorial $=70 \\%$; $a x i a l=30 \\%$\n4-17\nPROBLEM (a) $2.0 \\mathrm{~kJ} / \\mathrm{mol}$ (axial Cl) (b) $11.4 \\mathrm{~kJ} / \\mathrm{mol}\\left(\\mathrm{axial} \\mathrm{CH}_{3}\\right)$ (c) $2.0 \\mathrm{~kJ} / \\mathrm{mol}$ (axial Br)\n4-18 (d) $8.0 \\mathrm{~kJ} / \\mathrm{mol}\\left(\\right.$ axial $\\left.\\mathrm{CH}_{2} \\mathrm{CH}_{3}\\right)$\nPROBLEM\n\n\n1-Chloro-2,4-dimethylcyclohexane\n(less stable chair form)\nPROBLEM trans-Decalin is more stable because it has no 1,3-diaxial interactions.\n4-20\nPROBLEM Both ring-fusions are trans.\n4-21\nChapter 5\nPROBLEM Chiral: screw, shoe\n\n5-1\nPROBLEM (a)\n5-2\n\n(b)\n\n(c)\n\n\nPROBLEM\n5-3\n\nand\n\n\nPROBLEM (a)\n5-4\n\n(b)\n\n\nPROBLEM Levorotatory\n5-5\nPROBLEM $+16.1^{\\circ}$\n5-6\nPROBLEM (a) -Br\n(b) -Br (c) $-\\mathrm{CH}_{2} \\mathrm{CH}_{3}$\n(d) -OH\n(e) $-\\mathrm{CH}_{2} \\mathrm{OH}$ (f) $-\\mathrm{CH}=\\mathrm{O}$"}
{"id": 2094, "contents": "PROBLEM - \n5-7\nPROBLEM (a) $-\\mathrm{OH},-\\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{OH},-\\mathrm{CH}_{2} \\mathrm{CH}_{3},-\\mathrm{H}$ (b) $-\\mathrm{OH},-\\mathrm{CO}_{2} \\mathrm{CH}_{3},-\\mathrm{CO}_{2} \\mathrm{H},-\\mathrm{CH}_{2} \\mathrm{OH}$\n5-8 (c) $-\\mathrm{NH}_{2},-\\mathrm{CN},-\\mathrm{CH}_{2} \\mathrm{NHCH}_{3},-\\mathrm{CH}_{2} \\mathrm{NH}_{2}$ (d) $-\\mathrm{SSCH}_{3},-\\mathrm{SH},-\\mathrm{CH}_{2} \\mathrm{SCH}_{3},-\\mathrm{CH}_{3}$\nPROBLEM (a) $S$ (b) $R$ (c) $S$\n5-9\nPROBLEM (a) $S$ (b) $S$ (c) $R$\n5-10\nPROBLEM\n5-11\n\n\nPROBLEM $S$\n5-12\nPROBLEM Compound (a) is D-erythrose 4-phosphate, (d) is its enantiomer, and (b) and (c) are diastereomers.\n5-13\nPROBLEM Five chirality centers and $2^{5}=32$ stereoisomers\n5-14\nPROBLEM $S, S$\n5-15\nPROBLEM Compounds (a) and (d) are meso.\n5-16\nPROBLEM Compounds (a) and (c) have meso forms.\n5-17\nPROBLEM\n5-18\n\n\nPROBLEM Two diastereomeric salts: ( $R$ )-lactic acid plus ( $S$ )-1-phenylethylamine and ( $S$ )-lactic acid plus 5-20 (S)-1-phenylethylamine\n\nPROBLEM (a) Constitutional isomers (b) Diastereomers\n5-21\nPROBLEM (a)\n5-22\n\n(b) pro- $\\mathrm{C} \\longrightarrow$ pro-S\n\nPROBLEM (a)\n5-23\n\n(b)"}
{"id": 2095, "contents": "PROBLEM - \n(b) pro- $\\mathrm{C} \\longrightarrow$ pro-S\n\nPROBLEM (a)\n5-23\n\n(b)\n\n\nPROBLEM (S)-Lactate\n5-24\nPROBLEM The -OH adds to the Re face of C 2 , and -H adds to the Re face of C 3 . The overall addition has anti\n5-25 stereochemistry."}
{"id": 2096, "contents": "Chapter 6 - \nPROBLEM (a) Substitution (b) Elimination (c) Addition\n6-1\nPROBLEM (a) Carbon is electrophilic. (b) Sulfur is nucleophilic. (c) Nitrogens are nucleophilic.\n6-2 (d) Oxygen is nucleophilic; carbon is electrophilic.\nPROBLEM\n6-3\nPROBLEM Bromocyclohexane; chlorocyclohexane\n6-4\nPROBLEM\n6-5\n\n\nThe mechanism is shown in Figure 6.4\nPROBLEM (a)\n6-6\n\n(b)\n\n(c)\n\n\nPROBLEM\n6-7\n\n\nPROBLEM 1-Chloro-2-methylpentane, 2-chloro-2-methylpentane, 3-chloro-2-methylpentane, 6-8 2-chloro-4-methylpentane, 1-chloro-4-methylpentane\nPROBLEM\n\n\n\nChapter 7\nPROBLEM (a) 1 (b) 2 (c) 2\n7-1\nPROBLEM (a) 5 (b) 5 (c) 3 (d) 1 (e) 6 (f) 5\n7-2\nPROBLEM $\\mathrm{C}_{16} \\mathrm{H}_{13} \\mathrm{ClN}_{2} \\mathrm{O}$\n7-3\nPROBLEM (a) 3,4,4-Trimethyl-1-pentene (b) 3-Methyl-3-hexene (c) 4,7-Dimethyl-2,5-octadiene\n$7-4$ (d) 6-Ethyl-7-methyl-4-nonene\n7-4 (d) 6-Ethyl-7-methyl-4-nonene\nPROBLEM (a)\n7-5\n\n(b)\n\n(c)\n\n(d)"}
{"id": 2097, "contents": "Chapter 6 - \n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM (a) 1,2-Dimethylcyclohexene\n(b) 4,4-Dimethylcycloheptene (c) 3-Isopropylcyclopentene 7-6\nPROBLEM (a) 2,5,5-Trimethylhex-2-ene (b) 2,3-Dimethylcyclohexa-1,3-diene\n7-7\nPROBLEM\n7-8\n\n\nPROBLEM Compounds (c), (e), and (f) have cis-trans isomers.\n7-9\nPROBLEM (a) cis-4,5-Dimethyl-2-hexene (b) trans-6-Methyl-3-heptene 7-10\nPROBLEM\n(a) $-\\mathrm{CH}_{3}$\n(b) -Cl\n(c) $-\\mathrm{CH}=\\mathrm{CH}_{2}$\n(d) $-\\mathrm{OCH}_{3}$\n(e) $-\\mathrm{CH}=\\mathrm{O}$ (f) $-\\mathrm{CH}=\\mathrm{O}$\n\n7-11\nPROBLEM (a) $-\\mathrm{Cl},-\\mathrm{OH},-\\mathrm{CH}_{3},-\\mathrm{H}$ (b) $-\\mathrm{CH}_{2} \\mathrm{OH},-\\mathrm{CH}=\\mathrm{CH}_{2},-\\mathrm{CH}_{2} \\mathrm{CH}_{3},-\\mathrm{CH}_{3}$\n7-12 (c) $-\\mathrm{CO}_{2} \\mathrm{H},-\\mathrm{CH}_{2} \\mathrm{OH},-\\mathrm{C} \\equiv \\mathrm{N},-\\mathrm{CH}_{2} \\mathrm{NH}_{2}$ (d) $-\\mathrm{CH}_{2} \\mathrm{OCH}_{3},-\\mathrm{C} \\equiv \\mathrm{N},-\\mathrm{C} \\equiv \\mathrm{CH},-\\mathrm{CH}_{2} \\mathrm{CH}_{3}$\nPROBLEM (a) $Z$ (b) $E$ (c) $Z$ (d) $E$\n7-13\nPROBLEM\n7-14\n\nz"}
{"id": 2098, "contents": "Chapter 6 - \nz\n\nPROBLEM (a) 2-Methylpropene is more stable than 1-butene.\n7-15 (b) trans-2-Hexene is more stable than cis-2-hexene.\n(c) 1-Methylcyclohexene is more stable than 3-methylcyclohexene.\n\nPROBLEM (a) Chlorocyclohexane (b) 2-Bromo-2-methylpentane (c) 4-Methyl-2-pentanol\n7-16 (d) 1-Bromo-1-methylcyclohexane\nPROBLEM (a) Cyclopentene (b) 1-Ethylcyclohexene or ethylidenecyclohexane (c) 3-Hexene\n7-17 (d) Vinylcyclohexane (cyclohexylethylene)\nPROBLEM (a)\n7-18\n\n(b)\n\n\nPROBLEM In the conformation shown, only the methyl- group $\\mathrm{C}-\\mathrm{H}$ that is parallel to the carbocation $p$ orbital\n7-19 can show hyperconjugation.\nPROBLEM The second step is exergonic; the transition state resembles the carbocation.\n7-20\nPROBLEM\n7-21"}
{"id": 2099, "contents": "Chapter 8 - \nPROBLEM 2-Methyl-2-butene and 2-methyl-1-butene\n8-1\nPROBLEM Five\n\n8-2\nPROBLEM trans-1,2-Dichloro-1,2-dimethylcyclohexane\n8-3\nPROBLEM\n8-4\n\n\nPROBLEM trans-2-Bromocyclopentanol\n8-5\nPROBLEM Markovnikov\n8-6\nPROBLEM (a) 2-Pentanol (b) 2-Methyl-2-pentanol\n8-7\nPROBLEM (a) Oxymercuration of 2-methyl-1-hexene or 2-methyl-2-hexene\n8-8 (b) Oxymercuration of cyclohexylethylene or hydroboration of ethylidenecyclohexane\nPROBLEM (a)\n\n(b)\n\n\nPROBLEM (a) 3-Methyl-1-butene (b) 2-Methyl-2-butene (c) Methylenecyclohexane\n$\\mathbf{8 - 1 0}$\nPROBLEM\n\n8-11\n\n\nand\n\n(b) 1,1-Dimethylcyclopentane (c) tert-Butylcyclohexane 8-12\nPROBLEM\n8-13\ncis-2,3-Epoxybutane\n\nPROBLEM (a) 1-Methylcyclohexene (b) 2-Methyl-2-pentene (c) 1,3-Butadiene 8-14\n\nPROBLEM 8-15\n\nPROBLEM (a) 2-Methylpropen\n(b) 3-Hexene\n\n8-16\nPROBLEM (a)\n8-17\n\n(b)\n\n\nPROBLEM (a) $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHOCH}_{3}$\n(b) $\\mathrm{ClCH}=\\mathrm{CHCl}$\n\n8-18\nPROBLEM\n8-19"}
{"id": 2100, "contents": "Chapter 8 - \n8-18\nPROBLEM\n8-19\n\n\nPROBLEM An optically inactive, non-50:50 mixture of two racemic pairs: $(2 R, 4 R)+(2 S, 4 S)$ and $(2 R, 4 S)+$\n8-20 $(2 S, 4 R)$\nPROBLEM Non-50:50 mixture of two racemic pairs: $(1 S, 3 R)+(1 R, 3 S)$ and $(1 S, 3 S)+(1 R, 3 R)$"}
{"id": 2101, "contents": "Chapter 9 - \nPROBLEM (a) 2,5-Dimethyl-3-hexyne (b) 3,3-Dimethyl-1-butyne (c) 3,3-Dimethyl-4-octyne\n9-1 (d) 2,5,5-Trimethyl-3-heptyne (e) 2,4-Octadiene-6-yne\nPROBLEM 1-Hexyne, 2-hexyne, 3-hexyne, 3-methyl-1-pentyne, 4-methyl-1-pentyne, 4-methyl-2-pentyne, 9-2 3,3-dimethyl-1-butyne\nPROBLEM (a) 1,1,2,2-Tetrachloropentane (b) 1-Bromo-1-cyclopentylethylene\n9-3 (c) 2-Bromo-2-heptene and 3-bromo-2-heptene\nPROBLEM (a) 4-Octanone (b) 2-Methyl-4-octanone and 7-methyl-4-octanone 9-4\nPROBLEM (a) 1-Pentyne (b) 2-Pentyne 9-5\nPROBLEM (a) $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{C} \\equiv \\mathrm{CH}$ (b) 2,5-Dimethyl-3-hexyne 9-6\nPROBLEM (a) Mercuric sulfate-catalyzed hydration of phenylacetylene\n9-7 (b) Hydroboration/oxidation of cyclopentylacetylene\nPROBLEM (a) Reduce 2-octyne with Li/ $\\mathrm{NH}_{3}$. (b) Reduce 3-heptyne with $\\mathrm{H}_{2}$ /Lindlar catalyst.\n9-8 (c) Reduce 3-methyl-1-pentyne.\nPROBLEM No: (a), (c), (d); yes: (b)\n9-9"}
{"id": 2102, "contents": "Chapter 9 - \n9-8 (c) Reduce 3-methyl-1-pentyne.\nPROBLEM No: (a), (c), (d); yes: (b)\n9-9\nPROBLEM (a) 1-Pentyne $+\\mathrm{CH}_{3} \\mathrm{I}$, or propyne $+\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{I}$ (b) 3-Methyl-1-butyne $+\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{I}$ 9-10 (c) Cyclohexylacetylene $+\\mathrm{CH}_{3} \\mathrm{I}$\nPROBLEM\n9-11\n$\\mathrm{CH}_{3} \\mathrm{C} \\equiv \\mathrm{CH} \\xrightarrow[\\text { 2. } \\mathrm{CH}_{3} \\mathrm{I}]{\\text { 1. } \\mathrm{NaNH}_{2}} \\quad \\mathrm{CH}_{3} \\mathrm{C} \\equiv \\mathrm{CCH}_{3} \\xrightarrow[\\begin{array}{c}\\text { Lindlar } \\\\ \\text { cat. }\\end{array}]{\\mathrm{H}_{2}}$ cis $-\\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{CHCH}_{3}$\nPROBLEM (a) $\\mathrm{KMnO}_{4}, \\mathrm{H}_{3} \\mathrm{O}^{+}$(b) $\\mathrm{H}_{2} /$ Lindlar (c) 1. $\\mathrm{H}_{2} /$ Lindlar; 2. HBr\n9-12 (d) 1. $\\mathrm{H}_{2} /$ Lindlar; 2. $\\mathrm{BH}_{3} ; 3 . \\mathrm{NaOH}, \\mathrm{H}_{2} \\mathrm{O}_{2}$ (e) 1. $\\mathrm{H}_{2} /$ Lindlar; $2 . \\mathrm{Cl}_{2}$ (f) $\\mathrm{O}_{3}$"}
{"id": 2103, "contents": "Chapter 9 - \nPROBLEM (a) 1. $\\mathrm{HC} \\equiv \\mathrm{CH}+\\mathrm{NaNH}_{2} ; 2 . \\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{6} \\mathrm{CH}_{2} \\mathrm{Br} ; 3.2 \\mathrm{H}_{2} / \\mathrm{Pd}$\n9-13 (b) $\\mathrm{HC} \\equiv \\mathrm{CH}+\\mathrm{NaNH}_{2} ; 2 .\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CCH}_{2} \\mathrm{CH}_{2} \\mathrm{I} ; 3.2 \\mathrm{H}_{2} / \\mathrm{Pd}$\n(c) 1. $\\mathrm{HC} \\equiv \\mathrm{CH}+\\mathrm{NaNH}_{2} ; 2 . \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{I} ; 3 . \\mathrm{BH}_{3} ; 4 . \\mathrm{H}_{2} \\mathrm{O}_{2}$\n(d) 1. $\\mathrm{HC} \\equiv \\mathrm{CH}+\\mathrm{NaNH}_{2} ; 2 . \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{I} ; 3 . \\mathrm{HgSO}_{4}, \\mathrm{H}_{3} \\mathrm{O}^{+}$"}
{"id": 2104, "contents": "Chapter 9 - \nChapter 10\nPROBLEM (a) 1-Iodobutane (b) 1-Chloro-3-methylbutane (c) 1,5-Dibromo-2,2-dimethylpentane\n10-1 (d) 1,3-Dichloro-3-methylbutane (e) 1-Chloro-3-ethyl-4-iodopentane\n(f) 2-Bromo-5-chlorohexane\n\nPROBLEM (a)\n10-2\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n\nPROBLEM Chiral: 1-chloro-2-methylpentane, 3-chloro-2-methylpentane, 2-chloro-4-methylpentane Achiral:\n10-3 2-chloro-2-methylpentane, 1-chloro-4-methylpentane\n\n\n\n\n$\\mathrm{ClCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}\\left(\\mathrm{CH}_{3}\\right)_{2}$\nPROBLEM 1-Chloro-2-methylbutane (29\\%), 1-chloro-3-methylbutane (14\\%), 2-chloro-2-methylbutane (24\\%),\n10-4 2-chloro-3-methylbutane (33\\%)\nPROBLEM\n10-5"}
{"id": 2105, "contents": "Chapter 9 - \nPROBLEM The intermediate allylic radical reacts at the more accessible site and gives the more highly 10-6 substituted double bond.\nPROBLEM (a) 3-Bromo-5-methylcycloheptene and 3-bromo-6-methylcycloheptene (b) Four products 10-7\nPROBLEM (a) 2-Methyl-2-propanol +HCl (b) 4-Methyl-2-pentanol + $\\mathrm{PBr}_{3}$\n10-8 (c) 5-Methyl-1-pentanol $+\\mathrm{PBr}_{3}$ (d) 3,3-Dimethyl-cyclopentanol + HF, pyridine\nPROBLEM Both reactions occur\n10-9\nPROBLEM React Grignard reagent with $\\mathrm{D}_{2} \\mathrm{O}$.\n10-10\nPROBLEM (a) 1. NBS; 2. $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CuLi}$ (b) 1. Li; 2. $\\mathrm{CuI} ; 3 . \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{Br}$\n10-11 (c) 1. $\\mathrm{BH}_{3} ; 2 . \\mathrm{H}_{2} \\mathrm{O}_{2}, \\mathrm{NaOH} ; 3 . \\mathrm{PBr}_{3} ; 4$. Li, then $\\mathrm{CuI} ; 5 . \\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{4} \\mathrm{Br}$\nPROBLEM (a)\n10-12"}
{"id": 2106, "contents": "Chapter 9 - \n(b)\n$\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{NH}_{2}<\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}<\\mathrm{CH}_{3} \\mathrm{C} \\equiv \\mathrm{N}$\nPROBLEM (a) Reduction (b) Neither\n10-13\nChapter 11\nPROBLEM ( $R$ )-1-Methylpentyl acetate, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{CH}\\left(\\mathrm{CH}_{3}\\right) \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$\n11-1\nPROBLEM (S)-2-Butanol\n11-2\nPROBLEM\n(S)-2-Bromo-4-methylpentane $\\longrightarrow$\n\n11-3\n\n\nPROBLEM (a) 1-Iodobutane (b) 1-Butanol (c) 1-Hexyne (d) Butylammonium bromide\n11-4\nPROBLEM\n(a) $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{~N}^{-}$\n(b) $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{~N}$\n(c) $\\mathrm{H}_{2} \\mathrm{~S}$\n\n11-5\nPROBLEM $\\mathrm{CH}_{3} \\mathrm{OTos}>\\mathrm{CH}_{3} \\mathrm{Br}>\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CHCl}>\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CCl}$\n11-6\nPROBLEM Similar to protic solvents\n11-7\nPROBLEM Racemic 1-ethyl-1-methylhexyl acetate\n\n11-8\nPROBLEM 90.1\\% racemization, 9.9\\% inversion\n11-9\nPROBLEM\n11-10"}
{"id": 2107, "contents": "Chapter 9 - \n11-8\nPROBLEM 90.1\\% racemization, 9.9\\% inversion\n11-9\nPROBLEM\n11-10\n\n\nPROBLEM $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}(\\mathrm{Br}) \\mathrm{CH}_{3}>\\mathrm{CH}_{3} \\mathrm{CH}(\\mathrm{Br}) \\mathrm{CH}_{3}>\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{Br}>\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHBr}$\n11-11\nPROBLEM The same allylic carbocation intermediate is formed.\n11-12\nPROBLEM (a) $\\mathrm{S}_{\\mathrm{N}} 1$ (b) $\\mathrm{S}_{\\mathrm{N}} 2$\n11-13\nPROBLEM\n11-14\n\n\nLinalyl diphosphate\n\n\nPROBLEM (a) Major: 2-methyl-2-pentene; minor: 4-methyl-2-pentene\n11-15 (b) Major: 2,3,5-trimethyl-2-hexene; minor: 2,3,5-trimethyl-3-hexene and 2-isopropyl-4-methyl-1-pentene\n(c) Major: ethylidenecyclohexane; minor: cyclohexylethylene\n\nPROBLEM\n(a) 1-Bromo-3,6-dimethylheptane (b) 4-Bromo-1,2-dimethylcyclopentane"}
{"id": 2108, "contents": "Chapter 9 - \nPROBLEM\n(a) 1-Bromo-3,6-dimethylheptane (b) 4-Bromo-1,2-dimethylcyclopentane\n\n11-16\nPROBLEM (Z)-1-Bromo-1,2-diphenylethylene\n11-17\nPROBLEM (Z)-3-Methyl-2-pentene\n11-18\nPROBLEM Cis isomer reacts faster because the bromine is axial.\n11-19\nPROBLEM (a) $\\mathrm{S}_{\\mathrm{N}} 2$ (b) E2 (c) $\\mathrm{S}_{\\mathrm{N}} 1$ (d) E1cB\n11-20\nChapter 12\nPROBLEM $\\mathrm{C}_{19} \\mathrm{H}_{28} \\mathrm{O}_{2}$\n12-1\nPROBLEM (a) 2-Methyl-2-pentene (b) 2-Hexene\n12-2\nPROBLEM (a) 43,71 (b) 82 (c) 58 (d) 86\n12-3\nPROBLEM $102\\left(\\mathrm{M}^{+}\\right)$, 84 (dehydration), 87 (alpha cleavage), 59 (alpha cleavage)\n12-4\nPROBLEM X-ray energy is higher; $\\lambda=9.0 \\times 10^{-6} \\mathrm{~m}$ is higher in energy."}
{"id": 2109, "contents": "12-5 - \nPROBLEM (a) $\\left.2.4 \\times 10^{6} \\mathrm{~kJ} / \\mathrm{mol}\\right]$ (b) $4.0 \\times 10^{4} \\mathrm{~kJ} / \\mathrm{mol}$ (c) $2.4 \\times 10^{3} \\mathrm{~kJ} / \\mathrm{mol}$ (d) $2.8 \\times 10^{2} \\mathrm{~kJ} / \\mathrm{mol}$\n$12-6$ (e) $6.0 \\mathrm{~kJ} / \\mathrm{mol}$ (f) $4.0 \\times 10^{-2} \\mathrm{~kJ} / \\mathrm{mol}$\nPROBLEM (a) Ketone or aldehyde (b) Nitro compound (c) Carboxylic acid\n12-7\nPROBLEM (a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$ has an -OH absorption. (b) 1-Hexene has a double-bond absorption.\n12-8 (c) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$ has a very broad -OH absorption.\nPROBLEM $1450-1600 \\mathrm{~cm}^{-1}$ : aromatic ring; $2100 \\mathrm{~cm}^{-1}: \\mathrm{C} \\equiv \\mathrm{C} ; 3300 \\mathrm{~cm}^{-1}: \\mathrm{C} \\equiv \\mathrm{C}-\\mathrm{H}$\n12-9\nPROBLEM (a) $1715,1640,1250 \\mathrm{~cm}^{-1}$ (b) $1730,2100,3300 \\mathrm{~cm}^{-1}$\n12-10 (c) $1720,2500-3100,3400-3650 \\mathrm{~cm}^{-1}$\nPROBLEM 1690, 1650, $2230 \\mathrm{~cm}^{-1}$\n12-11"}
{"id": 2110, "contents": "Chapter 13 - \nPROBLEM $7.5 \\times 10^{-5} \\mathrm{~kJ} / \\mathrm{mol}$ for ${ }^{19} \\mathrm{~F} ; 8.0 \\times 10^{-5} \\mathrm{~kJ} / \\mathrm{mol}$ for ${ }^{1} \\mathrm{H}$\n13-1\nPROBLEM $1.2 \\times 10^{-4} \\mathrm{~kJ} / \\mathrm{mol}$\n13-2\nPROBLEM The vinylic $\\mathrm{C}-\\mathrm{H}$ protons are nonequivalent.\n$13-3$ b\n\n\nPROBLEM (a) $7.27 \\delta$ (b) $3.05 \\delta$ (c) $3.46 \\delta$ (d) $5.30 \\delta$\n13-4\nPROBLEM (a) 420 Hz (b) $2.1 \\delta$ (c) 1050 Hz\n13-5\nPROBLEM\n(a) $1.43 \\delta$ (b) $2.17 \\delta$ (c) $7.37 \\delta$ (d) $5.30 \\delta$ (e) $9.70 \\delta$ (f) $2.12 \\delta$\n\n13-6\nPROBLEM There are seven kinds of protons labeled. The types and expected range of absorption of each\n13-7 follow. a: ether, 3.5-4.5 $\\delta$; b: aryl, 6.5-8.0 $\\delta$; $\\boldsymbol{c}$ : aryl, 6.5-8.0; $\\boldsymbol{d}$ : vinylic, 4.5-6.5 $\\delta$; e: vinylic, 4.5-6.5 $\\delta$; $\\boldsymbol{f}$ : alkyl (secondary), 1.2-1.6 $\\delta$; $\\boldsymbol{g}$ : alkyl (primary), 0.7-1.3 $\\delta$."}
{"id": 2111, "contents": "Chapter 13 - \nPROBLEM Two peaks; 3:2 ratio\n13-8\nPROBLEM (a) $-\\mathrm{CHBr}_{2}$, quartet; $-\\mathrm{CH}_{3}$, doublet (b) $\\mathrm{CH}_{3} \\mathrm{O}-$, singlet; $-\\mathrm{OCH}_{2}-$, triplet; $-\\mathrm{CH}_{2} \\mathrm{Br}$, triplet\n13-9 (c) $\\mathrm{ClCH}_{2}-$, triplet; $-\\mathrm{CH}_{2}{ }^{-}$, quintet\n(d) $\\mathrm{CH}_{3}{ }^{-}$, triplet; $-\\mathrm{CH}_{2}{ }^{-}$, quartet; $-\\mathrm{CH}-$, septet; $\\left(\\mathrm{CH}_{3}\\right)_{2}$, doublet\n(e) $\\mathrm{CH}_{3}-$, triplet; $-\\mathrm{CH}_{2}-$, quartet; - $\\mathrm{CH}-$, septet; $\\left(\\mathrm{CH}_{3}\\right)_{2}$, doublet\n(f) $=\\mathrm{CH}$, triplet, $-\\mathrm{CH}_{2}-$, doublet, aromatic $\\mathrm{C}-\\mathrm{H}$, two multiplets"}
{"id": 2112, "contents": "Chapter 13 - \nPROBLEM (a) $\\mathrm{CH}_{3} \\mathrm{OCH}_{3}$ (b) $\\mathrm{CH}_{3} \\mathrm{CH}(\\mathrm{Cl}) \\mathrm{CH}_{3}$ (c) $\\mathrm{ClCH}_{2} \\mathrm{CH}_{2} \\mathrm{OCH}_{2} \\mathrm{CH}_{2} \\mathrm{Cl}$\n13-10 (d) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{CH}_{3}$ or $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$\nPROBLEM $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OCH}_{2} \\mathrm{CH}_{3}$\n13-11\nPROBLEM (a) Enantiotopic (b) Diastereotopic (c) Diastereotopic (d) Diastereotopic\n13-12 (e) Diastereotopic (f) Homotopic\nPROBLEM (a) 2 (b) 4 (c) 3 (d) 4 (e) 5 (f) 3\n\n13-13\nPROBLEM 4\n13-14\nPROBLEM $J_{1-2}=16 \\mathrm{~Hz} ; J_{2-3}=8 \\mathrm{~Hz}$\n13-15"}
{"id": 2113, "contents": "Chapter 13 - \n13-13\nPROBLEM 4\n13-14\nPROBLEM $J_{1-2}=16 \\mathrm{~Hz} ; J_{2-3}=8 \\mathrm{~Hz}$\n13-15\n\n\nun\nPROBLEM 1-Chloro-1-methylcyclohexane has a singlet methyl absorption.\n13-16\nPROBLEM (a) 4 (b) 7 (c) 4 (d) 5 (e) 5 (f) 7\n13-17\nPROBLEM (a) 1,3-Dimethylcyclopentene (b) 2-Methylpentane (c) 1-Chloro-2-methylpropane\n13-18\nPROBLEM $-\\mathrm{CH}_{3}, 9.3 \\delta ;-\\mathrm{CH}_{2}-27.6 \\delta$; $\\mathrm{C}=\\mathrm{O}, 174.6 \\delta ;-\\mathrm{OCH}_{3}, 51.4 \\delta$\n13-19\nPROBLEM\n13-20\n\n\nPROBLEM\n\n\nPROBLEM\n13-22"}
{"id": 2114, "contents": "Chapter 13 - \nPROBLEM\n\n\nPROBLEM\n13-22\n\n\nPROBLEM A DEPT-90 spectrum would show two absorptions for the non-Markovnikov product ( $\\mathrm{RCH}=\\mathrm{CHBr}$ )\n13-23 but no absorptions for the Markovnikov product $\\left(\\mathrm{RBrC}=\\mathrm{CH}_{2}\\right)$.\nChapter 14\nPROBLEM Expected $\\Delta H^{\\circ}{ }_{\\text {hydrog }}$ for allen is $-252 \\mathrm{~kJ} / \\mathrm{mol}$. Allen is less stable than a nonconjugated dene,\n14-1 which is less stable than a conjugated dene.\nPROBLEM 1-Chloro-2-pentene, 3-chloro-1-pentene, 4-chloro-2-pentene\n14-2\nPROBLEM 4-Chloro-2-pentene predominates in both.\n14-3\nPROBLEM 1,2 Addition: 6-bromo-1,6-dimethylcyclohexene 1,4 Addition: 3-bromo-1,2-dimethylcyclohexene 14-4\nPROBLEM Interconversion occurs by $\\mathrm{S}_{\\mathrm{N}} 1$ dissociation to a common intermediate cation.\n14-5\nPROBLEM The double bond is more highly substituted.\n14-6\n\nPROBLEM\n\n\nPROBLEM Good dienophiles: (a), (d)\n14-8\nPROBLEM Compound (a) is $s$-cis. Compound (c) can rotate to $s$-cis.\n14-9\nPROBLEM\n14-10\n\n\nPROBLEM\n14-11\n\n\nPROBLEM\n\n\nPROBLEM $300-600 \\mathrm{~kJ} / \\mathrm{mol}$; UV energy is greater than IR or NMR energy.\n14-13\nPROBLEM $1.46 \\times 10^{-5} \\mathrm{M}$\n14-14\nPROBLEM All except (a) have UV absorptions.\n14-15"}
{"id": 2115, "contents": "Chapter 15 - \nPROBLEM (a) Meta (b) Para (c) Ortho\n15-1\nPROBLEM (a) $m$-Bromochlorobenzene (b) (3-Methylbutyl)benzene (c) $p$-Bromoaniline\n15-2 (d) 2,5-Dichlorotoluene (e) 1-Ethyl-2,4-dinitrobenzene (f) 1,2,3,5-Tetramethylbenzene\nPROBLEM (a)\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM Pyridine has an aromatic sextet of electrons.\n15-4\n\n\nPyridine\n\nPROBLEM Cyclodecapentaene is not flat because of steric interactions.\n15-5\nPROBLEM All $\\mathrm{C}-\\mathrm{C}$ bonds are equivalent; one resonance line in both ${ }^{1} \\mathrm{H}$ and ${ }^{13} \\mathrm{C}$ NMR spectra.\n15-6\nPROBLEM The cyclooctatetraenyl dianion is aromatic (ten $\\pi$ electrons) and flat.\n\n| 15-7 <br> PROBLEM <br> 15-8 <br> 15 | - |  |  |  |  |\n| ---: | :---: | :---: | :---: | :---: | :---: |\n|  |  | - | - | - | - |\n|  | $\\uparrow \\downarrow$ |  | $\\uparrow$ | $\\uparrow$ | $\\uparrow \\downarrow$ |"}
{"id": 2116, "contents": "PROBLEM - \n15-9\n\n\nFuran\n\nPROBLEM The thiazolium ring has six $\\pi$ electrons.\n15-10\n\n\nPROBLEM Yes, it's aromatic.\n15-11\n\n\nPROBLEM The three nitrogens in double bonds each contribute one; the remaining nitrogen contributes two. 15-12"}
{"id": 2117, "contents": "Chapter 16 - \nPROBLEM $o-$, $m$-, and $p$-Bromotoluene\n16-1\nPROBLEM\n16-2\n\n\nPROBLEM $o$-xylene: $2 ; p$-xylene: $1 ; m$-xylene: 3\n16-3\nPROBLEM $\\mathrm{D}^{+}$does electrophilic substitutions on the ring.\n16-4\nPROBLEM No rearrangement: (a), (b), (e)\n16-5\nPROBLEM tert-Butylbenzene\n16-6\nPROBLEM (a) $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CHCOCl}$ (b) PhCOCl\n16-7\nPROBLEM (a) Phenol $>$ Toluene $>$ Benzene $>$ Nitrobenzene\n16-8 (b) Phenol $>$ Benzene $>$ Chlorobenzene $>$ Benzoic acid\n(c) Aniline $>$ Benzene $>$ Bromobenzene $>$ Benzaldehyde\n\nPROBLEM (a) $o$ - and $p$-Bromonitrobenzene (b) $m$-Bromonitrobenzene (c) $o$ - and $p$-Chlorophenol\n16-9 (d) $o$ - and $p$-Bromoaniline\nPROBLEM Alkylbenzenes are more reactive than benzene itself, but acylbenzenes are less reactive. 16-10"}
{"id": 2118, "contents": "Chapter 16 - \nPROBLEM Toluene is more reactive; the trifluoromethyl group is electron-withdrawing. 16-11\nPROBLEM The nitrogen electrons are donated to the nearby carbonyl group by resonance and are less 16-12 available to the ring.\nPROBLEM The meta intermediate is most favored. 16-13\nPROBLEM\n(a) Ortho and para to $-\\mathrm{OCH}_{3}$\n(b) Ortho and para to $-\\mathrm{NH}_{2}$\n(c) Ortho and para to -Cl 16-14\nPROBLEM (a) Reaction occurs ortho and para to the $-\\mathrm{CH}_{3}$ group.\n16-15 (b) Reaction occurs ortho and para to the $-\\mathrm{OCH}_{3}$ group.\nPROBLEM The phenol is deprotonated by KOH to give an anion that carries out a nucleophilic acyl substitution 16-16 reaction on the fluoronitrobenzene.\nPROBLEM Only one benzyne intermediate can form from $p$-bromotoluene; two different benzyne 16-17 intermediates can form from $m$-bromotoluene.\nPROBLEM (a) m-Nitrobenzoic acid (b) p-tert-Butylbenzoic acid 16-18\nPROBLEM A benzyl radical is more stable than a primary alkyl radical by $52 \\mathrm{~kJ} / \\mathrm{mol}$ and is similar in stability 16-19 to an allyl radical.\nPROBLEM 1. $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{Cl}, \\mathrm{AlCl}_{3} ; 2$. NBS; 3. KOH , ethanol 16-20\nPROBLEM 1. $\\mathrm{PhCOCl}, \\mathrm{AlCl}_{3} ; 2 . \\mathrm{H}_{2} / \\mathrm{Pd}$ 16-21"}
{"id": 2119, "contents": "Chapter 16 - \nPROBLEM 1. $\\mathrm{PhCOCl}, \\mathrm{AlCl}_{3} ; 2 . \\mathrm{H}_{2} / \\mathrm{Pd}$ 16-21\nPROBLEM (a) 1. $\\mathrm{HNO}_{3}, \\mathrm{H}_{2} \\mathrm{SO}_{4} ; 2 . \\mathrm{Cl}_{2}, \\mathrm{FeCl}_{3}$ (b) 1. $\\mathrm{CH}_{3} \\mathrm{COCl}, \\mathrm{AlCl}_{3} ; 2 . \\mathrm{Cl}_{2}, \\mathrm{FeCl}_{3} ; 3 . \\mathrm{H} 2 / \\mathrm{Pd}$\n16-22 (c) 1. $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COCl}, \\mathrm{AlCl}_{3} ; 2 . \\mathrm{Cl}_{2}, \\mathrm{FeCl}_{3} ; 3 . \\mathrm{H}_{2} / \\mathrm{Pd} ; 4 . \\mathrm{HNO}_{3}, \\mathrm{H}_{2} \\mathrm{SO}_{4}$\n(d) 1. $\\mathrm{CH}_{3} \\mathrm{Cl}, \\mathrm{AlCl}_{3} ; 2 . \\mathrm{Br}_{2}, \\mathrm{FeBr}_{3} ; 3 . \\mathrm{SO}_{3}, \\mathrm{H}_{2} \\mathrm{SO}_{4}$"}
{"id": 2120, "contents": "Chapter 16 - \nPROBLEM (a) Friedel-Crafts acylation does not occur on a deactivated ring.\n16-23 (b) Rearrangement occurs during Friedel-Crafts alkylation with primary halides; chlorination occurs ortho to the alkyl group.\n\nChapter 17\nPROBLEM (a) 5-Methyl-2,4-hexanediol (b) 2-Methyl-4-phenyl-2-butanol (c) 4,4-Dimethylcyclohexanol\n17-1 (d) trans-2-Bromocyclopentanol (e) 4-Bromo-3-methylphenol (f) 2-Cyclopenten-1-ol\nPROBLEM\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n\nPROBLEM Hydrogen-bonding is more difficult in hindered alcohols.\n17-3\nPROBLEM (a) $\\mathrm{HC} \\equiv \\mathrm{CH}<\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CHOH}<\\mathrm{CH}_{3} \\mathrm{OH}<\\left(\\mathrm{CF}_{3}\\right)_{2} \\mathrm{CHOH}$\n17-4 (b) $p$-Methylphenol < Phenol < p-(Trifluoromethyl)phenol\n(c) Benzyl alcohol < Phenol < p-Hydroxybenzoic acid\n\nPROBLEM The electron-withdrawing nitro group stabilizes an alkoxide ion, but the electron-donating\n17-5 methoxyl group destabilizes the anion.\nPROBLEM (a) 2-Methyl-3-pentanol (b) 2-Methyl-4-phenyl-2-butanol (c) meso-5,6-Decanediol 17-6\nPROBLEM (a) $\\mathrm{NaBH}_{4}$ (b) $\\mathrm{LiAlH}_{4}$ (c) $\\mathrm{LiAlH}_{4}$"}
{"id": 2121, "contents": "Chapter 16 - \n17-7\nPROBLEM (a) Benzaldehyde or benzoic acid (or ester) (b) Acetophenone (c) Cyclohexanone\n17-8 (d) 2-Methylpropanal or 2-methylpropanoic acid (or ester)\nPROBLEM (a) 1-Methylcyclopentanol (b) 1,1-Diphenylethanol (c) 3-Methyl-3-hexanol\n17-9\nPROBLEM (a)\n17-10\n\nor\n\n\n2-Methyl-2-propanol\n\n(b)\n\n\n1-Methylcyclohexanol\n(c)\n\nor\n\nor\n\n(d)\n\nor\n\n\n(e)\n\n(f)"}
{"id": 2122, "contents": "Chapter 16 - \nor\n\n\n2-Methyl-2-propanol\n\n(b)\n\n\n1-Methylcyclohexanol\n(c)\n\nor\n\nor\n\n(d)\n\nor\n\n\n(e)\n\n(f)\n\n\nPROBLEM Cyclohexanone $+\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{MgBr}$ 17-11\nPROBLEM 1. $p$-TosCl, pyridine; 2. NaCN 17-12\nPROBLEM (a) 2-Methyl-2-pentene (b) 3-Methylcyclohexene (c) 1-Methylcyclohexene 17-13 (d) 2,3-Dimethyl-2-pentene (e) 2-Methyl-2-pentene\nPROBLEM (a) 1-Phenylethanol (b) 2-Methyl-1-propanol (c) Cyclopentanol 17-14\nPROBLEM (a) Hexanoic acid, hexanal (b) 2-Hexanone (c) Hexanoic acid, no reaction 17-15\nPROBLEM $\\mathrm{S}_{\\mathrm{N}} 2$ reaction of $\\mathrm{F}^{-}$on silicon with displacement of alkoxide ion. 17-16\nPROBLEM Protonation of 2-methylpropene gives the tert-butyl cation, which carries out an electrophilic 17-17 aromatic substitution reaction.\nPROBLEM Disappearance of -OH absorption; appearance of $\\mathrm{C}=\\mathrm{O}$ 17-18\nPROBLEM (a) Singlet (b) Doublet (c) Triplet (d) Doublet (e) Doublet (f) Singlet 17-19"}
{"id": 2123, "contents": "Chapter 18 - \nPROBLEM (a) Diisopropyl ether (b) Cyclopentyl propyl ether\n18-1 (c) $p$-Bromoanisole or 4-bromo-1-methoxybenzene\n(d) 1-Methoxycyclohexene\n(e) Ethyl isobutyl ether (f) Allyl vinyl ether\n\nPROBLEM A mixture of diethyl ether, dipropyl ether, and ethyl propyl ether is formed in a $1: 1: 2$ ratio. 18-2\nPROBLEM (a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{O}^{-}+\\mathrm{CH}_{3} \\mathrm{Br}$ (b) $\\mathrm{PhO}^{-}+\\mathrm{CH}_{3} \\mathrm{Br}$ (c) $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CHO}^{-}+\\mathrm{PhCH}_{2} \\mathrm{Br}$\n18-3 (d) $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CCH}_{2} \\mathrm{O}^{-}+\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{Br}$\nPROBLEM\n18-4\n\n\n\nPROBLEM (a) Either method (b) Williamson (c) Alkoxymercuration (d) Williamson 18-5\nPROBLEM (a) Bromoethane $>$ 2-Bromopropane $>$ Bromobenzene\n18-6 (b) Bromoethane $>$ Chloroethane $>$ 1-Iodopropene\n\nPROBLEM (a)\n18-7\n\n(b)\n\n\nPROBLEM Protonation of the oxygen atom, followed by E1 reaction\n18-8\nPROBLEM $\\mathrm{Br}^{-}$and $\\mathrm{I}^{-}$are better nucleophiles than $\\mathrm{Cl}^{-}$.\n18-9\nPROBLEM\n18-10\n\n\nPROBLEM Epoxidation of cis-2-butene yields cis-2,3-epoxybutane, while epoxidation of trans-2-butene yields 18-11 trans-2,3-epoxybutane.\nPROBLEM (a)\n\n(b)"}
{"id": 2124, "contents": "Chapter 18 - \n(b)\n\n18-12\n\n\n\nPROBLEM (a) 1-Methylcyclohexene $+\\mathrm{OsO}_{4}$; then $\\mathrm{NaHSO}_{3}$\n18-13 (b) 1-Methylcyclohexene + m-chloroperoxybenzoic acid, then $\\mathrm{H}_{3} \\mathrm{O}^{+}$\nPROBLEM (a)\n\n(b)\n\n(c)\n\n\nPROBLEM\n18-15\n\n\n\nPROBLEM (a) 2-Butanethiol (b) 2,2,6-Trimethyl-4-heptanethiol (c) 2-Cyclopentene-1-thiol\n18-16 (d) Ethyl isopropyl sulfide (e) o-Di(methylthio)benzene (f) 3-(Ethylthio)cyclohexanone\nPROBLEM (a) 1. $\\mathrm{LiAlH}_{4} ; 2 . \\mathrm{PBr}_{3} ; 3 .\\left(\\mathrm{H}_{2} \\mathrm{~N}\\right)_{2} \\mathrm{C}=\\mathrm{S} ; 4 . \\mathrm{H}_{2} \\mathrm{O}, \\mathrm{NaOH}$\n18-17 (b) 1. HBr ; 2. $\\left(\\mathrm{H}_{2} \\mathrm{~N}\\right)_{2} \\mathrm{C}=\\mathrm{S} ; 3 . \\mathrm{H}_{2} \\mathrm{O}, \\mathrm{NaOH}$\nPROBLEM 1,2-Epoxybutane\n18-18\nPROBLEM Acetyl chloride is more electrophilic than acetone.\n18-69\nPROBLEM\n18-70\n\n\nPROBLEM (a) Nucleophilic acyl substitution (b) Nucleophilic addition (c) Carbonyl condensation 18-71"}
{"id": 2125, "contents": "Chapter 19 - \nPROBLEM (a) 2-Methyl-3-pentanone (b) 3-Phenylpropanal (c) 2,6-Octanedione\n19-1 (d) trans-2-Methylcyclohexanecarbaldehyde (e) 4-Hexenal\n(f) cis-2,5-Dimethylcyclohexanone\n\nPROBLEM (a)\n19-2\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n\nPROBLEM (a) Dess-Martin periodinane (b) $1 . \\mathrm{O}_{3} ; 2 \\mathrm{Zn}$ (c) DIBAH\n19-3 (d) 1. $\\mathrm{BH}_{3}$, then $\\mathrm{H}_{2} \\mathrm{O}_{2}, \\mathrm{NaOH}$; 2. Dess-Martin periodinane\nPROBLEM (a) $\\mathrm{HgSO}_{4}, \\mathrm{H}_{3} \\mathrm{O}^{+}$(b) $1 . \\mathrm{CH}_{3} \\mathrm{COCl}, \\mathrm{AlCl}_{3} ; 2 . \\mathrm{Br}_{2}, \\mathrm{FeBr}_{3}$ (c) 1. $\\mathrm{Mg} ; 2 . \\mathrm{CH}_{3} \\mathrm{CHO} ; 3 . \\mathrm{H}_{3} \\mathrm{O}^{+} ; 4 . \\mathrm{CrO}_{3}$\n19-4 (d) 1. $\\mathrm{BH}_{3} ; 2 . \\mathrm{H}_{2} \\mathrm{O}_{2}, \\mathrm{NaOH} ; 3 . \\mathrm{CrO}_{3}$\nPROBLEM\n19-5\n\n\nPROBLEM The electron-withdrawing nitro group in $p$-nitrobenzaldehyde polarizes the carbonyl group.\n19-6\nPROBLEM $\\mathrm{CCl}_{3} \\mathrm{CH}(\\mathrm{OH})_{2}$\n19-7\nPROBLEM Labeled water adds reversibly to the carbonyl group.\n19-8\nPROBLEM The equilibrium is unfavorable for sterically hindered ketones.\n19-9\nPROBLEM\n19-10\n\nand"}
{"id": 2126, "contents": "Chapter 19 - \nand\n\n\nPROBLEM The steps are the exact reverse of the forward reaction shown in Figure 19.7.\n19-11\nPROBLEM\n19-12\n\n\nPROBLEM (a) $\\mathrm{H}_{2} / \\mathrm{Pd}$ (b) $\\mathrm{N}_{2} \\mathrm{H}_{4}, \\mathrm{KOH}$ (c) 1. $\\mathrm{H}_{2} / \\mathrm{Pd} ; 2 . \\mathrm{N}_{2} \\mathrm{H}_{4}, \\mathrm{KOH}$\n19-13\nPROBLEM The mechanism is identical to that between a ketone and 2 equivalents of a monoalcohol, shown in\n19-14 Figure 19.11.\nPROBLEM\n19-15\n\n\nPROBLEM\n(a) Cyclohexanone $+(\\mathrm{Ph})_{3} \\mathrm{P}=\\mathrm{CHCH}_{3}$\n(b) Cyclohexanecarbaldehyde $+(\\mathrm{Ph})_{3} \\mathrm{P}=\\mathrm{CH}_{2}$\n\n19-16\n(e) Acetone $+(\\mathrm{Ph})_{3} \\mathrm{P}=\\mathrm{CHCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$ (d) Acetone $+(\\mathrm{Ph})_{3} \\mathrm{P}=\\mathrm{CHPh}$\n(f) 2-Cyclohexenone $+(\\mathrm{Ph})_{3} \\mathrm{P}=\\mathrm{CH}_{2}$\n\nPROBLEM\n19-17\n\n$\\beta$-Carotene\nPROBLEM Intramolecular Cannizzaro reaction\n19-18\nPROBLEM Addition of the pro- $R$ hydrogen of NADH takes place on the Re face of pyruvate.\n19-19\n\nPROBLEM The -OH group adds to the Re face at C 2 , and -H adds to the Re face at C 3 , to yield $(2 R, 3 S)$-isocitrate.\n19-20\nPROBLEM\n19-21"}
{"id": 2127, "contents": "Chapter 19 - \nPROBLEM (a) 3-Buten-2-one $+\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2}\\right)_{2} \\mathrm{CuLi}$ (b) 3-Methyl-2-cyclohexenone $+\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CuLi}$\n19-22 (c) 4-tert-Butyl-2-cyclohexenone $+\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2}\\right)_{2} \\mathrm{CuLi}$ (d) Unsaturated ketone $+\\left(\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}\\right)_{2} \\mathrm{CuLi}$\nPROBLEM Look for appearance of either an alcohol or a saturated ketone in the product. 19-23\nPROBLEM (a) $1715 \\mathrm{~cm}^{-1}$ (b) $1685 \\mathrm{~cm}^{-1}$ (c) $1750 \\mathrm{~cm}^{-1}$ (d) $1705 \\mathrm{~cm}^{-1}$ (e) $1715 \\mathrm{~cm}^{-1}$ $\\mathbf{1 9 - 2 4}$ (f) $1705 \\mathrm{~cm}^{-1}$\nPROBLEM (a) Different peaks due to McLafferty rearrangement\n19-25 (b) Different peaks due to $\\alpha$ cleavage and McLafferty rearrangement\n(c) Different peaks due to McLafferty rearrangement\n\nPROBLEM IR: $1750 \\mathrm{~cm}^{-1}$; MS: 140, 84\n19-26"}
{"id": 2128, "contents": "Chapter 20 - \nPROBLEM (a) 3-Methylbutanoic acid (b) 4-Bromopentanoic acid (c) 2-Ethylpentanoic acid\n20-1 (d) cis-4-Hexenoic acid (e) 2,4-Dimethylpentanenitrile\n(f) cis-1,3-Cyclopentanedicarboxylic acid\n\nPROBLEM (a)\n20-2\n\n(c)\n\n(d)\n\n(e)\n\n(f) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}=\\mathrm{CHCN}$"}
{"id": 2129, "contents": "Chapter 20 - \nPROBLEM Dissolve the mixture in ether, extract with aqueous NaOH , separate and acidify the aqueous layer,\n20-3 and extract with ether.\nPROBLEM 43\\%\n20-4\nPROBLEM (a) $82 \\%$ dissociation (b) $73 \\%$ dissociation 20-5\nPROBLEM Lactic acid is stronger because of the inductive effect of the -OH group.\n20-6\nPROBLEM The dianion is destabilized by repulsion between charges.\n20-7\nPROBLEM More reactive\n20-8\nPROBLEM (a) p-Methylbenzoic acid < Benzoic acid < $p$-Chlorobenzoic acid\n20-9 (b) Acetic acid < Benzoic acid < $p$-Nitrobenzoic acid\nPROBLEM (a) 1. $\\mathrm{Mg} ; 2 . \\mathrm{CO}_{2} ; 3 . \\mathrm{H}_{3} \\mathrm{O}^{+}$(b) 1. $\\mathrm{Mg} ; 2 . \\mathrm{CO}_{2} ; 3 . \\mathrm{H}_{3} \\mathrm{O}^{+}$or 1. $\\mathrm{NaCN} ; 2 . \\mathrm{H}_{3} \\mathrm{O}^{+}$ 20-10\nPROBLEM 1. NaCN; 2. $\\mathrm{H}_{3} \\mathrm{O}^{+}$; 3. $\\mathrm{LiAlH}_{4}$ 20-11\nPROBLEM 1. $\\mathrm{PBr}_{3} ; 2 . \\mathrm{NaCN} ; 3 . \\mathrm{H}_{3} \\mathrm{O}^{+} ; 4 . \\mathrm{LiAlH}_{4}$\n20-12"}
{"id": 2130, "contents": "Chapter 20 - \n20-12\nPROBLEM (a) Propanenitrile $+\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{MgBr}$, then $\\mathrm{H}_{3} \\mathrm{O}^{+}$(b) $p$-Chlorobenzonitrile $+\\mathrm{CH}_{3} \\mathrm{MgBr}$, then $\\mathrm{H}_{3} \\mathrm{O}^{+}$ 20-13\nPROBLEM 1. NaCN ; 2. $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{MgBr}$, then $\\mathrm{H}_{3} \\mathrm{O}^{+}$ 20-14"}
{"id": 2131, "contents": "Chapter 20 - \nPROBLEM A carboxylic acid has a very broad -OH absorption at $2500-3300 \\mathrm{~cm}^{-1}$.\n20-15\nPROBLEM 4-Hydroxycyclohexanone: $\\mathrm{H}-\\mathrm{C}-\\mathrm{O}$ absorption near $4 \\delta$ in the ${ }^{1} \\mathrm{H}$ spectrum and $\\mathrm{C}=\\mathrm{O}$ absorption near 20-16 $210 \\delta$ in the ${ }^{13} \\mathrm{C}$ spectrum. Cyclopentanecarboxylic acid: $-\\mathrm{CO}_{2} \\mathrm{H}$ absorption near $12 \\delta$ in the ${ }^{1} \\mathrm{H}$ spectrum and $-\\mathrm{CO}_{2} \\mathrm{H}$ absorption near $170 \\delta$ in the ${ }^{13} \\mathrm{C}$ spectrum.\n\nChapter 21\nPROBLEM (a) 4-Methylpentanoyl chloride (b) Cyclohexylacetamide (c) Isopropyl 2-methylpropanoate\n21-1 (d) Benzoic anhydride\n(e) Isopropyl cyclopentanecarboxylate\n(f) Cyclopentyl 2-methylpropanoate (g) N -Methyl-4-pentenamide\n(h) (R)-2-Hydroxypropanoyl phosphate (i) Ethyl 2,3-dimethyl-2-butenethioate\n\nPROBLEM\n(a) $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CO}_{2} \\mathrm{C}_{6} \\mathrm{H}_{5}$\n(b)\n$\\mathrm{H}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CON}\\left(\\mathrm{CH}_{3}\\right) \\mathrm{CH}_{2} \\mathrm{CH}_{3}$\n(c) $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CHCH}_{2} \\mathrm{CH}\\left(\\mathrm{CH}_{3}\\right) \\mathrm{COCl}$\n\n21-2\n(d)\n\n(e)\n\n(f)\n\n(g)\n\n(h)"}
{"id": 2132, "contents": "Chapter 20 - \n21-2\n(d)\n\n(e)\n\n(f)\n\n(g)\n\n(h)\n\n\nPROBLEM\n21-3\n\n\nPROBLEM (a) Acetyl chloride $>$ Methyl acetate $>$ Acetamide\n21-4 (b) Hexafluoroisopropyl acetate $>2,2,2$-Trichloroethyl acetate $>$ Ethyl acetate\nPROBLEM\n(a) $\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-} \\mathrm{Na}^{+}$\n(b) $\\mathrm{CH}_{3} \\mathrm{CONH}_{2}$\n(c) $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{CH}_{3}+\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-} \\mathrm{Na}^{+}$\n(d) $\\mathrm{CH}_{3} \\mathrm{CONHCH}_{3}$\n\n21-5\nPROBLEM\n\n\nPROBLEM (a) Acetic acid + 1-butanol (b) Butanoic acid + methanol\n21-7 (c) Cyclopentanecarboxylic acid + isopropyl alcohol\nPROBLEM\n21-8"}
{"id": 2133, "contents": "Chapter 20 - \nPROBLEM (a) Propanoyl chloride + methanol (b) Acetyl chloride + ethanol\n21-9 (c) Benzoyl chloride + ethanol\nPROBLEM Benzoyl chloride + cyclohexanol\n21-10\nPROBLEM This is a typical nucleophilic acyl substitution reaction, with morpholine as the nucleophile and\n21-11 chloride as the leaving group.\nPROBLEM (a) Propanoyl chloride + methylamine (b) Benzoyl chloride + diethylamine\n21-12 (c) Propanoyl chloride + ammonia\nPROBLEM (a) Benzoyl chloride $+\\left[\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CH}\\right]_{2} \\mathrm{CuLi}$, or 2-methylpropanoyl chloride $+\\mathrm{Ph}_{2} \\mathrm{CuLi}$\n21-13 (b) 2-Propenoyl chloride $+\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2}\\right)_{2} \\mathrm{CuLi}$, or butanoyl chloride $+\\left(\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}\\right)_{2} \\mathrm{CuLi}$\nPROBLEM This is a typical nucleophilic acyl substitution reaction, with $p$-hydroxyaniline as the nucleophile"}
{"id": 2134, "contents": "Chapter 20 - \n21-14 and acetate ion as the leaving group.\nPROBLEM Monomethyl ester of benzene-1,2-dicarboxylic acid\n21-15\nPROBLEM Reaction of a carboxylic acid with an alkoxide ion gives the carboxylate ion.\n21-16\nPROBLEM LiAlH 4 gives $\\mathrm{HOCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}$; DIBAH gives $\\mathrm{HOCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CHO}$.\n21-17\nPROBLEM (a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}\\left(\\mathrm{CH}_{3}\\right) \\mathrm{CH}_{2} \\mathrm{OH}+\\mathrm{CH}_{3} \\mathrm{OH}$ (b) $\\mathrm{PhOH}+\\mathrm{PhCH}_{2} \\mathrm{OH}$\n21-18\nPROBLEM (a) Ethyl benzoate $+2 \\mathrm{CH}_{3} \\mathrm{MgBr}$ (b) Ethyl acetate +2 PhMgBr\n21-19 (c) Ethyl pentanoate $+2 \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{MgBr}$\nPROBLEM (a) $\\mathrm{H}_{2} \\mathrm{O}, \\mathrm{NaOH}$ (b) Benzoic acid $+\\mathrm{LiAlH}_{4}$ (c) $\\mathrm{LiAlH}_{4}$\n21-20\nPROBLEM 1. Mg; 2. $\\mathrm{CO}_{2}$, then $\\mathrm{H}_{3} \\mathrm{O}^{+}$; 3. $\\mathrm{SOCl}_{2}$; 4. $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NH}$; 5. $\\mathrm{LiAlH}_{4}$\n21-21\nPROBLEM\n21-22\n\n\n$\\downarrow$\n\n\nPROBLEM (a)\n21-23\n\n(b)\n\n(c)\n\n\nPROBLEM\n21-24"}
{"id": 2135, "contents": "Chapter 20 - \n$\\downarrow$\n\n\nPROBLEM (a)\n21-23\n\n(b)\n\n(c)\n\n\nPROBLEM\n21-24\n\n\nPROBLEM (a) Ester (b) Acid chloride (c) Carboxylic acid (d) Aliphatic ketone or cyclohexanone\n21-25\nPROBLEM (a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$ and other possibilities (b) $\\mathrm{CH}_{3} \\mathrm{CON}\\left(\\mathrm{CH}_{3}\\right)_{2}$\n21-26 (c) $\\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{CHCOCl}$ or $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{C}\\left(\\mathrm{CH}_{3}\\right) \\mathrm{COCl}$\n\nChapter 22\nPROBLEM (a)\n22-1\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n\nPROBLEM\n(a) 4\n(b) 3\n(c) 3\n(d) 2\n(e) $4 \\quad$ (f) 5\n\n22-2\nPROBLEM\n22-3\n\n\nPROBLEM Acid-catalyzed formation of an enol is followed by deuteronation of the enol double bond and\n22-4 dedeuteronation of oxygen.\nPROBLEM 1. $\\mathrm{Br}_{2}$; 2. Pyridine, heat\n22-5\nPROBLEM The intermediate $\\alpha$-bromo acid bromide undergoes a nucleophilic acyl substitution reaction with\n22-6 methanol to give an $\\alpha$-bromo ester.\nPROBLEM (a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CHO}$ (b) $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CCOCH}_{3}$\n(c) $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$ (d) $\\mathrm{PhCONH}_{2}$ (e) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CN}$"}
{"id": 2136, "contents": "Chapter 20 - \n22-7 (f) $\\mathrm{CH}_{3} \\mathrm{CON}\\left(\\mathrm{CH}_{3}\\right)_{2}$\nPROBLEM\n22-8 ${ }^{-}: \\mathrm{CH}_{2} \\mathrm{C} \\equiv \\mathrm{N}: \\longleftrightarrow \\mathrm{H}_{2} \\mathrm{C}=\\mathrm{C}=\\ddot{\\mathrm{N}}:^{-}$\nPROBLEM Acid is regenerated, but base is used stoichiometrically.\n22-9\nPROBLEM (a) 1. $\\mathrm{Na}^{+-} \\mathrm{OEt}$; 2. $\\mathrm{PhCH}_{2} \\mathrm{Br}$; 3. $\\mathrm{H}_{3} \\mathrm{O}^{+}$\n22-10 (b) 1. $\\mathrm{Na}^{+-}{ }^{-} \\mathrm{OEt}$; 2. $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{Br}$; 3. $\\mathrm{Na}^{+{ }^{-}} \\mathrm{OEt}$; 4. $\\mathrm{CH}_{3} \\mathrm{Br} ;$ 5. $\\mathrm{H}_{3} \\mathrm{O}^{+}$\n(c) 1. $\\mathrm{Na}^{+-} \\mathrm{OEt}$; 2. $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CHCH}_{2} \\mathrm{Br} ; 3 . \\mathrm{H}_{3} \\mathrm{O}^{+}$"}
{"id": 2137, "contents": "Chapter 20 - \nPROBLEM Malonic ester has only two acidic hydrogens to be replaced. 22-11\nPROBLEM 1. $\\mathrm{Na}^{+-}$OEt; 2. $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CHCH}_{2} \\mathrm{Br}$; 3. $\\mathrm{Na}^{+-}{ }^{-} \\mathrm{OEt}$; 4. $\\mathrm{CH}_{3} \\mathrm{Br} ; 5 . \\mathrm{H}_{3} \\mathrm{O}^{+}$ 22-12\nPROBLEM (a) $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CHCH}_{2} \\mathrm{Br}$ (b) $\\mathrm{PhCH}_{2} \\mathrm{CH}_{2} \\mathrm{Br}$ 22-13\nPROBLEM None can be prepared.\n22-14\nPROBLEM 1.2 $\\mathrm{Na}^{+-} \\mathrm{OEt}$; 2. $\\mathrm{BrCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{Br}$; 3. $\\mathrm{H}_{3} \\mathrm{O}^{+}$\n22-15\nPROBLEM (a) Alkylate phenylacetone with $\\mathrm{CH}_{3} \\mathrm{I}$ (b) Alkylate pentanenitrile with $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{I}$\n22-16 (c) Alkylate cyclohexanone with $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}_{2} \\mathrm{Br}$ (d) Alkylate cyclohexanone with excess $\\mathrm{CH}_{3} \\mathrm{I}$\n(e) Alkylate $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{COCH}_{2} \\mathrm{CH}_{3}$ with $\\mathrm{CH}_{3} \\mathrm{I}$ (f) Alkylate methyl 3-methylbutanoate with $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{I}$\n\nChapter 23"}
{"id": 2138, "contents": "Chapter 20 - \nChapter 23\n\nPROBLEM (a)\n23-1\n\n(b)\n\n(c)\n\n\nPROBLEM The reverse reaction is the exact opposite of the forward reaction shown in Figure 23.2.\n\n23-2\n\nPROBLEM (a)\n23-3\n\n(b)\n\n(c)\n\n\nPROBLEM\n23-4\n\n\nPROBLEM (a) Not an aldol product (b) 3-Pentanone\n23-5\nPROBLEM 1. NaOH; 2. $\\mathrm{LiAlH}_{4} ; 3 . \\mathrm{H}_{2} / \\mathrm{Pd}$\n23-6\nPROBLEM\n23-7\n\n\nPROBLEM\n(a) $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CHO}+\\mathrm{CH}_{3} \\mathrm{COCH}_{3}$\n(b) Not easily prepared\n(c) Not easily prepared\n\n23-8\nPROBLEM The $\\mathrm{CH}_{2}$ position between the two carbonyl groups is so acidic that it is completely deprotonated to\n23-9 give a stable enolate ion.\nPROBLEM\n23-10\n\n\nPROBLEM (a)\n23-11\n\n(b)\n\n(c)\n\n\nPROBLEM The cleavage reaction is the exact reverse of the forward reaction.\n23-12\nPROBLEM\n23-13\n\n\nPROBLEM\n23-14\n\n\nPROBLEM\n23-15\n\n\nPROBLEM (a)\n23-16\n\n(b) $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{CHCH}_{2} \\mathrm{CH}_{2} \\mathrm{CN}$\n(c)\n\n\nPROBLEM (a)\n23-17\n\n(b)\n\n\nPROBLEM $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COCH}=\\mathrm{CH}_{2}+\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{NO}_{2}$ 23-18\nPROBLEM (a)\n\n(b)\n\n(c)\n\n\nPROBLEM (a) Cyclopentanone enamine + propenenitrile\n23-20 (b) Cyclohexanone enamine + methyl propenoate\nPROBLEM\n23-21"}
{"id": 2139, "contents": "Chapter 20 - \n(b)\n\n(c)\n\n\nPROBLEM (a) Cyclopentanone enamine + propenenitrile\n23-20 (b) Cyclohexanone enamine + methyl propenoate\nPROBLEM\n23-21\n\n\nPROBLEM 2,5,5-Trimethyl-1,3-cyclohexanedione + 1-penten-3-one\n23-22"}
{"id": 2140, "contents": "Chapter 24 - \nPROBLEM (a) $N$-Methylethylamine (b) Tricyclohexylamine (c) $N$-Ethyl- $N$-methylcyclohexylamine\n24-1 (d) $N$-Methylpyrrolidine\n(e) Diisopropylamine (f) 1,3-Butanediamine\n\nPROBLEM\n(a) $\\left[\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CH}\\right]_{3} \\mathrm{~N}$\n(b) $\\left(\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}_{2}\\right)_{3} \\mathrm{~N}$\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n\nPROBLEM (a)\n\n(b)\n\n(c)\n\n(d)\n\n\nPROBLEM (a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{NH}_{2}$\n(b) NaOH\n(c) $\\mathrm{CH}_{3} \\mathrm{NHCH}_{3}$\n\n24-4\nPROBLEM Propylamine is stronger; benzylamine $\\mathrm{p} K_{\\mathrm{b}}=4.67$; propylamine $\\mathrm{p} K_{\\mathrm{b}}=3.29$\n24-5\nPROBLEM (a) $p$-Nitroaniline $<p$-Aminobenzaldehyde $<p$-Bromoaniline\n24-6 (b) $p$-Aminoacetophenone $<p$-Chloroaniline $<p$-Methylaniline\n(c) $p$-(Trifluoromethyl)aniline $<p$-(Fluoromethyl)aniline $<p$-Methylaniline"}
{"id": 2141, "contents": "Chapter 24 - \nPROBLEM Pyrimidine is essentially 100\\% neutral (unprotonated).\n24-7\nPROBLEM (a) Propanenitrile or propanamide (b) $N$-Propylpropanamide (c) Benzonitrile or benzamide\n24-8 (d) $N$-Phenylacetamide\nPROBLEM The reaction takes place by two nucleophilic acyl substitution reactions.\n24-9\nPROBLEM\n24-10\n\n\nPROBLEM (a) Ethylamine + acetone, or isopropylamine + acetaldehyde (b) Aniline + acetaldehyde\n24-11 (c) Cyclopentylamine + formaldehyde, or methylamine + cyclopentanone\nPROBLEM\n24-12\n\n\nPROBLEM (a) 4,4-Dimethylpentanamide or 4,4-dimethylpentanoyl azide\n24-13 (b) $p$-Methylbenzamide or $p$-methylbenzoyl azide\nPROBLEM (a) 3-Octene and 4-octene (b) Cyclohexene (c) 3-Heptene (d) Ethylene and cyclohexene 24-14\nPROBLEM $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{~N}\\left(\\mathrm{CH}_{3}\\right)_{2}$"}
{"id": 2142, "contents": "24-15 - \nPROBLEM 1. $\\mathrm{HNO}_{3}, \\mathrm{H}_{2} \\mathrm{SO}_{4} ; 2 . \\mathrm{H}_{2} / \\mathrm{PtO}_{2}$; 3. $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O} ; 4 . \\mathrm{HOSO}_{2} \\mathrm{Cl}$; 5. aminothiazole; 6. $\\mathrm{H}_{2} \\mathrm{O}, \\mathrm{NaOH}$ 24-16\nPROBLEM (a) 1. $\\mathrm{HNO}_{3}, \\mathrm{H}_{2} \\mathrm{SO}_{4} ; 2 . \\mathrm{H}_{2} / \\mathrm{PtO}_{2} ; 3.2 \\mathrm{CH}_{3} \\mathrm{Br}$\n24-17 (b) 1. $\\mathrm{HNO}_{3}, \\mathrm{H}_{2} \\mathrm{SO}_{4} ; 2 . \\mathrm{H}_{2} / \\mathrm{PtO}_{2}$; 3. $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O} ; 4 . \\mathrm{Cl}_{2} ; 5 . \\mathrm{H} 2 \\mathrm{O}, \\mathrm{NaOH}$\n(c) 1. $\\mathrm{HNO}_{3}, \\mathrm{H}_{2} \\mathrm{SO}_{4} ; 2 . \\mathrm{Cl}_{2}, \\mathrm{FeCl}_{3} ; 3 . \\mathrm{SnCl}_{2}$\n(d) 1. $\\mathrm{HNO}_{3}, \\mathrm{H}_{2} \\mathrm{SO}_{4} ; 2 . \\mathrm{H}_{2} / \\mathrm{PtO}_{2}$; 3. $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O} ; 4.2 \\mathrm{CH}_{3} \\mathrm{Cl}, \\mathrm{AlCl}_{3} ; 5 . \\mathrm{H}_{2} \\mathrm{O}, \\mathrm{NaOH}$"}
{"id": 2143, "contents": "24-15 - \nPROBLEM (a) 1. $\\mathrm{CH}_{3} \\mathrm{Cl}, \\mathrm{AlCl}_{3} ; 2 . \\mathrm{HNO}_{3}, \\mathrm{H}_{2} \\mathrm{SO}_{4} ; 3 . \\mathrm{SnCl}_{2} ; 4 . \\mathrm{NaNO}_{2}, \\mathrm{H}_{2} \\mathrm{SO}_{4} ; 5 . \\mathrm{CuBr} ; 6 . \\mathrm{KMnO}_{4}, \\mathrm{H}_{2} \\mathrm{O}$\n24-18 (b) 1. $\\mathrm{HNO}_{3}, \\mathrm{H}_{2} \\mathrm{SO}_{4} ; 2 . \\mathrm{Br}_{2}, \\mathrm{FeBr}_{3} ; 3 . \\mathrm{SnCl}_{2}, \\mathrm{H}_{3} \\mathrm{O}^{+} ; 4 . \\mathrm{NaNO}_{2}, \\mathrm{H}_{2} \\mathrm{SO}_{4} ; 5 . \\mathrm{CuCN} ; 6 . \\mathrm{H}_{3} \\mathrm{O}^{+}$\n(c) 1. $\\mathrm{HNO}_{3}, \\mathrm{H}_{2} \\mathrm{SO}_{4} ; 2 . \\mathrm{Cl}_{2}, \\mathrm{FeCl}_{3} ; 3 . \\mathrm{SnCl}_{2} ; 4 . \\mathrm{NaNO}_{2}, \\mathrm{H}_{2} \\mathrm{SO}_{4} ; 5 . \\mathrm{CuBr}$\n(d) 1. $\\mathrm{CH}_{3} \\mathrm{Cl}, \\mathrm{AlCl}_{3} ; 2 . \\mathrm{HNO}_{3}, \\mathrm{H}_{2} \\mathrm{SO}_{4} ; 3 . \\mathrm{SnCl}_{2} ; 4 . \\mathrm{NaNO}_{2}, \\mathrm{H}_{2} \\mathrm{SO}_{4} ; 5 . \\mathrm{CuCN} ; 6 . \\mathrm{H}_{3} \\mathrm{O}^{+}$"}
{"id": 2144, "contents": "24-15 - \n(e) 1. $\\mathrm{HNO}_{3}, \\mathrm{H}_{2} \\mathrm{SO}_{4} ; 2 . \\mathrm{H}_{2} / \\mathrm{PtO}_{2}$; 3. $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O}$; $4.2 \\mathrm{Br}_{2}$; 5. $\\mathrm{H}_{2} \\mathrm{O}, \\mathrm{NaOH} ; 6 . \\mathrm{NaNO}_{2}, \\mathrm{H}_{2} \\mathrm{SO}_{4}$; 7. CuBr"}
{"id": 2145, "contents": "24-15 - \nPROBLEM 1. $\\mathrm{HNO}_{3}, \\mathrm{H}_{2} \\mathrm{SO}_{4} ; 2 . \\mathrm{SnCl}_{2}$; 3a. 2 equiv. $\\mathrm{CH}_{3} \\mathrm{I} ; 3 \\mathrm{~b} . \\mathrm{NaNO}_{2}, \\mathrm{H}_{2} \\mathrm{SO}_{4} ; 4$. product of $3 \\mathrm{a}+$ product of 3 b 24-19\nPROBLEM\n24-20\n\n\nPROBLEM 4.1\\% protonated\n24-21\nPROBLEM Attack at C2:\n24-22\n\n\nUnfavorable\n\n\n\nPROBLEM The side-chain nitrogen is more basic than the ring nitrogen.\n\n24-23\nPROBLEM Reaction at C 2 is disfavored because the aromaticity of the benzene ring is lost. 24-24\n\n\n\nPROBLEM $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CCOCH}_{3} \\rightarrow\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CCH}\\left(\\mathrm{NH}_{2}\\right) \\mathrm{CH}_{3}$ 24-25\n\nChapter 25\nPROBLEM (a) Aldotetrose (b) Ketopentose (c) Ketohexose (d) Aldopentose\n25-1\nPROBLEM (a) $S$ (b) $R$ (c) $S$\n25-2\nPROBLEM A, B, and C are the same.\n25-3\nPROBLEM\n25-4\nPROBLEM\n25-5\n\n\nPROBLEM (a) L-Erythrose; $2 S, 3 S$ (b) D-Xylose; $2 R, 3 S, 4 R$ (c) D-Xylulose; $3 S, 4 R$\n25-6\nPROBLEM\n25-7\n\n\nPROBLEM (a)\n25-8\n\n(c)\n\n\nPROBLEM 16 D and 16 L aldoheptoses\n25-9\nPROBLEM\n25-10\n\n\nPROBLEM\n25-11\n\n\nPROBLEM\n25-12\n\n\n$\\alpha$-D-Fructofuranose"}
{"id": 2146, "contents": "PROBLEM - \n25-13\n\n$\\beta$-D-Galactopyranose\n\n$\\beta$-D-Mannopyranose\n\nPROBLEM\n25-14\nPROBLEM $\\alpha$-D-Allopyranose\n25-15\nPROBLEM\n25-16\n\n\nPROBLEM D-Galactitol has a plane of symmetry and is a meso compound, whereas D-glucitol is chiral.\n25-17\nPROBLEM The -CHO end of L-gulose corresponds to the $-\\mathrm{CH}_{2} \\mathrm{OH}$ end of D -glucose after reduction.\n25-18\nPROBLEM D-Allaric acid has a symmetry plane and is a meso compound, but D-glucaric acid is chiral.\n25-19\nPROBLEM D-Allose and D-galactose yield meso aldaric acids; the other six D-hexoses yield optically active\n25-20 aldaric acids.\nPROBLEM D-Allose + D-altrose\n25-21\nPROBLEM L-Xylose\n25-22\nPROBLEM D-Xylose and D-lyxose\n25-23"}
{"id": 2147, "contents": "PROBLEM - \n25-24\n\n\n\nPROBLEM (a) The hemiacetal ring is reduced. (b) The hemiacetal ring is oxidized.\n25-25 (c) All hydroxyl groups are acetylated.\nChapter 26\nPROBLEM Aromatic: Phe, Tyr, Trp, His; sulfur-containing: Cys, Met; alcohols: Ser, Thr; hydrocarbon side\n26-1 chains: Ala, Ile, Leu, Val, Phe\nPROBLEM The sulfur atom in the $-\\mathrm{CH}_{2} \\mathrm{SH}$ group of cysteine makes the side chain higher in ranking than the\n26-2 $-\\mathrm{CO}_{2} \\mathrm{H}$ group.\nPROBLEM\n26-3\n\n\nL-Threonine\n\n\n\nDiastereomers of L-threonine\n\nPROBLEM Net positive at $\\mathrm{pH}=5.3$; net negative at $\\mathrm{pH}=7.3$\n26-4\nPROBLEM (a) Start with 3-phenylpropanoic acid: 1. $\\mathrm{Br}_{2}, \\mathrm{PBr}_{3} ; 2 . \\mathrm{NH}_{3}$\n26-5 (b) Start with 3-methylbutanoic acid: 1. $\\mathrm{Br}_{2}, \\mathrm{PBr}_{3} ; 2 . \\mathrm{NH}_{3}$\nPROBLEM\n26-6\n(a) $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CHCH}_{2} \\mathrm{Br}$\n(b)\n\n(c)\n\n(d) $\\mathrm{CH}_{3} \\mathrm{SCH}_{2} \\mathrm{CH}_{2} \\mathrm{Br}$\n\nPROBLEM\n26-7\n\n\nPROBLEM Val-Tyr-Gly (VYG), Tyr-Gly-Val (YGV), Gly-Val-Tyr (GVY), Val-Gly-Tyr (VGY), Tyr-Val-Gly (YVG), Gly-\n26-8 Tyr-Val (GYV)\nPROBLEM\n26-9\n\n\nPROBLEM\n26-10\n\n\nPROBLEM\n26-11"}
{"id": 2148, "contents": "PROBLEM - \nPROBLEM\n26-10\n\n\nPROBLEM\n26-11\n\n\nPROBLEM Trypsin: Asp-Arg + Val-Tyr-Ile-His-Pro-Phe\n26-12 Chymotrypsin: Asp-Arg-Val-Tyr + Ile-His-Pro-Phe\nPROBLEM Methionine\n26-13\nPROBLEM\n26-14\n\n\nPROBLEM (a) Arg-Pro-Leu-Gly-Ile-Val (b) Val-Met-Trp-Asp-Val-Leu (VMWNVL)\n26-15\nPROBLEM This is a typical nucleophilic acyl substitution reaction, with the amine of the amino acid as the\n26-16 nucleophile and tert-butyl carbonate as the leaving group. The tert-butyl carbonate then loses $\\mathrm{CO}_{2}$ and gives tert-butoxide, which is protonated.\nPROBLEM (1) Protect the amino group of leucine.\n26-17 (2) Protect the carboxylic acid group of alanine.\n(3) Couple the protected amino acids with DCC.\n(4) Remove the leucine protecting group.\n(5) Remove the alanine protecting group.\n\nPROBLEM (a) Lyase (b) Hydrolase (c) Oxidoreductase\n26-18"}
{"id": 2149, "contents": "Chapter 27 - \nPROBLEM $\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{18} \\mathrm{CO}_{2} \\mathrm{CH}_{2}\\left(\\mathrm{CH}_{2}\\right)_{30} \\mathrm{CH}_{3}$\n27-1\nPROBLEM Glyceryl tripalmitate is higher melting.\n27-2\nPROBLEM $\\left[\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{7} \\mathrm{CH}=\\mathrm{CH}\\left(\\mathrm{CH}_{2}\\right)_{7} \\mathrm{CO}_{2}^{-}\\right] 2 \\mathrm{Mg}^{2+}$\n27-3\nPROBLEM Glyceryl dioleate monopalmitate $\\rightarrow$ glycerol +2 sodium oleate + sodium palmitate\n27-4\nPROBLEM\n27-5\n\n\nPROBLEM The pro- $S$ hydrogen is cis to the $-\\mathrm{CH}_{3}$ group; the pro- $R$ hydrogen is trans.\n27-6\n\nPROBLEM (a)\n27-7\n\n\n\n(b)\n\n\n\n\n\n\n$\\gamma$-Bisabolene\n\nPROBLEM (a)\n27-8\n\n(b)\n\n\nPROBLEM\n27-9\n\n\nPROBLEM Three methyl groups are removed, the side-chain double bond is reduced, and the double bond in $\\mathbf{2 7 - 1 0}$ the $B$ ring is migrated.\n\nChapter 28\n\n\nPROBLEM\n28-2\n\n\nPROBLEM (5') ACGGATTAGCC (3')\n28-3"}
{"id": 2150, "contents": "PROBLEM - \n28-4\n\n\nPROBLEM (3') CUAAUGGCAU (5')\n28-5\nPROBLEM (5') ACTCTGCGAA (3')\n28-6\nPROBLEM (a) GCU, GCC, GCA, GCG (b) UUU, UUC (c) UUA, UUG, CUU, CUC, CUA, CUG (d) UAU, UAC\n28-7\nPROBLEM\n28-8\nPROBLEM Leu-Met-Ala-Trp-Pro-Stop\n28-9\nPROBLEM (5') TTA-GGG-CCA-AGC-CAT-AAG (3')\n28-10\nPROBLEM The cleavage is an $\\mathrm{S}_{\\mathrm{N}} 1$ reaction that occurs by protonation of the oxygen atom followed by loss of\n28-11 the stable triarylmethyl carbocation.\nPROBLEM\n28-12"}
{"id": 2151, "contents": "Chapter 29 - \nPROBLEM $\\mathrm{HOCH}_{2} \\mathrm{CH}(\\mathrm{OH}) \\mathrm{CH}_{2} \\mathrm{OH}+\\mathrm{ATP} \\rightarrow \\mathrm{HOCH}_{2} \\mathrm{CH}(\\mathrm{OH}) \\mathrm{CH}_{2} \\mathrm{OPO}_{3}{ }^{2-}+\\mathrm{ADP}$\n29-1\nPROBLEM Caprylyl CoA $\\rightarrow$ Hexanoyl CoA $\\rightarrow$ Butyryl CoA $\\rightarrow 2$ Acetyl CoA 29-2\nPROBLEM (a) 8 acetyl CoA; 7 passages (b) 10 acetyl CoA; 9 passages\n29-3\nPROBLEM The dehydration is an E1cB reaction.\n29-4\nPROBLEM At C2, C4, C6, C8, and so forth\n29-5\nPROBLEM The Si face\n29-6\nPROBLEM Steps 7 and 10\n29-7\nPROBLEM Steps 1, 3: Phosphate transfers; steps 2, 5, 8: isomerizations; step 4: retro-aldol reaction; step 29-8 5: oxidation and nucleophilic acyl substitution; steps 7, 10: phosphate transfers; step 9: E1cB dehydration\nPROBLEM C 1 and C 6 of glucose become $-\\mathrm{CH}_{3}$ groups; C 3 and C 4 become $\\mathrm{CO}_{2}$. 29-9\nPROBLEM Citrate and isocitrate 29-10\nPROBLEM E1cB elimination of water, followed by conjugate addition 29-11\nPROBLEM pro- $R$; anti geometry"}
{"id": 2152, "contents": "Chapter 29 - \nPROBLEM The reaction occurs by two sequential nucleophilic acyl substitutions, the first by a cysteine residue\n29-13 in the enzyme, with phosphate as leaving group, and the second by hydride donation from NADH, with the cysteine residue as leaving group.\nPROBLEM Initial imine formation between PMP and $\\alpha$-ketoglutarate is followed by double-bond 29-14 rearrangement to an isomeric imine and hydrolysis.\nPROBLEM $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CHCH}_{2} \\mathrm{COCO}_{2}{ }^{-}$\n29-15\nPROBLEM Asparagine\n29-16\nChapter 30\nPROBLEM Ethylene: $\\psi_{1}$ is the HOMO and $\\psi_{2}{ }^{*}$ is the LUMO in the ground state; $\\psi_{2}{ }^{*}$ is the HOMO and there is no\n30-1 LUMO in the excited state. 1,3-Butadiene: $\\psi_{2}$ is the HOMO and $\\psi_{3}{ }^{*}$ is the LUMO in the ground state; $\\psi_{3}{ }^{*}$ is the HOMO and $\\psi_{4}{ }^{*}$ is the LUMO in the excited state.\nPROBLEM Disrotatory: cis-5,6-dimethyl-1,3-cyclohexadiene;\n30-2 conrotatory: trans-5,6-dimethyl-1,3-cyclohexadiene. Disrotatory closure occurs.\nPROBLEM The more stable of two allowed products is formed.\n30-3\nPROBLEM trans-5,6-Dimethyl-1,3-cyclohexadiene; cis-5,6-dimethyl-1,3-cyclohexadiene 30-4\nPROBLEM cis-3,6-Dimethylcyclohexene; trans-3,6-dimethylcyclohexene 30-5\nPROBLEM A [6 + 4] suprafacial cycloaddition 30-6\nPROBLEM An antarafacial [1,7] sigmatropic rearrangement\n30-7\nPROBLEM\n30-8"}
{"id": 2153, "contents": "Chapter 29 - \nPROBLEM A series of [1,5] hydrogen shifts occur. 30-9\nPROBLEM o-(1-Methylallyl)phenol\n30-10\nPROBLEM Claisen rearrangement is followed by a Cope rearrangement. 30-11\nPROBLEM (a) Conrotatory (b) Disrotatory (c) Suprafacial (d) Antarafacial (e) Suprafacial 30-12"}
{"id": 2154, "contents": "Chapter 31 - \nPROBLEM $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCO}_{2} \\mathrm{CH}_{3}<\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCl}<\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}_{3}<\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}-\\mathrm{C}_{6} \\mathrm{H}_{5}$ 31-1\nPROBLEM $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHCH}_{3}<\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHC}_{6} \\mathrm{H}_{5}<\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CHC} \\equiv \\mathrm{N}$ 31-2\nPROBLEM The intermediate is a resonance-stabilized benzylic carbanion, $\\mathrm{Ph}-\\overline{\\ddot{\\mathrm{C}}} \\mathrm{HR}$ 31-3\nPROBLEM The polymer has no chirality centers. 31-4\nPROBLEM The polymers are racemic and have no optical rotation. 31-5"}
{"id": 2155, "contents": "PROBLEM <br> 31-6 <br> - \nPROBLEM\n31-7\nPolybutadiene chain\n\n\nPolystyrene chain\nPROBLEM\n31-8"}
{"id": 2156, "contents": "PROBLEM - \n31-9\n\n\n\nPROBLEM Vestenamer: ADMET polymerization of 1,9-decadiene or ROMP of cyclooctene; Norsorex: ROMP of 31-10 norbornene."}
{"id": 2157, "contents": "PROBLEM - \n31-12\n\n\n\n\n\nare a class of saturated hydrocarbons with the general formula $\\mathrm{C}_{n} \\mathrm{H}_{2 n+2}$. They contain no functional groups, are relatively inert, and can be either straight-chain (normal) or branched. Alkanes are named by a series of IUPAC rules of nomenclature. Compounds that have the same chemical formula but different"}
{"id": 2158, "contents": "1. CHAPTER 1 Essential Ideas - \nFigure 1.1 Chemical substances and processes are essential for our existence, providing sustenance, keeping us clean and healthy, fabricating electronic devices, enabling transportation, and much more. (credit \"left\": modification of work by \"vxla\"/Flickr; credit \"left middle\": modification of work by \"the Italian voice\"/Flickr; credit \"right middle\": modification of work by Jason Trim; credit \"right\": modification of work by \"gosheshe\"/Flickr)"}
{"id": 2159, "contents": "2. CHAPTER OUTLINE - 2.1. Chemistry in Context\n1.2 Phases and Classification of Matter\n1.3 Physical and Chemical Properties"}
{"id": 2160, "contents": "2. CHAPTER OUTLINE - 2.2. Measurements\n1.5 Measurement Uncertainty, Accuracy, and Precision\n1.6 Mathematical Treatment of Measurement Results\n\nINTRODUCTION Your alarm goes off and, after hitting \"snooze\" once or twice, you pry yourself out of bed. You make a cup of coffee to help you get going, and then you shower, get dressed, eat breakfast, and check your phone for messages. On your way to school, you stop to fill your car's gas tank, almost making you late for the first day of chemistry class. As you find a seat in the classroom, you read the question projected on the screen: \"Welcome to class! Why should we study chemistry?\"\n\nDo you have an answer? You may be studying chemistry because it fulfills an academic requirement, but if you consider your daily activities, you might find chemistry interesting for other reasons. Most everything you do and encounter during your day involves chemistry. Making coffee, cooking eggs, and toasting bread involve chemistry. The products you use-like soap and shampoo, the fabrics you wear, the electronics that keep you connected to your world, the gasoline that propels your car-all of these and more involve chemical substances and processes. Whether you are aware or not, chemistry is part of your everyday world. In this course, you will learn many of the essential principles underlying the chemistry of modern-day life."}
{"id": 2161, "contents": "3. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Outline the historical development of chemistry\n- Provide examples of the importance of chemistry in everyday life\n- Describe the scientific method\n- Differentiate among hypotheses, theories, and laws\n- Provide examples illustrating macroscopic, microscopic, and symbolic domains\n\nThroughout human history, people have tried to convert matter into more useful forms. Our Stone Age\nancestors chipped pieces of flint into useful tools and carved wood into statues and toys. These endeavors involved changing the shape of a substance without changing the substance itself. But as our knowledge increased, humans began to change the composition of the substances as well-clay was converted into pottery, hides were cured to make garments, copper ores were transformed into copper tools and weapons, and grain was made into bread.\n\nHumans began to practice chemistry when they learned to control fire and use it to cook, make pottery, and smelt metals. Subsequently, they began to separate and use specific components of matter. A variety of drugs such as aloe, myrrh, and opium were isolated from plants. Dyes, such as indigo and Tyrian purple, were extracted from plant and animal matter. Metals were combined to form alloys-for example, copper and tin were mixed together to make bronze-and more elaborate smelting techniques produced iron. Alkalis were extracted from ashes, and soaps were prepared by combining these alkalis with fats. Alcohol was produced by fermentation and purified by distillation.\n\nAttempts to understand the behavior of matter extend back for more than 2500 years. As early as the sixth century BC, Greek philosophers discussed a system in which water was the basis of all things. You may have heard of the Greek postulate that matter consists of four elements: earth, air, fire, and water. Subsequently, an amalgamation of chemical technologies and philosophical speculations was spread from Egypt, China, and the eastern Mediterranean by alchemists, who endeavored to transform \"base metals\" such as lead into \"noble metals\" like gold, and to create elixirs to cure disease and extend life (Figure 1.2)."}
{"id": 2162, "contents": "3. LEARNING OBJECTIVES - \nFIGURE 1.2 (a) This portrayal shows an alchemist's workshop circa 1580. Although alchemy made some useful contributions to how to manipulate matter, it was not scientific by modern standards. (b) While the equipment used by Alma Levant Hayden in this 1952 picture might not seem as sleek as you might find in a lab today, her approach was highly methodical and carefully recorded. A department head at the FDA, Hayden is most famous for exposing an aggressively marketed anti-cancer drug as nothing more than an unhelpful solution of common substances. (credit a: Chemical Heritage Foundation; b: NIH History Office)\n\nFrom alchemy came the historical progressions that led to modern chemistry: the isolation of drugs from natural sources, such as plants and animals. But while many of the substances extracted or processed from those natural sources were critical in the treatment of diseases, many were scarce. For example, progesterone, which is critical to women's health, became available as a medicine in 1935, but its animal sources produced extremely small quantities, limiting its availability and increasing its expense. Likewise, in the 1940s, cortisone came into use to treat arthritis and other disorders and injuries, but it took a 36-step process to synthesize. Chemist Percy Lavon Julian turned to a more plentiful source: soybeans. Previously, Julian had developed a lab to isolate soy protein, which was used in firefighting among other applications. He focused on using the soy sterols-substances mostly used in plant membranes-and was able to quickly produce progesterone and later testosterone and other hormones. He later developed a process to do the same for cortisone, and laid the groundwork for modern drug design. Since soybeans and similar plant sources were extremely plentiful, the drugs soon became widely available, saving many lives."}
{"id": 2163, "contents": "4. Chemistry: The Central Science - \nChemistry is sometimes referred to as \"the central science\" due to its interconnectedness with a vast array of other STEM disciplines (STEM stands for areas of study in the science, technology, engineering, and math fields). Chemistry and the language of chemists play vital roles in biology, medicine, materials science, forensics, environmental science, and many other fields (Figure 1.3). The basic principles of physics are essential for understanding many aspects of chemistry, and there is extensive overlap between many subdisciplines within the two fields, such as chemical physics and nuclear chemistry. Mathematics, computer science, and information theory provide important tools that help us calculate, interpret, describe, and generally make sense of the chemical world. Biology and chemistry converge in biochemistry, which is crucial to understanding the many complex factors and processes that keep living organisms (such as us) alive. Chemical engineering, materials science, and nanotechnology combine chemical principles and empirical findings to produce useful substances, ranging from gasoline to fabrics to electronics. Agriculture, food science, veterinary science, and brewing and wine making help provide sustenance in the form of food and drink to the world's population. Medicine, pharmacology, biotechnology, and botany identify and produce substances that help keep us healthy. Environmental science, geology, oceanography, and atmospheric science incorporate many chemical ideas to help us better understand and protect our physical world. Chemical ideas are used to help understand the universe in astronomy and cosmology.\n\n\nFIGURE 1.3 Knowledge of chemistry is central to understanding a wide range of scientific disciplines. This diagram shows just some of the interrelationships between chemistry and other fields."}
{"id": 2164, "contents": "4. Chemistry: The Central Science - \nFIGURE 1.3 Knowledge of chemistry is central to understanding a wide range of scientific disciplines. This diagram shows just some of the interrelationships between chemistry and other fields.\n\nWhat are some changes in matter that are essential to daily life? Digesting and assimilating food, synthesizing polymers that are used to make clothing, containers, cookware, and credit cards, and refining crude oil into gasoline and other products are just a few examples. As you proceed through this course, you will discover many different examples of changes in the composition and structure of matter, how to classify these changes and how they occurred, their causes, the changes in energy that accompany them, and the principles and laws involved. As you learn about these things, you will be learning chemistry, the study of the composition, properties, and interactions of matter. The practice of chemistry is not limited to chemistry books or laboratories: It happens whenever someone is involved in changes in matter or in conditions that may lead to such changes."}
{"id": 2165, "contents": "5. The Scientific Method - \nChemistry is a science based on observation and experimentation. Doing chemistry involves attempting to answer questions and explain observations in terms of the laws and theories of chemistry, using procedures that are accepted by the scientific community. There is no single route to answering a question or explaining an observation, but there is an aspect common to every approach: Each uses knowledge based on experiments that can be reproduced to verify the results. Some routes involve a hypothesis, a tentative explanation of\nobservations that acts as a guide for gathering and checking information. A hypothesis is tested by experimentation, calculation, and/or comparison with the experiments of others and then refined as needed.\n\nSome hypotheses are attempts to explain the behavior that is summarized in laws. The laws of science summarize a vast number of experimental observations, and describe or predict some facet of the natural world. If such a hypothesis turns out to be capable of explaining a large body of experimental data, it can reach the status of a theory. Scientific theories are well-substantiated, comprehensive, testable explanations of particular aspects of nature. Theories are accepted because they provide satisfactory explanations, but they can be modified if new data become available. The path of discovery that leads from question and observation to law or hypothesis to theory, combined with experimental verification of the hypothesis and any necessary modification of the theory, is called the scientific method (Figure 1.4).\n\n\nFIGURE 1.4 The scientific method follows a process similar to the one shown in this diagram. All the key components are shown, in roughly the right order. Scientific progress is seldom neat and clean: It requires open inquiry and the reworking of questions and ideas in response to findings."}
{"id": 2166, "contents": "6. The Domains of Chemistry - \nChemists study and describe the behavior of matter and energy in three different domains: macroscopic, microscopic, and symbolic. These domains provide different ways of considering and describing chemical behavior.\n\nMacro is a Greek word that means \"large.\" The macroscopic domain is familiar to us: It is the realm of everyday things that are large enough to be sensed directly by human sight or touch. In daily life, this includes the food you eat and the breeze you feel on your face. The macroscopic domain includes everyday and laboratory chemistry, where we observe and measure physical and chemical properties such as density, solubility, and flammability.\n\nMicro comes from Greek and means \"small.\" The microscopic domain of chemistry is often visited in the imagination. Some aspects of the microscopic domain are visible through standard optical microscopes, for example, many biological cells. More sophisticated instruments are capable of imaging even smaller entities such as molecules and atoms (see Figure 1.5 (b)).\n\nHowever, most of the subjects in the microscopic domain of chemistry are too small to be seen even with the most advanced microscopes and may only be pictured in the mind. Other components of the microscopic\ndomain include ions and electrons, protons and neutrons, and chemical bonds, each of which is far too small to see.\n\nThe symbolic domain contains the specialized language used to represent components of the macroscopic and microscopic domains. Chemical symbols (such as those used in the periodic table), chemical formulas, and chemical equations are part of the symbolic domain, as are graphs, drawings, and calculations. These symbols play an important role in chemistry because they help interpret the behavior of the macroscopic domain in terms of the components of the microscopic domain. One of the challenges for students learning chemistry is recognizing that the same symbols can represent different things in the macroscopic and microscopic domains, and one of the features that makes chemistry fascinating is the use of a domain that must be imagined to explain behavior in a domain that can be observed."}
{"id": 2167, "contents": "6. The Domains of Chemistry - \nA helpful way to understand the three domains is via the essential and ubiquitous substance of water. That water is a liquid at moderate temperatures, will freeze to form a solid at lower temperatures, and boil to form a gas at higher temperatures (Figure 1.5) are macroscopic observations. But some properties of water fall into the microscopic domain-what cannot be observed with the naked eye. The description of water as comprising two hydrogen atoms and one oxygen atom, and the explanation of freezing and boiling in terms of attractions between these molecules, is within the microscopic arena. The formula $\\mathrm{H}_{2} \\mathrm{O}$, which can describe water at either the macroscopic or microscopic levels, is an example of the symbolic domain. The abbreviations ( $g$ ) for gas, ( $S$ ) for solid, and ( $I$ ) for liquid are also symbolic.\n\n\nFIGURE 1.5 (a) Moisture in the air, icebergs, and the ocean represent water in the macroscopic domain. (b) At the molecular level (microscopic domain), gas molecules are far apart and disorganized, solid water molecules are close together and organized, and liquid molecules are close together and disorganized. (c) The formula $\\mathrm{H}_{2} \\mathrm{O}$ symbolizes water, and (g), (s), and ( $)$ symbolize its phases. Note that clouds are actually comprised of either very small liquid water droplets or solid water crystals; gaseous water in our atmosphere is not visible to the naked eye, although it may be sensed as humidity. (credit a: modification of work by \"Gorkaazk\"/Wikimedia Commons)"}
{"id": 2168, "contents": "7. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the basic properties of each physical state of matter: solid, liquid, and gas\n- Distinguish between mass and weight\n- Apply the law of conservation of matter\n- Classify matter as an element, compound, homogeneous mixture, or heterogeneous mixture with regard to its physical state and composition\n- Define and give examples of atoms and molecules\n\nMatter is defined as anything that occupies space and has mass, and it is all around us. Solids and liquids are more obviously matter: We can see that they take up space, and their weight tells us that they have mass. Gases are also matter; if gases did not take up space, a balloon would not inflate (increase its volume) when filled with gas.\n\nSolids, liquids, and gases are the three states of matter commonly found on earth (Figure 1.6). A solid is rigid and possesses a definite shape. A liquid flows and takes the shape of its container, except that it forms a flat or slightly curved upper surface when acted upon by gravity. (In zero gravity, liquids assume a spherical shape.) Both liquid and solid samples have volumes that are very nearly independent of pressure. A gas takes both the shape and volume of its container.\n\n\nFIGURE 1.6 The three most common states or phases of matter are solid, liquid, and gas.\nA fourth state of matter, plasma, occurs naturally in the interiors of stars. A plasma is a gaseous state of matter that contains appreciable numbers of electrically charged particles (Figure 1.7). The presence of these charged particles imparts unique properties to plasmas that justify their classification as a state of matter distinct from gases. In addition to stars, plasmas are found in some other high-temperature environments (both natural and man-made), such as lightning strikes, certain television screens, and specialized analytical instruments used to detect trace amounts of metals.\n\n\nFIGURE 1.7 A plasma torch can be used to cut metal. (credit: \"Hypertherm\"/Wikimedia Commons)"}
{"id": 2169, "contents": "8. LINK TO LEARNING - \nIn a tiny cell in a plasma television, the plasma emits ultraviolet light, which in turn causes the display at that location to appear a specific color. The composite of these tiny dots of color makes up the image that you see. Watch this video (http://openstax.org/1/16plasma) to learn more about plasma and the places you encounter it.\n\nSome samples of matter appear to have properties of solids, liquids, and/or gases at the same time. This can occur when the sample is composed of many small pieces. For example, we can pour sand as if it were a liquid because it is composed of many small grains of solid sand. Matter can also have properties of more than one state when it is a mixture, such as with clouds. Clouds appear to behave somewhat like gases, but they are actually mixtures of air (gas) and tiny particles of water (liquid or solid).\n\nThe mass of an object is a measure of the amount of matter in it. One way to measure an object's mass is to measure the force it takes to accelerate the object. It takes much more force to accelerate a car than a bicycle because the car has much more mass. A more common way to determine the mass of an object is to use a balance to compare its mass with a standard mass.\n\nAlthough weight is related to mass, it is not the same thing. Weight refers to the force that gravity exerts on an object. This force is directly proportional to the mass of the object. The weight of an object changes as the force of gravity changes, but its mass does not. An astronaut's mass does not change just because she goes to the moon. But her weight on the moon is only one-sixth her earth-bound weight because the moon's gravity is only one-sixth that of the earth's. She may feel \"weightless\" during her trip when she experiences negligible external forces (gravitational or any other), although she is, of course, never \"massless.\""}
{"id": 2170, "contents": "8. LINK TO LEARNING - \nThe law of conservation of matter summarizes many scientific observations about matter: It states that there is no detectable change in the total quantity of matter present when matter converts from one type to another (a chemical change) or changes among solid, liquid, or gaseous states (a physical change). Brewing beer and the operation of batteries provide examples of the conservation of matter (Figure 1.8). During the brewing of beer, the ingredients (water, yeast, grains, malt, hops, and sugar) are converted into beer (water, alcohol, carbonation, and flavoring substances) with no actual loss of substance. This is most clearly seen during the bottling process, when glucose turns into ethanol and carbon dioxide, and the total mass of the substances does not change. This can also be seen in a lead-acid car battery: The original substances (lead, lead oxide, and sulfuric acid), which are capable of producing electricity, are changed into other substances (lead sulfate and water) that do not produce electricity, with no change in the actual amount of matter.\n\n\nFIGURE 1.8 (a) The mass of beer precursor materials is the same as the mass of beer produced: Sugar has become alcohol and carbon dioxide. (b) The mass of the lead, lead oxide, and sulfuric acid consumed by the production of electricity is exactly equal to the mass of lead sulfate and water that is formed.\n\nAlthough this conservation law holds true for all conversions of matter, convincing examples are few and far between because, outside of the controlled conditions in a laboratory, we seldom collect all of the material that is produced during a particular conversion. For example, when you eat, digest, and assimilate food, all of the matter in the original food is preserved. But because some of the matter is incorporated into your body, and much is excreted as various types of waste, it is challenging to verify by measurement."}
{"id": 2171, "contents": "9. Classifying Matter - \nMatter can be classified into several categories. Two broad categories are mixtures and pure substances. A pure substance has a constant composition. All specimens of a pure substance have exactly the same makeup and properties. Any sample of sucrose (table sugar) consists of $42.1 \\%$ carbon, $6.5 \\%$ hydrogen, and $51.4 \\%$ oxygen by mass. Any sample of sucrose also has the same physical properties, such as melting point, color, and sweetness, regardless of the source from which it is isolated.\n\nPure substances may be divided into two classes: elements and compounds. Pure substances that cannot be broken down into simpler substances by chemical changes are called elements. Iron, silver, gold, aluminum, sulfur, oxygen, and copper are familiar examples of the more than 100 known elements, of which about 90 occur naturally on the earth, and two dozen or so have been created in laboratories.\n\nPure substances that are comprised of two or more elements are called compounds. Compounds may be broken down by chemical changes to yield either elements or other compounds, or both. Mercury(II) oxide, an orange, crystalline solid, can be broken down by heat into the elements mercury and oxygen (Figure 1.9). When heated in the absence of air, the compound sucrose is broken down into the element carbon and the compound water. (The initial stage of this process, when the sugar is turning brown, is known as caramelization-this is what imparts the characteristic sweet and nutty flavor to caramel apples, caramelized onions, and caramel). Silver(I) chloride is a white solid that can be broken down into its elements, silver and chlorine, by absorption of light. This property is the basis for the use of this compound in photographic films and photochromic eyeglasses (those with lenses that darken when exposed to light).\n\n\nFIGURE 1.9 (a) The compound mercury(II) oxide, (b) when heated, (c) decomposes into silvery droplets of liquid mercury and invisible oxygen gas. (credit: modification of work by Paul Flowers)"}
{"id": 2172, "contents": "10. LINK TO LEARNING - \nMany compounds break down when heated. This site (http://openstax.org/l/16mercury) shows the breakdown of mercury oxide, HgO. You can also view an example of the photochemical decomposition of silver chloride (http://openstax.org/l/16silvchloride) (AgCl), the basis of early photography.\n\nThe properties of combined elements are different from those in the free, or uncombined, state. For example, white crystalline sugar (sucrose) is a compound resulting from the chemical combination of the element carbon, which is a black solid in one of its uncombined forms, and the two elements hydrogen and oxygen, which are colorless gases when uncombined. Free sodium, an element that is a soft, shiny, metallic solid, and free chlorine, an element that is a yellow-green gas, combine to form sodium chloride (table salt), a compound that is a white, crystalline solid.\n\nA mixture is composed of two or more types of matter that can be present in varying amounts and can be separated by physical changes, such as evaporation (you will learn more about this later). A mixture with a composition that varies from point to point is called a heterogeneous mixture. Italian dressing is an example of a heterogeneous mixture (Figure 1.10). Its composition can vary because it may be prepared from varying amounts of oil, vinegar, and herbs. It is not the same from point to point throughout the mixture-one drop may be mostly vinegar, whereas a different drop may be mostly oil or herbs because the oil and vinegar separate and the herbs settle. Other examples of heterogeneous mixtures are chocolate chip cookies (we can see the separate bits of chocolate, nuts, and cookie dough) and granite (we can see the quartz, mica, feldspar, and more)."}
{"id": 2173, "contents": "10. LINK TO LEARNING - \nA homogeneous mixture, also called a solution, exhibits a uniform composition and appears visually the same throughout. An example of a solution is a sports drink, consisting of water, sugar, coloring, flavoring, and electrolytes mixed together uniformly (Figure 1.10). Each drop of a sports drink tastes the same because each drop contains the same amounts of water, sugar, and other components. Note that the composition of a sports drink can vary-it could be made with somewhat more or less sugar, flavoring, or other components, and still be a sports drink. Other examples of homogeneous mixtures include air, maple syrup, gasoline, and a solution of salt in water.\n\n\nFIGURE 1.10 (a) Oil and vinegar salad dressing is a heterogeneous mixture because its composition is not uniform throughout. (b) A commercial sports drink is a homogeneous mixture because its composition is uniform throughout. (credit a \"left\": modification of work by John Mayer; credit a \"right\": modification of work by Umberto Salvagnin; credit b \"left: modification of work by Jeff Bedford)\n\nAlthough there are just over 100 elements, tens of millions of chemical compounds result from different combinations of these elements. Each compound has a specific composition and possesses definite chemical and physical properties that distinguish it from all other compounds. And, of course, there are innumerable ways to combine elements and compounds to form different mixtures. A summary of how to distinguish between the various major classifications of matter is shown in (Figure 1.11).\n\n\nFIGURE 1.11 Depending on its properties, a given substance can be classified as a homogeneous mixture, a heterogeneous mixture, a compound, or an element.\n\nEleven elements make up about $99 \\%$ of the earth's crust and atmosphere (Table 1.1). Oxygen constitutes nearly one-half and silicon about one-quarter of the total quantity of these elements. A majority of elements on earth are found in chemical combinations with other elements; about one-quarter of the elements are also found in the free state.\n\nElemental Composition of Earth"}
{"id": 2174, "contents": "10. LINK TO LEARNING - \nElemental Composition of Earth\n\n| Element | Symbol |  | Percent Mass | Element | Symbol |  | Percent Mass |\n| :--- | :--- | :--- | :--- | :--- | :--- | :---: | :---: |\n| oxygen | O | 49.20 | chlorine | Cl | 0.19 |  |  |\n| silicon | Si | 25.67 | phosphorus | P | 0.11 |  |  |\n| aluminum | Al | 7.50 | manganese | Mn | 0.09 |  |  |\n| calcium | Ca | 3.39 | carbon | C | 0.08 |  |  |\n| sodium | Na | 2.63 | sulfur | S | 0.06 |  |  |\n| potassium | K | 2.40 | barium | Ba | 0.04 |  |  |\n| magnesium | Mg | 1.93 | fluorine | F | 0.03 |  |  |\n| hydrogen | H | 0.87 | strontium | Sr | 0.02 |  |  |\n| titanium | Ti | 0.58 | all others | - | 0.47 |  |  |\n\nTABLE 1.1"}
{"id": 2175, "contents": "11. Atoms and Molecules - \nAn atom is the smallest particle of an element that has the properties of that element and can enter into a chemical combination. Consider the element gold, for example. Imagine cutting a gold nugget in half, then cutting one of the halves in half, and repeating this process until a piece of gold remained that was so small that it could not be cut in half (regardless of how tiny your knife may be). This minimally sized piece of gold is an atom (from the Greek atomos, meaning \"indivisible\") (Figure 1.12). This atom would no longer be gold if it were divided any further.\n\n\nFIGURE 1.12 (a) This photograph shows a gold nugget. (b) A scanning-tunneling microscope (STM) can generate views of the surfaces of solids, such as this image of a gold crystal. Each sphere represents one gold atom. (credit a: modification of work by United States Geological Survey; credit b: modification of work by \"Erwinrossen\"/Wikimedia Commons)\n\nThe first suggestion that matter is composed of atoms is attributed to the Greek philosophers Leucippus and Democritus, who developed their ideas in the 5th century BCE. However, it was not until the early nineteenth century that John Dalton (1766-1844), a British schoolteacher with a keen interest in science, supported this hypothesis with quantitative measurements. Since that time, repeated experiments have confirmed many aspects of this hypothesis, and it has become one of the central theories of chemistry. Other aspects of Dalton's atomic theory are still used but with minor revisions (details of Dalton's theory are provided in the chapter on atoms and molecules)."}
{"id": 2176, "contents": "11. Atoms and Molecules - \nAn atom is so small that its size is difficult to imagine. One of the smallest things we can see with our unaided eye is a single thread of a spider web: These strands are about $1 / 10,000$ of a centimeter $(0.0001 \\mathrm{~cm})$ in diameter. Although the cross-section of one strand is almost impossible to see without a microscope, it is huge on an atomic scale. A single carbon atom in the web has a diameter of about 0.000000015 centimeter, and it would take about 7000 carbon atoms to span the diameter of the strand. To put this in perspective, if a carbon atom were the size of a dime, the cross-section of one strand would be larger than a football field, which would require about 150 million carbon atom \"dimes\" to cover it. (Figure 1.13) shows increasingly close microscopic and atomic-level views of ordinary cotton.\n\n\nFIGURE 1.13 These images provide an increasingly closer view: (a) a cotton boll, (b) a single cotton fiber viewed under an optical microscope (magnified 40 times), (c) an image of a cotton fiber obtained with an electron microscope (much higher magnification than with the optical microscope); and (d and e) atomic-level models of the fiber (spheres of different colors represent atoms of different elements). (credit c: modification of work by \"Featheredtar\"/Wikimedia Commons)\n\nAn atom is so light that its mass is also difficult to imagine. A billion lead atoms ( $1,000,000,000$ atoms) weigh about $3 \\times 10^{-13}$ grams, a mass that is far too light to be weighed on even the world's most sensitive balances. It would require over $300,000,000,000,000$ lead atoms ( 300 trillion, or $3 \\times 10^{14}$ ) to be weighed, and they would weigh only 0.0000001 gram."}
{"id": 2177, "contents": "11. Atoms and Molecules - \nIt is rare to find collections of individual atoms. Only a few elements, such as the gases helium, neon, and argon, consist of a collection of individual atoms that move about independently of one another. Other elements, such as the gases hydrogen, nitrogen, oxygen, and chlorine, are composed of units that consist of pairs of atoms (Figure 1.14). One form of the element phosphorus consists of units composed of four phosphorus atoms. The element sulfur exists in various forms, one of which consists of units composed of eight sulfur atoms. These units are called molecules. A molecule consists of two or more atoms joined by strong forces called chemical bonds. The atoms in a molecule move around as a unit, much like the cans of soda in a six-pack or a bunch of keys joined together on a single key ring. A molecule may consist of two or more identical atoms, as in the molecules found in the elements hydrogen, oxygen, and sulfur, or it may consist of two or more different atoms, as in the molecules found in water. Each water molecule is a unit that contains two hydrogen atoms and one oxygen atom. Each glucose molecule is a unit that contains 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. Like atoms, molecules are incredibly small and light. If an ordinary glass of water were enlarged to the size of the earth, the water molecules inside it would be about the size of golf balls.\n\n\nFIGURE 1.14 The elements hydrogen, oxygen, phosphorus, and sulfur form molecules consisting of two or more atoms of the same element. The compounds water, carbon dioxide, and glucose consist of combinations of atoms of different elements."}
{"id": 2178, "contents": "13. Decomposition of Water / Production of Hydrogen - \nWater consists of the elements hydrogen and oxygen combined in a 2 to 1 ratio. Water can be broken down\ninto hydrogen and oxygen gases by the addition of energy. One way to do this is with a battery or power supply, as shown in (Figure 1.15).\n\n\nFIGURE 1.15 The decomposition of water is shown at the macroscopic, microscopic, and symbolic levels. The battery provides an electric current (microscopic) that decomposes water. At the macroscopic level, the liquid separates into the gases hydrogen (on the left) and oxygen (on the right). Symbolically, this change is presented by showing how liquid $\\mathrm{H}_{2} \\mathrm{O}$ separates into $\\mathrm{H}_{2}$ and $\\mathrm{O}_{2}$ gases.\n\nThe breakdown of water involves a rearrangement of the atoms in water molecules into different molecules, each composed of two hydrogen atoms and two oxygen atoms, respectively. Two water molecules form one oxygen molecule and two hydrogen molecules. The representation for what occurs, $2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{H}_{2}(g)+\\mathrm{O}_{2}(g)$, will be explored in more depth in later chapters.\n\nThe two gases produced have distinctly different properties. Oxygen is not flammable but is required for combustion of a fuel, and hydrogen is highly flammable and a potent energy source. How might this knowledge be applied in our world? One application involves research into more fuel-efficient transportation. Fuel-cell vehicles (FCV) run on hydrogen instead of gasoline (Figure 1.16). They are more efficient than vehicles with internal combustion engines, are nonpolluting, and reduce greenhouse gas emissions, making us less dependent on fossil fuels. FCVs are not yet economically viable, however, and current hydrogen production depends on natural gas. If we can develop a process to economically decompose water, or produce hydrogen in another environmentally sound way, FCVs may be the way of the future.\n\n\nFIGURE 1.16 A fuel cell generates electrical energy from hydrogen and oxygen via an electrochemical process and produces only water as the waste product."}
{"id": 2179, "contents": "15. Chemistry of Cell Phones - \nImagine how different your life would be without cell phones (Figure 1.17) and other smart devices. Cell phones are made from numerous chemical substances, which are extracted, refined, purified, and assembled using an extensive and in-depth understanding of chemical principles. About $30 \\%$ of the elements that are found in nature are found within a typical smart phone. The case/body/frame consists of a combination of sturdy, durable polymers composed primarily of carbon, hydrogen, oxygen, and nitrogen [acrylonitrile butadiene styrene (ABS) and polycarbonate thermoplastics], and light, strong, structural metals, such as aluminum, magnesium, and iron. The display screen is made from a specially toughened glass (silica glass strengthened by the addition of aluminum, sodium, and potassium) and coated with a material to make it conductive (such as indium tin oxide). The circuit board uses a semiconductor material (usually silicon); commonly used metals like copper, tin, silver, and gold; and more unfamiliar elements such as yttrium, praseodymium, and gadolinium. The battery relies upon lithium ions and a variety of other materials, including iron, cobalt, copper, polyethylene oxide, and polyacrylonitrile.\n\n\nFIGURE 1.17 Almost one-third of naturally occurring elements are used to make a cell phone. (credit: modification of work by John Taylor)"}
{"id": 2180, "contents": "16. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Identify properties of and changes in matter as physical or chemical\n- Identify properties of matter as extensive or intensive\n\nThe characteristics that distinguish one substance from another are called properties. A physical property is a characteristic of matter that is not associated with a change in its chemical composition. Familiar examples of physical properties include density, color, hardness, melting and boiling points, and electrical conductivity. Some physical properties, such as density and color, may be observed without changing the physical state of the matter. Other physical properties, such as the melting temperature of iron or the freezing temperature of water, can only be observed as matter undergoes a physical change. A physical change is a change in the state or properties of matter without any accompanying change in the chemical identities of the substances contained in the matter. Physical changes are observed when wax melts, when sugar dissolves in coffee, and when steam condenses into liquid water (Figure 1.18). Other examples of physical changes include magnetizing and demagnetizing metals (as is done with common antitheft security tags) and grinding solids into powders (which can sometimes yield noticeable changes in color). In each of these examples, there is a change in the physical state, form, or properties of the substance, but no change in its chemical composition.\n\n(a)\n\n(b)\n\nFIGURE 1.18 (a) Wax undergoes a physical change when solid wax is heated and forms liquid wax. (b) Steam condensing inside a cooking pot is a physical change, as water vapor is changed into liquid water. (credit a: modification of work by \"95jb14\"/Wikimedia Commons; credit b: modification of work by \"mjneuby\"/Flickr)\n\nThe change of one type of matter into another type (or the inability to change) is a chemical property. Examples of chemical properties include flammability, toxicity, acidity, and many other types of reactivity. Iron, for example, combines with oxygen in the presence of water to form rust; chromium does not oxidize (Figure 1.19). Nitroglycerin is very dangerous because it explodes easily; neon poses almost no hazard because it is very unreactive."}
{"id": 2181, "contents": "16. LEARNING OBJECTIVES - \nFIGURE 1.19 (a) One of the chemical properties of iron is that it rusts; (b) one of the chemical properties of chromium is that it does not. (credit a: modification of work by Tony Hisgett; credit b: modification of work by \"Atoma\"/Wikimedia Commons)\n\nA chemical change always produces one or more types of matter that differ from the matter present before the change. The formation of rust is a chemical change because rust is a different kind of matter than the iron, oxygen, and water present before the rust formed. The explosion of nitroglycerin is a chemical change because the gases produced are very different kinds of matter from the original substance. Other examples of chemical changes include reactions that are performed in a lab (such as copper reacting with nitric acid), all forms of combustion (burning), and food being cooked, digested, or rotting (Figure 1.20).\n\n\nFIGURE 1.20 (a) Copper and nitric acid undergo a chemical change to form copper nitrate and brown, gaseous nitrogen dioxide. (b) During the combustion of a match, cellulose in the match and oxygen from the air undergo a chemical change to form carbon dioxide and water vapor. (c) Cooking red meat causes a number of chemical changes, including the oxidation of iron in myoglobin that results in the familiar red-to-brown color change. (d) A banana turning brown is a chemical change as new, darker (and less tasty) substances form. (credit b: modification of work by Jeff Turner; credit c: modification of work by Gloria Cabada-Leman; credit d: modification of work by Roberto Verzo)"}
{"id": 2182, "contents": "16. LEARNING OBJECTIVES - \nProperties of matter fall into one of two categories. If the property depends on the amount of matter present, it is an extensive property. The mass and volume of a substance are examples of extensive properties; for instance, a gallon of milk has a larger mass than a cup of milk. The value of an extensive property is directly proportional to the amount of matter in question. If the property of a sample of matter does not depend on the amount of matter present, it is an intensive property. Temperature is an example of an intensive property. If the gallon and cup of milk are each at $20^{\\circ} \\mathrm{C}$ (room temperature), when they are combined, the temperature remains at $20^{\\circ} \\mathrm{C}$. As another example, consider the distinct but related properties of heat and temperature. A drop of hot cooking oil spattered on your arm causes brief, minor discomfort, whereas a pot of hot oil yields severe burns. Both the drop and the pot of oil are at the same temperature (an intensive property), but the pot clearly contains much more heat (extensive property)."}
{"id": 2183, "contents": "18. Hazard Diamond - \nYou may have seen the symbol shown in Figure 1.21 on containers of chemicals in a laboratory or workplace. Sometimes called a \"fire diamond\" or \"hazard diamond,\" this chemical hazard diamond provides valuable information that briefly summarizes the various dangers of which to be aware when working with a particular substance.\n\n\nFIGURE 1.21 The National Fire Protection Agency (NFPA) hazard diamond summarizes the major hazards of a chemical substance.\n\nThe National Fire Protection Agency (NFPA) 704 Hazard Identification System was developed by NFPA to provide safety information about certain substances. The system details flammability, reactivity, health, and other hazards. Within the overall diamond symbol, the top (red) diamond specifies the level of fire hazard (temperature range for flash point). The blue (left) diamond indicates the level of health hazard. The yellow (right) diamond describes reactivity hazards, such as how readily the substance will undergo detonation or a violent chemical change. The white (bottom) diamond points out special hazards, such as if it is an oxidizer (which allows the substance to burn in the absence of air/oxygen), undergoes an unusual or dangerous reaction with water, is corrosive, acidic, alkaline, a biological hazard, radioactive, and so on. Each hazard is rated on a scale from 0 to 4 , with 0 being no hazard and 4 being extremely hazardous.\n\nWhile many elements differ dramatically in their chemical and physical properties, some elements have similar properties. For example, many elements conduct heat and electricity well, whereas others are poor conductors. These properties can be used to sort the elements into three classes: metals (elements that conduct well), nonmetals (elements that conduct poorly), and metalloids (elements that have intermediate conductivities).\n\nThe periodic table is a table of elements that places elements with similar properties close together (Figure 1.22). You will learn more about the periodic table as you continue your study of chemistry."}
{"id": 2184, "contents": "19. Periodic Table of the Elements - \nFIGURE 1.22 The periodic table shows how elements may be grouped according to certain similar properties. Note the background color denotes whether an element is a metal, metalloid, or nonmetal, whereas the element symbol color indicates whether it is a solid, liquid, or gas."}
{"id": 2185, "contents": "20. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Explain the process of measurement\n- Identify the three basic parts of a quantity\n- Describe the properties and units of length, mass, volume, density, temperature, and time\n- Perform basic unit calculations and conversions in the metric and other unit systems\n\nMeasurements provide much of the information that informs the hypotheses, theories, and laws describing the behavior of matter and energy in both the macroscopic and microscopic domains of chemistry. Every measurement provides three kinds of information: the size or magnitude of the measurement (a number); a standard of comparison for the measurement (a unit); and an indication of the uncertainty of the measurement. While the number and unit are explicitly represented when a quantity is written, the uncertainty is an aspect of the measurement result that is more implicitly represented and will be discussed later.\n\nThe number in the measurement can be represented in different ways, including decimal form and scientific notation. (Scientific notation is also known as exponential notation; a review of this topic can be found in Appendix B.) For example, the maximum takeoff weight of a Boeing 777-200ER airliner is 298,000 kilograms, which can also be written as $2.98 \\times 10^{5} \\mathrm{~kg}$. The mass of the average mosquito is about 0.0000025 kilograms,\nwhich can be written as $2.5 \\times 10^{-6} \\mathrm{~kg}$.\nUnits, such as liters, pounds, and centimeters, are standards of comparison for measurements. A 2-liter bottle of a soft drink contains a volume of beverage that is twice that of the accepted volume of 1 liter. The meat used to prepare a 0.25 -pound hamburger weighs one-fourth as much as the accepted weight of 1 pound. Without units, a number can be meaningless, confusing, or possibly life threatening. Suppose a doctor prescribes phenobarbital to control a patient's seizures and states a dosage of \" 100 \" without specifying units. Not only will this be confusing to the medical professional giving the dose, but the consequences can be dire: 100 mg given three times per day can be effective as an anticonvulsant, but a single dose of 100 g is more than 10 times the lethal amount."}
{"id": 2186, "contents": "20. LEARNING OBJECTIVES - \nThe measurement units for seven fundamental properties (\"base units\") are listed in Table 1.2. The standards for these units are fixed by international agreement, and they are called the International System of Units or SI Units (from the French, Le Syst\u00e8me International d'Unit\u00e9s). SI units have been used by the United States National Institute of Standards and Technology (NIST) since 1964. Units for other properties may be derived from these seven base units.\n\nBase Units of the SI System\n\n| Property Measured | Name of Unit | Symbol of Unit |\n| :--- | :--- | :--- |\n| length | meter | m |\n| mass | kilogram | kg |\n| time | second | s |\n| temperature | kelvin | K |\n| electric current | ampere | A |\n| amount of substance | mole | mol |\n| luminous intensity | candela | cd |\n\nTABLE 1.2\n\nEveryday measurement units are often defined as fractions or multiples of other units. Milk is commonly packaged in containers of 1 gallon ( 4 quarts), 1 quart ( 0.25 gallon), and one pint ( 0.5 quart). This same approach is used with SI units, but these fractions or multiples are always powers of 10. Fractional or multiple SI units are named using a prefix and the name of the base unit. For example, a length of 1000 meters is also called a kilometer because the prefix kilo means \"one thousand,\" which in scientific notation is $10^{3}(1$ kilometer $=1000 \\mathrm{~m}=10^{3} \\mathrm{~m}$ ). The prefixes used and the powers to which 10 are raised are listed in Table 1.3.\n\nCommon Unit Prefixes\n\n| Prefix | Symbol | Factor | Example |\n| :--- | :--- | :--- | :--- |\n| femto | f | $10^{-15}$ | 1 femtosecond $(\\mathrm{fs})=1 \\times 10^{-15} \\mathrm{~s}(0.000000000000001 \\mathrm{~s})$ |\n| pico | p | $10^{-12}$ | 1 picometer $(\\mathrm{pm})=1 \\times 10^{-12} \\mathrm{~m}(0.000000000001 \\mathrm{~m})$ |"}
{"id": 2187, "contents": "20. LEARNING OBJECTIVES - \nTABLE 1.3\n\n| Prefix | Symbol | Factor | Example |\n| :--- | :--- | :--- | :--- |\n| nano | n | $10^{-9}$ | 4 nanograms $(\\mathrm{ng})=4 \\times 10^{-9} \\mathrm{~g}(0.000000004 \\mathrm{~g})$ |\n| micro | $\\mu$ | $10^{-6}$ | 1 microliter $(\\mu \\mathrm{L})=1 \\times 10^{-6} \\mathrm{~L}(0.000001 \\mathrm{~L})$ |\n| milli | m | $10^{-3}$ | 2 millimoles $(\\mathrm{mmol})=2 \\times 10^{-3} \\mathrm{~mol}(0.002 \\mathrm{~mol})$ |\n| centi | c | $10^{-2}$ | 7 centimeters $(\\mathrm{cm})=7 \\times 10^{-2} \\mathrm{~m}(0.07 \\mathrm{~m})$ |\n| deci | d | $10^{-1}$ | 1 deciliter $(\\mathrm{dL})=1 \\times 10^{-1} \\mathrm{~L}(0.1 \\mathrm{~L})$ |\n| kilo | k | $10^{3}$ | 1 kilometer $(\\mathrm{km})=1 \\times 10^{3} \\mathrm{~m}(1000 \\mathrm{~m})$ |\n| mega | M | $10^{6}$ | 3 megahertz $(\\mathrm{MHz})=3 \\times 10^{6} \\mathrm{~Hz}(3,000,000 \\mathrm{~Hz})$ |\n| giga | G | $10^{9}$ | 8 gigayears $(\\mathrm{Gyr})=8 \\times 10^{9} \\mathrm{yr}(8,000,000,000 \\mathrm{yr})$ |\n| tera | T | $10^{12}$ | 5 terawatts $(\\mathrm{TW})=5 \\times 10^{12} \\mathrm{~W}(5,000,000,000,000 \\mathrm{~W})$ |\n\nTABLE 1.3"}
{"id": 2188, "contents": "21. LINK TO LEARNING - \nNeed a refresher or more practice with scientific notation? Visit this site (http://openstax.org/l/16notation) to go over the basics of scientific notation."}
{"id": 2189, "contents": "22. SI Base Units - \nThe initial units of the metric system, which eventually evolved into the SI system, were established in France during the French Revolution. The original standards for the meter and the kilogram were adopted there in 1799 and eventually by other countries. This section introduces four of the SI base units commonly used in chemistry. Other SI units will be introduced in subsequent chapters."}
{"id": 2190, "contents": "23. Length - \nThe standard unit of length in both the SI and original metric systems is the meter (m). A meter was originally specified as $1 / 10,000,000$ of the distance from the North Pole to the equator. It is now defined as the distance light in a vacuum travels in $1 / 299,792,458$ of a second. A meter is about 3 inches longer than a yard (Figure 1.23); one meter is about 39.37 inches or 1.094 yards. Longer distances are often reported in kilometers ( 1 km $=1000 \\mathrm{~m}=10^{3} \\mathrm{~m}$ ), whereas shorter distances can be reported in centimeters ( $1 \\mathrm{~cm}=0.01 \\mathrm{~m}=10^{-2} \\mathrm{~m}$ ) or millimeters $\\left(1 \\mathrm{~mm}=0.001 \\mathrm{~m}=10^{-3} \\mathrm{~m}\\right)$.\n\n\nFIGURE 1.23 The relative lengths of $1 \\mathrm{~m}, 1 \\mathrm{yd}, 1 \\mathrm{~cm}$, and 1 in . are shown (not actual size), as well as comparisons of 2.54 cm and 1 in ., and of 1 m and 1.094 yd ."}
{"id": 2191, "contents": "24. Mass - \nThe standard unit of mass in the SI system is the kilogram (kg). The kilogram was previously defined by the International Union of Pure and Applied Chemistry (IUPAC) as the mass of a specific reference object. This object was originally one liter of pure water, and more recently it was a metal cylinder made from a platinumiridium alloy with a height and diameter of 39 mm (Figure 1.24). In May 2019, this definition was changed to one that is based instead on precisely measured values of several fundamental physical constants. ${ }^{\\underline{1}}$. One kilogram is about 2.2 pounds. The gram (g) is exactly equal to $1 / 1000$ of the mass of the kilogram $\\left(10^{-3} \\mathrm{~kg}\\right)$.\n\n\nFIGURE 1.24 This replica prototype kilogram as previously defined is housed at the National Institute of Standards and Technology (NIST) in Maryland. (credit: National Institutes of Standards and Technology)"}
{"id": 2192, "contents": "25. Temperature - \nTemperature is an intensive property. The SI unit of temperature is the kelvin (K). The IUPAC convention is to use kelvin (all lowercase) for the word, K (uppercase) for the unit symbol, and neither the word \"degree\" nor the degree symbol $\\left({ }^{\\circ}\\right)$. The degree Celsius $\\left({ }^{\\circ} \\mathbf{C}\\right)$ is also allowed in the SI system, with both the word \"degree\" and the degree symbol used for Celsius measurements. Celsius degrees are the same magnitude as those of kelvin, but the two scales place their zeros in different places. Water freezes at $273.15 \\mathrm{~K}\\left(0^{\\circ} \\mathrm{C}\\right)$ and boils at 373.15 K $\\left(100^{\\circ} \\mathrm{C}\\right)$ by definition, and normal human body temperature is approximately $310 \\mathrm{~K}\\left(37^{\\circ} \\mathrm{C}\\right)$. The conversion\n\n1 For details see https://www.nist.gov/pml/weights-and-measures/si-units-mass\nbetween these two units and the Fahrenheit scale will be discussed later in this chapter.\nTime\nThe SI base unit of time is the second (s). Small and large time intervals can be expressed with the appropriate prefixes; for example, 3 microseconds $=0.000003 \\mathrm{~s}=3 \\times 10^{-6}$ and 5 megaseconds $=5,000,000 \\mathrm{~s}=5 \\times 10^{6} \\mathrm{~s}$. Alternatively, hours, days, and years can be used."}
{"id": 2193, "contents": "26. Derived SI Units - \nWe can derive many units from the seven SI base units. For example, we can use the base unit of length to define a unit of volume, and the base units of mass and length to define a unit of density.\n\nVolume\nVolume is the measure of the amount of space occupied by an object. The standard SI unit of volume is defined by the base unit of length (Figure 1.25). The standard volume is a cubic meter ( $\\mathbf{m}^{\\mathbf{3}}$ ), a cube with an edge length of exactly one meter. To dispense a cubic meter of water, we could build a cubic box with edge lengths of exactly one meter. This box would hold a cubic meter of water or any other substance.\n\nA more commonly used unit of volume is derived from the decimeter ( 0.1 m , or 10 cm ). A cube with edge lengths of exactly one decimeter contains a volume of one cubic decimeter ( $\\mathrm{dm}^{3}$ ). A liter ( $\\mathbf{L}$ ) is the more common name for the cubic decimeter. One liter is about 1.06 quarts.\n\nA cubic centimeter ( $\\mathbf{c m}^{\\mathbf{3}}$ ) is the volume of a cube with an edge length of exactly one centimeter. The abbreviation cc (for cubic centimeter) is often used by health professionals. A cubic centimeter is equivalent to a milliliter ( $\\mathbf{m L}$ ) and is $1 / 1000$ of a liter.\n\n\nFIGURE 1.25 (a) The relative volumes are shown for cubes of $1 \\mathrm{~m}^{3}, 1 \\mathrm{dm}^{3}(1 \\mathrm{~L})$, and $1 \\mathrm{~cm}^{3}(1 \\mathrm{~mL})$ (not to scale). (b) The diameter of a dime is compared relative to the edge length of a $1-\\mathrm{cm}^{3}(1-\\mathrm{mL}) \\mathrm{cube}$."}
{"id": 2194, "contents": "27. Density - \nWe use the mass and volume of a substance to determine its density. Thus, the units of density are defined by the base units of mass and length.\n\nThe density of a substance is the ratio of the mass of a sample of the substance to its volume. The SI unit for density is the kilogram per cubic meter $\\left(\\mathrm{kg} / \\mathrm{m}^{3}\\right)$. For many situations, however, this is an inconvenient unit, and we often use grams per cubic centimeter $\\left(\\mathrm{g} / \\mathrm{cm}^{3}\\right)$ for the densities of solids and liquids, and grams per liter $(\\mathrm{g} / \\mathrm{L})$ for gases. Although there are exceptions, most liquids and solids have densities that range from about 0.7 $\\mathrm{g} / \\mathrm{cm}^{3}$ (the density of gasoline) to $19 \\mathrm{~g} / \\mathrm{cm}^{3}$ (the density of gold). The density of air is about $1.2 \\mathrm{~g} / \\mathrm{L}$. Table 1.4\nshows the densities of some common substances.\n\nDensities of Common Substances"}
{"id": 2195, "contents": "27. Density - \n| Solids | Liquids | Gases (at $25^{\\circ} \\mathrm{C}$ and 1 atm ) |\n| :--- | :--- | :--- |\n| ice $\\left(\\right.$ at $\\left.0^{\\circ} \\mathrm{C}\\right) 0.92 \\mathrm{~g} / \\mathrm{cm}^{3}$ | water $1.0 \\mathrm{~g} / \\mathrm{cm}^{3}$ | dry air $1.20 \\mathrm{~g} / \\mathrm{L}$ |\n| oak (wood) $0.60-0.90 \\mathrm{~g} / \\mathrm{cm}^{3}$ | ethanol $0.79 \\mathrm{~g} / \\mathrm{cm}^{3}$ | oxygen $1.31 \\mathrm{~g} / \\mathrm{L}$ |\n| iron $7.9 \\mathrm{~g} / \\mathrm{cm}^{3}$ | acetone $0.79 \\mathrm{~g} / \\mathrm{cm}^{3}$ | nitrogen $1.14 \\mathrm{~g} / \\mathrm{L}$ |\n| copper $9.0 \\mathrm{~g} / \\mathrm{cm}^{3}$ | glycerin $1.26 \\mathrm{~g} / \\mathrm{cm}^{3}$ | carbon dioxide $1.80 \\mathrm{~g} / \\mathrm{L}$ |\n| lead $11.3 \\mathrm{~g} / \\mathrm{cm}^{3}$ | olive oil $0.92 \\mathrm{~g} / \\mathrm{cm}^{3}$ | helium $0.16 \\mathrm{~g} / \\mathrm{L}$ |\n| silver $10.5 \\mathrm{~g} / \\mathrm{cm}^{3}$ | gasoline $0.70-0.77 \\mathrm{~g} / \\mathrm{cm}^{3}$ | neon $0.83 \\mathrm{~g} / \\mathrm{L}$ |"}
{"id": 2196, "contents": "27. Density - \n| gold $19.3 \\mathrm{~g} / \\mathrm{cm}^{3}$ | mercury $13.6 \\mathrm{~g} / \\mathrm{cm}^{3}$ | radon $9.1 \\mathrm{~g} / \\mathrm{L}$ |"}
{"id": 2197, "contents": "27. Density - \nTABLE 1.4\n\nWhile there are many ways to determine the density of an object, perhaps the most straightforward method involves separately finding the mass and volume of the object, and then dividing the mass of the sample by its volume. In the following example, the mass is found directly by weighing, but the volume is found indirectly through length measurements.\n\n$$\n\\text { density }=\\frac{\\text { mass }}{\\text { volume }}\n$$"}
{"id": 2198, "contents": "29. Calculation of Density - \nGold-in bricks, bars, and coins-has been a form of currency for centuries. In order to swindle people into paying for a brick of gold without actually investing in a brick of gold, people have considered filling the centers of hollow gold bricks with lead to fool buyers into thinking that the entire brick is gold. It does not work: Lead is a dense substance, but its density is not as great as that of gold, $19.3 \\mathrm{~g} / \\mathrm{cm}^{3}$. What is the density of lead if a cube of lead has an edge length of 2.00 cm and a mass of 90.7 g ?"}
{"id": 2199, "contents": "30. Solution - \nThe density of a substance can be calculated by dividing its mass by its volume. The volume of a cube is calculated by cubing the edge length.\n\n$$\n\\begin{aligned}\n& \\text { volume of lead cube }=2.00 \\mathrm{~cm} \\times 2.00 \\mathrm{~cm} \\times 2.00 \\mathrm{~cm}=8.00 \\mathrm{~cm}^{3} \\\\\n& \\qquad \\text { density }=\\frac{\\text { mass }}{\\text { volume }}=\\frac{90.7 \\mathrm{~g}}{8.00 \\mathrm{~cm}^{3}}=11.3 \\mathrm{~g} / \\mathrm{cm}^{3}\n\\end{aligned}\n$$\n\n(We will discuss the reason for rounding to the first decimal place in the next section.)"}
{"id": 2200, "contents": "31. Check Your Learning - \n(a) To three decimal places, what is the volume of a cube $\\left(\\mathrm{cm}^{3}\\right)$ with an edge length of 0.843 cm ?\n(b) If the cube in part (a) is copper and has a mass of 5.34 g , what is the density of copper to two decimal places?"}
{"id": 2201, "contents": "32. Answer: - \n(a) $0.599 \\mathrm{~cm}^{3}$; (b) $8.91 \\mathrm{~g} / \\mathrm{cm}^{3}$"}
{"id": 2202, "contents": "33. LINK TO LEARNING - \nTo learn more about the relationship between mass, volume, and density, use this interactive simulator (http://openstax.org/l/16phetmasvolden) to explore the density of different materials."}
{"id": 2203, "contents": "35. Using Displacement of Water to Determine Density - \nThis exercise uses a simulation (http://openstax.org/l/16phetmasvolden) to illustrate an alternative approach to the determination of density that involves measuring the object's volume via displacement of water. Use the simulator to determine the densities iron and wood."}
{"id": 2204, "contents": "36. Solution - \nClick the \"turn fluid into water\" button in the simulator to adjust the density of liquid in the beaker to $1.00 \\mathrm{~g} /$ mL . Remove the red block from the beaker and note the volume of water is 25.5 mL . Select the iron sample by clicking \"iron\" in the table of materials at the bottom of the screen, place the iron block on the balance pan, and observe its mass is 31.48 g . Transfer the iron block to the beaker and notice that it sinks, displacing a volume of water equal to its own volume and causing the water level to rise to 29.5 mL . The volume of the iron block is therefore:\n\n$$\nv_{\\text {iron }}=29.5 \\mathrm{~mL}-25.5 \\mathrm{~mL}=4.0 \\mathrm{~mL}\n$$\n\nThe density of the iron is then calculated to be:\n\n$$\n\\text { density }=\\frac{\\text { mass }}{\\text { volume }}=\\frac{31.48 \\mathrm{~g}}{4.0 \\mathrm{~mL}}=7.9 \\mathrm{~g} / \\mathrm{mL}\n$$\n\nRemove the iron block from the beaker, change the block material to wood, and then repeat the mass and volume measurements. Unlike iron, the wood block does not sink in the water but instead floats on the water's surface. To measure its volume, drag it beneath the water's surface so that it is fully submerged.\n\n$$\n\\text { density }=\\frac{\\text { mass }}{\\text { volume }}=\\frac{1.95 \\mathrm{~g}}{3.0 \\mathrm{~mL}}=0.65 \\mathrm{~g} / \\mathrm{mL}\n$$\n\nNote: The sink versus float behavior illustrated in this example demonstrates the property of \"buoyancy\" (see end of chapter Exercise 1.42 and Exercise 1.43)."}
{"id": 2205, "contents": "37. Check Your Learning - \nFollowing the water displacement approach, use the simulator to measure the density of the foam sample."}
{"id": 2206, "contents": "38. Answer: - \n$0.230 \\mathrm{~g} / \\mathrm{mL}$"}
{"id": 2207, "contents": "39. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Define accuracy and precision\n- Distinguish exact and uncertain numbers\n- Correctly represent uncertainty in quantities using significant figures\n- Apply proper rounding rules to computed quantities\n\nCounting is the only type of measurement that is free from uncertainty, provided the number of objects being\ncounted does not change while the counting process is underway. The result of such a counting measurement is an example of an exact number. By counting the eggs in a carton, one can determine exactly how many eggs the carton contains. The numbers of defined quantities are also exact. By definition, 1 foot is exactly 12 inches, 1 inch is exactly 2.54 centimeters, and 1 gram is exactly 0.001 kilogram. Quantities derived from measurements other than counting, however, are uncertain to varying extents due to practical limitations of the measurement process used."}
{"id": 2208, "contents": "40. Significant Figures in Measurement - \nThe numbers of measured quantities, unlike defined or directly counted quantities, are not exact. To measure the volume of liquid in a graduated cylinder, you should make a reading at the bottom of the meniscus, the lowest point on the curved surface of the liquid.\n\n\nFIGURE 1.26 To measure the volume of liquid in this graduated cylinder, you must mentally subdivide the distance between the 21 and 22 mL marks into tenths of a milliliter, and then make a reading (estimate) at the bottom of the meniscus.\n\nRefer to the illustration in Figure 1.26. The bottom of the meniscus in this case clearly lies between the 21 and 22 markings, meaning the liquid volume is certainly greater than 21 mL but less than 22 mL . The meniscus appears to be a bit closer to the $22-\\mathrm{mL}$ mark than to the $21-\\mathrm{mL}$ mark, and so a reasonable estimate of the liquid's volume would be 21.6 mL . In the number 21.6, then, the digits 2 and 1 are certain, but the 6 is an estimate. Some people might estimate the meniscus position to be equally distant from each of the markings and estimate the tenth-place digit as 5 , while others may think it to be even closer to the $22-\\mathrm{mL}$ mark and estimate this digit to be 7 . Note that it would be pointless to attempt to estimate a digit for the hundredths place, given that the tenths-place digit is uncertain. In general, numerical scales such as the one on this graduated cylinder will permit measurements to one-tenth of the smallest scale division. The scale in this case has $1-\\mathrm{mL}$ divisions, and so volumes may be measured to the nearest 0.1 mL ."}
{"id": 2209, "contents": "40. Significant Figures in Measurement - \nThis concept holds true for all measurements, even if you do not actively make an estimate. If you place a quarter on a standard electronic balance, you may obtain a reading of 6.72 g . The digits 6 and 7 are certain, and the 2 indicates that the mass of the quarter is likely between 6.71 and 6.73 grams. The quarter weighs about 6.72 grams, with a nominal uncertainty in the measurement of $\\pm 0.01$ gram. If the coin is weighed on a more sensitive balance, the mass might be 6.723 g . This means its mass lies between 6.722 and 6.724 grams, an uncertainty of 0.001 gram. Every measurement has some uncertainty, which depends on the device used\n(and the user's ability). All of the digits in a measurement, including the uncertain last digit, are called significant figures or significant digits. Note that zero may be a measured value; for example, if you stand on a scale that shows weight to the nearest pound and it shows \" 120 ,\" then the 1 (hundreds), 2 (tens) and 0 (ones) are all significant (measured) values.\n\nA measurement result is properly reported when its significant digits accurately represent the certainty of the measurement process. But what if you were analyzing a reported value and trying to determine what is significant and what is not? Well, for starters, all nonzero digits are significant, and it is only zeros that require some thought. We will use the terms \"leading,\" \"trailing,\" and \"captive\" for the zeros and will consider how to deal with them.\n\n\nStarting with the first nonzero digit on the left, count this digit and all remaining digits to the right. This is the number of significant figures in the measurement unless the last digit is a trailing zero lying to the left of the decimal point.\n\n\nCaptive zeros result from measurement and are therefore always significant. Leading zeros, however, are never significant-they merely tell us where the decimal point is located.\n\n\nThe leading zeros in this example are not significant. We could use exponential notation (as described in Appendix B) and express the number as $8.32407 \\times 10^{-3}$; then the number 8.32407 contains all of the significant figures, and $10^{-3}$ locates the decimal point."}
{"id": 2210, "contents": "40. Significant Figures in Measurement - \nThe number of significant figures is uncertain in a number that ends with a zero to the left of the decimal point location. The zeros in the measurement 1,300 grams could be significant or they could simply indicate where the decimal point is located. The ambiguity can be resolved with the use of exponential notation: $1.3 \\times 10^{3}$ (two significant figures), $1.30 \\times 10^{3}$ (three significant figures, if the tens place was measured), or $1.300 \\times 10^{3}$ (four significant figures, if the ones place was also measured). In cases where only the decimal-formatted number is available, it is prudent to assume that all trailing zeros are not significant.\n\n\nWhen determining significant figures, be sure to pay attention to reported values and think about the measurement and significant figures in terms of what is reasonable or likely when evaluating whether the\nvalue makes sense. For example, the official January 2014 census reported the resident population of the US as $317,297,725$. Do you think the US population was correctly determined to the reported nine significant figures, that is, to the exact number of people? People are constantly being born, dying, or moving into or out of the country, and assumptions are made to account for the large number of people who are not actually counted. Because of these uncertainties, it might be more reasonable to expect that we know the population to within perhaps a million or so, in which case the population should be reported as $3.17 \\times 10^{8}$ people."}
{"id": 2211, "contents": "41. Significant Figures in Calculations - \nA second important principle of uncertainty is that results calculated from a measurement are at least as uncertain as the measurement itself. Take the uncertainty in measurements into account to avoid misrepresenting the uncertainty in calculated results. One way to do this is to report the result of a calculation with the correct number of significant figures, which is determined by the following three rules for rounding numbers:\n\n1. When adding or subtracting numbers, round the result to the same number of decimal places as the number with the least number of decimal places (the least certain value in terms of addition and subtraction).\n2. When multiplying or dividing numbers, round the result to the same number of digits as the number with the least number of significant figures (the least certain value in terms of multiplication and division).\n3. If the digit to be dropped (the one immediately to the right of the digit to be retained) is less than 5, \"round down\" and leave the retained digit unchanged; if it is more than 5, \"round up\" and increase the retained digit by 1 . If the dropped digit is 5 , and it's either the last digit in the number or it's followed only by zeros, round up or down, whichever yields an even value for the retained digit. If any nonzero digits follow the dropped 5 , round up. (The last part of this rule may strike you as a bit odd, but it's based on reliable statistics and is aimed at avoiding any bias when dropping the digit \" 5 ,\" since it is equally close to both possible values of the retained digit.)\n\nThe following examples illustrate the application of this rule in rounding a few different numbers to three significant figures:\n\n- 0.028675 rounds \"up\" to 0.0287 (the dropped digit, 7 , is greater than 5 )\n- 18.3384 rounds \"down\" to 18.3 (the dropped digit, 3 , is less than 5)\n- 6.8752 rounds \"up\" to 6.88 (the dropped digit is 5 , and a nonzero digit follows it)\n- 92.85 rounds \"down\" to 92.8 (the dropped digit is 5 , and the retained digit is even)\n\nLet's work through these rules with a few examples."}
{"id": 2212, "contents": "43. Rounding Numbers - \nRound the following to the indicated number of significant figures:\n(a) 31.57 (to two significant figures)\n(b) 8.1649 (to three significant figures)\n(c) 0.051065 (to four significant figures)\n(d) 0.90275 (to four significant figures)"}
{"id": 2213, "contents": "44. Solution - \n(a) 31.57 rounds \"up\" to 32 (the dropped digit is 5 , and the retained digit is even)\n(b) 8.1649 rounds \"down\" to 8.16 (the dropped digit, 4 , is less than 5)\n(c) 0.051065 rounds \"down\" to 0.05106 (the dropped digit is 5 , and the retained digit is even)\n(d) 0.90275 rounds \"up\" to 0.9028 (the dropped digit is 5 , and the retained digit is even)"}
{"id": 2214, "contents": "45. Check Your Learning - \nRound the following to the indicated number of significant figures:\n(a) 0.424 (to two significant figures)\n(b) 0.0038661 (to three significant figures)\n(c) 421.25 (to four significant figures)\n(d) 28,683.5 (to five significant figures)"}
{"id": 2215, "contents": "46. Answer: - \n(a) 0.42; (b) 0.00387; (c) 421.2; (d) 28,684"}
{"id": 2216, "contents": "48. Addition and Subtraction with Significant Figures - \nRule: When adding or subtracting numbers, round the result to the same number of decimal places as the number with the fewest decimal places (i.e., the least certain value in terms of addition and subtraction).\n(a) Add 1.0023 g and 4.383 g .\n(b) Subtract 421.23 g from 486 g ."}
{"id": 2217, "contents": "49. Solution - \n(a)\n\n$$\n\\begin{array}{r}\n1.0023 \\mathrm{~g} \\\\\n+4.383 \\mathrm{~g} \\\\\n\\hline 5.3853 \\mathrm{~g}\n\\end{array}\n$$\n\nAnswer is 5.385 g (round to the thousandths place; three decimal places)\n(b)\n\n$$\n\\begin{array}{r}\n486 \\mathrm{~g} \\\\\n-421.23 \\mathrm{~g} \\\\\n\\hline 64.77 \\mathrm{~g}\n\\end{array}\n$$\n\nAnswer is 65 g (round to the ones place; no decimal places)\n\n\nCheck Your Learning\n(a) Add 2.334 mL and 0.31 mL .\n(b) Subtract 55.8752 m from 56.533 m ."}
{"id": 2218, "contents": "50. Answer: - \n(a) 2.64 mL ; (b) 0.658 m"}
{"id": 2219, "contents": "52. Multiplication and Division with Significant Figures - \nRule: When multiplying or dividing numbers, round the result to the same number of digits as the number with the fewest significant figures (the least certain value in terms of multiplication and division).\n(a) Multiply 0.6238 cm by 6.6 cm .\n(b) Divide 421.23 g by 486 mL ."}
{"id": 2220, "contents": "53. Solution - \n(a) $0.6238 \\mathrm{~cm} \\times 6.6 \\mathrm{~cm}=4.11708 \\mathrm{~cm}^{2} \\longrightarrow$ result is $4.1 \\mathrm{~cm}^{2}$ (round to two significant figures)\nfour significant figures $\\times$ two significant figures $\\longrightarrow$ two significant figures answer\n$\\frac{421.23 \\mathrm{~g}}{486 \\mathrm{~mL}}=0.866728 \\ldots \\mathrm{~g} / \\mathrm{mL} \\longrightarrow$ result is $0.867 \\mathrm{~g} / \\mathrm{mL}$ (round to three significant figures)\n(b)\n$\\frac{\\text { five significant figures }}{\\text { three significant figures }} \\longrightarrow$ three significant figures answer\nCheck Your Learning\n(a) Multiply 2.334 cm and 0.320 cm .\n(b) Divide 55.8752 m by 56.53 s ."}
{"id": 2221, "contents": "54. Answer: - \n(a) $0.747 \\mathrm{~cm}^{2}$ (b) $0.9884 \\mathrm{~m} / \\mathrm{s}$\n\nIn the midst of all these technicalities, it is important to keep in mind the reason for these rules about significant figures and rounding-to correctly represent the certainty of the values reported and to ensure that a calculated result is not represented as being more certain than the least certain value used in the calculation."}
{"id": 2222, "contents": "56. Calculation with Significant Figures - \nOne common bathtub is 13.44 dm long, 5.920 dm wide, and 2.54 dm deep. Assume that the tub is rectangular and calculate its approximate volume in liters."}
{"id": 2223, "contents": "57. Solution - \n$$\n\\begin{aligned}\nV & =l \\times w \\times d \\\\\n& =13.44 \\mathrm{dm} \\times 5.920 \\mathrm{dm} \\times 2.54 \\mathrm{dm} \\\\\n& =202.09459 \\ldots \\mathrm{dm}^{3}(\\text { value from calculator) } \\\\\n& =202 \\mathrm{dm}^{3}, \\text { or } 202 \\mathrm{~L} \\text { (answer rounded to three significant figures) }\n\\end{aligned}\n$$"}
{"id": 2224, "contents": "58. Check Your Learning - \nWhat is the density of a liquid with a mass of 31.1415 g and a volume of $30.13 \\mathrm{~cm}^{3}$ ?"}
{"id": 2225, "contents": "59. Answer: - \n$1.034 \\mathrm{~g} / \\mathrm{mL}$"}
{"id": 2226, "contents": "61. Experimental Determination of Density Using Water Displacement - \nA piece of rebar is weighed and then submerged in a graduated cylinder partially filled with water, with results as shown.\n\n(a) Use these values to determine the density of this piece of rebar.\n(b) Rebar is mostly iron. Does your result in (a) support this statement? How?"}
{"id": 2227, "contents": "62. Solution - \nThe volume of the piece of rebar is equal to the volume of the water displaced:\n\n$$\n\\text { volume }=22.4 \\mathrm{~mL}-13.5 \\mathrm{~mL}=8.9 \\mathrm{~mL}=8.9 \\mathrm{~cm}^{3}\n$$\n\n(rounded to the nearest 0.1 mL , per the rule for addition and subtraction)\nThe density is the mass-to-volume ratio:\n\n$$\n\\text { density }=\\frac{\\text { mass }}{\\text { volume }}=\\frac{69.658 \\mathrm{~g}}{8.9 \\mathrm{~cm}^{3}}=7.8 \\mathrm{~g} / \\mathrm{cm}^{3}\n$$\n\n(rounded to two significant figures, per the rule for multiplication and division)\nFrom Table 1.4, the density of iron is $7.9 \\mathrm{~g} / \\mathrm{cm}^{3}$, very close to that of rebar, which lends some support to the fact that rebar is mostly iron."}
{"id": 2228, "contents": "63. Check Your Learning - \nAn irregularly shaped piece of a shiny yellowish material is weighed and then submerged in a graduated cylinder, with results as shown.\n\n(a) Use these values to determine the density of this material.\n(b) Do you have any reasonable guesses as to the identity of this material? Explain your reasoning."}
{"id": 2229, "contents": "64. Answer: - \n(a) $19 \\mathrm{~g} / \\mathrm{cm}^{3}$; (b) It is likely gold; the right appearance for gold and very close to the density given for gold in Table 1.4."}
{"id": 2230, "contents": "65. Accuracy and Precision - \nScientists typically make repeated measurements of a quantity to ensure the quality of their findings and to evaluate both the precision and the accuracy of their results. Measurements are said to be precise if they yield very similar results when repeated in the same manner. A measurement is considered accurate if it yields a result that is very close to the true or accepted value. Precise values agree with each other; accurate values agree with a true value. These characterizations can be extended to other contexts, such as the results of an archery competition (Figure 1.27).\n\n(a)\n\n\nPrecise, not accurate\n(b)\n\n\nNot accurate, not precise\n(c)\n\nFIGURE 1.27 (a) These arrows are close to both the bull's eye and one another, so they are both accurate and precise. (b) These arrows are close to one another but not on target, so they are precise but not accurate. (c) These arrows are neither on target nor close to one another, so they are neither accurate nor precise.\n\nSuppose a quality control chemist at a pharmaceutical company is tasked with checking the accuracy and precision of three different machines that are meant to dispense 10 ounces ( 296 mL ) of cough syrup into storage bottles. She proceeds to use each machine to fill five bottles and then carefully determines the actual volume dispensed, obtaining the results tabulated in Table 1.5.\n\nVolume (mL) of Cough Medicine Delivered by 10-oz (296 mL) Dispensers\n\n| Dispenser \\#1 | Dispenser \\#2 | Dispenser \\#3 |\n| :--- | :--- | :--- |\n| 283.3 | 298.3 | 296.1 |\n| 284.1 | 294.2 | 295.9 |\n| 283.9 | 296.0 | 296.1 |\n| 284.0 | 297.8 | 296.0 |\n| 284.1 | 293.9 | 296.1 |\n\nTABLE 1.5"}
{"id": 2231, "contents": "65. Accuracy and Precision - \nTABLE 1.5\n\nConsidering these results, she will report that dispenser \\#1 is precise (values all close to one another, within a few tenths of a milliliter) but not accurate (none of the values are close to the target value of 296 mL , each being more than 10 mL too low). Results for dispenser \\#2 represent improved accuracy (each volume is less than 3 mL away from 296 mL ) but worse precision (volumes vary by more than 4 mL ). Finally, she can report that dispenser \\#3 is working well, dispensing cough syrup both accurately (all volumes within 0.1 mL of the target volume) and precisely (volumes differing from each other by no more than 0.2 mL )."}
{"id": 2232, "contents": "66. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Explain the dimensional analysis (factor label) approach to mathematical calculations involving quantities\n- Use dimensional analysis to carry out unit conversions for a given property and computations involving two or more properties\n\nIt is often the case that a quantity of interest may not be easy (or even possible) to measure directly but instead must be calculated from other directly measured properties and appropriate mathematical relationships. For example, consider measuring the average speed of an athlete running sprints. This is typically accomplished by measuring the time required for the athlete to run from the starting line to the finish line, and the distance between these two lines, and then computing speed from the equation that relates these three properties:\n\n$$\n\\text { speed }=\\frac{\\text { distance }}{\\text { time }}\n$$\n\nAn Olympic-quality sprinter can run 100 m in approximately 10 s , corresponding to an average speed of\n\n$$\n\\frac{100 \\mathrm{~m}}{10 \\mathrm{~s}}=10 \\mathrm{~m} / \\mathrm{s}\n$$\n\nNote that this simple arithmetic involves dividing the numbers of each measured quantity to yield the number of the computed quantity $(100 / 10=10)$ and likewise dividing the units of each measured quantity to yield the unit of the computed quantity ( $\\mathrm{m} / \\mathrm{s}=\\mathrm{m} / \\mathrm{s}$ ). Now, consider using this same relation to predict the time required for a person running at this speed to travel a distance of 25 m . The same relation among the three properties is used, but in this case, the two quantities provided are a speed ( $10 \\mathrm{~m} / \\mathrm{s}$ ) and a distance ( 25 m ). To yield the sought property, time, the equation must be rearranged appropriately:\n\n$$\n\\text { time }=\\frac{\\text { distance }}{\\text { speed }}\n$$\n\nThe time can then be computed as:\n\n$$\n\\frac{25 \\mathrm{~m}}{10 \\mathrm{~m} / \\mathrm{s}}=2.5 \\mathrm{~s}\n$$"}
{"id": 2233, "contents": "66. LEARNING OBJECTIVES - \nThe time can then be computed as:\n\n$$\n\\frac{25 \\mathrm{~m}}{10 \\mathrm{~m} / \\mathrm{s}}=2.5 \\mathrm{~s}\n$$\n\nAgain, arithmetic on the numbers $(25 / 10=2.5)$ was accompanied by the same arithmetic on the units $(\\mathrm{m} /(\\mathrm{m} / \\mathrm{s})$ $=\\mathrm{s}$ ) to yield the number and unit of the result, 2.5 s . Note that, just as for numbers, when a unit is divided by an identical unit (in this case, $\\mathrm{m} / \\mathrm{m}$ ), the result is \" 1 \"-or, as commonly phrased, the units \"cancel.\"\n\nThese calculations are examples of a versatile mathematical approach known as dimensional analysis (or the factor-label method). Dimensional analysis is based on this premise: the units of quantities must be subjected to the same mathematical operations as their associated numbers. This method can be applied to computations ranging from simple unit conversions to more complex, multi-step calculations involving several different quantities."}
{"id": 2234, "contents": "67. Conversion Factors and Dimensional Analysis - \nA ratio of two equivalent quantities expressed with different measurement units can be used as a unit conversion factor. For example, the lengths of 2.54 cm and 1 in . are equivalent (by definition), and so a unit conversion factor may be derived from the ratio,\n\n$$\n\\frac{2.54 \\mathrm{~cm}}{1 \\mathrm{in} .}(2.54 \\mathrm{~cm}=1 \\mathrm{in} .) \\text { or } 2.54 \\frac{\\mathrm{~cm}}{\\mathrm{in} .}\n$$\n\nSeveral other commonly used conversion factors are given in Table 1.6.\nCommon Conversion Factors\n\n| Length | Volume | Mass |\n| :--- | :--- | :--- |\n| $1 \\mathrm{~m}=1.0936 \\mathrm{yd}$ | $1 \\mathrm{~L}=1.0567 \\mathrm{qt}$ | $1 \\mathrm{~kg}=2.2046 \\mathrm{lb}$ |\n| $1 \\mathrm{in} .=2.54 \\mathrm{~cm}$ (exact) | $1 \\mathrm{qt}=0.94635 \\mathrm{~L}$ | $1 \\mathrm{lb}=453.59 \\mathrm{~g}$ |\n| $1 \\mathrm{~km}=0.62137 \\mathrm{mi}$ | $1 \\mathrm{ft}^{3}=28.317 \\mathrm{~L}$ | 1 (avoirdupois) oz $=28.349 \\mathrm{~g}$ |\n| $1 \\mathrm{mi}=1609.3 \\mathrm{~m}$ | $1 \\mathrm{tbsp}=14.787 \\mathrm{~mL}$ | 1 (troy) oz $=31.103 \\mathrm{~g}$ |\n\nTABLE 1.6\n\nWhen a quantity (such as distance in inches) is multiplied by an appropriate unit conversion factor, the quantity is converted to an equivalent value with different units (such as distance in centimeters). For example, a basketball player's vertical jump of 34 inches can be converted to centimeters by:\n\n$$\n34 \\mathrm{in} \\times \\frac{2.54 \\mathrm{~cm}}{1 \\mathrm{in} \\cdot}=86 \\mathrm{~cm}\n$$"}
{"id": 2235, "contents": "67. Conversion Factors and Dimensional Analysis - \n$$\n34 \\mathrm{in} \\times \\frac{2.54 \\mathrm{~cm}}{1 \\mathrm{in} \\cdot}=86 \\mathrm{~cm}\n$$\n\nSince this simple arithmetic involves quantities, the premise of dimensional analysis requires that we multiply both numbers and units. The numbers of these two quantities are multiplied to yield the number of the product quantity, 86 , whereas the units are multiplied to yield $\\frac{\\mathrm{in} . \\times \\mathrm{cm}}{\\mathrm{in} \\text {. }}$. Just as for numbers, a ratio of identical units is also numerically equal to one, $\\frac{\\mathrm{in} .}{\\mathrm{in} .}=1$, and the unit product thus simplifies to cm . (When identical units divide to yield a factor of 1, they are said to \"cancel.\") Dimensional analysis may be used to confirm the proper application of unit conversion factors as demonstrated in the following example."}
{"id": 2236, "contents": "69. Using a Unit Conversion Factor - \nThe mass of a competition frisbee is 125 g . Convert its mass to ounces using the unit conversion factor derived\nfrom the relationship $1 \\mathrm{oz}=28.349 \\mathrm{~g}$ (Table 1.6)."}
{"id": 2237, "contents": "70. Solution - \nGiven the conversion factor, the mass in ounces may be derived using an equation similar to the one used for converting length from inches to centimeters.\n\n$$\nx \\mathrm{oz}=125 \\mathrm{~g} \\times \\text { unit conversion factor }\n$$\n\nThe unit conversion factor may be represented as:\n\n$$\n\\frac{1 \\mathrm{oz}}{28.349 \\mathrm{~g}} \\text { and } \\frac{28.349 \\mathrm{~g}}{1 \\mathrm{oz}}\n$$\n\nThe correct unit conversion factor is the ratio that cancels the units of grams and leaves ounces.\n\n$$\n\\begin{aligned}\nx \\mathrm{oz} & =125 \\frac{\\mathrm{~g}}{\\mathrm{o}} \\times \\frac{1 \\mathrm{oz}}{28.349 \\frac{\\mathrm{~g}}{\\mathrm{o}}} \\\\\n& =\\left(\\frac{125}{28.349}\\right) \\mathrm{oz} \\\\\n& =4.41 \\mathrm{oz} \\text { (three significant figures) }\n\\end{aligned}\n$$"}
{"id": 2238, "contents": "71. Check Your Learning - \nConvert a volume of 9.345 qt to liters."}
{"id": 2239, "contents": "72. Answer: - \n8.844 L\n\nBeyond simple unit conversions, the factor-label method can be used to solve more complex problems involving computations. Regardless of the details, the basic approach is the same-all the factors involved in the calculation must be appropriately oriented to ensure that their labels (units) will appropriately cancel and/ or combine to yield the desired unit in the result. As your study of chemistry continues, you will encounter many opportunities to apply this approach."}
{"id": 2240, "contents": "74. Computing Quantities from Measurement Results and Known Mathematical Relations - \nWhat is the density of common antifreeze in units of $\\mathrm{g} / \\mathrm{mL}$ ? A 4.00-qt sample of the antifreeze weighs 9.26 lb ."}
{"id": 2241, "contents": "75. Solution - \nSince density $=\\frac{\\text { mass }}{\\text { volume }}$, we need to divide the mass in grams by the volume in milliliters. In general: the number of units of $B=$ the number of units of $A \\times$ unit conversion factor. The necessary conversion factors are given in Table 1.6: $1 \\mathrm{lb}=453.59 \\mathrm{~g} ; 1 \\mathrm{~L}=1.0567 \\mathrm{qt} ; 1 \\mathrm{~L}=1,000 \\mathrm{~mL}$. Mass may be converted from pounds to grams as follows:\n\n$$\n9.26 \\mathrm{H} \\times \\frac{453.59 \\mathrm{~g}}{1 \\mathrm{H}}=4.20 \\times 10^{3} \\mathrm{~g}\n$$\n\nVolume may be converted from quarts to milliliters via two steps:\nStep 1. Convert quarts to liters.\n\n$$\n4.00 \\mathrm{qt} \\times \\frac{1 \\mathrm{~L}}{1.0567 \\mathrm{qt}}=3.78 \\mathrm{~L}\n$$\n\nStep 2. Convert liters to milliliters.\n\n$$\n3.78 \\mathrm{\\leftarrow} \\times \\frac{1000 \\mathrm{~mL}}{1 \\mathrm{\u0141}}=3.78 \\times 10^{3} \\mathrm{~mL}\n$$\n\nThen,\n\n$$\n\\text { density }=\\frac{4.20 \\times 10^{3} \\mathrm{~g}}{3.78 \\times 10^{3} \\mathrm{~mL}}=1.11 \\mathrm{~g} / \\mathrm{mL}\n$$\n\nAlternatively, the calculation could be set up in a way that uses three unit conversion factors sequentially as follows:"}
{"id": 2242, "contents": "75. Solution - \nAlternatively, the calculation could be set up in a way that uses three unit conversion factors sequentially as follows:\n\n$$\n\\frac{9.26 \\mathrm{~b}}{4.00 \\mathrm{qt}} \\times \\frac{453.59 \\mathrm{~g}}{1 \\mathrm{Hb}} \\times \\frac{1.0567 \\mathrm{q}}{1 \\mathrm{t}} \\times \\frac{1 \\mathrm{~L}}{1000 \\mathrm{~mL}}=1.11 \\mathrm{~g} / \\mathrm{mL}\n$$"}
{"id": 2243, "contents": "76. Check Your Learning - \nWhat is the volume in liters of 1.000 oz , given that $1 \\mathrm{~L}=1.0567 \\mathrm{qt}$ and $1 \\mathrm{qt}=32 \\mathrm{oz}$ (exactly)?"}
{"id": 2244, "contents": "77. Answer: - \n$2.956 \\times 10^{-2} \\mathrm{~L}$"}
{"id": 2245, "contents": "79. Computing Quantities from Measurement Results and Known Mathematical Relations - \nWhile being driven from Philadelphia to Atlanta, a distance of about 1250 km , a 2014 Lamborghini Aventador Roadster uses 213 L gasoline.\n(a) What (average) fuel economy, in miles per gallon, did the Roadster get during this trip?\n(b) If gasoline costs $\\$ 3.80$ per gallon, what was the fuel cost for this trip?"}
{"id": 2246, "contents": "80. Solution - \n(a) First convert distance from kilometers to miles:\n\n$$\n1250 \\mathrm{~km} \\times \\frac{0.62137 \\mathrm{mi}}{1 \\mathrm{kmt}}=777 \\mathrm{mi}\n$$\n\nand then convert volume from liters to gallons:\n\n$$\n213 \\mathrm{~L} \\times \\frac{1.0567 \\mathrm{qt}}{1 \\mathrm{t}} \\times \\frac{1 \\mathrm{gal}}{4 \\mathrm{qt}}=56.3 \\mathrm{gal}\n$$\n\nFinally,\n\n$$\n\\text { (average) } \\text { mileage }=\\frac{777 \\mathrm{mi}}{56.3 \\mathrm{gal}}=13.8 \\text { miles } / \\text { gallon }=13.8 \\mathrm{mpg}\n$$\n\nAlternatively, the calculation could be set up in a way that uses all the conversion factors sequentially, as follows:\n\n$$\n\\frac{1250 \\mathrm{~km}}{213 \\mathrm{~L}} \\times \\frac{0.62137 \\mathrm{mi}}{1 \\mathrm{kmt}} \\times \\frac{1 \\mathrm{~L}}{1.0567 \\mathrm{qt}} \\times \\frac{4 \\mathrm{qt}}{1 \\mathrm{gal}}=13.8 \\mathrm{mpg}\n$$\n\n(b) Using the previously calculated volume in gallons, we find:\n\n$$\n56.3 \\text { sat } \\times \\frac{\\$ 3.80}{1-\\text { gat }}=\\$ 214\n$$"}
{"id": 2247, "contents": "81. Check Your Learning - \nA Toyota Prius Hybrid uses 59.7 L gasoline to drive from San Francisco to Seattle, a distance of 1300 km (two significant digits).\n(a) What (average) fuel economy, in miles per gallon, did the Prius get during this trip?\n(b) If gasoline costs $\\$ 3.90$ per gallon, what was the fuel cost for this trip?"}
{"id": 2248, "contents": "82. Answer: - \n(a) 51 mpg ; (b) $\\$ 62$"}
{"id": 2249, "contents": "83. Conversion of Temperature Units - \nWe use the word temperature to refer to the hotness or coldness of a substance. One way we measure a change in temperature is to use the fact that most substances expand when their temperature increases and contract when their temperature decreases. The liquid in a common glass thermometer changes its volume as the temperature changes, and the position of the trapped liquid's surface along a printed scale may be used as a measure of temperature.\n\nTemperature scales are defined relative to selected reference temperatures: Two of the most commonly used are the freezing and boiling temperatures of water at a specified atmospheric pressure. On the Celsius scale, 0 ${ }^{\\circ} \\mathrm{C}$ is defined as the freezing temperature of water and $100^{\\circ} \\mathrm{C}$ as the boiling temperature of water. The space between the two temperatures is divided into 100 equal intervals, which we call degrees. On the Fahrenheit scale, the freezing point of water is defined as $32^{\\circ} \\mathrm{F}$ and the boiling temperature as $212^{\\circ} \\mathrm{F}$. The space between these two points on a Fahrenheit thermometer is divided into 180 equal parts (degrees).\n\nDefining the Celsius and Fahrenheit temperature scales as described in the previous paragraph results in a slightly more complex relationship between temperature values on these two scales than for different units of measure for other properties. Most measurement units for a given property are directly proportional to one another ( $\\mathrm{y}=\\mathrm{mx}$ ). Using familiar length units as one example:\n\n$$\n\\text { length in feet }=\\left(\\frac{1 \\mathrm{ft}}{12 \\mathrm{in} .}\\right) \\times \\text { length in inches }\n$$\n\nwhere $\\mathrm{y}=$ length in feet, $\\mathrm{x}=$ length in inches, and the proportionality constant, m , is the conversion factor. The Celsius and Fahrenheit temperature scales, however, do not share a common zero point, and so the relationship between these two scales is a linear one rather than a proportional one ( $y=m x+b$ ). Consequently, converting a temperature from one of these scales into the other requires more than simple multiplication by a conversion factor, $m$, it also must take into account differences in the scales' zero points (b)."}
{"id": 2250, "contents": "83. Conversion of Temperature Units - \nThe linear equation relating Celsius and Fahrenheit temperatures is easily derived from the two temperatures used to define each scale. Representing the Celsius temperature as $x$ and the Fahrenheit temperature as $y$, the slope, $m$, is computed to be:\n\n$$\nm=\\frac{\\Delta y}{\\Delta x}=\\frac{212^{\\circ} \\mathrm{F}-32^{\\circ} \\mathrm{F}}{100^{\\circ} \\mathrm{C}-0^{\\circ} \\mathrm{C}}=\\frac{180^{\\circ} \\mathrm{F}}{100^{\\circ} \\mathrm{C}}=\\frac{9^{\\circ} \\mathrm{F}}{5^{\\circ} \\mathrm{C}}\n$$\n\nThe $y$-intercept of the equation, $b$, is then calculated using either of the equivalent temperature pairs, $\\left(100^{\\circ} \\mathrm{C}\\right.$, $212^{\\circ} \\mathrm{F}$ ) or ( $0^{\\circ} \\mathrm{C}, 32^{\\circ} \\mathrm{F}$ ), as:\n\n$$\nb=y-m x=32^{\\circ} \\mathrm{F}-\\frac{9^{\\circ} \\mathrm{F}}{5^{\\circ} \\mathrm{C}} \\times 0^{\\circ} \\mathrm{C}=32^{\\circ} \\mathrm{F}\n$$\n\nThe equation relating the temperature ( $T$ ) scales is then:\n\n$$\nT_{{ }^{\\circ} \\mathrm{F}}=\\left(\\frac{9^{\\circ} \\mathrm{F}}{5^{\\circ} \\mathrm{C}} \\times T_{{ }^{\\circ} \\mathrm{C}}\\right)+32{ }^{\\circ} \\mathrm{F}\n$$\n\nAn abbreviated form of this equation that omits the measurement units is:\n\n$$\nT_{{ }^{\\circ} \\mathrm{F}}=\\left(\\frac{9}{5} \\times T_{{ }^{\\circ} \\mathrm{C}}\\right)+32\n$$\n\nRearrangement of this equation yields the form useful for converting from Fahrenheit to Celsius:"}
{"id": 2251, "contents": "83. Conversion of Temperature Units - \nRearrangement of this equation yields the form useful for converting from Fahrenheit to Celsius:\n\n$$\nT^{\\circ} \\mathrm{C}=\\frac{5}{9}\\left(T_{{ }^{\\circ} \\mathrm{F}}-32\\right)\n$$\n\nAs mentioned earlier in this chapter, the SI unit of temperature is the kelvin (K). Unlike the Celsius and Fahrenheit scales, the kelvin scale is an absolute temperature scale in which 0 (zero) K corresponds to the lowest temperature that can theoretically be achieved. Since the kelvin temperature scale is absolute, a degree symbol is not included in the unit abbreviation, K. The early 19th-century discovery of the relationship between a gas's volume and temperature suggested that the volume of a gas would be zero at $-273.15{ }^{\\circ} \\mathrm{C}$. In 1848, British physicist William Thompson, who later adopted the title of Lord Kelvin, proposed an absolute temperature scale based on this concept (further treatment of this topic is provided in this text's chapter on gases).\n\nThe freezing temperature of water on this scale is 273.15 K and its boiling temperature is 373.15 K . Notice the numerical difference in these two reference temperatures is 100 , the same as for the Celsius scale, and so the linear relation between these two temperature scales will exhibit a slope of $1 \\frac{\\mathrm{~K}}{{ }^{\\circ} \\mathrm{C}}$. Following the same approach, the equations for converting between the kelvin and Celsius temperature scales are derived to be:\n\n$$\n\\begin{aligned}\n& T_{\\mathrm{K}}=T_{{ }^{\\circ} \\mathrm{C}}+273.15 \\\\\n& T^{{ }^{\\circ} \\mathrm{C}}=T_{\\mathrm{K}}-273.15\n\\end{aligned}\n$$\n\nThe 273.15 in these equations has been determined experimentally, so it is not exact. Figure 1.28 shows the relationship among the three temperature scales."}
{"id": 2252, "contents": "83. Conversion of Temperature Units - \nThe 273.15 in these equations has been determined experimentally, so it is not exact. Figure 1.28 shows the relationship among the three temperature scales.\n\n\nFIGURE 1.28 The Fahrenheit, Celsius, and kelvin temperature scales are compared.\nAlthough the kelvin (absolute) temperature scale is the official SI temperature scale, Celsius is commonly used in many scientific contexts and is the scale of choice for nonscience contexts in almost all areas of the world. Very few countries (the U.S. and its territories, the Bahamas, Belize, Cayman Islands, and Palau) still use Fahrenheit for weather, medicine, and cooking."}
{"id": 2253, "contents": "85. Conversion from Celsius - \nNormal body temperature has been commonly accepted as $37.0^{\\circ} \\mathrm{C}$ (although it varies depending on time of day and method of measurement, as well as among individuals). What is this temperature on the kelvin scale and on the Fahrenheit scale?"}
{"id": 2254, "contents": "86. Solution - \n$$\n\\begin{gathered}\n\\mathrm{K}={ }^{\\circ} \\mathrm{C}+273.15=37.0+273.2=310.2 \\mathrm{~K} \\\\\n{ }^{\\circ} \\mathrm{F}=\\frac{9}{5}^{\\circ} \\mathrm{C}+32.0=\\left(\\frac{9}{5} \\times 37.0\\right)+32.0=66.6+32.0=98.6^{\\circ} \\mathrm{F}\n\\end{gathered}\n$$"}
{"id": 2255, "contents": "87. Check Your Learning - \nConvert $80.92^{\\circ} \\mathrm{C}$ to K and ${ }^{\\circ} \\mathrm{F}$."}
{"id": 2256, "contents": "88. Answer: - \n$354.07 \\mathrm{~K}, 177.7^{\\circ} \\mathrm{F}$"}
{"id": 2257, "contents": "90. Conversion from Fahrenheit - \nBaking a ready-made pizza calls for an oven temperature of $450^{\\circ} \\mathrm{F}$. If you are in Europe, and your oven thermometer uses the Celsius scale, what is the setting? What is the kelvin temperature?"}
{"id": 2258, "contents": "91. Solution - \n$$\n\\begin{gathered}\n{ }^{\\circ} \\mathrm{C}=\\frac{5}{9}\\left({ }^{\\circ} \\mathrm{F}-32\\right)=\\frac{5}{9}(450-32)=\\frac{5}{9} \\times 418=232{ }^{\\circ} \\mathrm{C} \\longrightarrow \\text { set oven to } 230{ }^{\\circ} \\mathrm{C} \\quad \\text { (two significant figures) } \\\\\n\\mathrm{K}={ }^{\\circ} \\mathrm{C}+273.15=230+273=503 \\mathrm{~K} \\longrightarrow 5.0 \\times 10^{2} \\mathrm{~K} \\quad \\text { (two significant figures) }\n\\end{gathered}\n$$"}
{"id": 2259, "contents": "92. Check Your Learning - \nConvert $50^{\\circ} \\mathrm{F}$ to ${ }^{\\circ} \\mathrm{C}$ and K ."}
{"id": 2260, "contents": "93. Answer: - \n$10^{\\circ} \\mathrm{C}, 280 \\mathrm{~K}$"}
{"id": 2261, "contents": "94. Key Terms - \naccuracy how closely a measurement aligns with a correct value\natom smallest particle of an element that can enter into a chemical combination\nCelsius ( ${ }^{\\circ} \\mathbf{C}$ ) unit of temperature; water freezes at 0 ${ }^{\\circ} \\mathrm{C}$ and boils at $100^{\\circ} \\mathrm{C}$ on this scale\nchemical change change producing a different kind of matter from the original kind of matter\nchemical property behavior that is related to the change of one kind of matter into another kind of matter\nchemistry study of the composition, properties, and interactions of matter\ncompound pure substance that can be decomposed into two or more elements\ncubic centimeter ( $\\mathbf{c m}^{3}$ or cc) volume of a cube with an edge length of exactly 1 cm\ncubic meter ( $\\mathbf{m}^{\\mathbf{3}}$ ) SI unit of volume\ndensity ratio of mass to volume for a substance or object\ndimensional analysis (also, factor-label method) versatile mathematical approach that can be applied to computations ranging from simple unit conversions to more complex, multi-step calculations involving several different quantities\nelement substance that is composed of a single type of atom; a substance that cannot be decomposed by a chemical change\nexact number number derived by counting or by definition\nextensive property property of a substance that depends on the amount of the substance\nFahrenheit unit of temperature; water freezes at $32^{\\circ} \\mathrm{F}$ and boils at $212^{\\circ} \\mathrm{F}$ on this scale\ngas state in which matter has neither definite volume nor shape\nheterogeneous mixture combination of substances with a composition that varies from point to point\nhomogeneous mixture (also, solution) combination of substances with a composition that is uniform throughout\nhypothesis tentative explanation of observations that acts as a guide for gathering and checking information\nintensive property property of a substance that is independent of the amount of the substance\nkelvin (K) SI unit of temperature; 273.15 $\\mathrm{K}=0^{\\circ} \\mathrm{C}$\nkilogram (kg) standard SI unit of mass; $1 \\mathrm{~kg}=$ approximately 2.2 pounds\nlaw statement that summarizes a vast number of experimental observations, and describes or\npredicts some aspect of the natural world"}
{"id": 2262, "contents": "94. Key Terms - \nlaw statement that summarizes a vast number of experimental observations, and describes or\npredicts some aspect of the natural world\nlaw of conservation of matter when matter converts from one type to another or changes form, there is no detectable change in the total amount of matter present\nlength measure of one dimension of an object\nliquid state of matter that has a definite volume but indefinite shape\nliter (L) (also, cubic decimeter) unit of volume; 1 L $=1,000 \\mathrm{~cm}^{3}$\nmacroscopic domain realm of everyday things that are large enough to sense directly by human sight and touch\nmass fundamental property indicating amount of matter\nmatter anything that occupies space and has mass\nmeter (m) standard metric and SI unit of length; 1 $\\mathrm{m}=$ approximately 1.094 yards\nmicroscopic domain realm of things that are much too small to be sensed directly\nmilliliter (mL) $1 / 1,000$ of a liter; equal to $1 \\mathrm{~cm}^{3}$\nmixture matter that can be separated into its components by physical means\nmolecule bonded collection of two or more atoms of the same or different elements\nphysical change change in the state or properties of matter that does not involve a change in its chemical composition\nphysical property characteristic of matter that is not associated with any change in its chemical composition\nplasma gaseous state of matter containing a large number of electrically charged atoms and/or molecules\nprecision how closely a measurement matches the same measurement when repeated\npure substance homogeneous substance that has a constant composition\nrounding procedure used to ensure that calculated results properly reflect the uncertainty in the measurements used in the calculation\nscientific method path of discovery that leads from question and observation to law or hypothesis to theory, combined with experimental verification of the hypothesis and any necessary modification of the theory\nsecond (s) SI unit of time\nSI units (International System of Units)\nstandards fixed by international agreement in the International System of Units (Le Syst\u00e8me International d'Unit\u00e9s)\nsignificant figures (also, significant digits) all of\nthe measured digits in a determination, including the uncertain last digit\nsolid state of matter that is rigid, has a definite shape, and has a fairly constant volume"}
{"id": 2263, "contents": "94. Key Terms - \nsignificant figures (also, significant digits) all of\nthe measured digits in a determination, including the uncertain last digit\nsolid state of matter that is rigid, has a definite shape, and has a fairly constant volume\nsymbolic domain specialized language used to represent components of the macroscopic and microscopic domains, such as chemical symbols, chemical formulas, chemical equations, graphs, drawings, and calculations\ntemperature intensive property representing the hotness or coldness of matter"}
{"id": 2264, "contents": "95. Key Equations - \ndensity $=\\frac{\\text { mass }}{\\text { volume }}$\n$T^{\\circ} \\mathrm{C}=\\frac{5}{9} \\times\\left(T_{{ }^{\\circ}} \\mathrm{F}-32\\right)$\n$T^{\\circ} \\mathrm{F}=\\left(\\frac{9}{5} \\times T^{\\circ} \\mathrm{C}\\right)+32$\n$T_{\\mathrm{K}}={ }^{\\circ} \\mathrm{C}+273.15$\n$T^{\\circ} \\mathrm{C}=\\mathrm{K}-273.15$"}
{"id": 2265, "contents": "96. Summary - 96.1. Chemistry in Context\nChemistry deals with the composition, structure, and properties of matter, and the ways by which various forms of matter may be interconverted. Thus, it occupies a central place in the study and practice of science and technology. Chemists use the scientific method to perform experiments, pose hypotheses, and formulate laws and develop theories, so that they can better understand the behavior of the natural world. To do so, they operate in the macroscopic, microscopic, and symbolic domains. Chemists measure, analyze, purify, and synthesize a wide variety of substances that are important to our lives."}
{"id": 2266, "contents": "96. Summary - 96.2. Phases and Classification of Matter\nMatter is anything that occupies space and has mass. The basic building block of matter is the atom, the smallest unit of an element that can enter into combinations with atoms of the same or other elements. In many substances, atoms are combined into molecules. On earth, matter commonly exists in three states: solids, of fixed shape and volume; liquids, of variable shape but fixed volume; and gases, of variable shape and volume. Under hightemperature conditions, matter also can exist as a plasma. Most matter is a mixture: It is composed of two or more types of matter that can be present in varying amounts and can be separated by physical means. Heterogeneous mixtures vary in\ntheory well-substantiated, comprehensive, testable explanation of a particular aspect of nature\nuncertainty estimate of amount by which measurement differs from true value\nunit standard of comparison for measurements\nunit conversion factor ratio of equivalent quantities expressed with different units; used to convert from one unit to a different unit volume amount of space occupied by an object weight force that gravity exerts on an object\ncomposition from point to point; homogeneous mixtures have the same composition from point to point. Pure substances consist of only one type of matter. A pure substance can be an element, which consists of only one type of atom and cannot be broken down by a chemical change, or a compound, which consists of two or more types of atoms."}
{"id": 2267, "contents": "96. Summary - 96.3. Physical and Chemical Properties\nAll substances have distinct physical and chemical properties, and may undergo physical or chemical changes. Physical properties, such as hardness and boiling point, and physical changes, such as melting or freezing, do not involve a change in the composition of matter. Chemical properties, such flammability and acidity, and chemical changes, such as rusting, involve production of matter that differs from that present beforehand.\n\nMeasurable properties fall into one of two categories. Extensive properties depend on the amount of matter present, for example, the mass of gold. Intensive properties do not depend on the amount of matter present, for example, the density of gold. Heat is an example of an extensive property, and temperature is an example of an intensive property."}
{"id": 2268, "contents": "96. Summary - 96.4. Measurements\nMeasurements provide quantitative information that is critical in studying and practicing chemistry. Each measurement has an amount, a unit for comparison,\nand an uncertainty. Measurements can be represented in either decimal or scientific notation. Scientists primarily use SI (International System) units such as meters, seconds, and kilograms, as well as derived units, such as liters (for volume) and $\\mathrm{g} / \\mathrm{cm}^{3}$ (for density). In many cases, it is convenient to use prefixes that yield fractional and multiple units, such as microseconds ( $10^{-6}$ seconds) and megahertz ( $10^{6}$ hertz), respectively."}
{"id": 2269, "contents": "96. Summary - 96.5. Measurement Uncertainty, Accuracy, and Precision\nQuantities can be defined or measured. Measured quantities have an associated uncertainty that is represented by the number of significant figures in the quantity's number. The uncertainty of a calculated quantity depends on the uncertainties in the quantities used in the calculation and is\nreflected in how the value is rounded. Quantities are characterized with regard to accuracy (closeness to a true or accepted value) and precision (variation among replicate measurement results)."}
{"id": 2270, "contents": "96. Summary - 96.6. Mathematical Treatment of Measurement Results\nMeasurements are made using a variety of units. It is often useful or necessary to convert a measured quantity from one unit into another. These conversions are accomplished using unit conversion factors, which are derived by simple applications of a mathematical approach called the factor-label method or dimensional analysis. This strategy is also employed to calculate sought quantities using measured quantities and appropriate mathematical relations."}
{"id": 2271, "contents": "97. Exercises - 97.1. Chemistry in Context\n1. Explain how you could experimentally determine whether the outside temperature is higher or lower than $0^{\\circ} \\mathrm{C}\\left(32^{\\circ} \\mathrm{F}\\right)$ without using a thermometer.\n2. Identify each of the following statements as being most similar to a hypothesis, a law, or a theory. Explain your reasoning.\n(a) Falling barometric pressure precedes the onset of bad weather.\n(b) All life on earth has evolved from a common, primitive organism through the process of natural selection.\n(c) My truck's gas mileage has dropped significantly, probably because it's due for a tune-up.\n3. Identify each of the following statements as being most similar to a hypothesis, a law, or a theory. Explain your reasoning.\n(a) The pressure of a sample of gas is directly proportional to the temperature of the gas.\n(b) Matter consists of tiny particles that can combine in specific ratios to form substances with specific properties.\n(c) At a higher temperature, solids (such as salt or sugar) will dissolve better in water.\n4. Identify each of the underlined items as a part of either the macroscopic domain, the microscopic domain, or the symbolic domain of chemistry. For any in the symbolic domain, indicate whether they are symbols for a macroscopic or a microscopic feature.\n(a) The mass of a lead pipe is 14 lb .\n(b) The mass of a certain chlorine atom is 35 amu .\n(c) A bottle with a label that reads Al contains aluminum metal.\n(d) Al is the symbol for an aluminum atom.\n5. Identify each of the underlined items as a part of either the macroscopic domain, the microscopic domain, or the symbolic domain of chemistry. For those in the symbolic domain, indicate whether they are symbols for a macroscopic or a microscopic feature.\n(a) A certain molecule contains one $\\underline{H}$ atom and one Cl atom.\n(b) Copper wire has a density of about $8 \\mathrm{~g} / \\mathrm{cm}^{3}$.\n(c) The bottle contains 15 grams of Ni powder.\n(d) A sulfur molecule is composed of eight sulfur atoms."}
{"id": 2272, "contents": "97. Exercises - 97.1. Chemistry in Context\n(c) The bottle contains 15 grams of Ni powder.\n(d) A sulfur molecule is composed of eight sulfur atoms.\n6. According to one theory, the pressure of a gas increases as its volume decreases because the molecules in the gas have to move a shorter distance to hit the walls of the container. Does this theory follow a macroscopic or microscopic description of chemical behavior? Explain your answer.\n7. The amount of heat required to melt 2 lbs of ice is twice the amount of heat required to melt 1 lb of ice. Is this observation a macroscopic or microscopic description of chemical behavior? Explain your answer."}
{"id": 2273, "contents": "97. Exercises - 97.2. Phases and Classification of Matter\n8. Why is an object's mass, rather than its weight, used to indicate the amount of matter it contains?\n9. What properties distinguish solids from liquids? Liquids from gases? Solids from gases?\n10. How does a heterogeneous mixture differ from a homogeneous mixture? How are they similar?\n11. How does a homogeneous mixture differ from a pure substance? How are they similar?\n12. How does an element differ from a compound? How are they similar?\n13. How do molecules of elements and molecules of compounds differ? In what ways are they similar?\n14. How does an atom differ from a molecule? In what ways are they similar?\n15. Many of the items you purchase are mixtures of pure compounds. Select three of these commercial products and prepare a list of the ingredients that are pure compounds.\n16. Classify each of the following as an element, a compound, or a mixture:\n(a) copper\n(b) water\n(c) nitrogen\n(d) sulfur\n(e) air\n(f) sucrose\n(g) a substance composed of molecules each of which contains two iodine atoms\n(h) gasoline\n17. Classify each of the following as an element, a compound, or a mixture:\n(a) iron\n(b) oxygen\n(c) mercury oxide\n(d) pancake syrup\n(e) carbon dioxide\n(f) a substance composed of molecules each of which contains one hydrogen atom and one chlorine atom\n(g) baking soda\n(h) baking powder\n18. A sulfur atom and a sulfur molecule are not identical. What is the difference?\n19. How are the molecules in oxygen gas, the molecules in hydrogen gas, and water molecules similar? How do they differ?\n20. Why are astronauts in space said to be \"weightless,\" but not \"massless\"?\n21. Prepare a list of the principal chemicals consumed and produced during the operation of an automobile.\n22. Matter is everywhere around us. Make a list by name of fifteen different kinds of matter that you encounter every day. Your list should include (and label at least one example of each) the following: a solid, a liquid, a gas, an element, a compound, a homogenous mixture, a heterogeneous mixture, and a pure substance."}
{"id": 2274, "contents": "97. Exercises - 97.2. Phases and Classification of Matter\n23. When elemental iron corrodes it combines with oxygen in the air to ultimately form red brown iron(III) oxide called rust. (a) If a shiny iron nail with an initial mass of 23.2 g is weighed after being coated in a layer of rust, would you expect the mass to have increased, decreased, or remained the same? Explain. (b) If the mass of the iron nail increases to 24.1 g , what mass of oxygen combined with the iron?\n24. As stated in the text, convincing examples that demonstrate the law of conservation of matter outside of the laboratory are few and far between. Indicate whether the mass would increase, decrease, or stay the same for the following scenarios where chemical reactions take place:\n(a) Exactly one pound of bread dough is placed in a baking tin. The dough is cooked in an oven at $350^{\\circ} \\mathrm{F}$ releasing a wonderful aroma of freshly baked bread during the cooking process. Is the mass of the baked loaf less than, greater than, or the same as the one pound of original dough? Explain.\n(b) When magnesium burns in air a white flaky ash of magnesium oxide is produced. Is the mass of magnesium oxide less than, greater than, or the same as the original piece of magnesium? Explain.\n(c) Antoine Lavoisier, the French scientist credited with first stating the law of conservation of matter, heated a mixture of tin and air in a sealed flask to produce tin oxide. Did the mass of the sealed flask and contents decrease, increase, or remain the same after the heating?\n25. Yeast converts glucose to ethanol and carbon dioxide during anaerobic fermentation as depicted in the simple chemical equation here:"}
{"id": 2275, "contents": "97. Exercises - 97.2. Phases and Classification of Matter\n$$\n\\text { glucose } \\longrightarrow \\text { ethanol + carbon dioxide }\n$$\n\n(a) If 200.0 g of glucose is fully converted, what will be the total mass of ethanol and carbon dioxide produced?\n(b) If the fermentation is carried out in an open container, would you expect the mass of the container and contents after fermentation to be less than, greater than, or the same as the mass of the container and contents before fermentation? Explain.\n(c) If 97.7 g of carbon dioxide is produced, what mass of ethanol is produced?"}
{"id": 2276, "contents": "97. Exercises - 97.3. Physical and Chemical Properties\n26. Classify the six underlined properties in the following paragraph as chemical or physical: Fluorine is a pale yellow gas that reacts with most substances. The free element melts at $-220^{\\circ} \\mathrm{C}$ and boils at $-188^{\\circ} \\mathrm{C}$. Finely divided metals burn in fluorine with a bright flame. Nineteen grams of fluorine will react with 1.0 gram of hydrogen.\n27. Classify each of the following changes as physical or chemical:\n(a) condensation of steam\n(b) burning of gasoline\n(c) souring of milk\n(d) dissolving of sugar in water\n(e) melting of gold\n28. Classify each of the following changes as physical or chemical:\n(a) coal burning\n(b) ice melting\n(c) mixing chocolate syrup with milk\n(d) explosion of a firecracker\n(e) magnetizing of a screwdriver\n29. The volume of a sample of oxygen gas changed from 10 mL to 11 mL as the temperature changed. Is this a chemical or physical change?\n30. A 2.0-liter volume of hydrogen gas combined with 1.0 liter of oxygen gas to produce 2.0 liters of water vapor. Does oxygen undergo a chemical or physical change?\n31. Explain the difference between extensive properties and intensive properties.\n32. Identify the following properties as either extensive or intensive.\n(a) volume\n(b) temperature\n(c) humidity\n(d) heat\n(e) boiling point\n33. The density (d) of a substance is an intensive property that is defined as the ratio of its mass (m) to its volume (V).\n\n$$\n\\text { density }=\\frac{\\text { mass }}{\\text { volume }} \\quad \\mathrm{d}=\\frac{\\mathrm{m}}{\\mathrm{~V}}\n$$\n\nConsidering that mass and volume are both extensive properties, explain why their ratio, density, is intensive."}
{"id": 2277, "contents": "97. Exercises - 97.4. Measurements\n34. Is one liter about an ounce, a pint, a quart, or a gallon?\n35. Is a meter about an inch, a foot, a yard, or a mile?\n36. Indicate the SI base units or derived units that are appropriate for the following measurements:\n(a) the length of a marathon race ( 26 miles 385 yards)\n(b) the mass of an automobile\n(c) the volume of a swimming pool\n(d) the speed of an airplane\n(e) the density of gold\n(f) the area of a football field\n(g) the maximum temperature at the South Pole on April 1, 1913\n37. Indicate the SI base units or derived units that are appropriate for the following measurements:\n(a) the mass of the moon\n(b) the distance from Dallas to Oklahoma City\n(c) the speed of sound\n(d) the density of air\n(e) the temperature at which alcohol boils\n(f) the area of the state of Delaware\n(g) the volume of a flu shot or a measles vaccination\n38. Give the name and symbol of the prefixes used with SI units to indicate multiplication by the following exact quantities.\n(a) $10^{3}$\n(b) $10^{-2}$\n(c) 0.1\n(d) $10^{-3}$\n(e) $1,000,000$\n(f) 0.000001\n39. Give the name of the prefix and the quantity indicated by the following symbols that are used with SI base units.\n(a) c\n(b) d\n(c) G\n(d) k\n(e) m\n(f) n\n(g) p\n(h) T\n40. A large piece of jewelry has a mass of 132.6 g . A graduated cylinder initially contains 48.6 mL water. When the jewelry is submerged in the graduated cylinder, the total volume increases to 61.2 mL .\n(a) Determine the density of this piece of jewelry.\n(b) Assuming that the jewelry is made from only one substance, what substance is it likely to be? Explain.\n41. Visit this density simulation (http://openstax.org/l/16phetmasvolden) and click the \"turn fluid into water\" button to adjust the density of liquid in the beaker to $1.00 \\mathrm{~g} / \\mathrm{mL}$."}
{"id": 2278, "contents": "97. Exercises - 97.4. Measurements\n(a) Use the water displacement approach to measure the mass and volume of the unknown material (select the green block with question marks).\n(b) Use the measured mass and volume data from step (a) to calculate the density of the unknown material.\n(c) Link out to the link provided.\n(d) Assuming this material is a copper-containing gemstone, identify its three most likely identities by comparing the measured density to the values tabulated at this gemstone density guide\n(https://www.ajsgem.com/articles/gemstone-density-definitive-guide.html).\n(e) How are mass and density related for blocks of the same volume?\n42. Visit this density simulation (http://openstax.org/l/16phetmasvolden) and click the \"reset\" button to ensure all simulator parameters are at their default values.\n(a) Use the water displacement approach to measure the mass and volume of the red block.\n(b) Use the measured mass and volume data from step (a) to calculate the density of the red block.\n(c) Use the vertical green slide control to adjust the fluid density to values well above, then well below, and finally nearly equal to the density of the red block, reporting your observations.\n43. Visit this density simulation (http://openstax.org/l/16phetmasvolden) and click the \"turn fluid into water\" button to adjust the density of liquid in the beaker to $1.00 \\mathrm{~g} / \\mathrm{mL}$. Change the block material to foam, and then wait patiently until the foam block stops bobbing up and down in the water.\n(a) The foam block should be floating on the surface of the water (that is, only partially submerged). What is the volume of water displaced?\n(b) Use the water volume from part (a) and the density of water $(1.00 \\mathrm{~g} / \\mathrm{mL})$ to calculate the mass of water displaced.\n(c) Remove and weigh the foam block. How does the block's mass compare to the mass of displaced water from part (b)?"}
{"id": 2279, "contents": "97. Exercises - 97.5. Measurement Uncertainty, Accuracy, and Precision\n44. Express each of the following numbers in scientific notation with correct significant figures:\n(a) 711.0\n(b) 0.239\n(c) 90743\n(d) 134.2\n(e) 0.05499\n(f) 10000.0\n(g) 0.000000738592\n45. Express each of the following numbers in exponential notation with correct significant figures:\n(a) 704\n(b) 0.03344\n(c) 547.9\n(d) 22086\n(e) 1000.00\n(f) 0.0000000651\n(g) 0.007157\n46. Indicate whether each of the following can be determined exactly or must be measured with some degree of uncertainty:\n(a) the number of eggs in a basket\n(b) the mass of a dozen eggs\n(c) the number of gallons of gasoline necessary to fill an automobile gas tank\n(d) the number of cm in 2 m\n(e) the mass of a textbook\n(f) the time required to drive from San Francisco to Kansas City at an average speed of $53 \\mathrm{mi} / \\mathrm{h}$\n47. Indicate whether each of the following can be determined exactly or must be measured with some degree of uncertainty:\n(a) the number of seconds in an hour\n(b) the number of pages in this book\n(c) the number of grams in your weight\n(d) the number of grams in 3 kilograms\n(e) the volume of water you drink in one day\n(f) the distance from San Francisco to Kansas City\n48. How many significant figures are contained in each of the following measurements?\n(a) 38.7 g\n(b) $2 \\times 10^{18} \\mathrm{~m}$\n(c) $3,486,002 \\mathrm{~kg}$\n(d) $9.74150 \\times 10^{-4} \\mathrm{~J}$\n(e) $0.0613 \\mathrm{~cm}^{3}$\n(f) 17.0 kg\n(g) $0.01400 \\mathrm{~g} / \\mathrm{mL}$\n49. How many significant figures are contained in each of the following measurements?\n(a) 53 cm"}
{"id": 2280, "contents": "97. Exercises - 97.5. Measurement Uncertainty, Accuracy, and Precision\n(g) $0.01400 \\mathrm{~g} / \\mathrm{mL}$\n49. How many significant figures are contained in each of the following measurements?\n(a) 53 cm\n(b) $2.05 \\times 10^{8} \\mathrm{~m}$\n(c) $86,002 \\mathrm{~J}$\n(d) $9.740 \\times 10^{4} \\mathrm{~m} / \\mathrm{s}$\n(e) $10.0613 \\mathrm{~m}^{3}$\n(f) $0.17 \\mathrm{~g} / \\mathrm{mL}$\n(g) 0.88400 s\n50. The following quantities were reported on the labels of commercial products. Determine the number of significant figures in each.\n(a) 0.0055 g active ingredients\n(b) 12 tablets\n(c) $3 \\%$ hydrogen peroxide\n(d) 5.5 ounces\n(e) 473 mL\n(f) $1.75 \\%$ bismuth\n(g) $0.001 \\%$ phosphoric acid\n(h) $99.80 \\%$ inert ingredients\n51. Round off each of the following numbers to two significant figures:\n(a) 0.436\n(b) 9.000\n(c) 27.2\n(d) 135\n(e) $1.497 \\times 10^{-3}$\n(f) 0.445\n52. Round off each of the following numbers to two significant figures:\n(a) 517\n(b) 86.3\n(c) $6.382 \\times 10^{3}$\n(d) 5.0008\n(e) 22.497\n(f) 0.885\n53. Perform the following calculations and report each answer with the correct number of significant figures.\n(a) $628 \\times 342$\n(b) $\\left(5.63 \\times 10^{2}\\right) \\times\\left(7.4 \\times 10^{3}\\right)$\n(c) $\\frac{28.0}{13.483}$\n(d) $8119 \\times 0.000023$\n(e) $14.98+27,340+84.7593$\n(f) $42.7+0.259$"}
{"id": 2281, "contents": "97. Exercises - 97.5. Measurement Uncertainty, Accuracy, and Precision\n(d) $8119 \\times 0.000023$\n(e) $14.98+27,340+84.7593$\n(f) $42.7+0.259$\n54. Perform the following calculations and report each answer with the correct number of significant figures.\n(a) $62.8 \\times 34$\n(b) $0.147+0.0066+0.012$\n(c) $38 \\times 95 \\times 1.792$\n(d) $15-0.15-0.6155$\n(e) $8.78 \\times\\left(\\frac{0.0500}{0.478}\\right)$\n(f) $140+7.68+0.014$\n(g) $28.7-0.0483$\n(h) $\\frac{(88.5-87.57)}{45.13}$\n55. Consider the results of the archery contest shown in this figure.\n(a) Which archer is most precise?\n(b) Which archer is most accurate?\n(c) Who is both least precise and least accurate?"}
{"id": 2282, "contents": "97. Exercises - 97.5. Measurement Uncertainty, Accuracy, and Precision\nArcher W\n\n\nArcher X\n\n\nArcher Y\n\n\nArcher Z\n56. Classify the following sets of measurements as accurate, precise, both, or neither.\n(a) Checking for consistency in the weight of chocolate chip cookies: $17.27 \\mathrm{~g}, 13.05 \\mathrm{~g}, 19.46 \\mathrm{~g}, 16.92 \\mathrm{~g}$\n(b) Testing the volume of a batch of $25-\\mathrm{mL}$ pipettes: $27.02 \\mathrm{~mL}, 26.99 \\mathrm{~mL}, 26.97 \\mathrm{~mL}, 27.01 \\mathrm{~mL}$\n(c) Determining the purity of gold: $99.9999 \\%, 99.9998 \\%, 99.9998 \\%, 99.9999 \\%$"}
{"id": 2283, "contents": "97. Exercises - 97.6. Mathematical Treatment of Measurement Results\n57. Write conversion factors (as ratios) for the number of:\n(a) yards in 1 meter\n(b) liters in 1 liquid quart\n(c) pounds in 1 kilogram\n58. Write conversion factors (as ratios) for the number of:\n(a) kilometers in 1 mile\n(b) liters in 1 cubic foot\n(c) grams in 1 ounce\n59. The label on a soft drink bottle gives the volume in two units: 2.0 L and 67.6 fl oz . Use this information to derive a conversion factor between the English and metric units. How many significant figures can you justify in your conversion factor?\n60. The label on a box of cereal gives the mass of cereal in two units: 978 grams and 34.5 oz . Use this information to find a conversion factor between the English and metric units. How many significant figures can you justify in your conversion factor?\n61. Soccer is played with a round ball having a circumference between 27 and 28 in . and a weight between 14 and 16 oz . What are these specifications in units of centimeters and grams?\n62. A woman's basketball has a circumference between 28.5 and 29.0 inches and a maximum weight of 20 ounces (two significant figures). What are these specifications in units of centimeters and grams?\n63. How many milliliters of a soft drink are contained in a $12.0-\\mathrm{oz}$ can?\n64. A barrel of oil is exactly 42 gal. How many liters of oil are in a barrel?\n65. The diameter of a red blood cell is about $3 \\times 10^{-4}$ in. What is its diameter in centimeters?\n66. The distance between the centers of the two oxygen atoms in an oxygen molecule is $1.21 \\times 10^{-8} \\mathrm{~cm}$. What is this distance in inches?\n67. Is a 197-lb weight lifter light enough to compete in a class limited to those weighing 90 kg or less?\n68. A very good 197-lb weight lifter lifted 192 kg in a move called the clean and jerk. What was the mass of the weight lifted in pounds?"}
{"id": 2284, "contents": "97. Exercises - 97.6. Mathematical Treatment of Measurement Results\n68. A very good 197-lb weight lifter lifted 192 kg in a move called the clean and jerk. What was the mass of the weight lifted in pounds?\n69. Many medical laboratory tests are run using $5.0 \\mu \\mathrm{~L}$ blood serum. What is this volume in milliliters?\n70. If an aspirin tablet contains 325 mg aspirin, how many grams of aspirin does it contain?\n71. Use scientific (exponential) notation to express the following quantities in terms of the SI base units in Table 1.2:\n(a) 0.13 g\n(b) 232 Gg\n(c) 5.23 pm\n(d) 86.3 mg\n(e) 37.6 cm\n(f) $54 \\mu \\mathrm{~m}$\n(g) 1 Ts\n(h) 27 ps\n(i) 0.15 mK\n72. Complete the following conversions between SI units.\n(a) $612 \\mathrm{~g}=$ $\\qquad$ mg\n(b) $8.160 \\mathrm{~m}=$ $\\qquad$ cm\n(c) $3779 \\mu \\mathrm{~g}=$ $\\qquad$ g\n(d) $781 \\mathrm{~mL}=$ $\\qquad$\n(e) $4.18 \\mathrm{~kg}=$ $\\qquad$ g\n(f) $27.8 \\mathrm{~m}=$ $\\qquad$ km\n(g) $0.13 \\mathrm{~mL}=$ $\\qquad$ L\n(h) $1738 \\mathrm{~km}=$ $\\qquad$ m\n(i) $1.9 \\mathrm{Gg}=$ $\\qquad$ g\n73. Gasoline is sold by the liter in many countries. How many liters are required to fill a 12.0 -gal gas tank?\n74. Milk is sold by the liter in many countries. What is the volume of exactly $1 / 2$ gal of milk in liters?\n75. A long ton is defined as exactly 2240 lb . What is this mass in kilograms?\n76. Make the conversion indicated in each of the following:"}
{"id": 2285, "contents": "97. Exercises - 97.6. Mathematical Treatment of Measurement Results\n75. A long ton is defined as exactly 2240 lb . What is this mass in kilograms?\n76. Make the conversion indicated in each of the following:\n(a) the men's world record long jump, $29 \\mathrm{ft} 4 \\frac{1}{4} \\mathrm{in}$., to meters\n(b) the greatest depth of the ocean, about 6.5 mi , to kilometers\n(c) the area of the state of Oregon, $96,981 \\mathrm{mi}^{2}$, to square kilometers\n(d) the volume of 1 gill (exactly 4 oz ) to milliliters\n(e) the estimated volume of the oceans, $330,000,000 \\mathrm{mi}^{3}$, to cubic kilometers.\n(f) the mass of a $3525-1 \\mathrm{lb}$ car to kilograms\n(g) the mass of a $2.3-\\mathrm{oz}$ egg to grams\n77. Make the conversion indicated in each of the following:\n(a) the length of a soccer field, 120 m (three significant figures), to feet\n(b) the height of Mt. Kilimanjaro, at 19,565 ft, the highest mountain in Africa, to kilometers\n(c) the area of an 8.5- $\\times 11$-inch sheet of paper in $\\mathrm{cm}^{2}$\n(d) the displacement volume of an automobile engine, $161 \\mathrm{in}^{3}$, to liters\n(e) the estimated mass of the atmosphere, $5.6 \\times 10^{15}$ tons, to kilograms\n(f) the mass of a bushel of rye, 32.0 lb , to kilograms\n(g) the mass of a 5.00-grain aspirin tablet to milligrams ( 1 grain $=0.00229 \\mathrm{oz}$ )\n78. Many chemistry conferences have held a $50-\\mathrm{Trillion}$ Angstrom Run (two significant figures). How long is this run in kilometers and in miles? $\\left(1 \\AA=1 \\times 10^{-10} \\mathrm{~m}\\right)$"}
{"id": 2286, "contents": "97. Exercises - 97.6. Mathematical Treatment of Measurement Results\n79. A chemist's 50 -Trillion Angstrom Run (see Exercise 1.78) would be an archeologist's 10,900 cubit run. How long is one cubit in meters and in feet? ( $1 \\AA=1 \\times 10^{-8} \\mathrm{~cm}$ )\n80. The gas tank of a certain luxury automobile holds 22.3 gallons according to the owner's manual. If the density of gasoline is $0.8206 \\mathrm{~g} / \\mathrm{mL}$, determine the mass in kilograms and pounds of the fuel in a full tank.\n81. As an instructor is preparing for an experiment, he requires 225 g phosphoric acid. The only container readily available is a $150-\\mathrm{mL}$ Erlenmeyer flask. Is it large enough to contain the acid, whose density is 1.83 $\\mathrm{g} / \\mathrm{mL}$ ?\n82. To prepare for a laboratory period, a student lab assistant needs 125 g of a compound. A bottle containing $1 / 4 \\mathrm{lb}$ is available. Did the student have enough of the compound?\n83. A chemistry student is 159 cm tall and weighs 45.8 kg . What is her height in inches and weight in pounds?\n84. In a recent Grand Prix, the winner completed the race with an average speed of $229.8 \\mathrm{~km} / \\mathrm{h}$. What was his speed in miles per hour, meters per second, and feet per second?\n85. Solve these problems about lumber dimensions.\n(a) To describe to a European how houses are constructed in the US, the dimensions of \"two-by-four\" lumber must be converted into metric units. The thickness $\\times$ width $\\times$ length dimensions are $1.50 \\mathrm{in} . \\times$ 3.50 in . $\\times 8.00 \\mathrm{ft}$ in the US. What are the dimensions in $\\mathrm{cm} \\times \\mathrm{cm} \\times \\mathrm{m}$ ?\n(b) This lumber can be used as vertical studs, which are typically placed 16.0 in. apart. What is that distance in centimeters?"}
{"id": 2287, "contents": "97. Exercises - 97.6. Mathematical Treatment of Measurement Results\n(b) This lumber can be used as vertical studs, which are typically placed 16.0 in. apart. What is that distance in centimeters?\n86. The mercury content of a stream was believed to be above the minimum considered safe -1 part per billion (ppb) by weight. An analysis indicated that the concentration was 0.68 parts per billion. What quantity of mercury in grams was present in 15.0 L of the water, the density of which is $0.998 \\mathrm{~g} / \\mathrm{ml}$ ? $\\left(1 \\mathrm{ppb} \\mathrm{Hg}=\\frac{1 \\mathrm{ng} \\mathrm{Hg}}{1 \\mathrm{~g} \\text { water }}\\right)$\n87. Calculate the density of aluminum if $27.6 \\mathrm{~cm}^{3}$ has a mass of 74.6 g .\n88. Osmium is one of the densest elements known. What is its density if 2.72 g has a volume of $0.121 \\mathrm{~cm}^{3}$ ?\n89. Calculate these masses.\n(a) What is the mass of $6.00 \\mathrm{~cm}^{3}$ of mercury, density $=13.5939 \\mathrm{~g} / \\mathrm{cm}^{3}$ ?\n(b) What is the mass of 25.0 mL octane, density $=0.702 \\mathrm{~g} / \\mathrm{cm}^{3}$ ?\n90. Calculate these masses.\n(a) What is the mass of $4.00 \\mathrm{~cm}^{3}$ of sodium, density $=0.97 \\mathrm{~g} / \\mathrm{cm}^{3}$ ?\n(b) What is the mass of 125 mL gaseous chlorine, density $=3.16 \\mathrm{~g} / \\mathrm{L}$ ?\n91. Calculate these volumes.\n(a) What is the volume of 25 g iodine, density $=4.93 \\mathrm{~g} / \\mathrm{cm}^{3}$ ?\n(b) What is the volume of 3.28 g gaseous hydrogen, density $=0.089 \\mathrm{~g} / \\mathrm{L}$ ?"}
{"id": 2288, "contents": "97. Exercises - 97.6. Mathematical Treatment of Measurement Results\n(b) What is the volume of 3.28 g gaseous hydrogen, density $=0.089 \\mathrm{~g} / \\mathrm{L}$ ?\n92. Calculate these volumes.\n(a) What is the volume of 11.3 g graphite, density $=2.25 \\mathrm{~g} / \\mathrm{cm}^{3}$ ?\n(b) What is the volume of 39.657 g bromine, density $=2.928 \\mathrm{~g} / \\mathrm{cm}^{3}$ ?\n93. Convert the boiling temperature of gold, $2966^{\\circ} \\mathrm{C}$, into degrees Fahrenheit and kelvin.\n94. Convert the temperature of scalding water, $54^{\\circ} \\mathrm{C}$, into degrees Fahrenheit and kelvin.\n95. Convert the temperature of the coldest area in a freezer, $-10^{\\circ} \\mathrm{F}$, to degrees Celsius and kelvin.\n96. Convert the temperature of dry ice, $-77^{\\circ} \\mathrm{C}$, into degrees Fahrenheit and kelvin.\n97. Convert the boiling temperature of liquid ammonia, $-28.1^{\\circ} \\mathrm{F}$, into degrees Celsius and kelvin.\n98. The label on a pressurized can of spray disinfectant warns against heating the can above $130^{\\circ} \\mathrm{F}$. What are the corresponding temperatures on the Celsius and kelvin temperature scales?\n99. The weather in Europe was unusually warm during the summer of 1995. The TV news reported temperatures as high as $45^{\\circ} \\mathrm{C}$. What was the temperature on the Fahrenheit scale?"}
{"id": 2289, "contents": "97. Exercises - 97.6. Mathematical Treatment of Measurement Results\n60 1\u2022Exercises\n\n\n\nFigure 2.1 Analysis of molecules in an exhaled breath can provide valuable information, leading to early diagnosis of diseases or detection of environmental exposure to harmful substances. (credit: modification of work by Paul Flowers)"}
{"id": 2290, "contents": "98. CHAPTER OUTLINE - 98.1. Early Ideas in Atomic Theory\n2.2 Evolution of Atomic Theory\n2.3 Atomic Structure and Symbolism\n2.4 Chemical Formulas\n\nINTRODUCTION Lung diseases and lung cancers are among the world's most devastating illnesses partly due to delayed detection and diagnosis. Most noninvasive screening procedures aren't reliable, and patients often resist more accurate methods due to discomfort with the procedures or with the potential danger that the procedures cause. But what if you could be accurately diagnosed through a simple breath test?\n\nEarly detection of biomarkers, substances that indicate an organism's disease or physiological state, could allow diagnosis and treatment before a condition becomes serious or irreversible. Recent studies have shown that your exhaled breath can contain molecules that may be biomarkers for recent exposure to environmental contaminants or for pathological conditions ranging from asthma to lung cancer. Scientists are working to develop biomarker \"fingerprints\" that could be used to diagnose a specific disease based on the amounts and identities of certain molecules in a patient's exhaled breath. In Sangeeta Bhatia's lab at MIT, a team used substances that react specifically inside diseased lung tissue; the products of the reactions will be present as biomarkers that can be identified through mass spectrometry (an analytical method discussed later in the chapter). A potential application would allow patients with early symptoms to inhale or ingest a \"sensor\" substance, and, minutes later, to breathe into a detector for diagnosis. Similar research by scientists such as Laura L\u00f3pez-S\u00e1nchez has provided similar processes for lung cancer. An essential concept underlying this goal is that of a molecule's identity, which is determined by the numbers and types of atoms it contains, and how\nthey are bonded together. This chapter will describe some of the fundamental chemical principles related to the composition of matter, including those central to the concept of molecular identity."}
{"id": 2291, "contents": "99. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- State the postulates of Dalton's atomic theory\n- Use postulates of Dalton's atomic theory to explain the laws of definite and multiple proportions\n\nThe earliest recorded discussion of the basic structure of matter comes from ancient Greek philosophers, the scientists of their day. In the fifth century BC, Leucippus and Democritus argued that all matter was composed of small, finite particles that they called atomos, a term derived from the Greek word for \"indivisible.\" They thought of atoms as moving particles that differed in shape and size, and which could join together. Later, Aristotle and others came to the conclusion that matter consisted of various combinations of the four \"elements\"-fire, earth, air, and water-and could be infinitely divided. Interestingly, these philosophers thought about atoms and \"elements\" as philosophical concepts, but apparently never considered performing experiments to test their ideas.\n\nThe Aristotelian view of the composition of matter held sway for over two thousand years, until English schoolteacher John Dalton helped to revolutionize chemistry with his hypothesis that the behavior of matter could be explained using an atomic theory. First published in 1807, many of Dalton's hypotheses about the microscopic features of matter are still valid in modern atomic theory. Here are the postulates of Dalton's atomic theory.\n\n1. Matter is composed of exceedingly small particles called atoms. An atom is the smallest unit of an element that can participate in a chemical change.\n2. An element consists of only one type of atom, which has a mass that is characteristic of the element and is the same for all atoms of that element (Figure 2.2). A macroscopic sample of an element contains an incredibly large number of atoms, all of which have identical chemical properties."}
{"id": 2292, "contents": "99. LEARNING OBJECTIVES - \nFIGURE 2.2 A pre-1982 copper penny (left) contains approximately $3 \\times 10^{22}$ copper atoms (several dozen are represented as brown spheres at the right), each of which has the same chemical properties. (credit: modification of work by \"slgckgc\"/Flickr)\n3. Atoms of one element differ in properties from atoms of all other elements.\n4. A compound consists of atoms of two or more elements combined in a small, whole-number ratio. In a given compound, the numbers of atoms of each of its elements are always present in the same ratio (Figure 2.3).\n\n\nFIGURE 2.3 Copper(II) oxide, a powdery, black compound, results from the combination of two types of atoms-copper (brown spheres) and oxygen (red spheres)-in a 1:1 ratio. (credit: modification of work by \"Chemicalinterest\"/Wikimedia Commons)\n5. Atoms are neither created nor destroyed during a chemical change, but are instead rearranged to yield substances that are different from those present before the change (Figure 2.4).\n\n\nThe elements copper and oxygen\n\n\nThe compound copper(II) oxide\n\nFIGURE 2.4 When the elements copper (a shiny, red-brown solid, shown here as brown spheres) and oxygen (a clear and colorless gas, shown here as red spheres) react, their atoms rearrange to form a compound containing copper and oxygen (a powdery, black solid). (credit copper: modification of work by http://images-ofelements.com/copper.php)\n\nDalton's atomic theory provides a microscopic explanation of the many macroscopic properties of matter that you've learned about. For example, if an element such as copper consists of only one kind of atom, then it cannot be broken down into simpler substances, that is, into substances composed of fewer types of atoms. And if atoms are neither created nor destroyed during a chemical change, then the total mass of matter present when matter changes from one type to another will remain constant (the law of conservation of matter)."}
{"id": 2293, "contents": "101. Testing Dalton's Atomic Theory - \nIn the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another element. If the spheres touch, they are part of a single unit of a compound. Does the following chemical change represented by these symbols violate any of the ideas of Dalton's atomic theory? If so, which one?"}
{"id": 2294, "contents": "102. Solution - \nThe starting materials consist of two green spheres and two purple spheres. The products consist of only one green sphere and one purple sphere. This violates Dalton's postulate that atoms are neither created nor destroyed during a chemical change, but are merely redistributed. (In this case, atoms appear to have been destroyed.)"}
{"id": 2295, "contents": "103. Check Your Learning - \nIn the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another element. If the spheres touch, they are part of a single unit of a compound. Does the following chemical change represented by these symbols violate any of the ideas of Dalton's atomic theory? If so, which one?"}
{"id": 2296, "contents": "104. Answer: - \nThe starting materials consist of four green spheres and two purple spheres. The products consist of four green spheres and two purple spheres. This does not violate any of Dalton's postulates: Atoms are neither created nor destroyed, but are redistributed in small, whole-number ratios.\n\nDalton knew of the experiments of French chemist Joseph Proust, who demonstrated that all samples of a pure compound contain the same elements in the same proportion by mass. This statement is known as the law of definite proportions or the law of constant composition. The suggestion that the numbers of atoms of the elements in a given compound always exist in the same ratio is consistent with these observations. For example, when different samples of isooctane (a component of gasoline and one of the standards used in the octane rating system) are analyzed, they are found to have a carbon-to-hydrogen mass ratio of 5.33:1, as shown in Table 2.1.\n\nConstant Composition of Isooctane\n\n| Sample | Carbon | Hydrogen |\n| :---: | :---: | :---: | Mass Ratio.\n\nTABLE 2.1\n\nIt is worth noting that although all samples of a particular compound have the same mass ratio, the converse is not true in general. That is, samples that have the same mass ratio are not necessarily the same substance. For example, there are many compounds other than isooctane that also have a carbon-to-hydrogen mass ratio of 5.33:1.00.\n\nDalton also used data from Proust, as well as results from his own experiments, to formulate another interesting law. The law of multiple proportions states that when two elements react to form more than one compound, a fixed mass of one element will react with masses of the other element in a ratio of small, whole\nnumbers. For example, copper and chlorine can form a green, crystalline solid with a mass ratio of 0.558 g chlorine to 1 g copper, as well as a brown crystalline solid with a mass ratio of 1.116 g chlorine to 1 g copper. These ratios by themselves may not seem particularly interesting or informative; however, if we take a ratio of these ratios, we obtain a useful and possibly surprising result: a small, whole-number ratio."}
{"id": 2297, "contents": "104. Answer: - \n$$\n\\frac{\\frac{1.116 \\mathrm{~g} \\mathrm{Cl}}{1 \\mathrm{~g} \\mathrm{Cu}}}{\\frac{0.558 \\mathrm{~g} \\mathrm{Cl}}{1 \\mathrm{~g} \\mathrm{Cu}}}=\\frac{2}{1}\n$$\n\nThis 2-to-1 ratio means that the brown compound has twice the amount of chlorine per amount of copper as the green compound.\n\nThis can be explained by atomic theory if the copper-to-chlorine ratio in the brown compound is 1 copper atom to 2 chlorine atoms, and the ratio in the green compound is 1 copper atom to 1 chlorine atom. The ratio of chlorine atoms (and thus the ratio of their masses) is therefore 2 to 1 (Figure 2.5).\n\n\nFIGURE 2.5 Compared to the copper chlorine compound in (a), where copper is represented by brown spheres and chlorine by green spheres, the copper chlorine compound in (b) has twice as many chlorine atoms per copper atom. (credit a: modification of work by \"Benjah-bmm27\"/Wikimedia Commons; credit b: modification of work by \"Walkerma\"/Wikimedia Commons)"}
{"id": 2298, "contents": "106. Laws of Definite and Multiple Proportions - \nA sample of compound A (a clear, colorless gas) is analyzed and found to contain 4.27 g carbon and 5.69 g oxygen. A sample of compound B (also a clear, colorless gas) is analyzed and found to contain 5.19 g carbon and 13.84 g oxygen. Are these data an example of the law of definite proportions, the law of multiple proportions, or neither? What do these data tell you about substances A and B?"}
{"id": 2299, "contents": "107. Solution - \nIn compound A , the mass ratio of oxygen to carbon is:\n\n$$\n\\frac{1.33 \\mathrm{~g} \\mathrm{O}}{1 \\mathrm{~g} \\mathrm{C}}\n$$\n\nIn compound $B$, the mass ratio of oxygen to carbon is:\n\n$$\n\\frac{2.67 \\mathrm{~g} \\mathrm{O}}{1 \\mathrm{~g} \\mathrm{C}}\n$$\n\nThe ratio of these ratios is:\n\n$$\n\\frac{\\frac{1.33 \\mathrm{~g} \\mathrm{O}}{1 \\mathrm{~g} \\mathrm{C}}}{\\frac{2.67 \\mathrm{~g} \\mathrm{O}}{1 \\mathrm{~g} \\mathrm{C}}}=\\frac{1}{2}\n$$\n\nThis supports the law of multiple proportions. This means that A and B are different compounds, with A having one-half as much oxygen per amount of carbon (or twice as much carbon per amount of oxygen) as B. A possible pair of compounds that would fit this relationship would be $\\mathrm{A}=\\mathrm{CO}$ and $\\mathrm{B}=\\mathrm{CO}_{2}$."}
{"id": 2300, "contents": "108. Check Your Learning - \nA sample of compound $X$ (a clear, colorless, combustible liquid with a noticeable odor) is analyzed and found to contain 14.13 g carbon and 2.96 g hydrogen. A sample of compound Y (a clear, colorless, combustible liquid with a noticeable odor that is slightly different from X's odor) is analyzed and found to contain 19.91 g carbon and 3.34 g hydrogen. Are these data an example of the law of definite proportions, the law of multiple proportions, or neither? What do these data tell you about substances X and Y ?"}
{"id": 2301, "contents": "109. Answer: - \nIn compound $X$, the mass ratio of carbon to hydrogen is $\\frac{14.13 \\mathrm{~g} \\mathrm{C}}{2.96 \\mathrm{~g} \\mathrm{H}}$. In compound $Y$, the mass ratio of carbon to\nnumber ratio supports the law of multiple proportions. This means that $X$ and $Y$ are different compounds."}
{"id": 2302, "contents": "110. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Outline milestones in the development of modern atomic theory\n- Summarize and interpret the results of the experiments of Thomson, Millikan, and Rutherford\n- Describe the three subatomic particles that compose atoms\n- Define isotopes and give examples for several elements\n\nIf matter is composed of atoms, what are atoms composed of? Are they the smallest particles, or is there something smaller? In the late 1800 s , a number of scientists interested in questions like these investigated the electrical discharges that could be produced in low-pressure gases, with the most significant discovery made by English physicist J. J. Thomson using a cathode ray tube. This apparatus consisted of a sealed glass tube from which almost all the air had been removed; the tube contained two metal electrodes. When high voltage was applied across the electrodes, a visible beam called a cathode ray appeared between them. This beam was deflected toward the positive charge and away from the negative charge, and was produced in the same way with identical properties when different metals were used for the electrodes. In similar experiments, the ray was simultaneously deflected by an applied magnetic field, and measurements of the extent of deflection and the magnetic field strength allowed Thomson to calculate the charge-to-mass ratio of the cathode ray particles. The results of these measurements indicated that these particles were much lighter than atoms (Figure 2.6).\n\n\nFIGURE 2.6 (a) J. J. Thomson produced a visible beam in a cathode ray tube. (b) This is an early cathode ray tube, invented in 1897 by Ferdinand Braun. (c) In the cathode ray, the beam (shown in yellow) comes from the cathode and is accelerated past the anode toward a fluorescent scale at the end of the tube. Simultaneous deflections by applied electric and magnetic fields permitted Thomson to calculate the mass-to-charge ratio of the particles composing the cathode ray. (credit a: modification of work by Nobel Foundation; credit b: modification of work by Eugen Nesper; credit c: modification of work by \"Kurzon\"/Wikimedia Commons)"}
{"id": 2303, "contents": "110. LEARNING OBJECTIVES - \nBased on his observations, here is what Thomson proposed and why: The particles are attracted by positive (+) charges and repelled by negative (-) charges, so they must be negatively charged (like charges repel and unlike charges attract); they are less massive than atoms and indistinguishable, regardless of the source material, so they must be fundamental, subatomic constituents of all atoms. Although controversial at the time, Thomson's idea was gradually accepted, and his cathode ray particle is what we now call an electron, a negatively charged, subatomic particle with a mass more than one thousand-times less that of an atom. The term \"electron\" was coined in 1891 by Irish physicist George Stoney, from \"electric ion.\""}
{"id": 2304, "contents": "111. LINK TO LEARNING - \nClick here (http://openstax.org/l/16JJThomson) to hear Thomson describe his discovery in his own voice.\n\nIn 1909, more information about the electron was uncovered by American physicist Robert A. Millikan via his \"oil drop\" experiments. Millikan created microscopic oil droplets, which could be electrically charged by friction as they formed or by using X-rays. These droplets initially fell due to gravity, but their downward progress could be slowed or even reversed by an electric field lower in the apparatus. By adjusting the electric field strength and making careful measurements and appropriate calculations, Millikan was able to determine the charge on individual drops (Figure 2.7).\n\n\nFIGURE 2.7 Millikan's experiment measured the charge of individual oil drops. The tabulated data are examples of a few possible values.\n\nLooking at the charge data that Millikan gathered, you may have recognized that the charge of an oil droplet is always a multiple of a specific charge, $1.6 \\times 10^{-19} \\mathrm{C}$. Millikan concluded that this value must therefore be a fundamental charge-the charge of a single electron-with his measured charges due to an excess of one electron ( 1 times $1.6 \\times 10^{-19} \\mathrm{C}$ ), two electrons ( 2 times $1.6 \\times 10^{-19} \\mathrm{C}$ ), three electrons ( 3 times $1.6 \\times 10^{-19} \\mathrm{C}$ ), and so on, on a given oil droplet. Since the charge of an electron was now known due to Millikan's research, and the charge-to-mass ratio was already known due to Thomson's research ( $1.759 \\times 10^{11} \\mathrm{C} / \\mathrm{kg}$ ), it only required a simple calculation to determine the mass of the electron as well.\n\n$$\n\\text { Mass of electron }=1.602 \\times 10^{-19} \\mathrm{C} \\times \\frac{1 \\mathrm{~kg}}{1.759 \\times 10^{11} \\mathrm{C}}=9.107 \\times 10^{-31} \\mathrm{~kg}\n$$"}
{"id": 2305, "contents": "111. LINK TO LEARNING - \nScientists had now established that the atom was not indivisible as Dalton had believed, and due to the work of Thomson, Millikan, and others, the charge and mass of the negative, subatomic particles-the electrons-were known. However, the positively charged part of an atom was not yet well understood. In 1904, Thomson proposed the \"plum pudding\" model of atoms, which described a positively charged mass with an equal amount of negative charge in the form of electrons embedded in it, since all atoms are electrically neutral. A competing model had been proposed in 1903 by Hantaro Nagaoka, who postulated a Saturn-like atom, consisting of a positively charged sphere surrounded by a halo of electrons (Figure 2.8).\n\n\nFIGURE 2.8 (a) Thomson suggested that atoms resembled plum pudding, an English dessert consisting of moist cake with embedded raisins (\"plums\"). (b) Nagaoka proposed that atoms resembled the planet Saturn, with a ring of\nelectrons surrounding a positive \"planet.\" (credit a: modification of work by \"Man vyi\"/Wikimedia Commons; credit b: modification of work by \"NASA\"/Wikimedia Commons)"}
{"id": 2306, "contents": "111. LINK TO LEARNING - \nThe next major development in understanding the atom came from Ernest Rutherford, a physicist from New Zealand who largely spent his scientific career in Canada and England. He performed a series of experiments using a beam of high-speed, positively charged alpha particles ( $\\alpha$ particles) that were produced by the radioactive decay of radium; $\\alpha$ particles consist of two protons and two neutrons (you will learn more about radioactive decay in the chapter on nuclear chemistry). Rutherford and his colleagues Hans Geiger (later famous for the Geiger counter) and Ernest Marsden aimed a beam of $\\alpha$ particles, the source of which was embedded in a lead block to absorb most of the radiation, at a very thin piece of gold foil and examined the resultant scattering of the $\\alpha$ particles using a luminescent screen that glowed briefly where hit by an $\\alpha$ particle.\nWhat did they discover? Most particles passed right through the foil without being deflected at all. However, some were diverted slightly, and a very small number were deflected almost straight back toward the source (Figure 2.9). Rutherford described finding these results: \"It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you., ${ }^{\\underline{1}}$\n\n\nFIGURE 2.9 Geiger and Rutherford fired $\\alpha$ particles at a piece of gold foil and detected where those particles went, as shown in this schematic diagram of their experiment. Most of the particles passed straight through the foil, but a few were deflected slightly and a very small number were significantly deflected.\n\nHere is what Rutherford deduced: Because most of the fast-moving $\\alpha$ particles passed through the gold atoms undeflected, they must have traveled through essentially empty space inside the atom. Alpha particles are positively charged, so deflections arose when they encountered another positive charge (like charges repel each other). Since like charges repel one another, the few positively charged $\\alpha$ particles that changed paths abruptly must have hit, or closely approached, another body that also had a highly concentrated, positive charge. Since the deflections occurred a small fraction of the time, this charge only occupied a small amount of the space in the gold foil. Analyzing a series of such experiments in detail, Rutherford drew two conclusions:"}
{"id": 2307, "contents": "111. LINK TO LEARNING - \n1. The volume occupied by an atom must consist of a large amount of empty space.\n2. A small, relatively heavy, positively charged body, the nucleus, must be at the center of each atom."}
{"id": 2308, "contents": "112. LINK TO LEARNING - \nView this simulation (http://openstax.org/l/16Rutherford) of the Rutherford gold foil experiment. Adjust the slit width to produce a narrower or broader beam of $\\alpha$ particles to see how that affects the scattering pattern.\n\n[^0]This analysis led Rutherford to propose a model in which an atom consists of a very small, positively charged nucleus, in which most of the mass of the atom is concentrated, surrounded by the negatively charged electrons, so that the atom is electrically neutral (Figure 2.10). After many more experiments, Rutherford also discovered that the nuclei of other elements contain the hydrogen nucleus as a \"building block,\" and he named this more fundamental particle the proton, the positively charged, subatomic particle found in the nucleus. With one addition, which you will learn next, this nuclear model of the atom, proposed over a century ago, is still used today.\n\n\nFIGURE 2.10 The $\\alpha$ particles are deflected only when they collide with or pass close to the much heavier, positively charged gold nucleus. Because the nucleus is very small compared to the size of an atom, very few $\\alpha$ particles are deflected. Most pass through the relatively large region occupied by electrons, which are too light to deflect the rapidly moving particles."}
{"id": 2309, "contents": "113. LINK TO LEARNING - \nThe Rutherford Scattering simulation (http://openstax.org/1/16PhetScatter) allows you to investigate the differences between a \"plum pudding\" atom and a Rutherford atom by firing $\\alpha$ particles at each type of atom.\n\nAnother important finding was the discovery of isotopes. During the early 1900s, scientists identified several substances that appeared to be new elements, isolating them from radioactive ores. For example, a \"new element\" produced by the radioactive decay of thorium was initially given the name mesothorium. However, a more detailed analysis showed that mesothorium was chemically identical to radium (another decay product), despite having a different atomic mass. This result, along with similar findings for other elements, led the English chemist Frederick Soddy to realize that an element could have types of atoms with different masses that were chemically indistinguishable. These different types are called isotopes-atoms of the same element that differ in mass. Soddy was awarded the Nobel Prize in Chemistry in 1921 for this discovery.\n\nOne puzzle remained: The nucleus was known to contain almost all of the mass of an atom, with the number of protons only providing half, or less, of that mass. Different proposals were made to explain what constituted the remaining mass, including the existence of neutral particles in the nucleus. As you might expect, detecting uncharged particles is very challenging, and it was not until 1932 that James Chadwick found evidence of neutrons, uncharged, subatomic particles with a mass approximately the same as that of protons. The existence of the neutron also explained isotopes: They differ in mass because they have different numbers of\nneutrons, but they are chemically identical because they have the same number of protons. This will be explained in more detail later."}
{"id": 2310, "contents": "114. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Write and interpret symbols that depict the atomic number, mass number, and charge of an atom or ion\n- Define the atomic mass unit and average atomic mass\n- Calculate average atomic mass and isotopic abundance\n\nThe development of modern atomic theory revealed much about the inner structure of atoms. It was learned that an atom contains a very small nucleus composed of positively charged protons and uncharged neutrons, surrounded by a much larger volume of space containing negatively charged electrons. The nucleus contains the majority of an atom's mass because protons and neutrons are much heavier than electrons, whereas electrons occupy almost all of an atom's volume. The diameter of an atom is on the order of $10^{-10} \\mathrm{~m}$, whereas the diameter of the nucleus is roughly $10^{-15} \\mathrm{~m}$-about 100,000 times smaller. For a perspective about their relative sizes, consider this: If the nucleus were the size of a blueberry, the atom would be about the size of a football stadium (Figure 2.11).\n\n\nFIGURE 2.11 If an atom could be expanded to the size of a football stadium, the nucleus would be the size of a single blueberry. (credit middle: modification of work by \"babyknight\"/Wikimedia Commons; credit right: modification of work by Paxson Woelber)"}
{"id": 2311, "contents": "114. LEARNING OBJECTIVES - \nAtoms-and the protons, neutrons, and electrons that compose them-are extremely small. For example, a carbon atom weighs less than $2 \\times 10^{-23} \\mathrm{~g}$, and an electron has a charge of less than $2 \\times 10^{-19} \\mathrm{C}$ (coulomb). When describing the properties of tiny objects such as atoms, we use appropriately small units of measure, such as the atomic mass unit (amu) and the fundamental unit of charge (e). The amu was originally defined based on hydrogen, the lightest element, then later in terms of oxygen. Since 1961, it has been defined with regard to the most abundant isotope of carbon, atoms of which are assigned masses of exactly 12 amu . (This isotope is known as \"carbon-12\" as will be discussed later in this module.) Thus, one amu is exactly $\\frac{1}{12}$ of the mass of one carbon-12 atom: $1 \\mathrm{amu}=1.6605 \\times 10^{-24} \\mathrm{~g}$. (The Dalton (Da) and the unified atomic mass unit (u) are alternative units that are equivalent to the amu.) The fundamental unit of charge (also called the elementary charge) equals the magnitude of the charge of an electron (e) with $\\mathrm{e}=1.602 \\times 10^{-19} \\mathrm{C}$."}
{"id": 2312, "contents": "114. LEARNING OBJECTIVES - \nA proton has a mass of 1.0073 amu and a charge of $1+$. A neutron is a slightly heavier particle with a mass 1.0087 amu and a charge of zero; as its name suggests, it is neutral. The electron has a charge of 1 - and is a much lighter particle with a mass of about 0.00055 amu (it would take about 1800 electrons to equal the mass of one proton). The properties of these fundamental particles are summarized in Table 2.2. (An observant student might notice that the sum of an atom's subatomic particles does not equal the atom's actual mass: The total mass of six protons, six neutrons, and six electrons is 12.0993 amu , slightly larger than 12.00 amu . This \"missing\" mass is known as the mass defect, and you will learn about it in the chapter on nuclear chemistry.)\n\nProperties of Subatomic Particles\n\n| Name | Location | Charge (C) | Unit Charge | Mass (amu) | Mass (g) |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| electron | outside nucleus | $-1.602 \\times 10^{-19}$ | $1-$ | 0.00055 | $0.00091 \\times 10^{-24}$ |\n| proton | nucleus | $1.602 \\times 10^{-19}$ | $1+$ | 1.00727 | $1.67262 \\times 10^{-24}$ |\n| neutron | nucleus | 0 | 0 | 1.00866 | $1.67493 \\times 10^{-24}$ |\n\nTABLE 2.2"}
{"id": 2313, "contents": "114. LEARNING OBJECTIVES - \nTABLE 2.2\n\nThe number of protons in the nucleus of an atom is its atomic number (Z). This is the defining trait of an element: Its value determines the identity of the atom. For example, any atom that contains six protons is the element carbon and has the atomic number 6, regardless of how many neutrons or electrons it may have. A neutral atom must contain the same number of positive and negative charges, so the number of protons equals the number of electrons. Therefore, the atomic number also indicates the number of electrons in an atom. The total number of protons and neutrons in an atom is called its mass number (A). The number of neutrons is therefore the difference between the mass number and the atomic number: $\\mathrm{A}-\\mathrm{Z}=$ number of neutrons.\n\n$$\n\\begin{aligned}\n\\text { atomic number }(\\mathrm{Z}) & =\\text { number of protons } \\\\\n\\text { mass number }(\\mathrm{A}) & =\\text { number of protons+ number of neutrons } \\\\\n\\mathrm{A}-\\mathrm{Z} & =\\text { number of neutrons }\n\\end{aligned}\n$$\n\nAtoms are electrically neutral if they contain the same number of positively charged protons and negatively charged electrons. When the numbers of these subatomic particles are not equal, the atom is electrically charged and is called an ion. The charge of an atom is defined as follows:\n\nAtomic charge $=$ number of protons - number of electrons\nAs will be discussed in more detail, atoms (and molecules) typically acquire charge by gaining or losing electrons. An atom that gains one or more electrons will exhibit a negative charge and is called an anion. Positively charged atoms called cations are formed when an atom loses one or more electrons. For example, a neutral sodium atom $(Z=11)$ has 11 electrons. If this atom loses one electron, it will become a cation with a 1+ charge ( $11-10=1+$ ). A neutral oxygen atom $(Z=8)$ has eight electrons, and if it gains two electrons it will become an anion with a $2-$ charge ( $8-10=2-$ )."}
{"id": 2314, "contents": "116. Composition of an Atom - \nIodine is an essential trace element in our diet; it is needed to produce thyroid hormone. Insufficient iodine in the diet can lead to the development of a goiter, an enlargement of the thyroid gland (Figure 2.12).\n\n\nFIGURE 2.12 (a) Insufficient iodine in the diet can cause an enlargement of the thyroid gland called a goiter. (b) The addition of small amounts of iodine to salt, which prevents the formation of goiters, has helped eliminate this concern in the US where salt consumption is high. (credit a: modification of work by \"Almazi\"/Wikimedia Commons; credit b: modification of work by Mike Mozart)\n\nThe addition of small amounts of iodine to table salt (iodized salt) has essentially eliminated this health concern in the United States, but as much as $40 \\%$ of the world's population is still at risk of iodine deficiency. The iodine atoms are added as anions, and each has a 1-charge and a mass number of 127. Determine the numbers of protons, neutrons, and electrons in one of these iodine anions."}
{"id": 2315, "contents": "117. Solution - \nThe atomic number of iodine (53) tells us that a neutral iodine atom contains 53 protons in its nucleus and 53 electrons outside its nucleus. Because the sum of the numbers of protons and neutrons equals the mass number, 127 , the number of neutrons is $74(127-53=74)$. Since the iodine is added as a 1 - anion, the number of electrons is 54 [53-(1-)=54]."}
{"id": 2316, "contents": "118. Check Your Learning - \nAn ion of platinum has a mass number of 195 and contains 74 electrons. How many protons and neutrons does it contain, and what is its charge?"}
{"id": 2317, "contents": "119. Answer: - \n78 protons; 117 neutrons; charge is 4+"}
{"id": 2318, "contents": "120. Chemical Symbols - \nA chemical symbol is an abbreviation that we use to indicate an element or an atom of an element. For example, the symbol for mercury is Hg (Figure 2.13). We use the same symbol to indicate one atom of mercury (microscopic domain) or to label a container of many atoms of the element mercury (macroscopic domain).\n\n\nFIGURE 2.13 The symbol Hg represents the element mercury regardless of the amount; it could represent one atom of mercury or a large amount of mercury.\n\nThe symbols for several common elements and their atoms are listed in Table 2.3. Some symbols are derived from the common name of the element; others are abbreviations of the name in another language. Most symbols have one or two letters, but three-letter symbols have been used to describe some elements that have atomic numbers greater than 112 . To avoid confusion with other notations, only the first letter of a symbol is capitalized. For example, Co is the symbol for the element cobalt, but CO is the notation for the compound carbon monoxide, which contains atoms of the elements carbon (C) and oxygen (O). All known elements and their symbols are in the periodic table in Figure 3.37 (also found in Figure A1)."}
{"id": 2319, "contents": "121. Some Common Elements and Their Symbols - \n| Element | Symbol | Element | Symbol |\n| :--- | :--- | :--- | :--- |\n| aluminum | Al | iron | Fe (from ferrum) |\n| bromine | Br | lead | Pb (from plumbum) |\n| calcium | Ca | magnesium | Mg |\n| carbon | C | mercury | Hg (from hydrargyrum) |\n| chlorine | Cl | nitrogen | N |\n| chromium | Cr | oxygen | O |\n| cobalt | Co | potassium | K (from kalium) |\n| copper | Cu (from cuprum) | silicon | Si |\n| fluorine | F | silver | Ag (from argentum) |\n| gold | Au (from aurum) | sodium | Na (from natrium) |\n| helium | He | sulfur | S |\n\nTABLE 2.3\n\n| Element | Symbol | Element | Symbol |\n| :--- | :--- | :--- | :--- |\n| hydrogen | H | tin | Sn (from stannum) |\n| iodine | I | zinc | Zn |\n\nTABLE 2.3\n\nTraditionally, the discoverer (or discoverers) of a new element names the element. However, until the name is recognized by the International Union of Pure and Applied Chemistry (IUPAC), the recommended name of the new element is based on the Latin word(s) for its atomic number. For example, element 106 was called unnilhexium (Unh), element 107 was called unnilseptium (Uns), and element 108 was called unniloctium (Uno) for several years. These elements are now named after scientists (or occasionally locations); for example, element 106 is now known as seaborgium (Sg) in honor of Glenn Seaborg, a Nobel Prize winner who was active in the discovery of several heavy elements. Element 109 was named in honor of Lise Meitner, who discovered nuclear fission, a phenomenon that would have world-changing impacts; Meitner also contributed to the discovery of some major isotopes, discussed immediately below."}
{"id": 2320, "contents": "122. (2) LINK TO LEARNING - \nVisit this site (http://openstax.org/l/16IUPAC) to learn more about IUPAC, the International Union of Pure and Applied Chemistry, and explore its periodic table."}
{"id": 2321, "contents": "123. Isotopes - \nThe symbol for a specific isotope of any element is written by placing the mass number as a superscript to the left of the element symbol (Figure 2.14). The atomic number is sometimes written as a subscript preceding the symbol, but since this number defines the element's identity, as does its symbol, it is often omitted. For example, magnesium exists as a mixture of three isotopes, each with an atomic number of 12 and with mass numbers of 24,25 , and 26 , respectively. These isotopes can be identified as ${ }^{24} \\mathrm{Mg}$, ${ }^{25} \\mathrm{Mg}$, and ${ }^{26} \\mathrm{Mg}$. These isotope symbols are read as \"element, mass number\" and can be symbolized consistent with this reading. For instance, ${ }^{24} \\mathrm{Mg}$ is read as \"magnesium 24 ,\" and can be written as \"magnesium- 24 \" or \" $\\mathrm{Mg}-24$.\" ${ }^{25} \\mathrm{Mg}$ is read as \"magnesium 25,\" and can be written as \"magnesium-25\" or \"Mg-25.\" All magnesium atoms have 12 protons in their nucleus. They differ only because a ${ }^{24} \\mathrm{Mg}$ atom has 12 neutrons in its nucleus, a ${ }^{25} \\mathrm{Mg}$ atom has 13 neutrons, and a ${ }^{26} \\mathrm{Mg}$ has 14 neutrons.\n\n\nFIGURE 2.14 The symbol for an atom indicates the element via its usual two-letter symbol, the mass number as a left superscript, the atomic number as a left subscript (sometimes omitted), and the charge as a right superscript."}
{"id": 2322, "contents": "123. Isotopes - \nInformation about the naturally occurring isotopes of elements with atomic numbers 1 through 10 is given in Table 2.4. Note that in addition to standard names and symbols, the isotopes of hydrogen are often referred to using common names and accompanying symbols. Hydrogen- 2 , symbolized ${ }^{2} \\mathrm{H}$, is also called deuterium and sometimes symbolized D. Hydrogen-3, symbolized ${ }^{3} \\mathrm{H}$, is also called tritium and sometimes symbolized T.\n\nNuclear Compositions of Atoms of the Very Light Elements"}
{"id": 2323, "contents": "123. Isotopes - \n| Element | Symbol | Atomic <br> Number | Number of Protons | Number of Neutrons | Mass <br> (amu) | \\% Natural <br> Abundance |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| hydrogen | ${ }_{1}^{1} \\mathrm{H}$ (protium) | 1 | 1 | 0 | 1.0078 | 99.989 |\n|  | ${ }_{1}^{2} \\mathrm{H}$ (deuterium) | 1 | 1 | 1 | 2.0141 | 0.0115 |\n|  | ${ }_{1}^{3} \\mathrm{H}$ (tritium) | 1 | 1 | 2 | 3.01605 | - (trace) |\n| helium | ${ }_{2}^{3} \\mathrm{He}$ | 2 | 2 | 1 | 3.01603 | 0.00013 |\n|  | ${ }_{2}^{4} \\mathrm{He}$ | 2 | 2 | 2 | 4.0026 | 100 |\n| lithium | ${ }_{3}^{6} \\mathrm{Li}$ | 3 | 3 | 3 | 6.0151 | 7.59 |\n|  | ${ }_{3}^{7} \\mathrm{Li}$ | 3 | 3 | 4 | 7.0160 | 92.41 |\n| beryllium | ${ }_{4}^{9} \\mathrm{Be}$ | 4 | 4 | 5 | 9.0122 | 100 |\n| boron | ${ }_{5}^{10} \\mathrm{~B}$ | 5 | 5 | 5 | 10.0129 | 19.9 |\n|  | ${ }_{5}^{11} \\mathrm{~B}$ | 5 | 5 | 6 | 11.0093 | 80.1 |"}
{"id": 2324, "contents": "123. Isotopes - \n|  | ${ }_{5}^{11} \\mathrm{~B}$ | 5 | 5 | 6 | 11.0093 | 80.1 |\n| carbon | ${ }_{6}^{12} \\mathrm{C}$ | 6 | 6 | 6 | 12.0000 | 98.89 |\n|  | ${ }_{6}^{13} \\mathrm{C}$ | 6 | 6 | 7 | 13.0034 | 1.11 |\n|  | ${ }_{6}^{14} \\mathrm{C}$ | 6 | 6 | 8 | 14.0032 | - (trace) |\n| nitrogen | ${ }_{7}^{14} \\mathrm{~N}$ | 7 | 7 | 7 | 14.0031 | 99.63 |\n|  | ${ }_{7}^{15} \\mathrm{~N}$ | 7 | 7 | 8 | 15.0001 | 0.37 |\n| oxygen | ${ }_{8}^{16} \\mathrm{O}$ | 8 | 8 | 8 | 15.9949 | 99.757 |\n|  | ${ }_{8}^{17} \\mathrm{O}$ | 8 | 8 | 9 | 16.9991 | 0.038 |\n|  | ${ }_{8}^{18} \\mathrm{O}$ | 8 | 8 | 10 | 17.9992 | 0.205 |\n| fluorine | ${ }_{9}^{19} \\mathrm{~F}$ | 9 | 9 | 10 | 18.9984 | 100 |\n| neon | ${ }_{10}^{20} \\mathrm{Ne}$ | 10 | 10 | 10 | 19.9924 | 90.48 |"}
{"id": 2325, "contents": "123. Isotopes - \nTABLE 2.4\n\n| Element | Symbol | Atomic <br> Number | Number of <br> Protons | Number of <br> Neutrons | Mass <br> (amu) | \\% Natural <br> Abundance |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 21 <br> 10 | 10 | 10 | 11 | 20.9938 | 0.27 |\n|  | ${ }_{10}^{22} \\mathrm{Ne}$ | 10 | 10 | 12 | 21.9914 | 9.25 |\n\nTABLE 2.4"}
{"id": 2326, "contents": "124. LINK TO LEARNING - \nUse this Build an Atom simulator (http://openstax.org/l/16PhetAtomBld) to build atoms of the first 10 elements, see which isotopes exist, check nuclear stability, and gain experience with isotope symbols."}
{"id": 2327, "contents": "125. Atomic Mass - \nBecause each proton and each neutron contribute approximately one amu to the mass of an atom, and each electron contributes far less, the atomic mass of a single atom is approximately equal to its mass number (a whole number). However, the average masses of atoms of most elements are not whole numbers because most elements exist naturally as mixtures of two or more isotopes.\n\nThe mass of an element shown in a periodic table or listed in a table of atomic masses is a weighted, average mass of all the isotopes present in a naturally occurring sample of that element. This is equal to the sum of each individual isotope's mass multiplied by its fractional abundance.\n\n$$\n\\text { average mass }=\\sum_{i}(\\text { fractional abundance } \\times \\text { isotopic mass })_{i}\n$$\n\nFor example, the element boron is composed of two isotopes: About $19.9 \\%$ of all boron atoms are ${ }^{10} \\mathrm{~B}$ with a mass of 10.0129 amu , and the remaining $80.1 \\%$ are ${ }^{11} \\mathrm{~B}$ with a mass of 11.0093 amu . The average atomic mass for boron is calculated to be:\n\n$$\n\\begin{aligned}\n\\text { boron average mass } & =(0.199 \\times 10.0129 \\mathrm{amu})+(0.801 \\times 11.0093 \\mathrm{amu}) \\\\\n& =1.99 \\mathrm{amu}+8.82 \\mathrm{amu} \\\\\n& =10.81 \\mathrm{amu}\n\\end{aligned}\n$$\n\nIt is important to understand that no single boron atom weighs exactly $10.8 \\mathrm{amu} ; 10.8 \\mathrm{amu}$ is the average mass of all boron atoms, and individual boron atoms weigh either approximately 10 amu or 11 amu ."}
{"id": 2328, "contents": "127. Calculation of Average Atomic Mass - \nA meteorite found in central Indiana contains traces of the noble gas neon picked up from the solar wind during the meteorite's trip through the solar system. Analysis of a sample of the gas showed that it consisted of $91.84 \\%{ }^{20} \\mathrm{Ne}$ (mass 19.9924 amu ), $0.47 \\%{ }^{21} \\mathrm{Ne}$ (mass 20.9940 amu ), and $7.69 \\%{ }^{22} \\mathrm{Ne}$ (mass 21.9914 amu ). What is the average mass of the neon in the solar wind?"}
{"id": 2329, "contents": "128. Solution - \n$$\n\\begin{aligned}\n\\text { average mass } & =(0.9184 \\times 19.9924 \\mathrm{amu})+(0.0047 \\times 20.9940 \\mathrm{amu})+(0.0769 \\times 21.9914 \\mathrm{amu}) \\\\\n& =(18.36+0.099+1.69) \\mathrm{amu} \\\\\n& =20.15 \\mathrm{amu}\n\\end{aligned}\n$$\n\nThe average mass of a neon atom in the solar wind is 20.15 amu . (The average mass of a terrestrial neon atom is 20.1796 amu . This result demonstrates that we may find slight differences in the natural abundance of\nisotopes, depending on their origin.)"}
{"id": 2330, "contents": "129. Check Your Learning - \nA sample of magnesium is found to contain $78.70 \\%$ of ${ }^{24} \\mathrm{Mg}$ atoms (mass 23.98 amu ), $10.13 \\%$ of ${ }^{25} \\mathrm{Mg}$ atoms (mass 24.99 amu ), and $11.17 \\%$ of ${ }^{26} \\mathrm{Mg}$ atoms (mass 25.98 amu ). Calculate the average mass of a Mg atom."}
{"id": 2331, "contents": "130. Answer: - \n24.31 amu\n\nWe can also do variations of this type of calculation, as shown in the next example."}
{"id": 2332, "contents": "132. Calculation of Percent Abundance - \nNaturally occurring chlorine consists of ${ }^{35} \\mathrm{Cl}$ (mass 34.96885 amu ) and ${ }^{37} \\mathrm{Cl}$ (mass 36.96590 amu ), with an average mass of 35.453 amu . What is the percent composition of Cl in terms of these two isotopes?"}
{"id": 2333, "contents": "133. Solution - \nThe average mass of chlorine is the fraction that is ${ }^{35} \\mathrm{Cl}$ times the mass of ${ }^{35} \\mathrm{Cl}$ plus the fraction that is ${ }^{37} \\mathrm{Cl}$ times the mass of ${ }^{37} \\mathrm{Cl}$.\n\n$$\n\\text { average mass }=\\left(\\text { fraction of }{ }^{35} \\mathrm{Cl} \\times \\text { mass of }{ }^{35} \\mathrm{Cl}\\right)+\\left(\\text { fraction of }{ }^{37} \\mathrm{Cl} \\times \\text { mass of }{ }^{37} \\mathrm{Cl}\\right)\n$$\n\nIf we let $x$ represent the fraction that is ${ }^{35} \\mathrm{Cl}$, then the fraction that is ${ }^{37} \\mathrm{Cl}$ is represented by $1.00-x$.\n(The fraction that is ${ }^{35} \\mathrm{Cl}$ + the fraction that is ${ }^{37} \\mathrm{Cl}$ must add up to 1 , so the fraction of ${ }^{37} \\mathrm{Cl}$ must equal 1.00 - the fraction of ${ }^{35} \\mathrm{Cl}$.)\n\nSubstituting this into the average mass equation, we have:\n\n$$\n\\begin{aligned}\n35.453 \\mathrm{amu} & =(x \\times 34.96885 \\mathrm{amu})+[(1.00-x) \\times 36.96590 \\mathrm{amu}] \\\\\n35.453 & =34.96885 x+36.96590-36.96590 x \\\\\n1.99705 x & =1.513 \\\\\nx & =\\frac{1.513}{1.99705}=0.7576\n\\end{aligned}\n$$\n\nSo solving yields: $x=0.7576$, which means that $1.00-0.7576=0.2424$. Therefore, chlorine consists of $75.76 \\%$ ${ }^{35} \\mathrm{Cl}$ and $24.24 \\%{ }^{37} \\mathrm{Cl}$."}
{"id": 2334, "contents": "134. Check Your Learning - \nNaturally occurring copper consists of ${ }^{63} \\mathrm{Cu}$ (mass 62.9296 amu ) and ${ }^{65} \\mathrm{Cu}$ (mass 64.9278 amu ), with an average mass of 63.546 amu . What is the percent composition of Cu in terms of these two isotopes?"}
{"id": 2335, "contents": "135. Answer: - \n69.15\\% Cu-63 and 30.85\\% Cu-65"}
{"id": 2336, "contents": "136. LINK TO LEARNING - \nVisit this site (http://openstax.org/l/16PhetAtomMass) to make mixtures of the main isotopes of the first 18 elements, gain experience with average atomic mass, and check naturally occurring isotope ratios using the Isotopes and Atomic Mass simulation.\n\nAs you will learn, isotopes are important in nature and especially in human understanding of science and medicine. Let's consider just one natural, stable isotope: Oxygen-18, which is noted in the table above and is referred to as one of the environmental isotopes. It is important in paleoclimatology, for example, because scientists can use the ratio between Oxygen-18 and Oxygen-16 in an ice core to determine the temperature of\nprecipitation over time. Oxygen-18 was also critical to the discovery of metabolic pathways and the mechanisms of enzymes. Mildred Cohn pioneered the usage of these isotopes to act as tracers, so that researchers could follow their path through reactions and gain a better understanding of what is happening. One of her first discoveries provided insight into the phosphorylation of glucose that takes place in mitochondria. And the methods of using isotopes for this research contributed to entire fields of study."}
{"id": 2337, "contents": "136. LINK TO LEARNING - \nThe occurrence and natural abundances of isotopes can be experimentally determined using an instrument called a mass spectrometer. Mass spectrometry (MS) is widely used in chemistry, forensics, medicine, environmental science, and many other fields to analyze and help identify the substances in a sample of material. In a typical mass spectrometer (Figure 2.15), the sample is vaporized and exposed to a high-energy electron beam that causes the sample's atoms (or molecules) to become electrically charged, typically by losing one or more electrons. These cations then pass through a (variable) electric or magnetic field that deflects each cation's path to an extent that depends on both its mass and charge (similar to how the path of a large steel ball rolling past a magnet is deflected to a lesser extent that that of a small steel ball). The ions are detected, and a plot of the relative number of ions generated versus their mass-to-charge ratios (a mass spectrum) is made. The height of each vertical feature or peak in a mass spectrum is proportional to the fraction of cations with the specified mass-to-charge ratio. Since its initial use during the development of modern atomic theory, MS has evolved to become a powerful tool for chemical analysis in a wide range of applications.\n\n\nFIGURE 2.15 Analysis of zirconium in a mass spectrometer produces a mass spectrum with peaks showing the different isotopes of Zr ."}
{"id": 2338, "contents": "137. LINK TO LEARNING - \nSee an animation (http://openstax.org/l/16MassSpec) that explains mass spectrometry. Watch this video (http://openstax.org/l/16RSChemistry) from the Royal Society for Chemistry for a brief description of the rudiments of mass spectrometry."}
{"id": 2339, "contents": "138. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Symbolize the composition of molecules using molecular formulas and empirical formulas\n- Represent the bonding arrangement of atoms within molecules using structural formulas\n- Define the amount unit mole and the related quantity Avogadro's number\n- Explain the relation between mass, moles, and numbers of atoms or molecules and perform calculations deriving these quantities from one another"}
{"id": 2340, "contents": "139. Molecular and Empirical Formulas - \nA molecular formula is a representation of a molecule that uses chemical symbols to indicate the types of atoms followed by subscripts to show the number of atoms of each type in the molecule. (A subscript is used only when more than one atom of a given type is present.) Molecular formulas are also used as abbreviations for the names of compounds.\n\nThe structural formula for a compound gives the same information as its molecular formula (the types and numbers of atoms in the molecule) but also shows how the atoms are connected in the molecule. The structural formula for methane contains symbols for one C atom and four H atoms, indicating the number of atoms in the molecule (Figure 2.16). The lines represent bonds that hold the atoms together. (A chemical bond is an attraction between atoms or ions that holds them together in a molecule or a crystal.) We will discuss chemical bonds and see how to predict the arrangement of atoms in a molecule later. For now, simply know that the lines are an indication of how the atoms are connected in a molecule. A ball-and-stick model shows the geometric arrangement of the atoms with atomic sizes not to scale, and a space-filling model shows the relative sizes of the atoms.\n\n(a)\n\n(b)\n\n(c)\n\n(d)\n\nFIGURE 2.16 A methane molecule can be represented as (a) a molecular formula, (b) a structural formula, (c) a ball-and-stick model, and (d) a space-filling model. Carbon and hydrogen atoms are represented by black and white spheres, respectively."}
{"id": 2341, "contents": "139. Molecular and Empirical Formulas - \nAlthough many elements consist of discrete, individual atoms, some exist as molecules made up of two or more atoms of the element chemically bonded together. For example, most samples of the elements hydrogen, oxygen, and nitrogen are composed of molecules that contain two atoms each (called diatomic molecules) and thus have the molecular formulas $\\mathrm{H}_{2}, \\mathrm{O}_{2}$, and $\\mathrm{N}_{2}$, respectively. Other elements commonly found as diatomic molecules are fluorine $\\left(\\mathrm{F}_{2}\\right)$, chlorine $\\left(\\mathrm{Cl}_{2}\\right)$, bromine $\\left(\\mathrm{Br}_{2}\\right)$, and iodine $\\left(\\mathrm{I}_{2}\\right)$. The most common form of the element sulfur is composed of molecules that consist of eight atoms of sulfur; its molecular formula is $\\mathrm{S}_{8}$ (Figure 2.17).\n\n(a)\n\n(b)\n\n(c)\n\nFIGURE 2.17 A molecule of sulfur is composed of eight sulfur atoms and is therefore written as $\\mathrm{S}_{8}$. It can be represented as (a) a structural formula, (b) a ball-and-stick model, and (c) a space-filling model. Sulfur atoms are represented by yellow spheres.\n\nIt is important to note that a subscript following a symbol and a number in front of a symbol do not represent the same thing; for example, $\\mathrm{H}_{2}$ and 2 H represent distinctly different species. $\\mathrm{H}_{2}$ is a molecular formula; it represents a diatomic molecule of hydrogen, consisting of two atoms of the element that are chemically bonded together. The expression 2 H , on the other hand, indicates two separate hydrogen atoms that are not combined as a unit. The expression $2 \\mathrm{H}_{2}$ represents two molecules of diatomic hydrogen (Figure 2.18)."}
{"id": 2342, "contents": "139. Molecular and Empirical Formulas - \nFIGURE 2.18 The symbols $\\mathrm{H}, 2 \\mathrm{H}, \\mathrm{H}_{2}$, and $2 \\mathrm{H}_{2}$ represent very different entities.\nCompounds are formed when two or more elements chemically combine, resulting in the formation of bonds. For example, hydrogen and oxygen can react to form water, and sodium and chlorine can react to form table salt. We sometimes describe the composition of these compounds with an empirical formula, which indicates the types of atoms present and the simplest whole-number ratio of the number of atoms (or ions) in the compound. For example, titanium dioxide (used as pigment in white paint and in the thick, white, blocking type of sunscreen) has an empirical formula of $\\mathrm{TiO}_{2}$. This identifies the elements titanium ( Ti ) and oxygen (O) as the constituents of titanium dioxide, and indicates the presence of twice as many atoms of the element oxygen as atoms of the element titanium (Figure 2.19).\n\n\nFIGURE 2.19 (a) The white compound titanium dioxide provides effective protection from the sun. (b) A crystal of titanium dioxide, $\\mathrm{TiO}_{2}$, contains titanium and oxygen in a ratio of 1 to 2 . The titanium atoms are gray and the oxygen atoms are red. (credit a: modification of work by \"osseous\"/Flickr)\n\nAs discussed previously, we can describe a compound with a molecular formula, in which the subscripts indicate the actual numbers of atoms of each element in a molecule of the compound. In many cases, the molecular formula of a substance is derived from experimental determination of both its empirical formula and its molecular mass (the sum of atomic masses for all atoms composing the molecule). For example, it can be determined experimentally that benzene contains two elements, carbon (C) and hydrogen (H), and that for every carbon atom in benzene, there is one hydrogen atom. Thus, the empirical formula is CH. An experimental determination of the molecular mass reveals that a molecule of benzene contains six carbon atoms and six hydrogen atoms, so the molecular formula for benzene is $\\mathrm{C}_{6} \\mathrm{H}_{6}$ (Figure 2.20).\n\n(a)\n\n(b)\n\n(c)\n\n(d)"}
{"id": 2343, "contents": "139. Molecular and Empirical Formulas - \n(a)\n\n(b)\n\n(c)\n\n(d)\n\nFIGURE 2.20 Benzene, $\\mathrm{C}_{6} \\mathrm{H}_{6}$, is produced during oil refining and has many industrial uses. A benzene molecule can be represented as (a) a structural formula, (b) a ball-and-stick model, and (c) a space-filling model. (d) Benzene is a clear liquid. (credit d: modification of work by Sahar Atwa)\n\nIf we know a compound's formula, we can easily determine the empirical formula. (This is somewhat of an academic exercise; the reverse chronology is generally followed in actual practice.) For example, the molecular formula for acetic acid, the component that gives vinegar its sharp taste, is $\\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}_{2}$. This formula indicates that a molecule of acetic acid (Figure 2.21) contains two carbon atoms, four hydrogen atoms, and two oxygen atoms. The ratio of atoms is $2: 4: 2$. Dividing by the lowest common denominator (2) gives the simplest, whole-number ratio of atoms, 1:2:1, so the empirical formula is $\\mathrm{CH}_{2} \\mathrm{O}$. Note that a molecular formula is always a wholenumber multiple of an empirical formula.\n\n(a)\n\n(b)\n\n(c)\n\nFIGURE 2.21 (a) Vinegar contains acetic acid, $\\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}_{2}$, which has an empirical formula of $\\mathrm{CH}_{2} \\mathrm{O}$. It can be represented as (b) a structural formula and (c) as a ball-and-stick model. (credit a: modification of work by \"HomeSpot HQ\"/Flickr)"}
{"id": 2344, "contents": "141. Empirical and Molecular Formulas - \nMolecules of glucose (blood sugar) contain 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. What are the molecular and empirical formulas of glucose?"}
{"id": 2345, "contents": "142. Solution - \nThe molecular formula is $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$ because one molecule actually contains $6 \\mathrm{C}, 12 \\mathrm{H}$, and 6 O atoms. The simplest whole-number ratio of C to H to O atoms in glucose is $1: 2: 1$, so the empirical formula is $\\mathrm{CH}_{2} \\mathrm{O}$."}
{"id": 2346, "contents": "143. Check Your Learning - \nA molecule of metaldehyde (a pesticide used for snails and slugs) contains 8 carbon atoms, 16 hydrogen atoms, and 4 oxygen atoms. What are the molecular and empirical formulas of metaldehyde?"}
{"id": 2347, "contents": "144. Answer: - \nMolecular formula, $\\mathrm{C}_{8} \\mathrm{H}_{16} \\mathrm{O}_{4}$; empirical formula, $\\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}$"}
{"id": 2348, "contents": "145. LINK TO LEARNING - \nYou can explore molecule building (http://openstax.org/l/16molbuilding) using an online simulation."}
{"id": 2349, "contents": "147. Lee Cronin - \nWhat is it that chemists do? According to Lee Cronin (Figure 2.22), chemists make very complicated molecules by \"chopping up\" small molecules and \"reverse engineering\" them. He wonders if we could \"make a really cool universal chemistry set\" by what he calls \"app-ing\" chemistry. Could we \"app\" chemistry?\n\nIn a 2012 TED talk, Lee describes one fascinating possibility: combining a collection of chemical \"inks\" with a 3D printer capable of fabricating a reaction apparatus (tiny test tubes, beakers, and the like) to fashion a \"universal toolkit of chemistry.\" This toolkit could be used to create custom-tailored drugs to fight a new superbug or to \"print\" medicine personally configured to your genetic makeup, environment, and health situation. Says Cronin, \"What Apple did for music, I'd like to do for the discovery and distribution of prescription drugs.\" ${ }^{2}$ View his full talk (http://openstax.org/l/16LeeCronin) at the TED website.\n\n\nFIGURE 2.22 Chemist Lee Cronin has been named one of the UK's 10 most inspirational scientists. The youngest chair at the University of Glasgow, Lee runs a large research group, collaborates with many scientists worldwide, has published over 250 papers in top scientific journals, and has given more than 150 invited talks. His research focuses on complex chemical systems and their potential to transform technology, but also branches into nanoscience, solar fuels, synthetic biology, and even artificial life and evolution. (credit: image courtesy of Lee Cronin)\n\nIt is important to be aware that it may be possible for the same atoms to be arranged in different ways: Compounds with the same molecular formula may have different atom-to-atom bonding and therefore different structures. For example, could there be another compound with the same formula as acetic acid, $\\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}_{2}$ ? And if so, what would be the structure of its molecules?"}
{"id": 2350, "contents": "147. Lee Cronin - \nIf you predict that another compound with the formula $\\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}_{2}$ could exist, then you demonstrated good chemical insight and are correct. Two C atoms, four H atoms, and two O atoms can also be arranged to form a methyl formate, which is used in manufacturing, as an insecticide, and for quick-drying finishes. Methyl formate molecules have one of the oxygen atoms between the two carbon atoms, differing from the arrangement in acetic acid molecules. Acetic acid and methyl formate are examples of isomers-compounds with the same chemical formula but different molecular structures (Figure 2.23). Note that this small difference in the arrangement of the atoms has a major effect on their respective chemical properties. You would certainly not want to use a solution of methyl formate as a substitute for a solution of acetic acid (vinegar) when you make salad dressing.\n\n\nAcetic acid $\\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}_{2}$\n(a)\n\n\nMethyl formate $\\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}_{2}$\n(b)\n\nFIGURE 2.23 Molecules of (a) acetic acid and methyl formate (b) are structural isomers; they have the same formula $\\left(\\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}_{2}\\right)$ but different structures (and therefore different chemical properties).\n\nMany types of isomers exist (Figure 2.24). Acetic acid and methyl formate are structural isomers, compounds in which the molecules differ in how the atoms are connected to each other. There are also various types of spatial isomers, in which the relative orientations of the atoms in space can be different. For example, the compound carvone (found in caraway seeds, spearmint, and mandarin orange peels) consists of two isomers that are mirror images of each other. $S$-(+)-carvone smells like caraway, and $R$-(-)-carvone smells like spearmint.\n\n\n\n(+)-Carvone $\\mathrm{C}_{10} \\mathrm{H}_{14} \\mathrm{O}$"}
{"id": 2351, "contents": "147. Lee Cronin - \n(+)-Carvone $\\mathrm{C}_{10} \\mathrm{H}_{14} \\mathrm{O}$\n\n\n\n\nFIGURE 2.24 Molecules of carvone are spatial isomers; they only differ in the relative orientations of the atoms in space. (credit bottom left: modification of work by \"Miansari66\"/Wikimedia Commons; credit bottom right: modification of work by Forest \\& Kim Starr)"}
{"id": 2352, "contents": "148. LINK TO LEARNING - \nSelect this link (http://openstax.org/l/16isomers) to view an explanation of isomers, spatial isomers, and why they have different smells (select the video titled \"Mirror Molecule: Carvone\")."}
{"id": 2353, "contents": "149. The Mole - \nThe identity of a substance is defined not only by the types of atoms or ions it contains, but by the quantity of each type of atom or ion. For example, water, $\\mathrm{H}_{2} \\mathrm{O}$, and hydrogen peroxide, $\\mathrm{H}_{2} \\mathrm{O}_{2}$, are alike in that their respective molecules are composed of hydrogen and oxygen atoms. However, because a hydrogen peroxide molecule contains two oxygen atoms, as opposed to the water molecule, which has only one, the two\nsubstances exhibit very different properties. Today, we possess sophisticated instruments that allow the direct measurement of these defining microscopic traits; however, the same traits were originally derived from the measurement of macroscopic properties (the masses and volumes of bulk quantities of matter) using relatively simple tools (balances and volumetric glassware). This experimental approach required the introduction of a new unit for amount of substances, the mole, which remains indispensable in modern chemical science.\n\nThe mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a sample of matter. One Latin connotation for the word \"mole\" is \"large mass\" or \"bulk,\" which is consistent with its use as the name for this unit. The mole provides a link between an easily measured macroscopic property, bulk mass, and an extremely important fundamental property, number of atoms, molecules, and so forth. A mole of substance is that amount in which there are $6.02214076 \\times$ $10^{23}$ discrete entities (atoms or molecules). This large number is a fundamental constant known as Avogadro's number ( $\\mathbf{N}_{\\mathbf{A}}$ ) or the Avogadro constant in honor of Italian scientist Amedeo Avogadro. This constant is properly reported with an explicit unit of \"per mole,\" a conveniently rounded version being $6.022 \\times 10^{23} / \\mathrm{mol}$."}
{"id": 2354, "contents": "149. The Mole - \nConsistent with its definition as an amount unit, 1 mole of any element contains the same number of atoms as 1 mole of any other element. The masses of 1 mole of different elements, however, are different, since the masses of the individual atoms are drastically different. The molar mass of an element (or compound) is the mass in grams of 1 mole of that substance, a property expressed in units of grams per mole (g/mol) (see Figure 2.25).\n\n\nFIGURE 2.25 Each sample contains $6.022 \\times 10^{23}$ atoms -1.00 mol of atoms. From left to right (top row): 65.4 g zinc, 12.0 g carbon, 24.3 g magnesium, and 63.5 g copper. From left to right (bottom row): 32.1 g sulfur, 28.1 g silicon, 207 g lead, and 118.7 g tin. (credit: modification of work by Mark Ott)\n\nThe molar mass of any substance is numerically equivalent to its atomic or formula weight in amu. Per the amu definition, a single ${ }^{12} \\mathrm{C}$ atom weighs 12 amu (its atomic mass is 12 amu ). A mole of ${ }^{12} \\mathrm{C}$ atoms weighs 12 g (its molar mass is $12 \\mathrm{~g} / \\mathrm{mol}$ ). This relationship holds for all elements, since their atomic masses are measured relative to that of the amu-reference substance, ${ }^{12}$ C. Extending this principle, the molar mass of a compound in grams is likewise numerically equivalent to its formula mass in amu (Figure 2.26)."}
{"id": 2355, "contents": "149. The Mole - \nFIGURE 2.26 Each sample contains $6.02 \\times 10^{23}$ molecules or formula units-1.00 mol of the compound or element. Clock-wise from the upper left: 130.2 g of $\\mathrm{C}_{8} \\mathrm{H}_{17} \\mathrm{OH}$ (1-octanol, formula mass 130.2 amu ), 454.4 g of $\\mathrm{HgI}_{2}$ (mercury(II) iodide, formula mass 454.4 amu ), 32.0 g of $\\mathrm{CH}_{3} \\mathrm{OH}$ (methanol, formula mass 32.0 amu ) and 256.5 g of $\\mathrm{S}_{8}$ (sulfur, formula mass 256.5 amu ). (credit: Sahar Atwa)\n\n| Element | Average Atomic Mass (amu) | Molar Mass (g/mol) | Atoms/Mole |\n| :---: | :---: | :---: | :---: |\n| C | 12.01 | 12.01 | $6.022 \\times 10^{23}$ |\n| H | 1.008 | 1.008 | $6.022 \\times 10^{23}$ |\n| O | 16.00 | 16.00 | $6.022 \\times 10^{23}$ |\n| Na | 22.99 | 22.99 | $6.022 \\times 10^{23}$ |\n| Cl | 35.45 | 35.45 | $6.022 \\times 10^{23}$ |\n\nWhile atomic mass and molar mass are numerically equivalent, keep in mind that they are vastly different in terms of scale, as represented by the vast difference in the magnitudes of their respective units (amu versus g). To appreciate the enormity of the mole, consider a small drop of water weighing about 0.03 g (see Figure 2.27). Although this represents just a tiny fraction of 1 mole of water $(\\sim 18 \\mathrm{~g})$, it contains more water molecules than can be clearly imagined. If the molecules were distributed equally among the roughly seven billion people on earth, each person would receive more than 100 billion molecules."}
{"id": 2356, "contents": "149. The Mole - \nFIGURE 2.27 The number of molecules in a single droplet of water is roughly 100 billion times greater than the number of people on earth. (credit: \"tanakawho\"/Wikimedia commons)"}
{"id": 2357, "contents": "150. LINK TO LEARNING - \nThe mole is used in chemistry to represent $6.022 \\times 10^{23}$ of something, but it can be difficult to conceptualize such a large number. Watch this video (http://openstax.org/l/16molevideo) and then complete the \"Think\" questions that follow. Explore more about the mole by reviewing the information under \"Dig Deeper.\"\n\nThe relationships between formula mass, the mole, and Avogadro's number can be applied to compute various quantities that describe the composition of substances and compounds. For example, if we know the mass and chemical composition of a substance, we can determine the number of moles and calculate number of atoms or molecules in the sample. Likewise, if we know the number of moles of a substance, we can derive the number of atoms or molecules and calculate the substance's mass."}
{"id": 2358, "contents": "152. Deriving Moles from Grams for an Element - \nAccording to nutritional guidelines from the US Department of Agriculture, the estimated average requirement for dietary potassium is 4.7 g . What is the estimated average requirement of potassium in moles?"}
{"id": 2359, "contents": "153. Solution - \nThe mass of $K$ is provided, and the corresponding amount of $K$ in moles is requested. Referring to the periodic table, the atomic mass of K is 39.10 amu , and so its molar mass is $39.10 \\mathrm{~g} / \\mathrm{mol}$. The given mass of $\\mathrm{K}(4.7 \\mathrm{~g})$ is a bit more than one-tenth the molar mass ( 39.10 g ), so a reasonable \"ballpark\" estimate of the number of moles would be slightly greater than 0.1 mol .\n\nThe molar amount of a substance may be calculated by dividing its mass (g) by its molar mass (g/mol):\n\n\nThe factor-label method supports this mathematical approach since the unit \"g\" cancels and the answer has units of \"mol:\"\n\n$$\n4.7 \\mathrm{gK}\\left(\\frac{\\mathrm{~mol} \\mathrm{~K}}{39.10 \\mathrm{~g} \\mathrm{~K}}\\right)=0.12 \\mathrm{~mol} \\mathrm{~K}\n$$\n\nThe calculated magnitude $(0.12 \\mathrm{~mol} \\mathrm{~K})$ is consistent with our ballpark expectation, since it is a bit greater than 0.1 mol."}
{"id": 2360, "contents": "154. Check Your Learning - \nBeryllium is a light metal used to fabricate transparent X-ray windows for medical imaging instruments. How many moles of Be are in a thin-foil window weighing 3.24 g ?"}
{"id": 2361, "contents": "155. Answer: - \n0.360 mol"}
{"id": 2362, "contents": "157. Deriving Grams from Moles for an Element - \nA liter of air contains $9.2 \\times 10^{-4} \\mathrm{~mol}$ argon. What is the mass of Ar in a liter of air?"}
{"id": 2363, "contents": "158. Solution - \nThe molar amount of Ar is provided and must be used to derive the corresponding mass in grams. Since the amount of Ar is less than 1 mole, the mass will be less than the mass of 1 mole of Ar , approximately 40 g . The molar amount in question is approximately one-one thousandth $\\left(\\sim 10^{-3}\\right)$ of a mole, and so the corresponding mass should be roughly one-one thousandth of the molar mass ( $\\sim 0.04 \\mathrm{~g}$ ):\n\n\nIn this case, logic dictates (and the factor-label method supports) multiplying the provided amount (mol) by the molar mass (g/mol):\n\n$$\n9.2 \\times 10^{-4} \\text { mol Ar }\\left(\\frac{39.95 \\mathrm{~g} \\mathrm{Ar}}{\\text { mol Ar }}\\right)=0.037 \\mathrm{~g} \\mathrm{Ar}\n$$\n\nThe result is in agreement with our expectations, around 0.04 g Ar .\nCheck Your Learning\nWhat is the mass of 2.561 mol of gold?"}
{"id": 2364, "contents": "159. Answer: - \n504.4 g"}
{"id": 2365, "contents": "161. Deriving Number of Atoms from Mass for an Element - \nCopper is commonly used to fabricate electrical wire (Figure 2.28). How many copper atoms are in 5.00 g of copper wire?\n\n\nFIGURE 2.28 Copper wire is composed of many, many atoms of Cu. (credit: Emilian Robert Vicol)"}
{"id": 2366, "contents": "162. Solution - \nThe number of Cu atoms in the wire may be conveniently derived from its mass by a two-step computation: first calculating the molar amount of Cu , and then using Avogadro's number ( $N_{A}$ ) to convert this molar amount to number of Cu atoms:\n\n\nConsidering that the provided sample mass ( 5.00 g ) is a little less than one-tenth the mass of 1 mole of $\\mathrm{Cu}(\\sim 64$ g ), a reasonable estimate for the number of atoms in the sample would be on the order of one-tenth $N_{A}$, or approximately $10^{22} \\mathrm{Cu}$ atoms. Carrying out the two-step computation yields:\n\n$$\n5.00 \\mathrm{~g} \\mathrm{Cu}\\left(\\frac{\\mathrm{molCu}}{63.55-\\mathrm{g} \\mathrm{Cu}}\\right)\\left(\\frac{6.022 \\times 10^{23} \\mathrm{Cu} \\text { atoms }}{\\mathrm{molCu}}\\right)=4.74 \\times 10^{22} \\mathrm{Cu} \\text { atoms }\n$$\n\nThe factor-label method yields the desired cancellation of units, and the computed result is on the order of $10^{22}$ as expected."}
{"id": 2367, "contents": "163. Check Your Learning - \nA prospector panning for gold in a river collects 15.00 g of pure gold. How many Au atoms are in this quantity of gold?"}
{"id": 2368, "contents": "164. Answer: - \n$4.586 \\times 10^{22} \\mathrm{Au}$ atoms"}
{"id": 2369, "contents": "166. Deriving Moles from Grams for a Compound - \nOur bodies synthesize protein from amino acids. One of these amino acids is glycine, which has the molecular formula $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{O}_{2} \\mathrm{~N}$. How many moles of glycine molecules are contained in 28.35 g of glycine?"}
{"id": 2370, "contents": "167. Solution - \nWe can derive the number of moles of a compound from its mass following the same procedure we used for an element in Example 2.7:\n\n\nThe molar mass of glycine is required for this calculation, and it is computed in the same fashion as its molecular mass. One mole of glycine, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{O}_{2} \\mathrm{~N}$, contains 2 moles of carbon, 5 moles of hydrogen, 2 moles of oxygen, and 1 mole of nitrogen:\n\n| Element | Quantity (mol element/ mol compound) |  | Molar mass (g/mol element) |  | Subtotal (g/mol compound) |  |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| C | 2 | $\\times$ | 12.01 | $=$ | 24.02 |  |\n| H | 5 | $\\times$ | 1.008 | $=$ | 5.040 |  |\n| 0 | 2 | $\\times$ | 16.00 | $=$ | 32.00 |  |\n| N | 1 | $\\times$ | 14.007 | $=$ | 14.007 |  |\n| Molecular mass (g/mol compound) |  |  |  |  | 75.07 |  |\n\nThe provided mass of glycine ( $\\sim 28 \\mathrm{~g}$ ) is a bit more than one-third the molar mass ( $\\sim 75 \\mathrm{~g} / \\mathrm{mol}$ ), so we would expect the computed result to be a bit greater than one-third of a mole ( $\\sim 0.33 \\mathrm{~mol}$ ). Dividing the compound's mass by its molar mass yields:\n\n$$\n28.35 \\text { golycine- }\\left(\\frac{\\text { mol glycine }}{75.07 \\text {-g glyeine }}\\right)=0.378 \\text { mol glycine }\n$$\n\nThis result is consistent with our rough estimate."}
{"id": 2371, "contents": "168. Check Your Learning - \nHow many moles of sucrose, $\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}$, are in a 25 -g sample of sucrose?"}
{"id": 2372, "contents": "169. Answer: - \n0.073 mol"}
{"id": 2373, "contents": "171. Deriving Grams from Moles for a Compound - \nVitamin C is a covalent compound with the molecular formula $\\mathrm{C}_{6} \\mathrm{H}_{8} \\mathrm{O}_{6}$. The recommended daily dietary allowance of vitamin $C$ for children aged $4-8$ years is $1.42 \\times 10^{-4} \\mathrm{~mol}$. What is the mass of this allowance in grams?"}
{"id": 2374, "contents": "172. Solution - \nAs for elements, the mass of a compound can be derived from its molar amount as shown:\n\n\nThe molar mass for this compound is computed to be $176.124 \\mathrm{~g} / \\mathrm{mol}$. The given number of moles is a very small fraction of a mole ( $\\sim 10^{-4}$ or one-ten thousandth); therefore, we would expect the corresponding mass to be about one-ten thousandth of the molar mass ( $\\sim 0.02 \\mathrm{~g}$ ). Performing the calculation, we get:\n\n$$\n1.42 \\times 10^{-4} \\text { mol vitamin } \\mathrm{C}\\left(\\frac{176.124 \\mathrm{~g} \\text { vitamin } \\mathrm{C}}{\\text { mot vitamin } \\mathrm{C}}\\right)=0.0250 \\mathrm{~g} \\text { vitamin } \\mathrm{C}\n$$\n\nThis is consistent with the anticipated result.\nCheck Your Learning\nWhat is the mass of 0.443 mol of hydrazine, $\\mathrm{N}_{2} \\mathrm{H}_{4}$ ?"}
{"id": 2375, "contents": "173. Answer: - \n14.2 g"}
{"id": 2376, "contents": "175. Deriving the Number of Atoms and Molecules from the Mass of a Compound - \nA packet of an artificial sweetener contains 40.0 mg of saccharin $\\left(\\mathrm{C}_{7} \\mathrm{H}_{5} \\mathrm{NO}_{3} \\mathrm{~S}\\right)$, which has the structural formula:\n\n\nGiven that saccharin has a molar mass of $183.18 \\mathrm{~g} / \\mathrm{mol}$, how many saccharin molecules are in a 40.0-mg ( $0.0400-\\mathrm{g}$ ) sample of saccharin? How many carbon atoms are in the same sample?"}
{"id": 2377, "contents": "176. Solution - \nThe number of molecules in a given mass of compound is computed by first deriving the number of moles, as demonstrated in Example 2.10, and then multiplying by Avogadro's number:\n\n\nUsing the provided mass and molar mass for saccharin yields:\n\n$$\n\\begin{aligned}\n& 0.0400-{ }_{8}^{8} \\mathrm{C}_{7} \\mathrm{H}_{5} \\mathrm{NO}_{3} \\mathrm{~S}-\\left(\\frac{\\mathrm{molC}_{7} \\mathrm{H}_{5} \\mathrm{NO}_{3} \\mathrm{~S}}{183.18-\\mathrm{C}_{7} \\mathrm{H}_{5} \\mathrm{NO}_{3} \\mathrm{~S}}\\right)\\left(\\frac{6.022 \\times 10^{23} \\mathrm{C}_{7} \\mathrm{H}_{5} \\mathrm{NO}_{3} \\mathrm{~S} \\text { molecules }}{1-\\mathrm{mol}_{7} \\mathrm{H}_{5} \\mathrm{NO}_{3} \\mathrm{~S}}\\right) \\\\\n& =1.31 \\times 10^{20} \\mathrm{C}_{7} \\mathrm{H}_{5} \\mathrm{NO}_{3} \\mathrm{~S} \\text { molecules }\n\\end{aligned}\n$$\n\nThe compound's formula shows that each molecule contains seven carbon atoms, and so the number of C atoms in the provided sample is:\n\n$$\n1.31 \\times 10^{20} \\mathrm{C}_{7} \\mathrm{H}_{5} \\mathrm{NO}_{3} \\mathrm{~S} \\text { molecules }\\left(\\frac{7 \\mathrm{C} \\text { atoms }}{1 \\mathrm{C}_{7} \\mathrm{H}_{5} \\mathrm{NO}_{3} \\mathrm{~S} \\text { molecule }}\\right)=9.17 \\times 10^{20} \\mathrm{C} \\text { atoms }\n$$"}
{"id": 2378, "contents": "177. Check Your Learning - \nHow many $\\mathrm{C}_{4} \\mathrm{H}_{10}$ molecules are contained in 9.213 g of this compound? How many hydrogen atoms?"}
{"id": 2379, "contents": "178. Answer: - \n$9.545 \\times 10^{22}$ molecules $\\mathrm{C}_{4} \\mathrm{H}_{10} ; 9.545 \\times 10^{23}$ atoms H"}
{"id": 2380, "contents": "180. Counting Neurotransmitter Molecules in the Brain - \nThe brain is the control center of the central nervous system (Figure 2.29). It sends and receives signals to and from muscles and other internal organs to monitor and control their functions; it processes stimuli detected by sensory organs to guide interactions with the external world; and it houses the complex physiological processes that give rise to our intellect and emotions. The broad field of neuroscience spans all aspects of the structure and function of the central nervous system, including research on the anatomy and physiology of the brain. Great progress has been made in brain research over the past few decades, and the BRAIN Initiative, a federal initiative announced in 2013, aims to accelerate and capitalize on these advances through the concerted efforts of various industrial, academic, and government agencies (see BRAIN Initiative website (https://openstax.org/l/16BRAINInit) for details).\n\n\nFIGURE 2.29 (a) A typical human brain weighs about 1.5 kg and occupies a volume of roughly 1.1 L. (b) Information is transmitted in brain tissue and throughout the central nervous system by specialized cells called neurons (micrograph shows cells at $1600 \\times$ magnification).\n\nSpecialized cells called neurons transmit information between different parts of the central nervous system by way of electrical and chemical signals. Chemical signaling occurs at the interface between different neurons when one of the cells releases molecules (called neurotransmitters) that diffuse across the small gap between the cells (called the synapse) and bind to the surface of the other cell. These neurotransmitter molecules are stored in small intracellular structures called vesicles that fuse to the cell membrane and then break open to release their contents when the neuron is appropriately stimulated. This process is called exocytosis (see Figure 2.30). One neurotransmitter that has been very extensively studied is dopamine, $\\mathrm{C}_{8} \\mathrm{H}_{11} \\mathrm{NO}_{2}$. Dopamine is involved in various neurological processes that impact a wide variety of human behaviors. Dysfunctions in the dopamine systems of the brain underlie serious neurological diseases such as Parkinson's and schizophrenia."}
{"id": 2381, "contents": "180. Counting Neurotransmitter Molecules in the Brain - \nFIGURE 2.30 (a) Chemical signals are transmitted from neurons to other cells by the release of neurotransmitter molecules into the small gaps (synapses) between the cells. (b) Dopamine, $\\mathrm{C}_{8} \\mathrm{H}_{11} \\mathrm{NO}_{2}$, is a neurotransmitter involved in a number of neurological processes.\n\nOne important aspect of the complex processes related to dopamine signaling is the number of neurotransmitter molecules released during exocytosis. Since this number is a central factor in determining neurological response (and subsequent human thought and action), it is important to know how this number changes with certain controlled stimulations, such as the administration of drugs. It is also important to understand the mechanism responsible for any changes in the number of neurotransmitter molecules released-for example, some dysfunction in exocytosis, a change in the number of vesicles in the neuron, or a change in the number of neurotransmitter molecules in each vesicle.\n\nSignificant progress has been made recently in directly measuring the number of dopamine molecules stored in individual vesicles and the amount actually released when the vesicle undergoes exocytosis. Using miniaturized probes that can selectively detect dopamine molecules in very small amounts, scientists have determined that the vesicles of a certain type of mouse brain neuron contain an average of 30,000 dopamine molecules per vesicle (about $5 \\times 10^{-20} \\mathrm{~mol}$ or 50 zmol ). Analysis of these neurons from mice subjected to various drug therapies shows significant changes in the average number of dopamine molecules contained in individual vesicles, increasing or decreasing by up to three-fold, depending on the specific drug used. These studies also indicate that not all of the dopamine in a given vesicle is released during exocytosis, suggesting that it may be possible to regulate the fraction released using pharmaceutical therapies. ${ }^{-3}$\n\n[^1]"}
{"id": 2382, "contents": "181. Key Terms - \nalpha particle ( $\\alpha$ particle) positively charged particle consisting of two protons and two neutrons\nanion negatively charged atom or molecule (contains more electrons than protons)\natomic mass average mass of atoms of an element, expressed in amu\natomic mass unit (amu) (also, unified atomic mass unit, $u$, or Dalton, Da) unit of mass equal to $\\frac{1}{12}$ of the mass of a ${ }^{12} \\mathrm{C}$ atom\natomic number (Z) number of protons in the nucleus of an atom\ncation positively charged atom or molecule (contains fewer electrons than protons)\nchemical symbol one-, two-, or three-letter abbreviation used to represent an element or its atoms\nDalton (Da) alternative unit equivalent to the atomic mass unit\nDalton's atomic theory set of postulates that established the fundamental properties of atoms\nelectron negatively charged, subatomic particle of relatively low mass located outside the nucleus\nempirical formula formula showing the composition of a compound given as the simplest whole-number ratio of atoms\nfundamental unit of charge (also called the elementary charge) equals the magnitude of the charge of an electron (e) with $\\mathrm{e}=1.602 \\times 10^{-19} \\mathrm{C}$\nion electrically charged atom or molecule (contains unequal numbers of protons and electrons)\nisomers compounds with the same chemical formula but different structures\nisotopes atoms that contain the same number of protons but different numbers of neutrons\nlaw of constant composition (also, law of definite proportions) all samples of a pure compound contain the same elements in the same proportions by mass\nlaw of definite proportions (also, law of constant composition) all samples of a pure compound contain the same elements in the same proportions by mass\nlaw of multiple proportions when two elements react to form more than one compound, a fixed mass of one element will react with masses of the other element in a ratio of small whole numbers\nmass number (A) sum of the numbers of neutrons and protons in the nucleus of an atom\nmolecular formula formula indicating the composition of a molecule of a compound and giving the actual number of atoms of each element in a molecule of the compound.\nneutron uncharged, subatomic particle located in the nucleus\nnucleus massive, positively charged center of an atom made up of protons and neutrons"}
{"id": 2383, "contents": "181. Key Terms - \nneutron uncharged, subatomic particle located in the nucleus\nnucleus massive, positively charged center of an atom made up of protons and neutrons\nproton positively charged, subatomic particle located in the nucleus\nspatial isomers compounds in which the relative orientations of the atoms in space differ\nstructural formula shows the atoms in a molecule and how they are connected\nstructural isomer one of two substances that have the same molecular formula but different physical and chemical properties because their atoms are bonded differently\nunified atomic mass unit (u) alternative unit equivalent to the atomic mass unit"}
{"id": 2384, "contents": "182. Key Equations - \naverage mass $=\\sum_{i}(\\text { fractional abundance } \\times \\text { isotopic mass })_{i}$"}
{"id": 2385, "contents": "183. Summary - 183.1. Early Ideas in Atomic Theory\nThe ancient Greeks proposed that matter consists of extremely small particles called atoms. Dalton postulated that each element has a characteristic type of atom that differs in properties from atoms of all other elements, and that atoms of different elements can combine in fixed, small, wholenumber ratios to form compounds. Samples of a particular compound all have the same elemental\nproportions by mass. When two elements form different compounds, a given mass of one element will combine with masses of the other element in a small, whole-number ratio. During any chemical change, atoms are neither created nor destroyed."}
{"id": 2386, "contents": "183. Summary - 183.2. Evolution of Atomic Theory\nAlthough no one has actually seen the inside of an atom, experiments have demonstrated much about\natomic structure. Thomson's cathode ray tube showed that atoms contain small, negatively charged particles called electrons. Millikan discovered that there is a fundamental electric charge-the charge of an electron. Rutherford's gold foil experiment showed that atoms have a small, dense, positively charged nucleus; the positively charged particles within the nucleus are called protons. Chadwick discovered that the nucleus also contains neutral particles called neutrons. Soddy demonstrated that atoms of the same element can differ in mass; these are called isotopes."}
{"id": 2387, "contents": "183. Summary - 183.3. Atomic Structure and Symbolism\nAn atom consists of a small, positively charged nucleus surrounded by electrons. The nucleus contains protons and neutrons; its diameter is about 100,000 times smaller than that of the atom. The mass of one atom is usually expressed in atomic mass units (amu), which is referred to as the atomic mass. An amu is defined as exactly $\\frac{1}{12}$ of the mass of a carbon-12 atom and is equal to $1.6605 \\times 10^{-24} \\mathrm{~g}$.\n\nProtons are relatively heavy particles with a charge of $1+$ and a mass of 1.0073 amu . Neutrons are relatively heavy particles with no charge and a mass of 1.0087 amu . Electrons are light particles with a charge of 1 - and a mass of 0.00055 amu . The number of protons in the nucleus is called the atomic number ( Z ) and is the property that defines an atom's elemental identity. The sum of the numbers of protons and neutrons in the nucleus is called the mass number and, expressed in amu, is\napproximately equal to the mass of the atom. An atom is neutral when it contains equal numbers of electrons and protons.\n\nIsotopes of an element are atoms with the same atomic number but different mass numbers; isotopes of an element, therefore, differ from each other only in the number of neutrons within the nucleus. When a naturally occurring element is composed of several isotopes, the atomic mass of the element represents the average of the masses of the isotopes involved. A chemical symbol identifies the atoms in a substance using symbols, which are one-, two-, or three-letter abbreviations for the atoms."}
{"id": 2388, "contents": "183. Summary - 183.4. Chemical Formulas\nA molecular formula uses chemical symbols and subscripts to indicate the exact numbers of different atoms in a molecule or compound. An empirical formula gives the simplest, whole-number ratio of atoms in a compound. A structural formula indicates the bonding arrangement of the atoms in the molecule. Ball-and-stick and space-filling models show the geometric arrangement of atoms in a molecule. Isomers are compounds with the same molecular formula but different arrangements of atoms. A convenient amount unit for expressing very large numbers of atoms or molecules is the mole. Experimental measurements have determined the number of entities composing 1 mole of substance to be $6.022 \\times 10^{23}$, a quantity called Avogadro's number. The mass in grams of 1 mole of substance is its molar mass."}
{"id": 2389, "contents": "184. Exercises - 184.1. Early Ideas in Atomic Theory\n1. In the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another element. If the spheres of different elements touch, they are part of a single unit of a compound. The following chemical change represented by these spheres may violate one of the ideas of Dalton's atomic theory. Which one?\n\n2. Which postulate of Dalton's theory is consistent with the following observation concerning the weights of reactants and products? When 100 grams of solid calcium carbonate is heated, 44 grams of carbon dioxide and 56 grams of calcium oxide are produced.\n3. Identify the postulate of Dalton's theory that is violated by the following observations: $59.95 \\%$ of one sample of titanium dioxide is titanium; $60.10 \\%$ of a different sample of titanium dioxide is titanium.\n4. Samples of compound $X, Y$, and $Z$ are analyzed, with results shown here.\n\n| Compound | Description | Mass of Carbon | Mass of Hydrogen |\n| :---: | :---: | :---: | :---: |\n| X | clear, colorless, liquid <br> with strong odor | 1.776 g | 0.148 g |\n| Y | clear, colorless, liquid <br> with strong odor | 1.974 g | 0.329 g |\n| Z | clear, colorless, liquid <br> with strong odor | 7.812 g | 0.651 g |\n\nDo these data provide example(s) of the law of definite proportions, the law of multiple proportions, neither, or both? What do these data tell you about compounds X, Y, and Z?"}
{"id": 2390, "contents": "184. Exercises - 184.2. Evolution of Atomic Theory\n5. The existence of isotopes violates one of the original ideas of Dalton's atomic theory. Which one?\n6. How are electrons and protons similar? How are they different?\n7. How are protons and neutrons similar? How are they different?\n8. Predict and test the behavior of $\\alpha$ particles fired at a \"plum pudding\" model atom.\n(a) Predict the paths taken by $\\alpha$ particles that are fired at atoms with a Thomson's plum pudding model structure. Explain why you expect the $\\alpha$ particles to take these paths.\n(b) If $\\alpha$ particles of higher energy than those in (a) are fired at plum pudding atoms, predict how their paths will differ from the lower-energy $\\alpha$ particle paths. Explain your reasoning.\n(c) Now test your predictions from (a) and (b). Open the Rutherford Scattering simulation\n(http://openstax.org/l/16PhetScatter) and select the \"Plum Pudding Atom\" tab. Set \"Alpha Particles\nEnergy\" to \"min,\" and select \"show traces.\" Click on the gun to start firing $\\alpha$ particles. Does this match your prediction from (a)? If not, explain why the actual path would be that shown in the simulation. Hit the pause button, or \"Reset All.\" Set \"Alpha Particles Energy\" to \"max,\" and start firing $\\alpha$ particles. Does this match your prediction from (b)? If not, explain the effect of increased energy on the actual paths as shown in the simulation.\n9. Predict and test the behavior of $\\alpha$ particles fired at a Rutherford atom model.\n(a) Predict the paths taken by $\\alpha$ particles that are fired at atoms with a Rutherford atom model structure. Explain why you expect the $\\alpha$ particles to take these paths.\n(b) If $\\alpha$ particles of higher energy than those in (a) are fired at Rutherford atoms, predict how their paths will differ from the lower-energy $\\alpha$ particle paths. Explain your reasoning.\n(c) Predict how the paths taken by the $\\alpha$ particles will differ if they are fired at Rutherford atoms of elements other than gold. What factor do you expect to cause this difference in paths, and why?"}
{"id": 2391, "contents": "184. Exercises - 184.2. Evolution of Atomic Theory\n(c) Predict how the paths taken by the $\\alpha$ particles will differ if they are fired at Rutherford atoms of elements other than gold. What factor do you expect to cause this difference in paths, and why?\n(d) Now test your predictions from (a), (b), and (c). Open the Rutherford Scattering simulation (http://openstax.org/l/16PhetScatter) and select the \"Rutherford Atom\" tab. Due to the scale of the simulation, it is best to start with a small nucleus, so select \" 20 \" for both protons and neutrons, \"min\" for energy, show traces, and then start firing $\\alpha$ particles. Does this match your prediction from (a)? If not, explain why the actual path would be that shown in the simulation. Pause or reset, set energy to \"max,\" and start firing $\\alpha$ particles. Does this match your prediction from (b)? If not, explain the effect of increased energy on the actual path as shown in the simulation. Pause or reset, select \" 40 \" for both protons and neutrons, \"min\" for energy, show traces, and fire away. Does this match your prediction from (c)? If not, explain why the actual path would be that shown in the simulation. Repeat this with larger numbers of protons and neutrons. What generalization can you make regarding the type of atom and effect on the path of $\\alpha$ particles? Be clear and specific."}
{"id": 2392, "contents": "184. Exercises - 184.3. Atomic Structure and Symbolism\n10. In what way are isotopes of a given element always different? In what way(s) are they always the same?\n11. Write the symbol for each of the following ions:\n(a) the ion with a $1+$ charge, atomic number 55 , and mass number 133\n(b) the ion with 54 electrons, 53 protons, and 74 neutrons\n(c) the ion with atomic number 15, mass number 31, and a 3 - charge\n(d) the ion with 24 electrons, 30 neutrons, and a $3+$ charge\n12. Write the symbol for each of the following ions:\n(a) the ion with a $3+$ charge, 28 electrons, and a mass number of 71\n(b) the ion with 36 electrons, 35 protons, and 45 neutrons\n(c) the ion with 86 electrons, 142 neutrons, and a 4+ charge\n(d) the ion with a $2+$ charge, atomic number 38 , and mass number 87\n13. Open the Build an Atom simulation (http://openstax.org/l/16PhetAtomBld) and click on the Atom icon.\n(a) Pick any one of the first 10 elements that you would like to build and state its symbol.\n(b) Drag protons, neutrons, and electrons onto the atom template to make an atom of your element. State the numbers of protons, neutrons, and electrons in your atom, as well as the net charge and mass number.\n(c) Click on \"Net Charge\" and \"Mass Number,\" check your answers to (b), and correct, if needed.\n(d) Predict whether your atom will be stable or unstable. State your reasoning.\n(e) Check the \"Stable/Unstable\" box. Was your answer to (d) correct? If not, first predict what you can do to make a stable atom of your element, and then do it and see if it works. Explain your reasoning.\n14. Open the Build an Atom simulation (http://openstax.org/l/16PhetAtomBld)\n(a) Drag protons, neutrons, and electrons onto the atom template to make a neutral atom of Oxygen-16 and give the isotope symbol for this atom."}
{"id": 2393, "contents": "184. Exercises - 184.3. Atomic Structure and Symbolism\n(a) Drag protons, neutrons, and electrons onto the atom template to make a neutral atom of Oxygen-16 and give the isotope symbol for this atom.\n(b) Now add two more electrons to make an ion and give the symbol for the ion you have created.\n15. Open the Build an Atom simulation (http://openstax.org/l/16PhetAtomBld)\n(a) Drag protons, neutrons, and electrons onto the atom template to make a neutral atom of Lithium-6 and give the isotope symbol for this atom.\n(b) Now remove one electron to make an ion and give the symbol for the ion you have created.\n16. Determine the number of protons, neutrons, and electrons in the following isotopes that are used in medical diagnoses:\n(a) atomic number 9, mass number 18, charge of 1-\n(b) atomic number 43 , mass number 99 , charge of $7+$\n(c) atomic number 53, atomic mass number 131, charge of $1-$\n(d) atomic number 81, atomic mass number 201, charge of $1+$\n(e) Name the elements in parts (a), (b), (c), and (d).\n17. The following are properties of isotopes of two elements that are essential in our diet. Determine the number of protons, neutrons and electrons in each and name them.\n(a) atomic number 26 , mass number 58 , charge of $2+$\n(b) atomic number 53 , mass number 127, charge of 1-\n18. Give the number of protons, electrons, and neutrons in neutral atoms of each of the following isotopes:\n(a) ${ }_{5}^{10} \\mathrm{~B}$\n(b) ${ }_{80}^{199} \\mathrm{Hg}$\n(c) ${ }_{29}^{63} \\mathrm{Cu}$\n(d) ${ }_{6}^{13} \\mathrm{C}$\n(e) ${ }_{34}^{77} \\mathrm{Se}$\n19. Give the number of protons, electrons, and neutrons in neutral atoms of each of the following isotopes:\n(a) ${ }_{3}^{7} \\mathrm{Li}$"}
{"id": 2394, "contents": "184. Exercises - 184.3. Atomic Structure and Symbolism\n19. Give the number of protons, electrons, and neutrons in neutral atoms of each of the following isotopes:\n(a) ${ }_{3}^{7} \\mathrm{Li}$\n(b) ${ }_{52}^{125} \\mathrm{Te}$\n(c) ${ }_{47}^{109} \\mathrm{Ag}$\n(d) ${ }_{7}^{15} \\mathrm{~N}$\n(e) ${ }_{15}^{31} \\mathrm{P}$\n20. Click on the site (http://openstax.org/l/16PhetAtomMass) and select the \"Mix Isotopes\" tab, hide the \"Percent Composition\" and \"Average Atomic Mass\" boxes, and then select the element boron.\n(a) Write the symbols of the isotopes of boron that are shown as naturally occurring in significant amounts.\n(b) Predict the relative amounts (percentages) of these boron isotopes found in nature. Explain the reasoning behind your choice.\n(c) Add isotopes to the black box to make a mixture that matches your prediction in (b). You may drag isotopes from their bins or click on \"More\" and then move the sliders to the appropriate amounts.\n(d) Reveal the \"Percent Composition\" and \"Average Atomic Mass\" boxes. How well does your mixture match with your prediction? If necessary, adjust the isotope amounts to match your prediction.\n(e) Select \"Nature's\" mix of isotopes and compare it to your prediction. How well does your prediction compare with the naturally occurring mixture? Explain. If necessary, adjust your amounts to make them match \"Nature's\" amounts as closely as possible.\n21. Repeat Exercise 2.20 using an element that has three naturally occurring isotopes.\n22. An element has the following natural abundances and isotopic masses: $90.92 \\%$ abundance with 19.99 amu, $0.26 \\%$ abundance with 20.99 amu , and $8.82 \\%$ abundance with 21.99 amu . Calculate the average atomic mass of this element."}
{"id": 2395, "contents": "184. Exercises - 184.3. Atomic Structure and Symbolism\n23. Average atomic masses listed by IUPAC are based on a study of experimental results. Bromine has two isotopes, ${ }^{79} \\mathrm{Br}$ and ${ }^{81} \\mathrm{Br}$, whose masses ( 78.9183 and 80.9163 amu , respectively) and abundances ( $50.69 \\%$ and $49.31 \\%$, respectively) were determined in earlier experiments. Calculate the average atomic mass of bromine based on these experiments.\n24. Variations in average atomic mass may be observed for elements obtained from different sources. Lithium provides an example of this. The isotopic composition of lithium from naturally occurring minerals is $7.5 \\%{ }^{6} \\mathrm{Li}$ and $92.5 \\%{ }^{7} \\mathrm{Li}$, which have masses of 6.01512 amu and 7.01600 amu , respectively. A commercial source of lithium, recycled from a military source, was $3.75 \\%{ }^{6} \\mathrm{Li}$ (and the rest ${ }^{7} \\mathrm{Li}$ ). Calculate the average atomic mass values for each of these two sources.\n25. The average atomic masses of some elements may vary, depending upon the sources of their ores. Naturally occurring boron consists of two isotopes with accurately known masses $\\left({ }^{10} \\mathrm{~B}, 10.0129 \\mathrm{amu}\\right.$ and ${ }^{11} \\mathrm{~B}, 11.00931 \\mathrm{amu}$ ). The actual atomic mass of boron can vary from 10.807 to 10.819 , depending on whether the mineral source is from Turkey or the United States. Calculate the percent abundances leading to the two values of the average atomic masses of boron from these two countries.\n26. The ${ }^{18} \\mathrm{O}:{ }^{16} \\mathrm{O}$ abundance ratio in some meteorites is greater than that used to calculate the average atomic mass of oxygen on earth. Is the average mass of an oxygen atom in these meteorites greater than, less than, or equal to that of a terrestrial oxygen atom?"}
{"id": 2396, "contents": "184. Exercises - 184.4. Chemical Formulas\n27. Explain why the symbol for an atom of the element oxygen and the formula for a molecule of oxygen differ.\n28. Explain why the symbol for the element sulfur and the formula for a molecule of sulfur differ.\n29. Write the molecular and empirical formulas of the following compounds:\n(a)\n\n$$\n\\mathrm{O}=\\mathrm{C}=\\mathrm{O}\n$$\n\n(b)\n\n$$\n\\mathrm{H}-\\mathrm{C} \\equiv \\mathrm{C}-\\mathrm{H}\n$$\n\n(c)\n\n(d)\n\n30. Write the molecular and empirical formulas of the following compounds:\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n31. Determine the empirical formulas for the following compounds:\n(a) caffeine, $\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2}$\n(b) sucrose, $\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}$\n(c) hydrogen peroxide, $\\mathrm{H}_{2} \\mathrm{O}_{2}$\n(d) glucose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$\n(e) ascorbic acid (vitamin C), $\\mathrm{C}_{6} \\mathrm{H}_{8} \\mathrm{O}_{6}$\n32. Determine the empirical formulas for the following compounds:\n(a) acetic acid, $\\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}_{2}$\n(b) citric acid, $\\mathrm{C}_{6} \\mathrm{H}_{8} \\mathrm{O}_{7}$\n(c) hydrazine, $\\mathrm{N}_{2} \\mathrm{H}_{4}$\n(d) nicotine, $\\mathrm{C}_{10} \\mathrm{H}_{14} \\mathrm{~N}_{2}$\n(e) butane, $\\mathrm{C}_{4} \\mathrm{H}_{10}$\n33. Write the empirical formulas for the following compounds:\n(a)\n\n(b)"}
{"id": 2397, "contents": "184. Exercises - 184.4. Chemical Formulas\n34. Open the Build a Molecule simulation (http://openstax.org/l/16molbuilding) and select the \"Larger Molecules\" tab. Select an appropriate atom's \"Kit\" to build a molecule with two carbon and six hydrogen atoms. Drag atoms into the space above the \"Kit\" to make a molecule. A name will appear when you have made an actual molecule that exists (even if it is not the one you want). You can use the scissors tool to separate atoms if you would like to change the connections. Click on \"3D\" to see the molecule, and look at both the space-filling and ball-and-stick possibilities.\n(a) Draw the structural formula of this molecule and state its name.\n(b) Can you arrange these atoms in any way to make a different compound?\n35. Use the Build a Molecule simulation (http://openstax.org/l/16molbuilding) to repeat Exercise 2.34, but build a molecule with two carbons, six hydrogens, and one oxygen.\n(a) Draw the structural formula of this molecule and state its name.\n(b) Can you arrange these atoms to make a different molecule? If so, draw its structural formula and state its name.\n(c) How are the molecules drawn in (a) and (b) the same? How do they differ? What are they called (the type of relationship between these molecules, not their names)?\n36. Use the Build a Molecule simulation (http://openstax.org/l/16molbuilding) to repeat Exercise 2.34, but build a molecule with three carbons, seven hydrogens, and one chlorine.\n(a) Draw the structural formula of this molecule and state its name.\n(b) Can you arrange these atoms to make a different molecule? If so, draw its structural formula and state its name.\n(c) How are the molecules drawn in (a) and (b) the same? How do they differ? What are they called (the type of relationship between these molecules, not their names)?\n37. Write a sentence that describes how to determine the number of moles of a compound in a known mass of the compound if we know its molecular formula."}
{"id": 2398, "contents": "184. Exercises - 184.4. Chemical Formulas\n37. Write a sentence that describes how to determine the number of moles of a compound in a known mass of the compound if we know its molecular formula.\n38. Compare 1 mole of $\\mathrm{H}_{2}, 1$ mole of $\\mathrm{O}_{2}$, and 1 mole of $\\mathrm{F}_{2}$.\n(a) Which has the largest number of molecules? Explain why.\n(b) Which has the greatest mass? Explain why.\n39. Which contains the greatest mass of oxygen: 0.75 mol of ethanol $\\left(\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}\\right), 0.60 \\mathrm{~mol}$ of formic acid $\\left(\\mathrm{HCO}_{2} \\mathrm{H}\\right)$, or 1.0 mol of water $\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)$ ? Explain why.\n40. Which contains the greatest number of moles of oxygen atoms: 1 mol of ethanol $\\left(\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}\\right), 1 \\mathrm{~mol}$ of formic acid $\\left(\\mathrm{HCO}_{2} \\mathrm{H}\\right)$, or 1 mol of water $\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)$ ? Explain why.\n41. How are the molecular mass and the molar mass of a compound similar and how are they different?\n42. Calculate the molar mass of each of the following compounds:\n(a) hydrogen fluoride, HF\n(b) ammonia, $\\mathrm{NH}_{3}$\n(c) nitric acid, $\\mathrm{HNO}_{3}$\n(d) silver sulfate, $\\mathrm{Ag}_{2} \\mathrm{SO}_{4}$\n(e) boric acid, $\\mathrm{B}(\\mathrm{OH})_{3}$\n43. Calculate the molar mass of each of the following:\n(a) $\\mathrm{S}_{8}$\n(b) $\\mathrm{C}_{5} \\mathrm{H}_{12}$\n(c) $\\mathrm{Sc}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}$\n(d) $\\mathrm{CH}_{3} \\mathrm{COCH}_{3}$ (acetone)"}
{"id": 2399, "contents": "184. Exercises - 184.4. Chemical Formulas\n(c) $\\mathrm{Sc}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}$\n(d) $\\mathrm{CH}_{3} \\mathrm{COCH}_{3}$ (acetone)\n(e) $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$ (glucose)\n44. Calculate the empirical or molecular formula mass and the molar mass of each of the following minerals:\n(a) limestone, $\\mathrm{CaCO}_{3}$\n(b) halite, NaCl\n(c) beryl, $\\mathrm{Be}_{3} \\mathrm{Al}_{2} \\mathrm{Si}_{6} \\mathrm{O}_{18}$\n(d) malachite, $\\mathrm{Cu}_{2}(\\mathrm{OH})_{2} \\mathrm{CO}_{3}$\n(e) turquoise, $\\mathrm{CuAl}_{6}\\left(\\mathrm{PO}_{4}\\right)_{4}(\\mathrm{OH})_{8}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{4}$\n45. Calculate the molar mass of each of the following:\n(a) the anesthetic halothane, $\\mathrm{C}_{2} \\mathrm{HBrClF}_{3}$\n(b) the herbicide paraquat, $\\mathrm{C}_{12} \\mathrm{H}_{14} \\mathrm{~N}_{2} \\mathrm{Cl}_{2}$\n(c) caffeine, $\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2}$\n(d) urea, $\\mathrm{CO}\\left(\\mathrm{NH}_{2}\\right)_{2}$\n(e) a typical soap, $\\mathrm{C}_{17} \\mathrm{H}_{35} \\mathrm{CO}_{2} \\mathrm{Na}$\n46. Determine the number of moles of compound and the number of moles of each type of atom in each of the following:\n(a) 25.0 g of propylene, $\\mathrm{C}_{3} \\mathrm{H}_{6}$"}
{"id": 2400, "contents": "184. Exercises - 184.4. Chemical Formulas\n(a) 25.0 g of propylene, $\\mathrm{C}_{3} \\mathrm{H}_{6}$\n(b) $3.06 \\times 10^{-3} \\mathrm{~g}$ of the amino acid glycine, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{NO}_{2}$\n(c) 25 lb of the herbicide Treflan, $\\mathrm{C}_{13} \\mathrm{H}_{16} \\mathrm{~N}_{2} \\mathrm{O}_{4} \\mathrm{~F}(1 \\mathrm{lb}=454 \\mathrm{~g})$\n(d) 0.125 kg of the insecticide Paris Green, $\\mathrm{Cu}_{4}\\left(\\mathrm{AsO}_{3}\\right)_{2}\\left(\\mathrm{CH}_{3} \\mathrm{CO}_{2}\\right)_{2}$\n(e) 325 mg of aspirin, $\\mathrm{C}_{6} \\mathrm{H}_{4}\\left(\\mathrm{CO}_{2} \\mathrm{H}\\right)\\left(\\mathrm{CO}_{2} \\mathrm{CH}_{3}\\right)$\n47. Determine the mass of each of the following:\n(a) 0.0146 mol KOH\n(b) 10.2 mol ethane, $\\mathrm{C}_{2} \\mathrm{H}_{6}$\n(c) $1.6 \\times 10^{-3} \\mathrm{~mol} \\mathrm{Na}_{2} \\mathrm{SO}_{4}$\n(d) $6.854 \\times 10^{3} \\mathrm{~mol}$ glucose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$\n(e) $2.86 \\mathrm{~mol} \\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6} \\mathrm{Cl}_{3}$\n48. Determine the number of moles of the compound and determine the number of moles of each type of atom in each of the following:\n(a) 2.12 g of potassium bromide, KBr\n(b) 0.1488 g of phosphoric acid, $\\mathrm{H}_{3} \\mathrm{PO}_{4}$"}
{"id": 2401, "contents": "184. Exercises - 184.4. Chemical Formulas\n(a) 2.12 g of potassium bromide, KBr\n(b) 0.1488 g of phosphoric acid, $\\mathrm{H}_{3} \\mathrm{PO}_{4}$\n(c) 23 kg of calcium carbonate, $\\mathrm{CaCO}_{3}$\n(d) 78.452 g of aluminum sulfate, $\\mathrm{Al}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}$\n(e) 0.1250 mg of caffeine, $\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2}$\n49. Determine the mass of each of the following:\n(a) 2.345 mol LiCl\n(b) 0.0872 mol acetylene, $\\mathrm{C}_{2} \\mathrm{H}_{2}$\n(c) $3.3 \\times 10^{-2} \\mathrm{~mol} \\mathrm{Na}_{2} \\mathrm{CO}_{3}$\n(d) $1.23 \\times 10^{3}$ mol fructose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$\n(e) $0.5758 \\mathrm{~mol} \\mathrm{FeSO}_{4}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{7}$\n50. The approximate minimum daily dietary requirement of the amino acid leucine, $\\mathrm{C}_{6} \\mathrm{H}_{13} \\mathrm{NO}_{2}$, is 1.1 g . What is this requirement in moles?\n51. Determine the mass in grams of each of the following:\n(a) 0.600 mol of oxygen atoms\n(b) 0.600 mol of oxygen molecules, $\\mathrm{O}_{2}$\n(c) 0.600 mol of ozone molecules, $\\mathrm{O}_{3}$\n52. A $55-\\mathrm{kg}$ woman has $7.5 \\times 10^{-3} \\mathrm{~mol}$ of hemoglobin (molar mass $=64,456 \\mathrm{~g} / \\mathrm{mol}$ ) in her blood. How many hemoglobin molecules is this? What is this quantity in grams?"}
{"id": 2402, "contents": "184. Exercises - 184.4. Chemical Formulas\n53. Determine the number of atoms and the mass of zirconium, silicon, and oxygen found in 0.3384 mol of zircon, $\\mathrm{ZrSiO}_{4}$, a semiprecious stone.\n54. Determine which of the following contains the greatest mass of hydrogen: 1 mol of $\\mathrm{CH}_{4}, 0.6 \\mathrm{~mol}^{\\text {of }} \\mathrm{C}_{6} \\mathrm{H}_{6}$, or 0.4 mol of $\\mathrm{C}_{3} \\mathrm{H}_{8}$.\n55. Determine which of the following contains the greatest mass of aluminum: 122 g of $\\mathrm{AlPO}_{4}, 266 \\mathrm{~g} \\mathrm{of} \\mathrm{Al}_{2} \\mathrm{Cl}_{6}$, or 225 g of $\\mathrm{Al}_{2} \\mathrm{~S}_{3}$.\n56. Diamond is one form of elemental carbon. An engagement ring contains a diamond weighing 1.25 carats ( 1 carat $=200 \\mathrm{mg}$ ). How many atoms are present in the diamond?\n57. The Cullinan diamond was the largest natural diamond ever found (January 25, 1905). It weighed 3104 carats ( 1 carat $=200 \\mathrm{mg}$ ). How many carbon atoms were present in the stone?\n58. One 55 -gram serving of a particular cereal supplies 270 mg of sodium, $11 \\%$ of the recommended daily allowance. How many moles and atoms of sodium are in the recommended daily allowance?\n59. A certain nut crunch cereal contains 11.0 grams of sugar (sucrose, $\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}$ ) per serving size of 60.0 grams. How many servings of this cereal must be eaten to consume 0.0278 moles of sugar?\n60. A tube of toothpaste contains 0.76 g of sodium monofluorophosphate $\\left(\\mathrm{Na}_{2} \\mathrm{PO}_{3} \\mathrm{~F}\\right)$ in 100 mL .\n(a) What mass of fluorine atoms in mg was present?"}
{"id": 2403, "contents": "184. Exercises - 184.4. Chemical Formulas\n(a) What mass of fluorine atoms in mg was present?\n(b) How many fluorine atoms were present?\n61. Which of the following represents the least number of molecules?\n(a) 20.0 g of $\\mathrm{H}_{2} \\mathrm{O}(18.02 \\mathrm{~g} / \\mathrm{mol})$\n(b) 77.0 g of $\\mathrm{CH}_{4}(16.06 \\mathrm{~g} / \\mathrm{mol})$\n(c) 68.0 g of $\\mathrm{C}_{3} \\mathrm{H}_{6}(42.08 \\mathrm{~g} / \\mathrm{mol})$\n(d) 100.0 g of $\\mathrm{N}_{2} \\mathrm{O}(44.02 \\mathrm{~g} / \\mathrm{mol})$\n(e) 84.0 g of $\\mathrm{HF}(20.01 \\mathrm{~g} / \\mathrm{mol})$"}
{"id": 2404, "contents": "185. CHAPTER 3 <br> Electronic Structure and Periodic Properties of Elements - \nFigure 3.1 The Crab Nebula consists of remnants of a supernova (the explosion of a star). NASA's Hubble Space Telescope produced this composite image. Measurements of the emitted light wavelengths enabled astronomers to identify the elements in the nebula, determining that it contains specific ions including $\\mathrm{S}^{+}$(green filaments) and $\\mathrm{O}^{2+}$ (red filaments). (credit: modification of work by NASA and ESA)"}
{"id": 2405, "contents": "186. CHAPTER OUTLINE - 186.2. The Bohr Model\n3.3 Development of Quantum Theory\n3.4 Electronic Structure of Atoms (Electron Configurations)\n3.5 Periodic Variations in Element Properties\n3.6 The Periodic Table\n3.7 Ionic and Molecular Compounds\n\nINTRODUCTION In 1054, Chinese astronomers recorded the appearance of a \"guest star\" in the sky, visible even during the day, which then disappeared slowly over the next two years. The sudden appearance was due to a supernova explosion, which was much brighter than the original star. Even though this supernova was observed almost a millennium ago, the remaining Crab Nebula (Figure 3.1) continues to release energy today. It emits not only visible light but also infrared light, X-rays, and other forms of electromagnetic radiation. The nebula emits both continuous spectra (the blue-white glow) and atomic emission spectra (the colored\nfilaments). In this chapter, we will discuss light and other forms of electromagnetic radiation and how they are related to the electronic structure of atoms. We will also see how this radiation can be used to identify elements, even from thousands of light years away."}
{"id": 2406, "contents": "187. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Explain the basic behavior of waves, including travelling waves and standing waves\n- Describe the wave nature of light\n- Use appropriate equations to calculate related light-wave properties such as period, frequency, wavelength, and energy\n- Distinguish between line and continuous emission spectra\n- Describe the particle nature of light\n\nThe nature of light has been a subject of inquiry since antiquity. In the seventeenth century, Isaac Newton performed experiments with lenses and prisms and was able to demonstrate that white light consists of the individual colors of the rainbow combined together. Newton explained his optics findings in terms of a \"corpuscular\" view of light, in which light was composed of streams of extremely tiny particles travelling at high speeds according to Newton's laws of motion. Others in the seventeenth century, such as Christiaan Huygens, had shown that optical phenomena such as reflection and refraction could be equally well explained in terms of light as waves travelling at high speed through a medium called \"luminiferous aether\" that was thought to permeate all space. Early in the nineteenth century, Thomas Young demonstrated that light passing through narrow, closely spaced slits produced interference patterns that could not be explained in terms of Newtonian particles but could be easily explained in terms of waves. Later in the nineteenth century, after James Clerk Maxwell developed his theory of electromagnetic radiation and showed that light was the visible part of a vast spectrum of electromagnetic waves, the particle view of light became thoroughly discredited. By the end of the nineteenth century, scientists viewed the physical universe as roughly comprising two separate domains: matter composed of particles moving according to Newton's laws of motion, and electromagnetic radiation consisting of waves governed by Maxwell's equations. Today, these domains are referred to as classical mechanics and classical electrodynamics (or classical electromagnetism). Although there were a few physical phenomena that could not be explained within this framework, scientists at that time were so confident of the overall soundness of this framework that they viewed these aberrations as puzzling paradoxes that would ultimately be resolved somehow within this framework. As we shall see, these paradoxes led to a contemporary framework that intimately connects particles and waves at a fundamental level called waveparticle duality, which has superseded the classical view."}
{"id": 2407, "contents": "187. LEARNING OBJECTIVES - \nVisible light and other forms of electromagnetic radiation play important roles in chemistry, since they can be used to infer the energies of electrons within atoms and molecules. Much of modern technology is based on electromagnetic radiation. For example, radio waves from a mobile phone, X-rays used by dentists, the energy used to cook food in your microwave, the radiant heat from red-hot objects, and the light from your television screen are forms of electromagnetic radiation that all exhibit wavelike behavior."}
{"id": 2408, "contents": "188. Waves - \nA wave is an oscillation or periodic movement that can transport energy from one point in space to another. Common examples of waves are all around us. Shaking the end of a rope transfers energy from your hand to the other end of the rope, dropping a pebble into a pond causes waves to ripple outward along the water's surface, and the expansion of air that accompanies a lightning strike generates sound waves (thunder) that can travel outward for several miles. In each of these cases, kinetic energy is transferred through matter (the rope, water, or air) while the matter remains essentially in place. An insightful example of a wave occurs in sports stadiums when fans in a narrow region of seats rise simultaneously and stand with their arms raised up for a few seconds before sitting down again while the fans in neighboring sections likewise stand up and sit down in sequence. While this wave can quickly encircle a large stadium in a few seconds, none of the fans actually travel with the wave-they all stay in or above their seats.\n\nWaves need not be restricted to travel through matter. As Maxwell showed, electromagnetic waves consist of an electric field oscillating in step with a perpendicular magnetic field, both of which are perpendicular to the direction of travel. These waves can travel through a vacuum at a constant speed of $2.998 \\times 10^{8} \\mathrm{~m} / \\mathrm{s}$, the speed of light (denoted by $c$ )."}
{"id": 2409, "contents": "188. Waves - \nAll waves, including forms of electromagnetic radiation, are characterized by, a wavelength (denoted by $\\lambda$, the lowercase Greek letter lambda), a frequency (denoted by $\\nu$, the lowercase Greek letter nu), and an amplitude. As can be seen in Figure 3.2, the wavelength is the distance between two consecutive peaks or troughs in a wave (measured in meters in the SI system). Electromagnetic waves have wavelengths that fall within an enormous range-wavelengths of kilometers $\\left(10^{3} \\mathrm{~m}\\right)$ to picometers $\\left(10^{-12} \\mathrm{~m}\\right)$ have been observed. The frequency is the number of wave cycles that pass a specified point in space in a specified amount of time (in the SI system, this is measured in seconds). A cycle corresponds to one complete wavelength. The unit for frequency, expressed as cycles per second $\\left[\\mathrm{s}^{-1}\\right]$, is the hertz (Hz). Common multiples of this unit are megahertz, $\\left(1 \\mathrm{MHz}=1 \\times 10^{6} \\mathrm{~Hz}\\right)$ and gigahertz $\\left(1 \\mathrm{GHz}=1 \\times 10^{9} \\mathrm{~Hz}\\right)$. The amplitude corresponds to the magnitude of the wave's displacement and so, in Figure 3.2, this corresponds to one-half the height between the peaks and troughs. The amplitude is related to the intensity of the wave, which for light is the brightness, and for sound is the loudness.\n\n\n\nFIGURE 3.2 One-dimensional sinusoidal waves show the relationship among wavelength, frequency, and speed. The wave with the shortest wavelength has the highest frequency. Amplitude is one-half the height of the wave from peak to trough.\n\nThe product of a wave's wavelength $(\\lambda)$ and its frequency $(\\nu), \\lambda \\nu$, is the speed of the wave. Thus, for electromagnetic radiation in a vacuum, speed is equal to the fundamental constant, $c$ :\n\n$$\nc=2.998 \\times 10^{8} \\mathrm{~ms}^{-1}=\\lambda v\n$$"}
{"id": 2410, "contents": "188. Waves - \n$$\nc=2.998 \\times 10^{8} \\mathrm{~ms}^{-1}=\\lambda v\n$$\n\nWavelength and frequency are inversely proportional: As the wavelength increases, the frequency decreases. The inverse proportionality is illustrated in Figure 3.3. This figure also shows the electromagnetic spectrum, the range of all types of electromagnetic radiation. Each of the various colors of visible light has specific frequencies and wavelengths associated with them, and you can see that visible light makes up only a small portion of the electromagnetic spectrum. Because the technologies developed to work in various parts of the electromagnetic spectrum are different, for reasons of convenience and historical legacies, different units are typically used for different parts of the spectrum. For example, radio waves are usually specified as frequencies (typically in units of MHz ), while the visible region is usually specified in wavelengths (typically in units of $n m$ or angstroms).\n\n\nFIGURE 3.3 Portions of the electromagnetic spectrum are shown in order of decreasing frequency and increasing wavelength. (credit \"Cosmic ray\": modification of work by NASA; credit \"PET scan\": modification of work by the National Institute of Health; credit \"X-ray\": modification of work by Dr. Jochen Lengerke; credit \"Dental curing\": modification of work by the Department of the Navy; credit \"Night vision\": modification of work by the Department of the Army; credit \"Remote\": modification of work by Emilian Robert Vicol; credit \"Cell phone\": modification of work by Brett Jordan; credit \"Microwave oven\": modification of work by Billy Mabray; credit \"Ultrasound\": modification of work by Jane Whitney; credit \"AM radio\": modification of work by Dave Clausen)"}
{"id": 2411, "contents": "190. Determining the Frequency and Wavelength of Radiation - \nA sodium streetlight gives off yellow light that has a wavelength of $589 \\mathrm{~nm}\\left(1 \\mathrm{~nm}=1 \\times 10^{-9} \\mathrm{~m}\\right)$. What is the frequency of this light?"}
{"id": 2412, "contents": "191. Solution - \nWe can rearrange the equation $c=\\lambda \\nu$ to solve for the frequency:\n\n$$\n\\nu=\\frac{c}{\\lambda}\n$$\n\nSince $c$ is expressed in meters per second, we must also convert 589 nm to meters.\n\n$$\nv=\\left(\\frac{2.998 \\times 10^{8}-\\mathrm{m}^{-1}}{589 \\mathrm{~nm}}\\right)\\left(\\frac{1 \\times 10^{9} \\mathrm{~nm}}{1 \\mathrm{~m}}\\right)=5.09 \\times 10^{14} \\mathrm{~s}^{-1}\n$$"}
{"id": 2413, "contents": "192. Check Your Learning - \nOne of the frequencies used to transmit and receive cellular telephone signals in the United States is 850 MHz . What is the wavelength in meters of these radio waves?"}
{"id": 2414, "contents": "193. Answer: - \n$0.353 \\mathrm{~m}=35.3 \\mathrm{~cm}$"}
{"id": 2415, "contents": "194. Chemistry in Everyday Life - \nWireless Communication\n\n\nFIGURE 3.4 Radio and cell towers are typically used to transmit long-wavelength electromagnetic radiation. Increasingly, cell towers are designed to blend in with the landscape, as with the Tucson, Arizona, cell tower (right) disguised as a palm tree. (credit left: modification of work by Sir Mildred Pierce; credit middle: modification of work by M.O. Stevens)\n\nMany valuable technologies operate in the radio ( $3 \\mathrm{kHz}-300 \\mathrm{GHz}$ ) frequency region of the electromagnetic spectrum. At the low frequency (low energy, long wavelength) end of this region are AM (amplitude modulation) radio signals ( $540-2830 \\mathrm{kHz}$ ) that can travel long distances. FM (frequency modulation) radio signals are used at higher frequencies ( $87.5-108.0 \\mathrm{MHz}$ ). In AM radio, the information is transmitted by varying the amplitude of the wave (Figure 3.5). In FM radio, by contrast, the amplitude is constant and the instantaneous frequency varies.\n\n\nFIGURE 3.5 This schematic depicts how amplitude modulation (AM) and frequency modulation (FM) can be used to transmit a radio wave.\n\nOther technologies also operate in the radio-wave portion of the electromagnetic spectrum. For example, 4G cellular telephone signals are approximately 880 MHz , while Global Positioning System (GPS) signals operate at 1.228 and 1.575 GHz , local area wireless technology (Wi-Fi) networks operate at 2.4 to 5 GHz , and highway toll sensors operate at 5.8 GHz . The frequencies associated with these applications are convenient because such waves tend not to be absorbed much by common building materials."}
{"id": 2416, "contents": "194. Chemistry in Everyday Life - \nOne particularly characteristic phenomenon of waves results when two or more waves come into contact: They interfere with each other. Figure 3.6 shows the interference patterns that arise when light passes through narrow slits closely spaced about a wavelength apart. The fringe patterns produced depend on the wavelength, with the fringes being more closely spaced for shorter wavelength light passing through a given set of slits. When the light passes through the two slits, each slit effectively acts as a new source, resulting in two closely spaced waves coming into contact at the detector (the camera in this case). The dark regions in Figure 3.6 correspond to regions where the peaks for the wave from one slit happen to coincide with the troughs for the wave from the other slit (destructive interference), while the brightest regions correspond to the regions where the peaks for the two waves (or their two troughs) happen to coincide (constructive interference). Likewise, when two stones are tossed close together into a pond, interference patterns are visible in the interactions between the waves produced by the stones. Such interference patterns cannot be explained by particles moving according to the laws of classical mechanics.\n\n\nFIGURE 3.6 Interference fringe patterns are shown for light passing through two closely spaced, narrow slits. The spacing of the fringes depends on the wavelength, with the fringes being more closely spaced for the shorterwavelength blue light. (credit: PASCO)"}
{"id": 2417, "contents": "196. Dorothy Crowfoot Hodgkin - \nX-rays exhibit wavelengths of approximately $0.01-10 \\mathrm{~nm}$. Since these wavelengths are comparable to the spaces between atoms in a crystalline solid, X-rays are scattered when they pass through crystals. The scattered rays undergo constructive and destructive interference that creates a specific diffraction pattern that may be measured and used to precisely determine the positions of atoms within the crystal. This phenomenon of X-ray diffraction is the basis for very powerful techniques enabling the determination of molecular structure. One of the pioneers who applied this powerful technology to important biochemical substances was Dorothy Crowfoot Hodgkin.\n\nBorn in Cairo, Egypt, in 1910 to British parents, Dorothy's fascination with chemistry was fostered early in her life. At age 11 she was enrolled in a prestigious English grammar school where she was one of just two girls allowed to study chemistry. On her 16th birthday, her mother, Molly, gifted her a book on X-ray crystallography, which had a profound impact on the trajectory of her career. She studied chemistry at Oxford University, graduating with first-class honors in 1932 and directly entering Cambridge University to pursue a doctoral degree. At Cambridge, Dorothy recognized the promise of X-ray crystallography for protein structure determinations, conducting research that earned her a PhD in 1937. Over the course of a very productive career, Dr. Hodgkin was credited with determining structures for several important biomolecules, including cholesterol iodide, penicillin, and vitamin B12. In recognition of her achievements in the use of X-ray techniques to elucidate the structures of biochemical substances, she was awarded the 1964 Nobel Prize in Chemistry. In 1969, she led a team of scientists who deduced the structure of insulin, facilitating the mass production of this hormone and greatly advancing the treatment of diabetic patients worldwide. Dr. Hodgkin continued working with the international scientific community, earning numerous distinctions and awards prior to her death in 1993."}
{"id": 2418, "contents": "196. Dorothy Crowfoot Hodgkin - \nNot all waves are travelling waves. Standing waves (also known as stationary waves) remain constrained within some region of space. As we shall see, standing waves play an important role in our understanding of the electronic structure of atoms and molecules. The simplest example of a standing wave is a onedimensional wave associated with a vibrating string that is held fixed at its two end points. Figure 3.7 shows the four lowest-energy standing waves (the fundamental wave and the lowest three harmonics) for a vibrating string at a particular amplitude. Although the string's motion lies mostly within a plane, the wave itself is considered to be one dimensional, since it lies along the length of the string. The motion of string segments in a direction perpendicular to the string length generates the waves and so the amplitude of the waves is visible as the maximum displacement of the curves seen in Figure 3.7. The key observation from the figure is that only those waves having an integer number, n, of half-wavelengths between the end points can form. A system with fixed end points such as this restricts the number and type of the possible waveforms. This is an example of quantization, in which only discrete values from a more general set of continuous values of some property are observed. Another important observation is that the harmonic waves (those waves displaying more than one-half wavelength) all have one or more points between the two end points that are not in motion. These special points are nodes. The energies of the standing waves with a given amplitude in a vibrating string increase with the number of half-wavelengths $n$. Since the number of nodes is $n-1$, the energy can also be said to depend on the number of nodes, generally increasing as the number of nodes increases.\n\n\nFIGURE 3.7 A vibrating string shows some one-dimensional standing waves. Since the two end points of the string are held fixed, only waves having an integer number of half-wavelengths can form. The points on the string between the end points that are not moving are called the nodes."}
{"id": 2419, "contents": "196. Dorothy Crowfoot Hodgkin - \nAn example of two-dimensional standing waves is shown in Figure 3.8, which shows the vibrational patterns on a flat surface. Although the vibrational amplitudes cannot be seen like they could in the vibrating string, the nodes have been made visible by sprinkling the drum surface with a powder that collects on the areas of the surface that have minimal displacement. For one-dimensional standing waves, the nodes were points on the line, but for two-dimensional standing waves, the nodes are lines on the surface (for three-dimensional standing waves, the nodes are two-dimensional surfaces within the three-dimensional volume).\n\n\nFIGURE 3.8 Two-dimensional standing waves can be visualized on a vibrating surface. The surface has been sprinkled with a powder that collects near the nodal lines. There are two types of nodes visible: radial nodes\n(circles) and angular nodes (radii)."}
{"id": 2420, "contents": "197. LINK TO LEARNING - \nYou can watch the formation of various radial nodes here (http://openstax.org/l/16radnodes) as singer Imogen Heap projects her voice across a kettle drum."}
{"id": 2421, "contents": "198. Blackbody Radiation and the Ultraviolet Catastrophe - \nThe last few decades of the nineteenth century witnessed intense research activity in commercializing newly discovered electric lighting. This required obtaining a better understanding of the distributions of light emitted from various sources being considered. Artificial lighting is usually designed to mimic natural sunlight within the limitations of the underlying technology. Such lighting consists of a range of broadly distributed frequencies that form a continuous spectrum. Figure 3.9 shows the wavelength distribution for sunlight. The most intense radiation is in the visible region, with the intensity dropping off rapidly for shorter wavelength ultraviolet (UV) light, and more slowly for longer wavelength infrared (IR) light.\n\n\nFIGURE 3.9 The spectral distribution (light intensity vs. wavelength) of sunlight reaches the Earth's atmosphere as UV light, visible light, and IR light. The unabsorbed sunlight at the top of the atmosphere has a distribution that approximately matches the theoretical distribution of a blackbody at $5250^{\\circ} \\mathrm{C}$, represented by the blue curve. (credit: modification of work by American Society for Testing and Materials (ASTM) Terrestrial Reference Spectra for Photovoltaic Performance Evaluation)"}
{"id": 2422, "contents": "198. Blackbody Radiation and the Ultraviolet Catastrophe - \nIn Figure 3.9, the solar distribution is compared to a representative distribution, called a blackbody spectrum, that corresponds to a temperature of $5250^{\\circ} \\mathrm{C}$. The blackbody spectrum matches the solar spectrum quite well. A blackbody is a convenient, ideal emitter that approximates the behavior of many materials when heated. It is \"ideal\" in the same sense that an ideal gas is a convenient, simple representation of real gases that works well, provided that the pressure is not too high nor the temperature too low. A good approximation of a blackbody that can be used to observe blackbody radiation is a metal oven that can be heated to very high temperatures. The oven has a small hole allowing for the light being emitted within the oven to be observed\nwith a spectrometer so that the wavelengths and their intensities can be measured. Figure 3.10 shows the resulting curves for some representative temperatures. Each distribution depends only on a single parameter: the temperature. The maxima in the blackbody curves, $\\lambda_{\\max }$, shift to shorter wavelengths as the temperature increases, reflecting the observation that metals being heated to high temperatures begin to glow a darker red that becomes brighter as the temperature increases, eventually becoming white hot at very high temperatures as the intensities of all of the visible wavelengths become appreciable. This common observation was at the heart of the first paradox that showed the fundamental limitations of classical physics that we will examine."}
{"id": 2423, "contents": "198. Blackbody Radiation and the Ultraviolet Catastrophe - \nPhysicists derived mathematical expressions for the blackbody curves using well-accepted concepts from the theories of classical mechanics and classical electromagnetism. The theoretical expressions as functions of temperature fit the observed experimental blackbody curves well at longer wavelengths, but showed significant discrepancies at shorter wavelengths. Not only did the theoretical curves not show a peak, they absurdly showed the intensity becoming infinitely large as the wavelength became smaller, which would imply that everyday objects at room temperature should be emitting large amounts of UV light. This became known as the \"ultraviolet catastrophe\" because no one could find any problems with the theoretical treatment that could lead to such unrealistic short-wavelength behavior. Finally, around 1900, Max Planck derived a theoretical expression for blackbody radiation that fit the experimental observations exactly (within experimental error). Planck developed his theoretical treatment by extending the earlier work that had been based on the premise that the atoms composing the oven vibrated at increasing frequencies (or decreasing wavelengths) as the temperature increased, with these vibrations being the source of the emitted electromagnetic radiation. But where the earlier treatments had allowed the vibrating atoms to have any energy values obtained from a continuous set of energies (perfectly reasonable, according to classical physics), Planck found that by restricting the vibrational energies to discrete values for each frequency, he could derive an expression for blackbody radiation that correctly had the intensity dropping rapidly for the short wavelengths in the UV region.\n\n$$\nE=n h \\nu, n=1,2,3, \\ldots\n$$\n\nThe quantity $h$ is a constant now known as Planck's constant, in his honor. Although Planck was pleased he had resolved the blackbody radiation paradox, he was disturbed that to do so, he needed to assume the vibrating atoms required quantized energies, which he was unable to explain. The value of Planck's constant is very small, $6.626 \\times 10^{-34}$ joule seconds ( J s), which helps explain why energy quantization had not been observed previously in macroscopic phenomena.\n\n\nFIGURE 3.10 Blackbody spectral distribution curves are shown for some representative temperatures."}
{"id": 2424, "contents": "199. The Photoelectric Effect - \nThe next paradox in the classical theory to be resolved concerned the photoelectric effect (Figure 3.11). It had been observed that electrons could be ejected from the clean surface of a metal when light having a frequency greater than some threshold frequency was shone on it. Surprisingly, the kinetic energy of the ejected electrons did not depend on the brightness of the light, but increased with increasing frequency of the light. Since the electrons in the metal had a certain amount of binding energy keeping them there, the incident light needed to have more energy to free the electrons. According to classical wave theory, a wave's energy depends on its intensity (which depends on its amplitude), not its frequency. One part of these observations was that the number of electrons ejected within in a given time period was seen to increase as the brightness increased. In 1905, Albert Einstein was able to resolve the paradox by incorporating Planck's quantization findings into the discredited particle view of light (Einstein actually won his Nobel prize for this work, and not for his theories of relativity for which he is most famous).\n\nEinstein argued that the quantized energies that Planck had postulated in his treatment of blackbody radiation could be applied to the light in the photoelectric effect so that the light striking the metal surface should not be viewed as a wave, but instead as a stream of particles (later called photons) whose energy depended on their frequency, according to Planck's formula, $E=h \\nu$ (or, in terms of wavelength using $c=\\nu \\lambda, E=\\frac{h c}{\\lambda}$ ). Electrons were ejected when hit by photons having sufficient energy (a frequency greater than the threshold). The greater the frequency, the greater the kinetic energy imparted to the escaping electrons by the collisions. Processes that increase the energy of an atom involve the absorption of light and are called endothermic. Processes that decrease the energy involve emission of light and are called exothermic. Einstein also argued that the light intensity did not depend on the amplitude of the incoming wave, but instead corresponded to the number of photons striking the surface within a given time period. This explains why the number of ejected electrons increased with increasing brightness, since the greater the number of incoming photons, the greater the likelihood that they would collide with some of the electrons."}
{"id": 2425, "contents": "199. The Photoelectric Effect - \nWith Einstein's findings, the nature of light took on a new air of mystery. Although many light phenomena could be explained either in terms of waves or particles, certain phenomena, such as the interference patterns obtained when light passed through a double slit, were completely contrary to a particle view of light, while other phenomena, such as the photoelectric effect, were completely contrary to a wave view of light. Somehow, at a deep fundamental level still not fully understood, light is both wavelike and particle-like. This is known as wave-particle duality.\n\n$$\nE=h \\nu\n$$\n\n\n\nFIGURE 3.11 Photons with low frequencies do not have enough energy to cause electrons to be ejected via the photoelectric effect. For any frequency of light above the threshold frequency, the kinetic energy of an ejected electron will increase linearly with the energy of the incoming photon."}
{"id": 2426, "contents": "201. Calculating the Energy of Radiation - \nWhen we see light from a neon sign, we are observing radiation from excited neon atoms. If this radiation has a wavelength of 640 nm , what is the energy of the photon being emitted?"}
{"id": 2427, "contents": "202. Solution - \nWe use the part of Planck's equation that includes the wavelength, $\\lambda$, and convert units of nanometers to meters so that the units of $\\lambda$ and $c$ are the same.\n\n$$\n\\begin{gathered}\nE=\\frac{h c}{\\lambda} \\\\\nE=\\frac{\\left(6.626 \\times 10^{-34} \\mathrm{~J}-\\right)\\left(2.998 \\times 10^{8} \\mathrm{~m}^{-1}\\right)}{(640 \\mathrm{~mm})\\left(\\frac{1 \\mathrm{~m}}{10^{9} \\mathrm{~mm}}\\right)} \\\\\nE=3.10 \\times 10^{-19} \\mathrm{~J}\n\\end{gathered}\n$$"}
{"id": 2428, "contents": "203. Check Your Learning - \nThe microwaves in an oven are of a specific frequency that will heat the water molecules contained in food. (This is why most plastics and glass do not become hot in a microwave oven-they do not contain water molecules.) This frequency is about $3 \\times 10^{9} \\mathrm{~Hz}$. What is the energy of one photon in these microwaves?"}
{"id": 2429, "contents": "204. Answer: - \n$2 \\times 10^{-24} \\mathrm{~J}$"}
{"id": 2430, "contents": "205. LINK TO LEARNING - \nUse this simulation program (http://openstax.org/l/16photelec) to experiment with the photoelectric effect to see how intensity, frequency, type of metal, and other factors influence the ejected photons."}
{"id": 2431, "contents": "207. Photoelectric Effect - \nIdentify which of the following statements are false and, where necessary, change the italicized word or phrase to make them true, consistent with Einstein's explanation of the photoelectric effect.\n(a) Increasing the brightness of incoming light increases the kinetic energy of the ejected electrons.\n(b) Increasing the wavelength of incoming light increases the kinetic energy of the ejected electrons.\n(c) Increasing the brightness of incoming light increases the number of ejected electrons.\n(d) Increasing the frequency of incoming light can increase the number of ejected electrons."}
{"id": 2432, "contents": "208. Solution - \n(a) False. Increasing the brightness of incoming light has no effect on the kinetic energy of the ejected electrons. Only energy, not the number or amplitude, of the photons influences the kinetic energy of the electrons.\n(b) False. Increasing the frequency of incoming light increases the kinetic energy of the ejected electrons. Frequency is proportional to energy and inversely proportional to wavelength. Frequencies above the threshold value transfer the excess energy into the kinetic energy of the electrons.\n(c) True. Because the number of collisions with photons increases with brighter light, the number of ejected electrons increases.\n(d) True with regard to the threshold energy binding the electrons to the metal. Below this threshold, electrons are not emitted and above it they are. Once over the threshold value, further increasing the frequency does not increase the number of ejected electrons"}
{"id": 2433, "contents": "209. Check Your Learning - \nCalculate the threshold energy in $\\mathrm{kJ} / \\mathrm{mol}$ of electrons in aluminum, given that the lowest frequency photon for which the photoelectric effect is observed is $9.87 \\times 10^{14} \\mathrm{~Hz}$."}
{"id": 2434, "contents": "210. Answer: - \n394"}
{"id": 2435, "contents": "211. Line Spectra - \nAnother paradox within the classical electromagnetic theory that scientists in the late nineteenth century struggled with concerned the light emitted from atoms and molecules. When solids, liquids, or condensed gases are heated sufficiently, they radiate some of the excess energy as light. Photons produced in this manner have a range of energies, and thereby produce a continuous spectrum in which an unbroken series of wavelengths is present. Most of the light generated from stars (including our sun) is produced in this fashion. You can see all the visible wavelengths of light present in sunlight by using a prism to separate them. As can be seen in Figure 3.9, sunlight also contains UV light (shorter wavelengths) and IR light (longer wavelengths) that can be detected using instruments but that are invisible to the human eye. Incandescent (glowing) solids such as tungsten filaments in incandescent lights also give off light that contains all wavelengths of visible light. These continuous spectra can often be approximated by blackbody radiation curves at some appropriate temperature, such as those shown in Figure 3.10.\n\nIn contrast to continuous spectra, light can also occur as discrete or line spectra having very narrow line widths interspersed throughout the spectral regions such as those shown in Figure 3.13. Exciting a gas at low partial pressure using an electrical current, or heating it, will produce line spectra. Fluorescent light bulbs and neon signs operate in this way (Figure 3.12). Each element displays its own characteristic set of lines, as do molecules, although their spectra are generally much more complicated.\n\n\nFIGURE 3.12 Neon signs operate by exciting a gas at low partial pressure using an electrical current. This sign shows the elaborate artistic effects that can be achieved. (credit: Dave Shaver)\n\nEach emission line consists of a single wavelength of light, which implies that the light emitted by a gas consists of a set of discrete energies. For example, when an electric discharge passes through a tube containing hydrogen gas at low pressure, the $\\mathrm{H}_{2}$ molecules are broken apart into separate H atoms and we see a blue-pink color. Passing the light through a prism produces a line spectrum, indicating that this light is composed of photons of four visible wavelengths, as shown in Figure 3.13."}
{"id": 2436, "contents": "211. Line Spectra - \nFIGURE 3.13 Compare the two types of emission spectra: continuous spectrum of white light (top) and the line spectra of the light from excited sodium, hydrogen, calcium, and mercury atoms.\n\nThe origin of discrete spectra in atoms and molecules was extremely puzzling to scientists in the late nineteenth century, since according to classical electromagnetic theory, only continuous spectra should be observed. Even more puzzling, in 1885, Johann Balmer was able to derive an empirical equation that related the four visible wavelengths of light emitted by hydrogen atoms to whole integers. That equation is the following one, in which $k$ is a constant:\n\n$$\n\\frac{1}{\\lambda}=k\\left(\\frac{1}{4}-\\frac{1}{n^{2}}\\right), n=3,4,5,6\n$$\n\nOther discrete lines for the hydrogen atom were found in the UV and IR regions. Johannes Rydberg generalized Balmer's work and developed an empirical formula that predicted all of hydrogen's emission lines, not just those restricted to the visible range, where, $n_{1}$ and $n_{2}$ are integers, $n_{1}<n_{2}$, and $R_{\\infty}$ is the Rydberg constant $\\left(1.097 \\times 10^{7} \\mathrm{~m}^{-1}\\right)$.\n\n$$\n\\frac{1}{\\lambda}=R_{\\infty}\\left(\\frac{1}{n_{1}^{2}}-\\frac{1}{n_{2}^{2}}\\right)\n$$\n\nEven in the late nineteenth century, spectroscopy was a very precise science, and so the wavelengths of hydrogen were measured to very high accuracy, which implied that the Rydberg constant could be determined very precisely as well. That such a simple formula as the Rydberg formula could account for such precise measurements seemed astounding at the time, but it was the eventual explanation for emission spectra by Neils Bohr in 1913 that ultimately convinced scientists to abandon classical physics and spurred the development of modern quantum mechanics."}
{"id": 2437, "contents": "212. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the Bohr model of the hydrogen atom\n- Use the Rydberg equation to calculate energies of light emitted or absorbed by hydrogen atoms\n\nFollowing the work of Ernest Rutherford and his colleagues in the early twentieth century, the picture of atoms consisting of tiny dense nuclei surrounded by lighter and even tinier electrons continually moving about the nucleus was well established. This picture was called the planetary model, since it pictured the atom as a miniature \"solar system\" with the electrons orbiting the nucleus like planets orbiting the sun. The simplest atom is hydrogen, consisting of a single proton as the nucleus about which a single electron moves. The electrostatic force attracting the electron to the proton depends only on the distance between the two particles. This classical mechanics description of the atom is incomplete, however, since an electron moving in an elliptical orbit would be accelerating (by changing direction) and, according to classical electromagnetism, it should continuously emit electromagnetic radiation. This loss in orbital energy should result in the electron's orbit getting continually smaller until it spirals into the nucleus, implying that atoms are inherently unstable.\n\nIn 1913, Niels Bohr attempted to resolve the atomic paradox by ignoring classical electromagnetism's prediction that the orbiting electron in hydrogen would continuously emit light. Instead, he incorporated into the classical mechanics description of the atom Planck's ideas of quantization and Einstein's finding that light consists of photons whose energy is proportional to their frequency. Bohr assumed that the electron orbiting the nucleus would not normally emit any radiation (the stationary state hypothesis), but it would emit or absorb a photon if it moved to a different orbit. The energy absorbed or emitted would reflect differences in the orbital energies according to this equation:\n\n$$\n|\\Delta E|=\\left|E_{\\mathrm{f}}-E_{\\mathrm{i}}\\right|=h \\nu=\\frac{h c}{\\lambda}\n$$\n\nIn this equation, $h$ is Planck's constant and $E_{i}$ and $E_{f}$ are the initial and final orbital energies, respectively. The absolute value of the energy difference is used, since frequencies and wavelengths are always positive. Instead of allowing for continuous values of energy, Bohr assumed the energies of these electron orbitals were quantized:"}
{"id": 2438, "contents": "212. LEARNING OBJECTIVES - \n$$\nE_{n}=-\\frac{k}{n^{2}}, n=1,2,3, \\ldots\n$$\n\nIn this expression, $k$ is a constant comprising fundamental constants such as the electron mass and charge and Planck's constant. Inserting the expression for the orbit energies into the equation for $\\Delta E$ gives\n\n$$\n\\Delta E=k\\left(\\frac{1}{n_{1}^{2}}-\\frac{1}{n_{2}^{2}}\\right)=\\frac{h c}{\\lambda}\n$$\n\nor\n\n$$\n\\frac{1}{\\lambda}=\\frac{k}{h c}\\left(\\frac{1}{n_{1}^{2}}-\\frac{1}{n_{2}^{2}}\\right)\n$$\n\nwhich is identical to the Rydberg equation in which $R_{\\infty}=\\frac{k}{h c}$. When Bohr calculated his theoretical value for the Rydberg constant, $R_{\\infty}$, and compared it with the experimentally accepted value, he got excellent agreement. Since the Rydberg constant was one of the most precisely measured constants at that time, this level of agreement was astonishing and meant that Bohr's model was taken seriously, despite the many assumptions that Bohr needed to derive it."}
{"id": 2439, "contents": "212. LEARNING OBJECTIVES - \nThe lowest few energy levels are shown in Figure 3.14. One of the fundamental laws of physics is that matter is most stable with the lowest possible energy. Thus, the electron in a hydrogen atom usually moves in the $n=1$ orbit, the orbit in which it has the lowest energy. When the electron is in this lowest energy orbit, the atom is\nsaid to be in its ground electronic state (or simply ground state). If the atom receives energy from an outside source, it is possible for the electron to move to an orbit with a higher $n$ value and the atom is now in an excited electronic state (or simply an excited state) with a higher energy. When an electron transitions from an excited state (higher energy orbit) to a less excited state, or ground state, the difference in energy is emitted as a photon. Similarly, if a photon is absorbed by an atom, the energy of the photon moves an electron from a lower energy orbit up to a more excited one. We can relate the energy of electrons in atoms to what we learned previously about energy. The law of conservation of energy says that we can neither create nor destroy energy. Thus, if a certain amount of external energy is required to excite an electron from one energy level to another, that same amount of energy will be liberated when the electron returns to its initial state (Figure 3.15).\n\nSince Bohr's model involved only a single electron, it could also be applied to the single electron ions $\\mathrm{He}^{+}, \\mathrm{Li}^{2+}$, $\\mathrm{Be}^{3+}$, and so forth, which differ from hydrogen only in their nuclear charges, and so one-electron atoms and ions are collectively referred to as hydrogen-like atoms. The energy expression for hydrogen-like atoms is a generalization of the hydrogen atom energy, in which $Z$ is the nuclear charge ( +1 for hydrogen, +2 for $\\mathrm{He},+3$ for Li , and so on) and $k$ has a value of $2.179 \\times 10^{-18} \\mathrm{~J}$.\n\n$$\nE_{n}=-\\frac{k Z^{2}}{n^{2}}\n$$"}
{"id": 2440, "contents": "212. LEARNING OBJECTIVES - \n$$\nE_{n}=-\\frac{k Z^{2}}{n^{2}}\n$$\n\nThe sizes of the circular orbits for hydrogen-like atoms are given in terms of their radii by the following expression, in which $\\mathrm{a}_{0}$ is a constant called the Bohr radius, with a value of $5.292 \\times 10^{-11} \\mathrm{~m}$ :\n\n$$\nr=\\frac{n^{2}}{Z} a_{0}\n$$\n\nThe equation also shows us that as the electron's energy increases (as $n$ increases), the electron is found at greater distances from the nucleus. This is implied by the inverse dependence of electrostatic attraction on distance, since, as the electron moves away from the nucleus, the electrostatic attraction between it and the nucleus decreases and it is held less tightly in the atom. Note that as $n$ gets larger and the orbits get larger, their energies get closer to zero, and so the limits $n \\longrightarrow \\infty$ and $r \\longrightarrow \\infty$ imply that $E=0$ corresponds to the ionization limit where the electron is completely removed from the nucleus. Thus, for hydrogen in the ground state $n=1$, the ionization energy would be:\n\n$$\n\\Delta E=E_{n \\rightarrow \\infty}-E_{1}=0+k=k\n$$\n\nWith three extremely puzzling paradoxes now solved (blackbody radiation, the photoelectric effect, and the hydrogen atom), and all involving Planck's constant in a fundamental manner, it became clear to most physicists at that time that the classical theories that worked so well in the macroscopic world were fundamentally flawed and could not be extended down into the microscopic domain of atoms and molecules. Unfortunately, despite Bohr's remarkable achievement in deriving a theoretical expression for the Rydberg constant, he was unable to extend his theory to the next simplest atom, He, which only has two electrons. Bohr's model was severely flawed, since it was still based on the classical mechanics notion of precise orbits, a concept that was later found to be untenable in the microscopic domain, when a proper model of quantum mechanics was developed to supersede classical mechanics.\n\n\nFIGURE 3.14 Quantum numbers and energy levels in a hydrogen atom. The more negative the calculated value, the lower the energy."}
{"id": 2441, "contents": "214. Calculating the Energy of an Electron in a Bohr Orbit - \nEarly researchers were very excited when they were able to predict the energy of an electron at a particular distance from the nucleus in a hydrogen atom. If a spark promotes the electron in a hydrogen atom into an orbit with $n=3$, what is the calculated energy, in joules, of the electron?"}
{"id": 2442, "contents": "215. Solution - \nThe energy of the electron is given by this equation:\n\n$$\nE=\\frac{-k Z^{2}}{n^{2}}\n$$\n\nThe atomic number, $Z$, of hydrogen is $1 ; k=2.179 \\times 10^{-18} \\mathrm{~J}$; and the electron is characterized by an $n$ value of 3. Thus,\n\n$$\nE=\\frac{-\\left(2.179 \\times 10^{-18} \\mathrm{~J}\\right) \\times(1)^{2}}{(3)^{2}}=-2.421 \\times 10^{-19} \\mathrm{~J}\n$$"}
{"id": 2443, "contents": "216. Check Your Learning - \nThe electron in Figure 3.15 is promoted even further to an orbit with $n=6$. What is its new energy?"}
{"id": 2444, "contents": "217. Answer: - \n$$\n-6.053 \\times 10^{-20} \\mathrm{~J}\n$$\n\n\n\nFIGURE 3.15 The horizontal lines show the relative energy of orbits in the Bohr model of the hydrogen atom, and the vertical arrows depict the energy of photons absorbed (left) or emitted (right) as electrons move between these orbits."}
{"id": 2445, "contents": "219. Calculating the Energy and Wavelength of Electron Transitions in a One-electron (Bohr) System - \nWhat is the energy (in joules) and the wavelength (in meters) of the line in the spectrum of hydrogen that represents the movement of an electron from Bohr orbit with $n=4$ to the orbit with $n=6$ ? In what part of the electromagnetic spectrum do we find this radiation?"}
{"id": 2446, "contents": "220. Solution - \nIn this case, the electron starts out with $n=4$, so $n_{1}=4$. It comes to rest in the $n=6$ orbit, so $n_{2}=6$. The difference in energy between the two states is given by this expression:\n\n$$\n\\begin{gathered}\n\\Delta E=E_{1}-E_{2}=2.179 \\times 10^{-18}\\left(\\frac{1}{n_{1}^{2}}-\\frac{1}{n_{2}^{2}}\\right) \\\\\n\\Delta E=2.179 \\times 10^{-18}\\left(\\frac{1}{4^{2}}-\\frac{1}{6^{2}}\\right) \\mathrm{J} \\\\\n\\Delta E=2.179 \\times 10^{-18}\\left(\\frac{1}{16}-\\frac{1}{36}\\right) \\mathrm{J} \\\\\n\\Delta E=7.566 \\times 10^{-20} \\mathrm{~J}\n\\end{gathered}\n$$\n\nThis energy difference is positive, indicating a photon enters the system (is absorbed) to excite the electron\nfrom the $n=4$ orbit up to the $n=6$ orbit. The wavelength of a photon with this energy is found by the expression $E=\\frac{h c}{\\lambda}$. Rearrangement gives:\n\n$$\n\\begin{aligned}\n& \\qquad \\lambda=\\frac{h c}{E} \\\\\n& =\\left(6.626 \\times 10^{-34} \\mathrm{Js}\\right) \\times \\frac{2.998 \\times 10^{8} \\mathrm{~m}^{-1}}{7.566 \\times 10^{-20} \\mathrm{f}} \\\\\n& =2.626 \\times 10^{-6} \\mathrm{~m}\n\\end{aligned}\n$$\n\nFrom the illustration of the electromagnetic spectrum in Electromagnetic Energy, we can see that this wavelength is found in the infrared portion of the electromagnetic spectrum."}
{"id": 2447, "contents": "221. Check Your Learning - \nWhat is the energy in joules and the wavelength in meters of the photon produced when an electron falls from the $n=5$ to the $n=3$ level in a $\\mathrm{He}^{+}$ion $\\left(Z=2\\right.$ for $\\left.\\mathrm{He}^{+}\\right)$?"}
{"id": 2448, "contents": "222. Answer: - \n$6.198 \\times 10^{-19} \\mathrm{~J} ; 3.205 \\times 10^{-7} \\mathrm{~m}$\n\nBohr's model of the hydrogen atom provides insight into the behavior of matter at the microscopic level, but it does not account for electron-electron interactions in atoms with more than one electron. It does introduce several important features of all models used to describe the distribution of electrons in an atom. These features include the following:\n\n- The energies of electrons (energy levels) in an atom are quantized, described by quantum numbers: integer numbers having only specific allowed value and used to characterize the arrangement of electrons in an atom.\n- An electron's energy increases with increasing distance from the nucleus.\n- The discrete energies (lines) in the spectra of the elements result from quantized electronic energies.\n\nOf these features, the most important is the postulate of quantized energy levels for an electron in an atom. As a consequence, the model laid the foundation for the quantum mechanical model of the atom. Bohr won a Nobel Prize in Physics for his contributions to our understanding of the structure of atoms and how that is related to line spectra emissions."}
{"id": 2449, "contents": "223. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Extend the concept of wave-particle duality that was observed in electromagnetic radiation to matter as well\n- Understand the general idea of the quantum mechanical description of electrons in an atom, and that it uses the notion of three-dimensional wave functions, or orbitals, that define the distribution of probability to find an electron in a particular part of space\n- List and describe traits of the four quantum numbers that form the basis for completely specifying the state of an electron in an atom\n\nBohr's model explained the experimental data for the hydrogen atom and was widely accepted, but it also raised many questions. Why did electrons orbit at only fixed distances defined by a single quantum number $n$ $=1,2,3$, and so on, but never in between? Why did the model work so well describing hydrogen and oneelectron ions, but could not correctly predict the emission spectrum for helium or any larger atoms? To answer these questions, scientists needed to completely revise the way they thought about matter."}
{"id": 2450, "contents": "224. Behavior in the Microscopic World - \nWe know how matter behaves in the macroscopic world-objects that are large enough to be seen by the naked eye follow the rules of classical physics. A billiard ball moving on a table will behave like a particle: It will continue in a straight line unless it collides with another ball or the table cushion, or is acted on by some other force (such as friction). The ball has a well-defined position and velocity (or a well-defined momentum, $p=m v$, defined by mass $m$ and velocity $v$ ) at any given moment. In other words, the ball is moving in a classical trajectory. This is the typical behavior of a classical object.\n\nWhen waves interact with each other, they show interference patterns that are not displayed by macroscopic particles such as the billiard ball. For example, interacting waves on the surface of water can produce interference patterns similar to those shown on Figure 3.16. This is a case of wave behavior on the macroscopic scale, and it is clear that particles and waves are very different phenomena in the macroscopic realm.\n\n\nFIGURE 3.16 An interference pattern on the water surface is formed by interacting waves. The waves are caused by reflection of water from the rocks. (credit: modification of work by Sukanto Debnath)\n\nAs technological improvements allowed scientists to probe the microscopic world in greater detail, it became increasingly clear by the 1920s that very small pieces of matter follow a different set of rules from those we observe for large objects. The unquestionable separation of waves and particles was no longer the case for the microscopic world.\n\nOne of the first people to pay attention to the special behavior of the microscopic world was Louis de Broglie. He asked the question: If electromagnetic radiation can have particle-like character, can electrons and other submicroscopic particles exhibit wavelike character? In his 1925 doctoral dissertation, de Broglie extended the wave-particle duality of light that Einstein used to resolve the photoelectric-effect paradox to material particles. He predicted that a particle with mass $m$ and velocity $v$ (that is, with linear momentum $p$ ) should also exhibit the behavior of a wave with a wavelength value $\\lambda$, given by this expression in which $h$ is the familiar Planck's constant:\n\n$$\n\\lambda=\\frac{h}{m v}=\\frac{h}{p}\n$$"}
{"id": 2451, "contents": "224. Behavior in the Microscopic World - \n$$\n\\lambda=\\frac{h}{m v}=\\frac{h}{p}\n$$\n\nThis is called the de Broglie wavelength. Unlike the other values of $\\lambda$ discussed in this chapter, the de Broglie wavelength is a characteristic of particles and other bodies, not electromagnetic radiation (note that this equation involves velocity [ $\\mathrm{v}, \\mathrm{m} / \\mathrm{s}$ ], not frequency [ $\\nu, \\mathrm{Hz}$ ]. Although these two symbols appear nearly identical, they mean very different things). Where Bohr had postulated the electron as being a particle orbiting the nucleus in quantized orbits, de Broglie argued that Bohr's assumption of quantization can be explained if the electron is considered not as a particle, but rather as a circular standing wave such that only an integer number of wavelengths could fit exactly within the orbit (Figure 3.17).\n\n\nFIGURE 3.17 If an electron is viewed as a wave circling around the nucleus, an integer number of wavelengths must fit into the orbit for this standing wave behavior to be possible.\n\nFor a circular orbit of radius $r$, the circumference is $2 \\pi r$, and so de Broglie's condition is:\n\n$$\n2 \\pi r=n \\lambda, n=1,2,3, \\ldots\n$$"}
{"id": 2452, "contents": "224. Behavior in the Microscopic World - \n$$\n2 \\pi r=n \\lambda, n=1,2,3, \\ldots\n$$\n\nShortly after de Broglie proposed the wave nature of matter, two scientists at Bell Laboratories, C. J. Davisson and L. H. Germer, demonstrated experimentally that electrons can exhibit wavelike behavior by showing an interference pattern for electrons travelling through a regular atomic pattern in a crystal. The regularly spaced atomic layers served as slits, as used in other interference experiments. Since the spacing between the layers serving as slits needs to be similar in size to the wavelength of the tested wave for an interference pattern to form, Davisson and Germer used a crystalline nickel target for their \"slits,\" since the spacing of the atoms within the lattice was approximately the same as the de Broglie wavelengths of the electrons that they used. Figure 3.18 shows an interference pattern. It is strikingly similar to the interference patterns for light shown in Electromagnetic Energy for light passing through two closely spaced, narrow slits. The wave-particle duality of matter can be seen in Figure 3.18 by observing what happens if electron collisions are recorded over a long period of time. Initially, when only a few electrons have been recorded, they show clear particle-like behavior, having arrived in small localized packets that appear to be random. As more and more electrons arrived and were recorded, a clear interference pattern that is the hallmark of wavelike behavior emerged. Thus, it appears that while electrons are small localized particles, their motion does not follow the equations of motion implied by classical mechanics, but instead it is governed by some type of a wave equation. Thus the wave-particle duality first observed with photons is actually a fundamental behavior intrinsic to all quantum particles.\n\n\nFIGURE 3.18 (a) The interference pattern for electrons passing through very closely spaced slits demonstrates that quantum particles such as electrons can exhibit wavelike behavior. (b) The experimental results illustrated here demonstrate the wave-particle duality in electrons."}
{"id": 2453, "contents": "225. LINK TO LEARNING - \nView the Dr. Quantum - Double Slit Experiment cartoon (http://openstax.org/l/16duality) for an easy-tounderstand description of wave-particle duality and the associated experiments."}
{"id": 2454, "contents": "227. Calculating the Wavelength of a Particle - \nIf an electron travels at a velocity of $1.000 \\times 10^{7} \\mathrm{~m} \\mathrm{~s}^{-1}$ and has a mass of $9.109 \\times 10^{-28} \\mathrm{~g}$, what is its wavelength?"}
{"id": 2455, "contents": "228. Solution - \nWe can use de Broglie's equation to solve this problem, but we first must do a unit conversion of Planck's constant. You learned earlier that $1 \\mathrm{~J}=1 \\mathrm{~kg} \\mathrm{~m}^{2} / \\mathrm{s}^{2}$. Thus, we can write $h=6.626 \\times 10^{-34} \\mathrm{~J} \\mathrm{~s}$ as $6.626 \\times 10^{-34} \\mathrm{~kg}$ $\\mathrm{m}^{2} / \\mathrm{s}$.\n\n$$\n\\lambda=\\frac{h}{m v}\n$$\n\n$$\n\\begin{aligned}\n& =\\frac{6.626 \\times 10^{-34} \\mathrm{~kg} \\mathrm{~m}^{2} / \\mathrm{s}}{\\left(9.109 \\times 10^{-31} \\mathrm{~kg}\\right)\\left(1.000 \\times 10^{7} \\mathrm{~m} / \\mathrm{s}\\right)} \\\\\n& =7.274 \\times 10^{-11} \\mathrm{~m}\n\\end{aligned}\n$$\n\nThis is a small value, but it is significantly larger than the size of an electron in the classical (particle) view. This size is the same order of magnitude as the size of an atom. This means that electron wavelike behavior is going to be noticeable in an atom."}
{"id": 2456, "contents": "229. Check Your Learning - \nCalculate the wavelength of a softball with a mass of 100 g traveling at a velocity of $35 \\mathrm{~m} \\mathrm{~s}^{-1}$, assuming that it can be modeled as a single particle."}
{"id": 2457, "contents": "230. Answer: - \n$1.9 \\times 10^{-34} \\mathrm{~m}$.\nWe never think of a thrown softball having a wavelength, since this wavelength is so small it is impossible for our senses or any known instrument to detect (strictly speaking, the wavelength of a real baseball would correspond to the wavelengths of its constituent atoms and molecules, which, while much larger than this value, would still be microscopically tiny). The de Broglie wavelength is only appreciable for matter that has a very small mass and/or a very high velocity.\n\nWerner Heisenberg considered the limits of how accurately we can measure properties of an electron or other microscopic particles. He determined that there is a fundamental limit to how accurately one can measure both a particle's position and its momentum simultaneously. The more accurately we measure the momentum of a particle, the less accurately we can determine its position at that time, and vice versa. This is summed up in what we now call the Heisenberg uncertainty principle: It is fundamentally impossible to determine simultaneously and exactly both the momentum and the position of a particle. For a particle of mass $m$ moving with velocity $v_{x}$ in the $x$ direction (or equivalently with momentum $p_{x}$ ), the product of the uncertainty in the position, $\\Delta x$, and the uncertainty in the momentum, $\\Delta p_{x}$, must be greater than or equal to $\\frac{\\hbar}{2}$ (where $\\hbar=\\frac{h}{2 \\pi}$, the value of Planck's constant divided by $2 \\pi$ ).\n\n$$\n\\Delta x \\times \\Delta p_{x}=(\\Delta x)(m \\Delta v) \\geq \\frac{\\hbar}{2}\n$$\n\nThis equation allows us to calculate the limit to how precisely we can know both the simultaneous position of an object and its momentum. For example, if we improve our measurement of an electron's position so that the uncertainty in the position ( $\\Delta x$ ) has a value of, say, $1 \\mathrm{pm}\\left(10^{-12} \\mathrm{~m}\\right.$, about $1 \\%$ of the diameter of a hydrogen atom), then our determination of its momentum must have an uncertainty with a value of at least"}
{"id": 2458, "contents": "230. Answer: - \n$$\n\\left[\\Delta p=m \\Delta v=\\frac{\\hbar}{(2 \\Delta x)}\\right]=\\frac{\\left(1.055 \\times 10^{-34} \\mathrm{~kg} \\mathrm{~m}^{2} / \\mathrm{s}\\right)}{\\left(2 \\times 1 \\times 10^{-12} \\mathrm{~m}\\right)}=5 \\times 10^{-23} \\mathrm{~kg} \\mathrm{~m} / \\mathrm{s} .\n$$\n\nThe value of $\\hbar$ is not large, so the uncertainty in the position or momentum of a macroscopic object like a baseball is too insignificant to observe. However, the mass of a microscopic object such as an electron is small enough that the uncertainty can be large and significant.\n\nIt should be noted that Heisenberg's uncertainty principle is not just limited to uncertainties in position and momentum, but it also links other dynamical variables. For example, when an atom absorbs a photon and makes a transition from one energy state to another, the uncertainty in the energy and the uncertainty in the time required for the transition are similarly related, as $\\Delta E \\Delta t \\geq \\frac{\\hbar}{2}$.\nHeisenberg's principle imposes ultimate limits on what is knowable in science. The uncertainty principle can\nbe shown to be a consequence of wave-particle duality, which lies at the heart of what distinguishes modern quantum theory from classical mechanics."}
{"id": 2459, "contents": "231. LINK TO LEARNING - \nRead this article (http://openstax.org/l/16uncertainty) that describes a recent macroscopic demonstration of the uncertainty principle applied to microscopic objects."}
{"id": 2460, "contents": "232. The Quantum-Mechanical Model of an Atom - \nShortly after de Broglie published his ideas that the electron in a hydrogen atom could be better thought of as being a circular standing wave instead of a particle moving in quantized circular orbits, Erwin Schr\u00f6dinger extended de Broglie's work by deriving what is today known as the Schr\u00f6dinger equation. When Schr\u00f6dinger applied his equation to hydrogen-like atoms, he was able to reproduce Bohr's expression for the energy and, thus, the Rydberg formula governing hydrogen spectra. Schr\u00f6dinger described electrons as three-dimensional stationary waves, or wavefunctions, represented by the Greek letter psi, $\\psi$. A few years later, Max Born proposed an interpretation of the wavefunction $\\psi$ that is still accepted today: Electrons are still particles, and so the waves represented by $\\psi$ are not physical waves but, instead, are complex probability amplitudes. The square of the magnitude of a wavefunction $|\\psi|^{2}$ describes the probability of the quantum particle being present near a certain location in space. This means that wavefunctions can be used to determine the distribution of the electron's density with respect to the nucleus in an atom. In the most general form, the Schr\u00f6dinger equation can be written as:\n\n$$\n\\widehat{H} \\psi=E \\psi\n$$\n\n$\\widehat{H}$ is the Hamiltonian operator, a set of mathematical operations representing the total energy of the quantum particle (such as an electron in an atom), $\\psi$ is the wavefunction of this particle that can be used to find the special distribution of the probability of finding the particle, and $E$ is the actual value of the total energy of the particle.\n\nSchr\u00f6dinger's work, as well as that of Heisenberg and many other scientists following in their footsteps, is generally referred to as quantum mechanics."}
{"id": 2461, "contents": "233. LINK TO LEARNING - \nYou may also have heard of Schr\u00f6dinger because of his famous thought experiment. This story (http://openstax.org/l/16superpos) explains the concepts of superposition and entanglement as related to a cat in a box with poison."}
{"id": 2462, "contents": "234. Understanding Quantum Theory of Electrons in Atoms - \nThe goal of this section is to understand the electron orbitals (location of electrons in atoms), their different energies, and other properties. The use of quantum theory provides the best understanding to these topics. This knowledge is a precursor to chemical bonding.\n\nAs was described previously, electrons in atoms can exist only on discrete energy levels but not between them. It is said that the energy of an electron in an atom is quantized, that is, it can be equal only to certain specific values and can jump from one energy level to another but not transition smoothly or stay between these levels.\n\nThe energy levels are labeled with an $n$ value, where $n=1,2,3, \\ldots$. Generally speaking, the energy of an electron in an atom is greater for greater values of $n$. This number, $n$, is referred to as the principal quantum number. The principal quantum number defines the location of the energy level. It is essentially the same concept as the $n$ in the Bohr atom description. Another name for the principal quantum number is the shell number. The shells of an atom can be thought of concentric circles radiating out from the nucleus. The electrons that belong to a specific shell are most likely to be found within the corresponding circular area. The further we proceed from the nucleus, the higher the shell number, and so the higher the energy level (Figure 3.19). The positively charged protons in the nucleus stabilize the electronic orbitals by electrostatic attraction\nbetween the positive charges of the protons and the negative charges of the electrons. So the further away the electron is from the nucleus, the greater the energy it has.\n\n\nFIGURE 3.19 Different shells are numbered by principal quantum numbers.\nThis quantum mechanical model for where electrons reside in an atom can be used to look at electronic transitions, the events when an electron moves from one energy level to another. If the transition is to a higher energy level, energy is absorbed, and the energy change has a positive value. To obtain the amount of energy necessary for the transition to a higher energy level, a photon is absorbed by the atom. A transition to a lower energy level involves a release of energy, and the energy change is negative. This process is accompanied by emission of a photon by the atom. The following equation summarizes these relationships and is based on the hydrogen atom:"}
{"id": 2463, "contents": "234. Understanding Quantum Theory of Electrons in Atoms - \n$$\n\\begin{gathered}\n\\Delta E=E_{\\text {final }}-E_{\\text {initial }} \\\\\n=-2.18 \\times 10^{-18}\\left(\\frac{1}{n_{\\mathrm{f}}^{2}}-\\frac{1}{n_{\\mathrm{i}}^{2}}\\right) \\mathrm{J}\n\\end{gathered}\n$$\n\nThe values $n_{\\mathrm{f}}$ and $n_{\\mathrm{i}}$ are the final and initial energy states of the electron. Example 3.5 in the previous section of the chapter demonstrates calculations of such energy changes.\n\nThe principal quantum number is one of three quantum numbers used to characterize an orbital. An atomic orbital is a general region in an atom within which an electron is most probable to reside. The quantum mechanical model specifies the probability of finding an electron in the three-dimensional space around the nucleus and is based on solutions of the Schr\u00f6dinger equation. In addition, the principal quantum number defines the energy of an electron in a hydrogen or hydrogen-like atom or an ion (an atom or an ion with only one electron) and the general region in which discrete energy levels of electrons in a multi-electron atoms and ions are located.\n\nAnother quantum number is $l$, the secondary (angular momentum) quantum number. It is an integer that may take the values, $l=0,1,2, \\ldots, n-1$. This means that an orbital with $n=1$ can have only one value of $l, l=0$, whereas $n=2$ permits $l=0$ and $l=1$, and so on. Whereas the principal quantum number, $n$, defines the general size and energy of the orbital, the secondary quantum number 1 specifies the shape of the orbital. Orbitals with the same value of $l$ define a subshell."}
{"id": 2464, "contents": "234. Understanding Quantum Theory of Electrons in Atoms - \nOrbitals with $l=0$ are called $\\boldsymbol{s}$ orbitals and they make up the $s$ subshells. The value $l=1$ corresponds to the $p$ orbitals. For a given $n, \\boldsymbol{p}$ orbitals constitute a $p$ subshell (e.g., $3 p$ if $n=3$ ). The orbitals with $l=2$ are called the $\\boldsymbol{d}$ orbitals, followed by the $f$-, $g$-, and $h$-orbitals for $l=3,4$, and 5 .\n\nThere are certain distances from the nucleus at which the probability density of finding an electron located at a particular orbital is zero. In other words, the value of the wavefunction $\\psi$ is zero at this distance for this orbital. Such a value of radius $r$ is called a radial node. The number of radial nodes in an orbital is $n-1-1$.\n\n\nFIGURE 3.20 The graphs show the probability ( $y$ axis) of finding an electron for the $1 s, 2 s, 3 s$ orbitals as a function of distance from the nucleus.\n\nConsider the examples in Figure 3.20. The orbitals depicted are of the $s$ type, thus $l=0$ for all of them. It can be seen from the graphs of the probability densities that there are $1-0-1=0$ places where the density is zero (nodes) for $1 s(n=1), 2-0-1=1$ node for $2 s$, and $3-0-1=2$ nodes for the $3 s$ orbitals.\n\nThe $s$ subshell electron density distribution is spherical and the $p$ subshell has a dumbbell shape. The $d$ and $\\boldsymbol{f}$ orbitals are more complex. These shapes represent the three-dimensional regions within which the electron is likely to be found."}
{"id": 2465, "contents": "234. Understanding Quantum Theory of Electrons in Atoms - \nFIGURE 3.21 Shapes of $s, p, d$, and $f$ orbitals.\nThe magnetic quantum number, $m_{l}$, specifies the relative spatial orientation of a particular orbital. Generally speaking, $m_{l}$ can be equal to $-l,-(l-1), \\ldots, 0, \\ldots,(l-1), l$. The total number of possible orbitals with the same value of $l$ (that is, in the same subshell) is $2 l+1$. Thus, there is one $s$-orbital in an $s$ subshell $(l=0)$, there are three $p$-orbitals in a $p$ subshell $(l=1)$, five $d$-orbitals in a $d$ subshell $(l=2)$, seven $f$-orbitals in an $f$ subshell $(l=$ 3 ), and so forth. The principal quantum number defines the general value of the electronic energy. The angular momentum quantum number determines the shape of the orbital. And the magnetic quantum number\nspecifies orientation of the orbital in space, as can be seen in Figure 3.21."}
{"id": 2466, "contents": "234. Understanding Quantum Theory of Electrons in Atoms - \nFIGURE 3.22 The chart shows the energies of electron orbitals in a multi-electron atom.\nFigure 3.22 illustrates the energy levels for various orbitals. The number before the orbital name (such as $2 s$, $3 p$, and so forth) stands for the principal quantum number, $n$. The letter in the orbital name defines the subshell with a specific angular momentum quantum number $l=0$ for $s$ orbitals, 1 for $p$ orbitals, 2 for $d$ orbitals. Finally, there are more than one possible orbitals for $l \\geq 1$, each corresponding to a specific value of $m_{l}$. In the case of a hydrogen atom or a one-electron ion (such as $\\mathrm{He}^{+}, \\mathrm{Li}^{2+}$, and so on), energies of all the orbitals with the same $n$ are the same. This is called a degeneracy, and the energy levels for the same principal quantum number, $n$, are called degenerate orbitals. However, in atoms with more than one electron, this degeneracy is eliminated by the electron-electron interactions, and orbitals that belong to different subshells have different energies, as shown on Figure 3.22. Orbitals within the same subshell are still degenerate and have the same energy.\nWhile the three quantum numbers discussed in the previous paragraphs work well for describing electron orbitals, some experiments showed that they were not sufficient to explain all observed results. It was demonstrated in the 1920s that when hydrogen-line spectra are examined at extremely high resolution, some lines are actually not single peaks but, rather, pairs of closely spaced lines. This is the so-called fine structure of the spectrum, and it implies that there are additional small differences in energies of electrons even when they are located in the same orbital. These observations led Samuel Goudsmit and George Uhlenbeck to propose that electrons have a fourth quantum number. They called this the spin quantum number, or $\\boldsymbol{m}_{\\boldsymbol{s}}$."}
{"id": 2467, "contents": "234. Understanding Quantum Theory of Electrons in Atoms - \nThe other three quantum numbers, $n, l$, and $m_{l}$, are properties of specific atomic orbitals that also define in what part of the space an electron is most likely to be located. Orbitals are a result of solving the Schr\u00f6dinger equation for electrons in atoms. The electron spin is a different kind of property. It is a completely quantum phenomenon with no analogues in the classical realm. In addition, it cannot be derived from solving the Schr\u00f6dinger equation and is not related to the normal spatial coordinates (such as the Cartesian $x, y$, and $z$ ). Electron spin describes an intrinsic electron \"rotation\" or \"spinning.\" Each electron acts as a tiny magnet or a tiny rotating object with an angular momentum, or as a loop with an electric current, even though this rotation or current cannot be observed in terms of spatial coordinates.\n\nThe magnitude of the overall electron spin can only have one value, and an electron can only \"spin\" in one of two quantized states. One is termed the $\\alpha$ state, with the $z$ component of the spin being in the positive direction of the $z$ axis. This corresponds to the spin quantum number $m_{s}=\\frac{1}{2}$. The other is called the $\\beta$ state, with the $z$ component of the spin being negative and $m_{s}=-\\frac{1}{2}$. Any electron, regardless of the atomic orbital it is located in, can only have one of those two values of the spin quantum number. The energies of electrons having $m_{s}=-\\frac{1}{2}$ and $m_{s}=\\frac{1}{2}$ are different if an external magnetic field is applied."}
{"id": 2468, "contents": "234. Understanding Quantum Theory of Electrons in Atoms - \nFIGURE 3.23 Electrons with spin values $\\pm \\frac{1}{2}$ in an external magnetic field.\nFigure 3.23 illustrates this phenomenon. An electron acts like a tiny magnet. Its moment is directed up (in the positive direction of the $z$ axis) for the $\\frac{1}{2}$ spin quantum number and down (in the negative $z$ direction) for the spin quantum number of $-\\frac{1}{2}$. A magnet has a lower energy if its magnetic moment is aligned with the external magnetic field (the left electron on Figure 3.23) and a higher energy for the magnetic moment being opposite to the applied field. This is why an electron with $m_{s}=\\frac{1}{2}$ has a slightly lower energy in an external field in the positive $z$ direction, and an electron with $m_{s}=-\\frac{1}{2}$ has a slightly higher energy in the same field. This is true even for an electron occupying the same orbital in an atom. A spectral line corresponding to a transition for electrons from the same orbital but with different spin quantum numbers has two possible values of energy; thus, the line in the spectrum will show a fine structure splitting."}
{"id": 2469, "contents": "235. The Pauli Exclusion Principle - \nAn electron in an atom is completely described by four quantum numbers: $n, l, m_{l}$, and $m_{s}$. The first three quantum numbers define the orbital and the fourth quantum number describes the intrinsic electron property called spin. An Austrian physicist Wolfgang Pauli formulated a general principle that gives the last piece of information that we need to understand the general behavior of electrons in atoms. The Pauli exclusion principle can be formulated as follows: No two electrons in the same atom can have exactly the same set of all the four quantum numbers. What this means is that two electrons can share the same orbital (the same set of the quantum numbers $n, l$, and $m_{l}$ ) only if their spin quantum numbers $m_{s}$ have different values. Since the spin quantum number can only have two values $\\left( \\pm \\frac{1}{2}\\right)$, no more than two electrons can occupy the same orbital (and if two electrons are located in the same orbital, they must have opposite spins). Therefore, any atomic orbital can be populated by only zero, one, or two electrons.\n\nThe properties and meaning of the quantum numbers of electrons in atoms are briefly summarized in Table 3.1.\n\nQuantum Numbers, Their Properties, and Significance\n\n|  | Symbol | Allowed <br> values | Physical meaning |\n| :--- | :--- | :--- | :--- |\n| principal quantum number | $n$ | $1,2,3,4$, <br> $\\ldots$. | shell, the general region for the value of energy <br> for an electron on the orbital |\n| angular momentum or <br> azimuthal quantum number | 1 | $0 \\leq 1 \\leq n$ <br> -1 | subshell, the shape of the orbital |\n| magnetic quantum number | $m_{l}$ | $-1 \\leq m_{l} \\leq$ <br> $l$ | orientation of the orbital |\n\nQuantum Numbers, Their Properties, and Significance\n\n|  | Symbol | Allowed <br> values | Physical meaning |\n| :--- | :--- | :--- | :--- |\n| spin quantum number | $m_{S}$ | $\\frac{1}{2},-\\frac{1}{2}$ | direction of the intrinsic quantum \"spinning\" of <br> the electron |"}
{"id": 2470, "contents": "235. The Pauli Exclusion Principle - \nTABLE 3.1"}
{"id": 2471, "contents": "237. Working with Shells and Subshells - \nIndicate the number of subshells, the number of orbitals in each subshell, and the values of 1 and $m_{l}$ for the orbitals in the $n=4$ shell of an atom."}
{"id": 2472, "contents": "238. Solution - \nFor $n=4, l$ can have values of $0,1,2$, and 3 . Thus, $s, p, d$, and $f$ subshells are found in the $n=4$ shell of an atom. For $l=0$ (the $s$ subshell), $m_{l}$ can only be 0 . Thus, there is only one $4 s$ orbital. For $l=1$ ( $p$-type orbitals), $m$ can have values of $-1,0,+1$, so we find three $4 p$ orbitals. For $l=2$ ( $d$-type orbitals), $m_{l}$ can have values of $-2,-1,0$, $+1,+2$, so we have five $4 d$ orbitals. When $l=3$ ( $f$-type orbitals), $m_{l}$ can have values of $-3,-2,-1,0,+1,+2,+3$, and we can have seven $4 f$ orbitals. Thus, we find a total of 16 orbitals in the $n=4$ shell of an atom."}
{"id": 2473, "contents": "239. Check Your Learning - \nIdentify the subshell in which electrons with the following quantum numbers are found: (a) $n=3,1=1$; (b) $n=$ $5, l=3$; (c) $n=2, l=0$."}
{"id": 2474, "contents": "240. Answer: - \n(a) $3 p$ (b) $5 f$ (c) 2 s"}
{"id": 2475, "contents": "242. Maximum Number of Electrons - \nCalculate the maximum number of electrons that can occupy a shell with (a) $n=2$, (b) $n=5$, and (c) $n$ as a variable. Note you are only looking at the orbitals with the specified $n$ value, not those at lower energies."}
{"id": 2476, "contents": "243. Solution - \n(a) When $n=2$, there are four orbitals (a single $2 s$ orbital, and three orbitals labeled $2 p$ ). These four orbitals can contain eight electrons.\n(b) When $n=5$, there are five subshells of orbitals that we need to sum:\n\n1 orbital labeled $5 s$\n3 orbitals labeled 5p\n5 orbitals labeled $5 d$\n7 orbitals labeled $5 f$\n+9 orbitals labeled $5 g$\n25 orbitals total\n\nAgain, each orbital holds two electrons, so 50 electrons can fit in this shell.\n(c) The number of orbitals in any shell $n$ will equal $n^{2}$. There can be up to two electrons in each orbital, so the maximum number of electrons will be $2 \\times n^{2}$."}
{"id": 2477, "contents": "244. Check Your Learning - \nIf a shell contains a maximum of 32 electrons, what is the principal quantum number, $n$ ?"}
{"id": 2478, "contents": "245. Answer: - \n$n=4$"}
{"id": 2479, "contents": "247. Working with Quantum Numbers - \nComplete the following table for atomic orbitals:\n\n| Orbital | $\\boldsymbol{n}$ | $\\boldsymbol{I}$ | $\\boldsymbol{m}_{\\boldsymbol{l}}$ degeneracy | Radial nodes (no.) |\n| :--- | :--- | :--- | :--- | :--- |\n| $4 f$ |  |  |  |  |\n|  | 4 | 1 |  |  |\n|  | 7 |  | 7 | 3 |\n| $5 d$ |  |  |  |  |"}
{"id": 2480, "contents": "248. Solution - \nThe table can be completed using the following rules:\n\n- The orbital designation is $n l$, where $l=0,1,2,3,4,5, \\ldots$ is mapped to the letter sequence $s, p, d, f, g, h, \\ldots$,\n- The $m_{l}$ degeneracy is the number of orbitals within an 1 subshell, and so is $2 l+1$ (there is one $s$ orbital, three $p$ orbitals, five $d$ orbitals, seven $f$ orbitals, and so forth).\n- The number of radial nodes is equal to $n-1-1$.\n\n| Orbital | $\\boldsymbol{n}$ | $\\boldsymbol{l}$ | $\\boldsymbol{m}_{\\boldsymbol{l}}$ degeneracy | Radial nodes (no.) |\n| :--- | :--- | :--- | :--- | :--- |\n| $4 f$ | 4 | 3 | 7 | 0 |\n| $4 p$ | 4 | 1 | 3 | 2 |\n| $7 f$ | 7 | 3 | 7 | 3 |\n| $5 d$ | 5 | 2 | 5 | 2 |"}
{"id": 2481, "contents": "249. Check Your Learning - \nHow many orbitals have $l=2$ and $n=3$ ?"}
{"id": 2482, "contents": "250. Answer: - \nThe five degenerate $3 d$ orbitals"}
{"id": 2483, "contents": "251. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Derive the predicted ground-state electron configurations of atoms\n- Identify and explain exceptions to predicted electron configurations for atoms and ions\n- Relate electron configurations to element classifications in the periodic table\n\nHaving introduced the basics of atomic structure and quantum mechanics, we can use our understanding of quantum numbers to determine how atomic orbitals relate to one another. This allows us to determine which orbitals are occupied by electrons in each atom. The specific arrangement of electrons in orbitals of an atom determines many of the chemical properties of that atom."}
{"id": 2484, "contents": "252. Orbital Energies and Atomic Structure - \nThe energy of atomic orbitals increases as the principal quantum number, $n$, increases. In any atom with two or more electrons, the repulsion between the electrons makes energies of subshells with different values of 1 differ so that the energy of the orbitals increases within a shell in the order $s<p<d<f$. Figure 3.24 depicts how these two trends in increasing energy relate. The $1 s$ orbital at the bottom of the diagram is the orbital with electrons of lowest energy. The energy increases as we move up to the $2 s$ and then $2 p, 3 s$, and $3 p$ orbitals, showing that the increasing $n$ value has more influence on energy than the increasing $l$ value for small atoms. However, this pattern does not hold for larger atoms. The $3 d$ orbital is higher in energy than the $4 s$ orbital. Such overlaps continue to occur frequently as we move up the chart.\n\n\nFIGURE 3.24 Generalized energy-level diagram for atomic orbitals in an atom with two or more electrons (not to scale)."}
{"id": 2485, "contents": "252. Orbital Energies and Atomic Structure - \nFIGURE 3.24 Generalized energy-level diagram for atomic orbitals in an atom with two or more electrons (not to scale).\n\nElectrons in successive atoms on the periodic table tend to fill low-energy orbitals first. Thus, many students find it confusing that, for example, the $5 p$ orbitals fill immediately after the $4 d$, and immediately before the $6 s$. The filling order is based on observed experimental results, and has been confirmed by theoretical calculations. As the principal quantum number, $n$, increases, the size of the orbital increases and the electrons spend more time farther from the nucleus. Thus, the attraction to the nucleus is weaker and the energy associated with the orbital is higher (less stabilized). But this is not the only effect we have to take into account. Within each shell, as the value of 1 increases, the electrons are less penetrating (meaning there is less electron density found close to the nucleus), in the order $s>p>d>f$. Electrons that are closer to the nucleus slightly repel electrons that are farther out, offsetting the more dominant electron-nucleus attractions slightly (recall\nthat all electrons have -1 charges, but nuclei have $+Z$ charges). This phenomenon is called shielding and will be discussed in more detail in the next section. Electrons in orbitals that experience more shielding are less stabilized and thus higher in energy. For small orbitals ( $1 s$ through $3 p$ ), the increase in energy due to $n$ is more significant than the increase due to $l$; however, for larger orbitals the two trends are comparable and cannot be simply predicted. We will discuss methods for remembering the observed order.\n\nThe arrangement of electrons in the orbitals of an atom is called the electron configuration of the atom. We describe an electron configuration with a symbol that contains three pieces of information (Figure 3.25):\n\n1. The number of the principal quantum shell, $n$,\n2. The letter that designates the orbital type (the subshell, $I$ ), and\n3. A superscript number that designates the number of electrons in that particular subshell."}
{"id": 2486, "contents": "252. Orbital Energies and Atomic Structure - \nFor example, the notation $2 p^{4}$ (read \"two-p-four\") indicates four electrons in a $p$ subshell $(l=1)$ with a principal quantum number ( $n$ ) of 2 . The notation $3 d^{8}$ (read \"three-d-eight\") indicates eight electrons in the $d$ subshell (i.e., $l=2$ ) of the principal shell for which $n=3$.\n\n\nFIGURE 3.25 The diagram of an electron configuration specifies the subshell ( $n$ and $l$ value, with letter symbol) and superscript number of electrons."}
{"id": 2487, "contents": "253. The Aufbau Principle - \nTo determine the electron configuration for any particular atom, we can \"build\" the structures in the order of atomic numbers. Beginning with hydrogen, and continuing across the periods of the periodic table, we add one proton at a time to the nucleus and one electron to the proper subshell until we have described the electron configurations of all the elements. This procedure is called the Aufbau principle, from the German word Aufbau (\"to build up\"). Each added electron occupies the subshell of lowest energy available (in the order shown in Figure 3.24), subject to the limitations imposed by the allowed quantum numbers according to the Pauli exclusion principle. Electrons enter higher-energy subshells only after lower-energy subshells have been filled to capacity. Figure 3.26 illustrates the traditional way to remember the filling order for atomic orbitals. Since the arrangement of the periodic table is based on the electron configurations, Figure 3.27 provides an alternative method for determining the electron configuration. The filling order simply begins at hydrogen and includes each subshell as you proceed in increasing $Z$ order. For example, after filling the $3 p$ block up to Ar, we see the orbital will be $4 \\mathrm{~s}(\\mathrm{~K}, \\mathrm{Ca})$, followed by the $3 d$ orbitals.\n\n\nFIGURE 3.26 This diagram depicts the energy order for atomic orbitals and is useful for deriving ground-state electron configurations.\n\n\nFIGURE 3.27 This partial periodic table shows electron configurations for the valence subshells of atoms. By \"building up\" from hydrogen, this table can be used to determine the electron configuration for atoms of most elements in the periodic table. (Electron configurations of the lanthanides and actinides are not accurately predicted by this simple approach. See Figure 3.29"}
{"id": 2488, "contents": "253. The Aufbau Principle - \nWe will now construct the ground-state electron configuration and orbital diagram for a selection of atoms in the first and second periods of the periodic table. Orbital diagrams are pictorial representations of the electron configuration, showing the individual orbitals and the pairing arrangement of electrons. We start with a single hydrogen atom (atomic number 1), which consists of one proton and one electron. Referring to Figure 3.26 or Figure 3.27, we would expect to find the electron in the $1 s$ orbital. By convention, the $m_{s}=+\\frac{1}{2}$ value is usually filled first. The electron configuration and the orbital diagram are:\n\n\nFollowing hydrogen is the noble gas helium, which has an atomic number of 2 . The helium atom contains two protons and two electrons. The first electron has the same four quantum numbers as the hydrogen atom electron ( $n=1, l=0, m_{l}=0, m_{s}=+\\frac{1}{2}$ ). The second electron also goes into the $1 s$ orbital and fills that orbital. The second electron has the same $n, l$, and $m_{l}$ quantum numbers, but must have the opposite spin quantum number, $m_{s}=-\\frac{1}{2}$. This is in accord with the Pauli exclusion principle: No two electrons in the same atom can have the same set of four quantum numbers. For orbital diagrams, this means two arrows go in each box (representing two electrons in each orbital) and the arrows must point in opposite directions (representing paired spins). The electron configuration and orbital diagram of helium are:\n\n\nThe $n=1$ shell is completely filled in a helium atom.\n\nThe next atom is the alkali metal lithium with an atomic number of 3 . The first two electrons in lithium fill the $1 s$ orbital and have the same sets of four quantum numbers as the two electrons in helium. The remaining electron must occupy the orbital of next lowest energy, the $2 s$ orbital (Figure 3.26 or Figure 3.27). Thus, the electron configuration and orbital diagram of lithium are:\n\n\nAn atom of the alkaline earth metal beryllium, with an atomic number of 4 , contains four protons in the nucleus and four electrons surrounding the nucleus. The fourth electron fills the remaining space in the $2 s$ orbital."}
{"id": 2489, "contents": "253. The Aufbau Principle - \nAn atom of boron (atomic number 5) contains five electrons. The $n=1$ shell is filled with two electrons and three electrons will occupy the $n=2$ shell. Because any $s$ subshell can contain only two electrons, the fifth electron must occupy the next energy level, which will be a $2 p$ orbital. There are three degenerate $2 p$ orbitals ( $m_{l}=-1,0,+1$ ) and the electron can occupy any one of these $p$ orbitals. When drawing orbital diagrams, we include empty boxes to depict any empty orbitals in the same subshell that we are filling.\n\n\nCarbon (atomic number 6) has six electrons. Four of them fill the $1 s$ and $2 s$ orbitals. The remaining two electrons occupy the $2 p$ subshell. We now have a choice of filling one of the $2 p$ orbitals and pairing the electrons or of leaving the electrons unpaired in two different, but degenerate, $p$ orbitals. The orbitals are filled as described by Hund's rule: the lowest-energy configuration for an atom with electrons within a set of degenerate orbitals is that having the maximum number of unpaired electrons. Thus, the two electrons in the carbon $2 p$ orbitals have identical $n, l$, and $m_{s}$ quantum numbers and differ in their $m_{l}$ quantum number (in accord with the Pauli exclusion principle). The electron configuration and orbital diagram for carbon are:\n\n\nNitrogen (atomic number 7) fills the $1 s$ and $2 s$ subshells and has one electron in each of the three $2 p$ orbitals, in accordance with Hund's rule. These three electrons have unpaired spins. Oxygen (atomic number 8) has a pair of electrons in any one of the $2 p$ orbitals (the electrons have opposite spins) and a single electron in each of the other two. Fluorine (atomic number 9) has only one $2 p$ orbital containing an unpaired electron. All of the electrons in the noble gas neon (atomic number 10) are paired, and all of the orbitals in the $n=1$ and the $n=2$ shells are filled. The electron configurations and orbital diagrams of these four elements are:"}
{"id": 2490, "contents": "253. The Aufbau Principle - \nThe alkali metal sodium (atomic number 11) has one more electron than the neon atom. This electron must go into the lowest-energy subshell available, the $3 s$ orbital, giving a $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$ configuration. The electrons occupying the outermost shell orbital(s) (highest value of $n$ ) are called valence electrons, and those occupying the inner shell orbitals are called core electrons (Figure 3.28). Since the core electron shells correspond to noble gas electron configurations, we can abbreviate electron configurations by writing the noble gas that matches the core electron configuration, along with the valence electrons in a condensed format. For our sodium example, the symbol [ Ne ] represents core electrons, $\\left(1 s^{2} 2 s^{2} 2 p^{6}\\right)$ and our abbreviated or condensed configuration is [ Ne$] 3 s^{1}$.\n\n\nAbbreviation $[\\mathrm{Ne}] 3 s^{1}$\n\nFIGURE 3.28 A core-abbreviated electron configuration (right) replaces the core electrons with the noble gas symbol whose configuration matches the core electron configuration of the other element.\n\nSimilarly, the abbreviated configuration of lithium can be represented as [He] $2 s^{1}$, where [He] represents the configuration of the helium atom, which is identical to that of the filled inner shell of lithium. Writing the configurations in this way emphasizes the similarity of the configurations of lithium and sodium. Both atoms, which are in the alkali metal family, have only one electron in a valence $s$ subshell outside a filled set of inner shells.\n\n$$\n\\begin{aligned}\n& \\mathrm{Li}:[\\mathrm{He}] 2 s^{1} \\\\\n& \\mathrm{Na}:[\\mathrm{Ne}] 3 s^{1}\n\\end{aligned}\n$$"}
{"id": 2491, "contents": "253. The Aufbau Principle - \nThe alkaline earth metal magnesium (atomic number 12), with its 12 electrons in a [ Ne$] 3 s^{2}$ configuration, is analogous to its family member beryllium, [ He$] 2 s^{2}$. Both atoms have a filled $s$ subshell outside their filled inner shells. Aluminum (atomic number 13), with 13 electrons and the electron configuration $[\\mathrm{Ne}] 3 s^{2} 3 p^{1}$, is analogous to its family member boron, [He] $2 s^{2} 2 p^{1}$.\n\nThe electron configurations of silicon (14 electrons), phosphorus (15 electrons), sulfur ( 16 electrons), chlorine (17 electrons), and argon (18 electrons) are analogous in the electron configurations of their outer shells to their corresponding family members carbon, nitrogen, oxygen, fluorine, and neon, respectively, except that the principal quantum number of the outer shell of the heavier elements has increased by one to $n=3$. Figure 3.29 shows the lowest energy, or ground-state, electron configuration for these elements as well as that for atoms of each of the known elements."}
{"id": 2492, "contents": "254. Electron Configuration Table - \nFIGURE 3.29 This version of the periodic table shows the outer-shell electron configuration of each element. Note that down each group, the configuration is often similar."}
{"id": 2493, "contents": "254. Electron Configuration Table - \nWhen we come to the next element in the periodic table, the alkali metal potassium (atomic number 19), we might expect that we would begin to add electrons to the $3 d$ subshell. However, all available chemical and physical evidence indicates that potassium is like lithium and sodium, and that the next electron is not added to the $3 d$ level but is, instead, added to the $4 s$ level (Figure 3.29). As discussed previously, the $3 d$ orbital with no radial nodes is higher in energy because it is less penetrating and more shielded from the nucleus than the $4 s$, which has three radial nodes. Thus, potassium has an electron configuration of [Ar]4s ${ }^{1}$. Hence, potassium corresponds to Li and Na in its valence shell configuration. The next electron is added to complete the $4 S$ subshell and calcium has an electron configuration of $[\\mathrm{Ar}] 4 s^{2}$. This gives calcium an outer-shell electron configuration corresponding to that of beryllium and magnesium."}
{"id": 2494, "contents": "254. Electron Configuration Table - \nBeginning with the transition metal scandium (atomic number 21), additional electrons are added successively to the $3 d$ subshell. This subshell is filled to its capacity with 10 electrons (remember that for $1=2$ [ $d$ orbitals], there are $2 l+1=5$ values of $m_{l}$, meaning that there are five $d$ orbitals that have a combined capacity of 10 electrons). The $4 p$ subshell fills next. Note that for three series of elements, scandium (Sc) through copper ( Cu ), yttrium ( Y ) through silver ( Ag ), and lutetium ( Lu ) through gold ( Au ), a total of 10 d electrons are successively added to the $(n-1)$ shell next to the $n$ shell to bring that $(n-1)$ shell from 8 to 18 electrons. For two series, lanthanum (La) through lutetium (Lu) and actinium (Ac) through lawrencium (Lr), 14 $f$ electrons ( $l=3,2 l+1=7 m_{l}$ values; thus, seven orbitals with a combined capacity of 14 electrons) are successively added to the $(n-2)$ shell to bring that shell from 18 electrons to a total of 32 electrons."}
{"id": 2495, "contents": "256. Quantum Numbers and Electron Configurations - \nWhat is the electron configuration and orbital diagram for a phosphorus atom? What are the four quantum numbers for the last electron added?"}
{"id": 2496, "contents": "257. Solution - \nThe atomic number of phosphorus is 15 . Thus, a phosphorus atom contains 15 electrons. The order of filling of the energy levels is $1 s, 2 s, 2 p, 3 s, 3 p, 4 s, \\ldots$ The 15 electrons of the phosphorus atom will fill up to the $3 p$ orbital, which will contain three electrons:\n\n\nThe last electron added is a $3 p$ electron. Therefore, $n=3$ and, for a $p$-type orbital, $l=1$. The $m_{l}$ value could be $-1,0$, or +1 . The three $p$ orbitals are degenerate, so any of these $m_{l}$ values is correct. For unpaired electrons, convention assigns the value of $+\\frac{1}{2}$ for the spin quantum number; thus, $m_{s}=+\\frac{1}{2}$."}
{"id": 2497, "contents": "258. Check Your Learning - \nIdentify the atoms from the electron configurations given:\n(a) $[\\mathrm{Ar}] 4 s^{2} 3 d^{5}$\n(b) $[\\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{6}$"}
{"id": 2498, "contents": "259. Answer: - \n(a) Mn (b) Xe\n\nThe periodic table can be a powerful tool in predicting the electron configuration of an element. However, we do find exceptions to the order of filling of orbitals that are shown in Figure 3.26 or Figure 3.27. For instance, the electron configurations (shown in Figure 3.29) of the transition metals chromium (Cr; atomic number 24) and copper ( Cu ; atomic number 29), among others, are not those we would expect. In general, such exceptions involve subshells with very similar energy, and small effects can lead to changes in the order of filling.\n\nIn the case of Cr and Cu , we find that half-filled and completely filled subshells apparently represent conditions of preferred stability. This stability is such that an electron shifts from the $4 s$ into the $3 d$ orbital to gain the extra stability of a half-filled $3 d$ subshell (in Cr ) or a filled $3 d$ subshell (in Cu ). Other exceptions also occur. For example, niobium ( Nb , atomic number 41) is predicted to have the electron configuration $[\\mathrm{Kr}] 5 s^{2} 4 d^{3}$. Experimentally, we observe that its ground-state electron configuration is actually $[\\mathrm{Kr}] 5 s^{1} 4 d^{4}$. We can rationalize this observation by saying that the electron-electron repulsions experienced by pairing the electrons in the $5 s$ orbital are larger than the gap in energy between the $5 s$ and $4 d$ orbitals. There is no simple method to predict the exceptions for atoms where the magnitude of the repulsions between electrons is greater than the small differences in energy between subshells."}
{"id": 2499, "contents": "260. Electron Configurations and the Periodic Table - \nAs described earlier, the periodic table arranges atoms based on increasing atomic number so that elements with the same chemical properties recur periodically. When their electron configurations are added to the table (Figure 3.29), we also see a periodic recurrence of similar electron configurations in the outer shells of these elements. Because they are in the outer shells of an atom, valence electrons play the most important role in chemical reactions. The outer electrons have the highest energy of the electrons in an atom and are more easily lost or shared than the core electrons. Valence electrons are also the determining factor in some physical properties of the elements.\n\nElements in any one group (or column) have the same number of valence electrons; the alkali metals lithium\nand sodium each have only one valence electron, the alkaline earth metals beryllium and magnesium each have two, and the halogens fluorine and chlorine each have seven valence electrons. The similarity in chemical properties among elements of the same group occurs because they have the same number of valence electrons. It is the loss, gain, or sharing of valence electrons that defines how elements react.\n\nIt is important to remember that the periodic table was developed on the basis of the chemical behavior of the elements, well before any idea of their atomic structure was available. Now we can understand why the periodic table has the arrangement it has-the arrangement puts elements whose atoms have the same number of valence electrons in the same group. This arrangement is emphasized in Figure 3.29, which shows in periodic-table form the electron configuration of the last subshell to be filled by the Aufbau principle. The colored sections of Figure 3.29 show the three categories of elements classified by the orbitals being filled: main group, transition, and inner transition elements. These classifications determine which orbitals are counted in the valence shell, or highest energy level orbitals of an atom."}
{"id": 2500, "contents": "260. Electron Configurations and the Periodic Table - \n1. Main group elements (sometimes called representative elements) are those in which the last electron added enters an $s$ or a $p$ orbital in the outermost shell, shown in blue and red in Figure 3.29. This category includes all the nonmetallic elements, as well as many metals and the metalloids. The valence electrons for main group elements are those with the highest $n$ level. For example, gallium ( Ga , atomic number 31) has the electron configuration $[\\operatorname{Ar}] 4 s^{2} 3 d^{10} 4 p^{1}$, which contains three valence electrons (underlined). The completely filled $d$ orbitals count as core, not valence, electrons.\n2. Transition elements or transition metals. These are metallic elements in which the last electron added enters a $d$ orbital. The valence electrons (those added after the last noble gas configuration) in these elements include the $n s$ and $(n-1) d$ electrons. The official IUPAC definition of transition elements specifies those with partially filled $d$ orbitals. Thus, the elements with completely filled orbitals (Zn, Cd, Hg , as well as $\\mathrm{Cu}, \\mathrm{Ag}$, and Au in Figure 3.29) are not technically transition elements. However, the term is frequently used to refer to the entire $d$ block (colored yellow in Figure 3.29), and we will adopt this usage in this textbook.\n3. Inner transition elements are metallic elements in which the last electron added occupies an $f$ orbital. They are shown in green in Figure 3.29. The valence shells of the inner transition elements consist of the $(n-2) f$, the $(n-1) d$, and the $n s$ subshells. There are two inner transition series:\na. The lanthanide series: lanthanum (La) through lutetium (Lu)\nb. The actinide series: actinium (Ac) through lawrencium (Lr)\n\nLanthanum and actinium, because of their similarities to the other members of the series, are included and used to name the series, even though they are transition metals with no $f$ electrons."}
{"id": 2501, "contents": "261. Electron Configurations of Ions - \nIons are formed when atoms gain or lose electrons. A cation (positively charged ion) forms when one or more electrons are removed from a parent atom. For main group elements, the electrons that were added last are the first electrons removed. For transition metals and inner transition metals, however, electrons in the $s$ orbital are easier to remove than the $d$ or $f$ electrons, and so the highest $n s$ electrons are lost, and then the $(n-1) d$ or ( $n-2$ )felectrons are removed. An anion (negatively charged ion) forms when one or more electrons are added to a parent atom. The added electrons fill in the order predicted by the Aufbau principle."}
{"id": 2502, "contents": "263. Predicting Electron Configurations of Ions - \nWhat is the electron configuration of:\n(a) $\\mathrm{Na}^{+}$\n(b) $\\mathrm{P}^{3-}$\n(c) $\\mathrm{Al}^{2+}$\n(d) $\\mathrm{Fe}^{2+}$\n(e) $\\mathrm{Sm}^{3+}$"}
{"id": 2503, "contents": "264. Solution - \nFirst, write out the electron configuration for each parent atom. We have chosen to show the full, unabbreviated configurations to provide more practice for students who want it, but listing the coreabbreviated electron configurations is also acceptable."}
{"id": 2504, "contents": "264. Solution - \nNext, determine whether an electron is gained or lost. Remember electrons are negatively charged, so ions with a positive charge have lost an electron. For main group elements, the last orbital gains or loses the electron. For transition metals, the last $s$ orbital loses an electron before the $d$ orbitals.\n(a) Na: $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$. Sodium cation loses one electron, so $\\mathrm{Na}^{+}: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}=\\mathrm{Na}^{+}: 1 s^{2} 2 s^{2} 2 p^{6}$.\n(b) P: $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{3}$. Phosphorus trianion gains three electrons, so $\\mathrm{P}^{3-}: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$.\n(c) Al: $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1}$. Aluminum dication loses two electrons $\\mathrm{Al}^{2+}: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1}=$\n$\\mathrm{Al}^{2+}: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$."}
{"id": 2505, "contents": "264. Solution - \n$\\mathrm{Al}^{2+}: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$.\n(d) Fe: $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{6}$. Iron(II) loses two electrons and, since it is a transition metal, they are removed from the $4 s$ orbital $\\mathrm{Fe}^{2+}: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{6}=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6}$.\n(e). Sm: $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{2} 4 d^{10} 5 p^{6} 6 s^{2} 4 f^{6}$. Samarium trication loses three electrons. The first two will be lost from the $6 s$ orbital, and the final one is removed from the $4 f$ orbital. $\\mathrm{Sm}^{3+}$ :\n$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{2} 4 d^{10} 5 p^{6} 6 s^{2} 4 f^{6}=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{2} 4 d^{10} 5 p^{6} 4 f^{5}$.\nCheck Your Learning"}
{"id": 2506, "contents": "264. Solution - \nCheck Your Learning\nWhich ion with a +2 charge has the electron configuration $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 4 d^{5}$ ? Which ion with a +3 charge has this configuration?"}
{"id": 2507, "contents": "265. Answer: - \n$\\mathrm{Tc}^{2+}, \\mathrm{Ru}^{3+}$"}
{"id": 2508, "contents": "266. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe and explain the observed trends in atomic size, ionization energy, and electron affinity of the elements\n\nThe elements in groups (vertical columns) of the periodic table exhibit similar chemical behavior. This similarity occurs because the members of a group have the same number and distribution of electrons in their valence shells. However, there are also other patterns in chemical properties on the periodic table. For example, as we move down a group, the metallic character of the atoms increases. Oxygen, at the top of group 16 (6A), is a colorless gas; in the middle of the group, selenium is a semiconducting solid; and, toward the bottom, polonium is a silver-grey solid that conducts electricity.\n\nAs we go across a period from left to right, we add a proton to the nucleus and an electron to the valence shell with each successive element. As we go down the elements in a group, the number of electrons in the valence shell remains constant, but the principal quantum number increases by one each time. An understanding of the electronic structure of the elements allows us to examine some of the properties that govern their chemical behavior. These properties vary periodically as the electronic structure of the elements changes. They are (1) size (radius) of atoms and ions, (2) ionization energies, and (3) electron affinities."}
{"id": 2509, "contents": "267. LINK TO LEARNING - \nExplore visualizations (http://openstax.org/l/16pertrends) of the periodic trends discussed in this section (and many more trends). With just a few clicks, you can create three-dimensional versions of the periodic table showing atomic size or graphs of ionization energies from all measured elements."}
{"id": 2510, "contents": "268. Variation in Covalent Radius - \nThe quantum mechanical picture makes it difficult to establish a definite size of an atom. However, there are several practical ways to define the radius of atoms and, thus, to determine their relative sizes that give roughly similar values. We will use the covalent radius (Figure 3.30), which is defined as one-half the distance between the nuclei of two identical atoms when they are joined by a covalent bond (this measurement is possible because atoms within molecules still retain much of their atomic identity). We know that as we scan down a group, the principal quantum number, $n$, increases by one for each element. Thus, the electrons are being added to a region of space that is increasingly distant from the nucleus. Consequently, the size of the atom (and its covalent radius) must increase as we increase the distance of the outermost electrons from the nucleus. This trend is illustrated for the covalent radii of the halogens in Table 3.2 and Figure 3.30. The trends for the entire periodic table can be seen in Figure 3.30.\n\nCovalent Radii of the Halogen Group Elements\n\n| Atom | Covalent radius (pm) | Nuclear charge |\n| :--- | :--- | :--- |\n| F | 64 | +9 |\n| Cl | 99 | +17 |\n| Br | 114 | +35 |\n| I | 133 | +53 |\n| At | 148 | +85 |\n\nTABLE 3.2\n\n\nFIGURE 3.30 (a) The radius of an atom is defined as one-half the distance between the nuclei in a molecule consisting of two identical atoms joined by a covalent bond. The atomic radius for the halogens increases down the group as $n$ increases. (b) Covalent radii of the elements are shown to scale. The general trend is that radii increase down a group and decrease across a period.\n\n\nFIGURE 3.31 Within each period, the trend in atomic radius decreases as $Z$ increases; for example, from K to Kr . Within each group (e.g., the alkali metals shown in purple), the trend is that atomic radius increases as $Z$ increases."}
{"id": 2511, "contents": "268. Variation in Covalent Radius - \nAs shown in Figure 3.31, as we move across a period from left to right, we generally find that each element has a smaller covalent radius than the element preceding it. This might seem counterintuitive because it implies that atoms with more electrons have a smaller atomic radius. This can be explained with the concept of effective nuclear charge, $Z_{\\text {eff }}$. This is the pull exerted on a specific electron by the nucleus, taking into account any electron-electron repulsions. For hydrogen, there is only one electron and so the nuclear charge ( $Z$ ) and the effective nuclear charge ( $Z_{\\text {eff }}$ ) are equal. For all other atoms, the inner electrons partially shield the outer electrons from the pull of the nucleus, and thus:\n\n$$\nZ_{\\mathrm{eff}}=Z-\\text { shielding }\n$$\n\nShielding is determined by the probability of another electron being between the electron of interest and the nucleus, as well as by the electron-electron repulsions the electron of interest encounters. Core electrons are adept at shielding, while electrons in the same valence shell do not block the nuclear attraction experienced by each other as efficiently. Thus, each time we move from one element to the next across a period, $Z$ increases by one, but the shielding increases only slightly. Thus, $Z_{\\text {eff }}$ increases as we move from left to right across a period. The stronger pull (higher effective nuclear charge) experienced by electrons on the right side of the periodic table draws them closer to the nucleus, making the covalent radii smaller.\n\nThus, as we would expect, the outermost or valence electrons are easiest to remove because they have the highest energies, are shielded more, and are farthest from the nucleus. As a general rule, when the representative elements form cations, they do so by the loss of the $n s$ or $n p$ electrons that were added last in the Aufbau process. The transition elements, on the other hand, lose the $n s$ electrons before they begin to lose the ( $n-1$ )d electrons, even though the $n s$ electrons are added first, according to the Aufbau principle."}
{"id": 2512, "contents": "270. Sorting Atomic Radii - \nPredict the order of increasing covalent radius for $\\mathrm{Ge}, \\mathrm{Fl}, \\mathrm{Br}, \\mathrm{Kr}$."}
{"id": 2513, "contents": "271. Solution - \nRadius increases as we move down a group, so $\\mathrm{Ge}<\\mathrm{Fl}$ (Note: Fl is the symbol for flerovium, element 114, NOT fluorine). Radius decreases as we move across a period, so $\\mathrm{Kr}<\\mathrm{Br}<\\mathrm{Ge}$. Putting the trends together, we obtain $\\mathrm{Kr}<\\mathrm{Br}<\\mathrm{Ge}<\\mathrm{Fl}$."}
{"id": 2514, "contents": "272. Check Your Learning - \nGive an example of an atom whose size is smaller than fluorine."}
{"id": 2515, "contents": "273. Answer: - \nNe or He"}
{"id": 2516, "contents": "274. Variation in Ionic Radii - \nIonic radius is the measure used to describe the size of an ion. A cation always has fewer electrons and the same number of protons as the parent atom; it is smaller than the atom from which it is derived (Figure 3.32). For example, the covalent radius of an aluminum atom $\\left(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1}\\right)$ is 118 pm , whereas the ionic radius of an $\\mathrm{Al}^{3+}\\left(1 s^{2} 2 s^{2} 2 p^{6}\\right)$ is 68 pm . As electrons are removed from the outer valence shell, the remaining core electrons occupying smaller shells experience a greater effective nuclear charge $Z_{\\text {eff }}$ (as discussed) and are drawn even closer to the nucleus.\n\n\nFIGURE 3.32 The radius for a cation is smaller than the parent atom (Al), due to the lost electrons; the radius for an anion is larger than the parent ( S ), due to the gained electrons.\n\nCations with larger charges are smaller than cations with smaller charges (e.g., $\\mathrm{V}^{2+}$ has an ionic radius of 79 pm , while that of $\\mathrm{V}^{3+}$ is 64 pm ). Proceeding down the groups of the periodic table, we find that cations of successive elements with the same charge generally have larger radii, corresponding to an increase in the principal quantum number, $n$."}
{"id": 2517, "contents": "274. Variation in Ionic Radii - \nAn anion (negative ion) is formed by the addition of one or more electrons to the valence shell of an atom. This results in a greater repulsion among the electrons and a decrease in $Z_{\\text {eff }}$ per electron. Both effects (the increased number of electrons and the decreased $Z_{\\text {eff }}$ ) cause the radius of an anion to be larger than that of the parent atom (Figure 3.32). For example, a sulfur atom ([Ne]3s ${ }^{2} 3 p^{4}$ ) has a covalent radius of 104 pm , whereas the ionic radius of the sulfide anion ( $[\\mathrm{Ne}] 3 s^{2} 3 p^{6}$ ) is 170 pm . For consecutive elements proceeding down any group, anions have larger principal quantum numbers and, thus, larger radii.\n\nAtoms and ions that have the same electron configuration are said to be isoelectronic. Examples of isoelectronic species are $\\mathrm{N}^{3-}, \\mathrm{O}^{2-}, \\mathrm{F}^{-}, \\mathrm{Ne}, \\mathrm{Na}^{+}, \\mathrm{Mg}^{2+}$, and $\\mathrm{Al}^{3+}\\left(1 s^{2} 2 s^{2} 2 p^{6}\\right)$. Another isoelectronic series is $\\mathrm{P}^{3-}$, $\\mathrm{S}^{2-}, \\mathrm{Cl}^{-}, \\mathrm{Ar}, \\mathrm{K}^{+}, \\mathrm{Ca}^{2+}$, and $\\mathrm{Sc}^{3+}\\left([\\mathrm{Ne}] 3 s^{2} 3 p^{6}\\right)$. For atoms or ions that are isoelectronic, the number of protons determines the size. The greater the nuclear charge, the smaller the radius in a series of isoelectronic ions and atoms."}
{"id": 2518, "contents": "275. Variation in Ionization Energies - \nThe amount of energy required to remove the most loosely bound electron from a gaseous atom in its ground state is called its first ionization energy ( $\\mathrm{IE}_{1}$ ). The first ionization energy for an element, X , is the energy required to form a cation with +1 charge:\n\n$$\n\\mathrm{X}(g) \\longrightarrow \\mathrm{X}^{+}(g)+\\mathrm{e}^{-} \\quad \\mathrm{IE}_{1}\n$$\n\nThe energy required to remove the second most loosely bound electron is called the second ionization energy ( $\\mathrm{IE}_{2}$ ).\n\n$$\n\\mathrm{X}^{+}(g) \\longrightarrow \\mathrm{X}^{2+}(g)+\\mathrm{e}^{-} \\quad \\mathrm{IE}_{2}\n$$\n\nThe energy required to remove the third electron is the third ionization energy, and so on. Energy is always required to remove electrons from atoms or ions, so ionization processes are endothermic and IE values are always positive. For larger atoms, the most loosely bound electron is located farther from the nucleus and so is easier to remove. Thus, as size (atomic radius) increases, the ionization energy should decrease. Relating this logic to what we have just learned about radii, we would expect first ionization energies to decrease down a group and to increase across a period."}
{"id": 2519, "contents": "275. Variation in Ionization Energies - \nFigure 3.33 graphs the relationship between the first ionization energy and the atomic number of several elements. The values of first ionization energy for the elements are given in Figure 3.34. Within a period, the $\\mathrm{IE}_{1}$ generally increases with increasing $Z$. Down a group, the $\\mathrm{IE}_{1}$ value generally decreases with increasing $Z$. There are some systematic deviations from this trend, however. Note that the ionization energy of boron (atomic number 5) is less than that of beryllium (atomic number 4) even though the nuclear charge of boron is greater by one proton. This can be explained because the energy of the subshells increases as $l$ increases, due to penetration and shielding (as discussed previously in this chapter). Within any one shell, the $s$ electrons are lower in energy than the $p$ electrons. This means that an $s$ electron is harder to remove from an atom than a $p$ electron in the same shell. The electron removed during the ionization of beryllium ( $[\\mathrm{He}] 2 s^{2}$ ) is an $s$ electron, whereas the electron removed during the ionization of boron ( $[\\mathrm{He}] 2 s^{2} 2 p^{1}$ ) is a $p$ electron; this results in a lower first ionization energy for boron, even though its nuclear charge is greater by one proton. Thus, we see a small deviation from the predicted trend occurring each time a new subshell begins.\n\n\nFIGURE 3.33 The first ionization energy of the elements in the first five periods are plotted against their atomic number.\n\nFirst Ionization Energies of Some Elements (kJ/mol)"}
{"id": 2520, "contents": "275. Variation in Ionization Energies - \n| 1 | 2 |  |  |  |  |  |  |  |  |  |  | 13 | 14 | 15 | 16 | 17 | 18 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $1 \\begin{gathered} \\mathbf{H} \\\\ 1310 \\end{gathered}$ |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |  | $\\begin{array}{\\|c\\|} \\hline \\mathbf{H e} \\\\ 2370 \\\\ \\hline \\end{array}$ |\n| $2 \\begin{gathered} \\mathbf{L i} \\\\ 520 \\end{gathered}$ | $\\begin{array}{\\|c\\|} \\hline \\mathrm{Be} \\\\ 900 \\end{array}$ |  |  |  |  |  |  |  |  |  |  | $\\begin{gathered} \\text { B } \\\\ 800 \\\\ \\hline \\end{gathered}$ | $\\begin{array}{\\|c\\|} \\hline \\mathbf{C} \\\\ 1090 \\\\ \\hline \\end{array}$ | $\\begin{array}{\\|c\\|} \\hline \\mathbf{N} \\\\ 1400 \\end{array}$ | $\\begin{array}{\\|c\\|} \\hline \\mathbf{o} \\\\ 1310 \\\\ \\hline \\end{array}$ | $\\begin{array}{\\|c\\|} \\hline F \\\\ 1680 \\end{array}$ | $\\begin{array}{\\|c\\|} \\hline \\mathrm{Ne} \\\\ 2080 \\\\ \\hline \\end{array}$ |"}
{"id": 2521, "contents": "275. Variation in Ionization Energies - \n| $3 \\begin{gathered} \\mathrm{Na} \\\\ 490 \\end{gathered}$ | $\\begin{array}{r} \\mathbf{M g} \\\\ 730 \\\\ \\hline \\end{array}$ | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | $\\begin{gathered} \\text { AI } \\\\ 580 \\\\ \\hline \\end{gathered}$ | $\\begin{array}{\\|c\\|} \\hline \\mathbf{S i} \\\\ 780 \\\\ \\hline \\end{array}$ | $\\begin{array}{\\|c\\|} \\hline \\mathbf{P} \\\\ 1060 \\\\ \\hline \\end{array}$ | $\\begin{gathered} \\mathbf{S} \\\\ 1000 \\\\ \\hline \\end{gathered}$ | $\\begin{gathered} \\mathrm{Cl} \\\\ 1250 \\\\ \\hline \\end{gathered}$ | $\\begin{array}{\\|c\\|} \\hline \\text { Ar } \\\\ 1520 \\\\ \\hline \\end{array}$ |"}
{"id": 2522, "contents": "275. Variation in Ionization Energies - \n| $4 \\begin{gathered} K \\\\ 420 \\end{gathered}$ | $\\begin{array}{r} \\mathrm{Ca} \\\\ 590 \\\\ \\hline \\end{array}$ | $\\begin{gathered} \\mathbf{S c} \\\\ 630 \\\\ \\hline \\end{gathered}$ | $\\begin{gathered} \\mathrm{Ti} \\\\ 660 \\\\ \\hline \\end{gathered}$ | $\\begin{gathered} \\text { v } \\\\ 650 \\end{gathered}$ | $\\begin{gathered} \\mathrm{Cr} \\\\ 660 \\\\ \\hline \\end{gathered}$ | $\\begin{array}{r} \\mathbf{M n} \\\\ 710 \\\\ \\hline \\end{array}$ | $\\begin{gathered} \\mathrm{Fe} \\\\ 760 \\\\ \\hline \\end{gathered}$ | $\\begin{gathered} \\text { Co } \\\\ 760 \\\\ \\hline \\end{gathered}$ | $\\begin{gathered} \\mathbf{N i} \\\\ 730 \\end{gathered}$ | $\\begin{array}{r} \\mathrm{Cu} \\\\ 740 \\end{array}$ | $\\begin{gathered} \\mathrm{Zn} \\\\ 910 \\end{gathered}$ | $\\begin{array}{r} \\mathbf{G a} \\\\ 580 \\\\ \\hline \\end{array}$ | $\\begin{gathered} \\text { Ge } \\\\ 780 \\end{gathered}$ | $\\begin{gathered} \\text { As } \\\\ 960 \\end{gathered}$ | $\\begin{gathered} \\mathrm{Se} \\\\ 950 \\\\ \\hline \\end{gathered}$ | $\\begin{gathered} \\mathrm{Br} \\\\ 1140 \\end{gathered}$ | $\\begin{gathered} \\mathrm{Kr} \\\\ 1350 \\end{gathered}$ |"}
{"id": 2523, "contents": "275. Variation in Ionization Energies - \n| $5 \\begin{gathered} \\mathbf{R b} \\\\ 400 \\end{gathered}$ | $\\begin{array}{r} \\mathrm{Sr} \\\\ 550 \\\\ \\hline \\end{array}$ | $\\begin{array}{\\|c} \\mathbf{Y} \\\\ 620 \\\\ \\hline \\end{array}$ | $\\begin{gathered} \\mathrm{Zr} \\\\ 660 \\\\ \\hline \\end{gathered}$ | $\\begin{gathered} \\mathrm{Nb} \\\\ 670 \\\\ \\hline \\end{gathered}$ | $\\begin{array}{r} \\text { Mo } \\\\ 680 \\\\ \\hline \\end{array}$ | $\\begin{gathered} \\text { Tc } \\\\ 700 \\\\ \\hline \\end{gathered}$ | $\\begin{array}{r} \\mathrm{Ru} \\\\ 710 \\\\ \\hline \\end{array}$ | $\\begin{array}{r\\|} \\hline \\text { Rh } \\\\ 720 \\\\ \\hline \\end{array}$ | $\\begin{gathered} \\text { Pd } \\\\ 800 \\\\ \\hline \\end{gathered}$ | $\\begin{array}{r} \\mathrm{Ag} \\\\ 730 \\\\ \\hline \\end{array}$ | $\\begin{array}{r} \\text { Cd } \\\\ 870 \\\\ \\hline \\end{array}$ | $\\begin{gathered} \\text { In } \\\\ 560 \\\\ \\hline \\end{gathered}$ | $\\begin{array}{r} \\text { Sn } \\\\ 700 \\\\ \\hline \\end{array}$ | $\\begin{array}{r} \\text { Sb } \\\\ 830 \\\\ \\hline \\end{array}$ | $\\begin{gathered} \\mathrm{Te} \\\\ 870 \\\\ \\hline \\end{gathered}$ | $\\begin{gathered} \\hline 1 \\\\ 1010 \\\\ \\hline \\end{gathered}$ | $\\begin{array}{\\|c\\|} \\hline \\mathbf{X e} \\\\ 1170 \\\\ \\hline \\end{array}$ |"}
{"id": 2524, "contents": "275. Variation in Ionization Energies - \n| $6 \\begin{gathered} \\text { Cs } \\\\ 380 \\end{gathered}$ | $\\begin{array}{r} \\mathrm{Ba} \\\\ 500 \\\\ \\hline \\end{array}$ | $\\begin{array}{\\|c} \\mathrm{La} \\\\ 540 \\\\ \\hline \\end{array}$ | $\\begin{gathered} \\mathrm{Hf} \\\\ 700 \\end{gathered}$ | $\\begin{gathered} \\mathrm{Ta} \\\\ 760 \\end{gathered}$ | $\\begin{gathered} \\mathbf{w} \\\\ 770 \\end{gathered}$ | $\\begin{gathered} \\mathrm{Re} \\\\ 760 \\end{gathered}$ | $\\begin{gathered} \\text { Os } \\\\ 840 \\end{gathered}$ | $\\begin{gathered} \\text { Ir } \\\\ 890 \\\\ \\hline \\end{gathered}$ | $\\begin{array}{r} \\text { Pt } \\\\ 870 \\\\ \\hline \\end{array}$ | $\\begin{array}{r} \\mathrm{Au} \\\\ 890 \\\\ \\hline \\end{array}$ | $\\begin{gathered} \\mathrm{Hg} \\\\ 1000 \\\\ \\hline \\end{gathered}$ | $\\begin{array}{r} \\mathrm{TI} \\\\ 590 \\\\ \\hline \\end{array}$ | $\\begin{gathered} \\text { Pb } \\\\ 710 \\\\ \\hline \\end{gathered}$ | $\\begin{gathered} \\text { Bi } \\\\ 800 \\end{gathered}$ | $\\begin{gathered} \\text { Po } \\\\ 810 \\end{gathered}$ | At | $\\begin{gathered} \\mathrm{Rn} \\\\ 1030 \\\\ \\hline \\end{gathered}$ |"}
{"id": 2525, "contents": "275. Variation in Ionization Energies - \n7 | Fr | Ra |\n| :---: | :---: |\n| $\\ldots$ | 510 |\n\nFIGURE 3.34 This version of the periodic table shows the first ionization energy ( $\\mathrm{IE}_{1}$ ), in $\\mathrm{kJ} / \\mathrm{mol}$, of selected elements.\n\nAnother deviation occurs as orbitals become more than one-half filled. The first ionization energy for oxygen is slightly less than that for nitrogen, despite the trend in increasing $I E_{1}$ values across a period. Looking at the orbital diagram of oxygen, we can see that removing one electron will eliminate the electron-electron repulsion caused by pairing the electrons in the $2 p$ orbital and will result in a half-filled orbital (which is energetically favorable). Analogous changes occur in succeeding periods (note the dip for sulfur after phosphorus in Figure 3.34).\n\n\nRemoving an electron from a cation is more difficult than removing an electron from a neutral atom because of the greater electrostatic attraction to the cation. Likewise, removing an electron from a cation with a higher positive charge is more difficult than removing an electron from an ion with a lower charge. Thus, successive ionization energies for one element always increase. As seen in Table 3.3, there is a large increase in the ionization energies for each element. This jump corresponds to removal of the core electrons, which are harder to remove than the valence electrons. For example, Sc and Ga both have three valence electrons, so the rapid increase in ionization energy occurs after the third ionization.\n\nSuccessive Ionization Energies for Selected Elements (kJ/mol)"}
{"id": 2526, "contents": "275. Variation in Ionization Energies - \nSuccessive Ionization Energies for Selected Elements (kJ/mol)\n\n| Element | $\\mathrm{IE}_{1}$ | $\\mathrm{IE}_{2}$ | $\\mathrm{IE}_{3}$ | $\\mathrm{IE}_{4}$ | $\\mathrm{IE}_{5}$ | $\\mathrm{IE}_{6}$ | $\\mathrm{IE}_{7}$ |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :--- |\n| K | 418.8 | 3051.8 | 4419.6 | 5876.9 | 7975.5 | 9590.6 | 11343 |\n| Ca | 589.8 | 1145.4 | 4912.4 | 6490.6 | 8153.0 | 10495.7 | 12272.9 |\n| Sc | 633.1 | 1235.0 | 2388.7 | 7090.6 | 8842.9 | 10679.0 | 13315.0 |\n| Ga | 578.8 | 1979.4 | 2964.6 | 6180 | 8298.7 | 10873.9 | 13594.8 |\n\nTABLE 3.3\n\n| Element | $\\mathbf{I E}_{\\mathbf{1}}$ | $\\mathbf{I E}_{\\mathbf{2}}$ | $\\mathbf{I E}_{\\mathbf{3}}$ | $\\mathbf{I E}_{\\mathbf{4}}$ | $\\mathbf{I E}_{\\mathbf{5}}$ | $\\mathbf{I E}_{6}$ | $\\mathbf{I E}_{\\mathbf{7}}$ |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| Ge | 762.2 | 1537.5 | 3302.1 | 4410.6 | 9021.4 | Not available | Not available |\n| As | 944.5 | 1793.6 | 2735.5 | 4836.8 | 6042.9 | 12311.5 | Not available |"}
{"id": 2527, "contents": "275. Variation in Ionization Energies - \nTABLE 3.3"}
{"id": 2528, "contents": "277. Ranking Ionization Energies - \nPredict the order of increasing energy for the following processes: $\\mathrm{IE}_{1}$ for $\\mathrm{Al}, \\mathrm{IE}_{1}$ for $\\mathrm{Tl}, \\mathrm{IE}_{2}$ for $\\mathrm{Na}, \\mathrm{IE}_{3}$ for Al ."}
{"id": 2529, "contents": "278. Solution - \nRemoving the $6 p^{1}$ electron from Tl is easier than removing the $3 p^{1}$ electron from Al because the higher $n$ orbital is farther from the nucleus, so $\\mathrm{IE}_{1}(\\mathrm{Tl})<\\mathrm{IE}_{1}(\\mathrm{Al})$. Ionizing the third electron from\n$\\mathrm{Al} \\quad\\left(\\mathrm{Al}^{2+} \\longrightarrow \\mathrm{Al}^{3+}+\\mathrm{e}^{-}\\right)$requires more energy because the cation $\\mathrm{Al}^{2+}$ exerts a stronger pull on the electron than the neutral Al atom, so $\\mathrm{IE}_{1}(\\mathrm{Al})<\\mathrm{IE}_{3}(\\mathrm{Al})$. The second ionization energy for sodium removes a core electron, which is a much higher energy process than removing valence electrons. Putting this all together, we obtain: $\\mathrm{IE}_{1}(\\mathrm{Tl})<\\mathrm{IE}_{1}(\\mathrm{Al})<\\mathrm{IE}_{3}(\\mathrm{Al})<\\mathrm{IE}_{2}(\\mathrm{Na})$."}
{"id": 2530, "contents": "279. Check Your Learning - \nWhich has the lowest value for $\\mathrm{IE}_{1}$ : $\\mathrm{O}, \\mathrm{Po}, \\mathrm{Pb}$, or Ba ?"}
{"id": 2531, "contents": "280. Answer: - \nBa"}
{"id": 2532, "contents": "281. Variation in Electron Affinities - \nThe electron affinity (EA) is the energy change for the process of adding an electron to a gaseous atom to form an anion (negative ion).\n\n$$\n\\mathrm{X}(g)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{X}^{-}(g) \\quad \\mathrm{EA}_{1}\n$$\n\nThis process can be either endothermic or exothermic, depending on the element. The EA of some of the elements is given in Figure 3.35. You can see that many of these elements have negative values of EA, which means that energy is released when the gaseous atom accepts an electron. However, for some elements, energy is required for the atom to become negatively charged and the value of their EA is positive. Just as with ionization energy, subsequent EA values are associated with forming ions with more charge. The second EA is the energy associated with adding an electron to an anion to form a -2 ion, and so on.\n\nAs we might predict, it becomes easier to add an electron across a series of atoms as the effective nuclear charge of the atoms increases. We find, as we go from left to right across a period, EAs tend to become more negative. The exceptions found among the elements of group 2 (2A), group 15 ( 5 A ), and group 18 (8A) can be understood based on the electronic structure of these groups. The noble gases, group 18 (8A), have a completely filled shell and the incoming electron must be added to a higher $n$ level, which is more difficult to do. Group $2(2 \\mathrm{~A})$ has a filled $n s$ subshell, and so the next electron added goes into the higher energy $n p$, so, again, the observed EA value is not as the trend would predict. Finally, group 15 (5A) has a half-filled $n p$ subshell and the next electron must be paired with an existing $n p$ electron. In all of these cases, the initial relative stability of the electron configuration disrupts the trend in EA."}
{"id": 2533, "contents": "281. Variation in Electron Affinities - \nWe also might expect the atom at the top of each group to have the most negative EA; their first ionization potentials suggest that these atoms have the largest effective nuclear charges. However, as we move down a group, we see that the second element in the group most often has the most negative EA. This can be attributed\nto the small size of the $n=2$ shell and the resulting large electron-electron repulsions. For example, chlorine, with an EA value of $-348 \\mathrm{~kJ} / \\mathrm{mol}$, has the highest value of any element in the periodic table. The EA of fluorine is $-322 \\mathrm{~kJ} / \\mathrm{mol}$. When we add an electron to a fluorine atom to form a fluoride anion ( $\\mathrm{F}^{-}$), we add an electron to the $n=2$ shell. The electron is attracted to the nucleus, but there is also significant repulsion from the other electrons already present in this small valence shell. The chlorine atom has the same electron configuration in the valence shell, but because the entering electron is going into the $n=3$ shell, it occupies a considerably larger region of space and the electron-electron repulsions are reduced. The entering electron does not experience as much repulsion and the chlorine atom accepts an additional electron more readily, resulting in a more negative EA.\n\n\nFIGURE 3.35 This version of the periodic table displays the electron affinity values (in $\\mathrm{kJ} / \\mathrm{mol}$ ) for selected elements.\n\nThe properties discussed in this section (size of atoms and ions, effective nuclear charge, ionization energies, and electron affinities) are central to understanding chemical reactivity. For example, because fluorine has an energetically favorable EA and a large energy barrier to ionization (IE), it is much easier to form fluorine anions than cations. Metallic properties including conductivity and malleability (the ability to be formed into sheets) depend on having electrons that can be removed easily. Thus, metallic character increases as we move down a group and decreases across a period in the same trend observed for atomic size because it is easier to remove an electron that is farther away from the nucleus."}
{"id": 2534, "contents": "282. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- State the periodic law and explain the organization of elements in the periodic table\n- Predict the general properties of elements based on their location within the periodic table\n- Identify metals, nonmetals, and metalloids by their properties and/or location on the periodic table\n\nAs early chemists worked to purify ores and discovered more elements, they realized that various elements could be grouped together by their similar chemical behaviors. One such grouping includes lithium (Li), sodium ( Na ), and potassium (K): These elements all are shiny, conduct heat and electricity well, and have similar chemical properties. A second grouping includes calcium (Ca), strontium (Sr), and barium (Ba), which also are shiny, good conductors of heat and electricity, and have chemical properties in common. However, the specific properties of these two groupings are notably different from each other. For example: $\\mathrm{Li}, \\mathrm{Na}$, and K are much more reactive than are $\\mathrm{Ca}, \\mathrm{Sr}$, and Ba ; $\\mathrm{Li}, \\mathrm{Na}$, and K form compounds with oxygen in a ratio of two of their atoms to one oxygen atom, whereas $\\mathrm{Ca}, \\mathrm{Sr}$, and Ba form compounds with one of their atoms to one oxygen\natom. Fluorine ( F ), chlorine ( Cl ), bromine ( Br ), and iodine (I) also exhibit similar properties to each other, but these properties are drastically different from those of any of the elements above.\n\nDimitri Mendeleev in Russia (1869) and Lothar Meyer in Germany (1870) independently recognized that there was a periodic relationship among the properties of the elements known at that time. Both published tables with the elements arranged according to increasing atomic mass. But Mendeleev went one step further than Meyer: He used his table to predict the existence of elements that would have the properties similar to aluminum and silicon, but were yet unknown. The discoveries of gallium (1875) and germanium (1886) provided great support for Mendeleev's work. Although Mendeleev and Meyer had a long dispute over priority, Mendeleev's contributions to the development of the periodic table are now more widely recognized (Figure 3.36)."}
{"id": 2535, "contents": "282. LEARNING OBJECTIVES - \nFIGURE 3.36 (a) Dimitri Mendeleev is widely credited with creating (b) the first periodic table of the elements. (credit a: modification of work by Serge Lachinov; credit b: modification of work by \"Den fj\u00e4ttrade ankan\"/Wikimedia Commons)\n\nBy the twentieth century, it became apparent that the periodic relationship involved atomic numbers rather than atomic masses. The modern statement of this relationship, the periodic law, is as follows: the properties of the elements are periodic functions of their atomic numbers. A modern periodic table arranges the elements in increasing order of their atomic numbers and groups atoms with similar properties in the same vertical column (Figure 3.37). Each box represents an element and contains its atomic number, symbol, average atomic mass, and (sometimes) name. The elements are arranged in seven horizontal rows, called periods or series, and 18 vertical columns, called groups. Groups are labeled at the top of each column. In the United States, the labels traditionally were numerals with capital letters. However, IUPAC recommends that the numbers 1 through 18 be used, and these labels are more common. For the table to fit on a single page, parts of two of the rows, a total of 14 columns, are usually written below the main body of the table."}
{"id": 2536, "contents": "283. Periodic Table of the Elements - \nFIGURE 3.37 Elements in the periodic table are organized according to their properties.\nEven after the periodic nature of elements and the table itself were widely accepted, gaps remained. Mendeleev had predicted, and others including Henry Moseley had later confirmed, that there should be elements below Manganese in Group 7. German chemists Ida Tacke and Walter Noddack set out to find the elements, a quest being pursued by scientists around the world. Their method was unique in that they did not only consider the properties of manganese, but also the elements horizontally adjacent to the missing elements 43 and 75 on the table. Thus, by investigating ores containing minerals of ruthenium (Ru), tungsten (W), osmium (Os), and so on, they were able to identify naturally occurring elements that helped complete the table. Rhenium, one of their discoveries, was one of the last natural elements to be discovered and is the last stable element to be discovered. (Francium, the last natural element to be discovered, was identified by Marguerite Perey in 1939.)\n\nMany elements differ dramatically in their chemical and physical properties, but some elements are similar in their behaviors. For example, many elements appear shiny, are malleable (able to be deformed without breaking) and ductile (can be drawn into wires), and conduct heat and electricity well. Other elements are not shiny, malleable, or ductile, and are poor conductors of heat and electricity. We can sort the elements into large classes with common properties: metals (elements that are shiny, malleable, good conductors of heat and electricity-shaded yellow); nonmetals (elements that appear dull, poor conductors of heat and electricity-shaded green); and metalloids (elements that conduct heat and electricity moderately well, and possess some properties of metals and some properties of nonmetals-shaded purple)."}
{"id": 2537, "contents": "283. Periodic Table of the Elements - \nThe elements can also be classified into the main-group elements (or representative elements) in the columns labeled 1, 2, and 13-18; the transition metals in the columns labeled $3-12 ;^{\\frac{1}{n}}$ and inner transition metals in the two rows at the bottom of the table (the top-row elements are called lanthanides and the bottomrow elements are actinides; Figure 3.38). The elements can be subdivided further by more specific properties, such as the composition of the compounds they form. For example, the elements in group 1 (the first column) form compounds that consist of one atom of the element and one atom of hydrogen. These elements (except hydrogen) are known as alkali metals, and they all have similar chemical properties. The elements in group 2 (the second column) form compounds consisting of one atom of the element and two atoms of hydrogen: These are called alkaline earth metals, with similar properties among members of that group. Other groups with specific names are the pnictogens (group 15), chalcogens (group 16), halogens (group 17), and the noble gases (group 18, also known as inert gases). The groups can also be referred to by the first element of the group: For example, the chalcogens can be called the oxygen group or oxygen family. Hydrogen is a unique, nonmetallic element with properties similar to both group 1 and group 17 elements. For that reason, hydrogen may be shown at the top of both groups, or by itself.\n\n\nFIGURE 3.38 The periodic table organizes elements with similar properties into groups."}
{"id": 2538, "contents": "284. LINK TO LEARNING - \nClick on this link (http://openstax.org/l/16Periodic) for an interactive periodic table, which you can use to explore the properties of the elements (includes podcasts and videos of each element). You may also want to try this one (http://openstax.org/1/16Periodic2) that shows photos of all the elements."}
{"id": 2539, "contents": "286. Naming Groups of Elements - \nAtoms of each of the following elements are essential for life. Give the group name for the following elements:\n(a) chlorine\n\n^2 calcium\n(c) sodium\n(d) sulfur"}
{"id": 2540, "contents": "287. Solution - \nThe family names are as follows:\n(a) halogen\n(b) alkaline earth metal\n(c) alkali metal\n(d) chalcogen"}
{"id": 2541, "contents": "288. Check Your Learning - \nGive the group name for each of the following elements:\n(a) krypton\n(b) selenium\n(c) barium\n(d) lithium"}
{"id": 2542, "contents": "289. Answer: - \n(a) noble gas; (b) chalcogen; (c) alkaline earth metal; (d) alkali metal\n\nAs you will learn in your further study of chemistry, elements in groups often behave in a somewhat similar manner. This is partly due to the number of electrons in their outer shell and their similar readiness to bond. These shared properties can have far-ranging implications in nature, science, and medicine. For example, when Gertrude Elion and George Hitchens were investigating ways to interrupt cell and virus replication to fight diseases, they utilized the similarity between sulfur and oxygen (both in Group 16) and their capacity to bond in similar ways. Elion focused on purines, which are key components of DNA and which contain oxygen. She found that by introducing sulfur-based compounds (called purine analogues) that mimic the structure of purines, molecules within DNA would bond to the analogues rather than the \"regular\" DNA purine. With the normal DNA bonding and structure altered, Elion successfully interrupted cell replication. At its core, the strategy worked because of the similarity between sulfur and oxygen. Her discovery led directly to important treatments for leukemia. Overall, Elion's work with George Hitchens not only led to more treatments, but also changed the entire methodology of drug development. By using specific elements and compounds to target specific aspects of tumor cells, viruses, and bacteria, they laid the groundwork for many of today's most common and important medicines, used to help millions of people each year. They were awarded the Nobel Prize in 1988.\n\nIn studying the periodic table, you might have noticed something about the atomic masses of some of the elements. Element 43 (technetium), element 61 (promethium), and most of the elements with atomic number 84 (polonium) and higher have their atomic mass given in square brackets. This is done for elements that consist entirely of unstable, radioactive isotopes (you will learn more about radioactivity in the nuclear chemistry chapter). An average atomic weight cannot be determined for these elements because their radioisotopes may vary significantly in relative abundance, depending on the source, or may not even exist in nature. The number in square brackets is the atomic mass number (and approximate atomic mass) of the most stable isotope of that element."}
{"id": 2543, "contents": "290. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Define ionic and molecular (covalent) compounds\n- Predict the type of compound formed from elements based on their location within the periodic table\n- Determine formulas for simple ionic compounds\n\nIn ordinary chemical reactions, the nucleus of each atom (and thus the identity of the element) remains unchanged. Electrons, however, can be added to atoms by transfer from other atoms, lost by transfer to other atoms, or shared with other atoms. The transfer and sharing of electrons among atoms govern the chemistry of the elements. During the formation of some compounds, atoms gain or lose electrons, and form electrically charged particles called ions (Figure 3.39).\n\n\nFIGURE 3.39 (a) A sodium atom ( Na ) has equal numbers of protons and electrons (11) and is uncharged. (b) A sodium cation $\\left(\\mathrm{Na}^{+}\\right)$has lost an electron, so it has one more proton (11) than electrons (10), giving it an overall positive charge, signified by a superscripted plus sign.\n\nYou can use the periodic table to predict whether an atom will form an anion or a cation, and you can often predict the charge of the resulting ion. Atoms of many main-group metals lose enough electrons to leave them with the same number of electrons as an atom of the preceding noble gas. To illustrate, an atom of an alkali metal (group 1) loses one electron and forms a cation with a 1+ charge; an alkaline earth metal (group 2) loses two electrons and forms a cation with a $2+$ charge, and so on. For example, a neutral calcium atom, with 20 protons and 20 electrons, readily loses two electrons. This results in a cation with 20 protons, 18 electrons, and a $2+$ charge. It has the same number of electrons as atoms of the preceding noble gas, argon, and is symbolized $\\mathrm{Ca}^{2+}$. The name of a metal ion is the same as the name of the metal atom from which it forms, $\\mathrm{so}_{\\mathrm{Ca}}{ }^{2+}$ is called a calcium ion."}
{"id": 2544, "contents": "290. LEARNING OBJECTIVES - \nWhen atoms of nonmetal elements form ions, they generally gain enough electrons to give them the same number of electrons as an atom of the next noble gas in the periodic table. Atoms of group 17 gain one electron and form anions with a 1- charge; atoms of group 16 gain two electrons and form ions with a 2- charge, and so on. For example, the neutral bromine atom, with 35 protons and 35 electrons, can gain one electron to provide it with 36 electrons. This results in an anion with 35 protons, 36 electrons, and a 1 - charge. It has the same number of electrons as atoms of the next noble gas, krypton, and is symbolized $\\mathrm{Br}^{-}$. (A discussion of the theory supporting the favored status of noble gas electron numbers reflected in these predictive rules for ion formation is provided in a later chapter of this text.)\n\nNote the usefulness of the periodic table in predicting likely ion formation and charge (Figure 3.40). Moving from the far left to the right on the periodic table, main-group elements tend to form cations with a charge equal to the group number. That is, group 1 elements form $1+$ ions; group 2 elements form $2+$ ions, and so on. Moving from the far right to the left on the periodic table, elements often form anions with a negative charge\nequal to the number of groups moved left from the noble gases. For example, group 17 elements (one group left of the noble gases) form 1-ions; group 16 elements (two groups left) form 2-ions, and so on. This trend can be used as a guide in many cases, but its predictive value decreases when moving toward the center of the periodic table. In fact, transition metals and some other metals often exhibit variable charges that are not predictable by their location in the table. For example, copper can form ions with a $1+$ or $2+$ charge, and iron can form ions with a $2+$ or $3+$ charge.\n\n\nFIGURE 3.40 Some elements exhibit a regular pattern of ionic charge when they form ions."}
{"id": 2545, "contents": "292. Composition of Ions - \nAn ion found in some compounds used as antiperspirants contains 13 protons and 10 electrons. What is its symbol?"}
{"id": 2546, "contents": "293. Solution - \nBecause the number of protons remains unchanged when an atom forms an ion, the atomic number of the element must be 13 . Knowing this lets us use the periodic table to identify the element as Al (aluminum). The Al atom has lost three electrons and thus has three more positive charges (13) than it has electrons (10). This is the aluminum cation, $\\mathrm{Al}^{3+}$."}
{"id": 2547, "contents": "294. Check Your Learning - \nGive the symbol and name for the ion with 34 protons and 36 electrons."}
{"id": 2548, "contents": "295. Answer: - \n$\\mathrm{Se}^{2-}$, the selenide ion"}
{"id": 2549, "contents": "297. Formation of Ions - \nMagnesium and nitrogen react to form an ionic compound. Predict which forms an anion, which forms a cation, and the charges of each ion. Write the symbol for each ion and name them."}
{"id": 2550, "contents": "298. Solution - \nMagnesium's position in the periodic table (group 2) tells us that it is a metal. Metals form positive ions (cations). A magnesium atom must lose two electrons to have the same number electrons as an atom of the previous noble gas, neon. Thus, a magnesium atom will form a cation with two fewer electrons than protons and a charge of $2+$. The symbol for the ion is $\\mathrm{Mg}^{2+}$, and it is called a magnesium ion.\n\nNitrogen's position in the periodic table (group 15) reveals that it is a nonmetal. Nonmetals form negative ions (anions). A nitrogen atom must gain three electrons to have the same number of electrons as an atom of the following noble gas, neon. Thus, a nitrogen atom will form an anion with three more electrons than protons and a charge of $3-$. The symbol for the ion is $\\mathrm{N}^{3-}$, and it is called a nitride ion."}
{"id": 2551, "contents": "299. Check Your Learning - \nAluminum and carbon react to form an ionic compound. Predict which forms an anion, which forms a cation, and the charges of each ion. Write the symbol for each ion and name them."}
{"id": 2552, "contents": "300. Answer: - \nAl will form a cation with a charge of $3+: \\mathrm{Al}^{3+}$, an aluminum ion. Carbon will form an anion with a charge of 4-: $\\mathrm{C}^{4-}$, a carbide ion.\n\nThe ions that we have discussed so far are called monatomic ions, that is, they are ions formed from only one atom. We also find many polyatomic ions. These ions, which act as discrete units, are electrically charged molecules (a group of bonded atoms with an overall charge). Some of the more important polyatomic ions are listed in Table 3.4. Oxyanions are polyatomic ions that contain one or more oxygen atoms. At this point in your study of chemistry, you should memorize the names, formulas, and charges of the most common polyatomic ions. Because you will use them repeatedly, they will soon become familiar.\n\nCommon Polyatomic Ions\n\n| Name | Formula | Related Acid | Formula |\n| :--- | :--- | :--- | :--- |\n| ammonium | $\\mathrm{NH}_{4}{ }^{+}$ |  |  |\n| hydronium | $\\mathrm{H}_{3} \\mathrm{O}^{+}$ |  |  |\n| peroxide | $\\mathrm{O}_{2}{ }^{2-}$ |  |  |\n| hydroxide | $\\mathrm{OH}^{-}$ |  | $\\mathrm{CH}_{3} \\mathrm{COOH}$ |\n| acetate | $\\mathrm{CH}_{3} \\mathrm{COO}^{-}$ | acetic acid | $\\mathrm{HCN}^{-}$ |\n| cyanide | $\\mathrm{N}_{3}^{-}$ | hydrocyanic acid | $\\mathrm{HCN}_{3}$ |\n| azide | $\\mathrm{CO}_{3}{ }^{2-}$ | carbonic acid | $\\mathrm{H}_{2} \\mathrm{CO} 3_{3}$ |\n| carbonate |  |  |  |\n\nTABLE 3.4"}
{"id": 2553, "contents": "300. Answer: - \n| Name | Formula | Related Acid | Formula |\n| :---: | :---: | :---: | :---: |\n| bicarbonate | $\\mathrm{HCO}_{3}{ }^{-}$ |  |  |\n| nitrate | $\\mathrm{NO}_{3}{ }^{-}$ | nitric acid | $\\mathrm{HNO}_{3}$ |\n| nitrite | $\\mathrm{NO}_{2}{ }^{-}$ | nitrous acid | $\\mathrm{HNO}_{2}$ |\n| sulfate | $\\mathrm{SO}_{4}{ }^{2-}$ | sulfuric acid | $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ |\n| hydrogen sulfate | $\\mathrm{HSO}_{4}{ }^{-}$ |  |  |\n| sulfite | $\\mathrm{SO}_{3}{ }^{2-}$ | sulfurous acid | $\\mathrm{H}_{2} \\mathrm{SO}_{3}$ |\n| hydrogen sulfite | $\\mathrm{HSO}_{3}{ }^{-}$ |  |  |\n| phosphate | $\\mathrm{PO}_{4}{ }^{3-}$ | phosphoric acid | $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ |\n| hydrogen phosphate | $\\mathrm{HPO}_{4}{ }^{2-}$ |  |  |\n| dihydrogen phosphate | $\\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{-}$ |  |  |\n| perchlorate | $\\mathrm{ClO}_{4}{ }^{-}$ | perchloric acid | $\\mathrm{HClO}_{4}$ |\n| chlorate | $\\mathrm{ClO}_{3}{ }^{-}$ | chloric acid | $\\mathrm{HClO}_{3}$ |\n| chlorite | $\\mathrm{ClO}_{2}{ }^{-}$ | chlorous acid | $\\mathrm{HClO}_{2}$ |\n| hypochlorite | $\\mathrm{ClO}^{-}$ | hypochlorous acid | HClO |\n| chromate | $\\mathrm{CrO}_{4}{ }^{2-}$ | chromic acid | $\\mathrm{H}_{2} \\mathrm{CrO}_{4}$ |"}
{"id": 2554, "contents": "300. Answer: - \n| chromate | $\\mathrm{CrO}_{4}{ }^{2-}$ | chromic acid | $\\mathrm{H}_{2} \\mathrm{CrO}_{4}$ |\n| dichromate | $\\mathrm{Cr}_{2} \\mathrm{O}_{7}{ }^{2-}$ | dichromic acid | $\\mathrm{H}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$ |\n| permanganate | $\\mathrm{MnO}_{4}{ }^{-}$ | permanganic acid | $\\mathrm{HMnO}_{4}$ |"}
{"id": 2555, "contents": "300. Answer: - \nTABLE 3.4\n\nNote that there is a system for naming some polyatomic ions; -ate and -ite are suffixes designating polyatomic ions containing more or fewer oxygen atoms. Per- (short for \"hyper\") and hypo- (meaning \"under\") are prefixes meaning more oxygen atoms than -ate and fewer oxygen atoms than -ite, respectively. For example, perchlorate is $\\mathrm{ClO}_{4}^{-}$, chlorate is $\\mathrm{ClO}_{3}{ }^{-}$, chlorite is $\\mathrm{ClO}_{2}{ }^{-}$and hypochlorite is $\\mathrm{ClO}^{-}$. Unfortunately, the number of oxygen atoms corresponding to a given suffix or prefix is not consistent; for example, nitrate is $\\mathrm{NO}_{3}{ }^{-}$while sulfate is $\\mathrm{SO}_{4}{ }^{2-}$. This will be covered in more detail later in the module on nomenclature.\n\nThe nature of the attractive forces that hold atoms or ions together within a compound is the basis for classifying chemical bonding. When electrons are transferred and ions form, ionic bonds result. Ionic bonds are electrostatic forces of attraction, that is, the attractive forces experienced between objects of opposite electrical charge (in this case, cations and anions). When electrons are \"shared\" and molecules form, covalent bonds result. Covalent bonds are the attractive forces between the positively charged nuclei of the bonded atoms and one or more pairs of electrons that are located between the atoms. Compounds are classified as ionic or molecular (covalent) on the basis of the bonds present in them."}
{"id": 2556, "contents": "301. Ionic Compounds - \nWhen an element composed of atoms that readily lose electrons (a metal) reacts with an element composed of atoms that readily gain electrons (a nonmetal), a transfer of electrons usually occurs, producing ions. The compound formed by this transfer is stabilized by the electrostatic attractions (ionic bonds) between the ions of opposite charge present in the compound. For example, when each sodium atom in a sample of sodium metal (group 1) gives up one electron to form a sodium cation, $\\mathrm{Na}^{+}$, and each chlorine atom in a sample of chlorine gas (group 17) accepts one electron to form a chloride anion, $\\mathrm{Cl}^{-}$, the resulting compound, NaCl , is composed of sodium ions and chloride ions in the ratio of one $\\mathrm{Na}^{+}$ion for each $\\mathrm{Cl}^{-}$ion. Similarly, each calcium atom (group 2) can give up two electrons and transfer one to each of two chlorine atoms to form $\\mathrm{CaCl}_{2}$, which is composed of $\\mathrm{Ca}^{2+}$ and $\\mathrm{Cl}^{-}$ions in the ratio of one $\\mathrm{Ca}^{2+}$ ion to two $\\mathrm{Cl}^{-}$ions.\n\nA compound that contains ions and is held together by ionic bonds is called an ionic compound. The periodic table can help us recognize many of the compounds that are ionic: When a metal is combined with one or more nonmetals, the compound is usually ionic. This guideline works well for predicting ionic compound formation for most of the compounds typically encountered in an introductory chemistry course. However, it is not always true (for example, aluminum chloride, $\\mathrm{AlCl}_{3}$, is not ionic)."}
{"id": 2557, "contents": "301. Ionic Compounds - \nYou can often recognize ionic compounds because of their properties. Ionic compounds are solids that typically melt at high temperatures and boil at even higher temperatures. For example, sodium chloride melts at $801{ }^{\\circ} \\mathrm{C}$ and boils at $1413^{\\circ} \\mathrm{C}$. (As a comparison, the molecular compound water melts at $0^{\\circ} \\mathrm{C}$ and boils at 100 ${ }^{\\circ} \\mathrm{C}$.) In solid form, an ionic compound is not electrically conductive because its ions are unable to flow (\"electricity\" is the flow of charged particles). When molten, however, it can conduct electricity because its ions are able to move freely through the liquid (Figure 3.41).\n\n\nFIGURE 3.41 Sodium chloride melts at $801{ }^{\\circ} \\mathrm{C}$ and conducts electricity when molten. (credit: modification of work by Mark Blaser and Matt Evans)"}
{"id": 2558, "contents": "302. LINK TO LEARNING - \nWatch this video (http://openstax.org/l/16moltensalt) to see a mixture of salts melt and conduct electricity.\n\nIn every ionic compound, the total number of positive charges of the cations equals the total number of negative charges of the anions. Thus, ionic compounds are electrically neutral overall, even though they contain positive and negative ions. We can use this observation to help us write the formula of an ionic compound. The formula of an ionic compound must have a ratio of ions such that the numbers of positive and negative charges are equal.\ncations, $\\mathrm{Al}^{3+}$, and oxygen anions, $\\mathrm{O}^{2-}$. What is the formula of this compound?\n\n\nFIGURE 3.42 Although pure aluminum oxide is colorless, trace amounts of iron and titanium give blue sapphire its characteristic color. (credit: modification of work by Stanislav Doronenko)"}
{"id": 2559, "contents": "303. Solution - \nBecause the ionic compound must be electrically neutral, it must have the same number of positive and negative charges. Two aluminum ions, each with a charge of $3+$, would give us six positive charges, and three oxide ions, each with a charge of $2-$, would give us six negative charges. The formula would be $\\mathrm{Al}_{2} \\mathrm{O}_{3}$."}
{"id": 2560, "contents": "304. Check Your Learning - \nPredict the formula of the ionic compound formed between the sodium cation, $\\mathrm{Na}^{+}$, and the sulfide anion, $\\mathrm{S}^{2-}$."}
{"id": 2561, "contents": "305. Answer: - \n$\\mathrm{Na}_{2} \\mathrm{~S}$\n\nMany ionic compounds contain polyatomic ions (Table 3.4) as the cation, the anion, or both. As with simple ionic compounds, these compounds must also be electrically neutral, so their formulas can be predicted by treating the polyatomic ions as discrete units. We use parentheses in a formula to indicate a group of atoms that behave as a unit. For example, the formula for calcium phosphate, one of the minerals in our bones, is $\\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$. This formula indicates that there are three calcium ions $\\left(\\mathrm{Ca}^{2+}\\right)$ for every two phosphate $\\left(\\mathrm{PO}_{4}{ }^{3-}\\right)$ groups. The $\\mathrm{PO}_{4}{ }^{3-}$ groups are discrete units, each consisting of one phosphorus atom and four oxygen atoms, and having an overall charge of $3-$. The compound is electrically neutral, and its formula shows a total count of three Ca , two P , and eight O atoms."}
{"id": 2562, "contents": "307. Predicting the Formula of a Compound with a Polyatomic Anion - \nBaking powder contains calcium dihydrogen phosphate, an ionic compound composed of the ions $\\mathrm{Ca}^{2+}$ and $\\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{-}$. What is the formula of this compound?"}
{"id": 2563, "contents": "308. Solution - \nThe positive and negative charges must balance, and this ionic compound must be electrically neutral. Thus, we must have two negative charges to balance the $2+$ charge of the calcium ion. This requires a ratio of one $\\mathrm{Ca}^{2+}$ ion to two $\\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{-}$ions. We designate this by enclosing the formula for the dihydrogen phosphate ion in parentheses and adding a subscript 2. The formula is $\\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$."}
{"id": 2564, "contents": "309. Check Your Learning - \nPredict the formula of the ionic compound formed between the lithium ion and the peroxide ion, $\\mathrm{O}_{2}{ }^{2-}$ (Hint: Use the periodic table to predict the sign and the charge on the lithium ion.)"}
{"id": 2565, "contents": "310. Answer: - \n$\\mathrm{Li}_{2} \\mathrm{O}_{2}$\n\nBecause an ionic compound is not made up of single, discrete molecules, it may not be properly symbolized using a molecular formula. Instead, ionic compounds must be symbolized by a formula indicating the relative\nnumbers of its constituent ions. For compounds containing only monatomic ions (such as NaCl ) and for many compounds containing polyatomic ions (such as $\\mathrm{CaSO}_{4}$ ), these formulas are just the empirical formulas introduced earlier. However, the formulas for some ionic compounds containing polyatomic ions are not empirical formulas. For example, the ionic compound sodium oxalate is comprised of $\\mathrm{Na}^{+}$and $\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}$ ions combined in a $2: 1$ ratio, and its formula is written as $\\mathrm{Na}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$. The subscripts in this formula are not the smallest-possible whole numbers, as each can be divided by 2 to yield the empirical formula, $\\mathrm{NaCO}_{2}$. This is not the accepted formula for sodium oxalate, however, as it does not accurately represent the compound's polyatomic anion, $\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}$."}
{"id": 2566, "contents": "311. Molecular Compounds - \nMany compounds do not contain ions but instead consist solely of discrete, neutral molecules. These molecular compounds (covalent compounds) result when atoms share, rather than transfer (gain or lose), electrons. Covalent bonding is an important and extensive concept in chemistry, and it will be treated in considerable detail in a later chapter of this text. We can often identify molecular compounds on the basis of their physical properties. Under normal conditions, molecular compounds often exist as gases, low-boiling liquids, and low-melting solids, although many important exceptions exist.\n\nWhereas ionic compounds are usually formed when a metal and a nonmetal combine, covalent compounds are usually formed by a combination of nonmetals. Thus, the periodic table can help us recognize many of the compounds that are covalent. While we can use the positions of a compound's elements in the periodic table to predict whether it is ionic or covalent at this point in our study of chemistry, you should be aware that this is a very simplistic approach that does not account for a number of interesting exceptions. Shades of gray exist between ionic and molecular compounds, and you'll learn more about those later."}
{"id": 2567, "contents": "313. Predicting the Type of Bonding in Compounds - \nPredict whether the following compounds are ionic or molecular:\n(a) KI, the compound used as a source of iodine in table salt\n(b) $\\mathrm{H}_{2} \\mathrm{O}_{2}$, the bleach and disinfectant hydrogen peroxide\n(c) $\\mathrm{CHCl}_{3}$, the anesthetic chloroform\n(d) $\\mathrm{Li}_{2} \\mathrm{CO}_{3}$, a source of lithium in antidepressants"}
{"id": 2568, "contents": "314. Solution - \n(a) Potassium (group 1) is a metal, and iodine (group 17) is a nonmetal; KI is predicted to be ionic.\n(b) Hydrogen (group 1) is a nonmetal, and oxygen (group 16) is a nonmetal; $\\mathrm{H}_{2} \\mathrm{O}_{2}$ is predicted to be molecular.\n(c) Carbon (group 14) is a nonmetal, hydrogen (group 1) is a nonmetal, and chlorine (group 17) is a nonmetal; $\\mathrm{CHCl}_{3}$ is predicted to be molecular.\n(d) Lithium (group 1) is a metal, and carbonate is a polyatomic ion; $\\mathrm{Li}_{2} \\mathrm{CO}_{3}$ is predicted to be ionic."}
{"id": 2569, "contents": "315. Check Your Learning - \nUsing the periodic table, predict whether the following compounds are ionic or covalent:\n(a) $\\mathrm{SO}_{2}$\n(b) $\\mathrm{CaF}_{2}$\n(c) $\\mathrm{N}_{2} \\mathrm{H}_{4}$\n(d) $\\mathrm{Al}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}$\n\nAnswer:\n(a) molecular; (b) ionic; (c) molecular; (d) ionic"}
{"id": 2570, "contents": "316. Key Terms - \nactinide inner transition metal in the bottom of the bottom two rows of the periodic table\nalkali metal element in group 1\nalkaline earth metal element in group 2\namplitude extent of the displacement caused by a wave\natomic orbital mathematical function that describes the behavior of an electron in an atom (also called the wavefunction)\nAufbau principle procedure in which the electron configuration of the elements is determined by \"building\" them in order of atomic numbers, adding one proton to the nucleus and one electron to the proper subshell at a time\nblackbody idealized perfect absorber of all incident electromagnetic radiation; such bodies emit electromagnetic radiation in characteristic continuous spectra called blackbody radiation\nBohr's model of the hydrogen atom structural model in which an electron moves around the nucleus only in circular orbits, each with a specific allowed radius\nchalcogen element in group 16\ncontinuous spectrum electromagnetic radiation given off in an unbroken series of wavelengths (e.g., white light from the sun)\ncore electron electron in an atom that occupies the orbitals of the inner shells\ncovalent bond attractive force between the nuclei of a molecule's atoms and pairs of electrons between the atoms\ncovalent compound (also, molecular compound) composed of molecules formed by atoms of two or more different elements\ncovalent radius one-half the distance between the nuclei of two identical atoms when they are joined by a covalent bond\n$\\boldsymbol{d}$ orbital region of space with high electron density that is either four lobed or contains a dumbbell and torus shape; describes orbitals with $l=2$.\ndegenerate orbitals orbitals that have the same energy\neffective nuclear charge charge that leads to the Coulomb force exerted by the nucleus on an electron, calculated as the nuclear charge minus shielding\nelectromagnetic radiation energy transmitted by waves that have an electric-field component and a magnetic-field component\nelectromagnetic spectrum range of energies that electromagnetic radiation can comprise, including radio, microwaves, infrared, visible,\nultraviolet, X-rays, and gamma rays\nelectron affinity energy change associated with addition of an electron to a gaseous atom or ion\nelectron configuration listing that identifies the electron occupancy of an atom's shells and subshells\nelectron density a measure of the probability of locating an electron in a particular region of space, it is equal to the squared absolute value of the wave function $\\psi$"}
{"id": 2571, "contents": "316. Key Terms - \nelectron density a measure of the probability of locating an electron in a particular region of space, it is equal to the squared absolute value of the wave function $\\psi$\nendothermic processes that increase the energy of an atom and involve the absorption of light\nexcited state state having an energy greater than the ground-state energy\nexothermic processes that decrease the energy of an atom and involve the emission of light\nforbital multilobed region of space with high electron density, describes orbitals with $1=3$\nfrequency ( $\\boldsymbol{\\nu}$ ) number of wave cycles (peaks or troughs) that pass a specified point in space per unit time\nground state state in which the electrons in an atom, ion, or molecule have the lowest energy possible\ngroup vertical column of the periodic table\nhalogen element in group 17\nHeisenberg uncertainty principle rule stating that it is impossible to exactly determine both certain conjugate dynamical properties such as the momentum and the position of a particle at the same time. The uncertainty principle is a consequence of quantum particles exhibiting wave-particle duality\nhertz (Hz) the unit of frequency, which is the number of cycles per second, $\\mathrm{s}^{-1}$\nHund's rule every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin\ninert gas (also, noble gas) element in group 18\ninner transition metal (also, lanthanide or actinide) element in the bottom two rows; if in the first row, also called lanthanide, or if in the second row, also called actinide\nintensity property of wave-propagated energy related to the amplitude of the wave, such as brightness of light or loudness of sound\ninterference pattern pattern typically consisting of alternating bright and dark fringes; it results from constructive and destructive interference of waves\nionic bond electrostatic forces of attraction\nbetween the oppositely charged ions of an ionic compound\nionic compound compound composed of cations and anions combined in ratios, yielding an electrically neutral substance\nionization energy energy required to remove an electron from a gaseous atom or ion\nisoelectronic group of ions or atoms that have identical electron configurations\nlanthanide inner transition metal in the top of the bottom two rows of the periodic table"}
{"id": 2572, "contents": "316. Key Terms - \nisoelectronic group of ions or atoms that have identical electron configurations\nlanthanide inner transition metal in the top of the bottom two rows of the periodic table\nline spectrum electromagnetic radiation emitted at discrete wavelengths by a specific atom (or atoms) in an excited state\nmagnetic quantum number ( $\\boldsymbol{m}_{\\boldsymbol{l}}$ ) quantum number signifying the orientation of an atomic orbital around the nucleus\nmain-group element (also, representative element) element in groups 1, 2, and 13-18\nmetal element that is shiny, malleable, good conductor of heat and electricity\nmetalloid element that conducts heat and electricity moderately well, and possesses some properties of metals and some properties of nonmetals\nmolecular compound (also, covalent compound) composed of molecules formed by atoms of two or more different elements\nmonatomic ion ion composed of a single atom\nnoble gas (also, inert gas) element in group 18\nnode any point of a standing wave with zero amplitude\nnonmetal element that appears dull, poor conductor of heat and electricity\norbital diagram pictorial representation of the electron configuration showing each orbital as a box and each electron as an arrow\noxyanion polyatomic anion composed of a central atom bonded to oxygen atoms\n$\\boldsymbol{p}$ orbital dumbbell-shaped region of space with high electron density, describes orbitals with $l=1$\nPauli exclusion principle specifies that no two electrons in an atom can have the same value for all four quantum numbers\nperiod (also, series) horizontal row of the periodic table\nperiodic law properties of the elements are periodic function of their atomic numbers\nperiodic table table of the elements that places elements with similar chemical properties close together\nphoton smallest possible packet of electromagnetic radiation, a particle of light\npnictogen element in group 15\npolyatomic ion ion composed of more than one atom\nprincipal quantum number ( $n$ ) quantum number specifying the shell an electron occupies in an atom\nquantization limitation of some property to specific discrete values, not continuous\nquantum mechanics field of study that includes quantization of energy, wave-particle duality, and the Heisenberg uncertainty principle to describe matter\nquantum number number having only specific allowed values and used to characterize the arrangement of electrons in an atom"}
{"id": 2573, "contents": "316. Key Terms - \nquantum number number having only specific allowed values and used to characterize the arrangement of electrons in an atom\nrepresentative element (also, main-group element) element in columns 1, 2, and 12-18\n$\\boldsymbol{s}$ orbital spherical region of space with high electron density, describes orbitals with $1=0$\nsecondary (angular momentum) quantum number (1) quantum number distinguishing the different shapes of orbitals; it is also a measure of the orbital angular momentum\nseries (also, period) horizontal row of the period table\nshell atomic orbitals with the same principal quantum number, $n$\nspin quantum number ( $\\boldsymbol{m}_{\\boldsymbol{s}}$ ) number specifying the electron spin direction, either $+\\frac{1}{2}$ or $-\\frac{1}{2}$\nstanding wave (also, stationary wave) localized wave phenomenon characterized by discrete wavelengths determined by the boundary conditions used to generate the waves; standing waves are inherently quantized\nsubshell atomic orbitals with the same values of $n$ and 1\ntransition metal element in groups 3-12 (more strictly defined, 3-11; see chapter on transition metals and coordination chemistry)\nvalence electrons electrons in the high energy outer shell(s) of an atom\nvalence shell high energy outer shell(s) of an atom\nwave oscillation of a property over time or space; can transport energy from one point to another\nwave-particle duality observation that elementary particles can exhibit both wave-like and particlelike properties\nwavefunction ( $\\boldsymbol{\\Psi}$ ) mathematical description of an atomic orbital that describes the shape of the orbital; it can be used to calculate the probability of finding the electron at any given location in the orbital, as well as dynamical variables such as the energy and the angular momentum\nwavelength ( $\\lambda$ ) distance between two consecutive"}
{"id": 2574, "contents": "317. Key Equations - \n$c=\\lambda \\nu$\n$E=h \\nu=\\frac{h c}{\\lambda}$, where $h=6.626 \\times 10^{-34} \\mathrm{~J} \\mathrm{~S}$\n$\\frac{1}{\\lambda}=R_{\\infty}\\left(\\frac{1}{n_{1}^{2}}-\\frac{1}{n_{2}^{2}}\\right)$\n$E_{n}=-\\frac{k Z^{2}}{n^{2}}, n=1,2,3, \\ldots$\n$\\Delta E=k Z^{2}\\left(\\frac{1}{n_{1}^{2}}-\\frac{1}{n_{2}^{2}}\\right)$\n$r=\\frac{n^{2}}{Z} a_{0}$"}
{"id": 2575, "contents": "318. Summary - 318.1. Electromagnetic Energy\nLight and other forms of electromagnetic radiation move through a vacuum with a constant speed, $c$, of $2.998 \\times 10^{8} \\mathrm{~m} \\mathrm{~s}^{-1}$. This radiation shows wavelike behavior, which can be characterized by a frequency, $\\nu$, and a wavelength, $\\lambda$, such that $c=\\lambda \\nu$. Light is an example of a travelling wave. Other important wave phenomena include standing waves, periodic oscillations, and vibrations. Standing waves exhibit quantization, since their wavelengths are limited to discrete integer multiples of some characteristic lengths. Electromagnetic radiation that passes through two closely spaced narrow slits having dimensions roughly similar to the wavelength will show an interference pattern that is a result of constructive and destructive interference of the waves. Electromagnetic radiation also demonstrates properties of particles called photons. The energy of a photon is related to the frequency (or alternatively, the wavelength) of the radiation as $E=h \\nu$ (or $E=\\frac{h c}{\\lambda}$ ), where $h$ is Planck's constant. That light demonstrates both wavelike and particle-like behavior is known as wave-particle duality. All forms of electromagnetic radiation share these properties, although various forms including X-rays, visible light, microwaves, and radio waves interact differently with matter and have very different practical applications. Electromagnetic radiation can be generated by exciting matter to higher energies, such as by heating it. The emitted light can be either continuous (incandescent sources like the sun) or discrete (from specific types of excited atoms). Continuous spectra often have distributions that can be approximated as blackbody radiation at some appropriate temperature. The line spectrum of hydrogen can be obtained by passing the light from\npeaks or troughs in a wave\npeaks or troughs in a wave\nan electrified tube of hydrogen gas through a prism. This line spectrum was simple enough that an empirical formula called the Rydberg formula could be derived from the spectrum. Three historically important paradoxes from the late 19th and early 20th centuries that could not be explained within the existing framework of classical mechanics and classical electromagnetism were the blackbody problem, the photoelectric effect, and the discrete spectra of atoms. The resolution of these paradoxes ultimately led to quantum theories that superseded the classical theories."}
{"id": 2576, "contents": "318. Summary - 318.2. The Bohr Model\nBohr incorporated Planck's and Einstein's quantization ideas into a model of the hydrogen atom that resolved the paradox of atom stability and discrete spectra. The Bohr model of the hydrogen atom explains the connection between the quantization of photons and the quantized emission from atoms. Bohr described the hydrogen atom in terms of an electron moving in a circular orbit about a nucleus. He postulated that the electron was restricted to certain orbits characterized by discrete energies. Transitions between these allowed orbits result in the absorption or emission of photons. When an electron moves from a higher-energy orbit to a more stable one, energy is emitted in the form of a photon. To move an electron from a stable orbit to a more excited one, a photon of energy must be absorbed. Using the Bohr model, we can calculate the energy of an electron and the radius of its orbit in any one-electron system."}
{"id": 2577, "contents": "318. Summary - 318.3. Development of Quantum Theory\nMacroscopic objects act as particles. Microscopic\nobjects (such as electrons) have properties of both a particle and a wave. Their exact trajectories cannot be determined. The quantum mechanical model of atoms describes the three-dimensional position of the electron in a probabilistic manner according to a mathematical function called a wavefunction, often denoted as $\\psi$. Atomic wavefunctions are also called orbitals. The squared magnitude of the wavefunction describes the distribution of the probability of finding the electron in a particular region in space. Therefore, atomic orbitals describe the areas in an atom where electrons are most likely to be found.\n\nAn atomic orbital is characterized by three quantum numbers. The principal quantum number, $n$, can be any positive integer. The general region for value of energy of the orbital and the average distance of an electron from the nucleus are related to $n$. Orbitals having the same value of $n$ are said to be in the same shell. The secondary (angular momentum) quantum number, $l$, can have any integer value from 0 to $n-1$. This quantum number describes the shape or type of the orbital. Orbitals with the same principal quantum number and the same $l$ value belong to the same subshell. The magnetic quantum number, $m_{l}$, with $2 l+1$ values ranging from $-l$ to $+l$, describes the orientation of the orbital in space. In addition, each electron has a spin quantum number, $m_{s}$, that can be equal to $\\pm \\frac{1}{2}$. No two electrons in the same atom can have the same set of values for all the four quantum numbers."}
{"id": 2578, "contents": "318. Summary - 318.4. Electronic Structure of Atoms (Electron Configurations)\nThe relative energy of the subshells determine the order in which atomic orbitals are filled ( $1 s, 2 s, 2 p$, $3 s, 3 p, 4 s, 3 d, 4 p$, and so on). Electron configurations and orbital diagrams can be determined by applying the Pauli exclusion principle (no two electrons can have the same set of four quantum numbers) and Hund's rule (whenever possible, electrons retain unpaired spins in degenerate orbitals).\n\nElectrons in the outermost orbitals, called valence electrons, are responsible for most of the chemical behavior of elements. In the periodic table, elements with analogous valence electron configurations usually occur within the same group. There are some exceptions to the predicted filling order, particularly when half-filled or completely filled orbitals can be formed. The periodic table can be divided into three categories based on the orbital in which the last electron to be added is placed: main group elements ( $s$ and $p$ orbitals), transition\nelements ( $d$ orbitals), and inner transition elements (forbitals)."}
{"id": 2579, "contents": "318. Summary - 318.5. Periodic Variations in Element Properties\nElectron configurations allow us to understand many periodic trends. Covalent radius increases as we move down a group because the $n$ level (orbital size) increases. Covalent radius mostly decreases as we move left to right across a period because the effective nuclear charge experienced by the electrons increases, and the electrons are pulled in tighter to the nucleus. Anionic radii are larger than the parent atom, while cationic radii are smaller, because the number of valence electrons has changed while the nuclear charge has remained constant. Ionization energy (the energy associated with forming a cation) decreases down a group and mostly increases across a period because it is easier to remove an electron from a larger, higher energy orbital. Electron affinity (the energy associated with forming an anion) is more favorable (exothermic) when electrons are placed into lower energy orbitals, closer to the nucleus. Therefore, electron affinity becomes increasingly negative as we move left to right across the periodic table and decreases as we move down a group. For both IE and electron affinity data, there are exceptions to the trends when dealing with completely filled or half-filled subshells."}
{"id": 2580, "contents": "318. Summary - 318.6. The Periodic Table\nThe discovery of the periodic recurrence of similar properties among the elements led to the formulation of the periodic table, in which the elements are arranged in order of increasing atomic number in rows known as periods and columns known as groups. Elements in the same group of the periodic table have similar chemical properties. Elements can be classified as metals, metalloids, and nonmetals, or as a main-group elements, transition metals, and inner transition metals. Groups are numbered 1-18 from left to right. The elements in group 1 are known as the alkali metals; those in group 2 are the alkaline earth metals; those in 15 are the pnictogens; those in 16 are the chalcogens; those in 17 are the halogens; and those in 18 are the noble gases."}
{"id": 2581, "contents": "318. Summary - 318.7. Ionic and Molecular Compounds\nMetals (particularly those in groups 1 and 2) tend to lose the number of electrons that would leave them with the same number of electrons as in the\npreceding noble gas in the periodic table. By this means, a positively charged ion is formed. Similarly, nonmetals (especially those in groups 16 and 17, and, to a lesser extent, those in Group 15) can gain the number of electrons needed to provide atoms with the same number of electrons as in the next noble gas in the periodic table. Thus, nonmetals tend to form negative ions. Positively charged ions are called cations, and negatively charged ions are called anions. Ions can be either monatomic (containing only one atom) or polyatomic\n(containing more than one atom).\nCompounds that contain ions are called ionic compounds. Ionic compounds generally form from metals and nonmetals. Compounds that do not contain ions, but instead consist of atoms bonded tightly together in molecules (uncharged groups of atoms that behave as a single unit), are called covalent compounds. Covalent compounds usually form from two nonmetals."}
{"id": 2582, "contents": "319. Exercises - 319.1. Electromagnetic Energy\n1. The light produced by a red neon sign is due to the emission of light by excited neon atoms. Qualitatively describe the spectrum produced by passing light from a neon lamp through a prism.\n2. An FM radio station found at 103.1 on the FM dial broadcasts at a frequency of $1.031 \\times 10^{8} \\mathrm{~s}^{-1}(103.1$ MHz ). What is the wavelength of these radio waves in meters?\n3. $\\mathrm{FM}-95$, an FM radio station, broadcasts at a frequency of $9.51 \\times 10^{7} \\mathrm{~s}^{-1}(95.1 \\mathrm{MHz})$. What is the wavelength of these radio waves in meters?\n4. A bright violet line occurs at 435.8 nm in the emission spectrum of mercury vapor. What amount of energy, in joules, must be released by an electron in a mercury atom to produce a photon of this light?\n5. Light with a wavelength of 614.5 nm looks orange. What is the energy, in joules, per photon of this orange light? What is the energy in $\\mathrm{eV}\\left(1 \\mathrm{eV}=1.602 \\times 10^{-19} \\mathrm{~J}\\right)$ ?\n6. Heated lithium atoms emit photons of light with an energy of $2.961 \\times 10^{-19} \\mathrm{~J}$. Calculate the frequency and wavelength of one of these photons. What is the total energy in 1 mole of these photons? What is the color of the emitted light?\n7. A photon of light produced by a surgical laser has an energy of $3.027 \\times 10^{-19} \\mathrm{~J}$. Calculate the frequency and wavelength of the photon. What is the total energy in 1 mole of photons? What is the color of the emitted light?\n8. When rubidium ions are heated to a high temperature, two lines are observed in its line spectrum at wavelengths (a) $7.9 \\times 10^{-7} \\mathrm{~m}$ and (b) $4.2 \\times 10^{-7} \\mathrm{~m}$. What are the frequencies of the two lines? What color do we see when we heat a rubidium compound?"}
{"id": 2583, "contents": "319. Exercises - 319.1. Electromagnetic Energy\n9. The emission spectrum of cesium contains two lines whose frequencies are (a) $3.45 \\times 10^{14} \\mathrm{~Hz}$ and (b) 6.53 $\\times 10^{14} \\mathrm{~Hz}$. What are the wavelengths and energies per photon of the two lines? What color are the lines?\n10. Photons of infrared radiation are responsible for much of the warmth we feel when holding our hands before a fire. These photons will also warm other objects. How many infrared photons with a wavelength of $1.5 \\times 10^{-6} \\mathrm{~m}$ must be absorbed by the water to warm a cup of water $(175 \\mathrm{~g})$ from $25.0^{\\circ} \\mathrm{C}$ to $40^{\\circ} \\mathrm{C}$ ?\n11. One of the radiographic devices used in a dentist's office emits an X-ray of wavelength $2.090 \\times 10^{-11} \\mathrm{~m}$. What is the energy, in joules, and frequency of this X-ray?\n12. The eyes of certain reptiles pass a single visual signal to the brain when the visual receptors are struck by photons of a wavelength of 850 nm . If a total energy of $3.15 \\times 10^{-14} \\mathrm{~J}$ is required to trip the signal, what is the minimum number of photons that must strike the receptor?\n13. RGB color television and computer displays use cathode ray tubes that produce colors by mixing red, green, and blue light. If we look at the screen with a magnifying glass, we can see individual dots turn on and off as the colors change. Using a spectrum of visible light, determine the approximate wavelength of each of these colors. What is the frequency and energy of a photon of each of these colors?\n14. Answer the following questions about a Blu-ray laser:\n(a) The laser on a Blu-ray player has a wavelength of 405 nm . In what region of the electromagnetic spectrum is this radiation? What is its frequency?\n(b) A Blu-ray laser has a power of 5 milliwatts ( 1 watt $=1 \\mathrm{~J} \\mathrm{~s}^{-1}$ ). How many photons of light are produced by the laser in 1 hour?"}
{"id": 2584, "contents": "319. Exercises - 319.1. Electromagnetic Energy\n(c) The ideal resolution of a player using a laser (such as a Blu-ray player), which determines how close together data can be stored on a compact disk, is determined using the following formula: Resolution = $0.60(\\lambda / N A)$, where $\\lambda$ is the wavelength of the laser and NA is the numerical aperture. Numerical aperture is a measure of the size of the spot of light on the disk; the larger the NA, the smaller the spot. In a typical Blu-ray system, NA $=0.95$. If the $405-\\mathrm{nm}$ laser is used in a Blu-ray player, what is the closest that information can be stored on a Blu-ray disk?\n(d) The data density of a Blu-ray disk using a $405-\\mathrm{nm}$ laser is $1.5 \\times 10^{7} \\mathrm{bits} \\mathrm{mm}^{-2}$. Disks have an outside diameter of 120 mm and a hole of $15-\\mathrm{mm}$ diameter. How many data bits can be contained on the disk? If a Blu-ray disk can hold 9,400,000 pages of text, how many data bits are needed for a typed page? (Hint: Determine the area of the disk that is available to hold data. The area inside a circle is given by $\\mathrm{A}=\\pi r^{2}$, where the radius $r$ is one-half of the diameter.)\n15. What is the threshold frequency for sodium metal if a photon with frequency $6.66 \\times 10^{14} \\mathrm{~s}^{-1}$ ejects an electron with $7.74 \\times 10^{-20} \\mathrm{~J}$ kinetic energy? Will the photoelectric effect be observed if sodium is exposed to orange light?"}
{"id": 2585, "contents": "319. Exercises - 319.2. The Bohr Model\n16. Why is the electron in a Bohr hydrogen atom bound less tightly when it has a quantum number of 3 than when it has a quantum number of 1 ?\n17. What does it mean to say that the energy of the electrons in an atom is quantized?\n18. Using the Bohr model, determine the energy, in joules, necessary to ionize a ground-state hydrogen atom. Show your calculations.\n19. The electron volt (eV) is a convenient unit of energy for expressing atomic-scale energies. It is the amount of energy that an electron gains when subjected to a potential of 1 volt; $1 \\mathrm{eV}=1.602 \\times 10^{-19} \\mathrm{~J}$. Using the Bohr model, determine the energy, in electron volts, of the photon produced when an electron in a hydrogen atom moves from the orbit with $n=5$ to the orbit with $n=2$. Show your calculations.\n20. Using the Bohr model, determine the lowest possible energy, in joules, for the electron in the $\\mathrm{Li}^{2+}$ ion.\n21. Using the Bohr model, determine the lowest possible energy for the electron in the $\\mathrm{He}^{+}$ion.\n22. Using the Bohr model, determine the energy of an electron with $n=6$ in a hydrogen atom.\n23. Using the Bohr model, determine the energy of an electron with $n=8$ in a hydrogen atom.\n24. How far from the nucleus in angstroms ( 1 angstrom $=1 \\times 10^{-10} \\mathrm{~m}$ ) is the electron in a hydrogen atom if it has an energy of $-8.72 \\times 10^{-20} \\mathrm{~J}$ ?\n25. What is the radius, in angstroms, of the orbital of an electron with $n=8$ in a hydrogen atom?\n26. Using the Bohr model, determine the energy in joules of the photon produced when an electron in a $\\mathrm{He}^{+}$ ion moves from the orbit with $n=5$ to the orbit with $n=2$."}
{"id": 2586, "contents": "319. Exercises - 319.2. The Bohr Model\n27. Using the Bohr model, determine the energy in joules of the photon produced when an electron in a $\\mathrm{Li}^{2+}$ ion moves from the orbit with $n=2$ to the orbit with $n=1$.\n28. Consider a large number of hydrogen atoms with electrons randomly distributed in the $n=1,2,3$, and 4 orbits.\n(a) How many different wavelengths of light are emitted by these atoms as the electrons fall into lowerenergy orbits?\n(b) Calculate the lowest and highest energies of light produced by the transitions described in part (a).\n(c) Calculate the frequencies and wavelengths of the light produced by the transitions described in part (b).\n29. How are the Bohr model and the Rutherford model of the atom similar? How are they different?\n30. The spectra of hydrogen and of calcium are shown here."}
{"id": 2587, "contents": "319. Exercises - 319.2. The Bohr Model\nWhat causes the lines in these spectra? Why are the colors of the lines different? Suggest a reason for the observation that the spectrum of calcium is more complicated than the spectrum of hydrogen."}
{"id": 2588, "contents": "319. Exercises - 319.3. Development of Quantum Theory\n31. How are the Bohr model and the quantum mechanical model of the hydrogen atom similar? How are they different?\n32. What are the allowed values for each of the four quantum numbers: $n, l, m_{l}$, and $m_{s}$ ?\n33. Describe the properties of an electron associated with each of the following four quantum numbers: $n, l$, $m_{l}$, and $m_{s}$.\n34. Answer the following questions:\n(a) Without using quantum numbers, describe the differences between the shells, subshells, and orbitals of an atom.\n(b) How do the quantum numbers of the shells, subshells, and orbitals of an atom differ?\n35. Identify the subshell in which electrons with the following quantum numbers are found:\n(a) $n=2, l=1$\n(b) $n=4, l=2$\n(c) $n=6, l=0$\n36. Which of the subshells described in the previous question contain degenerate orbitals? How many degenerate orbitals are in each?\n37. Identify the subshell in which electrons with the following quantum numbers are found:\n(a) $n=3, l=2$\n(b) $n=1, l=0$\n(c) $n=4, l=3$\n38. Which of the subshells described in the previous question contain degenerate orbitals? How many degenerate orbitals are in each?\n39. Sketch the boundary surface of a $d_{x^{2}-y^{2}}$ and a $p_{y}$ orbital. Be sure to show and label the axes.\n40. Sketch the $p_{X}$ and $d_{X Z}$ orbitals. Be sure to show and label the coordinates.\n41. Consider the orbitals shown here in outline.\n\n(x)\n\n(y)"}
{"id": 2589, "contents": "319. Exercises - 319.3. Development of Quantum Theory\n(x)\n\n(y)\n\n(z)\n(a) What is the maximum number of electrons contained in an orbital of type (x)? Of type (y)? Of type (z)?\n(b) How many orbitals of type (x) are found in a shell with $n=2$ ? How many of type (y)? How many of type (z)?\n(c) Write a set of quantum numbers for an electron in an orbital of type (x) in a shell with $n=4$. Of an orbital of type (y) in a shell with $n=2$. Of an orbital of type (z) in a shell with $n=3$.\n(d) What is the smallest possible $n$ value for an orbital of type (x)? Of type (y)? Of type (z)?\n(e) What are the possible 1 and $m_{l}$ values for an orbital of type (x)? Of type (y)? Of type (z)?\n42. State the Heisenberg uncertainty principle. Describe briefly what the principle implies.\n43. How many electrons could be held in the second shell of an atom if the spin quantum number $m_{S}$ could have three values instead of just two? (Hint: Consider the Pauli exclusion principle.)\n44. Which of the following equations describe particle-like behavior? Which describe wavelike behavior? Do any involve both types of behavior? Describe the reasons for your choices.\n(a) $c=\\lambda \\nu$\n(b) $E=\\frac{m v^{2}}{2}$\n(c) $r=\\frac{n^{2} a_{0}}{Z}$\n(d) $E=h \\nu$\n(e) $\\lambda=\\frac{h}{m v}$\n45. Write a set of quantum numbers for each of the electrons with an $n$ of 4 in a Se atom."}
{"id": 2590, "contents": "319. Exercises - 319.4. Electronic Structure of Atoms (Electron Configurations)\n46. Read the labels of several commercial products and identify monatomic ions of at least four transition elements contained in the products. Write the complete electron configurations of these cations.\n47. Read the labels of several commercial products and identify monatomic ions of at least six main group elements contained in the products. Write the complete electron configurations of these cations and anions.\n48. Using complete subshell notation (not abbreviations, $1 s^{2} 2 s^{2} 2 p^{6}$, and so forth), predict the electron configuration of each of the following atoms:\n(a) C\n(b) P\n(c) V\n(d) Sb\n(e) Sm\n49. Using complete subshell notation $\\left(1 s^{2} 2 s^{2} 2 p^{6}\\right.$, and so forth), predict the electron configuration of each of the following atoms:\n(a) N\n(b) Si\n(c) Fe\n(d) Te\n(e) Tb\n50. Is $1 s^{2} 2 s^{2} 2 p^{6}$ the symbol for a macroscopic property or a microscopic property of an element? Explain your answer.\n51. What additional information do we need to answer the question \"Which ion has the electron configuration $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6 \"}$ ?\n52. Draw the orbital diagram for the valence shell of each of the following atoms:\n(a) C\n(b) P\n(c) V\n(d) Sb\n(e) Ru\n53. Use an orbital diagram to describe the electron configuration of the valence shell of each of the following atoms:\n(a) N\n(b) Si\n(c) Fe\n(d) Te\n(e) Mo\n54. Using complete subshell notation $\\left(1 s^{2} 2 s^{2} 2 p^{6}\\right.$, and so forth), predict the electron configurations of the following ions.\n(a) $\\mathrm{N}^{3-}$\n(b) $\\mathrm{Ca}^{2+}$\n(c) $\\mathrm{S}^{-}$\n(d) $\\mathrm{Cs}^{2+}$\n(e) $\\mathrm{Cr}^{2+}$"}
{"id": 2591, "contents": "319. Exercises - 319.4. Electronic Structure of Atoms (Electron Configurations)\n(b) $\\mathrm{Ca}^{2+}$\n(c) $\\mathrm{S}^{-}$\n(d) $\\mathrm{Cs}^{2+}$\n(e) $\\mathrm{Cr}^{2+}$\n(f) $\\mathrm{Gd}^{3+}$\n55. Which atom has the electron configuration $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{2} 4 d^{2}$ ?\n56. Which atom has the electron configuration $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}$ ?\n57. Which ion with a +1 charge has the electron configuration $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6}$ ? Which ion with a -2 charge has this configuration?\n58. Which of the following atoms contains only three valence electrons: $\\mathrm{Li}, \\mathrm{B}, \\mathrm{N}, \\mathrm{F}, \\mathrm{Ne}$ ?\n59. Which of the following has two unpaired electrons?\n(a) Mg\n(b) Si\n(c) S\n(d) Both Mg and S\n(e) Both Si and S.\n60. Which atom would be expected to have a half-filled $6 p$ subshell?\n61. Which atom would be expected to have a half-filled $4 s$ subshell?\n62. In one area of Australia, the cattle did not thrive despite the presence of suitable forage. An investigation showed the cause to be the absence of sufficient cobalt in the soil. Cobalt forms cations in two oxidation states, $\\mathrm{Co}^{2+}$ and $\\mathrm{Co}^{3+}$. Write the electron structure of the two cations."}
{"id": 2592, "contents": "319. Exercises - 319.4. Electronic Structure of Atoms (Electron Configurations)\n63. Thallium was used as a poison in the Agatha Christie mystery story \"The Pale Horse.\" Thallium has two possible cationic forms, +1 and +3 . The +1 compounds are the more stable. Write the electron structure of the +1 cation of thallium.\n64. Write the electron configurations for the following atoms or ions:\n(a) $\\mathrm{B}^{3+}$\n(b) $\\mathrm{O}^{-}$\n(c) $\\mathrm{Cl}^{3+}$\n(d) $\\mathrm{Ca}^{2+}$\n(e) Ti\n65. Cobalt-60 and iodine-131 are radioactive isotopes commonly used in nuclear medicine. How many protons, neutrons, and electrons are in atoms of these isotopes? Write the complete electron configuration for each isotope.\n66. Write a set of quantum numbers for each of the electrons with an $n$ of 3 in a Sc atom."}
{"id": 2593, "contents": "319. Exercises - 319.5. Periodic Variations in Element Properties\n67. Based on their positions in the periodic table, predict which has the smallest atomic radius: $\\mathrm{Mg}, \\mathrm{Sr}, \\mathrm{Si}, \\mathrm{Cl}$, I.\n68. Based on their positions in the periodic table, predict which has the largest atomic radius: Li, Rb, N, F, I.\n69. Based on their positions in the periodic table, predict which has the largest first ionization energy: $\\mathrm{Mg}, \\mathrm{Ba}$, $\\mathrm{B}, \\mathrm{O}, \\mathrm{Te}$.\n70. Based on their positions in the periodic table, predict which has the smallest first ionization energy: Li, Cs, N, F, I.\n71. Based on their positions in the periodic table, rank the following atoms in order of increasing first ionization energy: $\\mathrm{F}, \\mathrm{Li}, \\mathrm{N}, \\mathrm{Rb}$\n72. Based on their positions in the periodic table, rank the following atoms in order of increasing first ionization energy: $\\mathrm{Mg}, \\mathrm{O}, \\mathrm{S}, \\mathrm{Si}$\n73. Atoms of which group in the periodic table have a valence shell electron configuration of $n s^{2} n p^{3}$ ?\n74. Atoms of which group in the periodic table have a valence shell electron configuration of $n s^{2}$ ?\n75. Based on their positions in the periodic table, list the following atoms in order of increasing radius: Mg , $\\mathrm{Ca}, \\mathrm{Rb}, \\mathrm{Cs}$.\n76. Based on their positions in the periodic table, list the following atoms in order of increasing radius: $\\mathrm{Sr}, \\mathrm{Ca}$, Si, Cl.\n77. Based on their positions in the periodic table, list the following ions in order of increasing radius: $\\mathrm{K}^{+}, \\mathrm{Ca}^{2+}$, $\\mathrm{Al}^{3+}, \\mathrm{Si}^{4+}$.\n78. List the following ions in order of increasing radius: $\\mathrm{Li}^{+}, \\mathrm{Mg}^{2+}, \\mathrm{Br}^{-}, \\mathrm{Te}^{2-}$."}
{"id": 2594, "contents": "319. Exercises - 319.5. Periodic Variations in Element Properties\n78. List the following ions in order of increasing radius: $\\mathrm{Li}^{+}, \\mathrm{Mg}^{2+}, \\mathrm{Br}^{-}, \\mathrm{Te}^{2-}$.\n79. Which atom and/or ion is (are) isoelectronic with $\\mathrm{Br}^{+}: \\mathrm{Se}^{2+}, \\mathrm{Se}, \\mathrm{As}^{-}, \\mathrm{Kr}, \\mathrm{Ga}^{3+}, \\mathrm{Cl}^{-}$?\n80. Which of the following atoms and ions is (are) isoelectronic with $\\mathrm{S}^{2+}: \\mathrm{Si}^{4+}, \\mathrm{Cl}^{3+}, \\mathrm{Ar}, \\mathrm{As}^{3+}, \\mathrm{Si}, \\mathrm{Al}^{3+}$ ?\n81. Compare both the numbers of protons and electrons present in each to rank the following ions in order of increasing radius: $\\mathrm{As}^{3-}, \\mathrm{Br}^{-}, \\mathrm{K}^{+}, \\mathrm{Mg}^{2+}$.\n82. Of the five elements $\\mathrm{Al}, \\mathrm{Cl}, \\mathrm{I}, \\mathrm{Na}, \\mathrm{Rb}$, which has the most exothermic reaction? (E represents an atom.) What name is given to the energy for the reaction? Hint: Note the process depicted does not correspond to electron affinity.)\n$\\mathrm{E}^{+}(g)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{E}(g)$\n83. Of the five elements $\\mathrm{Sn}, \\mathrm{Si}, \\mathrm{Sb}, \\mathrm{O}, \\mathrm{Te}$, which has the most endothermic reaction? (E represents an atom.) What name is given to the energy for the reaction? $\\mathrm{E}(g) \\longrightarrow \\mathrm{E}^{+}(g)+\\mathrm{e}^{-}$"}
{"id": 2595, "contents": "319. Exercises - 319.5. Periodic Variations in Element Properties\n84. The ionic radii of the ions $\\mathrm{S}^{2-}, \\mathrm{Cl}^{-}$, and $\\mathrm{K}^{+}$are $184,181,138 \\mathrm{pm}$ respectively. Explain why these ions have different sizes even though they contain the same number of electrons.\n85. Which main group atom would be expected to have the lowest second ionization energy?\n86. Explain why Al is a member of group 13 rather than group 3 ?"}
{"id": 2596, "contents": "319. Exercises - 319.6. The Periodic Table\n87. Using the periodic table, classify each of the following elements as a metal or a nonmetal, and then further classify each as a main-group (representative) element, transition metal, or inner transition metal:\n(a) uranium\n(b) bromine\n(c) strontium\n(d) neon\n(e) gold\n(f) americium\n(g) rhodium\n(h) sulfur\n(i) carbon\n(j) potassium\n88. Using the periodic table, classify each of the following elements as a metal or a nonmetal, and then further classify each as a main-group (representative) element, transition metal, or inner transition metal:\n(a) cobalt\n(b) europium\n(c) iodine\n(d) indium\n(e) lithium\n(f) oxygen\n(g) cadmium\n(h) terbium\n(i) rhenium\n89. Using the periodic table, identify the lightest member of each of the following groups:\n(a) noble gases\n(b) alkaline earth metals\n(c) alkali metals\n(d) chalcogens\n90. Using the periodic table, identify the heaviest member of each of the following groups:\n(a) alkali metals\n(b) chalcogens\n(c) noble gases\n(d) alkaline earth metals\n91. Use the periodic table to give the name and symbol for each of the following elements:\n(a) the noble gas in the same period as germanium\n(b) the alkaline earth metal in the same period as selenium\n(c) the halogen in the same period as lithium\n(d) the chalcogen in the same period as cadmium\n92. Use the periodic table to give the name and symbol for each of the following elements:\n(a) the halogen in the same period as the alkali metal with 11 protons\n(b) the alkaline earth metal in the same period with the neutral noble gas with 18 electrons\n(c) the noble gas in the same row as an isotope with 30 neutrons and 25 protons\n(d) the noble gas in the same period as gold\n93. Write a symbol for each of the following neutral isotopes. Include the atomic number and mass number for each.\n(a) the alkali metal with 11 protons and a mass number of 23"}
{"id": 2597, "contents": "319. Exercises - 319.6. The Periodic Table\n93. Write a symbol for each of the following neutral isotopes. Include the atomic number and mass number for each.\n(a) the alkali metal with 11 protons and a mass number of 23\n(b) the noble gas element with 75 neutrons in its nucleus and 54 electrons in the neutral atom\n(c) the isotope with 33 protons and 40 neutrons in its nucleus\n(d) the alkaline earth metal with 88 electrons and 138 neutrons\n94. Write a symbol for each of the following neutral isotopes. Include the atomic number and mass number for each.\n(a) the chalcogen with a mass number of 125\n(b) the halogen whose longest-lived isotope is radioactive\n(c) the noble gas, used in lighting, with 10 electrons and 10 neutrons\n(d) the lightest alkali metal with three neutrons"}
{"id": 2598, "contents": "319. Exercises - 319.7. Ionic and Molecular Compounds\n95. Using the periodic table, predict whether the following chlorides are ionic or covalent: $\\mathrm{KCl}, \\mathrm{NCl}_{3}, \\mathrm{ICl}$, $\\mathrm{MgCl}_{2}, \\mathrm{PCl}_{5}$, and $\\mathrm{CCl}_{4}$.\n96. Using the periodic table, predict whether the following chlorides are ionic or covalent: $\\mathrm{SiCl}_{4}, \\mathrm{PCl}_{3}, \\mathrm{CaCl}_{2}$, $\\mathrm{CsCl}, \\mathrm{CuCl}_{2}$, and $\\mathrm{CrCl}_{3}$.\n97. For each of the following compounds, state whether it is ionic or covalent. If it is ionic, write the symbols for the ions involved:\n(a) $\\mathrm{NF}_{3}$\n(b) BaO ,\n(c) $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{CO}_{3}$\n(d) $\\mathrm{Sr}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$\n(e) IBr\n(f) $\\mathrm{Na}_{2} \\mathrm{O}$\n98. For each of the following compounds, state whether it is ionic or covalent, and if it is ionic, write the symbols for the ions involved:\n(a) $\\mathrm{KClO}_{4}$\n(b) $\\mathrm{Mg}\\left(\\mathrm{C}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}\\right)_{2}$\n(c) $\\mathrm{H}_{2} \\mathrm{~S}$\n(d) $\\mathrm{Ag}_{2} \\mathrm{~S}$\n(e) $\\mathrm{N}_{2} \\mathrm{Cl}_{4}$\n(f) $\\mathrm{Co}\\left(\\mathrm{NO}_{3}\\right)_{2}$\n99. For each of the following pairs of ions, write the symbol for the formula of the compound they will form.\n(a) $\\mathrm{Ca}^{2+}, \\mathrm{S}^{2-}$\n(b) $\\mathrm{NH}_{4}{ }^{+}, \\mathrm{SO}_{4}{ }^{2-}$"}
{"id": 2599, "contents": "319. Exercises - 319.7. Ionic and Molecular Compounds\n(a) $\\mathrm{Ca}^{2+}, \\mathrm{S}^{2-}$\n(b) $\\mathrm{NH}_{4}{ }^{+}, \\mathrm{SO}_{4}{ }^{2-}$\n(c) $\\mathrm{Al}^{3+}, \\mathrm{Br}^{-}$\n(d) $\\mathrm{Na}^{+}, \\mathrm{HPO}_{4}{ }^{2-}$\n(e) $\\mathrm{Mg}^{2+}, \\mathrm{PO}_{4}{ }^{3-}$\n100. For each of the following pairs of ions, write the symbol for the formula of the compound they will form.\n(a) $\\mathrm{K}^{+}, \\mathrm{O}^{2-}$\n(b) $\\mathrm{NH}_{4}{ }^{+}, \\mathrm{PO}_{4}{ }^{3-}$\n(c) $\\mathrm{Al}^{3+}, \\mathrm{O}^{2-}$\n(d) $\\mathrm{Na}^{+}, \\mathrm{CO}_{3}{ }^{2-}$\n(e) $\\mathrm{Ba}^{2+}, \\mathrm{PO}_{4}{ }^{3-}$"}
{"id": 2600, "contents": "319. Exercises - 319.7. Ionic and Molecular Compounds\n174 3\u2022Exercises"}
{"id": 2601, "contents": "320. CHAPTER 4 <br> Chemical Bonding and Molecular Geometry - \nFigure 4.1 Nicknamed \"buckyballs,\" buckminsterfullerene molecules ( $\\mathrm{C}_{60}$ ) contain only carbon atoms (left) arranged to form a geometric framework of hexagons and pentagons, similar to the pattern on a soccer ball (center). This molecular structure is named after architect R. Buckminster Fuller, whose innovative designs combined simple geometric shapes to create large, strong structures such as this weather radar dome near Tucson, Arizona (right). (credit middle: modification of work by \"Petey21\"/Wikimedia Commons; credit right: modification of work by Bill Morrow)"}
{"id": 2602, "contents": "321. CHAPTER OUTLINE - 321.1. Ionic Bonding\n4.2 Covalent Bonding\n4.3 Chemical Nomenclature\n4.4 Lewis Symbols and Structures\n4.5 Formal Charges and Resonance\n4.6 Molecular Structure and Polarity\n\nINTRODUCTION It has long been known that pure carbon occurs in different forms (allotropes) including graphite and diamonds. But it was not until 1985 that a new form of carbon was recognized:\nbuckminsterfullerene. This molecule was named after the architect and inventor R. Buckminster Fuller (1895-1983), whose signature architectural design was the geodesic dome, characterized by a lattice shell structure supporting a spherical surface. Experimental evidence revealed the formula, $\\mathrm{C}_{60}$, and then scientists determined how 60 carbon atoms could form one symmetric, stable molecule. They were guided by bonding theory-the topic of this chapter-which explains how individual atoms connect to form more complex structures."}
{"id": 2603, "contents": "322. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Explain the formation of cations, anions, and ionic compounds\n- Predict the charge of common metallic and nonmetallic elements, and write their electron configurations\n\nAs you have learned, ions are atoms or molecules bearing an electrical charge. A cation (a positive ion) forms when a neutral atom loses one or more electrons from its valence shell, and an anion (a negative ion) forms when a neutral atom gains one or more electrons in its valence shell.\n\nCompounds composed of ions are called ionic compounds (or salts), and their constituent ions are held together by ionic bonds: electrostatic forces of attraction between oppositely charged cations and anions. The properties of ionic compounds shed some light on the nature of ionic bonds. Ionic solids exhibit a crystalline structure and tend to be rigid and brittle; they also tend to have high melting and boiling points, which suggests that ionic bonds are very strong. Ionic solids are also poor conductors of electricity for the same reason-the strength of ionic bonds prevents ions from moving freely in the solid state. Most ionic solids, however, dissolve readily in water. Once dissolved or melted, ionic compounds are excellent conductors of electricity and heat because the ions can move about freely.\n\nNeutral atoms and their associated ions have very different physical and chemical properties. Sodium atoms form sodium metal, a soft, silvery-white metal that burns vigorously in air and reacts explosively with water. Chlorine atoms form chlorine gas, $\\mathrm{Cl}_{2}$, a yellow-green gas that is extremely corrosive to most metals and very poisonous to animals and plants. The vigorous reaction between the elements sodium and chlorine forms the white, crystalline compound sodium chloride, common table salt, which contains sodium cations and chloride anions (Figure 4.2). The compound composed of these ions exhibits properties entirely different from the properties of the elements sodium and chlorine. Chlorine is poisonous, but sodium chloride is essential to life; sodium atoms react vigorously with water, but sodium chloride simply dissolves in water."}
{"id": 2604, "contents": "322. LEARNING OBJECTIVES - \nFIGURE 4.2 (a) Sodium is a soft metal that must be stored in mineral oil to prevent reaction with air or water. (b) Chlorine is a pale yellow-green gas. (c) When combined, they form white crystals of sodium chloride (table salt). (credit a: modification of work by \"Jurii\"/Wikimedia Commons)"}
{"id": 2605, "contents": "323. The Formation of Ionic Compounds - \nBinary ionic compounds are composed of just two elements: a metal (which forms the cations) and a nonmetal (which forms the anions). For example, NaCl is a binary ionic compound. We can think about the formation of such compounds in terms of the periodic properties of the elements. Many metallic elements have relatively low ionization potentials and lose electrons easily. These elements lie to the left in a period or near the bottom of a group on the periodic table. Nonmetal atoms have relatively high electron affinities and thus readily gain electrons lost by metal atoms, thereby filling their valence shells. Nonmetallic elements are found in the upper-right corner of the periodic table.\n\nAs all substances must be electrically neutral, the total number of positive charges on the cations of an ionic compound must equal the total number of negative charges on its anions. The formula of an ionic compound represents the simplest ratio of the numbers of ions necessary to give identical numbers of positive and negative charges. For example, the formula for aluminum oxide, $\\mathrm{Al}_{2} \\mathrm{O}_{3}$, indicates that this ionic compound contains two aluminum cations, $\\mathrm{Al}^{3+}$, for every three oxide anions, $\\mathrm{O}^{2-}$ [thus, $(2 \\times+3)+(3 \\times-2)=0$ ].\n\nIt is important to note, however, that the formula for an ionic compound does not represent the physical arrangement of its ions. It is incorrect to refer to a sodium chloride ( NaCl ) \"molecule\" because there is not a\nsingle ionic bond, per se, between any specific pair of sodium and chloride ions. The attractive forces between ions are isotropic-the same in all directions-meaning that any particular ion is equally attracted to all of the nearby ions of opposite charge. This results in the ions arranging themselves into a tightly bound, threedimensional lattice structure. Sodium chloride, for example, consists of a regular arrangement of equal numbers of $\\mathrm{Na}^{+}$cations and $\\mathrm{Cl}^{-}$anions (Figure 4.3)."}
{"id": 2606, "contents": "323. The Formation of Ionic Compounds - \nFIGURE 4.3 The atoms in sodium chloride (common table salt) are arranged to (a) maximize opposite charges interacting. The smaller spheres represent sodium ions, the larger ones represent chloride ions. In the expanded view (b), the geometry can be seen more clearly. Note that each ion is \"bonded\" to all of the surrounding ions-six in this case.\n\nThe strong electrostatic attraction between $\\mathrm{Na}^{+}$and $\\mathrm{Cl}^{-}$ions holds them tightly together in solid NaCl . It requires 769 kJ of energy to dissociate one mole of solid NaCl into separate gaseous $\\mathrm{Na}^{+}$and $\\mathrm{Cl}^{-}$ions:\n\n$$\n\\mathrm{NaCl}(s) \\longrightarrow \\mathrm{Na}^{+}(g)+\\mathrm{Cl}^{-}(g) \\quad \\Delta H=769 \\mathrm{~kJ}\n$$"}
{"id": 2607, "contents": "324. Electronic Structures of Cations - \nWhen forming a cation, an atom of a main group element tends to lose all of its valence electrons, thus assuming the electronic structure of the noble gas that precedes it in the periodic table. For groups 1 (the alkali metals) and 2 (the alkaline earth metals), the group numbers are equal to the numbers of valence shell electrons and, consequently, to the charges of the cations formed from atoms of these elements when all valence shell electrons are removed. For example, calcium is a group 2 element whose neutral atoms have 20 electrons and a ground state electron configuration of $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2}$. When a Ca atom loses both of its valence electrons, the result is a cation with 18 electrons, a $2+$ charge, and an electron configuration of $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$. The $\\mathrm{Ca}^{2+}$ ion is therefore isoelectronic with the noble gas Ar.\n\nFor groups 13-17, the group numbers exceed the number of valence electrons by 10 (accounting for the possibility of full $d$ subshells in atoms of elements in the fourth and greater periods). Thus, the charge of a cation formed by the loss of all valence electrons is equal to the group number minus 10. For example, aluminum (in group 13) forms $3+$ ions $\\left(\\mathrm{Al}^{3+}\\right)$."}
{"id": 2608, "contents": "324. Electronic Structures of Cations - \nExceptions to the expected behavior involve elements toward the bottom of the groups. In addition to the expected ions $\\mathrm{Tl}^{3+}, \\mathrm{Sn}^{4+}, \\mathrm{Pb}^{4+}$, and $\\mathrm{Bi}^{5+}$, a partial loss of these atoms' valence shell electrons can also lead to the formation of $\\mathrm{Tl}^{+}, \\mathrm{Sn}^{2+}, \\mathrm{Pb}^{2+}$, and $\\mathrm{Bi}^{3+}$ ions. The formation of these $1+, 2+$, and $3+$ cations is ascribed to the inert pair effect, which reflects the relatively low energy of the valence $s$-electron pair for atoms of the heavy elements of groups 13,14 , and 15 . Mercury (group 12) also exhibits an unexpected behavior: it forms a diatomic ion, $\\mathrm{Hg}_{2}{ }^{2+}$ (an ion formed from two mercury atoms, with an $\\mathrm{Hg}-\\mathrm{Hg}$ bond), in addition to the expected monatomic ion $\\mathrm{Hg}^{2+}$ (formed from only one mercury atom)."}
{"id": 2609, "contents": "324. Electronic Structures of Cations - \nTransition and inner transition metal elements behave differently than main group elements. Most transition metal cations have $2+$ or $3+$ charges that result from the loss of their outermost $s$ electron(s) first, sometimes followed by the loss of one or two $d$ electrons from the next-to-outermost shell. For example, iron $\\left(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6} 4 s^{2}\\right)$ forms the ion $\\mathrm{Fe}^{2+}\\left(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6}\\right)$ by the loss of the $4 s$ electrons and the ion $\\mathrm{Fe}^{3+}\\left(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{5}\\right)$ by the loss of the $4 s$ electron and one of the $3 d$ electrons. Although the $d$ orbitals of\nthe transition elements are-according to the Aufbau principle-the last to fill when building up electron configurations, the outermost $s$ electrons are the first to be lost when these atoms ionize. When the inner transition metals form ions, they usually have a $3+$ charge, resulting from the loss of their outermost $s$ electrons and a $d$ or $f$ electron."}
{"id": 2610, "contents": "326. Determining the Electronic Structures of Cations - \nThere are at least 14 elements categorized as \"essential trace elements\" for the human body. They are called \"essential\" because they are required for healthy bodily functions, \"trace\" because they are required only in small amounts, and \"elements\" in spite of the fact that they are really ions. Two of these essential trace elements, chromium and zinc, are required as $\\mathrm{Cr}^{3+}$ and $\\mathrm{Zn}^{2+}$. Write the electron configurations of these cations."}
{"id": 2611, "contents": "327. Solution - \nFirst, write the electron configuration for the neutral atoms:\n$\\mathrm{Zn}:[\\mathrm{Ar}] 3 \\mathrm{~d}^{10} 4 s^{2}$\nCr: $[\\mathrm{Ar}] 3 d^{5} 4 s^{1}$\nNext, remove electrons from the highest energy orbital. For the transition metals, electrons are removed from the $s$ orbital first and then from the $d$ orbital. For the $p$-block elements, electrons are removed from the $p$ orbitals and then from the $s$ orbital. Zinc is a member of group 12, so it should have a charge of $2+$, and thus loses only the two electrons in its $S$ orbital. Chromium is a transition element and should lose its $s$ electrons and then its $d$ electrons when forming a cation. Thus, we find the following electron configurations of the ions:\n$\\mathrm{Zn}^{2+}:[\\mathrm{Ar}] 3 d^{10}$\n$\\mathrm{Cr}^{3+}:[\\mathrm{Ar}] 3 d^{3}$"}
{"id": 2612, "contents": "328. Check Your Learning - \nPotassium and magnesium are required in our diet. Write the electron configurations of the ions expected from these elements."}
{"id": 2613, "contents": "329. Answer: - \n$\\mathrm{K}^{+}$: $[\\mathrm{Ar}], \\mathrm{Mg}^{2+}$ : $[\\mathrm{Ne}]$"}
{"id": 2614, "contents": "330. Electronic Structures of Anions - \nMost monatomic anions form when a neutral nonmetal atom gains enough electrons to completely fill its outer $s$ and $p$ orbitals, thereby reaching the electron configuration of the next noble gas. Thus, it is simple to determine the charge on such a negative ion: The charge is equal to the number of electrons that must be gained to fill the $s$ and $p$ orbitals of the parent atom. Oxygen, for example, has the electron configuration $1 s^{2} 2 s^{2} 2 p^{4}$, whereas the oxygen anion has the electron configuration of the noble gas neon ( Ne ), $1 s^{2} 2 s^{2} 2 p^{6}$. The two additional electrons required to fill the valence orbitals give the oxide ion the charge of $2-\\left(\\mathrm{O}^{2-}\\right)$."}
{"id": 2615, "contents": "332. Determining the Electronic Structure of Anions - \nSelenium and iodine are two essential trace elements that form anions. Write the electron configurations of the anions."}
{"id": 2616, "contents": "333. Solution - \n$\\mathrm{Se}^{2-}:[\\mathrm{Ar}] 3 d^{10} 4 s^{2} 4 p^{6}$\n$I^{-}:[\\mathrm{Kr}] 4 d^{10} 5 s^{2} 5 p^{6}$"}
{"id": 2617, "contents": "334. Check Your Learning - \nWrite the electron configurations of a phosphorus atom and its negative ion. Give the charge on the anion."}
{"id": 2618, "contents": "335. Answer: - \nP: [Ne] $3 s^{2} 3 p^{3} ; \\mathrm{P}^{3-}:[\\mathrm{Ne}] 3 s^{2} 3 p^{6}$"}
{"id": 2619, "contents": "335. Answer: - 335.1. Covalent Bonding\nLEARNING OBJECTIVES\nBy the end of this section, you will be able to:\n\n- Describe the formation of covalent bonds\n- Define electronegativity and assess the polarity of covalent bonds\n\nIonic bonding results from the electrostatic attraction of oppositely charged ions that are typically produced by the transfer of electrons between metallic and nonmetallic atoms. A different type of bonding results from the mutual attraction of atoms for a \"shared\" pair of electrons. Such bonds are called covalent bonds. Covalent bonds are formed between two atoms when both have similar tendencies to attract electrons to themselves (i.e., when both atoms have identical or fairly similar ionization energies and electron affinities). For example, two hydrogen atoms bond covalently to form an $\\mathrm{H}_{2}$ molecule; each hydrogen atom in the $\\mathrm{H}_{2}$ molecule has two electrons stabilizing it, giving each atom the same number of valence electrons as the noble gas He .\n\nCompounds that contain covalent bonds exhibit different physical properties than ionic compounds. Because the attraction between molecules, which are electrically neutral, is weaker than that between electrically charged ions, covalent compounds generally have much lower melting and boiling points than ionic compounds. In fact, many covalent compounds are liquids or gases at room temperature, and, in their solid states, they are typically much softer than ionic solids. Furthermore, whereas ionic compounds are good conductors of electricity when dissolved in water, most covalent compounds are insoluble in water; since they are electrically neutral, they are poor conductors of electricity in any state."}
{"id": 2620, "contents": "336. Formation of Covalent Bonds - \nNonmetal atoms frequently form covalent bonds with other nonmetal atoms. For example, the hydrogen molecule, $\\mathrm{H}_{2}$, contains a covalent bond between its two hydrogen atoms. Figure 4.4 illustrates why this bond is formed. Starting on the far right, we have two separate hydrogen atoms with a particular potential energy, indicated by the red line. Along the $x$-axis is the distance between the two atoms. As the two atoms approach each other (moving left along the $x$-axis), their valence orbitals (1s) begin to overlap. The single electrons on each hydrogen atom then interact with both atomic nuclei, occupying the space around both atoms. The strong attraction of each shared electron to both nuclei stabilizes the system, and the potential energy decreases as the bond distance decreases. If the atoms continue to approach each other, the positive charges in the two nuclei begin to repel each other, and the potential energy increases. The bond length is determined by the distance at which the lowest potential energy is achieved.\n\n\nFIGURE 4.4 The potential energy of two separate hydrogen atoms (right) decreases as they approach each other, and the single electrons on each atom are shared to form a covalent bond. The bond length is the internuclear distance at which the lowest potential energy is achieved.\n\nIt is essential to remember that energy must be added to break chemical bonds (an endothermic process), whereas forming chemical bonds releases energy (an exothermic process). In the case of $\\mathrm{H}_{2}$, the covalent bond is very strong; a large amount of energy, 436 kJ , must be added to break the bonds in one mole of hydrogen molecules and cause the atoms to separate:\n\n$$\n\\mathrm{H}_{2}(\\mathrm{~g}) \\longrightarrow 2 \\mathrm{H}(\\mathrm{~g}) \\quad \\Delta H=436 \\mathrm{~kJ}\n$$\n\nConversely, the same amount of energy is released when one mole of $\\mathrm{H}_{2}$ molecules forms from two moles of H atoms:\n\n$$\n2 \\mathrm{H}(\\mathrm{~g}) \\longrightarrow \\mathrm{H}_{2}(\\mathrm{~g}) \\quad \\Delta H=-436 \\mathrm{~kJ}\n$$"}
{"id": 2621, "contents": "337. Pure vs. Polar Covalent Bonds - \nIf the atoms that form a covalent bond are identical, as in $\\mathrm{H}_{2}, \\mathrm{Cl}_{2}$, and other diatomic molecules, then the electrons in the bond must be shared equally. We refer to this as a pure covalent bond. Electrons shared in\npure covalent bonds have an equal probability of being near each nucleus.\nIn the case of $\\mathrm{Cl}_{2}$, each atom starts off with seven valence electrons, and each Cl shares one electron with the other, forming one covalent bond:\n\n$$\n\\mathrm{Cl}+\\mathrm{Cl} \\longrightarrow \\mathrm{Cl}_{2}\n$$\n\nThe total number of electrons around each individual atom consists of six nonbonding electrons and two shared (i.e., bonding) electrons for eight total electrons, matching the number of valence electrons in the noble gas argon. Since the bonding atoms are identical, $\\mathrm{Cl}_{2}$ also features a pure covalent bond.\n\nWhen the atoms linked by a covalent bond are different, the bonding electrons are shared, but no longer equally. Instead, the bonding electrons are more attracted to one atom than the other, giving rise to a shift of electron density toward that atom. This unequal distribution of electrons is known as a polar covalent bond, characterized by a partial positive charge on one atom and a partial negative charge on the other. The atom that attracts the electrons more strongly acquires the partial negative charge and vice versa. For example, the electrons in the $\\mathrm{H}-\\mathrm{Cl}$ bond of a hydrogen chloride molecule spend more time near the chlorine atom than near the hydrogen atom. Thus, in an HCl molecule, the chlorine atom carries a partial negative charge and the hydrogen atom has a partial positive charge. Figure 4.5 shows the distribution of electrons in the $\\mathrm{H}-\\mathrm{Cl}$ bond. Note that the shaded area around Cl is much larger than it is around H . Compare this to Figure 4.4, which shows the even distribution of electrons in the $\\mathrm{H}_{2}$ nonpolar bond."}
{"id": 2622, "contents": "337. Pure vs. Polar Covalent Bonds - \nWe sometimes designate the positive and negative atoms in a polar covalent bond using a lowercase Greek letter \"delta,\" $\\delta$, with a plus sign or minus sign to indicate whether the atom has a partial positive charge ( $\\delta+$ ) or a partial negative charge ( $\\delta-$ ). This symbolism is shown for the $\\mathrm{H}-\\mathrm{Cl}$ molecule in Figure 4.5.\n\n(a)\n$\\delta+\\delta-$\n$\\mathrm{H}-\\mathrm{Cl}$\n(b)\n\nFIGURE 4.5 (a) The distribution of electron density in the HCl molecule is uneven. The electron density is greater around the chlorine nucleus. The small, black dots indicate the location of the hydrogen and chlorine nuclei in the molecule. (b) Symbols $\\delta+$ and $\\delta$ - indicate the polarity of the $\\mathrm{H}-\\mathrm{Cl}$ bond."}
{"id": 2623, "contents": "338. Electronegativity - \nWhether a bond is nonpolar or polar covalent is determined by a property of the bonding atoms called electronegativity. Electronegativity is a measure of the tendency of an atom to attract electrons (or electron density) towards itself. It determines how the shared electrons are distributed between the two atoms in a bond. The more strongly an atom attracts the electrons in its bonds, the larger its electronegativity. Electrons in a polar covalent bond are shifted toward the more electronegative atom; thus, the more electronegative atom is the one with the partial negative charge. The greater the difference in electronegativity, the more polarized the electron distribution and the larger the partial charges of the atoms.\n\nFigure 4.6 shows the electronegativity values of the elements as proposed by one of the most famous chemists of the twentieth century: Linus Pauling (Figure 4.7). In general, electronegativity increases from left to right across a period in the periodic table and decreases down a group. Thus, the nonmetals, which lie in the upper right, tend to have the highest electronegativities, with fluorine the most electronegative element of all (EN = 4.0). Metals tend to be less electronegative elements, and the group 1 metals have the lowest electronegativities. Note that noble gases are excluded from this figure because these atoms usually do not share electrons with others atoms since they have a full valence shell. (While noble gas compounds such as $\\mathrm{XeO}_{2}$ do exist, they can only be formed under extreme conditions, and thus they do not fit neatly into the general model of electronegativity.)\n\n\nFIGURE 4.6 The electronegativity values derived by Pauling follow predictable periodic trends, with the higher electronegativities toward the upper right of the periodic table."}
{"id": 2624, "contents": "339. Electronegativity versus Electron Affinity - \nWe must be careful not to confuse electronegativity and electron affinity. The electron affinity of an element is a measurable physical quantity, namely, the energy released or absorbed when an isolated gas-phase atom acquires an electron, measured in $\\mathrm{kJ} / \\mathrm{mol}$. Electronegativity, on the other hand, describes how tightly an atom attracts electrons in a bond. It is a dimensionless quantity that is calculated, not measured. Pauling derived the first electronegativity values by comparing the amounts of energy required to break different types of bonds. He chose an arbitrary relative scale ranging from 0 to 4 ."}
{"id": 2625, "contents": "341. Linus Pauling - \nLinus Pauling, shown in Figure 4.7, is the only person to have received two unshared (individual) Nobel Prizes: one for chemistry in 1954 for his work on the nature of chemical bonds and one for peace in 1962 for his opposition to weapons of mass destruction. He developed many of the theories and concepts that are foundational to our current understanding of chemistry, including electronegativity and resonance structures.\n\n\nFIGURE 4.7 Linus Pauling (1901-1994) made many important contributions to the field of chemistry. He was also a prominent activist, publicizing issues related to health and nuclear weapons.\n\nPauling also contributed to many other fields besides chemistry. His research on sickle cell anemia revealed the cause of the disease-the presence of a genetically inherited abnormal protein in the blood-and paved the way for the field of molecular genetics. His work was also pivotal in curbing the\ntesting of nuclear weapons; he proved that radioactive fallout from nuclear testing posed a public health risk."}
{"id": 2626, "contents": "342. Electronegativity and Bond Type - \nThe absolute value of the difference in electronegativity ( $\\Delta \\mathrm{EN}$ ) of two bonded atoms provides a rough measure of the polarity to be expected in the bond and, thus, the bond type. When the difference is very small or zero, the bond is covalent and nonpolar. When it is large, the bond is polar covalent or ionic. The absolute values of the electronegativity differences between the atoms in the bonds $\\mathrm{H}-\\mathrm{H}, \\mathrm{H}-\\mathrm{Cl}$, and $\\mathrm{Na}-\\mathrm{Cl}$ are 0 (nonpolar), 0.9 (polar covalent), and 2.1 (ionic), respectively. The degree to which electrons are shared between atoms varies from completely equal (pure covalent bonding) to not at all (ionic bonding). Figure 4.8 shows the relationship between electronegativity difference and bond type.\n\n\nFIGURE 4.8 As the electronegativity difference increases between two atoms, the bond becomes more ionic.\nA rough approximation of the electronegativity differences associated with covalent, polar covalent, and ionic bonds is shown in Figure 4.8. This table is just a general guide, however, with many exceptions. For example, the H and F atoms in HF have an electronegativity difference of 1.9, and the N and H atoms in $\\mathrm{NH}_{3}$ a difference of 0.9 , yet both of these compounds form bonds that are considered polar covalent. Likewise, the Na and Cl atoms in NaCl have an electronegativity difference of 2.1, and the Mn and I atoms in $\\mathrm{MnI}_{2}$ have a difference of 1.0, yet both of these substances form ionic compounds.\n\nThe best guide to the covalent or ionic character of a bond is to consider the types of atoms involved and their relative positions in the periodic table. Bonds between two nonmetals are generally covalent; bonding between a metal and a nonmetal is often ionic."}
{"id": 2627, "contents": "342. Electronegativity and Bond Type - \nSome compounds contain both covalent and ionic bonds. The atoms in polyatomic ions, such as $\\mathrm{OH}^{-}, \\mathrm{NO}_{3}{ }^{-}$, and $\\mathrm{NH}_{4}{ }^{+}$, are held together by polar covalent bonds. However, these polyatomic ions form ionic compounds by combining with ions of opposite charge. For example, potassium nitrate, $\\mathrm{KNO}_{3}$, contains the $\\mathrm{K}^{+}$cation and the polyatomic $\\mathrm{NO}_{3}{ }^{-}$anion. Thus, bonding in potassium nitrate is ionic, resulting from the electrostatic attraction between the ions $\\mathrm{K}^{+}$and $\\mathrm{NO}_{3}{ }^{-}$, as well as covalent between the nitrogen and oxygen atoms in $\\mathrm{NO}_{3}{ }^{-}$."}
{"id": 2628, "contents": "344. Electronegativity and Bond Polarity - \nBond polarities play an important role in determining the structure of proteins. Using the electronegativity values in Figure 4.6, arrange the following covalent bonds-all commonly found in amino acids-in order of increasing polarity. Then designate the positive and negative atoms using the symbols $\\delta+$ and $\\delta-$ :\n\nC-H, C-N, C-O, N-H, O-H, S-H"}
{"id": 2629, "contents": "345. Solution - \nThe polarity of these bonds increases as the absolute value of the electronegativity difference increases. The atom with the $\\delta$ - designation is the more electronegative of the two. Table 4.1 shows these bonds in order of increasing polarity.\n\nBond Polarity and Electronegativity Difference\n\n| Bond | $\\Delta \\mathrm{EN}$ | Polarity |\n| :---: | :---: | :---: |\n| C-H | 0.4 | $\\stackrel{\\delta-}{\\mathrm{C}}-\\stackrel{\\delta+}{\\mathrm{H}}$ |\n| S-H | 0.4 | $\\stackrel{\\delta-}{S-}-\\stackrel{\\delta+}{H}$ |\n| C-N | 0.5 | $\\stackrel{\\delta+}{\\mathrm{C}}-\\stackrel{\\delta-}{\\mathrm{N}}$ |\n| N-H | 0.9 | $\\stackrel{\\delta-}{\\mathrm{N}}-\\stackrel{\\delta+}{\\mathrm{H}}$ |\n| C-O | 1.0 | $\\stackrel{\\delta+}{\\mathrm{C}}-\\stackrel{\\delta-}{\\mathrm{O}}$ |\n| O-H | 1.4 | $\\stackrel{\\delta-}{\\mathrm{O}}-\\stackrel{\\delta+}{\\mathrm{H}}$ |\n\nTABLE 4.1"}
{"id": 2630, "contents": "346. Check Your Learning - \nSilicones are polymeric compounds containing, among others, the following types of covalent bonds: $\\mathrm{Si}-\\mathrm{O}$, $\\mathrm{Si}-\\mathrm{C}, \\mathrm{C}-\\mathrm{H}$, and $\\mathrm{C}-\\mathrm{C}$. Using the electronegativity values in Figure 4.6, arrange the bonds in order of increasing polarity and designate the positive and negative atoms using the symbols $\\delta+$ and $\\delta-$.\n\nAnswer:\n\n| Bond | Electronegativity Difference | Polarity |\n| :--- | :--- | :--- |\n| $\\mathrm{C}-\\mathrm{C}$ | 0.0 | nonpolar |\n| $\\mathrm{C}-\\mathrm{H}$ | 0.4 | $\\delta-\\stackrel{\\delta+}{\\mathrm{C}}-\\mathrm{H}$ |\n| $\\mathrm{Si}-\\mathrm{C}$ | 0.7 | $\\delta+\\delta-$ <br> $\\mathrm{Si}-\\mathrm{C}$ |\n| $\\mathrm{Si}-\\mathrm{O}$ | 1.7 | $\\delta+\\delta-$ <br> $\\mathrm{Si}-\\mathrm{O}$ |"}
{"id": 2631, "contents": "347. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Derive names for common types of inorganic compounds using a systematic approach\n\nNomenclature, a collection of rules for naming things, is important in science and in many other situations.\n\nThis module describes an approach that is used to name simple ionic and molecular compounds, such as $\\mathrm{NaCl}, \\mathrm{CaCO}_{3}$, and $\\mathrm{N}_{2} \\mathrm{O}_{4}$. The simplest of these are binary compounds, those containing only two elements, but we will also consider how to name ionic compounds containing polyatomic ions, and one specific, very important class of compounds known as acids (subsequent chapters in this text will focus on these compounds in great detail). We will limit our attention here to inorganic compounds, compounds that are composed principally of elements other than carbon, and will follow the nomenclature guidelines proposed by IUPAC. The rules for organic compounds, in which carbon is the principle element, will be treated in a later chapter on organic chemistry."}
{"id": 2632, "contents": "348. Ionic Compounds - \nTo name an inorganic compound, we need to consider the answers to several questions. First, is the compound ionic or molecular? If the compound is ionic, does the metal form ions of only one type (fixed charge) or more than one type (variable charge)? Are the ions monatomic or polyatomic? If the compound is molecular, does it contain hydrogen? If so, does it also contain oxygen? From the answers we derive, we place the compound in an appropriate category and then name it accordingly."}
{"id": 2633, "contents": "349. Compounds Containing Only Monatomic Ions - \nThe name of a binary compound containing monatomic ions consists of the name of the cation (the name of the metal) followed by the name of the anion (the name of the nonmetallic element with its ending replaced by the suffix -ide). Some examples are given in Table 4.2.\n\n| Names of Some Ionic Compounds |  |\n| :--- | :--- |\n| NaCl, sodium chloride | $\\mathrm{Na}_{2} \\mathrm{O}$, sodium oxide |\n| KBr , potassium bromide | CdS, cadmium sulfide |\n| $\\mathrm{CaI}_{2}$, calcium iodide | $\\mathrm{Mg}_{3} \\mathrm{~N}_{2}$, magnesium nitride |\n| CsF, cesium fluoride | $\\mathrm{Ca}_{3} \\mathrm{P}_{2}$, calcium phosphide |\n| LiCl, lithium chloride | $\\mathrm{Al}_{4} \\mathrm{C}_{3}$, aluminum carbide |\n\nTABLE 4.2"}
{"id": 2634, "contents": "350. Compounds Containing Polyatomic lons - \nCompounds containing polyatomic ions are named similarly to those containing only monatomic ions, i.e., by naming first the cation and then the anion. Examples are shown in Table 4.3.\n\n| Names of Some Polyatomic lonic Compounds |  |\n| :--- | :--- |\n| $\\mathrm{KC}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}$, potassium acetate | $\\mathrm{NH}_{4} \\mathrm{Cl}$, ammonium chloride |\n| $\\mathrm{NaHCO}_{3}$, sodium bicarbonate | $\\mathrm{CaSO}_{4}$, calcium sulfate |\n| $\\mathrm{Al}_{2}\\left(\\mathrm{CO}_{3}\\right)_{3}$, aluminum carbonate | $\\mathrm{Mg}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$, magnesium phosphate |\n\nTABLE 4.3"}
{"id": 2635, "contents": "352. Ionic Compounds in Your Cabinets - \nEvery day you encounter and use a large number of ionic compounds. Some of these compounds, where they are found, and what they are used for are listed in Table 4.4. Look at the label or ingredients list on the various products that you use during the next few days, and see if you run into any of those in this table, or find other ionic compounds that you could now name or write as a formula.\n\nEveryday Ionic Compounds\n\n| Ionic Compound | Use |\n| :---: | :---: |\n| NaCl , sodium chloride | ordinary table salt |\n| KI, potassium iodide | added to \"iodized\" salt for thyroid health |\n| NaF, sodium fluoride | ingredient in toothpaste |\n| $\\mathrm{NaHCO}_{3}$, sodium bicarbonate | baking soda; used in cooking (and as antacid) |\n| $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$, sodium carbonate | washing soda; used in cleaning agents |\n| NaOCl , sodium hypochlorite | active ingredient in household bleach |\n| $\\mathrm{CaCO}_{3}$ calcium carbonate | ingredient in antacids |\n| $\\mathrm{Mg}(\\mathrm{OH})_{2}$, magnesium hydroxide | ingredient in antacids |\n| $\\mathrm{Al}(\\mathrm{OH})_{3}$, aluminum hydroxide | ingredient in antacids |\n| NaOH , sodium hydroxide | lye; used as drain cleaner |\n| $\\mathrm{K}_{3} \\mathrm{PO}_{4}$, potassium phosphate | food additive (many purposes) |\n| $\\mathrm{MgSO}_{4}$, magnesium sulfate | added to purified water |\n| $\\mathrm{Na}_{2} \\mathrm{HPO}_{4}$, sodium hydrogen phosphate | anti-caking agent; used in powdered products |\n| $\\mathrm{Na}_{2} \\mathrm{SO}_{3}$, sodium sulfite | preservative |\n\nTABLE 4.4"}
{"id": 2636, "contents": "353. Compounds Containing a Metal Ion with a Variable Charge - \nMost of the transition metals and some main group metals can form two or more cations with different charges. Compounds of these metals with nonmetals are named with the same method as compounds in the first category, except the charge of the metal ion is specified by a Roman numeral in parentheses after the name of the metal. The charge of the metal ion is determined from the formula of the compound and the charge of the anion. For example, consider binary ionic compounds of iron and chlorine. Iron typically exhibits a charge of either 2+ or $3+$ (see Figure 3.40), and the two corresponding compound formulas are $\\mathrm{FeCl}_{2}$ and $\\mathrm{FeCl}_{3}$. The simplest name, \"iron chloride,\" will, in this case, be ambiguous, as it does not distinguish between\nthese two compounds. In cases like this, the charge of the metal ion is included as a Roman numeral in parentheses immediately following the metal name. These two compounds are then unambiguously named iron(II) chloride and iron(III) chloride, respectively. Other examples are provided in Table 4.5.\n\n| Some Ionic Compounds with Variably Charged Metal lons |  |\n| :--- | :--- |\n| Compound | Name |\n| $\\mathrm{FeCl}_{2}$ | iron(II) chloride |\n| $\\mathrm{FeCl}_{3}$ | iron(III) chloride |\n| $\\mathrm{Hg}_{2} \\mathrm{O}$ | mercury(I) oxide |\n| HgO | mercury(II) oxide |\n| $\\mathrm{SnF}_{2}$ | tin(II) fluoride |\n| $\\mathrm{SnF}_{4}$ | tin(IV) fluoride |\n\nTABLE 4.5"}
{"id": 2637, "contents": "353. Compounds Containing a Metal Ion with a Variable Charge - \nTABLE 4.5\n\nOut-of-date nomenclature used the suffixes -ic and -ous to designate metals with higher and lower charges, respectively: Iron(III) chloride, $\\mathrm{FeCl}_{3}$, was previously called ferric chloride, and iron(II) chloride, $\\mathrm{FeCl}_{2}$, was known as ferrous chloride. Though this naming convention has been largely abandoned by the scientific community, it remains in use by some segments of industry. For example, you may see the words stannous fluoride on a tube of toothpaste. This represents the formula $\\mathrm{SnF}_{2}$, which is more properly named tin(II) fluoride. The other fluoride of tin is $\\mathrm{SnF}_{4}$, which was previously called stannic fluoride but is now named tin(IV) fluoride."}
{"id": 2638, "contents": "354. Ionic Hydrates - \nIonic compounds that contain water molecules as integral components of their crystals are called hydrates. The name for an ionic hydrate is derived by adding a term to the name for the anhydrous (meaning \"not hydrated\") compound that indicates the number of water molecules associated with each formula unit of the compound. The added word begins with a Greek prefix denoting the number of water molecules (see Table 4.6) and ends with \"hydrate.\" For example, the anhydrous compound copper(II) sulfate also exists as a hydrate containing five water molecules and named copper(II) sulfate pentahydrate. Washing soda is the common name for a hydrate of sodium carbonate containing 10 water molecules; the systematic name is sodium carbonate decahydrate.\n\nFormulas for ionic hydrates are written by appending a vertically centered dot, a coefficient representing the number of water molecules, and the formula for water. The two examples mentioned in the previous paragraph are represented by the formulas\n\n> copper(II) sulfate pentahydrate $\\mathrm{CuSO}_{4} \\cdot 5 \\mathrm{H}_{2} \\mathrm{O}$\n> sodium carbonate decahydrate $\\mathrm{Na}_{2} \\mathrm{CO}_{3} \\cdot 10 \\mathrm{H}_{2} \\mathrm{O}$\n\nNomenclature Prefixes\n\n| Number |  | Prefix |  | Number |  | Prefix |\n| :--- | :--- | :--- | :--- | :--- | :---: | :---: |\n| 1 (sometimes omitted) | mono- | 6 | hexa- |  |  |  |\n| 2 | di- |  | 7 | hepta- |  |  |\n| 2 | tri- |  | 8 | octa- |  |  |\n| 3 | tetra- |  | 9 | nona- |  |  |\n| 4 | penta- |  | 10 | deca- |  |  |\n| 5 |  |  |  |  |  |  |\n\nTABLE 4.6"}
{"id": 2639, "contents": "356. Naming Ionic Compounds - \nName the following ionic compounds:\n(a) $\\mathrm{Fe}_{2} \\mathrm{~S}_{3}$\n(b) CuSe\n(c) GaN\n(d) $\\mathrm{MgSO}_{4} \\cdot 7 \\mathrm{H}_{2} \\mathrm{O}$\n(e) $\\mathrm{Ti}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}$"}
{"id": 2640, "contents": "357. Solution - \nThe anions in these compounds have a fixed negative charge ( $\\mathrm{S}^{2-}, \\mathrm{Se}^{2-}, \\mathrm{N}^{3-}$, and $\\mathrm{SO}_{4}{ }^{2-}$ ), and the compounds must be neutral. Because the total number of positive charges in each compound must equal the total number of negative charges, the positive ions must be $\\mathrm{Fe}^{3+}, \\mathrm{Cu}^{2+}, \\mathrm{Ga}^{3+}, \\mathrm{Mg}^{2+}$, and $\\mathrm{Ti}^{3+}$. These charges are used in the names of the metal ions:\n(a) iron(III) sulfide\n(b) copper(II) selenide\n(c) gallium(III) nitride\n(d) magnesium sulfate heptahydrate\n(e) titanium(III) sulfate"}
{"id": 2641, "contents": "358. Check Your Learning - \nWrite the formulas of the following ionic compounds:\n(a) chromium(III) phosphide\n(b) mercury(II) sulfide\n(c) manganese(II) phosphate\n(d) copper(I) oxide\n(e) iron(III) chloride dihydrate"}
{"id": 2642, "contents": "359. Answer: - \n(a) CrP ; (b) HgS ; (c) $\\mathrm{Mn}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$; (d) $\\mathrm{Cu}_{2} \\mathrm{O}$; (e) $\\mathrm{FeCl}_{3} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$"}
{"id": 2643, "contents": "361. Erin Brokovich and Chromium Contamination - \nIn the early 1990s, legal file clerk Erin Brockovich (Figure 4.9) discovered a high rate of serious illnesses in the small town of Hinckley, California. Her investigation eventually linked the illnesses to groundwater contaminated by $\\mathrm{Cr}(\\mathrm{VI})$ used by Pacific Gas \\& Electric (PG\\&E) to fight corrosion in a nearby natural gas pipeline. As dramatized in the film Erin Brokovich (for which Julia Roberts won an Oscar), Erin and lawyer Edward Masry sued PG\\&E for contaminating the water near Hinckley in 1993. The settlement they won in 1996-\\$333 million-was the largest amount ever awarded for a direct-action lawsuit in the US at that time.\n\n\nFIGURE 4.9 (a) Erin Brockovich found that $\\mathrm{Cr}(\\mathrm{VI})$, used by PG\\&E, had contaminated the Hinckley, California, water supply. (b) The $\\mathrm{Cr}(\\mathrm{VI})$ ion is often present in water as the polyatomic ions chromate, $\\mathrm{CrO}_{4}{ }^{2-}$ (left), and dichromate, $\\mathrm{Cr}_{2} \\mathrm{O}_{7}{ }^{2-}$ (right).\n\nChromium compounds are widely used in industry, such as for chrome plating, in dye-making, as preservatives, and to prevent corrosion in cooling tower water, as occurred near Hinckley. In the environment, chromium exists primarily in either the Cr (III) or Cr (VI) forms. Cr (III), an ingredient of many vitamin and nutritional supplements, forms compounds that are not very soluble in water, and it has low toxicity. But $\\mathrm{Cr}(\\mathrm{VI})$ is much more toxic and forms compounds that are reasonably soluble in water.\nExposure to small amounts of $\\mathrm{Cr}(\\mathrm{VI})$ can lead to damage of the respiratory, gastrointestinal, and immune systems, as well as the kidneys, liver, blood, and skin."}
{"id": 2644, "contents": "361. Erin Brokovich and Chromium Contamination - \nDespite cleanup efforts, $\\mathrm{Cr}(\\mathrm{VI})$ groundwater contamination remains a problem in Hinckley and other locations across the globe. A 2010 study by the Environmental Working Group found that of 35 US cities tested, 31 had higher levels of $\\mathrm{Cr}(\\mathrm{VI})$ in their tap water than the public health goal of 0.02 parts per billion set by the California Environmental Protection Agency."}
{"id": 2645, "contents": "362. Molecular (Covalent) Compounds - \nThe bonding characteristics of inorganic molecular compounds are different from ionic compounds, and they are named using a different system as well. The charges of cations and anions dictate their ratios in ionic compounds, so specifying the names of the ions provides sufficient information to determine chemical formulas. However, because covalent bonding allows for significant variation in the combination ratios of the atoms in a molecule, the names for molecular compounds must explicitly identify these ratios.\n\nCompounds Composed of Two Elements\nWhen two nonmetallic elements form a molecular compound, several combination ratios are often possible.\n\nFor example, carbon and oxygen can form the compounds CO and $\\mathrm{CO}_{2}$. Since these are different substances with different properties, they cannot both have the same name (they cannot both be called carbon oxide). To deal with this situation, we use a naming method that is somewhat similar to that used for ionic compounds, but with added prefixes to specify the numbers of atoms of each element. The name of the more metallic element (the one farther to the left and/or bottom of the periodic table) is first, followed by the name of the more nonmetallic element (the one farther to the right and/or top) with its ending changed to the suffix -ide. The numbers of atoms of each element are designated by the Greek prefixes shown in Table 4.6.\n\nWhen only one atom of the first element is present, the prefix mono- is usually deleted from that part. Thus, CO is named carbon monoxide, and $\\mathrm{CO}_{2}$ is called carbon dioxide. When two vowels are adjacent, the $a$ in the Greek prefix is usually dropped. Some other examples are shown in Table 4.7.\n\nNames of Some Molecular Compounds Composed of Two Elements"}
{"id": 2646, "contents": "362. Molecular (Covalent) Compounds - \nNames of Some Molecular Compounds Composed of Two Elements\n\n| Compound |  |  |  | Name |  | Compound |  | Name |\n| :--- | :--- | :--- | :--- | :--- | :---: | :---: | :---: | :---: |\n| $\\mathrm{SO}_{2}$ | sulfur dioxide |  | $\\mathrm{BCl}_{3}$ | boron trichloride |  |  |  |  |\n| $\\mathrm{SO}_{3}$ | sulfur trioxide |  | $\\mathrm{SF}_{6}$ | sulfur hexafluoride |  |  |  |  |\n| $\\mathrm{NO}_{2}$ | nitrogen dioxide |  | $\\mathrm{PF}_{5}$ | phosphorus pentafluoride |  |  |  |  |\n| $\\mathrm{N}_{2} \\mathrm{O}_{4}$ | dinitrogen tetroxide |  | $\\mathrm{P}_{4} \\mathrm{O}_{10}$ | tetraphosphorus decaoxide |  |  |  |  |\n| $\\mathrm{N}_{2} \\mathrm{O}_{5}$ | dinitrogen pentoxide |  | $\\mathrm{IF}_{7}$ | iodine heptafluoride |  |  |  |  |\n\nTABLE 4.7\n\nThere are a few common names that you will encounter as you continue your study of chemistry. For example, although NO is often called nitric oxide, its proper name is nitrogen monoxide. Similarly, $\\mathrm{N}_{2} \\mathrm{O}$ is known as nitrous oxide even though our rules would specify the name dinitrogen monoxide. (And $\\mathrm{H}_{2} \\mathrm{O}$ is usually called water, not dihydrogen monoxide.) You should commit to memory the common names of compounds as you encounter them."}
{"id": 2647, "contents": "364. Naming Covalent Compounds - \nName the following covalent compounds:\n(a) $\\mathrm{SF}_{6}$\n(b) $\\mathrm{N}_{2} \\mathrm{O}_{3}$\n(c) $\\mathrm{Cl}_{2} \\mathrm{O}_{7}$\n(d) $\\mathrm{P}_{4} \\mathrm{O}_{6}$"}
{"id": 2648, "contents": "365. Solution - \nBecause these compounds consist solely of nonmetals, we use prefixes to designate the number of atoms of each element:\n(a) sulfur hexafluoride\n(b) dinitrogen trioxide\n(c) dichlorine heptoxide\n(d) tetraphosphorus hexoxide"}
{"id": 2649, "contents": "366. Check Your Learning - \nWrite the formulas for the following compounds:\n(a) phosphorus pentachloride\n(b) dinitrogen monoxide\n(c) iodine heptafluoride\n(d) carbon tetrachloride"}
{"id": 2650, "contents": "367. Answer: - \n(a) $\\mathrm{PCl}_{5}$; (b) $\\mathrm{N}_{2} \\mathrm{O}$; (c) $\\mathrm{IF}_{7}$; (d) $\\mathrm{CCl}_{4}$"}
{"id": 2651, "contents": "368. LINK TO LEARNING - \nThe following website (http://openstax.org/l/16chemcompname) provides practice with naming chemical compounds and writing chemical formulas. You can choose binary, polyatomic, and variable charge ionic compounds, as well as molecular compounds."}
{"id": 2652, "contents": "369. Binary Acids - \nSome compounds containing hydrogen are members of an important class of substances known as acids. The chemistry of these compounds is explored in more detail in later chapters of this text, but for now, it will suffice to note that many acids release hydrogen ions, $\\mathrm{H}^{+}$, when dissolved in water. To denote this distinct chemical property, a mixture of water with an acid is given a name derived from the compound's name. If the compound is a binary acid (comprised of hydrogen and one other nonmetallic element):\n\n1. The word \"hydrogen\" is changed to the prefix hydro-\n2. The other nonmetallic element name is modified by adding the suffix -ic\n3. The word \"acid\" is added as a second word\n\nFor example, when the gas HCl (hydrogen chloride) is dissolved in water, the solution is called hydrochloric acid. Several other examples of this nomenclature are shown in Table 4.8.\n\nNames of Some Simple Acids\n\n| Name of Gas | Name of Acid |\n| :--- | :--- |\n| $\\mathrm{HF}(g)$, hydrogen fluoride | $\\mathrm{HF}(a q)$, hydrofluoric acid |\n| $\\mathrm{HCl}(g)$, hydrogen chloride | $\\mathrm{HCl}(a q)$, hydrochloric acid |\n| $\\mathrm{HBr}(g)$, hydrogen bromide | $\\mathrm{HBr}(\\mathrm{aq})$, hydrobromic acid |\n| $\\mathrm{HI}(g)$, hydrogen iodide | $\\mathrm{HI}(\\mathrm{aq})$, hydroiodic acid |\n| $\\mathrm{H}_{2} \\mathrm{~S}(g)$, hydrogen sulfide | $\\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{aq})$, hydrosulfuric acid |\n\nTABLE 4.8"}
{"id": 2653, "contents": "370. Oxyacids - \nMany compounds containing three or more elements (such as organic compounds or coordination compounds) are subject to specialized nomenclature rules that you will learn later. However, we will briefly\ndiscuss the important compounds known as oxyacids, compounds that contain hydrogen, oxygen, and at least one other element, and are bonded in such a way as to impart acidic properties to the compound (you will learn the details of this in a later chapter). Typical oxyacids consist of hydrogen combined with a polyatomic, oxygen-containing ion. To name oxyacids:\n\n1. Omit \"hydrogen\"\n2. Start with the root name of the anion\n3. Replace -ate with -ic, or -ite with -ous\n4. Add \"acid\"\n\nFor example, consider $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ (which you might be tempted to call \"hydrogen carbonate\"). To name this correctly, \"hydrogen\" is omitted; the -ate of carbonate is replace with -ic; and acid is added-so its name is carbonic acid. Other examples are given in Table 4.9. There are some exceptions to the general naming method (e.g., $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ is called sulfuric acid, not sulfic acid, and $\\mathrm{H}_{2} \\mathrm{SO}_{3}$ is sulfurous, not sulfous, acid).\n\nNames of Common Oxyacids\n\n| Formula | Anion Name | Acid Name |\n| :--- | :--- | :--- |\n| $\\mathrm{HC}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}$ | acetate | acetic acid |\n| $\\mathrm{HNO}_{3}$ | nitrate | nitric acid |\n| $\\mathrm{HNO}_{2}$ | nitrite | nitrous acid |\n| $\\mathrm{HClO}_{4}$ | perchlorate | perchloric acid |\n| $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ | carbonate | carbonic acid |\n| $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ | sulfate | sulfuric acid |\n| $\\mathrm{H}_{2} \\mathrm{SO}_{3}$ | sulfite | sulfurous acid |\n| $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ | phosphate | phosphoric acid |"}
{"id": 2654, "contents": "370. Oxyacids - \nTABLE 4.9"}
{"id": 2655, "contents": "371. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Write Lewis symbols for neutral atoms and ions\n- Draw Lewis structures depicting the bonding in simple molecules\n\nThus far in this chapter, we have discussed the various types of bonds that form between atoms and/or ions. In all cases, these bonds involve the sharing or transfer of valence shell electrons between atoms. In this section, we will explore the typical method for depicting valence shell electrons and chemical bonds, namely Lewis symbols and Lewis structures."}
{"id": 2656, "contents": "372. Lewis Symbols - \nWe use Lewis symbols to describe valence electron configurations of atoms and monatomic ions. A Lewis symbol consists of an elemental symbol surrounded by one dot for each of its valence electrons:\n\n- Ca\n\nFigure 4.10 shows the Lewis symbols for the elements of the third period of the periodic table.\n\n| Atoms | Electronic Configuration | Lewis Symbol |\n| :--- | :--- | :--- |\n| sodium | $[\\mathrm{Ne}] 3 s^{1}$ | $\\mathrm{Na} \\cdot$ |\n| magnesium | $[\\mathrm{Ne}] 3 s^{2}$ | $\\cdot \\mathrm{Mg} \\cdot$ |\n| aluminum | $[\\mathrm{Ne}] 3 s^{2} 3 p^{1}$ | $\\cdot \\dot{\\mathrm{Al} \\cdot} \\cdot$ |\n| silicon | $[\\mathrm{Ne}] 3 s^{2} 3 p^{2}$ | $\\cdot \\dot{\\mathrm{Si}} \\cdot$ |\n| phosphorus | $[\\mathrm{Ne}] 3 s^{2} 3 p^{3}$ | $\\ddot{\\mathrm{P}} \\cdot$ |\n| sulfur | $[\\mathrm{Ne}] 3 s^{2} 3 p^{4}$ | $\\ddot{\\mathrm{~S}} \\cdot$ |\n| chlorine | $[\\mathrm{Ne}] 3 s^{2} 3 p^{5}$ | $: \\ddot{\\mathrm{Cl}} \\cdot$ |\n| argon | $[\\mathrm{Ne}] 3 s^{2} 3 p^{6}$ | $: \\ddot{\\mathrm{Ar}}:$ |\n\nFIGURE 4.10 Lewis symbols illustrating the number of valence electrons for each element in the third period of the periodic table.\n\nLewis symbols can also be used to illustrate the formation of cations from atoms, as shown here for sodium and calcium:\n\\$\\underset{\\substack{sodium <br>\natom}}{\\mathrm{Na} \\cdot \\underset{\\substack{sodium <br>\n\ncation}}{\\mathrm{Na}^{+}}+\\mathrm{e}^{-}}\\$| calcium |\n| :---: |\n| atom |\\$>\\underset{\\substack{calcium <br>\n\ncation}}{\\mathrm{Ca}^{2+}}+\\mathrm{Ce}^{-}\\$"}
{"id": 2657, "contents": "372. Lewis Symbols - \ncation}}{\\mathrm{Ca}^{2+}}+\\mathrm{Ce}^{-}\\$\n\nLikewise, they can be used to show the formation of anions from atoms, as shown here for chlorine and sulfur:\n\n\nFigure 4.11 demonstrates the use of Lewis symbols to show the transfer of electrons during the formation of ionic compounds.\n\n| Metal |  | Nonmetal | Ionic Compound |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{Na} \\cdot$ <br> sodium atom | $+$ | $\\ddot{\\mathrm{Cl}}$ <br> chlorine atom | $\\longrightarrow \\quad \\mathrm{Na}^{+}[: \\stackrel{\\ddot{\\mathrm{C}} \\cdot:}{\\cdot .}]^{-}$ <br> sodium chloride (sodium ion and chloride ion) |\n| -Mg\u2022 <br> magnesium atom | + | $\\ddot{O}$ <br> oxygen atom | $\\longrightarrow \\quad \\mathrm{Mg}^{2+}[: \\ddot{\\mathrm{O}}:]^{2-}$ <br> magnesium oxide (magnesium ion and oxide ion) |\n| - Ca\u2022 <br> calcium atom | $+$ | $2: \\ddot{F}$ <br> fluorine atoms | $\\longrightarrow \\quad \\mathrm{Ca}^{2+}[: \\ddot{\\mathrm{F}}:]_{2}^{-}$ <br> calcium fluoride (calcium ion and two fluoride ions) |\n\nFIGURE 4.11 Cations are formed when atoms lose electrons, represented by fewer Lewis dots, whereas anions are formed by atoms gaining electrons. The total number of electrons does not change."}
{"id": 2658, "contents": "373. Lewis Structures - \nWe also use Lewis symbols to indicate the formation of covalent bonds, which are shown in Lewis structures, drawings that describe the bonding in molecules and polyatomic ions. For example, when two chlorine atoms form a chlorine molecule, they share one pair of electrons:\n\n\nThe Lewis structure indicates that each Cl atom has three pairs of electrons that are not used in bonding (called lone pairs) and one shared pair of electrons (written between the atoms). A dash (or line) is sometimes used to indicate a shared pair of electrons:\n\n\nA single shared pair of electrons is called a single bond. Each Cl atom interacts with eight valence electrons: the six in the lone pairs and the two in the single bond."}
{"id": 2659, "contents": "374. The Octet Rule - \nThe other halogen molecules ( $\\mathrm{F}_{2}, \\mathrm{Br}_{2}, \\mathrm{I}_{2}$, and $\\mathrm{At}_{2}$ ) form bonds like those in the chlorine molecule: one single bond between atoms and three lone pairs of electrons per atom. This allows each halogen atom to have a noble gas electron configuration. The tendency of main group atoms to form enough bonds to obtain eight valence electrons is known as the octet rule.\n\nThe number of bonds that an atom can form can often be predicted from the number of electrons needed to reach an octet (eight valence electrons); this is especially true of the nonmetals of the second period of the periodic table ( $\\mathrm{C}, \\mathrm{N}, \\mathrm{O}$, and F ). For example, each atom of a group 14 element has four electrons in its outermost shell and therefore requires four more electrons to reach an octet. These four electrons can be gained by forming four covalent bonds, as illustrated here for carbon in $\\mathrm{CCl}_{4}$ (carbon tetrachloride) and silicon in $\\mathrm{SiH}_{4}$ (silane). Because hydrogen only needs two electrons to fill its valence shell, it is an exception to the octet rule. The transition elements and inner transition elements also do not follow the octet rule:\n\n\nGroup 15 elements such as nitrogen have five valence electrons in the atomic Lewis symbol: one lone pair and three unpaired electrons. To obtain an octet, these atoms form three covalent bonds, as in $\\mathrm{NH}_{3}$ (ammonia). Oxygen and other atoms in group 16 obtain an octet by forming two covalent bonds:\n\n\nDouble and Triple Bonds\nAs previously mentioned, when a pair of atoms shares one pair of electrons, we call this a single bond. However, a pair of atoms may need to share more than one pair of electrons in order to achieve the requisite octet. A double bond forms when two pairs of electrons are shared between a pair of atoms, as between the carbon and oxygen atoms in $\\mathrm{CH}_{2} \\mathrm{O}$ (formaldehyde) and between the two carbon atoms in $\\mathrm{C}_{2} \\mathrm{H}_{4}$ (ethylene):"}
{"id": 2660, "contents": "374. The Octet Rule - \nA triple bond forms when three electron pairs are shared by a pair of atoms, as in carbon monoxide (CO) and the cyanide ion $\\left(\\mathrm{CN}^{-}\\right)$:\n\n```\n:C:::O: or :C\\equivO: :C:::N\\overline{* or :C\\equivN:}\n```\n\ncarbon monoxide cyanide ion"}
{"id": 2661, "contents": "375. Writing Lewis Structures with the Octet Rule - \nFor very simple molecules and molecular ions, we can write the Lewis structures by merely pairing up the unpaired electrons on the constituent atoms. See these examples:\n\n\nFor more complicated molecules and molecular ions, it is helpful to follow the step-by-step procedure outlined here:\n\n1. Determine the total number of valence (outer shell) electrons. For cations, subtract one electron for each positive charge. For anions, add one electron for each negative charge.\n2. Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom. (Generally, the least electronegative element should be placed in the center.) Connect each atom to the central atom with a single bond (one electron pair).\n3. Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen), completing an\noctet around each atom.\n4. Place all remaining electrons on the central atom.\n5. Rearrange the electrons of the outer atoms to make multiple bonds with the central atom in order to obtain octets wherever possible.\n\nLet us determine the Lewis structures of $\\mathrm{SiH}_{4}, \\mathrm{CHO}_{2}{ }^{-}, \\mathrm{NO}^{+}$, and $\\mathrm{OF}_{2}$ as examples in following this procedure:\n\n1. Determine the total number of valence (outer shell) electrons in the molecule or ion.\n\n- For a molecule, we add the number of valence electrons on each atom in the molecule:\n\n> | $\\mathrm{SiH}_{4}$ |\n| :--- |\n| $\\mathrm{Si}: 4$ valence electrons/atom $\\times 1$ atom $=4$ |\n| $+\\mathrm{H}: 1$ valence electron $/$ atom $\\times 4$ atoms $=4$ |\n\n$$\n=8 \\text { valence electrons }\n$$\n\n- For a negative ion, such as $\\mathrm{CHO}_{2}{ }^{-}$, we add the number of valence electrons on the atoms to the number of negative charges on the ion (one electron is gained for each single negative charge):"}
{"id": 2662, "contents": "375. Writing Lewis Structures with the Octet Rule - \n$$\n\\begin{aligned}\n& \\mathrm{CHO}_{2}{ }^{-} \\\\\n& \\mathrm{C}: 4 \\text { valence electrons/atom } \\times 1 \\text { atom }=4 \\\\\n& \\mathrm{H}: 1 \\text { valence electron/atom } \\times 1 \\text { atom }=1 \\\\\n& \\mathrm{O}: 6 \\text { valence electrons/atom } \\times 2 \\text { atoms }=12 \\\\\n& +\\quad 1 \\text { additional electron }=1 \\\\\n& \\hline\n\\end{aligned}\n$$\n\n- For a positive ion, such as $\\mathrm{NO}^{+}$, we add the number of valence electrons on the atoms in the ion and then subtract the number of positive charges on the ion (one electron is lost for each single positive charge) from the total number of valence electrons:\n\n$$\n\\begin{aligned}\n& \\mathrm{NO}^{+} \\\\\n& \\mathrm{N}: 5 \\text { valence electrons/atom } \\times 1 \\text { atom }=5 \\\\\n& \\\\\n& \\begin{aligned}\n\\mathrm{O}: 6 \\text { valence electron/atom } \\times 1 \\text { atom } & =6 \\\\\n+-1 \\text { electron (positive charge) } & =-1 \\\\\n\\hline & =10 \\text { valence electrons }\n\\end{aligned}\n\\end{aligned}\n$$\n\n- Since $\\mathrm{OF}_{2}$ is a neutral molecule, we simply add the number of valence electrons:\n\n$$\n\\begin{aligned}\n& \\mathrm{OF}_{2} \\\\\n& \\mathrm{O}: 6 \\text { valence electrons/atom } \\times 1 \\text { atom }=6 \\\\\n&+\\mathrm{F}: 7 \\text { valence electrons/atom } \\times 2 \\text { atoms }=14 \\\\\n& \\hline=20 \\text { valence electrons }\n\\end{aligned}\n$$\n\n2. Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom and connecting each atom to the central atom with a single (one electron pair) bond. (Note that we denote ions with brackets around the structure, indicating the charge outside the brackets:)"}
{"id": 2663, "contents": "375. Writing Lewis Structures with the Octet Rule - \nWhen several arrangements of atoms are possible, as for $\\mathrm{CHO}_{2}{ }^{-}$, we must use experimental evidence to choose the correct one. In general, the less electronegative elements are more likely to be central atoms. In $\\mathrm{CHO}_{2}{ }^{-}$, the less electronegative carbon atom occupies the central position with the oxygen and hydrogen atoms surrounding it. Other examples include P in $\\mathrm{POCl}_{3}, \\mathrm{~S}$ in $\\mathrm{SO}_{2}$, and Cl in $\\mathrm{ClO}_{4}{ }^{-}$. An exception is that hydrogen is almost never a central atom. As the most electronegative element, fluorine also cannot be a central atom.\n3. Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen) to complete their valence shells with an octet of electrons.\n\n- There are no remaining electrons on $\\mathrm{SiH}_{4}$, so it is unchanged:\n\n\n\n\n4. Place all remaining electrons on the central atom.\n\n- For $\\mathrm{SiH}_{4}, \\mathrm{CHO}_{2}{ }^{-}$, and $\\mathrm{NO}^{+}$, there are no remaining electrons; we already placed all of the electrons determined in Step 1.\n- For $\\mathrm{OF}_{2}$, we had 16 electrons remaining in Step 3, and we placed 12, leaving 4 to be placed on the central atom:\n\n\n5. Rearrange the electrons of the outer atoms to make multiple bonds with the central atom in order to obtain octets wherever possible.\n\n- $\\mathrm{SiH}_{4}$ : Si already has an octet, so nothing needs to be done.\n- $\\mathrm{CHO}_{2}{ }^{-}$: We have distributed the valence electrons as lone pairs on the oxygen atoms, but the carbon atom lacks an octet:\n\n- $\\mathrm{NO}^{+}$: For this ion, we added eight valence electrons, but neither atom has an octet. We cannot add any more electrons since we have already used the total that we found in Step 1, so we must move electrons to form a multiple bond:"}
{"id": 2664, "contents": "375. Writing Lewis Structures with the Octet Rule - \n$$\n[: \\ddot{N}-\\ddot{O}:]^{+} \\text {gives }[: \\ddot{N}=0:]^{+}\n$$\n\nThis still does not produce an octet, so we must move another pair, forming a triple bond:\n\n$$\n[: N \\equiv O:]^{+}\n$$\n\n- In $\\mathrm{OF}_{2}$, each atom has an octet as drawn, so nothing changes."}
{"id": 2665, "contents": "377. Writing Lewis Structures - \nNASA's Cassini-Huygens mission detected a large cloud of toxic hydrogen cyanide (HCN) on Titan, one of Saturn's moons. Titan also contains ethane $\\left(\\mathrm{H}_{3} \\mathrm{CCH}_{3}\\right)$, acetylene (HCCH), and ammonia $\\left(\\mathrm{NH}_{3}\\right)$. What are the Lewis structures of these molecules?"}
{"id": 2666, "contents": "378. Solution - \nStep 1. Calculate the number of valence electrons.\nHCN: $(1 \\times 1)+(4 \\times 1)+(5 \\times 1)=10$\n$\\mathrm{H}_{3} \\mathrm{CCH}_{3}:(1 \\times 3)+(2 \\times 4)+(1 \\times 3)=14$\nHCCH: $(1 \\times 1)+(2 \\times 4)+(1 \\times 1)=10$\n$\\mathrm{NH}_{3}:(5 \\times 1)+(3 \\times 1)=8$\nStep 2. Draw a skeleton and connect the atoms with single bonds. Remember that H is never a central atom:\n\n\nStep 3. Where needed, distribute electrons to the terminal atoms:\n\n\nHCN: six electrons placed on N\n$\\mathrm{H}_{3} \\mathrm{CCH}_{3}$ : no electrons remain\nHCCH: no terminal atoms capable of accepting electrons\n$\\mathrm{NH}_{3}$ : no terminal atoms capable of accepting electrons\nStep 4. Where needed, place remaining electrons on the central atom:\n\n\nHCN : no electrons remain\n$\\mathrm{H}_{3} \\mathrm{CCH}_{3}$ : no electrons remain\nHCCH: four electrons placed on carbon\n$\\mathrm{NH}_{3}$ : two electrons placed on nitrogen\nStep 5. Where needed, rearrange electrons to form multiple bonds in order to obtain an octet on each atom:\nHCN : form two more $\\mathrm{C}-\\mathrm{N}$ bonds\n$\\mathrm{H}_{3} \\mathrm{CCH}_{3}$ : all atoms have the correct number of electrons\nHCCH : form a triple bond between the two carbon atoms\n$\\mathrm{NH}_{3}$ : all atoms have the correct number of electrons"}
{"id": 2667, "contents": "379. Check Your Learning - \nBoth carbon monoxide, CO , and carbon dioxide, $\\mathrm{CO}_{2}$, are products of the combustion of fossil fuels. Both of these gases also cause problems: CO is toxic and $\\mathrm{CO}_{2}$ has been implicated in global climate change. What are the Lewis structures of these two molecules?"}
{"id": 2668, "contents": "380. Answer: - \n:c\u4e09o: : $\\ddot{\\mathrm{O}}=\\mathrm{c}=\\ddot{\\mathrm{o}}$ :"}
{"id": 2669, "contents": "382. Fullerene Chemistry - \nCarbon, in various forms and compounds, has been known since prehistoric times, . Soot has been used as a pigment (often called carbon black) for thousands of years. Charcoal, high in carbon content, has likewise been critical to human development. Carbon is the key additive to iron in the steelmaking process, and diamonds have a unique place in both culture and industry. With all this usage came significant study, particularly with the emergence of organic chemistry. And even with all the known forms and functions of the element, scientists began to uncover the potential for even more varied and extensive carbon structures.\n\nAs early as the 1960s, chemists began to observe complex carbon structures, but they had little evidence to support their concepts, or their work did not make it into the mainstream. Eiji Osawa predicted a spherical form based on observations of a similar structure, but his work was not widely known outside Japan. In a similar manner, the most comprehensive advance was likely computational chemist Elena Galpern's, who in 1973 predicted a highly stable, 60-carbon molecule; her work was also isolated to her native Russia. Still later, Harold Kroto, working with Canadian radio astronomers, sought to uncover the nature of long carbon chains that had been discovered in interstellar space.\n\nKroto sought to use a machine developed by Richard Smalley's team at Rice University to learn more about these structures. Together with Robert Curl, who had introduced them, and three graduate students-James Heath, Sean O'Brien, and Yuan Liu-they performed an intensive series of experiments that led to a major discovery.\n\nIn 1996, the Nobel Prize in Chemistry was awarded to Richard Smalley (Figure 4.12), Robert Curl, and Harold Kroto for their work in discovering a new form of carbon, the $\\mathrm{C}_{60}$ buckminsterfullerene molecule (Figure 4.1). An entire class of compounds, including spheres and tubes of various shapes, were discovered based on $\\mathrm{C}_{60}$. This type of molecule, called a fullerene, shows promise in a variety of applications. Because of their size and shape, fullerenes can encapsulate other molecules, so they have shown potential in various applications from hydrogen storage to targeted drug delivery systems. They also possess unique electronic and optical properties that have been put to good use in solar powered devices and chemical sensors."}
{"id": 2670, "contents": "382. Fullerene Chemistry - \nFIGURE 4.12 Richard Smalley (1943-2005), a professor of physics, chemistry, and astronomy at Rice University, was one of the leading advocates for fullerene chemistry. Upon his death in 2005, the US Senate honored him as the \"Father of Nanotechnology.\" (credit: United States Department of Energy)"}
{"id": 2671, "contents": "383. Exceptions to the Octet Rule - \nMany covalent molecules have central atoms that do not have eight electrons in their Lewis structures. These molecules fall into three categories:\n\n- Odd-electron molecules have an odd number of valence electrons, and therefore have an unpaired electron.\n- Electron-deficient molecules have a central atom that has fewer electrons than needed for a noble gas configuration.\n- Hypervalent molecules have a central atom that has more electrons than needed for a noble gas configuration."}
{"id": 2672, "contents": "384. Odd-electron Molecules - \nWe call molecules that contain an odd number of electrons free radicals. Nitric oxide, NO, is an example of an odd-electron molecule; it is produced in internal combustion engines when oxygen and nitrogen react at high temperatures.\n\nTo draw the Lewis structure for an odd-electron molecule like NO, we follow the same five steps we would for other molecules, but with a few minor changes:\n\n1. Determine the total number of valence (outer shell) electrons. The sum of the valence electrons is 5 (from $\\mathrm{N})+6($ from O$)=11$. The odd number immediately tells us that we have a free radical, so we know that not every atom can have eight electrons in its valence shell.\n2. Draw a skeleton structure of the molecule. We can easily draw a skeleton with an N-O single bond: N-O\n3. Distribute the remaining electrons as lone pairs on the terminal atoms. In this case, there is no central atom, so we distribute the electrons around both atoms. We give eight electrons to the more electronegative atom in these situations; thus oxygen has the filled valence shell:\n\n4. Place all remaining electrons on the central atom. Since there are no remaining electrons, this step does not apply.\n5. Rearrange the electrons to make multiple bonds with the central atom in order to obtain octets wherever possible. We know that an odd-electron molecule cannot have an octet for every atom, but we want to get each atom as close to an octet as possible. In this case, nitrogen has only five electrons around it. To move closer to an octet for nitrogen, we take one of the lone pairs from oxygen and use it to form a NO double bond. (We cannot take another lone pair of electrons on oxygen and form a triple bond because nitrogen would then have nine electrons:)\n$: \\dot{N}=\\ddot{O}$ :"}
{"id": 2673, "contents": "385. Electron-deficient Molecules - \nWe will also encounter a few molecules that contain central atoms that do not have a filled valence shell. Generally, these are molecules with central atoms from groups 2 and 13, outer atoms that are hydrogen, or other atoms that do not form multiple bonds. For example, in the Lewis structures of beryllium dihydride, $\\mathrm{BeH}_{2}$, and boron trifluoride, $\\mathrm{BF}_{3}$, the beryllium and boron atoms each have only four and six electrons, respectively. It is possible to draw a structure with a double bond between a boron atom and a fluorine atom in $\\mathrm{BF}_{3}$, satisfying the octet rule, but experimental evidence indicates the bond lengths are closer to that expected for B-F single bonds. This suggests the best Lewis structure has three B-F single bonds and an electron deficient boron. The reactivity of the compound is also consistent with an electron deficient boron. However, the $B-F$ bonds are slightly shorter than what is actually expected for $B-F$ single bonds, indicating that some double bond character is found in the actual molecule.\n\n\nAn atom like the boron atom in $\\mathrm{BF}_{3}$, which does not have eight electrons, is very reactive. It readily combines with a molecule containing an atom with a lone pair of electrons. For example, $\\mathrm{NH}_{3}$ reacts with $\\mathrm{BF}_{3}$ because the lone pair on nitrogen can be shared with the boron atom:"}
{"id": 2674, "contents": "386. Hypervalent Molecules - \nElements in the second period of the periodic table $(n=2)$ can accommodate only eight electrons in their valence shell orbitals because they have only four valence orbitals (one $2 s$ and three $2 p$ orbitals). Elements in the third and higher periods ( $n \\geq 3$ ) have more than four valence orbitals and can share more than four pairs of electrons with other atoms because they have empty $d$ orbitals in the same shell. Molecules formed from these elements are sometimes called hypervalent molecules. Figure 4.13 shows the Lewis structures for two hypervalent molecules, $\\mathrm{PCl}_{5}$ and $\\mathrm{SF}_{6}$.\n\n\n\nFIGURE 4.13 In $\\mathrm{PCl}_{5}$, the central atom phosphorus shares five pairs of electrons. In $\\mathrm{SF}_{6}$, sulfur shares six pairs of electrons.\n\nIn some hypervalent molecules, such as $\\mathrm{IF}_{5}$ and $\\mathrm{XeF}_{4}$, some of the electrons in the outer shell of the central atom are lone pairs:\n\n\n\nWhen we write the Lewis structures for these molecules, we find that we have electrons left over after filling the valence shells of the outer atoms with eight electrons. These additional electrons must be assigned to the central atom."}
{"id": 2675, "contents": "388. Writing Lewis Structures: Octet Rule Violations - \nXenon is a noble gas, but it forms a number of stable compounds. We examined $\\mathrm{XeF}_{4}$ earlier. What are the Lewis structures of $\\mathrm{XeF}_{2}$ and $\\mathrm{XeF}_{6}$ ?"}
{"id": 2676, "contents": "389. Solution - \nWe can draw the Lewis structure of any covalent molecule by following the six steps discussed earlier. In this case, we can condense the last few steps, since not all of them apply.\n\nStep 1. Calculate the number of valence electrons:\n$\\mathrm{XeF}_{2}: 8+(2 \\times 7)=22$\n$\\mathrm{XeF}_{6}: 8+(6 \\times 7)=50$\nStep 2. Draw a skeleton joining the atoms by single bonds. Xenon will be the central atom because fluorine cannot be a central atom:\n\n\nStep 3. Distribute the remaining electrons.\n$\\mathrm{XeF}_{2}$ : We place three lone pairs of electrons around each F atom, accounting for 12 electrons and giving\neach F atom 8 electrons. Thus, six electrons (three lone pairs) remain. These lone pairs must be placed on the Xe atom. This is acceptable because Xe atoms have empty valence shell $d$ orbitals and can accommodate more than eight electrons. The Lewis structure of $\\mathrm{XeF}_{2}$ shows two bonding pairs and three lone pairs of electrons around the Xe atom:\n\n$\\mathrm{XeF}_{6}$ : We place three lone pairs of electrons around each F atom, accounting for 36 electrons. Two electrons remain, and this lone pair is placed on the Xe atom:"}
{"id": 2677, "contents": "390. Check Your Learning - \nThe halogens form a class of compounds called the interhalogens, in which halogen atoms covalently bond to each other. Write the Lewis structures for the interhalogens $\\mathrm{BrCl}_{3}$ and $\\mathrm{ICl}_{4}{ }^{-}$."}
{"id": 2678, "contents": "392. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Compute formal charges for atoms in any Lewis structure\n- Use formal charges to identify the most reasonable Lewis structure for a given molecule\n- Explain the concept of resonance and draw Lewis structures representing resonance forms for a given molecule\n\nIn the previous section, we discussed how to write Lewis structures for molecules and polyatomic ions. As we have seen, however, in some cases, there is seemingly more than one valid structure for a molecule. We can use the concept of formal charges to help us predict the most appropriate Lewis structure when more than one is reasonable."}
{"id": 2679, "contents": "393. Calculating Formal Charge - \nThe formal charge of an atom in a molecule is the hypothetical charge the atom would have if we could redistribute the electrons in the bonds evenly between the atoms. Another way of saying this is that formal charge results when we take the number of valence electrons of a neutral atom, subtract the nonbonding electrons, and then subtract the number of bonds connected to that atom in the Lewis structure.\n\nThus, we calculate formal charge as follows:\n\n$$\n\\text { formal charge }=\\# \\text { valence shell electrons (free atom) }-\\# \\text { lone pair electrons }-\\frac{1}{2} \\# \\text { bonding electrons }\n$$\n\nWe can double-check formal charge calculations by determining the sum of the formal charges for the whole structure. The sum of the formal charges of all atoms in a molecule must be zero; the sum of the formal charges in an ion should equal the charge of the ion.\n\nWe must remember that the formal charge calculated for an atom is not the actual charge of the atom in the molecule. Formal charge is only a useful bookkeeping procedure; it does not indicate the presence of actual charges."}
{"id": 2680, "contents": "395. Calculating Formal Charge from Lewis Structures - \nAssign formal charges to each atom in the interhalogen ion $\\mathrm{ICl}_{4}{ }^{-}$."}
{"id": 2681, "contents": "396. Solution - \nStep 1. We divide the bonding electron pairs equally for all I-Cl bonds:\n\n\nStep 2. We assign lone pairs of electrons to their atoms. Each Cl atom now has seven electrons assigned to it, and the I atom has eight.\nStep 3. Subtract this number from the number of valence electrons for the neutral atom:\nI: $7-8=-1$\nCl: $7-7=0$\nThe sum of the formal charges of all the atoms equals -1 , which is identical to the charge of the ion ( -1 )."}
{"id": 2682, "contents": "397. Check Your Learning - \nCalculate the formal charge for each atom in the carbon monoxide molecule:\n\n```\n:C\\equivO:\n```"}
{"id": 2683, "contents": "398. Answer: - \nC $-1,0+1$"}
{"id": 2684, "contents": "400. Calculating Formal Charge from Lewis Structures - \nAssign formal charges to each atom in the interhalogen molecule $\\mathrm{BrCl}_{3}$."}
{"id": 2685, "contents": "401. Solution - \nStep 1. Assign one of the electrons in each $\\mathrm{Br}-\\mathrm{Cl}$ bond to the Br atom and one to the Cl atom in that bond:\n\n\nStep 2. Assign the lone pairs to their atom. Now each Cl atom has seven electrons and the Br atom has seven electrons.\nStep 3. Subtract this number from the number of valence electrons for the neutral atom. This gives the formal charge:\nBr: 7-7 = 0\nCl: 7-7=0\nAll atoms in $\\mathrm{BrCl}_{3}$ have a formal charge of zero, and the sum of the formal charges totals zero, as it must in a neutral molecule."}
{"id": 2686, "contents": "402. Check Your Learning - \nDetermine the formal charge for each atom in $\\mathrm{NCl}_{3}$."}
{"id": 2687, "contents": "403. Answer: - \nN : 0; all three Cl atoms: 0"}
{"id": 2688, "contents": "404. Using Formal Charge to Predict Molecular Structure - \nThe arrangement of atoms in a molecule or ion is called its molecular structure. In many cases, following the steps for writing Lewis structures may lead to more than one possible molecular structure-different multiple bond and lone-pair electron placements or different arrangements of atoms, for instance. A few guidelines involving formal charge can be helpful in deciding which of the possible structures is most likely for a particular molecule or ion:\n\n1. A molecular structure in which all formal charges are zero is preferable to one in which some formal charges are not zero.\n2. If the Lewis structure must have nonzero formal charges, the arrangement with the smallest nonzero formal charges is preferable.\n3. Lewis structures are preferable when adjacent formal charges are zero or of the opposite sign.\n4. When we must choose among several Lewis structures with similar distributions of formal charges, the structure with the negative formal charges on the more electronegative atoms is preferable.\n\nTo see how these guidelines apply, let us consider some possible structures for carbon dioxide, $\\mathrm{CO}_{2}$. We know from our previous discussion that the less electronegative atom typically occupies the central position, but formal charges allow us to understand why this occurs. We can draw three possibilities for the structure: carbon in the center and double bonds, carbon in the center with a single and triple bond, and oxygen in the center with double bonds:\n\n| $\\ddot{\\mathrm{O}}=\\mathrm{C}=\\ddot{\\mathrm{O}}$ | $: \\mathrm{O} \\equiv \\mathrm{C}-\\ddot{\\mathrm{O}}:$ | $\\ddot{\\mathrm{O}}=\\mathrm{O}=\\ddot{\\mathrm{C}}$ | Structure |  |  |  |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| 0 | 0 | 0 | +1 | 0 | -1 | 0 |\n| $\\cdots$ | +2 | -2 | Formal charge |  |  |  |\n\nComparing the three formal charges, we can definitively identify the structure on the left as preferable because it has only formal charges of zero (Guideline 1)."}
{"id": 2689, "contents": "404. Using Formal Charge to Predict Molecular Structure - \nComparing the three formal charges, we can definitively identify the structure on the left as preferable because it has only formal charges of zero (Guideline 1).\n\nAs another example, the thiocyanate ion, an ion formed from a carbon atom, a nitrogen atom, and a sulfur atom, could have three different molecular structures: $\\mathrm{NCS}^{-}, \\mathrm{CNS}^{-}$, or $\\mathrm{CSN}^{-}$. The formal charges present in each of these molecular structures can help us pick the most likely arrangement of atoms. Possible Lewis structures and the formal charges for each of the three possible structures for the thiocyanate ion are shown here:\n\n\nNote that the sum of the formal charges in each case is equal to the charge of the ion ( -1 ). However, the first arrangement of atoms is preferred because it has the lowest number of atoms with nonzero formal charges (Guideline 2). Also, it places the least electronegative atom in the center, and the negative charge on the more electronegative element (Guideline 4)."}
{"id": 2690, "contents": "406. Using Formal Charge to Determine Molecular Structure - \nNitrous oxide, $\\mathrm{N}_{2} \\mathrm{O}$, commonly known as laughing gas, is used as an anesthetic in minor surgeries, such as the routine extraction of wisdom teeth. Which is the more likely structure for nitrous oxide?"}
{"id": 2691, "contents": "407. Solution - \nDetermining formal charge yields the following:\n\n| $: N \\equiv N=\\ddot{O}:$ | $: \\ddot{N}=O=\\ddot{N}:$ |  |\n| ---: | :--- | :--- |\n| 0 | $+1-1$ | $-1+2-1$ |\n\nThe structure with a terminal oxygen atom best satisfies the criteria for the most stable distribution of formal charge:\n\n\nThe number of atoms with formal charges are minimized (Guideline 2), there is no formal charge with a magnitude greater than one (Guideline 2), the negative formal charge is on the more electronegative element (Guideline 4), and the less electronegative atom is in the center position."}
{"id": 2692, "contents": "408. Check Your Learning - \nWhich is the most likely molecular structure for the nitrite $\\left(\\mathrm{NO}_{2}{ }^{-}\\right)$ion?\n$[: \\ddot{N}=\\ddot{O}-\\ddot{O}:]^{-}$or $\\quad[: \\ddot{O}=\\ddot{N}-\\ddot{O}:]^{-}$\n\nAnswer:\n$\\mathrm{ONO}^{-}$"}
{"id": 2693, "contents": "409. Resonance - \nNotice that the more likely structure for the nitrite anion in Example 4.10 may actually be drawn in two different ways, distinguished by the locations of the $\\mathrm{N}-\\mathrm{O}$ and $\\mathrm{N}=\\mathrm{O}$ bonds:\n\n\n\nIf nitrite ions do indeed contain a single and a double bond, we would expect for the two bond lengths to be different. A double bond between two atoms is shorter (and stronger) than a single bond between the same two atoms. Experiments show, however, that both $\\mathrm{N}-\\mathrm{O}$ bonds in $\\mathrm{NO}_{2}{ }^{-}$have the same strength and length, and are identical in all other properties.\n\nIt is not possible to write a single Lewis structure for $\\mathrm{NO}_{2}{ }^{-}$in which nitrogen has an octet and both bonds are equivalent. Instead, we use the concept of resonance: if two or more Lewis structures with the same arrangement of atoms can be written for a molecule or ion, the actual distribution of electrons is an average of that shown by the various Lewis structures. The actual distribution of electrons in each of the nitrogen-oxygen bonds in $\\mathrm{NO}_{2}{ }^{-}$is the average of a double bond and a single bond. We call the individual Lewis structures resonance forms. The actual electronic structure of the molecule (the average of the resonance forms) is called a resonance hybrid of the individual resonance forms. A double-headed arrow between Lewis structures indicates that they are resonance forms."}
{"id": 2694, "contents": "409. Resonance - \nWe should remember that a molecule described as a resonance hybrid never possesses an electronic structure described by either resonance form. It does not fluctuate between resonance forms; rather, the actual electronic structure is always the average of that shown by all resonance forms. George Wheland, one of the pioneers of resonance theory, used a historical analogy to describe the relationship between resonance forms and resonance hybrids. A medieval traveler, having never before seen a rhinoceros, described it as a hybrid of a dragon and a unicorn because it had many properties in common with both. Just as a rhinoceros is neither a dragon sometimes nor a unicorn at other times, a resonance hybrid is neither of its resonance forms at any\ngiven time. Like a rhinoceros, it is a real entity that experimental evidence has shown to exist. It has some characteristics in common with its resonance forms, but the resonance forms themselves are convenient, imaginary images (like the unicorn and the dragon).\n\nThe carbonate anion, $\\mathrm{CO}_{3}{ }^{2-}$, provides a second example of resonance:\n\n\nOne oxygen atom must have a double bond to carbon to complete the octet on the central atom. All oxygen atoms, however, are equivalent, and the double bond could form from any one of the three atoms. This gives rise to three resonance forms of the carbonate ion. Because we can write three identical resonance structures, we know that the actual arrangement of electrons in the carbonate ion is the average of the three structures. Again, experiments show that all three $\\mathrm{C}-\\mathrm{O}$ bonds are exactly the same."}
{"id": 2695, "contents": "410. LINK TO LEARNING - \nUse this online quiz (http://openstax.org/l/16LewisMake) to practice your skills in drawing resonance structures and estimating formal charges."}
{"id": 2696, "contents": "411. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Predict the structures of small molecules using valence shell electron pair repulsion (VSEPR) theory\n- Explain the concepts of polar covalent bonds and molecular polarity\n- Assess the polarity of a molecule based on its bonding and structure\n\nThus far, we have used two-dimensional Lewis structures to represent molecules. However, molecular structure is actually three-dimensional, and it is important to be able to describe molecular bonds in terms of their distances, angles, and relative arrangements in space (Figure 4.14). A bond angle is the angle between any two bonds that include a common atom, usually measured in degrees. A bond distance (or bond length) is the distance between the nuclei of two bonded atoms along the straight line joining the nuclei. Bond distances are measured in \u00c5ngstroms ( $1 \\AA=10^{-10} \\mathrm{~m}$ ) or picometers ( $1 \\mathrm{pm}=10^{-12} \\mathrm{~m}, 100 \\mathrm{pm}=1 \\AA$ ).\n\n\nFIGURE 4.14 Bond distances (lengths) and angles are shown for the formaldehyde molecule, $\\mathrm{H}_{2} \\mathrm{CO}$."}
{"id": 2697, "contents": "412. VSEPR Theory - \nValence shell electron-pair repulsion theory (VSEPR theory) enables us to predict the molecular structure, including approximate bond angles around a central atom, of a molecule from an examination of the number of bonds and lone electron pairs in its Lewis structure. The VSEPR model assumes that electron pairs in the valence shell of a central atom will adopt an arrangement that minimizes repulsions between these electron pairs by maximizing the distance between them. The electrons in the valence shell of a central atom form\neither bonding pairs of electrons, located primarily between bonded atoms, or lone pairs. The electrostatic repulsion of these electrons is reduced when the various regions of high electron density assume positions as far from each other as possible.\n\nVSEPR theory predicts the arrangement of electron pairs around each central atom and, usually, the correct arrangement of atoms in a molecule. We should understand, however, that the theory only considers electronpair repulsions. Other interactions, such as nuclear-nuclear repulsions and nuclear-electron attractions, are also involved in the final arrangement that atoms adopt in a particular molecular structure.\n\nAs a simple example of VSEPR theory, let us predict the structure of a gaseous $\\mathrm{BeF}_{2}$ molecule. The Lewis structure of $\\mathrm{BeF}_{2}$ (Figure 4.15) shows only two electron pairs around the central beryllium atom. With two bonds and no lone pairs of electrons on the central atom, the bonds are as far apart as possible, and the electrostatic repulsion between these regions of high electron density is reduced to a minimum when they are on opposite sides of the central atom. The bond angle is $180^{\\circ}$ (Figure 4.15).\n\n$$\n\\because \\ddot{\\mathrm{F}} \\mathrm{Be}_{\\mathrm{Be}}^{180^{\\circ}}\n$$\n\nFIGURE 4.15 The $\\mathrm{BeF}_{2}$ molecule adopts a linear structure in which the two bonds are as far apart as possible, on opposite sides of the Be atom."}
{"id": 2698, "contents": "412. VSEPR Theory - \nFIGURE 4.15 The $\\mathrm{BeF}_{2}$ molecule adopts a linear structure in which the two bonds are as far apart as possible, on opposite sides of the Be atom.\n\nFigure 4.16 illustrates this and other electron-pair geometries that minimize the repulsions among regions of high electron density (bonds and/or lone pairs). Two regions of electron density around a central atom in a molecule form a linear geometry; three regions form a trigonal planar geometry; four regions form a tetrahedral geometry; five regions form a trigonal bipyramidal geometry; and six regions form an octahedral geometry.\n\n| Number of <br> regions | Two regions of <br> high electron <br> density (bonds <br> and/or <br> unshared <br> pairs) | Three regions <br> of high <br> electron <br> density (bonds <br> and/or <br> unshared <br> pairs) | Four regions <br> of high <br> electron <br> density (bonds <br> and/or <br> unshared <br> pairs) | Five regions of <br> high electron <br> density (bonds <br> and/or <br> unshared <br> pairs) | Six regions of <br> high electron <br> density (bonds <br> and/or <br> unshared <br> pairs) |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n| Spatial <br> arrangement |  |  |  |  |  |\n\nFIGURE 4.16 The basic electron-pair geometries predicted by VSEPR theory maximize the space around any region of electron density (bonds or lone pairs)."}
{"id": 2699, "contents": "413. Electron-pair Geometry versus Molecular Structure - \nIt is important to note that electron-pair geometry around a central atom is not the same thing as its molecular structure. The electron-pair geometries shown in Figure 4.16 describe all regions where electrons are located, bonds as well as lone pairs. Molecular structure describes the location of the atoms, not the electrons.\n\nWe differentiate between these two situations by naming the geometry that includes all electron pairs the electron-pair geometry. The structure that includes only the placement of the atoms in the molecule is called the molecular structure. The electron-pair geometries will be the same as the molecular structures when there are no lone electron pairs around the central atom, but they will be different when there are lone pairs present on the central atom.\n\nFor example, the methane molecule, $\\mathrm{CH}_{4}$, which is the major component of natural gas, has four bonding pairs of electrons around the central carbon atom; the electron-pair geometry is tetrahedral, as is the molecular structure (Figure 4.17). On the other hand, the ammonia molecule, $\\mathrm{NH}_{3}$, also has four electron pairs associated with the nitrogen atom, and thus has a tetrahedral electron-pair geometry. One of these regions, however, is a lone pair, which is not included in the molecular structure, and this lone pair influences the shape of the molecule (Figure 4.18).\n\n\nFIGURE 4.17 The molecular structure of the methane molecule, $\\mathrm{CH}_{4}$, is shown with a tetrahedral arrangement of the hydrogen atoms. VSEPR structures like this one are often drawn using the wedge and dash notation, in which solid lines represent bonds in the plane of the page, solid wedges represent bonds coming up out of the plane, and dashed lines represent bonds going down into the plane.\n\n(a)\n\n(b)\n\n(c)\n\nFIGURE 4.18 (a) The electron-pair geometry for the ammonia molecule is tetrahedral with one lone pair and three single bonds. (b) The trigonal pyramidal molecular structure is determined from the electron-pair geometry. (c) The actual bond angles deviate slightly from the idealized angles because the lone pair takes up a larger region of space than do the single bonds, causing the HNH angle to be slightly smaller than $109.5^{\\circ}$."}
{"id": 2700, "contents": "413. Electron-pair Geometry versus Molecular Structure - \nAs seen in Figure 4.18, small distortions from the ideal angles in Figure 4.16 can result from differences in repulsion between various regions of electron density. VSEPR theory predicts these distortions by establishing an order of repulsions and an order of the amount of space occupied by different kinds of electron pairs. The order of electron-pair repulsions from greatest to least repulsion is:"}
{"id": 2701, "contents": "414. lone pair-lone pair $>$ lone pair-bonding pair $>$ bonding pair-bonding pair - \nThis order of repulsions determines the amount of space occupied by different regions of electrons. A lone pair of electrons occupies a larger region of space than the electrons in a triple bond; in turn, electrons in a triple bond occupy more space than those in a double bond, and so on. The order of sizes from largest to smallest is:\n\n$$\n\\text { lone pair }>\\text { triple bond }>\\text { double bond }>\\text { single bond }\n$$\n\nConsider formaldehyde, $\\mathrm{H}_{2} \\mathrm{CO}$, which is used as a preservative for biological and anatomical specimens (Figure 4.14). This molecule has regions of high electron density that consist of two single bonds and one double bond. The basic geometry is trigonal planar with $120^{\\circ}$ bond angles, but we see that the double bond causes slightly larger angles $\\left(121^{\\circ}\\right)$, and the angle between the single bonds is slightly smaller ( $118^{\\circ}$ ).\n\nIn the ammonia molecule, the three hydrogen atoms attached to the central nitrogen are not arranged in a flat, trigonal planar molecular structure, but rather in a three-dimensional trigonal pyramid (Figure 4.18) with the nitrogen atom at the apex and the three hydrogen atoms forming the base. The ideal bond angles in a trigonal pyramid are based on the tetrahedral electron pair geometry. Again, there are slight deviations from the ideal because lone pairs occupy larger regions of space than do bonding electrons. The $\\mathrm{H}-\\mathrm{N}-\\mathrm{H}$ bond angles in $\\mathrm{NH}_{3}$ are slightly smaller than the $109.5^{\\circ}$ angle in a regular tetrahedron (Figure 4.16) because the lone pair-bonding pair repulsion is greater than the bonding pair-bonding pair repulsion (Figure 4.18). Figure 4.19 illustrates the ideal molecular structures, which are predicted based on the electron-pair geometries for various combinations of lone pairs and bonding pairs."}
{"id": 2702, "contents": "414. lone pair-lone pair $>$ lone pair-bonding pair $>$ bonding pair-bonding pair - \n| Number of electron regions | Electron region geometries: 0 lone pair | 1 lone pair | 2 lone pairs | 3 lone pairs | 4 lone pairs |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| 2 | $x \\overbrace{\\mathrm{E}}^{180^{\\circ}}$ <br> Linear |  |  |  |  |\n| 3 |  <br> Trigonal planar |  <br> Bent or angular |  |  |  |\n| 4 | <smiles>[Y]C([Y])([Y])C([Y])([Y])[Y10]</smiles> <br> Tetrahedral | Trigonal pyramid | Bent or angular |  |  |\n| 5 |  <br> Trigonal bipyramid |  <br> Sawhorse or seesaw |  <br> T-shape |  <br> Linear |  |\n| 6 |  <br> Octahedral |  |  <br> Square planar |  <br> T-shape | Linear |\n\nFIGURE 4.19 The molecular structures are identical to the electron-pair geometries when there are no lone pairs present (first column). For a particular number of electron pairs (row), the molecular structures for one or more lone pairs are determined based on modifications of the corresponding electron-pair geometry."}
{"id": 2703, "contents": "414. lone pair-lone pair $>$ lone pair-bonding pair $>$ bonding pair-bonding pair - \nAccording to VSEPR theory, the terminal atom locations (Xs in Figure 4.19) are equivalent within the linear, trigonal planar, and tetrahedral electron-pair geometries (the first three rows of the table). It does not matter which X is replaced with a lone pair because the molecules can be rotated to convert positions. For trigonal bipyramidal electron-pair geometries, however, there are two distinct $X$ positions, as shown in Figure 4.20: an axial position (if we hold a model of a trigonal bipyramid by the two axial positions, we have an axis around which we can rotate the model) and an equatorial position (three positions form an equator around the middle of the molecule). As shown in Figure 4.19, the axial position is surrounded by bond angles of $90^{\\circ}$, whereas the equatorial position has more space available because of the $120^{\\circ}$ bond angles. In a trigonal bipyramidal electron-pair geometry, lone pairs always occupy equatorial positions because these more spacious positions can more easily accommodate the larger lone pairs.\n\nTheoretically, we can come up with three possible arrangements for the three bonds and two lone pairs for the $\\mathrm{ClF}_{3}$ molecule (Figure 4.20). The stable structure is the one that puts the lone pairs in equatorial locations, giving a T-shaped molecular structure.\n\n\nFIGURE 4.20 (a) In a trigonal bipyramid, the two axial positions are located directly across from one another, whereas the three equatorial positions are located in a triangular arrangement. (b-d) The two lone pairs (red lines) in $\\mathrm{ClF}_{3}$ have several possible arrangements, but the T -shaped molecular structure (b) is the one actually observed, consistent with the larger lone pairs both occupying equatorial positions.\n\nWhen a central atom has two lone electron pairs and four bonding regions, we have an octahedral electronpair geometry. The two lone pairs are on opposite sides of the octahedron ( $180^{\\circ}$ apart), giving a square planar molecular structure that minimizes lone pair-lone pair repulsions (Figure 4.19)."}
{"id": 2704, "contents": "414. lone pair-lone pair $>$ lone pair-bonding pair $>$ bonding pair-bonding pair - \nPredicting Electron Pair Geometry and Molecular Structure\nThe following procedure uses VSEPR theory to determine the electron pair geometries and the molecular structures:\n\n1. Write the Lewis structure of the molecule or polyatomic ion.\n2. Count the number of regions of electron density (lone pairs and bonds) around the central atom. A single, double, or triple bond counts as one region of electron density.\n3. Identify the electron-pair geometry based on the number of regions of electron density: linear, trigonal planar, tetrahedral, trigonal bipyramidal, or octahedral (Figure 4.19, first column).\n4. Use the number of lone pairs to determine the molecular structure (Figure 4.19). If more than one arrangement of lone pairs and chemical bonds is possible, choose the one that will minimize repulsions, remembering that lone pairs occupy more space than multiple bonds, which occupy more space than single bonds. In trigonal bipyramidal arrangements, repulsion is minimized when every lone pair is in an equatorial position. In an octahedral arrangement with two lone pairs, repulsion is minimized when the lone pairs are on opposite sides of the central atom.\n\nThe following examples illustrate the use of VSEPR theory to predict the molecular structure of molecules or ions that have no lone pairs of electrons. In this case, the molecular structure is identical to the electron pair geometry."}
{"id": 2705, "contents": "416. Predicting Electron-pair Geometry and Molecular Structure: $\\mathbf{C O}_{\\mathbf{2}}$ and $\\mathbf{B C l}_{\\mathbf{3}}$ - \nPredict the electron-pair geometry and molecular structure for each of the following:\n(a) carbon dioxide, $\\mathrm{CO}_{2}$, a molecule produced by the combustion of fossil fuels\n(b) boron trichloride, $\\mathrm{BCl}_{3}$, an important industrial chemical"}
{"id": 2706, "contents": "417. Solution - \n(a) We write the Lewis structure of $\\mathrm{CO}_{2}$ as:\n\n\nThis shows us two regions of high electron density around the carbon atom-each double bond counts as one region, and there are no lone pairs on the carbon atom. Using VSEPR theory, we predict that the two regions of electron density arrange themselves on opposite sides of the central atom with a bond angle of $180^{\\circ}$. The\nelectron-pair geometry and molecular structure are identical, and $\\mathrm{CO}_{2}$ molecules are linear.\n(b) We write the Lewis structure of $\\mathrm{BCl}_{3}$ as:\n\n\nThus we see that $\\mathrm{BCl}_{3}$ contains three bonds, and there are no lone pairs of electrons on boron. The arrangement of three regions of high electron density gives a trigonal planar electron-pair geometry. The $\\mathrm{B}-\\mathrm{Cl}$ bonds lie in a plane with $120^{\\circ}$ angles between them. $\\mathrm{BCl}_{3}$ also has a trigonal planar molecular structure (Figure 4.21).\n\n\nFIGURE 4.21\nThe electron-pair geometry and molecular structure of $\\mathrm{BCl}_{3}$ are both trigonal planar. Note that the VSEPR geometry indicates the correct bond angles ( $120^{\\circ}$ ), unlike the Lewis structure shown above."}
{"id": 2707, "contents": "418. Check Your Learning - \nCarbonate, $\\mathrm{CO}_{3}{ }^{2-}$, is a common polyatomic ion found in various materials from eggshells to antacids. What are the electron-pair geometry and molecular structure of this polyatomic ion?"}
{"id": 2708, "contents": "419. Answer: - \nThe electron-pair geometry is trigonal planar and the molecular structure is trigonal planar. Due to resonance, all three $\\mathrm{C}-\\mathrm{O}$ bonds are identical. Whether they are single, double, or an average of the two, each bond counts as one region of electron density."}
{"id": 2709, "contents": "421. Predicting Electron-pair Geometry and Molecular Structure: Ammonium - \nTwo of the top 50 chemicals produced in the United States, ammonium nitrate and ammonium sulfate, both used as fertilizers, contain the ammonium ion. Predict the electron-pair geometry and molecular structure of the $\\mathrm{NH}_{4}{ }^{+}$cation."}
{"id": 2710, "contents": "422. Solution - \nWe write the Lewis structure of $\\mathrm{NH}_{4}{ }^{+}$as:\n\n\nWe can see that $\\mathrm{NH}_{4}{ }^{+}$contains four bonds from the nitrogen atom to hydrogen atoms and no lone pairs. We expect the four regions of high electron density to arrange themselves so that they point to the corners of a tetrahedron with the central nitrogen atom in the middle (Figure 4.19). Therefore, the electron pair geometry of $\\mathrm{NH}_{4}{ }^{+}$is tetrahedral, and the molecular structure is also tetrahedral (Figure 4.22).\n\n\nFIGURE 4.22 The ammonium ion displays a tetrahedral electron-pair geometry as well as a tetrahedral molecular structure.\n\nCheck Your Learning\nIdentify a molecule with trigonal bipyramidal molecular structure."}
{"id": 2711, "contents": "423. Answer: - \nAny molecule with five electron pairs around the central atoms including no lone pairs will be trigonal bipyramidal. $\\mathrm{PF}_{5}$ is a common example.\n\nThe next several examples illustrate the effect of lone pairs of electrons on molecular structure."}
{"id": 2712, "contents": "425. Predicting Electron-pair Geometry and Molecular Structure: Lone Pairs on the Central Atom - \nPredict the electron-pair geometry and molecular structure of a water molecule."}
{"id": 2713, "contents": "426. Solution - \nThe Lewis structure of $\\mathrm{H}_{2} \\mathrm{O}$ indicates that there are four regions of high electron density around the oxygen atom: two lone pairs and two chemical bonds:\n\n\nWe predict that these four regions are arranged in a tetrahedral fashion (Figure 4.23), as indicated in Figure 4.19. Thus, the electron-pair geometry is tetrahedral and the molecular structure is bent with an angle slightly less than $109.5^{\\circ}$. In fact, the bond angle is $104.5^{\\circ}$.\n\n(a)\n\n(b)\n\nFIGURE 4.23 (a) $\\mathrm{H}_{2} \\mathrm{O}$ has four regions of electron density around the central atom, so it has a tetrahedral electronpair geometry. (b) Two of the electron regions are lone pairs, so the molecular structure is bent."}
{"id": 2714, "contents": "427. Check Your Learning - \nThe hydronium ion, $\\mathrm{H}_{3} \\mathrm{O}^{+}$, forms when acids are dissolved in water. Predict the electron-pair geometry and molecular structure of this cation."}
{"id": 2715, "contents": "428. Answer: - \nelectron pair geometry: tetrahedral; molecular structure: trigonal pyramidal"}
{"id": 2716, "contents": "430. Predicting Electron-pair Geometry and Molecular Structure: $\\mathbf{S F}_{\\mathbf{4}}$ - \nSulfur tetrafluoride, $\\mathrm{SF}_{4}$, is extremely valuable for the preparation of fluorine-containing compounds used as herbicides (i.e., $\\mathrm{SF}_{4}$ is used as a fluorinating agent). Predict the electron-pair geometry and molecular structure of a $\\mathrm{SF}_{4}$ molecule."}
{"id": 2717, "contents": "431. Solution - \nThe Lewis structure of $\\mathrm{SF}_{4}$ indicates five regions of electron density around the sulfur atom: one lone pair and four bonding pairs:\n\n\nWe expect these five regions to adopt a trigonal bipyramidal electron-pair geometry. To minimize lone pair repulsions, the lone pair occupies one of the equatorial positions. The molecular structure (Figure 4.24) is that of a seesaw (Figure 4.19).\n\n\nFIGURE 4.24 (a) SF4 has a trigonal bipyramidal arrangement of the five regions of electron density. (b) One of the regions is a lone pair, which results in a seesaw-shaped molecular structure."}
{"id": 2718, "contents": "432. Check Your Learning - \nPredict the electron pair geometry and molecular structure for molecules of $\\mathrm{XeF}_{2}$."}
{"id": 2719, "contents": "433. Answer: - \nThe electron-pair geometry is trigonal bipyramidal. The molecular structure is linear."}
{"id": 2720, "contents": "435. Predicting Electron-pair Geometry and Molecular Structure: $\\mathrm{XeF}_{4}$ - \nOf all the noble gases, xenon is the most reactive, frequently reacting with elements such as oxygen and fluorine. Predict the electron-pair geometry and molecular structure of the $\\mathrm{XeF}_{4}$ molecule."}
{"id": 2721, "contents": "436. Solution - \nThe Lewis structure of $\\mathrm{XeF}_{4}$ indicates six regions of high electron density around the xenon atom: two lone pairs and four bonds:\n\n\nThese six regions adopt an octahedral arrangement (Figure 4.19), which is the electron-pair geometry. To minimize repulsions, the lone pairs should be on opposite sides of the central atom (Figure 4.25). The five atoms are all in the same plane and have a square planar molecular structure.\n\n\nFIGURE 4.25 (a) $\\mathrm{XeF}_{4}$ adopts an octahedral arrangement with two lone pairs (red lines) and four bonds in the electron-pair geometry. (b) The molecular structure is square planar with the lone pairs directly across from one another."}
{"id": 2722, "contents": "437. Check Your Learning - \nIn a certain molecule, the central atom has three lone pairs and two bonds. What will the electron pair geometry and molecular structure be?"}
{"id": 2723, "contents": "438. Answer: - \nelectron pair geometry: trigonal bipyramidal; molecular structure: linear"}
{"id": 2724, "contents": "439. Molecular Structure for Multicenter Molecules - \nWhen a molecule or polyatomic ion has only one central atom, the molecular structure completely describes the shape of the molecule. Larger molecules do not have a single central atom, but are connected by a chain of interior atoms that each possess a \"local\" geometry. The way these local structures are oriented with respect to each other also influences the molecular shape, but such considerations are largely beyond the scope of this introductory discussion. For our purposes, we will only focus on determining the local structures."}
{"id": 2725, "contents": "441. Predicting Structure in Multicenter Molecules - \nThe Lewis structure for the simplest amino acid, glycine, $\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$, is shown here. Predict the local geometry for the nitrogen atom, the two carbon atoms, and the oxygen atom with a hydrogen atom attached:"}
{"id": 2726, "contents": "442. Solution - \nConsider each central atom independently. The electron-pair geometries:\n\n- nitrogen--four regions of electron density; tetrahedral\n- carbon $\\left(\\mathrm{CH}_{2}\\right)$--four regions of electron density; tetrahedral\n- carbon $\\left(\\mathrm{CO}_{2}\\right)$-three regions of electron density; trigonal planar\n- oxygen ( OH )-four regions of electron density; tetrahedral\n\nThe local structures:\n\n- nitrogen--three bonds, one lone pair; trigonal pyramidal\n- carbon $\\left(\\mathrm{CH}_{2}\\right)$-four bonds, no lone pairs; tetrahedral\n- carbon $\\left(\\mathrm{CO}_{2}\\right)$-three bonds (double bond counts as one bond), no lone pairs; trigonal planar\n- oxygen ( OH )-two bonds, two lone pairs; bent (109 $)$"}
{"id": 2727, "contents": "443. Check Your Learning - \nAnother amino acid is alanine, which has the Lewis structure shown here. Predict the electron-pair geometry and local structure of the nitrogen atom, the three carbon atoms, and the oxygen atom with hydrogen attached:"}
{"id": 2728, "contents": "444. Answer: - \nelectron-pair geometries: nitrogen--tetrahedral; carbon ( $\\underline{\\mathrm{C}} \\mathbf{H}$ )-tetrahedral; carbon $\\left(\\mathrm{CH}_{3}\\right)$-tetrahedral; carbon $\\left(\\mathrm{CO}_{2}\\right)$-trigonal planar; oxygen ( OH )-tetrahedral; local structures: nitrogen-trigonal pyramidal; carbon $(\\mathrm{CH})$-tetrahedral; carbon $\\left(\\mathrm{CH}_{3}\\right)$-tetrahedral; carbon $\\left(\\mathrm{CO}_{2}\\right)$-trigonal planar; oxygen $(\\mathrm{OH})$-bent $\\left(109^{\\circ}\\right)$"}
{"id": 2729, "contents": "445. LINK TO LEARNING - \nThe molecular shape simulator (http://openstax.org/l/16MolecShape) lets you build various molecules and practice naming their electron-pair geometries and molecular structures."}
{"id": 2730, "contents": "447. Molecular Simulation - \nUsing molecular shape simulator (http://openstax.org/l/16MolecShape) allows us to control whether bond angles and/or lone pairs are displayed by checking or unchecking the boxes under \"Options\" on the right. We can also use the \"Name\" checkboxes at bottom-left to display or hide the electron pair geometry (called \"electron geometry\" in the simulator) and/or molecular structure (called \"molecular shape\" in the simulator).\n\nBuild the molecule HCN in the simulator based on the following Lewis structure:\n\n$$\n\\mathrm{H}-\\mathrm{C} \\equiv \\mathrm{~N}\n$$\n\nClick on each bond type or lone pair at right to add that group to the central atom. Once you have the complete molecule, rotate it to examine the predicted molecular structure. What molecular structure is this?"}
{"id": 2731, "contents": "448. Solution - \nThe molecular structure is linear."}
{"id": 2732, "contents": "449. Check Your Learning - \nBuild a more complex molecule in the simulator. Identify the electron-group geometry, molecular structure, and bond angles. Then try to find a chemical formula that would match the structure you have drawn."}
{"id": 2733, "contents": "450. Answer: - \nAnswers will vary. For example, an atom with four single bonds, a double bond, and a lone pair has an octahedral electron-group geometry and a square pyramidal molecular structure. $\\mathrm{XeOF}_{4}$ is a molecule that adopts this structure."}
{"id": 2734, "contents": "451. Molecular Polarity and Dipole Moment - \nAs discussed previously, polar covalent bonds connect two atoms with differing electronegativities, leaving one atom with a partial positive charge ( $\\delta^{+}$) and the other atom with a partial negative charge ( $\\delta^{-}$), as the electrons are pulled toward the more electronegative atom. This separation of charge gives rise to a bond dipole moment. The magnitude of a bond dipole moment is represented by the Greek letter mu( $\\mu$ ) and is given by the formula shown here, where Q is the magnitude of the partial charges (determined by the electronegativity difference) and $r$ is the distance between the charges:\n\n$$\n\\mu=\\mathrm{Qr}\n$$\n\nThis bond moment can be represented as a vector, a quantity having both direction and magnitude (Figure 4.26). Dipole vectors are shown as arrows pointing along the bond from the less electronegative atom toward the more electronegative atom. A small plus sign is drawn on the less electronegative end to indicate the partially positive end of the bond. The length of the arrow is proportional to the magnitude of the electronegativity difference between the two atoms.\n\n(a)\n\n(b)\n\nFIGURE 4.26 (a) There is a small difference in electronegativity between C and H , represented as a short vector. (b) The electronegativity difference between $B$ and $F$ is much larger, so the vector representing the bond moment is much longer.\n\nA whole molecule may also have a separation of charge, depending on its molecular structure and the polarity of each of its bonds. If such a charge separation exists, the molecule is said to be a polar molecule (or dipole); otherwise the molecule is said to be nonpolar. The dipole moment measures the extent of net charge separation in the molecule as a whole. We determine the dipole moment by adding the bond moments in three-dimensional space, taking into account the molecular structure."}
{"id": 2735, "contents": "451. Molecular Polarity and Dipole Moment - \nFor diatomic molecules, there is only one bond, so its bond dipole moment determines the molecular polarity. Homonuclear diatomic molecules such as $\\mathrm{Br}_{2}$ and $\\mathrm{N}_{2}$ have no difference in electronegativity, so their dipole moment is zero. For heteronuclear molecules such as CO, there is a small dipole moment. For HF, there is a larger dipole moment because there is a larger difference in electronegativity.\n\nWhen a molecule contains more than one bond, the geometry must be taken into account. If the bonds in a molecule are arranged such that their bond moments cancel (vector sum equals zero), then the molecule is nonpolar. This is the situation in $\\mathrm{CO}_{2}$ (Figure 4.27). Each of the bonds is polar, but the molecule as a whole is nonpolar. From the Lewis structure, and using VSEPR theory, we determine that the $\\mathrm{CO}_{2}$ molecule is linear with polar $\\mathrm{C}=\\mathrm{O}$ bonds on opposite sides of the carbon atom. The bond moments cancel because they are pointed in opposite directions. In the case of the water molecule (Figure 4.27), the Lewis structure again shows that there are two bonds to a central atom, and the electronegativity difference again shows that each of these bonds has a nonzero bond moment. In this case, however, the molecular structure is bent because of the lone pairs on O , and the two bond moments do not cancel. Therefore, water does have a net dipole moment and is a polar molecule (dipole).\n\n\nFIGURE 4.27 The overall dipole moment of a molecule depends on the individual bond dipole moments and how they are arranged. (a) Each CO bond has a bond dipole moment, but they point in opposite directions so that the net $\\mathrm{CO}_{2}$ molecule is nonpolar. (b) In contrast, water is polar because the OH bond moments do not cancel out.\n\nThe OCS molecule has a structure similar to $\\mathrm{CO}_{2}$, but a sulfur atom has replaced one of the oxygen atoms. To determine if this molecule is polar, we draw the molecular structure. VSEPR theory predicts a linear molecule:"}
{"id": 2736, "contents": "451. Molecular Polarity and Dipole Moment - \nThe C-O bond is considerably polar. Although C and $S$ have very similar electronegativity values, $S$ is slightly more electronegative than C , and so the $\\mathrm{C}-\\mathrm{S}$ bond is just slightly polar. Because oxygen is more electronegative than sulfur, the oxygen end of the molecule is the negative end.\n\nChloromethane, $\\mathrm{CH}_{3} \\mathrm{Cl}$, is a tetrahedral molecule with three slightly polar $\\mathrm{C}-\\mathrm{H}$ bonds and a more polar $\\mathrm{C}-\\mathrm{Cl}$ bond. The relative electronegativities of the bonded atoms is $\\mathrm{H}<\\mathrm{C}<\\mathrm{Cl}$, and so the bond moments all point toward the Cl end of the molecule and sum to yield a considerable dipole moment (the molecules are relatively polar).\n\n\nFor molecules of high symmetry such as $\\mathrm{BF}_{3}$ (trigonal planar), $\\mathrm{CH}_{4}$ (tetrahedral), $\\mathrm{PF}_{5}$ (trigonal bipyramidal), and $\\mathrm{SF}_{6}$ (octahedral), all the bonds are of identical polarity (same bond moment) and they are oriented in geometries that yield nonpolar molecules (dipole moment is zero). Molecules of less geometric symmetry, however, may be polar even when all bond moments are identical. For these molecules, the directions of the equal bond moments are such that they sum to give a nonzero dipole moment and a polar molecule. Examples of such molecules include hydrogen sulfide, $\\mathrm{H}_{2} \\mathrm{~S}$ (nonlinear), and ammonia, $\\mathrm{NH}_{3}$ (trigonal pyramidal).\n\n\n\nTo summarize, to be polar, a molecule must:\n\n1. Contain at least one polar covalent bond.\n2. Have a molecular structure such that the sum of the vectors of each bond dipole moment does not cancel."}
{"id": 2737, "contents": "452. Properties of Polar Molecules - \nPolar molecules tend to align when placed in an electric field with the positive end of the molecule oriented\ntoward the negative plate and the negative end toward the positive plate (Figure 4.28). We can use an electrically charged object to attract polar molecules, but nonpolar molecules are not attracted. Also, polar solvents are better at dissolving polar substances, and nonpolar solvents are better at dissolving nonpolar substances.\n\n\nFIGURE 4.28 (a) Molecules are always randomly distributed in the liquid state in the absence of an electric field. (b) When an electric field is applied, polar molecules like HF will align to the dipoles with the field direction."}
{"id": 2738, "contents": "453. LINK TO LEARNING - \nThe molecule polarity simulation (http://openstax.org/l/16MolecPolarity) provides many ways to explore dipole moments of bonds and molecules."}
{"id": 2739, "contents": "455. Polarity Simulations - \nOpen the molecule polarity simulation (http://openstax.org/l/16MolecPolarity) and select the \"Three Atoms\" tab at the top. This should display a molecule ABC with three electronegativity adjustors. You can display or hide the bond moments, molecular dipoles, and partial charges at the right. Turning on the Electric Field will show whether the molecule moves when exposed to a field, similar to Figure 4.28.\n\nUse the electronegativity controls to determine how the molecular dipole will look for the starting bent molecule if:\n(a) A and C are very electronegative and B is in the middle of the range.\n(b) A is very electronegative, and $B$ and $C$ are not."}
{"id": 2740, "contents": "456. Solution - \n(a) Molecular dipole moment points immediately between A and C.\n(b) Molecular dipole moment points along the A-B bond, toward A."}
{"id": 2741, "contents": "457. Check Your Learning - \nDetermine the partial charges that will give the largest possible bond dipoles."}
{"id": 2742, "contents": "458. Answer: - \nThe largest bond moments will occur with the largest partial charges. The two solutions above represent how unevenly the electrons are shared in the bond. The bond moments will be maximized when the electronegativity difference is greatest. The controls for A and C should be set to one extreme, and B should be set to the opposite extreme. Although the magnitude of the bond moment will not change based on whether B is the most electronegative or the least, the direction of the bond moment will."}
{"id": 2743, "contents": "459. Key Terms - \naxial position location in a trigonal bipyramidal geometry in which there is another atom at a $180^{\\circ}$ angle and the equatorial positions are at a $90^{\\circ}$ angle\nbinary acid compound that contains hydrogen and one other element, bonded in a way that imparts acidic properties to the compound (ability to release $\\mathrm{H}^{+}$ions when dissolved in water)\nbinary compound compound containing two different elements.\nbond angle angle between any two covalent bonds that share a common atom\nbond dipole moment separation of charge in a bond that depends on the difference in electronegativity and the bond distance represented by partial charges or a vector\nbond distance (also, bond length) distance between the nuclei of two bonded atoms\nbond length distance between the nuclei of two bonded atoms at which the lowest potential energy is achieved\ncovalent bond bond formed when electrons are shared between atoms\ndipole moment property of a molecule that describes the separation of charge determined by the sum of the individual bond moments based on the molecular structure\ndouble bond covalent bond in which two pairs of electrons are shared between two atoms\nelectron-pair geometry arrangement around a central atom of all regions of electron density (bonds, lone pairs, or unpaired electrons)\nelectronegativity tendency of an atom to attract electrons in a bond to itself\nequatorial position one of the three positions in a trigonal bipyramidal geometry with $120^{\\circ}$ angles between them; the axial positions are located at a $90^{\\circ}$ angle\nformal charge charge that would result on an atom by taking the number of valence electrons on the neutral atom and subtracting the nonbonding electrons and the number of bonds (one-half of the bonding electrons)\nfree radical molecule that contains an odd number of electrons\nhypervalent molecule molecule containing at least one main group element that has more than eight electrons in its valence shell\ninert pair effect tendency of heavy atoms to form ions in which their valence $s$ electrons are not lost\nionic bond strong electrostatic force of attraction between cations and anions in an ionic\ncompound\nLewis structure diagram showing lone pairs and bonding pairs of electrons in a molecule or an ion\nLewis symbol symbol for an element or monatomic ion that uses a dot to represent each valence electron in the element or ion"}
{"id": 2744, "contents": "459. Key Terms - \ncompound\nLewis structure diagram showing lone pairs and bonding pairs of electrons in a molecule or an ion\nLewis symbol symbol for an element or monatomic ion that uses a dot to represent each valence electron in the element or ion\nlinear shape in which two outside groups are placed on opposite sides of a central atom\nlone pair two (a pair of) valence electrons that are not used to form a covalent bond\nmolecular structure arrangement of atoms in a molecule or ion\nmolecular structure structure that includes only the placement of the atoms in the molecule\nnomenclature system of rules for naming objects of interest\noctahedral shape in which six outside groups are placed around a central atom such that a threedimensional shape is generated with four groups forming a square and the other two forming the apex of two pyramids, one above and one below the square plane\noctet rule guideline that states main group atoms will form structures in which eight valence electrons interact with each nucleus, counting bonding electrons as interacting with both atoms connected by the bond\noxyacid compound that contains hydrogen, oxygen, and one other element, bonded in a way that imparts acidic properties to the compound (ability to release $\\mathrm{H}^{+}$ions when dissolved in water)\npolar covalent bond covalent bond between atoms of different electronegativities; a covalent bond with a positive end and a negative end\npolar molecule (also, dipole) molecule with an overall dipole moment\npure covalent bond (also, nonpolar covalent bond) covalent bond between atoms of identical electronegativities\nresonance situation in which one Lewis structure is insufficient to describe the bonding in a molecule and the average of multiple structures is observed\nresonance forms two or more Lewis structures that have the same arrangement of atoms but different arrangements of electrons\nresonance hybrid average of the resonance forms shown by the individual Lewis structures\nsingle bond bond in which a single pair of electrons is shared between two atoms\ntetrahedral shape in which four outside groups are\nplaced around a central atom such that a threedimensional shape is generated with four corners and $109.5^{\\circ}$ angles between each pair and the central atom"}
{"id": 2745, "contents": "459. Key Terms - \ntetrahedral shape in which four outside groups are\nplaced around a central atom such that a threedimensional shape is generated with four corners and $109.5^{\\circ}$ angles between each pair and the central atom\ntrigonal bipyramidal shape in which five outside groups are placed around a central atom such that three form a flat triangle with $120^{\\circ}$ angles between each pair and the central atom, and the other two form the apex of two pyramids, one above and one below the triangular plane\ntrigonal planar shape in which three outside\ngroups are placed in a flat triangle around a central atom with $120^{\\circ}$ angles between each pair and the central atom\ntriple bond bond in which three pairs of electrons are shared between two atoms\nvalence shell electron-pair repulsion theory (VSEPR) theory used to predict the bond angles in a molecule based on positioning regions of high electron density as far apart as possible to minimize electrostatic repulsion\nvector quantity having magnitude and direction"}
{"id": 2746, "contents": "460. Key Equations - \nformal charge $=\\#$ valence shell electrons (free atom) $-\\#$ lone pair electrons $-\\frac{1}{2} \\#$ bonding electrons"}
{"id": 2747, "contents": "461. Summary - 461.1. Ionic Bonding\nAtoms gain or lose electrons to form ions with particularly stable electron configurations. The charges of cations formed by the representative metals may be determined readily because, with few exceptions, the electronic structures of these ions have either a noble gas configuration or a completely filled electron shell. The charges of anions formed by the nonmetals may also be readily determined because these ions form when nonmetal atoms gain enough electrons to fill their valence shells."}
{"id": 2748, "contents": "461. Summary - 461.2. Covalent Bonding\nCovalent bonds form when electrons are shared between atoms and are attracted by the nuclei of both atoms. In pure covalent bonds, the electrons are shared equally. In polar covalent bonds, the electrons are shared unequally, as one atom exerts a stronger force of attraction on the electrons than the other. The ability of an atom to attract a pair of electrons in a chemical bond is called its electronegativity. The difference in electronegativity between two atoms determines how polar a bond will be. In a diatomic molecule with two identical atoms, there is no difference in electronegativity, so the bond is nonpolar or pure covalent. When the electronegativity difference is very large, as is the case between metals and nonmetals, the bonding is characterized as ionic."}
{"id": 2749, "contents": "461. Summary - 461.3. Chemical Nomenclature\nChemists use nomenclature rules to clearly name compounds. Ionic and molecular compounds are named using somewhat-different methods. Binary ionic compounds typically consist of a metal and a\nnonmetal. The name of the metal is written first, followed by the name of the nonmetal with its ending changed to -ide. For example, $\\mathrm{K}_{2} \\mathrm{O}$ is called potassium oxide. If the metal can form ions with different charges, a Roman numeral in parentheses follows the name of the metal to specify its charge. Thus, $\\mathrm{FeCl}_{2}$ is iron(II) chloride and $\\mathrm{FeCl}_{3}$ is iron(III) chloride. Some compounds contain polyatomic ions; the names of common polyatomic ions should be memorized. Molecular compounds can form compounds with different ratios of their elements, so prefixes are used to specify the numbers of atoms of each element in a molecule of the compound. Examples include $\\mathrm{SF}_{6}$, sulfur hexafluoride, and $\\mathrm{N}_{2} \\mathrm{O}_{4}$, dinitrogen tetroxide. Acids are an important class of compounds containing hydrogen and having special nomenclature rules. Binary acids are named using the prefix hydro-, changing the -ide suffix to $-i c$, and adding \"acid;\" HCl is hydrochloric acid. Oxyacids are named by changing the ending of the anion (-ate to -ic and -ite to -ous), and adding \"acid;\" $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ is carbonic acid."}
{"id": 2750, "contents": "461. Summary - 461.4. Lewis Symbols and Structures\nValence electronic structures can be visualized by drawing Lewis symbols (for atoms and monatomic ions) and Lewis structures (for molecules and polyatomic ions). Lone pairs, unpaired electrons, and single, double, or triple bonds are used to indicate where the valence electrons are located around each atom in a Lewis structure. Most structures-especially those containing second row elements-obey the octet rule, in which every atom (except H ) is surrounded by eight electrons. Exceptions to the octet rule occur for odd-electron molecules (free radicals), electron-deficient\nmolecules, and hypervalent molecules."}
{"id": 2751, "contents": "461. Summary - 461.5. Formal Charges and Resonance\nIn a Lewis structure, formal charges can be assigned to each atom by treating each bond as if one-half of the electrons are assigned to each atom. These hypothetical formal charges are a guide to determining the most appropriate Lewis structure. A structure in which the formal charges are as close to zero as possible is preferred. Resonance occurs in cases where two or more Lewis structures with identical arrangements of atoms but different distributions of electrons can be written. The actual distribution of electrons (the resonance hybrid) is an average of the distribution indicated by the individual Lewis structures (the resonance forms)."}
{"id": 2752, "contents": "461. Summary - 461.6. Molecular Structure and Polarity\nVSEPR theory predicts the three-dimensional\narrangement of atoms in a molecule. It states that valence electrons will assume an electron-pair geometry that minimizes repulsions between areas of high electron density (bonds and/or lone pairs). Molecular structure, which refers only to the placement of atoms in a molecule and not the electrons, is equivalent to electron-pair geometry only when there are no lone electron pairs around the central atom. A dipole moment measures a separation of charge. For one bond, the bond dipole moment is determined by the difference in electronegativity between the two atoms. For a molecule, the overall dipole moment is determined by both the individual bond moments and how these dipoles are arranged in the molecular structure. Polar molecules (those with an appreciable dipole moment) interact with electric fields, whereas nonpolar molecules do not."}
{"id": 2753, "contents": "462. Exercises - 462.1. Ionic Bonding\n1. Does a cation gain protons to form a positive charge or does it lose electrons?\n2. Iron(III) sulfate $\\left[\\mathrm{Fe}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}\\right]$ is composed of $\\mathrm{Fe}^{3+}$ and $\\mathrm{SO}_{4}{ }^{2-}$ ions. Explain why a sample of iron(III) sulfate is uncharged.\n3. Which of the following atoms would be expected to form negative ions in binary ionic compounds and which would be expected to form positive ions: $\\mathrm{P}, \\mathrm{I}, \\mathrm{Mg}, \\mathrm{Cl}, \\mathrm{In}, \\mathrm{Cs}, \\mathrm{O}, \\mathrm{Pb}, \\mathrm{Co}$ ?\n4. Which of the following atoms would be expected to form negative ions in binary ionic compounds and which would be expected to form positive ions: $\\mathrm{Br}, \\mathrm{Ca}, \\mathrm{Na}, \\mathrm{N}, \\mathrm{F}, \\mathrm{Al}, \\mathrm{Sn}, \\mathrm{S}, \\mathrm{Cd}$ ?\n5. Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds:\n(a) P\n(b) Mg\n(c) Al\n(d) O\n(e) Cl\n(f) Cs\n6. Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds:\n(a) I\n(b) Sr\n(c) K\n(d) N\n(e) S\n(f) In\n7. Write the electron configuration for each of the following ions:\n(a) $\\mathrm{As}^{3-}$\n(b) $I^{-}$\n(c) $\\mathrm{Be}^{2+}$\n(d) $\\mathrm{Cd}^{2+}$\n(e) $\\mathrm{O}^{2-}$\n(f) $\\mathrm{Ga}^{3+}$\n(g) $\\mathrm{Li}^{+}$\n(h) $\\mathrm{N}^{3-}$\n(i) $\\mathrm{Sn}^{2+}$\n(j) $\\mathrm{Co}^{2+}$\n(k) $\\mathrm{Fe}^{2+}$\n(l) $\\mathrm{As}^{3+}$"}
{"id": 2754, "contents": "462. Exercises - 462.1. Ionic Bonding\n(i) $\\mathrm{Sn}^{2+}$\n(j) $\\mathrm{Co}^{2+}$\n(k) $\\mathrm{Fe}^{2+}$\n(l) $\\mathrm{As}^{3+}$\n8. Write the electron configuration for the monatomic ions formed from the following elements (which form the greatest concentration of monatomic ions in seawater):\n(a) Cl\n(b) Na\n(c) Mg\n(d) Ca\n(e) K\n(f) Br\n(g) Sr\n(h) F\n9. Write out the full electron configuration for each of the following atoms and for the monatomic ion found in binary ionic compounds containing the element:\n(a) Al\n(b) Br\n(c) Sr\n(d) Li\n(e) As\n(f) S\n10. From the labels of several commercial products, prepare a list of six ionic compounds in the products. For each compound, write the formula. (You may need to look up some formulas in a suitable reference.)"}
{"id": 2755, "contents": "462. Exercises - 462.2. Covalent Bonding\n11. Why is it incorrect to speak of a molecule of solid NaCl ?\n12. What information can you use to predict whether a bond between two atoms is covalent or ionic?\n13. Predict which of the following compounds are ionic and which are covalent, based on the location of their constituent atoms in the periodic table:\n(a) $\\mathrm{Cl}_{2} \\mathrm{CO}$\n(b) MnO\n(c) $\\mathrm{NCl}_{3}$\n(d) $\\mathrm{CoBr}_{2}$\n(e) $\\mathrm{K}_{2} \\mathrm{~S}$\n(f) CO\n(g) $\\mathrm{CaF}_{2}$\n(h) HI\n(i) CaO\n(j) IBr\n(k) $\\mathrm{CO}_{2}$\n14. Explain the difference between a nonpolar covalent bond, a polar covalent bond, and an ionic bond.\n15. From its position in the periodic table, determine which atom in each pair is more electronegative:\n(a) Br or Cl\n(b) N or O\n(c) S or O\n(d) P or S\n(e) Si or N\n(f) Ba or P\n(g) N or K\n16. From its position in the periodic table, determine which atom in each pair is more electronegative:\n(a) N or P\n(b) N or Ge\n(c) S or F\n(d) Cl or S\n(e) H or C\n(f) Se or P\n(g) C or Si\n17. From their positions in the periodic table, arrange the atoms in each of the following series in order of increasing electronegativity:\n(a) C, F, H, N, O\n(b) Br, Cl, F, H, I\n(c) F, H, O, P, S\n(d) Al, H, Na, O, P\n(e) Ba, H, N, O, As\n18. From their positions in the periodic table, arrange the atoms in each of the following series in order of increasing electronegativity:\n(a) As, H, N, P, Sb\n(b) Cl, H, P, S, Si"}
{"id": 2756, "contents": "462. Exercises - 462.2. Covalent Bonding\n(a) As, H, N, P, Sb\n(b) Cl, H, P, S, Si\n(c) $\\mathrm{Br}, \\mathrm{Cl}, \\mathrm{Ge}, \\mathrm{H}, \\mathrm{Sr}$\n(d) Ca, H, K, N, Si\n(e) $\\mathrm{Cl}, \\mathrm{Cs}, \\mathrm{Ge}, \\mathrm{H}, \\mathrm{Sr}$\n19. Which atoms can bond to sulfur so as to produce a positive partial charge on the sulfur atom?\n20. Which is the most polar bond?\n(a) $\\mathrm{C}-\\mathrm{C}$\n(b) $\\mathrm{C}-\\mathrm{H}$\n(c) $\\mathrm{N}-\\mathrm{H}$\n(d) $\\mathrm{O}-\\mathrm{H}$\n(e) $\\mathrm{Se}-\\mathrm{H}$\n21. Identify the more polar bond in each of the following pairs of bonds:\n(a) HF or HCl\n(b) NO or CO\n(c) SH or OH\n(d) PCl or SCl\n(e) CH or NH\n(f) SO or PO\n(g) CN or NN\n22. Which of the following molecules or ions contain polar bonds?\n(a) $\\mathrm{O}_{3}$\n(b) $\\mathrm{S}_{8}$\n(c) $\\mathrm{O}_{2}{ }^{2-}$\n(d) $\\mathrm{NO}_{3}{ }^{-}$\n(e) $\\mathrm{CO}_{2}$\n(f) $\\mathrm{H}_{2} \\mathrm{~S}$\n(g) $\\mathrm{BH}_{4}{ }^{-}$"}
{"id": 2757, "contents": "462. Exercises - 462.3. Chemical Nomenclature\n23. Name the following compounds:\n(a) CsCl\n(b) BaO\n(c) $\\mathrm{K}_{2} \\mathrm{~S}$\n(d) $\\mathrm{BeCl}_{2}$\n(e) HBr\n(f) $\\mathrm{AlF}_{3}$\n24. Name the following compounds:\n(a) NaF\n(b) $\\mathrm{Rb}_{2} \\mathrm{O}$\n(c) $\\mathrm{BCl}_{3}$\n(d) $\\mathrm{H}_{2} \\mathrm{Se}$\n(e) $\\mathrm{P}_{4} \\mathrm{O}_{6}$\n(f) $\\mathrm{ICl}_{3}$\n25. Write the formulas of the following compounds:\n(a) rubidium bromide\n(b) magnesium selenide\n(c) sodium oxide\n(d) calcium chloride\n(e) hydrogen fluoride\n(f) gallium phosphide\n(g) aluminum bromide\n(h) ammonium sulfate\n26. Write the formulas of the following compounds:\n(a) lithium carbonate\n(b) sodium perchlorate\n(c) barium hydroxide\n(d) ammonium carbonate\n(e) sulfuric acid\n(f) calcium acetate\n(g) magnesium phosphate\n(h) sodium sulfite\n27. Write the formulas of the following compounds:\n(a) chlorine dioxide\n(b) dinitrogen tetraoxide\n(c) potassium phosphide\n(d) silver(I) sulfide\n(e) aluminum fluoride trihydrate\n(f) silicon dioxide\n28. Write the formulas of the following compounds:\n(a) barium chloride\n(b) magnesium nitride\n(c) sulfur dioxide\n(d) nitrogen trichloride\n(e) dinitrogen trioxide\n(f) tin(IV) chloride\n29. Each of the following compounds contains a metal that can exhibit more than one ionic charge. Name these compounds:\n(a) $\\mathrm{Cr}_{2} \\mathrm{O}_{3}$\n(b) $\\mathrm{FeCl}_{2}$\n(c) $\\mathrm{CrO}_{3}$\n(d) $\\mathrm{TiCl}_{4}$\n(e) $\\mathrm{CoCl}_{2} \\cdot 6 \\mathrm{H}_{2} \\mathrm{O}$\n(f) $\\mathrm{MoS}_{2}$"}
{"id": 2758, "contents": "462. Exercises - 462.3. Chemical Nomenclature\n(d) $\\mathrm{TiCl}_{4}$\n(e) $\\mathrm{CoCl}_{2} \\cdot 6 \\mathrm{H}_{2} \\mathrm{O}$\n(f) $\\mathrm{MoS}_{2}$\n30. Each of the following compounds contains a metal that can exhibit more than one ionic charge. Name these compounds:\n(a) $\\mathrm{NiCO}_{3}$\n(b) $\\mathrm{MoO}_{3}$\n(c) $\\mathrm{Co}\\left(\\mathrm{NO}_{3}\\right)_{2}$\n(d) $\\mathrm{V}_{2} \\mathrm{O}_{5}$\n(e) $\\mathrm{MnO}_{2}$\n(f) $\\mathrm{Fe}_{2} \\mathrm{O}_{3}$\n31. The following ionic compounds are found in common household products. Write the formulas for each compound:\n(a) potassium phosphate\n(b) copper(II) sulfate\n(c) calcium chloride\n(d) titanium(IV) oxide\n(e) ammonium nitrate\n(f) sodium bisulfate (the common name for sodium hydrogen sulfate)\n32. The following ionic compounds are found in common household products. Name each of the compounds:\n(a) $\\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$\n(b) $\\mathrm{FeSO}_{4}$\n(c) $\\mathrm{CaCO}_{3}$\n(d) MgO\n(e) $\\mathrm{NaNO}_{2}$\n(f) KI\n33. What are the IUPAC names of the following compounds?\n(a) manganese dioxide\n(b) mercurous chloride $\\left(\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}\\right)$\n(c) ferric nitrate $\\left[\\mathrm{Fe}\\left(\\mathrm{NO}_{3}\\right)_{3}\\right]$\n(d) titanium tetrachloride\n(e) cupric bromide $\\left(\\mathrm{CuBr}_{2}\\right)$"}
{"id": 2759, "contents": "462. Exercises - 462.4. Lewis Symbols and Structures\n34. Write the Lewis symbols for each of the following ions:\n(a) $\\mathrm{As}^{3-}$\n(b) $\\mathrm{I}^{-}$\n(c) $\\mathrm{Be}^{2+}$\n(d) $\\mathrm{O}^{2-}$\n(e) $\\mathrm{Ga}^{3+}$\n(f) $\\mathrm{Li}^{+}$\n(g) $\\mathrm{N}^{3-}$\n35. Many monatomic ions are found in seawater, including the ions formed from the following list of elements. Write the Lewis symbols for the monatomic ions formed from the following elements:\n(a) Cl\n(b) Na\n(c) Mg\n(d) Ca\n(e) K\n(f) Br\n(g) Sr\n(h) F\n36. Write the Lewis symbols of the ions in each of the following ionic compounds and the Lewis symbols of the atom from which they are formed:\n(a) MgS\n(b) $\\mathrm{Al}_{2} \\mathrm{O}_{3}$\n(c) $\\mathrm{GaCl}_{3}$\n(d) $\\mathrm{K}_{2} \\mathrm{O}$\n(e) $\\mathrm{Li}_{3} \\mathrm{~N}$\n(f) KF\n37. In the Lewis structures listed here, M and X represent various elements in the third period of the periodic table. Write the formula of each compound using the chemical symbols of each element:\n(a)\n\n$$\n\\leftM^{2+}\\right^{2-}\n$$\n\n(b)\n\n$$\n\\leftM^{3+}\\right_{3}^{-}\n$$\n\n(c)\n\n$$\n\\left[M^{+}\\right]_{2}[: \\ddot{x}:]^{2-}\n$$\n\n(d)\n\n$$\n\\left[M^{3+}\\right]_{2}[: \\ddot{x}:]_{3}^{2-}\n$$"}
{"id": 2760, "contents": "462. Exercises - 462.4. Lewis Symbols and Structures\n38. Write the Lewis structure for the diatomic molecule $P_{2}$, an unstable form of phosphorus found in hightemperature phosphorus vapor.\n39. Write Lewis structures for the following:\n(a) $\\mathrm{H}_{2}$\n(b) HBr\n(c) $\\mathrm{PCl}_{3}$\n(d) $\\mathrm{SF}_{2}$\n(e) $\\mathrm{H}_{2} \\mathrm{CCH}_{2}$\n(f) HNNH\n(g) $\\mathrm{H}_{2} \\mathrm{CNH}$\n(h) $\\mathrm{NO}^{-}$\n(i) $\\mathrm{N}_{2}$\n(j) CO\n(k) $\\mathrm{CN}^{-}$\n40. Write Lewis structures for the following:\n(a) $\\mathrm{O}_{2}$\n(b) $\\mathrm{H}_{2} \\mathrm{CO}$\n(c) $\\mathrm{AsF}_{3}$\n(d) ClNO\n(e) $\\mathrm{SiCl}_{4}$\n(f) $\\mathrm{H}_{3} \\mathrm{O}^{+}$\n(g) $\\mathrm{NH}_{4}{ }^{+}$\n(h) $\\mathrm{BF}_{4}{ }^{-}$\n(i) HCCH\n(j) ClCN\n(k) $\\mathrm{C}_{2}{ }^{2+}$\n41. Write Lewis structures for the following:\n(a) $\\mathrm{ClF}_{3}$\n(b) $\\mathrm{PCl}_{5}$\n(c) $\\mathrm{BF}_{3}$\n(d) $\\mathrm{PF}_{6}{ }^{-}$\n42. Write Lewis structures for the following:\n(a) $\\mathrm{SeF}_{6}$\n(b) $\\mathrm{XeF}_{4}$\n(c) $\\mathrm{SeCl}_{3}{ }^{+}$\n(d) $\\mathrm{Cl}_{2} \\mathrm{BBCl}_{2}$ (contains a B-B bond)\n43. Write Lewis structures for:\n(a) $\\mathrm{PO}_{4}{ }^{3-}$\n(b) $\\mathrm{ICl}_{4}{ }^{-}$\n(c) $\\mathrm{SO}_{3}{ }^{2-}$\n(d) HONO"}
{"id": 2761, "contents": "462. Exercises - 462.4. Lewis Symbols and Structures\n(a) $\\mathrm{PO}_{4}{ }^{3-}$\n(b) $\\mathrm{ICl}_{4}{ }^{-}$\n(c) $\\mathrm{SO}_{3}{ }^{2-}$\n(d) HONO\n44. Correct the following statement: \"The bonds in solid $\\mathrm{PbCl}_{2}$ are ionic; the bond in a HCl molecule is covalent. Thus, all of the valence electrons in $\\mathrm{PbCl}_{2}$ are located on the $\\mathrm{Cl}^{-}$ions, and all of the valence electrons in a HCl molecule are shared between the H and Cl atoms.\"\n45. Write Lewis structures for the following molecules or ions:\n(a) $\\mathrm{SbH}_{3}$\n(b) $\\mathrm{XeF}_{2}$\n(c) $\\mathrm{Se}_{8}$ (a cyclic molecule with a ring of eight Se atoms)\n46. Methanol, $\\mathrm{H}_{3} \\mathrm{COH}$, is used as the fuel in some race cars. Ethanol, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$, is used extensively as motor fuel in Brazil. Both methanol and ethanol produce $\\mathrm{CO}_{2}$ and $\\mathrm{H}_{2} \\mathrm{O}$ when they burn. Write the chemical equations for these combustion reactions using Lewis structures instead of chemical formulas.\n47. Many planets in our solar system contain organic chemicals including methane $\\left(\\mathrm{CH}_{4}\\right)$ and traces of ethylene $\\left(\\mathrm{C}_{2} \\mathrm{H}_{4}\\right)$, ethane $\\left(\\mathrm{C}_{2} \\mathrm{H}_{6}\\right)$, propyne $\\left(\\mathrm{H}_{3} \\mathrm{CCCH}\\right)$, and diacetylene $(\\mathrm{HCCCCH})$. Write the Lewis structures for each of these molecules.\n48. Carbon tetrachloride was formerly used in fire extinguishers for electrical fires. It is no longer used for this purpose because of the formation of the toxic gas phosgene, $\\mathrm{Cl}_{2} \\mathrm{CO}$. Write the Lewis structures for carbon tetrachloride and phosgene."}
{"id": 2762, "contents": "462. Exercises - 462.4. Lewis Symbols and Structures\n49. Identify the atoms that correspond to each of the following electron configurations. Then, write the Lewis symbol for the common ion formed from each atom:\n(a) $1 s^{2} 2 s^{2} 2 p^{5}$\n(b) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}$\n(c) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10}$\n(d) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{4}$\n(e) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{1}$\n50. The arrangement of atoms in several biologically important molecules is given here. Complete the Lewis structures of these molecules by adding multiple bonds and lone pairs. Do not add any more atoms.\n(a) the amino acid serine:"}
{"id": 2763, "contents": "462. Exercises - 462.4. Lewis Symbols and Structures\n(b) urea:\n\n(c) pyruvic acid:\n\n(d) uracil:\n\n(e) carbonic acid:\n\n51. A compound with a molar mass of about $28 \\mathrm{~g} / \\mathrm{mol}$ contains $85.7 \\%$ carbon and $14.3 \\%$ hydrogen by mass. Write the Lewis structure for a molecule of the compound.\n52. A compound with a molar mass of about $42 \\mathrm{~g} / \\mathrm{mol}$ contains $85.7 \\%$ carbon and $14.3 \\%$ hydrogen by mass. Write the Lewis structure for a molecule of the compound.\n53. Two arrangements of atoms are possible for a compound with a molar mass of about $45 \\mathrm{~g} / \\mathrm{mol}$ that contains $52.2 \\% \\mathrm{C}, 13.1 \\% \\mathrm{H}$, and $34.7 \\%$ O by mass. Write the Lewis structures for the two molecules.\n54. How are single, double, and triple bonds similar? How do they differ?"}
{"id": 2764, "contents": "462. Exercises - 462.5. Formal Charges and Resonance\n55. Write resonance forms that describe the distribution of electrons in each of these molecules or ions.\n(a) selenium dioxide, OSeO\n(b) nitrate ion, $\\mathrm{NO}_{3}{ }^{-}$\n(c) nitric acid, $\\mathrm{HNO}_{3}$ ( N is bonded to an OH group and two O atoms)\n(d) benzene, $\\mathrm{C}_{6} \\mathrm{H}_{6}$ :\n\n(e) the formate ion:\n\n56. Write resonance forms that describe the distribution of electrons in each of these molecules or ions.\n(a) sulfur dioxide, $\\mathrm{SO}_{2}$\n(b) carbonate ion, $\\mathrm{CO}_{3}{ }^{2-}$\n(c) hydrogen carbonate ion, $\\mathrm{HCO}_{3}{ }^{-}$( C is bonded to an OH group and two O atoms)\n(d) pyridine:\n\n(e) the allyl ion:\n\n57. Write the resonance forms of ozone, $\\mathrm{O}_{3}$, the component of the upper atmosphere that protects the Earth from ultraviolet radiation.\n58. Sodium nitrite, which has been used to preserve bacon and other meats, is an ionic compound. Write the resonance forms of the nitrite ion, $\\mathrm{NO}_{2}{ }^{-}$.\n59. In terms of the bonds present, explain why acetic acid, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$, contains two distinct types of carbonoxygen bonds, whereas the acetate ion, formed by loss of a hydrogen ion from acetic acid, only contains one type of carbon-oxygen bond. The skeleton structures of these species are shown:"}
{"id": 2765, "contents": "462. Exercises - 462.5. Formal Charges and Resonance\n60. Write the Lewis structures for the following, and include resonance structures where appropriate. Indicate which has the strongest carbon-oxygen bond.\n(a) $\\mathrm{CO}_{2}$\n(b) CO\n61. Toothpastes containing sodium hydrogen carbonate (sodium bicarbonate) and hydrogen peroxide are widely used. Write Lewis structures for the hydrogen carbonate ion and hydrogen peroxide molecule, with resonance forms where appropriate.\n62. Determine the formal charge of each element in the following:\n(a) HCl\n(b) $\\mathrm{CF}_{4}$\n(c) $\\mathrm{PCl}_{3}$\n(d) $\\mathrm{PF}_{5}$\n63. Determine the formal charge of each element in the following:\n(a) $\\mathrm{H}_{3} \\mathrm{O}^{+}$\n(b) $\\mathrm{SO}_{4}{ }^{2-}$\n(c) $\\mathrm{NH}_{3}$\n(d) $\\mathrm{O}_{2}{ }^{2-}$\n(e) $\\mathrm{H}_{2} \\mathrm{O}_{2}$\n64. Calculate the formal charge of chlorine in the molecules $\\mathrm{Cl}_{2}, \\mathrm{BeCl}_{2}$, and $\\mathrm{ClF}_{5}$.\n65. Calculate the formal charge of each element in the following compounds and ions:\n(a) $\\mathrm{F}_{2} \\mathrm{CO}$\n(b) $\\mathrm{NO}^{-}$\n(c) $\\mathrm{BF}_{4}^{-}$\n(d) $\\mathrm{SnCl}_{3}-$\n(e) $\\mathrm{H}_{2} \\mathrm{CCH}_{2}$\n(f) $\\mathrm{ClF}_{3}$\n(g) $\\mathrm{SeF}_{6}$\n(h) $\\mathrm{PO}_{4}{ }^{3-}$\n66. Draw all possible resonance structures for each of these compounds. Determine the formal charge on each atom in each of the resonance structures:\n(a) $\\mathrm{O}_{3}$\n(b) $\\mathrm{SO}_{2}$\n(c) $\\mathrm{NO}_{2}{ }^{-}$\n(d) $\\mathrm{NO}_{3}{ }^{-}$\n67. Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in nitrosyl chloride: ClNO or ClON?"}
{"id": 2766, "contents": "462. Exercises - 462.5. Formal Charges and Resonance\n(d) $\\mathrm{NO}_{3}{ }^{-}$\n67. Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in nitrosyl chloride: ClNO or ClON?\n68. Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in hypochlorous acid: HOCl or OClH ?\n69. Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in sulfur dioxide: OSO or SOO?\n70. Draw the structure of hydroxylamine, $\\mathrm{H}_{3} \\mathrm{NO}$, and assign formal charges; look up the structure. Is the actual structure consistent with the formal charges?\n71. Iodine forms a series of fluorides (listed here). Write Lewis structures for each of the four compounds and determine the formal charge of the iodine atom in each molecule:\n(a) IF\n(b) $\\mathrm{IF}_{3}$\n(c) $\\mathrm{IF}_{5}$\n(d) $\\mathrm{IF}_{7}$\n72. Write the Lewis structure and chemical formula of the compound with a molar mass of about $70 \\mathrm{~g} / \\mathrm{mol}$ that contains $19.7 \\%$ nitrogen and $80.3 \\%$ fluorine by mass, and determine the formal charge of the atoms in this compound.\n73. Which of the following structures would we expect for nitrous acid? Determine the formal charges:"}
{"id": 2767, "contents": "462. Exercises - 462.5. Formal Charges and Resonance\n74. Sulfuric acid is the industrial chemical produced in greatest quantity worldwide. About 90 billion pounds are produced each year in the United States alone. Write the Lewis structure for sulfuric acid, $\\mathrm{H}_{2} \\mathrm{SO}_{4}$, which has two oxygen atoms and two OH groups bonded to the sulfur."}
{"id": 2768, "contents": "462. Exercises - 462.6. Molecular Structure and Polarity\n75. Explain why the HOH molecule is bent, whereas the HBeH molecule is linear.\n76. What feature of a Lewis structure can be used to tell if a molecule's (or ion's) electron-pair geometry and molecular structure will be identical?\n77. Explain the difference between electron-pair geometry and molecular structure.\n78. Why is the $\\mathrm{H}-\\mathrm{N}-\\mathrm{H}$ angle in $\\mathrm{NH}_{3}$ smaller than the $\\mathrm{H}-\\mathrm{C}-\\mathrm{H}$ bond angle in $\\mathrm{CH}_{4}$ ? Why is the $\\mathrm{H}-\\mathrm{N}-\\mathrm{H}$ angle in $\\mathrm{NH}_{4}{ }^{+}$identical to the $\\mathrm{H}-\\mathrm{C}-\\mathrm{H}$ bond angle in $\\mathrm{CH}_{4}$ ?\n79. Explain how a molecule that contains polar bonds can be nonpolar.\n80. As a general rule, $M X_{n}$ molecules (where $M$ represents a central atom and $X$ represents terminal atoms; $n$ $=2-5)$ are polar if there is one or more lone pairs of electrons on $M . \\mathrm{NH}_{3}(\\mathrm{M}=\\mathrm{N}, \\mathrm{X}=\\mathrm{H}, \\mathrm{n}=3)$ is an example. There are two molecular structures with lone pairs that are exceptions to this rule. What are they?\n81. Predict the electron pair geometry and the molecular structure of each of the following molecules or ions:\n(a) $\\mathrm{SF}_{6}$\n(b) $\\mathrm{PCl}_{5}$\n(c) $\\mathrm{BeH}_{2}$\n(d) $\\mathrm{CH}_{3}{ }^{+}$\n82. Identify the electron pair geometry and the molecular structure of each of the following molecules or ions:\n(a) $\\mathrm{IF}_{6}{ }^{+}$\n(b) $\\mathrm{CF}_{4}$\n(c) $\\mathrm{BF}_{3}$\n(d) $\\mathrm{SiF}_{5}{ }^{-}$\n(e) $\\mathrm{BeCl}_{2}$\n83. What are the electron-pair geometry and the molecular structure of each of the following molecules or ions?"}
{"id": 2769, "contents": "462. Exercises - 462.6. Molecular Structure and Polarity\n(d) $\\mathrm{SiF}_{5}{ }^{-}$\n(e) $\\mathrm{BeCl}_{2}$\n83. What are the electron-pair geometry and the molecular structure of each of the following molecules or ions?\n(a) $\\mathrm{ClF}_{5}$\n(b) $\\mathrm{ClO}_{2}-$\n(c) $\\mathrm{TeCl}_{4}{ }^{2-}$\n(d) $\\mathrm{PCl}_{3}$\n(e) $\\mathrm{SeF}_{4}$\n(f) $\\mathrm{PH}_{2}{ }^{-}$\n84. Predict the electron pair geometry and the molecular structure of each of the following ions:\n(a) $\\mathrm{H}_{3} \\mathrm{O}^{+}$\n(b) $\\mathrm{PCl}_{4}^{-}$\n(c) $\\mathrm{SnCl}_{3}+$\n(d) $\\mathrm{BrCl}_{4}^{-}$\n(e) $\\mathrm{ICl}_{3}$\n(f) $\\mathrm{XeF}_{4}$\n(g) $\\mathrm{SF}_{2}$\n85. Identify the electron pair geometry and the molecular structure of each of the following molecules:\n(a) ClNO ( N is the central atom)\n(b) $\\mathrm{CS}_{2}$\n(c) $\\mathrm{Cl}_{2} \\mathrm{CO}$ ( C is the central atom)\n(d) $\\mathrm{Cl}_{2} \\mathrm{SO}$ (S is the central atom)\n(e) $\\mathrm{SO}_{2} \\mathrm{~F}_{2}$ (S is the central atom)\n(f) $\\mathrm{XeO}_{2} \\mathrm{~F}_{2}$ (Xe is the central atom)\n(g) $\\mathrm{ClOF}_{2}+(\\mathrm{Cl}$ is the central atom $)$\n86. Predict the electron pair geometry and the molecular structure of each of the following:\n(a) $\\mathrm{IOF}_{5}$ ( I is the central atom)\n(b) $\\mathrm{POCl}_{3}$ ( P is the central atom)\n(c) $\\mathrm{Cl}_{2} \\mathrm{SeO}$ (Se is the central atom)\n(d) $\\mathrm{ClSO}^{+}$( S is the central atom)"}
{"id": 2770, "contents": "462. Exercises - 462.6. Molecular Structure and Polarity\n(c) $\\mathrm{Cl}_{2} \\mathrm{SeO}$ (Se is the central atom)\n(d) $\\mathrm{ClSO}^{+}$( S is the central atom)\n(e) $\\mathrm{F}_{2} \\mathrm{SO}$ ( S is the central atom)\n(f) $\\mathrm{NO}_{2}^{-}$\n(g) $\\mathrm{SiO}_{4}{ }^{4-}$\n87. Which of the following molecules and ions contain polar bonds? Which of these molecules and ions have dipole moments?\n(a) $\\mathrm{ClF}_{5}$\n(b) $\\mathrm{ClO}_{2}^{-}$\n(c) $\\mathrm{TeCl}_{4}{ }^{2-}$\n(d) $\\mathrm{PCl}_{3}$\n(e) $\\mathrm{SeF}_{4}$\n(f) $\\mathrm{PH}_{2}{ }^{-}$\n(g) $\\mathrm{XeF}_{2}$\n88. Which of these molecules and ions contain polar bonds? Which of these molecules and ions have dipole moments?\n(a) $\\mathrm{H}_{3} \\mathrm{O}^{+}$\n(b) $\\mathrm{PCl}_{4}^{-}$\n(c) $\\mathrm{SnCl}_{3}-$\n(d) $\\mathrm{BrCl}_{4}{ }^{-}$\n(e) $\\mathrm{ICl}_{3}$\n(f) $\\mathrm{XeF}_{4}$\n(g) $\\mathrm{SF}_{2}$\n89. Which of the following molecules have dipole moments?\n(a) $\\mathrm{CS}_{2}$\n(b) $\\mathrm{SeS}_{2}$\n(c) $\\mathrm{CCl}_{2} \\mathrm{~F}_{2}$\n(d) $\\mathrm{PCl}_{3}$ ( P is the central atom)\n(e) ClNO ( N is the central atom)\n90. Identify the molecules with a dipole moment:\n(a) $\\mathrm{SF}_{4}$\n(b) $\\mathrm{CF}_{4}$\n(c) $\\mathrm{Cl}_{2} \\mathrm{CCBr}_{2}$\n(d) $\\mathrm{CH}_{3} \\mathrm{Cl}$\n(e) $\\mathrm{H}_{2} \\mathrm{CO}$"}
{"id": 2771, "contents": "462. Exercises - 462.6. Molecular Structure and Polarity\n(c) $\\mathrm{Cl}_{2} \\mathrm{CCBr}_{2}$\n(d) $\\mathrm{CH}_{3} \\mathrm{Cl}$\n(e) $\\mathrm{H}_{2} \\mathrm{CO}$\n91. The molecule $\\mathrm{XF}_{3}$ has a dipole moment. Is X boron or phosphorus?\n92. The molecule $\\mathrm{XCl}_{2}$ has a dipole moment. Is X beryllium or sulfur?\n93. Is the $\\mathrm{Cl}_{2} \\mathrm{BBCl}_{2}$ molecule polar or nonpolar?\n94. There are three possible structures for $\\mathrm{PCl}_{2} \\mathrm{~F}_{3}$ with phosphorus as the central atom. Draw them and discuss how measurements of dipole moments could help distinguish among them.\n95. Describe the molecular structure around the indicated atom or atoms:\n(a) the sulfur atom in sulfuric acid, $\\mathrm{H}_{2} \\mathrm{SO}_{4}\\left[(\\mathrm{HO})_{2} \\mathrm{SO}_{2}\\right]$\n(b) the chlorine atom in chloric acid, $\\mathrm{HClO}_{3}\\left[\\mathrm{HOClO}_{2}\\right]$\n(c) the oxygen atom in hydrogen peroxide, HOOH\n(d) the nitrogen atom in nitric acid, $\\mathrm{HNO}_{3}\\left[\\mathrm{HONO}_{2}\\right]$\n(e) the oxygen atom in the OH group in nitric acid, $\\mathrm{HNO}_{3}\\left[\\mathrm{HONO}_{2}\\right]$\n(f) the central oxygen atom in the ozone molecule, $\\mathrm{O}_{3}$\n$(\\mathrm{g})$ each of the carbon atoms in propyne, $\\mathrm{CH}_{3} \\mathrm{CCH}$\n(h) the carbon atom in Freon, $\\mathrm{CCl}_{2} \\mathrm{~F}_{2}$\n(i) each of the carbon atoms in allene, $\\mathrm{H}_{2} \\mathrm{CCCH}_{2}$\n96. Draw the Lewis structures and predict the shape of each compound or ion:\n(a) $\\mathrm{CO}_{2}$\n(b) $\\mathrm{NO}_{2}{ }^{-}$\n(c) $\\mathrm{SO}_{3}$"}
{"id": 2772, "contents": "462. Exercises - 462.6. Molecular Structure and Polarity\n96. Draw the Lewis structures and predict the shape of each compound or ion:\n(a) $\\mathrm{CO}_{2}$\n(b) $\\mathrm{NO}_{2}{ }^{-}$\n(c) $\\mathrm{SO}_{3}$\n(d) $\\mathrm{SO}_{3}{ }^{2-}$\n97. A molecule with the formula $A B_{2}$, in which $A$ and $B$ represent different atoms, could have one of three different shapes. Sketch and name the three different shapes that this molecule might have. Give an example of a molecule or ion for each shape.\n98. A molecule with the formula $\\mathrm{AB}_{3}$, in which $A$ and $B$ represent different atoms, could have one of three different shapes. Sketch and name the three different shapes that this molecule might have. Give an example of a molecule or ion that has each shape.\n99. Draw the Lewis electron dot structures for these molecules, including resonance structures where appropriate:\n(a) $\\mathrm{CS}_{3}{ }^{2-}$\n(b) $\\mathrm{CS}_{2}$\n(c) CS\n(d) predict the molecular shapes for $\\mathrm{CS}_{3}{ }^{2-}$ and $\\mathrm{CS}_{2}$ and explain how you arrived at your predictions\n100. What is the molecular structure of the stable form of $\\mathrm{FNO}_{2}$ ? ( N is the central atom.)\n101. A compound with a molar mass of about $42 \\mathrm{~g} / \\mathrm{mol}$ contains $85.7 \\%$ carbon and $14.3 \\%$ hydrogen. What is its molecular structure?\n102. Use the simulation (http://openstax.org/l/16MolecPolarity) to perform the following exercises for a twoatom molecule:\n(a) Adjust the electronegativity value so the bond dipole is pointing toward B. Then determine what the electronegativity values must be to switch the dipole so that it points toward A.\n(b) With a partial positive charge on A, turn on the electric field and describe what happens.\n(c) With a small partial negative charge on A, turn on the electric field and describe what happens.\n(d) Reset all, and then with a large partial negative charge on A, turn on the electric field and describe what happens."}
{"id": 2773, "contents": "462. Exercises - 462.6. Molecular Structure and Polarity\n(c) With a small partial negative charge on A, turn on the electric field and describe what happens.\n(d) Reset all, and then with a large partial negative charge on A, turn on the electric field and describe what happens.\n103. Use the simulation (http://openstax.org/l/16MolecPolarity) to perform the following exercises for a real molecule. You may need to rotate the molecules in three dimensions to see certain dipoles.\n(a) Sketch the bond dipoles and molecular dipole (if any) for $\\mathrm{O}_{3}$. Explain your observations.\n(b) Look at the bond dipoles for $\\mathrm{NH}_{3}$. Use these dipoles to predict whether N or H is more electronegative.\n(c) Predict whether there should be a molecular dipole for $\\mathrm{NH}_{3}$ and, if so, in which direction it will point. Check the molecular dipole box to test your hypothesis.\n104. Use the Molecule Shape simulator (http://openstax.org/l/16MolecShape) to build a molecule. Starting with the central atom, click on the double bond to add one double bond. Then add one single bond and one lone pair. Rotate the molecule to observe the complete geometry. Name the electron group geometry and molecular structure and predict the bond angle. Then click the check boxes at the bottom and right of the simulator to check your answers.\n105. Use the Molecule Shape simulator (http://openstax.org/l/16MolecShape) to explore real molecules. On the Real Molecules tab, select $\\mathrm{H}_{2} \\mathrm{O}$. Switch between the \"real\" and \"model\" modes. Explain the difference observed.\n106. Use the Molecule Shape simulator (http://openstax.org/l/16MolecShape) to explore real molecules. On the Real Molecules tab, select \"model\" mode and $\\mathrm{S}_{2} \\mathrm{O}$. What is the model bond angle? Explain whether the \"real\" bond angle should be larger or smaller than the ideal model angle."}
{"id": 2774, "contents": "462. Exercises - 462.6. Molecular Structure and Polarity\n236 4\u2022Exercises"}
{"id": 2775, "contents": "463. CHAPTER 5 <br> Advanced Theories of Bonding - \nFigure 5.1 Oxygen molecules orient randomly most of the time, as shown in the top magnified view. However, when we pour liquid oxygen through a magnet, the molecules line up with the magnetic field, and the attraction allows them to stay suspended between the poles of the magnet where the magnetic field is strongest. Other diatomic molecules (like $\\mathrm{N}_{2}$ ) flow past the magnet. The detailed explanation of bonding described in this chapter allows us to understand this phenomenon. (credit: modification of work by Jefferson Lab)"}
{"id": 2776, "contents": "464. CHAPTER OUTLINE - \n5.1 Valence Bond Theory\n5.2 Hybrid Atomic Orbitals\n5.3 Multiple Bonds\n5.4 Molecular Orbital Theory\n\nINTRODUCTION We have examined the basic ideas of bonding, showing that atoms share electrons to form molecules with stable Lewis structures and that we can predict the shapes of those molecules by valence shell electron pair repulsion (VSEPR) theory. These ideas provide an important starting point for understanding chemical bonding. But these models sometimes fall short in their abilities to predict the behavior of real substances. How can we reconcile the geometries of $s, p$, and $d$ atomic orbitals with molecular shapes that show angles like $120^{\\circ}$ and $109.5^{\\circ}$ ? Furthermore, we know that electrons and magnetic behavior are related through electromagnetic fields. Both $\\mathrm{N}_{2}$ and $\\mathrm{O}_{2}$ have fairly similar Lewis structures that contain lone pairs of electrons.\n\n```\n:N\u4e09N: :\n```\n\nYet oxygen demonstrates very different magnetic behavior than nitrogen. We can pour liquid nitrogen through a magnetic field with no visible interactions, while liquid oxygen (shown in Figure 5.1) is attracted to the magnet and floats in the magnetic field. We need to understand the additional concepts of valence bond theory, orbital hybridization, and molecular orbital theory to understand these observations."}
{"id": 2777, "contents": "465. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the formation of covalent bonds in terms of atomic orbital overlap\n- Define and give examples of $\\sigma$ and $\\pi$ bonds\n\nAs we know, a scientific theory is a strongly supported explanation for observed natural laws or large bodies of experimental data. For a theory to be accepted, it must explain experimental data and be able to predict behavior. For example, VSEPR theory has gained widespread acceptance because it predicts threedimensional molecular shapes that are consistent with experimental data collected for thousands of different molecules. However, VSEPR theory does not provide an explanation of chemical bonding."}
{"id": 2778, "contents": "466. Atomic Orbital Overlap - \nThere are successful theories that describe the electronic structure of atoms. We can use quantum mechanics to predict the specific regions around an atom where electrons are likely to be located: A spherical shape for an $s$ orbital, a dumbbell shape for a $p$ orbital, and so forth. However, these predictions only describe the orbitals around free atoms. When atoms bond to form molecules, atomic orbitals are not sufficient to describe the regions where electrons will be located in the molecule. A more complete understanding of electron distributions requires a model that can account for the electronic structure of molecules. One popular theory holds that a covalent bond forms when a pair of electrons is shared by two atoms and is simultaneously attracted by the nuclei of both atoms. In the following sections, we will discuss how such bonds are described by valence bond theory and hybridization.\n\nValence bond theory describes a covalent bond as the overlap of half-filled atomic orbitals (each containing a single electron) that yield a pair of electrons shared between the two bonded atoms. We say that orbitals on two different atoms overlap when a portion of one orbital and a portion of a second orbital occupy the same region of space. According to valence bond theory, a covalent bond results when two conditions are met: (1) an orbital on one atom overlaps an orbital on a second atom and (2) the single electrons in each orbital combine to form an electron pair. The mutual attraction between this negatively charged electron pair and the two atoms' positively charged nuclei serves to physically link the two atoms through a force we define as a covalent bond. The strength of a covalent bond depends on the extent of overlap of the orbitals involved. Orbitals that overlap extensively form bonds that are stronger than those that have less overlap."}
{"id": 2779, "contents": "466. Atomic Orbital Overlap - \nThe energy of the system depends on how much the orbitals overlap. Figure 5.2 illustrates how the sum of the energies of two hydrogen atoms (the colored curve) changes as they approach each other. When the atoms are far apart there is no overlap, and by convention we set the sum of the energies at zero. As the atoms move together, their orbitals begin to overlap. Each electron begins to feel the attraction of the nucleus in the other atom. In addition, the electrons begin to repel each other, as do the nuclei. While the atoms are still widely separated, the attractions are slightly stronger than the repulsions, and the energy of the system decreases. (A bond begins to form.) As the atoms move closer together, the overlap increases, so the attraction of the nuclei for the electrons continues to increase (as do the repulsions among electrons and between the nuclei). At some specific distance between the atoms, which varies depending on the atoms involved, the energy reaches its lowest (most stable) value. This optimum distance between the two bonded nuclei is the bond distance between the two atoms. The bond is stable because at this point, the attractive and repulsive forces combine to create the lowest possible energy configuration. If the distance between the nuclei were to decrease further, the repulsions between nuclei and the repulsions as electrons are confined in closer proximity to each other would become stronger than the attractive forces. The energy of the system would then rise (making the system destabilized), as shown at the far left of Figure 5.2.\n\n\nFIGURE 5.2 (a) The interaction of two hydrogen atoms changes as a function of distance. (b) The energy of the system changes as the atoms interact. The lowest (most stable) energy occurs at a distance of 74 pm , which is the bond length observed for the $\\mathrm{H}_{2}$ molecule.\n\nIn addition to the distance between two orbitals, the orientation of orbitals also affects their overlap (other than for two $s$ orbitals, which are spherically symmetric). Greater overlap is possible when orbitals are oriented such that they overlap on a direct line between the two nuclei. Figure 5.3 illustrates this for two $p$ orbitals from different atoms; the overlap is greater when the orbitals overlap end to end rather than at an angle.\n\n(a)\n\n(b)"}
{"id": 2780, "contents": "466. Atomic Orbital Overlap - \n(a)\n\n(b)\n\nFIGURE 5.3 (a) The overlap of two $p$ orbitals is greatest when the orbitals are directed end to end. (b) Any other arrangement results in less overlap. The dots indicate the locations of the nuclei.\n\nThe overlap of two $s$ orbitals (as in $\\mathrm{H}_{2}$ ), the overlap of an $s$ orbital and a $p$ orbital (as in HCl ), and the end-to-end overlap of two $p$ orbitals (as in $\\mathrm{Cl}_{2}$ ) all produce sigma bonds ( $\\boldsymbol{\\sigma}$ bonds), as illustrated in Figure 5.4. A $\\sigma$ bond is a covalent bond in which the electron density is concentrated in the region along the internuclear axis; that is, a line between the nuclei would pass through the center of the overlap region. Single bonds in Lewis structures are described as $\\sigma$ bonds in valence bond theory.\n\n\nFIGURE 5.4 Sigma ( $\\sigma$ ) bonds form from the overlap of the following: (a) two $s$ orbitals, (b) an $s$ orbital and a $p$ orbital, and (c) two $p$ orbitals. The dots indicate the locations of the nuclei.\n\nA pi bond ( $\\boldsymbol{\\pi}$ bond) is a type of covalent bond that results from the side-by-side overlap of two $p$ orbitals, as illustrated in Figure 5.5. In a $\\pi$ bond, the regions of orbital overlap lie on opposite sides of the internuclear axis.\n\nAlong the axis itself, there is a node, that is, a plane with no probability of finding an electron.\n\n\nFIGURE 5.5 $\\mathrm{Pi}(\\pi)$ bonds form from the side-by-side overlap of two $p$ orbitals. The dots indicate the location of the nuclei."}
{"id": 2781, "contents": "466. Atomic Orbital Overlap - \nFIGURE 5.5 $\\mathrm{Pi}(\\pi)$ bonds form from the side-by-side overlap of two $p$ orbitals. The dots indicate the location of the nuclei.\n\nWhile all single bonds are $\\sigma$ bonds, multiple bonds consist of both $\\sigma$ and $\\pi$ bonds. As the Lewis structures below suggest, $\\mathrm{O}_{2}$ contains a double bond, and $\\mathrm{N}_{2}$ contains a triple bond. The double bond consists of one $\\sigma$ bond and one $\\pi$ bond, and the triple bond consists of one $\\sigma$ bond and two $\\pi$ bonds. Between any two atoms, the first bond formed will always be a $\\sigma$ bond, but there can only be one $\\sigma$ bond in any one location. In any multiple bond, there will be one $\\sigma$ bond, and the remaining one or two bonds will be $\\pi$ bonds. These bonds are described in more detail later in this chapter.\n\n\nOne $\\sigma$ bond\nNo $\\pi$ bonds\n\n$$\n: \\ddot{o}=\\ddot{\\mathrm{O}}:\n$$\n\nOne $\\sigma$ bond One $\\pi$ bond\n$: N \\equiv N:$"}
{"id": 2782, "contents": "467. One $\\sigma$ bond - \nTwo $\\pi$ bonds"}
{"id": 2783, "contents": "469. Counting $\\sigma$ and $\\pi$ Bonds - \nButadiene, $\\mathrm{C}_{4} \\mathrm{H}_{6}$, is used to make synthetic rubber. Identify the number of $\\sigma$ and $\\pi$ bonds contained in this molecule."}
{"id": 2784, "contents": "470. Solution - \nThere are six $\\sigma \\mathrm{C}-\\mathrm{H}$ bonds and one $\\sigma \\mathrm{C}-\\mathrm{C}$ bond, for a total of seven from the single bonds. There are two double bonds that each have a $\\pi$ bond in addition to the $\\sigma$ bond. This gives a total nine $\\sigma$ and two $\\pi$ bonds overall."}
{"id": 2785, "contents": "471. Check Your Learning - \nIdentify each illustration as depicting a $\\sigma$ or $\\pi$ bond:\n(a) side-by-side overlap of a $4 p$ and a $2 p$ orbital\n(b) end-to-end overlap of a $4 p$ and $4 p$ orbital\n(c) end-to-end overlap of a $4 p$ and a $2 p$ orbital\n\n(a)\n\n(b)\n\n(c)"}
{"id": 2786, "contents": "472. Answer: - \n(a) is a $\\pi$ bond with a node along the axis connecting the nuclei while (b) and (c) are $\\sigma$ bonds that overlap along the axis."}
{"id": 2787, "contents": "473. Dipole Moments and Ionic Character - \nNow that we have seen the importance of understanding the connection between the location of electrons in atoms and the properties of elements, we can expand our understanding of the connection between atoms. This will be an introduction to more advanced aspects of the chemical bond, which is the very heart of chemistry itself. With the sole exception of the noble gases, atoms by themselves do not possess the most stable possible electron configuration. That is where the concept of chemical bonding comes into its own: atoms can attain a stable configuration by exchanging electrons with another atom, resulting in the formation of ions.\nIons, in turn, can associate by charge - simple Coulombic attraction - resulting in the formation of compounds we call ionic compounds. We will look at the ionic nature of bonds first, from a simple positive-negative attraction standpoint. Just as important is that some atoms bond by sharing rather than exchanging electrons; the sharing of electrons gives rise to the covalent bond. To add just one more dimension, some chemical species are neither completely ionic nor completely covalent; these species possess a permanent dipole, and are classified as polar.\nIn your introductory physics course, you will likely discuss the concept of Coulombic interactions in much more rigorous detail than we will do here. We are interested primarily in the differences in properties between species that arise from their relative covalent, ionic, or polar nature - not in a rigorous model of those properties. We are concerned with the connection between potential energy and force and the relative separation (or lack of separation) between charges. We begin by defining the electric or Coulomb force as the product of the charges divided by the square of the distance between those charges:\n\n$$\nF=\\frac{Q_{1} Q_{2}}{d^{2}}\n$$\n\nHere, $Q$ is taken to be the fundamental constant of electron charge: one electron has a charge of $1.60218 \\times 10^{-19} \\mathrm{C}$. (We will work exclusively in the SI system, so distances will be measured in meters $(\\mathrm{m})$ ).\nAnd as you may recall, energy is force times distance, so\n\n$$\nE=\\frac{Q_{1} Q_{2}}{d}\n$$\n\nTo illustrate the trend in attractive force, we will consider first the attractive force between two ions of single charge separated by a distance of 2 d :"}
{"id": 2788, "contents": "473. Dipole Moments and Ionic Character - \n$$\nE=\\frac{Q_{1} Q_{2}}{d}\n$$\n\nTo illustrate the trend in attractive force, we will consider first the attractive force between two ions of single charge separated by a distance of 2 d :\n\n$$\nF=\\frac{(1)(-1)}{(2 d)^{2}}=-\\frac{1}{4 d^{2}}\n$$\n\nAnd then the attractive force between two ions of double charge separated by a distance $d$ :\n\n$$\nF=\\frac{(2)(-2)}{(d)^{2}}=-\\frac{4}{d^{2}}\n$$\n\nThe force of attraction increases with the charge and decreases with increased distance. If all matter were composed of ions that would be the end of the story, but it clearly is not. There are molecules for which the charge - either positive or negative - is permanently concentrated more on one atom than the other. Notice we say atom, because these compounds are not composed of ions but rather of atoms sharing electrons through covalent bonds."}
{"id": 2789, "contents": "474. Bond Dipole Moments - \nThe concept of a bond dipole moment allows us to examine the partial separation of charge between atoms. It is a simple model when applied to diatomic molecules, which will be more than sufficient for our purposes. The dipole moment of a bond is defined as the charge times the distance - charge once again being measured in multiples of the charge on an electron, or coulombs. The distance will always be in meters. Because we are considering very small charges and distances, and because it is the relative separation of charge rather than the actual value for it that we are interested in, we will introduce a new unit called the Debye, named after the physical chemist Peter Debye:\n\n$$\n1 \\text { Debye }(\\mathrm{D})=3.336 \\times 10^{-30} \\mathrm{C}-\\mathrm{m} \\quad \\mu=Q \\times d\n$$\n\nThe usefulness of the Debye unit will be shown by example:\nFor HCl , the bond dipole moment is known to be 1.08 D\nFor HI, the bond dipole moment is known to be 0.44 D\nComparing the two, we can see that HI is less polar than HCl , which is what we would expect based on electronegativity values.\n\nWe have now made a transition between the concept of an ionic compound and a partially ionic one. Of course, the partially ionic compound must also by definition be partially covalent."}
{"id": 2790, "contents": "475. Partial Ionic Character - \nThe concept of the bond dipole moment helps bridge the concepts of ionic and covalent bonding. Because there is a separation of charge that is less complete than it is in an ionic bond, we can refer to polar bonds as being partially ionic in nature. In contrast to sodium chloride, hydrogen chloride shows partial charges (indicated with a delta notation) on the hydrogen and chlorine. As you would expect from the electronegativity values, hydrogen carries a partial positive charge, while chlorine carries a partial negative charge. Where do these charges come from?\n\nIt is easy to come up with the partial charges by comparing the actual dipole moments (which can be obtained experimentally, using spectroscopy) with the dipole expected in the limiting case (that is, if we were to consider the molecule ionic). The actual dipole moment is 1.03 D ."}
{"id": 2791, "contents": "477. Finding the Partial Ionic Character - \nWhat are the partial charges of an HCl molecule, whose bond length is 0.127 nm ?"}
{"id": 2792, "contents": "478. Solution - \nThe bond dipole moment is $\\left(1.60218 \\times 10^{-19} \\mathrm{C}\\right)\\left(0.127 \\times 10^{-9} \\mathrm{~m}\\right)$ or $2.03 \\times 10^{-29} \\mathrm{C}-\\mathrm{m}$. Converted to D , this is $\\left(2.03 \\times 10^{-29} \\mathrm{C}-\\mathrm{m}\\right)\\left(\\frac{1 \\text { Debye }}{3.336 \\times 10^{-30} \\mathrm{C}-\\mathrm{m}}\\right)$ or 6.09 D . Were HCl completely ionic, this would be its molecular dipole moment. To get the partial ionic character, we divide the experimentally measured bond moment by this limiting value: \\%ionic character $\\left.=\\frac{\\mu_{\\exp }}{\\mu_{\\lim }} \\times 100 \\%=\\frac{(1.03}{(6.09} \\quad \\mathrm{D}\\right) ~ \\times 100 \\%=17 \\%$. This means the bond is about $17 \\%$ ionic - or, put another way, the positive charge in H is +0.17 and the partial negative charge on chlorine, -0.17."}
{"id": 2793, "contents": "479. Check Your Learning - \nRepeat the calculation for HI , which has a dipole moment of 0.42 D and a bond length of 0.161 nm .\nAnswer: Calculated 7.73, percent 5.43\n\nWhat does the result suggest about the relative polarity of the HI bond vs. that of the HCl bond? Does the calculated dipole and percent ionic character reconcile with the difference in electronegativity between Cl and I?\n\nThe electron configuration of an atom or ion is key to understanding the chemical behavior of an element. The atoms that make up the element combine in various ways, ranging from the mostly ionic ( NaCl ) to the partially ionic ( HCl ) to what we will call purely covalent. At the most fundamental level, all chemical bonds involve electrons, and a significant percentage of chemical and physical properties can be explained by considering the location and separation of charge in a species. By understanding the structure of matter at the atomic level, we can begin to build an understanding of the behavior of matter at both the microscopic and macroscopic levels.\n\nAn understanding of dipoles and partial ionic character is fundamental to understanding the interactions between particles, which we will examine in the chapter on liquids and solids. These intermolecular forces\nbecome important in the liquid and solid states of matter."}
{"id": 2794, "contents": "480. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Explain the concept of atomic orbital hybridization\n- Determine the hybrid orbitals associated with various molecular geometries\n\nThinking in terms of overlapping atomic orbitals is one way for us to explain how chemical bonds form in diatomic molecules. However, to understand how molecules with more than two atoms form stable bonds, we require a more detailed model. As an example, let us consider the water molecule, in which we have one oxygen atom bonding to two hydrogen atoms. Oxygen has the electron configuration $1 s^{2} 2 s^{2} 2 p^{4}$, with two unpaired electrons (one in each of the two $2 p$ orbitals). Valence bond theory would predict that the two $\\mathrm{O}-\\mathrm{H}$ bonds form from the overlap of these two $2 p$ orbitals with the $1 s$ orbitals of the hydrogen atoms. If this were the case, the bond angle would be $90^{\\circ}$, as shown in Figure 5.6 , because $p$ orbitals are perpendicular to each other. Experimental evidence shows that the bond angle is $104.5^{\\circ}$, not $90^{\\circ}$. The prediction of the valence bond theory model does not match the real-world observations of a water molecule; a different model is needed.\n\n\nFIGURE 5.6 The hypothetical overlap of two of the $2 p$ orbitals on an oxygen atom (red) with the 1 s orbitals of two hydrogen atoms (blue) would produce a bond angle of $90^{\\circ}$. This is not consistent with experimental evidence. ${ }^{1}$"}
{"id": 2795, "contents": "480. LEARNING OBJECTIVES - \nQuantum-mechanical calculations suggest why the observed bond angles in $\\mathrm{H}_{2} \\mathrm{O}$ differ from those predicted by the overlap of the $1 s$ orbital of the hydrogen atoms with the $2 p$ orbitals of the oxygen atom. The mathematical expression known as the wave function, $\\psi$, contains information about each orbital and the wavelike properties of electrons in an isolated atom. When atoms are bound together in a molecule, the wave functions combine to produce new mathematical descriptions that have different shapes. This process of combining the wave functions for atomic orbitals is called hybridization and is mathematically accomplished by the linear combination of atomic orbitals, LCAO, (a technique that we will encounter again later). The new orbitals that result are called hybrid orbitals. The valence orbitals in an isolated oxygen atom are a $2 s$ orbital and three $2 p$ orbitals. The valence orbitals in an oxygen atom in a water molecule differ; they consist of four equivalent hybrid orbitals that point approximately toward the corners of a tetrahedron (Figure 5.7). Consequently, the overlap of the O and H orbitals should result in a tetrahedral bond angle $\\left(109.5^{\\circ}\\right)$. The observed angle of $104.5^{\\circ}$ is experimental evidence for which quantum-mechanical calculations give a useful explanation: Valence bond theory must include a hybridization component to give accurate predictions.\n\n(a)\n\n(b)\n\nFIGURE 5.7 (a) A water molecule has four regions of electron density, so VSEPR theory predicts a tetrahedral arrangement of hybrid orbitals. (b) Two of the hybrid orbitals on oxygen contain lone pairs, and the other two overlap with the 1 s orbitals of hydrogen atoms to form the $\\mathrm{O}-\\mathrm{H}$ bonds in $\\mathrm{H}_{2} \\mathrm{O}$. This description is more consistent with the\n\n[^3]experimental structure.\nThe following ideas are important in understanding hybridization:"}
{"id": 2796, "contents": "480. LEARNING OBJECTIVES - \n[^3]experimental structure.\nThe following ideas are important in understanding hybridization:\n\n1. Hybrid orbitals do not exist in isolated atoms. They are formed only in covalently bonded atoms.\n2. Hybrid orbitals have shapes and orientations that are very different from those of the atomic orbitals in isolated atoms.\n3. A set of hybrid orbitals is generated by combining atomic orbitals. The number of hybrid orbitals in a set is equal to the number of atomic orbitals that were combined to produce the set.\n4. All orbitals in a set of hybrid orbitals are equivalent in shape and energy.\n5. The type of hybrid orbitals formed in a bonded atom depends on its electron-pair geometry as predicted by the VSEPR theory.\n6. Hybrid orbitals overlap to form $\\sigma$ bonds. Unhybridized orbitals overlap to form $\\pi$ bonds.\n\nIn the following sections, we shall discuss the common types of hybrid orbitals."}
{"id": 2797, "contents": "481. sp Hybridization - \nThe beryllium atom in a gaseous $\\mathrm{BeCl}_{2}$ molecule is an example of a central atom with no lone pairs of electrons in a linear arrangement of three atoms. There are two regions of valence electron density in the $\\mathrm{BeCl}_{2}$ molecule that correspond to the two covalent $\\mathrm{Be}-\\mathrm{Cl}$ bonds. To accommodate these two electron domains, two of the Be atom's four valence orbitals will mix to yield two hybrid orbitals. This hybridization process involves mixing of the valence $s$ orbital with one of the valence $p$ orbitals to yield two equivalent $\\boldsymbol{s p}$ hybrid orbitals that are oriented in a linear geometry (Figure 5.8). In this figure, the set of $s p$ orbitals appears similar in shape to the original $p$ orbital, but there is an important difference. The number of atomic orbitals combined always equals the number of hybrid orbitals formed. The $p$ orbital is one orbital that can hold up to two electrons. The sp set is two equivalent orbitals that point $180^{\\circ}$ from each other. The two electrons that were originally in the $s$ orbital are now distributed to the two $s p$ orbitals, which are half filled. In gaseous $\\mathrm{BeCl}_{2}$, these half-filled hybrid orbitals will overlap with orbitals from the chlorine atoms to form two identical $\\sigma$ bonds.\n\n\nGives a linear arrangement\n\n\nFIGURE 5.8 Hybridization of an $s$ orbital (blue) and a $p$ orbital (red) of the same atom produces two $s p$ hybrid orbitals (yellow). Each hybrid orbital is oriented primarily in just one direction. Note that each sp orbital contains one lobe that is significantly larger than the other. The set of two $s p$ orbitals are oriented at $180^{\\circ}$, which is consistent with the geometry for two domains."}
{"id": 2798, "contents": "481. sp Hybridization - \nWe illustrate the electronic differences in an isolated Be atom and in the bonded Be atom in the orbital energylevel diagram in Figure 5.9. These diagrams represent each orbital by a horizontal line (indicating its energy) and each electron by an arrow. Energy increases toward the top of the diagram. We use one upward arrow to indicate one electron in an orbital and two arrows (up and down) to indicate two electrons of opposite spin.\n\n\nFIGURE 5.9 This orbital energy-level diagram shows the $s p$ hybridized orbitals on Be in the linear $\\mathrm{BeCl}_{2}$ molecule. Each of the two $s p$ hybrid orbitals holds one electron and is thus half filled and available for bonding via overlap with a Cl 3 p orbital.\n\nWhen atomic orbitals hybridize, the valence electrons occupy the newly created orbitals. The Be atom had two valence electrons, so each of the $s p$ orbitals gets one of these electrons. Each of these electrons pairs up with the unpaired electron on a chlorine atom when a hybrid orbital and a chlorine orbital overlap during the formation of the $\\mathrm{Be}-\\mathrm{Cl}$ bonds.\n\nAny central atom surrounded by just two regions of valence electron density in a molecule will exhibit $s p$ hybridization. Other examples include the mercury atom in the linear $\\mathrm{HgCl}_{2}$ molecule, the zinc atom in $\\mathrm{Zn}\\left(\\mathrm{CH}_{3}\\right)_{2}$, which contains a linear $\\mathrm{C}-\\mathrm{Zn}-\\mathrm{C}$ arrangement, and the carbon atoms in HCCH and $\\mathrm{CO}_{2}$."}
{"id": 2799, "contents": "482. (8) LINK TO LEARNING - \nCheck out the University of Wisconsin-Oshkosh website (http://openstax.org/l/16hybridorbital) to learn about visualizing hybrid orbitals in three dimensions."}
{"id": 2800, "contents": "483. $s p^{2}$ Hybridization - \nThe valence orbitals of a central atom surrounded by three regions of electron density consist of a set of three $\\boldsymbol{s p}^{\\mathbf{2}}$ hybrid orbitals and one unhybridized $p$ orbital. This arrangement results from $s p^{2}$ hybridization, the mixing of one $s$ orbital and two $p$ orbitals to produce three identical hybrid orbitals oriented in a trigonal planar geometry (Figure 5.10).\n\n\nFIGURE 5.10 The hybridization of an $s$ orbital (blue) and two $p$ orbitals (red) produces three equivalent $s p^{2}$ hybridized orbitals (yellow) oriented at $120^{\\circ}$ with respect to each other. The remaining unhybridized $p$ orbital is not shown here, but is located along the $z$ axis.\n\nAlthough quantum mechanics yields the \"plump\" orbital lobes as depicted in Figure 5.10, sometimes for clarity these orbitals are drawn thinner and without the minor lobes, as in Figure 5.11, to avoid obscuring other features of a given illustration. We will use these \"thinner\" representations whenever the true view is too crowded to easily visualize.\n\n\nFIGURE 5.11 This alternate way of drawing the trigonal planar $s p^{2}$ hybrid orbitals is sometimes used in more crowded figures.\n\nThe observed structure of the borane molecule, $\\mathrm{BH}_{3}$, suggests $s p^{2}$ hybridization for boron in this compound. The molecule is trigonal planar, and the boron atom is involved in three bonds to hydrogen atoms (Figure 5.12). We can illustrate the comparison of orbitals and electron distribution in an isolated boron atom and in the bonded atom in $\\mathrm{BH}_{3}$ as shown in the orbital energy level diagram in Figure 5.13. We redistribute the three valence electrons of the boron atom in the three $s p^{2}$ hybrid orbitals, and each boron electron pairs with a hydrogen electron when $\\mathrm{B}-\\mathrm{H}$ bonds form.\n\n\nFIGURE $5.12 \\quad \\mathrm{BH}_{3}$ is an electron-deficient molecule with a trigonal planar structure."}
{"id": 2801, "contents": "483. $s p^{2}$ Hybridization - \nFIGURE $5.12 \\quad \\mathrm{BH}_{3}$ is an electron-deficient molecule with a trigonal planar structure.\n\n\nFIGURE 5.13 In an isolated $B$ atom, there are one $2 s$ and three $2 p$ valence orbitals. When boron is in a molecule with three regions of electron density, three of the orbitals hybridize and create a set of three $s p^{2}$ orbitals and one unhybridized $2 p$ orbital. The three half-filled hybrid orbitals each overlap with an orbital from a hydrogen atom to form three $\\sigma$ bonds in $\\mathrm{BH}_{3}$.\n\nAny central atom surrounded by three regions of electron density will exhibit $s p^{2}$ hybridization. This includes molecules with a lone pair on the central atom, such as ClNO (Figure 5.14), or molecules with two single bonds and a double bond connected to the central atom, as in formaldehyde, $\\mathrm{CH}_{2} \\mathrm{O}$, and ethene, $\\mathrm{H}_{2} \\mathrm{CCH}_{2}$.\n\n\n\n\nFIGURE 5.14 The central atom(s) in each of the structures shown contain three regions of electron density and are $s p^{2}$ hybridized. As we know from the discussion of VSEPR theory, a region of electron density contains all of the electrons that point in one direction. A lone pair, an unpaired electron, a single bond, or a multiple bond would each count as one region of electron density."}
{"id": 2802, "contents": "484. $s p^{3}$ Hybridization - \nThe valence orbitals of an atom surrounded by a tetrahedral arrangement of bonding pairs and lone pairs consist of a set of four $\\boldsymbol{s p}^{\\mathbf{3}}$ hybrid orbitals. The hybrids result from the mixing of one $s$ orbital and all three $p$ orbitals that produces four identical $s p^{3}$ hybrid orbitals (Figure 5.15). Each of these hybrid orbitals points toward a different corner of a tetrahedron.\n\n\nFIGURE 5.15 The hybridization of an $s$ orbital (blue) and three $p$ orbitals (red) produces four equivalent $s p^{3}$ hybridized orbitals (yellow) oriented at $109.5^{\\circ}$ with respect to each other.\n\nA molecule of methane, $\\mathrm{CH}_{4}$, consists of a carbon atom surrounded by four hydrogen atoms at the corners of a tetrahedron. The carbon atom in methane exhibits $s p^{3}$ hybridization. We illustrate the orbitals and electron distribution in an isolated carbon atom and in the bonded atom in $\\mathrm{CH}_{4}$ in Figure 5.16. The four valence electrons of the carbon atom are distributed equally in the hybrid orbitals, and each carbon electron pairs with a hydrogen electron when the $\\mathrm{C}-\\mathrm{H}$ bonds form.\n\n\nFIGURE 5.16 The four valence atomic orbitals from an isolated carbon atom all hybridize when the carbon bonds\nin a molecule like $\\mathrm{CH}_{4}$ with four regions of electron density. This creates four equivalent $s p^{3}$ hybridized orbitals. Overlap of each of the hybrid orbitals with a hydrogen orbital creates a $\\mathrm{C}-\\mathrm{H} \\sigma$ bond."}
{"id": 2803, "contents": "484. $s p^{3}$ Hybridization - \nIn a methane molecule, the $1 s$ orbital of each of the four hydrogen atoms overlaps with one of the four $s p^{3}$ orbitals of the carbon atom to form a sigma ( $\\sigma$ ) bond. This results in the formation of four strong, equivalent covalent bonds between the carbon atom and each of the hydrogen atoms to produce the methane molecule, $\\mathrm{CH}_{4}$.\nThe structure of ethane, $\\mathrm{C}_{2} \\mathrm{H}_{6}$, is similar to that of methane in that each carbon in ethane has four neighboring atoms arranged at the corners of a tetrahedron-three hydrogen atoms and one carbon atom (Figure 5.17). However, in ethane an $s p^{3}$ orbital of one carbon atom overlaps end to end with an $s p^{3}$ orbital of a second carbon atom to form a $\\sigma$ bond between the two carbon atoms. Each of the remaining $s p^{3}$ hybrid orbitals overlaps with an $s$ orbital of a hydrogen atom to form carbon-hydrogen $\\sigma$ bonds. The structure and overall outline of the bonding orbitals of ethane are shown in Figure 5.17. The orientation of the two $\\mathrm{CH}_{3}$ groups is not fixed relative to each other. Experimental evidence shows that rotation around $\\sigma$ bonds occurs easily.\n\n\nFIGURE 5.17 (a) In the ethane molecule, $\\mathrm{C}_{2} \\mathrm{H}_{6}$, each carbon has four $s p^{3}$ orbitals. (b) These four orbitals overlap to form seven $\\sigma$ bonds.\n\nAn $s p^{3}$ hybrid orbital can also hold a lone pair of electrons. For example, the nitrogen atom in ammonia is surrounded by three bonding pairs and a lone pair of electrons directed to the four corners of a tetrahedron. The nitrogen atom is $s p^{3}$ hybridized with one hybrid orbital occupied by the lone pair."}
{"id": 2804, "contents": "484. $s p^{3}$ Hybridization - \nThe molecular structure of water is consistent with a tetrahedral arrangement of two lone pairs and two bonding pairs of electrons. Thus we say that the oxygen atom is $s p^{3}$ hybridized, with two of the hybrid orbitals occupied by lone pairs and two by bonding pairs. Since lone pairs occupy more space than bonding pairs, structures that contain lone pairs have bond angles slightly distorted from the ideal. Perfect tetrahedra have angles of $109.5^{\\circ}$, but the observed angles in ammonia ( $107.3^{\\circ}$ ) and water ( $104.5^{\\circ}$ ) are slightly smaller. Other examples of $s p^{3}$ hybridization include $\\mathrm{CCl}_{4}, \\mathrm{PCl}_{3}$, and $\\mathrm{NCl}_{3}$.\n$s p^{3} d$ and $s p^{3} d^{2}$ Hybridization\nTo describe the five bonding orbitals in a trigonal bipyramidal arrangement, we must use five of the valence shell atomic orbitals (the $s$ orbital, the three $p$ orbitals, and one of the $d$ orbitals), which gives five $\\boldsymbol{s p}^{\\mathbf{3} \\boldsymbol{d}}$ hybrid orbitals. With an octahedral arrangement of six hybrid orbitals, we must use six valence shell atomic orbitals (the $s$ orbital, the three $p$ orbitals, and two of the $d$ orbitals in its valence shell), which gives six $\\boldsymbol{s p}^{3} \\boldsymbol{d}^{2}$ hybrid orbitals. These hybridizations are only possible for atoms that have $d$ orbitals in their valence subshells (that is, not those in the first or second period)."}
{"id": 2805, "contents": "484. $s p^{3}$ Hybridization - \nIn a molecule of phosphorus pentachloride, $\\mathrm{PCl}_{5}$, there are five $\\mathrm{P}-\\mathrm{Cl}$ bonds (thus five pairs of valence electrons around the phosphorus atom) directed toward the corners of a trigonal bipyramid. We use the $3 s$ orbital, the three $3 p$ orbitals, and one of the $3 d$ orbitals to form the set of five $s p^{3} d$ hybrid orbitals (Figure 5.19) that are involved in the $\\mathrm{P}-\\mathrm{Cl}$ bonds. Other atoms that exhibit $s p^{3} d$ hybridization include the sulfur atom in $\\mathrm{SF}_{4}$ and the chlorine atoms in $\\mathrm{ClF}_{3}$ and in $\\mathrm{ClF}_{4}{ }^{+}$. (The electrons on fluorine atoms are omitted for clarity.)"}
{"id": 2806, "contents": "484. $s p^{3}$ Hybridization - \nFIGURE 5.18 The three compounds pictured exhibit $s p^{3} d$ hybridization in the central atom and a trigonal bipyramid form. $\\mathrm{SF}_{4}$ and $\\mathrm{ClF}_{4}{ }^{+}$have one lone pair of electrons on the central atom, and $\\mathrm{ClF}_{3}$ has two lone pairs giving it the T -shape shown.\n\n(a)\n\n(b)\n\nFIGURE 5.19 (a) The five regions of electron density around phosphorus in $\\mathrm{PCl}_{5}$ require five hybrid $s p^{3} d$ orbitals. (b) These orbitals combine to form a trigonal bipyramidal structure with each large lobe of the hybrid orbital pointing at a vertex. As before, there are also small lobes pointing in the opposite direction for each orbital (not shown for clarity).\n\nThe sulfur atom in sulfur hexafluoride, $\\mathrm{SF}_{6}$, exhibits $s p^{3} d^{2}$ hybridization. A molecule of sulfur hexafluoride has six bonding pairs of electrons connecting six fluorine atoms to a single sulfur atom. There are no lone pairs of electrons on the central atom. To bond six fluorine atoms, the $3 s$ orbital, the three $3 p$ orbitals, and two of the $3 d$ orbitals form six equivalent $s p^{3} d^{2}$ hybrid orbitals, each directed toward a different corner of an octahedron. Other atoms that exhibit $s p^{3} d^{2}$ hybridization include the phosphorus atom in $\\mathrm{PCl}_{6}{ }^{-}$, the iodine atom in the interhalogens $\\mathrm{IF}_{6}{ }^{+}, \\mathrm{IF}_{5}, \\mathrm{ICl}_{4}{ }^{-}, \\mathrm{IF}_{4}{ }^{-}$and the xenon atom in $\\mathrm{XeF}_{4}$."}
{"id": 2807, "contents": "484. $s p^{3}$ Hybridization - \nFIGURE 5.20 (a) Sulfur hexafluoride, $\\mathrm{SF}_{6}$, has an octahedral structure that requires $s p^{3} d^{2}$ hybridization. (b) The six $s p^{3} d^{2}$ orbitals form an octahedral structure around sulfur. Again, the minor lobe of each orbital is not shown for clarity."}
{"id": 2808, "contents": "485. Assignment of Hybrid Orbitals to Central Atoms - \nThe hybridization of an atom is determined based on the number of regions of electron density that surround it. The geometrical arrangements characteristic of the various sets of hybrid orbitals are shown in Figure 5.21. These arrangements are identical to those of the electron-pair geometries predicted by VSEPR theory. VSEPR theory predicts the shapes of molecules, and hybrid orbital theory provides an explanation for how those shapes are formed. To find the hybridization of a central atom, we can use the following guidelines:\n\n1. Determine the Lewis structure of the molecule.\n2. Determine the number of regions of electron density around an atom using VSEPR theory, in which single bonds, multiple bonds, radicals, and lone pairs each count as one region.\n3. Assign the set of hybridized orbitals from Figure 5.21 that corresponds to this geometry.\n\n| Regions of Electron Density | Arrangement |  | Hybridization |  |\n| :---: | :---: | :---: | :---: | :---: |\n| 2 | ---------- | linear | $s p$ |  |\n| 3 |  | trigonal planar | $s p^{2}$ |  |\n| 4 |  | tetrahedral | $s p^{3}$ |  |\n| 5 |  | trigonal bipyramidal | $s p^{3} d$ |  |\n| 6 |  | octahedral | $s p^{3} d^{2}$ |  |\n\nFIGURE 5.21 The shapes of hybridized orbital sets are consistent with the electron-pair geometries. For example, an atom surrounded by three regions of electron density is $s p^{2}$ hybridized, and the three $s p^{2}$ orbitals are arranged in a trigonal planar fashion."}
{"id": 2809, "contents": "485. Assignment of Hybrid Orbitals to Central Atoms - \nIt is important to remember that hybridization was devised to rationalize experimentally observed molecular geometries. The model works well for molecules containing small central atoms, in which the valence electron pairs are close together in space. However, for larger central atoms, the valence-shell electron pairs are farther from the nucleus, and there are fewer repulsions. Their compounds exhibit structures that are often not consistent with VSEPR theory, and hybridized orbitals are not necessary to explain the observed data. For example, we have discussed the $\\mathrm{H}-\\mathrm{O}-\\mathrm{H}$ bond angle in $\\mathrm{H}_{2} \\mathrm{O}, 104.5^{\\circ}$, which is more consistent with $s p^{3}$ hybrid orbitals $\\left(109.5^{\\circ}\\right)$ on the central atom than with $2 p$ orbitals $\\left(90^{\\circ}\\right)$. Sulfur is in the same group as oxygen, and $\\mathrm{H}_{2} \\mathrm{~S}$ has a similar Lewis structure. However, it has a much smaller bond angle $\\left(92.1^{\\circ}\\right)$, which indicates much less hybridization on sulfur than oxygen. Continuing down the group, tellurium is even larger than sulfur, and for $\\mathrm{H}_{2} \\mathrm{Te}$, the observed bond angle $\\left(90^{\\circ}\\right)$ is consistent with overlap of the $5 p$ orbitals, without invoking hybridization. We invoke hybridization where it is necessary to explain the observed structures."}
{"id": 2810, "contents": "487. Assigning Hybridization - \nAmmonium sulfate is important as a fertilizer. What is the hybridization of the sulfur atom in the sulfate ion, $\\mathrm{SO}_{4}{ }^{2-}$ ?"}
{"id": 2811, "contents": "488. Solution - \nThe Lewis structure of sulfate shows there are four regions of electron density. The hybridization is $s p^{3}$."}
{"id": 2812, "contents": "489. Check Your Learning - \nWhat is the hybridization of the selenium atom in $\\mathrm{SeF}_{4}$ ?"}
{"id": 2813, "contents": "490. Answer: - \nThe selenium atom is $s p^{3} d$ hybridized."}
{"id": 2814, "contents": "492. Assigning Hybridization - \nUrea, $\\mathrm{NH}_{2} \\mathrm{C}(\\mathrm{O}) \\mathrm{NH}_{2}$, is sometimes used as a source of nitrogen in fertilizers. What is the hybridization of the carbon atom in urea?"}
{"id": 2815, "contents": "493. Solution - \nThe Lewis structure of urea is\n\n\nThe carbon atom is surrounded by three regions of electron density, positioned in a trigonal planar arrangement. The hybridization in a trigonal planar electron pair geometry is $s p^{2}$ (Figure 5.21), which is the hybridization of the carbon atom in urea."}
{"id": 2816, "contents": "494. Check Your Learning - \nAcetic acid, $\\mathrm{H}_{3} \\mathrm{CC}(\\mathrm{O}) \\mathrm{OH}$, is the molecule that gives vinegar its odor and sour taste. What is the hybridization of the two carbon atoms in acetic acid?"}
{"id": 2817, "contents": "495. Answer: - \n$\\mathrm{H}_{3} \\mathrm{C}, s p^{3} ; \\underline{\\mathrm{C}}(\\mathrm{O}) \\mathrm{OH}, s p^{2}$"}
{"id": 2818, "contents": "496. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe multiple covalent bonding in terms of atomic orbital overlap\n- Relate the concept of resonance to $\\pi$-bonding and electron delocalization\n\nThe hybrid orbital model appears to account well for the geometry of molecules involving single covalent bonds. Is it also capable of describing molecules containing double and triple bonds? We have already discussed that multiple bonds consist of $\\sigma$ and $\\pi$ bonds. Next we can consider how we visualize these components and how they relate to hybrid orbitals. The Lewis structure of ethene, $\\mathrm{C}_{2} \\mathrm{H}_{4}$, shows us that each carbon atom is surrounded by one other carbon atom and two hydrogen atoms.\n\n\nThe three bonding regions form a trigonal planar electron-pair geometry. Thus we expect the $\\sigma$ bonds from each carbon atom are formed using a set of $s p^{2}$ hybrid orbitals that result from hybridization of two of the $2 p$ orbitals and the $2 s$ orbital (Figure 5.22). These orbitals form the $\\mathrm{C}-\\mathrm{H}$ single bonds and the $\\sigma$ bond in the $\\mathrm{C}=\\mathrm{C}$ double bond (Figure 5.23). The $\\pi$ bond in the $\\mathbf{C}=\\mathbf{C}$ double bond results from the overlap of the third (remaining) $2 p$ orbital on each carbon atom that is not involved in hybridization. This unhybridized $p$ orbital (lobes shown in red and blue in Figure 5.23) is perpendicular to the plane of the $s p^{2}$ hybrid orbitals. Thus the unhybridized $2 p$ orbitals overlap in a side-by-side fashion, above and below the internuclear axis (Figure 5.23) and form a $\\pi$ bond.\n\n\nFIGURE 5.22 In ethene, each carbon atom is $s p^{2}$ hybridized, and the $s p^{2}$ orbitals and the $p$ orbital are singly occupied. The hybrid orbitals overlap to form $\\sigma$ bonds, while the $p$ orbitals on each carbon atom overlap to form $a \\pi$ bond.\n\n(a)\n\n(b)"}
{"id": 2819, "contents": "496. LEARNING OBJECTIVES - \n(a)\n\n(b)\n\nFIGURE 5.23 In the ethene molecule, $\\mathrm{C}_{2} \\mathrm{H}_{4}$, there are (a) five $\\sigma$ bonds. One $\\mathrm{C}-\\mathrm{C} \\sigma$ bond results from overlap of $s p^{2}$ hybrid orbitals on the carbon atom with one $s p^{2}$ hybrid orbital on the other carbon atom. Four $\\mathrm{C}-\\mathrm{H}$ bonds result from the overlap between the C atoms' $s p^{2}$ orbitals with $s$ orbitals on the hydrogen atoms. (b) The $\\pi$ bond is formed by the side-by-side overlap of the two unhybridized $p$ orbitals in the two carbon atoms. The two lobes of the $\\pi$ bond are above and below the plane of the $\\sigma$ system.\n\nIn an ethene molecule, the four hydrogen atoms and the two carbon atoms are all in the same plane. If the two planes of $s p^{2}$ hybrid orbitals tilted relative to each other, the $p$ orbitals would not be oriented to overlap efficiently to create the $\\pi$ bond. The planar configuration for the ethene molecule occurs because it is the most stable bonding arrangement. This is a significant difference between $\\sigma$ and $\\pi$ bonds; rotation around single ( $\\sigma$ ) bonds occurs easily because the end-to-end orbital overlap does not depend on the relative orientation of the orbitals on each atom in the bond. In other words, rotation around the internuclear axis does not change the extent to which the $\\sigma$ bonding orbitals overlap because the bonding electron density is symmetric about the axis. Rotation about the internuclear axis is much more difficult for multiple bonds; however, this would drastically alter the off-axis overlap of the $\\pi$ bonding orbitals, essentially breaking the $\\pi$ bond."}
{"id": 2820, "contents": "496. LEARNING OBJECTIVES - \nIn molecules with $s p$ hybrid orbitals, two unhybridized $p$ orbitals remain on the atom (Figure 5.24). We find this situation in acetylene, $\\mathrm{H}-\\mathrm{C} \\equiv \\mathrm{C}-\\mathrm{H}$, which is a linear molecule. The $s p$ hybrid orbitals of the two carbon atoms overlap end to end to form a $\\sigma$ bond between the carbon atoms (Figure 5.25 ). The remaining $s p$ orbitals form $\\sigma$ bonds with hydrogen atoms. The two unhybridized $p$ orbitals per carbon are positioned such that they overlap side by side and, hence, form two $\\pi$ bonds. The two carbon atoms of acetylene are thus bound together by one $\\sigma$ bond and two $\\pi$ bonds, giving a triple bond.\n\n\nFIGURE 5.24 Diagram of the two linear sp hybrid orbitals of a carbon atom, which lie in a straight line, and the two unhybridized $p$ orbitals at perpendicular angles.\n\n(a)\n\n(b)\n\nFIGURE 5.25 (a) In the acetylene molecule, $\\mathrm{C}_{2} \\mathrm{H}_{2}$, there are two $\\mathrm{C}-\\mathrm{H} \\sigma$ bonds and $\\mathrm{a} \\mathrm{C} \\equiv \\mathrm{C}$ triple bond involving one $\\mathrm{C}-\\mathrm{C} \\sigma$ bond and two $\\mathrm{C}-\\mathrm{C} \\pi$ bonds. The dashed lines, each connecting two lobes, indicate the side-by-side overlap of the four unhybridized $p$ orbitals. (b) This shows the overall outline of the bonds in $\\mathrm{C}_{2} \\mathrm{H}_{2}$. The two lobes of each of the $\\pi$ bonds are positioned across from each other around the line of the $\\mathrm{C}-\\mathrm{C} \\sigma$ bond."}
{"id": 2821, "contents": "496. LEARNING OBJECTIVES - \nHybridization involves only $\\sigma$ bonds, lone pairs of electrons, and single unpaired electrons (radicals). Structures that account for these features describe the correct hybridization of the atoms. However, many structures also include resonance forms. Remember that resonance forms occur when various arrangements of $\\pi$ bonds are possible. Since the arrangement of $\\pi$ bonds involves only the unhybridized orbitals, resonance does not influence the assignment of hybridization.\n\nFor example, molecule benzene has two resonance forms (Figure 5.26). We can use either of these forms to determine that each of the carbon atoms is bonded to three other atoms with no lone pairs, so the correct hybridization is $s p^{2}$. The electrons in the unhybridized $p$ orbitals form $\\pi$ bonds. Neither resonance structure completely describes the electrons in the $\\pi$ bonds. They are not located in one position or the other, but in reality are delocalized throughout the ring. Valence bond theory does not easily address delocalization. Bonding in molecules with resonance forms is better described by molecular orbital theory. (See the next module.)\n\n\nFIGURE 5.26 Each carbon atom in benzene, $\\mathrm{C}_{6} \\mathrm{H}_{6}$, is $s p^{2}$ hybridized, independently of which resonance form is considered. The electrons in the $\\pi$ bonds are not located in one set of $p$ orbitals or the other, but rather delocalized throughout the molecule."}
{"id": 2822, "contents": "498. Assignment of Hybridization Involving Resonance - \nSome acid rain results from the reaction of sulfur dioxide with atmospheric water vapor, followed by the formation of sulfuric acid. Sulfur dioxide, $\\mathrm{SO}_{2}$, is a major component of volcanic gases as well as a product of the combustion of sulfur-containing coal. What is the hybridization of the S atom in $\\mathrm{SO}_{2}$ ?"}
{"id": 2823, "contents": "499. Solution - \nThe resonance structures of $\\mathrm{SO}_{2}$ are\n\n\nThe sulfur atom is surrounded by two bonds and one lone pair of electrons in either resonance structure. Therefore, the electron-pair geometry is trigonal planar, and the hybridization of the sulfur atom is $s p^{2}$."}
{"id": 2824, "contents": "500. Check Your Learning - \nAnother acid in acid rain is nitric acid, $\\mathrm{HNO}_{3}$, which is produced by the reaction of nitrogen dioxide, $\\mathrm{NO}_{2}$, with atmospheric water vapor. What is the hybridization of the nitrogen atom in $\\mathrm{NO}_{2}$ ? (Note: the lone electron on nitrogen occupies a hybridized orbital just as a lone pair would.)"}
{"id": 2825, "contents": "501. Answer: - \n$s p^{2}$"}
{"id": 2826, "contents": "502. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Outline the basic quantum-mechanical approach to deriving molecular orbitals from atomic orbitals\n- Describe traits of bonding and antibonding molecular orbitals\n- Calculate bond orders based on molecular electron configurations\n- Write molecular electron configurations for first- and second-row diatomic molecules\n- Relate these electron configurations to the molecules' stabilities and magnetic properties\n\nFor almost every covalent molecule that exists, we can now draw the Lewis structure, predict the electron-pair geometry, predict the molecular geometry, and come close to predicting bond angles. However, one of the most important molecules we know, the oxygen molecule $\\mathrm{O}_{2}$, presents a problem with respect to its Lewis structure. We would write the following Lewis structure for $\\mathrm{O}_{2}$ :\n\n\nThis electronic structure adheres to all the rules governing Lewis theory. There is an $\\mathrm{O}=\\mathrm{O}$ double bond, and each oxygen atom has eight electrons around it. However, this picture is at odds with the magnetic behavior of oxygen. By itself, $\\mathrm{O}_{2}$ is not magnetic, but it is attracted to magnetic fields. Thus, when we pour liquid oxygen past a strong magnet, it collects between the poles of the magnet and defies gravity, as in Figure 5.1. Such attraction to a magnetic field is called paramagnetism, and it arises in molecules that have unpaired electrons. And yet, the Lewis structure of $\\mathrm{O}_{2}$ indicates that all electrons are paired. How do we account for this discrepancy?\n\nMagnetic susceptibility measures the force experienced by a substance in a magnetic field. When we compare the weight of a sample to the weight measured in a magnetic field (Figure 5.27), paramagnetic samples that are attracted to the magnet will appear heavier because of the force exerted by the magnetic field. We can calculate the number of unpaired electrons based on the increase in weight.\n\n\nFIGURE 5.27 A Gouy balance compares the mass of a sample in the presence of a magnetic field with the mass with the electromagnet turned off to determine the number of unpaired electrons in a sample."}
{"id": 2827, "contents": "502. LEARNING OBJECTIVES - \nFIGURE 5.27 A Gouy balance compares the mass of a sample in the presence of a magnetic field with the mass with the electromagnet turned off to determine the number of unpaired electrons in a sample.\n\nExperiments show that each $\\mathrm{O}_{2}$ molecule has two unpaired electrons. The Lewis-structure model does not predict the presence of these two unpaired electrons. Unlike oxygen, the apparent weight of most molecules decreases slightly in the presence of an inhomogeneous magnetic field. Materials in which all of the electrons are paired are diamagnetic and weakly repel a magnetic field. Paramagnetic and diamagnetic materials do not act as permanent magnets. Only in the presence of an applied magnetic field do they demonstrate attraction or repulsion."}
{"id": 2828, "contents": "503. LINK TO LEARNING - \nWater, like most molecules, contains all paired electrons. Living things contain a large percentage of water, so they demonstrate diamagnetic behavior. If you place a frog near a sufficiently large magnet, it will levitate. You can see videos (http://openstax.org/l/16diamagnetic) of diamagnetic floating frogs, strawberries, and more.\n\nMolecular orbital theory (MO theory) provides an explanation of chemical bonding that accounts for the paramagnetism of the oxygen molecule. It also explains the bonding in a number of other molecules, such as violations of the octet rule and more molecules with more complicated bonding (beyond the scope of this text) that are difficult to describe with Lewis structures. Additionally, it provides a model for describing the energies of electrons in a molecule and the probable location of these electrons. Unlike valence bond theory, which uses hybrid orbitals that are assigned to one specific atom, MO theory uses the combination of atomic orbitals to yield molecular orbitals that are delocalized over the entire molecule rather than being localized on its constituent atoms. MO theory also helps us understand why some substances are electrical conductors, others are semiconductors, and still others are insulators. Table 5.1 summarizes the main points of the two complementary bonding theories. Both theories provide different, useful ways of describing molecular structure.\n\n| Valence Bond Theory | Molecular Orbital Theory |\n| :--- | :--- |\n| considers bonds as localized between one pair of <br> atoms | considers electrons delocalized throughout the <br> entire molecule |\n| creates bonds from overlap of atomic orbitals ( $s, p, d . .)$. <br> and hybrid orbitals $\\left(s p, s p^{2}, s p^{3} \\ldots\\right)$ | combines atomic orbitals to form molecular <br> orbitals $\\left(\\sigma, \\sigma^{*}, \\pi, \\pi^{*}\\right)$ |\n| forms $\\sigma$ or $\\pi$ bonds | creates bonding and antibonding interactions <br> based on which orbitals are filled |\n| predicts molecular shape based on the number of <br> regions of electron density | predicts the arrangement of electrons in <br> molecules |\n| needs multiple structures to describe resonance |  |\n\nTABLE 5.1"}
{"id": 2829, "contents": "503. LINK TO LEARNING - \nTABLE 5.1\n\nMolecular orbital theory describes the distribution of electrons in molecules in much the same way that the distribution of electrons in atoms is described using atomic orbitals. Using quantum mechanics, the behavior of an electron in a molecule is still described by a wave function, $\\Psi$, analogous to the behavior in an atom. Just like electrons around isolated atoms, electrons around atoms in molecules are limited to discrete (quantized) energies. The region of space in which a valence electron in a molecule is likely to be found is called a molecular orbital ( $\\Psi^{2}$ ). Like an atomic orbital, a molecular orbital is full when it contains two electrons with opposite spin.\n\nWe will consider the molecular orbitals in molecules composed of two identical atoms ( $\\mathrm{H}_{2}$ or $\\mathrm{Cl}_{2}$, for example). Such molecules are called homonuclear diatomic molecules. In these diatomic molecules, several types of molecular orbitals occur.\n\nThe mathematical process of combining atomic orbitals to generate molecular orbitals is called the linear combination of atomic orbitals (LCAO). The wave function describes the wavelike properties of an electron. Molecular orbitals are combinations of atomic orbital wave functions. Combining waves can lead to constructive interference, in which peaks line up with peaks, or destructive interference, in which peaks line up with troughs (Figure 5.28). In orbitals, the waves are three dimensional, and they combine with in-phase waves producing regions with a higher probability of electron density and out-of-phase waves producing nodes, or regions of no electron density.\n\n\nFIGURE 5.28 (a) When in-phase waves combine, constructive interference produces a wave with greater amplitude. (b) When out-of-phase waves combine, destructive interference produces a wave with less (or no) amplitude."}
{"id": 2830, "contents": "503. LINK TO LEARNING - \nFIGURE 5.28 (a) When in-phase waves combine, constructive interference produces a wave with greater amplitude. (b) When out-of-phase waves combine, destructive interference produces a wave with less (or no) amplitude.\n\nThere are two types of molecular orbitals that can form from the overlap of two atomic $s$ orbitals on adjacent atoms. The two types are illustrated in Figure 5.29. The in-phase combination produces a lower energy $\\boldsymbol{\\sigma}_{\\boldsymbol{s}}$\nmolecular orbital (read as \"sigma-s\") in which most of the electron density is directly between the nuclei. The out-of-phase addition (which can also be thought of as subtracting the wave functions) produces a higher energy $\\sigma_{s}^{*}$ molecular orbital (read as \"sigma-s-star\") molecular orbital in which there is a node between the nuclei. The asterisk signifies that the orbital is an antibonding orbital. Electrons in a $\\sigma_{S}$ orbital are attracted by both nuclei at the same time and are more stable (of lower energy) than they would be in the isolated atoms. Adding electrons to these orbitals creates a force that holds the two nuclei together, so we call these orbitals bonding orbitals. Electrons in the $\\sigma_{s}^{*}$ orbitals are located well away from the region between the two nuclei. The attractive force between the nuclei and these electrons pulls the two nuclei apart. Hence, these orbitals are called antibonding orbitals. Electrons fill the lower-energy bonding orbital before the higher-energy antibonding orbital, just as they fill lower-energy atomic orbitals before they fill higher-energy atomic orbitals.\n\n\nFIGURE 5.29 Sigma ( $\\sigma$ ) and sigma-star ( $\\sigma^{*}$ ) molecular orbitals are formed by the combination of two $s$ atomic orbitals. The dots $(\\cdot)$ indicate the locations of nuclei."}
{"id": 2831, "contents": "504. LINK TO LEARNING - \nYou can watch animations (http://openstax.org/l/16molecorbital) visualizing the calculated atomic orbitals combining to form various molecular orbitals at the Orbitron website.\n\nIn $p$ orbitals, the wave function gives rise to two lobes with opposite phases, analogous to how a twodimensional wave has both parts above and below the average. We indicate the phases by shading the orbital lobes different colors. When orbital lobes of the same phase overlap, constructive wave interference increases the electron density. When regions of opposite phase overlap, the destructive wave interference decreases electron density and creates nodes. When $p$ orbitals overlap end to end, they create $\\sigma$ and $\\sigma^{*}$ orbitals (Figure 5.30). If two atoms are located along the $x$-axis in a Cartesian coordinate system, the two $p_{x}$ orbitals overlap end to end and form $\\sigma_{p x}$ (bonding) and $\\sigma_{p x}^{*}$ (antibonding) (read as \"sigma-p-x\" and \"sigma-p-x star,\" respectively). Just as with $s$-orbital overlap, the asterisk indicates the orbital with a node between the nuclei, which is a higher-energy, antibonding orbital."}
{"id": 2832, "contents": "504. LINK TO LEARNING - \nFIGURE 5.30 Combining wave functions of two $p$ atomic orbitals along the internuclear axis creates two molecular orbitals, $\\sigma_{p}$ and $\\sigma_{p}^{*}$.\nThe side-by-side overlap of two $p$ orbitals gives rise to a pi ( $\\boldsymbol{\\pi}$ ) bonding molecular orbital and a $\\boldsymbol{\\pi}^{*}$ *\nantibonding molecular orbital, as shown in Figure 5.31. In valence bond theory, we describe $\\pi$ bonds as containing a nodal plane containing the internuclear axis and perpendicular to the lobes of the $p$ orbitals, with electron density on either side of the node. In molecular orbital theory, we describe the $\\pi$ orbital by this same\nshape, and a $\\pi$ bond exists when this orbital contains electrons. Electrons in this orbital interact with both nuclei and help hold the two atoms together, making it a bonding orbital. For the out-of-phase combination, there are two nodal planes created, one along the internuclear axis and a perpendicular one between the nuclei.\n\n\nFIGURE 5.31 Side-by-side overlap of each two $p$ orbitals results in the formation of two $\\pi$ molecular orbitals. Combining the out-of-phase orbitals results in an antibonding molecular orbital with two nodes. One contains the internuclear axis, and one is perpendicular to the axis. Combining the in-phase orbitals results in a bonding orbital. There is a node (blue) containing the internuclear axis with the two lobes of the orbital located above and below this node."}
{"id": 2833, "contents": "504. LINK TO LEARNING - \nIn the molecular orbitals of diatomic molecules, each atom also has two sets of $p$ orbitals oriented side by side ( $p_{y}$ and $p_{z}$ ), so these four atomic orbitals combine pairwise to create two $\\pi$ orbitals and two $\\pi^{*}$ orbitals. The $\\pi_{p y}$ and $\\pi_{p y}^{*}$ orbitals are oriented at right angles to the $\\pi_{p z}$ and $\\pi_{p z}^{*}$ orbitals. Except for their orientation, the $\\pi_{p y}$ and $\\pi_{p z}$ orbitals are identical and have the same energy; they are degenerate orbitals. The $\\pi_{p y}^{*}$ and $\\pi_{p z}^{*}$ antibonding orbitals are also degenerate and identical except for their orientation. A total of six molecular orbitals results from the combination of the six atomic $p$ orbitals in two atoms: $\\sigma_{p x}$ and $\\sigma_{p x}^{*}, \\pi_{p y}$ and $\\pi_{p y}^{*}, \\pi_{p z}$ and $\\pi_{p z}^{*}$."}
{"id": 2834, "contents": "506. Molecular Orbitals - \nPredict what type (if any) of molecular orbital would result from adding the wave functions so each pair of orbitals shown overlap. The orbitals are all similar in energy."}
{"id": 2835, "contents": "507. Solution - \n(a) is an in-phase combination, resulting in a $\\sigma_{3 p}$ orbital\n(b) will not result in a new orbital because the in-phase component (bottom) and out-of-phase component (top) cancel out. Only orbitals with the correct alignment can combine.\n(c) is an out-of-phase combination, resulting in a $\\pi_{3 p}^{*}$ orbital."}
{"id": 2836, "contents": "508. Check Your Learning - \nLabel the molecular orbital shown as $\\sigma$ or $\\pi$, bonding or antibonding and indicate where the node occurs."}
{"id": 2837, "contents": "509. Answer: - \nThe orbital is located along the internuclear axis, so it is a $\\sigma$ orbital. There is a node bisecting the internuclear axis, so it is an antibonding orbital."}
{"id": 2838, "contents": "511. Walter Kohn: Nobel Laureate - \nWalter Kohn (Figure 5.32) is a theoretical physicist who studies the electronic structure of solids. His work combines the principles of quantum mechanics with advanced mathematical techniques. This technique, called density functional theory, makes it possible to compute properties of molecular orbitals, including their shape and energies. Kohn and mathematician John Pople were awarded the Nobel Prize in Chemistry in 1998 for their contributions to our understanding of electronic structure. Kohn also made significant contributions to the physics of semiconductors.\n\n\nFIGURE 5.32 Walter Kohn developed methods to describe molecular orbitals. (credit: image courtesy of Walter Kohn)\n\nKohn's biography has been remarkable outside the realm of physical chemistry as well. He was born in Austria, and during World War II he was part of the Kindertransport program that rescued 10,000 children from the Nazi regime. His summer jobs included discovering gold deposits in Canada and helping Polaroid explain how its instant film worked. Dr. Kohn passed away in 2016 at the age of 93."}
{"id": 2839, "contents": "513. Computational Chemistry in Drug Design - \nWhile the descriptions of bonding described in this chapter involve many theoretical concepts, they also have many practical, real-world applications. For example, drug design is an important field that uses our understanding of chemical bonding to develop pharmaceuticals. This interdisciplinary area of study uses biology (understanding diseases and how they operate) to identify specific targets, such as a binding site that is involved in a disease pathway. By modeling the structures of the binding site and potential drugs, computational chemists can predict which structures can fit together and how effectively they will bind (see Figure 5.33). Thousands of potential candidates can be narrowed down to a few of the most promising candidates. These candidate molecules are then carefully tested to determine side effects, how effectively they can be transported through the body, and other factors. Dozens of important new pharmaceuticals have been discovered with the aid of computational chemistry, and new research projects are underway.\n\n\nFIGURE 5.33 The molecule shown, HIV-1 protease, is an important target for pharmaceutical research. By designing molecules that bind to this protein, scientists are able to drastically inhibit the progress of the disease."}
{"id": 2840, "contents": "514. Molecular Orbital Energy Diagrams - \nThe relative energy levels of atomic and molecular orbitals are typically shown in a molecular orbital diagram (Figure 5.34). For a diatomic molecule, the atomic orbitals of one atom are shown on the left, and those of the other atom are shown on the right. Each horizontal line represents one orbital that can hold two electrons. The molecular orbitals formed by the combination of the atomic orbitals are shown in the center. Dashed lines show which of the atomic orbitals combine to form the molecular orbitals. For each pair of atomic orbitals that combine, one lower-energy (bonding) molecular orbital and one higher-energy (antibonding) orbital result. Thus we can see that combining the six $2 p$ atomic orbitals results in three bonding orbitals (one $\\sigma$ and two $\\pi$ ) and three antibonding orbitals (one $\\sigma^{*}$ and two $\\pi^{*}$ ).\n\nWe predict the distribution of electrons in these molecular orbitals by filling the orbitals in the same way that we fill atomic orbitals, by the Aufbau principle. Lower-energy orbitals fill first, electrons spread out among degenerate orbitals before pairing, and each orbital can hold a maximum of two electrons with opposite spins (Figure 5.34). Just as we write electron configurations for atoms, we can write the molecular electronic configuration by listing the orbitals with superscripts indicating the number of electrons present. For clarity, we place parentheses around molecular orbitals with the same energy. In this case, each orbital is at a different energy, so parentheses separate each orbital. Thus we would expect a diatomic molecule or ion containing seven electrons (such as $\\mathrm{Be}_{2}{ }^{+}$) would have the molecular electron configuration $\\left(\\sigma_{1 s}\\right)^{2}\\left(\\sigma_{1 s}^{*}\\right)^{2}\\left(\\sigma_{2 s}\\right)^{2}\\left(\\sigma_{2 s}^{*}\\right)^{1}$. It is common to omit the core electrons from molecular orbital diagrams and configurations and include only the valence electrons."}
{"id": 2841, "contents": "514. Molecular Orbital Energy Diagrams - \nFIGURE 5.34 This is the molecular orbital diagram for the homonuclear diatomic $\\mathrm{Be}_{2}{ }^{+}$, showing the molecular orbitals of the valence shell only. The molecular orbitals are filled in the same manner as atomic orbitals, using the Aufbau principle and Hund's rule."}
{"id": 2842, "contents": "515. Bond Order - \nThe filled molecular orbital diagram shows the number of electrons in both bonding and antibonding molecular orbitals. The net contribution of the electrons to the bond strength of a molecule is identified by determining the bond order that results from the filling of the molecular orbitals by electrons.\n\nWhen using Lewis structures to describe the distribution of electrons in molecules, we define bond order as the number of bonding pairs of electrons between two atoms. Thus a single bond has a bond order of 1 , a double bond has a bond order of 2 , and a triple bond has a bond order of 3 . We define bond order differently when we use the molecular orbital description of the distribution of electrons, but the resulting bond order is usually the same. The MO technique is more accurate and can handle cases when the Lewis structure method fails, but both methods describe the same phenomenon.\n\nIn the molecular orbital model, an electron contributes to a bonding interaction if it occupies a bonding orbital and it contributes to an antibonding interaction if it occupies an antibonding orbital. The bond order is calculated by subtracting the destabilizing (antibonding) electrons from the stabilizing (bonding) electrons. Since a bond consists of two electrons, we divide by two to get the bond order. We can determine bond order with the following equation:\n\n$$\n\\text { bond order }=\\frac{\\text { (number of bonding electrons) }- \\text { (number of antibonding electrons) }}{2}\n$$\n\nThe order of a covalent bond is a guide to its strength; a bond between two given atoms becomes stronger as the bond order increases. If the distribution of electrons in the molecular orbitals between two atoms is such that the resulting bond would have a bond order of zero, a stable bond does not form. We next look at some specific examples of MO diagrams and bond orders."}
{"id": 2843, "contents": "516. Bonding in Diatomic Molecules - \nA dihydrogen molecule $\\left(\\mathrm{H}_{2}\\right)$ forms from two hydrogen atoms. When the atomic orbitals of the two atoms combine, the electrons occupy the molecular orbital of lowest energy, the $\\sigma_{1 s}$ bonding orbital. A dihydrogen molecule, $\\mathrm{H}_{2}$, readily forms because the energy of a $\\mathrm{H}_{2}$ molecule is lower than that of two H atoms. The $\\sigma_{1 s}$ orbital that contains both electrons is lower in energy than either of the two 1 s atomic orbitals.\n\nA molecular orbital can hold two electrons, so both electrons in the $\\mathrm{H}_{2}$ molecule are in the $\\sigma_{1 s}$ bonding orbital; the electron configuration is $\\left(\\sigma_{1 s}\\right)^{2}$. We represent this configuration by a molecular orbital energy diagram (Figure 5.35) in which a single upward arrow indicates one electron in an orbital, and two (upward and downward) arrows indicate two electrons of opposite spin.\n\n\nFIGURE 5.35 The molecular orbital energy diagram predicts that $\\mathrm{H}_{2}$ will be a stable molecule with lower energy than the separated atoms.\n\nA dihydrogen molecule contains two bonding electrons and no antibonding electrons so we have\n\n$$\n\\text { bond order in } \\mathrm{H}_{2}=\\frac{(2-0)}{2}=1\n$$"}
{"id": 2844, "contents": "516. Bonding in Diatomic Molecules - \nA dihydrogen molecule contains two bonding electrons and no antibonding electrons so we have\n\n$$\n\\text { bond order in } \\mathrm{H}_{2}=\\frac{(2-0)}{2}=1\n$$\n\nBecause the bond order for the $\\mathrm{H}-\\mathrm{H}$ bond is equal to 1 , the bond is a single bond.\nA helium atom has two electrons, both of which are in its $1 s$ orbital. Two helium atoms do not combine to form a dihelium molecule, $\\mathrm{He}_{2}$, with four electrons, because the stabilizing effect of the two electrons in the lowerenergy bonding orbital would be offset by the destabilizing effect of the two electrons in the higher-energy antibonding molecular orbital. We would write the hypothetical electron configuration of $\\mathrm{He}_{2}$ as $\\left(\\sigma_{1 s}\\right)^{2}\\left(\\sigma_{1 s}^{*}\\right)^{2}$ as in Figure 5.36. The net energy change would be zero, so there is no driving force for helium atoms to form the diatomic molecule. In fact, helium exists as discrete atoms rather than as diatomic molecules. The bond order in a hypothetical dihelium molecule would be zero.\n\n$$\n\\text { bond order in } \\mathrm{He}_{2}=\\frac{(2-2)}{2}=0\n$$\n\nA bond order of zero indicates that no bond is formed between two atoms.\n\n\nFIGURE 5.36 The molecular orbital energy diagram predicts that $\\mathrm{He}_{2}$ will not be a stable molecule, since it has equal numbers of bonding and antibonding electrons.\n\nThe Diatomic Molecules of the Second Period\nEight possible homonuclear diatomic molecules might be formed by the atoms of the second period of the periodic table: $\\mathrm{Li}_{2}, \\mathrm{Be}_{2}, \\mathrm{~B}_{2}, \\mathrm{C}_{2}, \\mathrm{~N}_{2}, \\mathrm{O}_{2}, \\mathrm{~F}_{2}$, and $\\mathrm{Ne}_{2}$. However, we can predict that the $\\mathrm{Be}_{2}$ molecule and the $\\mathrm{Ne}_{2}$ molecule would not be stable. We can see this by a consideration of the molecular electron configurations (Table 5.2)."}
{"id": 2845, "contents": "516. Bonding in Diatomic Molecules - \nWe predict valence molecular orbital electron configurations just as we predict electron configurations of atoms. Valence electrons are assigned to valence molecular orbitals with the lowest possible energies.\n\nConsistent with Hund's rule, whenever there are two or more degenerate molecular orbitals, electrons fill each orbital of that type singly before any pairing of electrons takes place.\n\nAs we saw in valence bond theory, $\\sigma$ bonds are generally more stable than $\\pi$ bonds formed from degenerate atomic orbitals. Similarly, in molecular orbital theory, $\\sigma$ orbitals are usually more stable than $\\pi$ orbitals. However, this is not always the case. The MOs for the valence orbitals of the second period are shown in Figure 5.37. Looking at $\\mathrm{Ne}_{2}$ molecular orbitals, we see that the order is consistent with the generic diagram shown in the previous section. However, for atoms with three or fewer electrons in the $p$ orbitals (Li through N) we observe a different pattern, in which the $\\sigma_{p}$ orbital is higher in energy than the $\\pi_{p}$ set. Obtain the molecular orbital diagram for a homonuclear diatomic ion by adding or subtracting electrons from the diagram for the neutral molecule.\n\n\nFIGURE 5.37 This shows the MO diagrams for each homonuclear diatomic molecule in the second period. The orbital energies decrease across the period as the effective nuclear charge increases and atomic radius decreases. Between $\\mathrm{N}_{2}$ and $\\mathrm{O}_{2}$, the order of the orbitals changes.\n\nThis switch in orbital ordering occurs because of a phenomenon called s-p mixing. s-p mixing does not create new orbitals; it merely influences the energies of the existing molecular orbitals. The $\\sigma_{\\mathrm{s}}$ wavefunction mathematically combines with the $\\sigma_{p}$ wavefunction, with the result that the $\\sigma_{s}$ orbital becomes more stable, and the $\\sigma_{p}$ orbital becomes less stable (Figure 5.38). Similarly, the antibonding orbitals also undergo s-p mixing, with the $\\sigma_{\\mathrm{s}^{*}}$ becoming more stable and the $\\sigma_{\\mathrm{p}^{*}}$ becoming less stable."}
{"id": 2846, "contents": "516. Bonding in Diatomic Molecules - \nFIGURE 5.38 Without mixing, the MO pattern occurs as expected, with the $\\sigma_{p}$ orbital lower in energy than the $\\pi_{p}$ orbitals. When s-p mixing occurs, the orbitals shift as shown, with the $\\sigma_{p}$ orbital higher in energy than the $\\pi_{p}$ orbitals. $s$-p mixing occurs when the $s$ and $p$ orbitals have similar energies. The energy difference between $2 s$ and $2 p$ orbitals in $\\mathrm{O}, \\mathrm{F}$, and Ne is greater than that in Li, Be, B, C, and N. Because of this, $\\mathrm{O}_{2}, \\mathrm{~F}_{2}$, and $\\mathrm{Ne}_{2}$ exhibit negligible s-p mixing (not sufficient to change the energy ordering), and their MO diagrams follow the normal pattern, as shown in Figure 5.37. All of the other period 2 diatomic molecules do have s-p mixing, which leads to the pattern where the $\\sigma_{p}$ orbital is raised above the $\\pi_{p}$ set.\n\nUsing the MO diagrams shown in Figure 5.37, we can add in the electrons and determine the molecular electron configuration and bond order for each of the diatomic molecules. As shown in Table 5.2, $\\mathrm{Be}_{2}$ and $\\mathrm{Ne}_{2}$ molecules would have a bond order of 0 , and these molecules do not exist.\n\nElectron Configuration and Bond Order for Molecular Orbitals in Homonuclear Diatomic Molecules of Period Two Elements"}
{"id": 2847, "contents": "516. Bonding in Diatomic Molecules - \nElectron Configuration and Bond Order for Molecular Orbitals in Homonuclear Diatomic Molecules of Period Two Elements\n\n| Molecule | Electron Configuration | Bond Order |\n| :--- | :--- | :--- |\n| $\\mathrm{Li}_{2}$ | $\\left(\\sigma_{2 s}\\right)^{2}$ | 1 |\n| $\\mathrm{Be}_{2}$ (unstable) | $\\left(\\sigma_{2 s}\\right)^{2}\\left(\\sigma_{2 s}^{*}\\right)^{2}$ | 0 |\n| $\\mathrm{~B}_{2}$ | $\\left(\\sigma_{2 s}\\right)^{2}\\left(\\sigma_{2 s}^{*}\\right)^{2}\\left(\\pi_{2 p y}, \\pi_{2 p z}\\right)^{2}$ | 1 |\n| $\\mathrm{C}_{2}$ | $\\left(\\sigma_{2 s}\\right)^{2}\\left(\\sigma_{2 s}^{*}\\right)^{2}\\left(\\pi_{2 p y}, \\pi_{2 p z}\\right)^{4}$ | 2 |\n\nTABLE 5.2"}
{"id": 2848, "contents": "516. Bonding in Diatomic Molecules - \nTABLE 5.2\n\n| Molecule | Electron Configuration | Bond Order |\n| :--- | :--- | :--- |\n| $\\mathrm{N}_{2}$ | $\\left(\\sigma_{2 s}\\right)^{2}\\left(\\sigma_{2 s}^{*}\\right)^{2}\\left(\\pi_{2 p y}, \\pi_{2 p z}\\right)^{4}\\left(\\sigma_{2 p x}\\right)^{2}$ | 3 |\n| $\\mathrm{O}_{2}$ | $\\left(\\sigma_{2 s}\\right)^{2}\\left(\\sigma_{2 s}^{*}\\right)^{2}\\left(\\sigma_{2 p x}\\right)^{2}\\left(\\pi_{2 p y}, \\pi_{2 p z}\\right)^{4}\\left(\\pi_{2 p y}^{*}, \\pi_{2 p z}^{*}\\right)^{2}$ | 2 |\n| $\\mathrm{~F}_{2}$ | $\\left(\\sigma_{2 s}\\right)^{2}\\left(\\sigma_{2 s}^{*}\\right)^{2}\\left(\\sigma_{2 p x}\\right)^{2}\\left(\\pi_{2 p y}, \\pi_{2 p z}\\right)^{4}\\left(\\pi_{2 p y}^{*}, \\pi_{2 p z}^{*}\\right)^{4}$ | 1 |\n| $\\mathrm{Ne}_{2}$ (unstable) | $\\left(\\sigma_{2 s}\\right)^{2}\\left(\\sigma_{2 s}^{*}\\right)^{2}\\left(\\sigma_{2 p x}\\right)^{2}\\left(\\pi_{2 p y}, \\pi_{2 p z}\\right)^{4}\\left(\\pi_{2 p y}^{*}, \\pi_{2 p z}^{*}\\right)^{4}\\left(\\sigma_{2 p x}^{*}\\right)^{2}$ | 0 |\n\nTABLE 5.2"}
{"id": 2849, "contents": "516. Bonding in Diatomic Molecules - \nTABLE 5.2\n\nThe combination of two lithium atoms to form a lithium molecule, $\\mathrm{Li}_{2}$, is analogous to the formation of $\\mathrm{H}_{2}$, but the atomic orbitals involved are the valence $2 s$ orbitals. Each of the two lithium atoms has one valence electron. Hence, we have two valence electrons available for the $\\sigma_{2 s}$ bonding molecular orbital. Because both valence electrons would be in a bonding orbital, we would predict the $\\mathrm{Li}_{2}$ molecule to be stable. The molecule is, in fact, present in appreciable concentration in lithium vapor at temperatures near the boiling point of the element. All of the other molecules in Table 5.2 with a bond order greater than zero are also known.\n\nThe $\\mathrm{O}_{2}$ molecule has enough electrons to half fill the $\\left(\\pi_{2 p y}^{*}, \\pi_{2 p z}^{*}\\right)$ level. We expect the two electrons that occupy these two degenerate orbitals to be unpaired, and this molecular electronic configuration for $\\mathrm{O}_{2}$ is in accord with the fact that the oxygen molecule has two unpaired electrons (Figure 5.40). The presence of two unpaired electrons has proved to be difficult to explain using Lewis structures, but the molecular orbital theory explains it quite well. In fact, the unpaired electrons of the oxygen molecule provide a strong piece of support for the molecular orbital theory."}
{"id": 2850, "contents": "518. Band Theory - \nWhen two identical atomic orbitals on different atoms combine, two molecular orbitals result (see Figure 5.29). The bonding orbital is lower in energy than the original atomic orbitals because the atomic orbitals are inphase in the molecular orbital. The antibonding orbital is higher in energy than the original atomic orbitals because the atomic orbitals are out-of-phase.\n\nIn a solid, similar things happen, but on a much larger scale. Remember that even in a small sample there are a huge number of atoms (typically $>10^{23}$ atoms), and therefore a huge number of atomic orbitals that may be combined into molecular orbitals. When $N$ valence atomic orbitals, all of the same energy and each containing one (1) electron, are combined, $N / 2$ (filled) bonding orbitals and $N / 2$ (empty) antibonding orbitals will result. Each bonding orbital will show an energy lowering as the atomic orbitals are mostly in-phase, but each of the bonding orbitals will be a little different and have slightly different energies. The antibonding orbitals will show an increase in energy as the atomic orbitals are mostly out-of-phase, but each of the antibonding orbitals will also be a little different and have slightly different energies. The allowed energy levels for all the bonding orbitals are so close together that they form a band, called the valence band. Likewise, all the antibonding orbitals are very close together and form a band, called the conduction band. Figure 5.39 shows the bands for three important classes of materials: insulators, semiconductors, and conductors.\n\n\nFIGURE 5.39 Molecular orbitals in solids are so closely spaced that they are described as bands. The valence band is lower in energy and the conduction band is higher in energy. The type of solid is determined by the size of the \"band gap\" between the valence and conduction bands. Only a very small amount of energy is required to move electrons from the valance band to the conduction band in a conductor, and so they conduct electricity well. In an insulator, the band gap is large, so that very few electrons move, and they are poor conductors of electricity. Semiconductors are in between: they conduct electricity better than insulators, but not as well as conductors."}
{"id": 2851, "contents": "518. Band Theory - \nIn order to conduct electricity, electrons must move from the filled valence band to the empty conduction band where they can move throughout the solid. The size of the band gap, or the energy difference between the top of the valence band and the bottom of the conduction band, determines how easy it is to move electrons between the bands. Only a small amount of energy is required in a conductor because the band gap is very small. This small energy difference is \"easy\" to overcome, so they are good conductors of electricity. In an insulator, the band gap is so \"large\" that very few electrons move into the conduction band; as a result, insulators are poor conductors of electricity. Semiconductors conduct electricity when \"moderate\" amounts of energy are provided to move electrons out of the valence band and into the conduction band. Semiconductors, such as silicon, are found in many electronics.\n\nSemiconductors are used in devices such as computers, smartphones, and solar cells. Solar cells produce electricity when light provides the energy to move electrons out of the valence band. The electricity that is generated may then be used to power a light or tool, or it can be stored for later use by charging a battery. As of December 2014, up to $46 \\%$ of the energy in sunlight could be converted into electricity using solar cells."}
{"id": 2852, "contents": "519. EXAMPLE 5.7 - \nMolecular Orbital Diagrams, Bond Order, and Number of Unpaired Electrons\nDraw the molecular orbital diagram for the oxygen molecule, $\\mathrm{O}_{2}$. From this diagram, calculate the bond order for $\\mathrm{O}_{2}$. How does this diagram account for the paramagnetism of $\\mathrm{O}_{2}$ ?"}
{"id": 2853, "contents": "520. Solution - \nWe draw a molecular orbital energy diagram similar to that shown in Figure 5.37. Each oxygen atom contributes six electrons, so the diagram appears as shown in Figure 5.40.\n\n\nFIGURE 5.40 The molecular orbital energy diagram for $\\mathrm{O}_{2}$ predicts two unpaired electrons.\nWe calculate the bond order as\n\n$$\n\\mathrm{O}_{2}=\\frac{(8-4)}{2}=2\n$$\n\nOxygen's paramagnetism is explained by the presence of two unpaired electrons in the ( $\\Pi_{2 p y}, \\Pi_{2 p z}$ )* molecular orbitals."}
{"id": 2854, "contents": "521. Check Your Learning - \nThe main component of air is $N_{2}$. From the molecular orbital diagram of $N_{2}$, predict its bond order and whether it is diamagnetic or paramagnetic."}
{"id": 2855, "contents": "522. Answer: - \n$\\mathrm{N}_{2}$ has a bond order of 3 and is diamagnetic."}
{"id": 2856, "contents": "524. Ion Predictions with MO Diagrams - \nGive the molecular orbital configuration for the valence electrons in $\\mathrm{C}_{2}{ }^{2-}$. Will this ion be stable?"}
{"id": 2857, "contents": "525. Solution - \nLooking at the appropriate MO diagram, we see that the $\\pi$ orbitals are lower in energy than the $\\sigma_{p}$ orbital. The valence electron configuration for $\\mathrm{C}_{2}$ is $\\left(\\sigma_{2 s}\\right)^{2}\\left(\\sigma_{2 s}^{*}\\right)^{2}\\left(\\pi_{2 p y}, \\pi_{2 p z}\\right)^{4}$. Adding two more electrons to generate the $\\mathrm{C}_{2}{ }^{2-}$ anion will give a valence electron configuration of $\\left(\\sigma_{2 s}\\right)^{2}\\left(\\sigma_{2 s}^{*}\\right)^{2}\\left(\\pi_{2 p y}, \\pi_{2 p z}\\right)^{4}\\left(\\sigma_{2 p x}\\right)^{2}$. Since this has six more bonding electrons than antibonding, the bond order will be 3 , and the ion should be stable."}
{"id": 2858, "contents": "526. Check Your Learning - \nHow many unpaired electrons would be present on a $\\mathrm{Be}_{2}{ }^{2-}$ ion? Would it be paramagnetic or diamagnetic?"}
{"id": 2859, "contents": "527. Answer: - \ntwo, paramagnetic"}
{"id": 2860, "contents": "528. LINK TO LEARNING - \nCreating molecular orbital diagrams for molecules with more than two atoms relies on the same basic ideas as the diatomic examples presented here. However, with more atoms, computers are required to calculate how the atomic orbitals combine. See three-dimensional drawings (http://openstax.org/l/16orbitaldiag) of the molecular orbitals for $\\mathrm{C}_{6} \\mathrm{H}_{6}$."}
{"id": 2861, "contents": "529. Key Terms - \nantibonding orbital molecular orbital located outside of the region between two nuclei; electrons in an antibonding orbital destabilize the molecule\nbond order number of pairs of electrons between two atoms; it can be found by the number of bonds in a Lewis structure or by the difference between the number of bonding and antibonding electrons divided by two\nbonding orbital molecular orbital located between two nuclei; electrons in a bonding orbital stabilize a molecule\ndegenerate orbitals orbitals that have the same energy\ndiamagnetism phenomenon in which a material is not magnetic itself but is repelled by a magnetic field; it occurs when there are only paired electrons present\nhomonuclear diatomic molecule molecule consisting of two identical atoms\nhybrid orbital orbital created by combining atomic orbitals on a central atom\nhybridization model that describes the changes in the atomic orbitals of an atom when it forms a covalent compound\nlinear combination of atomic orbitals technique for combining atomic orbitals to create molecular orbitals\nmolecular orbital region of space in which an electron has a high probability of being found in a molecule\nmolecular orbital diagram visual representation of the relative energy levels of molecular orbitals\nmolecular orbital theory model that describes the behavior of electrons delocalized throughout a molecule in terms of the combination of atomic wave functions\nnode plane separating different lobes of orbitals, where the probability of finding an electron is zero\noverlap coexistence of orbitals from two different atoms sharing the same region of space, leading to the formation of a covalent bond\nparamagnetism phenomenon in which a material is not magnetic itself but is attracted to a magnetic field; it occurs when there are unpaired electrons present\npi bond ( $\\boldsymbol{\\pi}$ bond) covalent bond formed by side-byside overlap of atomic orbitals; the electron density is found on opposite sides of the internuclear axis\ns-p mixing change that causes $\\sigma_{p}$ orbitals to be less stable than $\\pi_{p}$ orbitals due to the mixing of $S$ and $p$-based molecular orbitals of similar energies.\nsigma bond ( $\\sigma$ bond) covalent bond formed by overlap of atomic orbitals along the internuclear axis\n$\\boldsymbol{s p}$ hybrid orbital one of a set of two orbitals with a linear arrangement that results from combining one $s$ and one $p$ orbital"}
{"id": 2862, "contents": "529. Key Terms - \n$\\boldsymbol{s p}$ hybrid orbital one of a set of two orbitals with a linear arrangement that results from combining one $s$ and one $p$ orbital\n$\\boldsymbol{s} \\boldsymbol{p}^{\\mathbf{2}}$ hybrid orbital one of a set of three orbitals with a trigonal planar arrangement that results from combining one $s$ and two $p$ orbitals\n$\\boldsymbol{s p} \\boldsymbol{p}^{\\mathbf{3}}$ hybrid orbital one of a set of four orbitals with a tetrahedral arrangement that results from combining one $s$ and three $p$ orbitals\n$\\boldsymbol{s p}^{\\mathbf{3}} \\boldsymbol{d}$ hybrid orbital one of a set of five orbitals with a trigonal bipyramidal arrangement that results from combining one $s$, three $p$, and one $d$ orbital\n$\\boldsymbol{s p}^{\\mathbf{3}} \\boldsymbol{d}^{\\mathbf{2}}$ hybrid orbital one of a set of six orbitals with an octahedral arrangement that results from combining one $s$, three $p$, and two $d$ orbitals\nvalence bond theory description of bonding that involves atomic orbitals overlapping to form $\\sigma$ or $\\pi$ bonds, within which pairs of electrons are shared\n$\\boldsymbol{\\pi}$ bonding orbital molecular orbital formed by side-by-side overlap of atomic orbitals, in which the electron density is found on opposite sides of the internuclear axis\n$\\boldsymbol{\\pi}$ * bonding orbital antibonding molecular orbital formed by out of phase side-by-side overlap of atomic orbitals, in which the electron density is found on both sides of the internuclear axis, and there is a node between the nuclei\n$\\boldsymbol{\\sigma}$ bonding orbital molecular orbital in which the electron density is found along the axis of the bond\n$\\boldsymbol{\\sigma}^{*}$ bonding orbital antibonding molecular orbital formed by out-of-phase overlap of atomic orbital along the axis of the bond, generating a node between the nuclei"}
{"id": 2863, "contents": "530. Key Equations - \nbond order $=\\frac{(\\text { number of bonding electron) }-(\\text { number of antibonding electrons })}{2}$"}
{"id": 2864, "contents": "531. Summary - 531.1. Valence Bond Theory\nValence bond theory describes bonding as a consequence of the overlap of two separate atomic orbitals on different atoms that creates a region with one pair of electrons shared between the two atoms. When the orbitals overlap along an axis containing the nuclei, they form a $\\sigma$ bond. When they overlap in a fashion that creates a node along this axis, they form a $\\pi$ bond. Dipole moments can be used to determine partial separations of charges between atoms."}
{"id": 2865, "contents": "531. Summary - 531.2. Hybrid Atomic Orbitals\nWe can use hybrid orbitals, which are mathematical combinations of some or all of the valence atomic orbitals, to describe the electron density around covalently bonded atoms. These hybrid orbitals either form sigma ( $\\sigma$ ) bonds directed toward other atoms of the molecule or contain lone pairs of electrons. We can determine the type of hybridization around a central atom from the geometry of the regions of electron density about it. Two such regions imply $s p$ hybridization; three, $s p^{2}$ hybridization; four, $s p^{3}$ hybridization; five, $s p^{3} d$ hybridization; and six, $s p^{3} d^{2}$ hybridization. $\\operatorname{Pi}(\\pi)$ bonds are formed from unhybridized atomic orbitals ( $p$ or $d$ orbitals)."}
{"id": 2866, "contents": "531. Summary - 531.3. Multiple Bonds\nMultiple bonds consist of a $\\sigma$ bond located along the axis between two atoms and one or two $\\pi$ bonds. The $\\sigma$ bonds are usually formed by the overlap of hybridized atomic orbitals, while the $\\pi$ bonds are formed by the side-by-side overlap of unhybridized orbitals. Resonance occurs when there are multiple unhybridized orbitals with the appropriate alignment to overlap, so the placement of $\\pi$ bonds\ncan vary."}
{"id": 2867, "contents": "531. Summary - 531.4. Molecular Orbital Theory\nMolecular orbital (MO) theory describes the behavior of electrons in a molecule in terms of combinations of the atomic wave functions. The resulting molecular orbitals may extend over all the atoms in the molecule. Bonding molecular orbitals are formed by in-phase combinations of atomic wave functions, and electrons in these orbitals stabilize a molecule. Antibonding molecular orbitals result from out-of-phase combinations of atomic wave functions and electrons in these orbitals make a molecule less stable. Molecular orbitals located along an internuclear axis are called $\\sigma$ MOs. They can be formed from $s$ orbitals or from $p$ orbitals oriented in an end-to-end fashion. Molecular orbitals formed from $p$ orbitals oriented in a side-by-side fashion have electron density on opposite sides of the internuclear axis and are called $\\pi$ orbitals.\n\nWe can describe the electronic structure of diatomic molecules by applying molecular orbital theory to the valence electrons of the atoms. Electrons fill molecular orbitals following the same rules that apply to filling atomic orbitals; Hund's rule and the Aufbau principle tell us that lower-energy orbitals will fill first, electrons will spread out before they pair up, and each orbital can hold a maximum of two electrons with opposite spins. Materials with unpaired electrons are paramagnetic and attracted to a magnetic field, while those with all-paired electrons are diamagnetic and repelled by a magnetic field. Correctly predicting the magnetic properties of molecules is in advantage of molecular orbital theory over Lewis structures and valence bond theory."}
{"id": 2868, "contents": "532. Exercises - 532.1. Valence Bond Theory\n1. Explain how $\\sigma$ and $\\pi$ bonds are similar and how they are different.\n2. Use valence bond theory to explain the bonding in $\\mathrm{F}_{2}$, HF , and ClBr . Sketch the overlap of the atomic orbitals involved in the bonds.\n3. Use valence bond theory to explain the bonding in $\\mathrm{O}_{2}$. Sketch the overlap of the atomic orbitals involved in the bonds in $\\mathrm{O}_{2}$.\n4. How many $\\sigma$ and $\\pi$ bonds are present in the molecule HCN?\n5. A friend tells you $N_{2}$ has three $\\pi$ bonds due to overlap of the three $p$-orbitals on each $N$ atom. Do you agree?\n6. Draw the Lewis structures for $\\mathrm{CO}_{2}$ and CO , and predict the number of $\\sigma$ and $\\pi$ bonds for each molecule.\n(a) $\\mathrm{CO}_{2}$\n(b) CO"}
{"id": 2869, "contents": "532. Exercises - 532.2. Hybrid Atomic Orbitals\n7. Why is the concept of hybridization required in valence bond theory?\n8. Give the shape that describes each hybrid orbital set:\n(a) $s p^{2}$\n(b) $s p^{3} d$\n(c) $s p$\n(d) $s p^{3} d^{2}$\n9. Explain why a carbon atom cannot form five bonds using $s p^{3} d$ hybrid orbitals.\n10. What is the hybridization of the central atom in each of the following?\n(a) $\\mathrm{BeH}_{2}$\n(b) $\\mathrm{SF}_{6}$\n(c) $\\mathrm{PO}_{4}{ }^{3-}$\n(d) $\\mathrm{PCl}_{5}$\n11. A molecule with the formula $A B_{3}$ could have one of four different shapes. Give the shape and the hybridization of the central A atom for each.\n12. Methionine, $\\mathrm{CH}_{3} \\mathrm{SCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}\\left(\\mathrm{NH}_{2}\\right) \\mathrm{CO}_{2} \\mathrm{H}$, is an amino acid found in proteins. The Lewis structure of this compound is shown below. What is the hybridization type of each carbon, oxygen, the nitrogen, and the sulfur?"}
{"id": 2870, "contents": "532. Exercises - 532.2. Hybrid Atomic Orbitals\n13. Sulfuric acid is manufactured by a series of reactions represented by the following equations:\n$\\mathrm{S}_{8}(\\mathrm{~s})+8 \\mathrm{O}_{2}(g) \\longrightarrow 8 \\mathrm{SO}_{2}(g)$\n$2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{SO}_{3}(g)$\n$\\mathrm{SO}_{3}(g)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{H}_{2} \\mathrm{SO}_{4}(l)$\nDraw a Lewis structure, predict the molecular geometry by VSEPR, and determine the hybridization of sulfur for the following:\n(a) circular $\\mathrm{S}_{8}$ molecule\n(b) $\\mathrm{SO}_{2}$ molecule\n(c) $\\mathrm{SO}_{3}$ molecule\n(d) $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ molecule (the hydrogen atoms are bonded to oxygen atoms)\n14. Two important industrial chemicals, ethene, $\\mathrm{C}_{2} \\mathrm{H}_{4}$, and propene, $\\mathrm{C}_{3} \\mathrm{H}_{6}$, are produced by the steam (or thermal) cracking process:\n\n$$\n2 \\mathrm{C}_{3} \\mathrm{H}_{8}(g) \\longrightarrow \\mathrm{C}_{2} \\mathrm{H}_{4}(g)+\\mathrm{C}_{3} \\mathrm{H}_{6}(g)+\\mathrm{CH}_{4}(g)+\\mathrm{H}_{2}(g)\n$$"}
{"id": 2871, "contents": "532. Exercises - 532.2. Hybrid Atomic Orbitals\nFor each of the four carbon compounds, do the following:\n(a) Draw a Lewis structure.\n(b) Predict the geometry about the carbon atom.\n(c) Determine the hybridization of each type of carbon atom.\n15. Analysis of a compound indicates that it contains $77.55 \\% \\mathrm{Xe}$ and $22.45 \\% \\mathrm{~F}$ by mass.\n(a) What is the empirical formula for this compound? (Assume this is also the molecular formula in responding to the remaining parts of this exercise).\n(b) Write a Lewis structure for the compound.\n(c) Predict the shape of the molecules of the compound.\n(d) What hybridization is consistent with the shape you predicted?\n16. Consider nitrous acid, $\\mathrm{HNO}_{2}$ (HONO).\n(a) Write a Lewis structure.\n(b) What are the electron pair and molecular geometries of the internal oxygen and nitrogen atoms in the $\\mathrm{HNO}_{2}$ molecule?\n(c) What is the hybridization on the internal oxygen and nitrogen atoms in $\\mathrm{HNO}_{2}$ ?\n17. Strike-anywhere matches contain a layer of $\\mathrm{KClO}_{3}$ and a layer of $\\mathrm{P}_{4} \\mathrm{~S}_{3}$. The heat produced by the friction of striking the match causes these two compounds to react vigorously, which sets fire to the wooden stem of the match. $\\mathrm{KClO}_{3}$ contains the $\\mathrm{ClO}_{3}{ }^{-}$ion. $\\mathrm{P}_{4} \\mathrm{~S}_{3}$ is an unusual molecule with the skeletal structure."}
{"id": 2872, "contents": "532. Exercises - 532.2. Hybrid Atomic Orbitals\n(a) Write Lewis structures for $\\mathrm{P}_{4} \\mathrm{~S}_{3}$ and the $\\mathrm{ClO}_{3}{ }^{-}$ion.\n(b) Describe the geometry about the P atoms, the S atom, and the Cl atom in these species.\n(c) Assign a hybridization to the P atoms, the S atom, and the Cl atom in these species.\n(d) Determine the oxidation states and formal charge of the atoms in $\\mathrm{P}_{4} \\mathrm{~S}_{3}$ and the $\\mathrm{ClO}_{3}{ }^{-}$ion.\n18. Identify the hybridization of each carbon atom in the following molecule. (The arrangement of atoms is given; you need to determine how many bonds connect each pair of atoms.)\n\n|  | H | H |  |  |  | H | H |  |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n|  | H | C | C | C | C | C | H |  |\n|  | H | H |  |  |  |  |  |  |\n\n19. Write Lewis structures for $\\mathrm{NF}_{3}$ and $\\mathrm{PF}_{5}$. On the basis of hybrid orbitals, explain the fact that $\\mathrm{NF}_{3}, \\mathrm{PF}_{3}$, and $\\mathrm{PF}_{5}$ are stable molecules, but $\\mathrm{NF}_{5}$ does not exist.\n20. In addition to $\\mathrm{NF}_{3}$, two other fluoro derivatives of nitrogen are known: $\\mathrm{N}_{2} \\mathrm{~F}_{4}$ and $\\mathrm{N}_{2} \\mathrm{~F}_{2}$. What shapes do you predict for these two molecules? What is the hybridization for the nitrogen in each molecule?"}
{"id": 2873, "contents": "532. Exercises - 532.3. Multiple Bonds\n21. The bond energy of a $\\mathrm{C}-\\mathrm{C}$ single bond averages $347 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$; that of a $\\mathrm{C} \\equiv \\mathrm{C}$ triple bond averages 839 kJ $\\mathrm{mol}^{-1}$. Explain why the triple bond is not three times as strong as a single bond.\n22. For the carbonate ion, $\\mathrm{CO}_{3}{ }^{2-}$, draw all of the resonance structures. Identify which orbitals overlap to create each bond.\n23. A useful solvent that will dissolve salts as well as organic compounds is the compound acetonitrile, $\\mathrm{H}_{3} \\mathrm{CCN}$. It is present in paint strippers.\n(a) Write the Lewis structure for acetonitrile, and indicate the direction of the dipole moment in the molecule.\n(b) Identify the hybrid orbitals used by the carbon atoms in the molecule to form $\\sigma$ bonds.\n(c) Describe the atomic orbitals that form the $\\pi$ bonds in the molecule. Note that it is not necessary to hybridize the nitrogen atom.\n24. For the molecule allene, $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{C}=\\mathrm{CH}_{2}$, give the hybridization of each carbon atom. Will the hydrogen atoms be in the same plane or perpendicular planes?\n25. Identify the hybridization of the central atom in each of the following molecules and ions that contain multiple bonds:\n(a) ClNO ( N is the central atom)\n(b) $\\mathrm{CS}_{2}$\n(c) $\\mathrm{Cl}_{2} \\mathrm{CO}$ ( C is the central atom)\n(d) $\\mathrm{Cl}_{2} \\mathrm{SO}$ ( S is the central atom)\n(e) $\\mathrm{SO}_{2} \\mathrm{~F}_{2}$ (S is the central atom)\n(f) $\\mathrm{XeO}_{2} \\mathrm{~F}_{2}$ ( Xe is the central atom)\n(g) $\\mathrm{ClOF}_{2}{ }^{+}$( Cl is the central atom)\n26. Describe the molecular geometry and hybridization of the $N, P$, or $S$ atoms in each of the following compounds."}
{"id": 2874, "contents": "532. Exercises - 532.3. Multiple Bonds\n26. Describe the molecular geometry and hybridization of the $N, P$, or $S$ atoms in each of the following compounds.\n(a) $\\mathrm{H}_{3} \\mathrm{PO}_{4}$, phosphoric acid, used in cola soft drinks\n(b) $\\mathrm{NH}_{4} \\mathrm{NO}_{3}$, ammonium nitrate, a fertilizer and explosive\n(c) $\\mathrm{S}_{2} \\mathrm{Cl}_{2}$, disulfur dichloride, used in vulcanizing rubber\n(d) $\\mathrm{K}_{4}\\left[\\mathrm{O}_{3} \\mathrm{POPO}_{3}\\right]$, potassium pyrophosphate, an ingredient in some toothpastes\n27. For each of the following molecules, indicate the hybridization requested and whether or not the electrons will be delocalized:\n(a) ozone $\\left(\\mathrm{O}_{3}\\right)$ central O hybridization\n(b) carbon dioxide $\\left(\\mathrm{CO}_{2}\\right)$ central C hybridization\n(c) nitrogen dioxide $\\left(\\mathrm{NO}_{2}\\right)$ central N hybridization\n(d) phosphate ion $\\left(\\mathrm{PO}_{4}{ }^{3-}\\right)$ central P hybridization\n28. For each of the following structures, determine the hybridization requested and whether the electrons will be delocalized:\n(a) Hybridization of each carbon"}
{"id": 2875, "contents": "532. Exercises - 532.3. Multiple Bonds\n(b) Hybridization of sulfur\n\n(c) All atoms\n\n29. Draw the orbital diagram for carbon in $\\mathrm{CO}_{2}$ showing how many carbon atom electrons are in each orbital."}
{"id": 2876, "contents": "532. Exercises - 532.4. Molecular Orbital Theory\n30. Sketch the distribution of electron density in the bonding and antibonding molecular orbitals formed from two $s$ orbitals and from two $p$ orbitals.\n31. How are the following similar, and how do they differ?\n(a) $\\sigma$ molecular orbitals and $\\pi$ molecular orbitals\n(b) $\\psi$ for an atomic orbital and $\\psi$ for a molecular orbital\n(c) bonding orbitals and antibonding orbitals\n32. If molecular orbitals are created by combining five atomic orbitals from atom $A$ and five atomic orbitals from atom B combine, how many molecular orbitals will result?\n33. Can a molecule with an odd number of electrons ever be diamagnetic? Explain why or why not.\n34. Can a molecule with an even number of electrons ever be paramagnetic? Explain why or why not.\n35. Why are bonding molecular orbitals lower in energy than the parent atomic orbitals?\n36. Calculate the bond order for an ion with this configuration:\n\n$$\n\\left(\\sigma_{2 s}\\right)^{2}\\left(\\sigma_{2 s}^{*}\\right)^{2}\\left(\\sigma_{2 p x}\\right)^{2}\\left(\\pi_{2 p y}, \\pi_{2 p z}\\right)^{4}\\left(\\pi_{2 p y}^{*}, \\pi_{2 p z}^{*}\\right)^{3}\n$$"}
{"id": 2877, "contents": "532. Exercises - 532.4. Molecular Orbital Theory\n37. Explain why an electron in the bonding molecular orbital in the $\\mathrm{H}_{2}$ molecule has a lower energy than an electron in the $1 s$ atomic orbital of either of the separated hydrogen atoms.\n38. Predict the valence electron molecular orbital configurations for the following, and state whether they will be stable or unstable ions.\n(a) $\\mathrm{Na}_{2}{ }^{2+}$\n(b) $\\mathrm{Mg}_{2}{ }^{2+}$\n(c) $\\mathrm{Al}_{2}{ }^{2+}$\n(d) $\\mathrm{Si}_{2}{ }^{2+}$\n(e) $\\mathrm{P}_{2}{ }^{2+}$\n(f) $\\mathrm{S}_{2}{ }^{2+}$\n(g) $\\mathrm{F}_{2}{ }^{2+}$\n(h) $\\mathrm{Ar}_{2}{ }^{2+}$\n39. Determine the bond order of each member of the following groups, and determine which member of each group is predicted by the molecular orbital model to have the strongest bond.\n(a) $\\mathrm{H}_{2}, \\mathrm{H}_{2}{ }^{+}, \\mathrm{H}_{2}^{-}$\n(b) $\\mathrm{O}_{2}, \\mathrm{O}_{2}{ }^{2+}, \\mathrm{O}_{2}{ }^{2-}$\n(c) $\\mathrm{Li}_{2}, \\mathrm{Be}_{2}{ }^{+}, \\mathrm{Be}_{2}$\n(d) $\\mathrm{F}_{2}, \\mathrm{~F}_{2}{ }^{+}, \\mathrm{F}_{2}^{-}$\n(e) $\\mathrm{N}_{2}, \\mathrm{~N}_{2}{ }^{+}, \\mathrm{N}_{2}{ }^{-}$\n40. For the first ionization energy for an $\\mathrm{N}_{2}$ molecule, what molecular orbital is the electron removed from?\n41. Compare the atomic and molecular orbital diagrams to identify the member of each of the following pairs that has the highest first ionization energy (the most tightly bound electron) in the gas phase:\n(a) H and $\\mathrm{H}_{2}$\n(b) N and $\\mathrm{N}_{2}$\n(c) O and $\\mathrm{O}_{2}$\n(d) C and $\\mathrm{C}_{2}$"}
{"id": 2878, "contents": "532. Exercises - 532.4. Molecular Orbital Theory\n(a) H and $\\mathrm{H}_{2}$\n(b) N and $\\mathrm{N}_{2}$\n(c) O and $\\mathrm{O}_{2}$\n(d) C and $\\mathrm{C}_{2}$\n(e) B and $\\mathrm{B}_{2}$\n42. Which of the period 2 homonuclear diatomic molecules are predicted to be paramagnetic?\n43. A friend tells you that the $2 s$ orbital for fluorine starts off at a much lower energy than the $2 s$ orbital for lithium, so the resulting $\\sigma_{2 s}$ molecular orbital in $\\mathrm{F}_{2}$ is more stable than in $\\mathrm{Li}_{2}$. Do you agree?\n44. True or false: Boron contains $2 s^{2} 2 p^{1}$ valence electrons, so only one $p$ orbital is needed to form molecular orbitals.\n45. What charge would be needed on $F_{2}$ to generate an ion with a bond order of 2 ?\n46. Predict whether the MO diagram for $S_{2}$ would show s-p mixing or not.\n47. Explain why $\\mathrm{N}_{2}{ }^{2+}$ is diamagnetic, while $\\mathrm{O}_{2}{ }^{4+}$, which has the same number of valence electrons, is paramagnetic.\n48. Using the MO diagrams, predict the bond order for the stronger bond in each pair:\n(a) $\\mathrm{B}_{2}$ or $\\mathrm{B}_{2}{ }^{+}$\n(b) $\\mathrm{F}_{2}$ or $\\mathrm{F}_{2}{ }^{+}$\n(c) $\\mathrm{O}_{2}$ or $\\mathrm{O}_{2}{ }^{2+}$\n(d) $\\mathrm{C}_{2}{ }^{+}$or $\\mathrm{C}_{2}{ }^{-}$"}
{"id": 2879, "contents": "533. CHAPTER 6 <br> Composition of Substances and Solutions - \nFigure 6.1 The water in a swimming pool is a complex mixture of substances whose relative amounts must be carefully maintained to ensure the health and comfort of people using the pool. (credit: modification of work by Vic Brincat)"}
{"id": 2880, "contents": "534. CHAPTER OUTLINE - 534.1. Formula Mass\n6.2 Determining Empirical and Molecular Formulas\n6.3 Molarity\n6.4 Other Units for Solution Concentrations\n\nINTRODUCTION Swimming pools have long been a popular means of recreation, exercise, and physical therapy. Since it is impractical to refill large pools with fresh water on a frequent basis, pool water is regularly treated with chemicals to prevent the growth of harmful bacteria and algae. Proper pool maintenance requires regular additions of various chemical compounds in carefully measured amounts. For example, the relative amount of calcium ion, $\\mathrm{Ca}^{2+}$, in the water should be maintained within certain limits to prevent eye irritation and avoid damage to the pool bed and plumbing. To maintain proper calcium levels, calcium cations are added to the water in the form of an ionic compound that also contains anions; thus, it is necessary to know both the relative amount of $\\mathrm{Ca}^{2+}$ in the compound and the volume of water in the pool in order to achieve the proper calcium level. Quantitative aspects of the composition of substances (such as the calcium-containing compound) and mixtures (such as the pool water) are the subject of this chapter."}
{"id": 2881, "contents": "535. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Calculate formula masses for covalent and ionic compounds\n\nMany argue that modern chemical science began when scientists started exploring the quantitative as well as the qualitative aspects of chemistry. For example, Dalton's atomic theory was an attempt to explain the results of measurements that allowed him to calculate the relative masses of elements combined in various compounds. Understanding the relationship between the masses of atoms and the chemical formulas of compounds allows us to quantitatively describe the composition of substances."}
{"id": 2882, "contents": "536. Formula Mass - \nAn earlier chapter of this text described the development of the atomic mass unit, the concept of average atomic masses, and the use of chemical formulas to represent the elemental makeup of substances. These ideas can be extended to calculate the formula mass of a substance by summing the average atomic masses of all the atoms represented in the substance's formula."}
{"id": 2883, "contents": "537. Formula Mass for Covalent Substances - \nFor covalent substances, the formula represents the numbers and types of atoms composing a single molecule of the substance; therefore, the formula mass may be correctly referred to as a molecular mass. Consider chloroform $\\left(\\mathrm{CHCl}_{3}\\right)$, a covalent compound once used as a surgical anesthetic and now primarily used in the production of the \"anti-stick\" polymer, Teflon. The molecular formula of chloroform indicates that a single molecule contains one carbon atom, one hydrogen atom, and three chlorine atoms. The average molecular mass of a chloroform molecule is therefore equal to the sum of the average atomic masses of these atoms. Figure 6.2 outlines the calculations used to derive the molecular mass of chloroform, which is 119.37 amu .\n\n| Element | Quantity | Average atomic <br> mass (amu) | Subtotal <br> $(\\mathbf{a m u})$ |  |  |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| C | 1 | $\\times$ | 12.01 | $=$ | 12.01 |\n| H | 1 | $\\times$ | 1.008 | $=$ | 1.008 |\n| Cl | 3 | $\\times$ | 35.45 | $=$ | 106.35 |\n| Molecular mass |  |  |  |  | 119.37 |\n\nFIGURE 6.2 The average mass of a chloroform molecule, $\\mathrm{CHCl}_{3}$, is 119.37 amu , which is the sum of the average atomic masses of each of its constituent atoms. The model shows the molecular structure of chloroform.\n\nLikewise, the molecular mass of an aspirin molecule, $\\mathrm{C}_{9} \\mathrm{H}_{8} \\mathrm{O}_{4}$, is the sum of the atomic masses of nine carbon atoms, eight hydrogen atoms, and four oxygen atoms, which amounts to 180.15 amu (Figure 6.3)."}
{"id": 2884, "contents": "537. Formula Mass for Covalent Substances - \n| Element | Quantity | Average atomic <br> mass (amu) | Subtotal <br> $(\\mathbf{a m u})$ |  |  |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| C | 9 | $\\times$ | 12.01 | $=$ | 108.09 |\n| H | 8 | $\\times$ | 1.008 | $=$ | 8.064 |\n| O | 4 | $\\times$ | 16.00 | $=$ | 64.00 |\n| Molecular mass |  |  |  |  | 180.15 |\n\nFIGURE 6.3 The average mass of an aspirin molecule is 180.15 amu . The model shows the molecular structure of aspirin, $\\mathrm{C}_{9} \\mathrm{H}_{8} \\mathrm{O}_{4}$."}
{"id": 2885, "contents": "539. Computing Molecular Mass for a Covalent Compound - \nIbuprofen, $\\mathrm{C}_{13} \\mathrm{H}_{18} \\mathrm{O}_{2}$, is a covalent compound and the active ingredient in several popular nonprescription pain medications, such as Advil and Motrin. What is the molecular mass (amu) for this compound?"}
{"id": 2886, "contents": "540. Solution - \nMolecules of this compound are comprised of 13 carbon atoms, 18 hydrogen atoms, and 2 oxygen atoms. Following the approach described above, the average molecular mass for this compound is therefore:\n\n| Element | Quantity | Average atomic <br> mass (amu) | Subtotal <br> (amu) |  |  |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| C | 13 | $\\times$ | 12.01 | $=$ | 156.13 |\n| H | 18 | $\\times$ | 1.008 | $=$ | 18.144 |\n| O | 2 | $\\times$ | 16.00 | $=$ | 32.00 |\n| Molecular mass |  |  |  |  | 206.27 |"}
{"id": 2887, "contents": "541. Check Your Learning - \nAcetaminophen, $\\mathrm{C}_{8} \\mathrm{H}_{9} \\mathrm{NO}_{2}$, is a covalent compound and the active ingredient in several popular nonprescription pain medications, such as Tylenol. What is the molecular mass (amu) for this compound?"}
{"id": 2888, "contents": "542. Answer: - \n151.16 amu"}
{"id": 2889, "contents": "543. Formula Mass for Ionic Compounds - \nIonic compounds are composed of discrete cations and anions combined in ratios to yield electrically neutral bulk matter. The formula mass for an ionic compound is calculated in the same way as the formula mass for covalent compounds: by summing the average atomic masses of all the atoms in the compound's formula. Keep in mind, however, that the formula for an ionic compound does not represent the composition of a discrete molecule, so it may not correctly be referred to as the \"molecular mass.\"\n\nAs an example, consider sodium chloride, NaCl , the chemical name for common table salt. Sodium chloride is an ionic compound composed of sodium cations, $\\mathrm{Na}^{+}$, and chloride anions, $\\mathrm{Cl}^{-}$, combined in a 1:1 ratio. The formula mass for this compound is computed as 58.44 amu (see Figure 6.4).\n\n| Element | Quantity |  | Average atomic <br> mass (amu) | Subtotal |  |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| Na | 1 | $\\times$ | 22.99 | $=$ | 22.99 |\n| Cl | 1 | $\\times$ | 35.45 | $=$ | 35.45 |\n|  |  |  |  |  |  |\n| Formula mass |  |  |  |  | 58.44 |\n\nFIGURE 6.4 Table salt, NaCl , contains an array of sodium and chloride ions combined in a 1:1 ratio. Its formula mass is 58.44 amu ."}
{"id": 2890, "contents": "543. Formula Mass for Ionic Compounds - \nFIGURE 6.4 Table salt, NaCl , contains an array of sodium and chloride ions combined in a 1:1 ratio. Its formula mass is 58.44 amu .\n\nNote that the average masses of neutral sodium and chlorine atoms were used in this computation, rather than the masses for sodium cations and chlorine anions. This approach is perfectly acceptable when computing the formula mass of an ionic compound. Even though a sodium cation has a slightly smaller mass than a sodium atom (since it is missing an electron), this difference will be offset by the fact that a chloride anion is slightly more massive than a chloride atom (due to the extra electron). Moreover, the mass of an electron is negligibly small with respect to the mass of a typical atom. Even when calculating the mass of an isolated ion, the missing\nor additional electrons can generally be ignored, since their contribution to the overall mass is negligible, reflected only in the nonsignificant digits that will be lost when the computed mass is properly rounded. The few exceptions to this guideline are very light ions derived from elements with precisely known atomic masses."}
{"id": 2891, "contents": "545. Computing Formula Mass for an Ionic Compound - \nAluminum sulfate, $\\mathrm{Al}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}$, is an ionic compound that is used in the manufacture of paper and in various water purification processes. What is the formula mass ( amu ) of this compound?"}
{"id": 2892, "contents": "546. Solution - \nThe formula for this compound indicates it contains $\\mathrm{Al}^{3+}$ and $\\mathrm{SO}_{4}{ }^{2-}$ ions combined in a $2: 3$ ratio. For purposes of computing a formula mass, it is helpful to rewrite the formula in the simpler format, $\\mathrm{Al}_{2} \\mathrm{~S}_{3} \\mathrm{O}_{12}$. Following the approach outlined above, the formula mass for this compound is calculated as follows:\n\n| Element | Quantity | Average atomic <br> mass (amu) | Subtotal <br> $(\\mathbf{a m u})$ |  |  |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| Al | 2 | $\\times$ | 26.98 | $=$ | 53.96 |\n| S | 3 | $\\times$ | 32.06 | $=$ | 96.18 |\n| O | 12 | $\\times$ | 16.00 | $=$ | 192.00 |\n| Molecular mass 342.14 |  |  |  |  |  |"}
{"id": 2893, "contents": "547. Check Your Learning - \nCalcium phosphate, $\\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$, is an ionic compound and a common anti-caking agent added to food products. What is the formula mass (amu) of calcium phosphate?"}
{"id": 2894, "contents": "548. Answer: - \n310.18 amu"}
{"id": 2895, "contents": "548. Answer: - 548.1. Determining Empirical and Molecular Formulas\nA previous chapter of this text discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, one may determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, these same principles will be applied to derive the chemical formulas of unknown substances from experimental mass measurements."}
{"id": 2896, "contents": "549. Percent Composition - \nThe elemental makeup of a compound defines its chemical identity, and chemical formulas are the most succinct way of representing this elemental makeup. When a compound's formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound's percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:\n\n$$\n\\begin{aligned}\n& \\% \\mathrm{H}=\\frac{\\text { mass } \\mathrm{H}}{\\text { mass compound }} \\times 100 \\% \\\\\n& \\% \\mathrm{C}=\\frac{\\text { mass } \\mathrm{C}}{\\text { mass compound }} \\times 100 \\%\n\\end{aligned}\n$$\n\nIf analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C , the percent composition would be calculated to be $25 \\% \\mathrm{H}$ and $75 \\% \\mathrm{C}$ :\n\n$$\n\\begin{aligned}\n& \\% \\mathrm{H}=\\frac{2.5 \\mathrm{~g} \\mathrm{H}}{10.0 \\mathrm{~g} \\text { compound }} \\times 100 \\%=25 \\% \\\\\n& \\% \\mathrm{C}=\\frac{7.5 \\mathrm{~g} \\mathrm{C}}{10.0 \\mathrm{~g} \\text { compound }} \\times 100 \\%=75 \\%\n\\end{aligned}\n$$"}
{"id": 2897, "contents": "551. Calculation of Percent Composition - \nAnalysis of a $12.04-\\mathrm{g}$ sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed it to contain $7.34 \\mathrm{~g} \\mathrm{C}, 1.85 \\mathrm{~g} \\mathrm{H}$, and 2.85 g N . What is the percent composition of this compound?"}
{"id": 2898, "contents": "552. Solution - \nTo calculate percent composition, divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:\n\n$$\n\\begin{aligned}\n& \\% \\mathrm{C}=\\frac{7.34 \\mathrm{~g} \\mathrm{C}}{12.04 \\mathrm{~g} \\text { compound }} \\times 100 \\%=61.0 \\% \\\\\n& \\% \\mathrm{H}=\\frac{1.85 \\mathrm{~g} \\mathrm{H}}{12.04 \\mathrm{~g} \\text { compound }} \\times 100 \\%=15.4 \\% \\\\\n& \\% \\mathrm{~N}=\\frac{2.85 \\mathrm{~g} \\mathrm{~N}}{12.04 \\mathrm{~g} \\text { compound }} \\times 100 \\%=23.7 \\%\n\\end{aligned}\n$$\n\nThe analysis results indicate that the compound is $61.0 \\% \\mathrm{C}, 15.4 \\% \\mathrm{H}$, and $23.7 \\% \\mathrm{~N}$ by mass."}
{"id": 2899, "contents": "553. Check Your Learning - \nA 24.81-g sample of a gaseous compound containing only carbon, oxygen, and chlorine is determined to contain $3.01 \\mathrm{~g} \\mathrm{C}, 4.00 \\mathrm{~g} \\mathrm{O}$, and 17.81 g Cl . What is this compound's percent composition?"}
{"id": 2900, "contents": "554. Answer: - \n12.1\\% C, 16.1\\% O, 71.79\\% Cl"}
{"id": 2901, "contents": "555. Determining Percent Composition from Molecular or Empirical Formulas - \nPercent composition is also useful for evaluating the relative abundance of a given element in different compounds of known formulas. As one example, consider the common nitrogen-containing fertilizers ammonia $\\left(\\mathrm{NH}_{3}\\right)$, ammonium nitrate $\\left(\\mathrm{NH}_{4} \\mathrm{NO}_{3}\\right)$, and urea $\\left(\\mathrm{CH}_{4} \\mathrm{~N}_{2} \\mathrm{O}\\right)$. The element nitrogen is the active ingredient for agricultural purposes, so the mass percentage of nitrogen in the compound is a practical and economic concern for consumers choosing among these fertilizers. For these sorts of applications, the percent composition of a compound is easily derived from its formula mass and the atomic masses of its constituent elements. A molecule of $\\mathrm{NH}_{3}$ contains one N atom weighing 14.01 amu and three H atoms weighing a total of (3 $\\times 1.008 \\mathrm{amu})=3.024 \\mathrm{amu}$. The formula mass of ammonia is therefore $(14.01 \\mathrm{amu}+3.024 \\mathrm{amu})=17.03 \\mathrm{amu}$, and its percent composition is:\n\n$$\n\\begin{aligned}\n& \\% \\mathrm{~N}=\\frac{14.01 \\mathrm{amu} \\mathrm{~N}}{17.03 \\mathrm{amu} \\mathrm{NH}} 3 \\mathrm{am} \\\\\n& \\% \\mathrm{H}=\\frac{3.024 \\mathrm{amu} \\mathrm{H}}{17.03 \\mathrm{amu} \\mathrm{NH}_{3} \\times 100 \\%=82.27 \\%}=17.76 \\%\n\\end{aligned}\n$$\n\nThis same approach may be taken considering a pair of molecules, a dozen molecules, or a mole of molecules, etc. The latter amount is most convenient and would simply involve the use of molar masses instead of atomic and formula masses, as demonstrated Example 6.4. As long as the molecular or empirical formula of the compound in question is known, the percent composition may be derived from the atomic or molar masses of the compound's elements."}
{"id": 2902, "contents": "557. Determining Percent Composition from a Molecular Formula - \nAspirin is a compound with the molecular formula $\\mathrm{C}_{9} \\mathrm{H}_{8} \\mathrm{O}_{4}$. What is its percent composition?"}
{"id": 2903, "contents": "558. Solution - \nTo calculate the percent composition, the masses of $\\mathrm{C}, \\mathrm{H}$, and O in a known mass of $\\mathrm{C}_{9} \\mathrm{H}_{8} \\mathrm{O}_{4}$ are needed. It is convenient to consider 1 mol of $\\mathrm{C}_{9} \\mathrm{H}_{8} \\mathrm{O}_{4}$ and use its molar mass ( $180.159 \\mathrm{~g} / \\mathrm{mole}$, determined from the chemical formula) to calculate the percentages of each of its elements:"}
{"id": 2904, "contents": "558. Solution - \n$$\n\\begin{aligned}\n& \\% \\mathrm{C}=\\frac{9 \\mathrm{~mol} \\mathrm{C} \\times \\text { molar mass } \\mathrm{C}}{\\mathrm{molar} \\text { mass } \\mathrm{C}_{9} \\mathrm{H}_{8} \\mathrm{O}_{4}} \\times 100=\\frac{9 \\times 12.01 \\mathrm{~g} / \\mathrm{mol}}{180.159 \\mathrm{~g} / \\mathrm{mol}} \\times 100=\\frac{108.09 \\mathrm{~g} / \\mathrm{mol}}{180.159 \\mathrm{~g} / \\mathrm{mol}} \\times 100 \\\\\n& \\% \\mathrm{C}=60.00 \\% \\mathrm{C} \\\\\n& \\% \\mathrm{H}=\\frac{8 \\mathrm{~mol} \\mathrm{H} \\times \\text { molar mass } \\mathrm{H}}{\\mathrm{molar} \\text { mass } \\mathrm{C}_{9} \\mathrm{H}_{8} \\mathrm{O}_{4}} \\times 100=\\frac{8 \\times 1.008 \\mathrm{~g} / \\mathrm{mol}}{180.159 \\mathrm{~g} / \\mathrm{mol}} \\times 100=\\frac{8.064 \\mathrm{~g} / \\mathrm{mol}}{180.159 \\mathrm{~g} / \\mathrm{mol}} \\times 100 \\\\\n& \\% \\mathrm{H}=4.476 \\% \\mathrm{H} \\\\\n& \\% \\mathrm{O}=\\frac{4 \\text { mol O } \\times \\text { molar mass } \\mathrm{O}}{\\text { molar mass } \\mathrm{C}_{9} \\mathrm{H}_{8} \\mathrm{O}_{4}} \\times 100=\\frac{4 \\times 16.00 \\mathrm{~g} / \\mathrm{mol}}{180.159 \\mathrm{~g} / \\mathrm{mol}} \\times 100=\\frac{64.00 \\mathrm{~g} / \\mathrm{mol}}{180.159 \\mathrm{~g} / \\mathrm{mol}} \\times 100 \\\\"}
{"id": 2905, "contents": "558. Solution - \n& \\% \\mathrm{O}=35.52 \\%\n\\end{aligned}\n$$"}
{"id": 2906, "contents": "558. Solution - \nNote that these percentages sum to equal $100.00 \\%$ when appropriately rounded.\nCheck Your Learning\nTo three significant digits, what is the mass percentage of iron in the compound $\\mathrm{Fe}_{2} \\mathrm{O}_{3}$ ?"}
{"id": 2907, "contents": "559. Answer: - \n69.9\\% Fe"}
{"id": 2908, "contents": "560. Determination of Empirical Formulas - \nAs previously mentioned, the most common approach to determining a compound's chemical formula is to first measure the masses of its constituent elements. However, keep in mind that chemical formulas represent the relative numbers, not masses, of atoms in the substance. Therefore, any experimentally derived data involving mass must be used to derive the corresponding numbers of atoms in the compound. This is accomplished using molar masses to convert the mass of each element to a number of moles. These molar amounts are used to compute whole-number ratios that can be used to derive the empirical formula of the substance. Consider a sample of compound determined to contain 1.71 g C and 0.287 g H . The corresponding numbers of atoms (in moles) are:\n\n$$\n\\begin{aligned}\n& 1.71 \\mathrm{~g} \\mathrm{C} \\times \\frac{1 \\mathrm{~mol} \\mathrm{C}}{12.01 \\mathrm{~g} \\mathrm{C}}=0.142 \\mathrm{~mol} \\mathrm{C} \\\\\n& 0.287 \\mathrm{~g} \\mathrm{H} \\times \\frac{1 \\mathrm{~mol} \\mathrm{H}}{1.008 \\mathrm{~g} \\mathrm{H}}=0.284 \\mathrm{~mol} \\mathrm{H}\n\\end{aligned}\n$$\n\nThus, this compound may be represented by the formula $\\mathrm{C}_{0.142} \\mathrm{H}_{0.284}$. Per convention, formulas contain wholenumber subscripts, which can be achieved by dividing each subscript by the smaller subscript:\n\n$$\n\\mathrm{C}_{\\frac{0.142}{0.142}} \\mathrm{H}_{\\frac{0.284}{0.142}} \\text { or } \\mathrm{CH}_{2}\n$$\n\n(Recall that subscripts of \" 1 \" are not written but rather assumed if no other number is present.)\nThe empirical formula for this compound is thus $\\mathrm{CH}_{2}$. This may or not be the compound's molecular formula\nas well; however, additional information is needed to make that determination (as discussed later in this section)."}
{"id": 2909, "contents": "560. Determination of Empirical Formulas - \nConsider as another example a sample of compound determined to contain 5.31 g Cl and 8.40 g O . Following the same approach yields a tentative empirical formula of:\n\n$$\n\\mathrm{Cl}_{0.150} \\mathrm{O}_{0.525}=\\mathrm{Cl}_{\\frac{0.150}{0.150}} \\mathrm{O}_{\\frac{0.525}{0.150}}=\\mathrm{ClO}_{3.5}\n$$\n\nIn this case, dividing by the smallest subscript still leaves us with a decimal subscript in the empirical formula. To convert this into a whole number, multiply each of the subscripts by two, retaining the same atom ratio and yielding $\\mathrm{Cl}_{2} \\mathrm{O}_{7}$ as the final empirical formula.\n\nIn summary, empirical formulas are derived from experimentally measured element masses by:\n\n1. Deriving the number of moles of each element from its mass\n2. Dividing each element's molar amount by the smallest molar amount to yield subscripts for a tentative empirical formula\n3. Multiplying all coefficients by an integer, if necessary, to ensure that the smallest whole-number ratio of subscripts is obtained\n\nFigure 6.5 outlines this procedure in flow chart fashion for a substance containing elements A and X.\n\n\nFIGURE 6.5 The empirical formula of a compound can be derived from the masses of all elements in the sample."}
{"id": 2910, "contents": "562. Determining a Compound's Empirical Formula from the Masses of Its Elements - \nA sample of the black mineral hematite (Figure 6.6), an oxide of iron found in many iron ores, contains 34.97 g of iron and 15.03 g of oxygen. What is the empirical formula of hematite?\n\n\nFIGURE 6.6 Hematite is an iron oxide that is used in jewelry. (credit: Mauro Cateb)"}
{"id": 2911, "contents": "563. Solution - \nThis problem provides the mass in grams of each element. Begin by finding the moles of each:\n\n$$\n\\begin{aligned}\n& 34.97 \\mathrm{~g} \\mathrm{Fe}\\left(\\frac{\\mathrm{~mol} \\mathrm{Fe}}{55.85 \\mathrm{~g}}\\right)=0.6261 \\mathrm{~mol} \\mathrm{Fe} \\\\\n& 15.03 \\mathrm{~g} \\mathrm{O}\\left(\\frac{\\mathrm{~mol} \\mathrm{O}}{16.00 \\mathrm{~g}}\\right)=0.9394 \\mathrm{~mol} \\mathrm{O}\n\\end{aligned}\n$$\n\nNext, derive the iron-to-oxygen molar ratio by dividing by the lesser number of moles:\n\n$$\n\\begin{aligned}\n& \\frac{0.6261}{0.6261}=1.000 \\mathrm{~mol} \\mathrm{Fe} \\\\\n& \\frac{0.9394}{0.6261}=1.500 \\mathrm{~mol} \\mathrm{O}\n\\end{aligned}\n$$\n\nThe ratio is 1.000 mol of iron to 1.500 mol of oxygen $\\left(\\mathrm{Fe}_{1} \\mathrm{O}_{1.5}\\right)$. Finally, multiply the ratio by two to get the smallest possible whole number subscripts while still maintaining the correct iron-to-oxygen ratio:\n\n$$\n2\\left(\\mathrm{Fe}_{1} \\mathrm{O}_{1.5}\\right)=\\mathrm{Fe}_{2} \\mathrm{O}_{3}\n$$\n\nThe empirical formula is $\\mathrm{Fe}_{2} \\mathrm{O}_{3}$."}
{"id": 2912, "contents": "564. Check Your Learning - \nWhat is the empirical formula of a compound if a sample contains 0.130 g of nitrogen and 0.370 g of oxygen?"}
{"id": 2913, "contents": "565. Answer: - \n$\\mathrm{N}_{2} \\mathrm{O}_{5}$"}
{"id": 2914, "contents": "566. LINK TO LEARNING - \nFor additional worked examples illustrating the derivation of empirical formulas, watch the brief video (http://openstax.org/l/16empforms) clip."}
{"id": 2915, "contents": "567. Deriving Empirical Formulas from Percent Composition - \nFinally, with regard to deriving empirical formulas, consider instances in which a compound's percent composition is available rather than the absolute masses of the compound's constituent elements. In such cases, the percent composition can be used to calculate the masses of elements present in any convenient mass of compound; these masses can then be used to derive the empirical formula in the usual fashion."}
{"id": 2916, "contents": "569. Determining an Empirical Formula from Percent Composition - \nThe bacterial fermentation of grain to produce ethanol forms a gas with a percent composition of 27.29\\% C and $72.71 \\% \\mathrm{O}$ (Figure 6.7). What is the empirical formula for this gas?\n\n\nFIGURE 6.7 An oxide of carbon is removed from these fermentation tanks through the large copper pipes at the top. (credit: \"Dual Freq\"/Wikimedia Commons)"}
{"id": 2917, "contents": "570. Solution - \nSince the scale for percentages is 100 , it is most convenient to calculate the mass of elements present in a sample weighing 100 g . The calculation is \"most convenient\" because, per the definition for percent composition, the mass of a given element in grams is numerically equivalent to the element's mass percentage. This numerical equivalence results from the definition of the \"percentage\" unit, whose name is derived from the Latin phrase per centum meaning \"by the hundred.\" Considering this definition, the mass percentages provided may be more conveniently expressed as fractions:\n\n$$\n\\begin{aligned}\n27.29 \\% \\mathrm{C} & =\\frac{27.29 \\mathrm{~g} \\mathrm{C}}{100 \\mathrm{~g} \\text { compound }} \\\\\n72.71 \\% \\mathrm{O} & =\\frac{72.71 \\mathrm{~g} \\mathrm{O}}{100 \\mathrm{~g} \\text { compound }}\n\\end{aligned}\n$$\n\nThe molar amounts of carbon and oxygen in a 100-g sample are calculated by dividing each element's mass by its molar mass:\n\n$$\n\\begin{aligned}\n& 27.29 \\mathrm{~g} \\mathrm{C}\\left(\\frac{\\mathrm{~mol} \\mathrm{C}}{12.01 \\mathrm{~g}}\\right)=2.272 \\mathrm{~mol} \\mathrm{C} \\\\\n& 72.71 \\mathrm{~g} \\mathrm{O}\\left(\\frac{\\mathrm{~mol} \\mathrm{O}}{16.00 \\mathrm{~g}}\\right)=4.544 \\mathrm{~mol} \\mathrm{O}\n\\end{aligned}\n$$\n\nCoefficients for the tentative empirical formula are derived by dividing each molar amount by the lesser of the two:\n\n$$\n\\begin{aligned}\n& \\frac{2.272 \\mathrm{~mol} \\mathrm{C}}{2.272}=1 \\\\\n& \\frac{4.544 \\mathrm{~mol} \\mathrm{O}}{2.272}=2\n\\end{aligned}\n$$\n\nSince the resulting ratio is one carbon to two oxygen atoms, the empirical formula is $\\mathrm{CO}_{2}$."}
{"id": 2918, "contents": "571. Check Your Learning - \nWhat is the empirical formula of a compound containing $40.0 \\% \\mathrm{C}, 6.71 \\% \\mathrm{H}$, and $53.28 \\% \\mathrm{O}$ ?"}
{"id": 2919, "contents": "572. Answer: - \n$\\mathrm{CH}_{2} \\mathrm{O}$"}
{"id": 2920, "contents": "573. Derivation of Molecular Formulas - \nRecall that empirical formulas are symbols representing the relative numbers of a compound's elements. Determining the absolute numbers of atoms that compose a single molecule of a covalent compound requires knowledge of both its empirical formula and its molecular mass or molar mass. These quantities may be determined experimentally by various measurement techniques. Molecular mass, for example, is often derived from the mass spectrum of the compound (see discussion of this technique in a previous chapter on atoms and molecules). Molar mass can be measured by a number of experimental methods, many of which will be introduced in later chapters of this text.\n\nMolecular formulas are derived by comparing the compound's molecular or molar mass to its empirical formula mass. As the name suggests, an empirical formula mass is the sum of the average atomic masses of all the atoms represented in an empirical formula. If the molecular (or molar) mass of the substance is known, it may be divided by the empirical formula mass to yield the number of empirical formula units per molecule (n):\n\n$$\n\\frac{\\text { molecular or molar mass }\\left(\\mathrm{amu} \\text { or } \\frac{\\mathrm{g}}{\\mathrm{~mol}}\\right)}{\\text { empirical formula mass }\\left(\\mathrm{amu} \\text { or } \\frac{\\mathrm{g}}{\\mathrm{~mol}}\\right)}=n \\text { formula units/molecule }\n$$\n\nThe molecular formula is then obtained by multiplying each subscript in the empirical formula by $n$, as shown by the generic empirical formula $\\mathrm{A}_{\\mathrm{x}} \\mathrm{B}_{\\mathrm{y}}$ :\n\n$$\n\\left(A_{x} B_{y}\\right)_{n}=A_{n x} B_{n y}\n$$\n\nFor example, consider a covalent compound whose empirical formula is determined to be $\\mathrm{CH}_{2} \\mathrm{O}$. The empirical formula mass for this compound is approximately 30 amu (the sum of 12 amu for one C atom, 2 amu for two H atoms, and 16 amu for one 0 atom). If the compound's molecular mass is determined to be 180 amu , this indicates that molecules of this compound contain six times the number of atoms represented in the empirical formula:"}
{"id": 2921, "contents": "573. Derivation of Molecular Formulas - \n$$\n\\frac{180 \\mathrm{amu} / \\mathrm{molecule}}{30 \\frac{\\mathrm{amu}}{\\text { formula unit }}}=6 \\text { formula units } / \\mathrm{molec} \\text {. }\n$$\n\nMolecules of this compound are then represented by molecular formulas whose subscripts are six times greater than those in the empirical formula:\n\n$$\n\\left(\\mathrm{CH}_{2} \\mathrm{O}\\right)_{6}=\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}\n$$\n\nNote that this same approach may be used when the molar mass ( $\\mathrm{g} / \\mathrm{mol}$ ) instead of the molecular mass (amu) is used. In this case, one mole of empirical formula units and molecules is considered, as opposed to single units and molecules."}
{"id": 2922, "contents": "575. Determination of the Molecular Formula for Nicotine - \nNicotine, an alkaloid in the nightshade family of plants that is mainly responsible for the addictive nature of cigarettes, contains $74.02 \\% \\mathrm{C}, 8.710 \\% \\mathrm{H}$, and $17.27 \\% \\mathrm{~N}$. If 40.57 g of nicotine contains 0.2500 mol nicotine, what is the molecular formula?"}
{"id": 2923, "contents": "576. Solution - \nDetermining the molecular formula from the provided data will require comparison of the compound's empirical formula mass to its molar mass. As the first step, use the percent composition to derive the compound's empirical formula. Assuming a convenient, a 100-g sample of nicotine yields the following molar amounts of its elements:\n\n$$\n\\begin{aligned}\n& (74.02 \\mathrm{~g} \\mathrm{C})\\left(\\frac{1 \\mathrm{~mol} \\mathrm{C}}{12.01 \\mathrm{~g} \\mathrm{C}}\\right)=6.163 \\mathrm{~mol} \\mathrm{C} \\\\\n& (8.710 \\mathrm{~g} \\mathrm{H})\\left(\\frac{1 \\mathrm{~mol} \\mathrm{H}}{1.01 \\mathrm{~g} \\mathrm{H}}\\right)=8.624 \\mathrm{~mol} \\mathrm{H} \\\\\n& (17.27 \\mathrm{~g} \\mathrm{~N})\\left(\\frac{1 \\mathrm{~mol} \\mathrm{~N}}{14.01 \\mathrm{~g} \\mathrm{~N}}\\right)=1.233 \\mathrm{~mol} \\mathrm{~N}\n\\end{aligned}\n$$\n\nNext, calculate the molar ratios of these elements relative to the least abundant element, N .\n\n$$\n\\begin{gathered}\n6.163 \\mathrm{~mol} \\mathrm{C} / 1.233 \\mathrm{~mol} \\mathrm{~N}=5 \\\\\n8.264 \\mathrm{~mol} \\mathrm{H} / 1.233 \\mathrm{~mol} \\mathrm{~N}=7 \\\\\n1.233 \\mathrm{~mol} \\mathrm{~N} / 1.233 \\mathrm{~mol} \\mathrm{~N}=1 \\\\\n\\frac{1.233}{1.233}=1.000 \\mathrm{~mol} \\mathrm{~N} \\\\\n\\frac{6.163}{1.233}=4.998 \\mathrm{~mol} \\mathrm{C} \\\\\n\\frac{8.624}{1.233}=6.994 \\mathrm{~mol} \\mathrm{H}\n\\end{gathered}\n$$"}
{"id": 2924, "contents": "576. Solution - \nThe C-to- N and H -to- N molar ratios are adequately close to whole numbers, and so the empirical formula is $\\mathrm{C}_{5} \\mathrm{H}_{7} \\mathrm{~N}$. The empirical formula mass for this compound is therefore $81.13 \\mathrm{amu} /$ formula unit, or $81.13 \\mathrm{~g} / \\mathrm{mol}$ formula unit.\n\nCalculate the molar mass for nicotine from the given mass and molar amount of compound:\n\n$$\n\\frac{40.57 \\mathrm{~g} \\text { nicotine }}{0.2500 \\mathrm{~mol} \\text { nicotine }}=\\frac{162.3 \\mathrm{~g}}{\\mathrm{~mol}}\n$$\n\nComparing the molar mass and empirical formula mass indicates that each nicotine molecule contains two formula units:\n\n$$\n\\frac{162.3 \\mathrm{~g} / \\mathrm{mol}}{81.13 \\frac{\\mathrm{~g}}{\\text { formula unit }}}=2 \\text { formula units } / \\text { molecule }\n$$\n\nFinally, derive the molecular formula for nicotine from the empirical formula by multiplying each subscript by two:\n\n$$\n\\left(\\mathrm{C}_{5} \\mathrm{H}_{7} \\mathrm{~N}\\right)_{2}=\\mathrm{C}_{10} \\mathrm{H}_{14} \\mathrm{~N}_{2}\n$$"}
{"id": 2925, "contents": "577. Check Your Learning - \nWhat is the molecular formula of a compound with a percent composition of $49.47 \\% \\mathrm{C}, 5.201 \\% \\mathrm{H}, 28.84 \\% \\mathrm{~N}$, and $16.48 \\% \\mathrm{O}$, and a molecular mass of 194.2 amu ?"}
{"id": 2926, "contents": "578. Answer: - \n$\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2}$"}
{"id": 2927, "contents": "579. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the fundamental properties of solutions\n- Calculate solution concentrations using molarity\n- Perform dilution calculations using the dilution equation\n\nPreceding sections of this chapter focused on the composition of substances: samples of matter that contain only one type of element or compound. However, mixtures-samples of matter containing two or more substances physically combined-are more commonly encountered in nature than are pure substances. Similar to a pure substance, the relative composition of a mixture plays an important role in determining its properties. The relative amount of oxygen in a planet's atmosphere determines its ability to sustain aerobic\nlife. The relative amounts of iron, carbon, nickel, and other elements in steel (a mixture known as an \"alloy\") determine its physical strength and resistance to corrosion. The relative amount of the active ingredient in a medicine determines its effectiveness in achieving the desired pharmacological effect. The relative amount of sugar in a beverage determines its sweetness (see Figure 6.8). This section will describe one of the most common ways in which the relative compositions of mixtures may be quantified.\n\n\nFIGURE 6.8 Sugar is one of many components in the complex mixture known as coffee. The amount of sugar in a given amount of coffee is an important determinant of the beverage's sweetness. (credit: Jane Whitney)"}
{"id": 2928, "contents": "580. Solutions - \nSolutions have previously been defined as homogeneous mixtures, meaning that the composition of the mixture (and therefore its properties) is uniform throughout its entire volume. Solutions occur frequently in nature and have also been implemented in many forms of manmade technology. A more thorough treatment of solution properties is provided in the chapter on solutions and colloids, but provided here is an introduction to some of the basic properties of solutions.\n\nThe relative amount of a given solution component is known as its concentration. Often, though not always, a solution contains one component with a concentration that is significantly greater than that of all other components. This component is called the solvent and may be viewed as the medium in which the other components are dispersed, or dissolved. Solutions in which water is the solvent are, of course, very common on our planet. A solution in which water is the solvent is called an aqueous solution.\n\nA solute is a component of a solution that is typically present at a much lower concentration than the solvent. Solute concentrations are often described with qualitative terms such as dilute (of relatively low concentration) and concentrated (of relatively high concentration).\n\nConcentrations may be quantitatively assessed using a wide variety of measurement units, each convenient for particular applications. Molarity ( $\\boldsymbol{M}$ ) is a useful concentration unit for many applications in chemistry. Molarity is defined as the number of moles of solute in exactly 1 liter ( 1 L ) of the solution:\n\n$$\nM=\\frac{\\text { mol solute }}{\\mathrm{L} \\text { solution }}\n$$"}
{"id": 2929, "contents": "582. Calculating Molar Concentrations - \nA $355-\\mathrm{mL}$ soft drink sample contains 0.133 mol of sucrose (table sugar). What is the molar concentration of sucrose in the beverage?"}
{"id": 2930, "contents": "583. Solution - \nSince the molar amount of solute and the volume of solution are both given, the molarity can be calculated using the definition of molarity. Per this definition, the solution volume must be converted from mL to L :\n\n$$\nM=\\frac{\\text { mol solute }}{\\mathrm{L} \\text { solution }}=\\frac{0.133 \\mathrm{~mol}}{355 \\mathrm{~mL} \\times \\frac{1 \\mathrm{~L}}{1000 \\mathrm{~mL}}}=0.375 \\mathrm{M}\n$$"}
{"id": 2931, "contents": "584. Check Your Learning - \nA teaspoon of table sugar contains about 0.01 mol sucrose. What is the molarity of sucrose if a teaspoon of sugar has been dissolved in a cup of tea with a volume of 200 mL ?"}
{"id": 2932, "contents": "587. Deriving Moles and Volumes from Molar Concentrations - \nHow much sugar (mol) is contained in a modest sip ( $\\sim 10 \\mathrm{~mL}$ ) of the soft drink from Example 6.8?"}
{"id": 2933, "contents": "588. Solution - \nRearrange the definition of molarity to isolate the quantity sought, moles of sugar, then substitute the value for molarity derived in Example 6.8, 0.375 M :\n\n$$\n\\begin{gathered}\n\\qquad \\begin{array}{c}\nM=\\frac{\\text { mol solute }}{\\mathrm{L} \\text { solution }} \\\\\n\\text { mol solute }\n\\end{array}=M \\times \\mathrm{L} \\text { solution } \\\\\n\\text { mol solute }=0.375 \\frac{\\mathrm{~mol} \\text { sugar }}{\\mathrm{L}} \\times\\left(10 \\mathrm{~mL} \\times \\frac{1 \\mathrm{~L}}{1000 \\mathrm{~mL}}\\right)=0.004 \\mathrm{~mol} \\text { sugar }\n\\end{gathered}\n$$"}
{"id": 2934, "contents": "589. Check Your Learning - \nWhat volume (mL) of the sweetened tea described in Example 6.8 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example?"}
{"id": 2935, "contents": "590. Answer: - \n80 mL"}
{"id": 2936, "contents": "592. Calculating Molar Concentrations from the Mass of Solute - \nDistilled white vinegar (Figure 6.9) is a solution of acetic acid, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$, in water. A 0.500 - L vinegar solution contains 25.2 g of acetic acid. What is the concentration of the acetic acid solution in units of molarity?\n\n\nFIGURE 6.9 Distilled white vinegar is a solution of acetic acid in water."}
{"id": 2937, "contents": "593. Solution - \nAs in previous examples, the definition of molarity is the primary equation used to calculate the quantity sought. Since the mass of solute is provided instead of its molar amount, use the solute's molar mass to obtain the amount of solute in moles:\n\n$$\n\\begin{gathered}\nM=\\frac{\\text { mol solute }}{\\mathrm{L} \\text { solution }=} \\frac{25.2 \\mathrm{~g} \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H} \\times \\frac{1 \\mathrm{~mol} \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}}{60.052 \\mathrm{~g} \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}}}{0.500 \\mathrm{~L} \\text { solution }}=0.839 \\mathrm{M} \\\\\n\\begin{array}{c}\nM=\\frac{\\text { mol solute }}{\\mathrm{L} \\text { solution }}=0.839 \\mathrm{M} \\\\\nM=\\frac{.0839 \\text { mol solute }}{1.00 \\mathrm{~L} \\text { solution }}\n\\end{array}\n\\end{gathered}\n$$"}
{"id": 2938, "contents": "594. Check Your Learning - \nCalculate the molarity of 6.52 g of $\\mathrm{CoCl}_{2}(128.9 \\mathrm{~g} / \\mathrm{mol})$ dissolved in an aqueous solution with a total volume of 75.0 mL ."}
{"id": 2939, "contents": "595. Answer: - \n0.674 M"}
{"id": 2940, "contents": "597. Determining the Mass of Solute in a Given Volume of Solution - \nHow many grams of NaCl are contained in 0.250 L of a $5.30-\\mathrm{M}$ solution?"}
{"id": 2941, "contents": "598. Solution - \nThe volume and molarity of the solution are specified, so the amount (mol) of solute is easily computed as demonstrated in Example 6.9:\n\n$$\n\\begin{gathered}\n\\qquad M=\\frac{\\text { mol solute }}{\\mathrm{L} \\text { solution }} \\\\\n\\text { mol solute }=M \\times \\mathrm{L} \\text { solution } \\\\\n\\text { mol solute }=5.30 \\frac{\\mathrm{~mol} \\mathrm{NaCl}}{\\mathrm{~L}} \\times 0.250 \\mathrm{~L}=1.325 \\mathrm{~mol} \\mathrm{NaCl}\n\\end{gathered}\n$$\n\nFinally, this molar amount is used to derive the mass of NaCl :\n\n$$\n1.325 \\mathrm{~mol} \\mathrm{NaCl} \\times \\frac{58.44 \\mathrm{~g} \\mathrm{NaCl}}{\\mathrm{~mol} \\mathrm{NaCl}}=77.4 \\mathrm{~g} \\mathrm{NaCl}\n$$"}
{"id": 2942, "contents": "599. Check Your Learning - \nHow many grams of $\\mathrm{CaCl}_{2}(110.98 \\mathrm{~g} / \\mathrm{mol})$ are contained in 250.0 mL of a $0.200-M$ solution of calcium chloride?"}
{"id": 2943, "contents": "600. Answer: - \n$5.55 \\mathrm{~g} \\mathrm{CaCl}_{2}$\n\nWhen performing calculations stepwise, as in Example 6.11, it is important to refrain from rounding any intermediate calculation results, which can lead to rounding errors in the final result. In Example 6.11, the molar amount of NaCl computed in the first step, 1.325 mol , would be properly rounded to 1.32 mol if it were to be reported; however, although the last digit (5) is not significant, it must be retained as a guard digit in the intermediate calculation. If the guard digit had not been retained, the final calculation for the mass of NaCl would have been 77.1 g , a difference of 0.3 g .\n\nIn addition to retaining a guard digit for intermediate calculations, rounding errors may also be avoided by performing computations in a single step (see Example 6.12). This eliminates intermediate steps so that only the final result is rounded."}
{"id": 2944, "contents": "602. Determining the Volume of Solution Containing a Given Mass of Solute - \nIn Example 6.10, the concentration of acetic acid in white vinegar was determined to be 0.839 M . What volume of vinegar contains 75.6 g of acetic acid?"}
{"id": 2945, "contents": "603. Solution - \nFirst, use the molar mass to calculate moles of acetic acid from the given mass:\n\n$$\n\\mathrm{g} \\text { solute } \\times \\frac{\\text { mol solute }}{\\mathrm{g} \\text { solute }}=\\text { mol solute }\n$$\n\nThen, use the molarity of the solution to calculate the volume of solution containing this molar amount of solute:\n\n$$\n\\text { mol solute } \\times \\frac{\\mathrm{L} \\text { solution }}{\\text { mol solute }}=\\mathrm{L} \\text { solution }\n$$\n\nCombining these two steps into one yields:\n\n$$\n\\begin{gathered}\n\\mathrm{g} \\text { solute } \\times \\frac{\\text { mol solute }}{\\mathrm{g} \\text { solute }} \\times \\frac{\\mathrm{L} \\text { solution }}{\\text { mol solute }}=\\mathrm{L} \\text { solution } \\\\\n75.6 \\mathrm{~g} \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\left(\\frac{\\mathrm{~mol} \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}}{60.05 \\mathrm{~g}}\\right)\\left(\\frac{\\mathrm{L} \\text { solution }}{0.839 \\mathrm{~mol} \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}}\\right)=1.50 \\mathrm{~L} \\text { solution }\n\\end{gathered}\n$$"}
{"id": 2946, "contents": "604. Check Your Learning - \nWhat volume of a $1.50-M \\mathrm{KBr}$ solution contains 66.0 g KBr ?"}
{"id": 2947, "contents": "605. Answer: - \n0.370 L"}
{"id": 2948, "contents": "606. Dilution of Solutions - \nDilution is the process whereby the concentration of a solution is lessened by the addition of solvent. For example, a glass of iced tea becomes increasingly diluted as the ice melts. The water from the melting ice increases the volume of the solvent (water) and the overall volume of the solution (iced tea), thereby reducing the relative concentrations of the solutes that give the beverage its taste (Figure 6.10).\n\n\nFIGURE 6.10 Both solutions contain the same mass of copper nitrate. The solution on the right is more dilute because the copper nitrate is dissolved in more solvent. (credit: Mark Ott)\n\nDilution is also a common means of preparing solutions of a desired concentration. By adding solvent to a measured portion of a more concentrated stock solution, a solution of lesser concentration may be prepared.\nFor example, commercial pesticides are typically sold as solutions in which the active ingredients are far more concentrated than is appropriate for their application. Before they can be used on crops, the pesticides must be diluted. This is also a very common practice for the preparation of a number of common laboratory reagents.\n\nA simple mathematical relationship can be used to relate the volumes and concentrations of a solution before and after the dilution process. According to the definition of molarity, the number of moles of solute in a solution $(n)$ is equal to the product of the solution's molarity $(M)$ and its volume in liters $(L)$ :\n\n$$\nn=M L\n$$\n\nExpressions like these may be written for a solution before and after it is diluted:\n\n$$\n\\begin{aligned}\n& n_{1}=M_{1} L_{1} \\\\\n& n_{2}=M_{2} L_{2}\n\\end{aligned}\n$$\n\nwhere the subscripts \" 1 \" and \" 2 \" refer to the solution before and after the dilution, respectively. Since the dilution process does not change the amount of solute in the solution, $n_{1}=n_{2}$. Thus, these two equations may be set equal to one another:\n\n$$\nM_{1} L_{1}=M_{2} L_{2}\n$$"}
{"id": 2949, "contents": "606. Dilution of Solutions - \n$$\nM_{1} L_{1}=M_{2} L_{2}\n$$\n\nThis relation is commonly referred to as the dilution equation. Although this equation uses molarity as the unit of concentration and liters as the unit of volume, other units of concentration and volume may be used as long as the units properly cancel per the factor-label method. Reflecting this versatility, the dilution equation is often written in the more general form:\n\n$$\nC_{1} V_{1}=C_{2} V_{2}\n$$\n\nwhere $C$ and $V$ are concentration and volume, respectively."}
{"id": 2950, "contents": "607. LINK TO LEARNING - \nUse the simulation (http://openstax.org/l/16Phetsolvents) to explore the relations between solute amount, solution volume, and concentration and to confirm the dilution equation."}
{"id": 2951, "contents": "609. Determining the Concentration of a Diluted Solution - \nIf 0.850 L of a $5.00-M$ solution of copper nitrate, $\\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}$, is diluted to a volume of 1.80 L by the addition of water, what is the molarity of the diluted solution?"}
{"id": 2952, "contents": "610. Solution - \nThe stock concentration, $C_{1}$, and volume, $V_{1}$, are provided as well as the volume of the diluted solution, $V_{2}$. Rearrange the dilution equation to isolate the unknown property, the concentration of the diluted solution, $C_{2}$ :\n\n$$\n\\begin{gathered}\nC_{1} V_{1}=C_{2} V_{2} \\\\\nC_{2}=\\frac{C_{1} V_{1}}{V_{2}}\n\\end{gathered}\n$$\n\nSince the stock solution is being diluted by more than two-fold (volume is increased from 0.85 L to 1.80 L ), the diluted solution's concentration is expected to be less than one-half 5 M . This ballpark estimate will be compared to the calculated result to check for any gross errors in computation (for example, such as an improper substitution of the given quantities). Substituting the given values for the terms on the right side of this equation yields:\n\n$$\nC_{2}=\\frac{0.850 \\mathrm{~L} \\times 5.00 \\frac{\\mathrm{~mol}}{\\mathrm{~L}}}{1.80 \\mathrm{~L}}=2.36 M\n$$\n\nThis result compares well to our ballpark estimate (it's a bit less than one-half the stock concentration, 5 M )."}
{"id": 2953, "contents": "611. Check Your Learning - \nWhat is the concentration of the solution that results from diluting 25.0 mL of a $2.04-\\mathrm{M}$ solution of $\\mathrm{CH}_{3} \\mathrm{OH}$ to 500.0 mL ?"}
{"id": 2954, "contents": "612. Answer: - \n$0.102 \\mathrm{MCH}_{3} \\mathrm{OH}$"}
{"id": 2955, "contents": "614. Volume of a Diluted Solution - \nWhat volume of 0.12 MHBr can be prepared from $11 \\mathrm{~mL}(0.011 \\mathrm{~L})$ of 0.45 M HBr ?"}
{"id": 2956, "contents": "615. Solution - \nProvided are the volume and concentration of a stock solution, $V_{1}$ and $C_{1}$, and the concentration of the resultant diluted solution, $C_{2}$. Find the volume of the diluted solution, $V_{2}$ by rearranging the dilution equation to isolate $V_{2}$ :\n\n$$\n\\begin{gathered}\nC_{1} V_{1}=C_{2} V_{2} \\\\\nV_{2}=\\frac{C_{1} V_{1}}{C_{2}}\n\\end{gathered}\n$$\n\nSince the diluted concentration $(0.12 M)$ is slightly more than one-fourth the original concentration $(0.45 M)$,\nthe volume of the diluted solution is expected to be roughly four times the original volume, or around 44 mL . Substituting the given values and solving for the unknown volume yields:\n\n$$\n\\begin{gathered}\nV_{2}=\\frac{(0.45 M)(0.011 \\mathrm{~L})}{(0.12 M)} \\\\\nV_{2}=0.041 \\mathrm{~L}\n\\end{gathered}\n$$\n\nThe volume of the $0.12-M$ solution is $0.041 \\mathrm{~L}(41 \\mathrm{~mL})$. The result is reasonable and compares well with the rough estimate."}
{"id": 2957, "contents": "616. Check Your Learning - \nA laboratory experiment calls for $0.125 \\mathrm{M} \\mathrm{HNO}_{3}$. What volume of $0.125 \\mathrm{M} \\mathrm{HNO}_{3}$ can be prepared from 0.250 L of $1.88 \\mathrm{M} \\mathrm{HNO}_{3}$ ?"}
{"id": 2958, "contents": "617. Answer: - \n3.76 L"}
{"id": 2959, "contents": "619. Volume of a Concentrated Solution Needed for Dilution - \nWhat volume of 1.59 M KOH is required to prepare 5.00 L of 0.100 M KOH ?"}
{"id": 2960, "contents": "620. Solution - \nGiven are the concentration of a stock solution, $C_{1}$, and the volume and concentration of the resultant diluted solution, $V_{2}$ and $C_{2}$. Find the volume of the stock solution, $V_{1}$ by rearranging the dilution equation to isolate $V_{1}$ :\n\n$$\n\\begin{gathered}\nC_{1} V_{1}=C_{2} V_{2} \\\\\nV_{1}=\\frac{C_{2} V_{2}}{C_{1}}\n\\end{gathered}\n$$\n\nSince the concentration of the diluted solution $0.100 M$ is roughly one-sixteenth that of the stock solution (1.59 $M$ ), the volume of the stock solution is expected to be about one-sixteenth that of the diluted solution, or around 0.3 liters. Substituting the given values and solving for the unknown volume yields:\n\n$$\n\\begin{gathered}\nV_{1}=\\frac{(0.100 M)(5.00 \\mathrm{~L})}{1.59 M} \\\\\nV_{1}=0.314 \\mathrm{~L}\n\\end{gathered}\n$$\n\nThus, 0.314 L of the $1.59-M$ solution is needed to prepare the desired solution. This result is consistent with the rough estimate."}
{"id": 2961, "contents": "621. Check Your Learning - \nWhat volume of a $0.575-M$ solution of glucose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$, can be prepared from 50.00 mL of a $3.00-\\mathrm{M}$ glucose solution?"}
{"id": 2962, "contents": "622. Answer: - \n0.261 L"}
{"id": 2963, "contents": "623. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Define the concentration units of mass percentage, volume percentage, mass-volume percentage, parts-permillion (ppm), and parts-per-billion (ppb)\n- Perform computations relating a solution's concentration and its components' volumes and/or masses using these units\n\nThe previous section introduced molarity, a very useful measurement unit for evaluating the concentration of solutions. However, molarity is only one measure of concentration. This section will describe some other units of concentration that are commonly used in various applications, either for convenience or by convention."}
{"id": 2964, "contents": "624. Mass Percentage - \nEarlier in this chapter, percent composition was introduced as a measure of the relative amount of a given element in a compound. Percentages are also commonly used to express the composition of mixtures, including solutions. The mass percentage of a solution component is defined as the ratio of the component's mass to the solution's mass, expressed as a percentage:\n\n$$\n\\text { mass percentage }=\\frac{\\text { mass of component }}{\\text { mass of solution }} \\times 100 \\%\n$$\n\nMass percentage is also referred to by similar names such as percent mass, percent weight, weight/weight percent, and other variations on this theme. The most common symbol for mass percentage is simply the percent sign, \\%, although more detailed symbols are often used including \\%mass, \\%weight, and (w/w)\\%. Use of these more detailed symbols can prevent confusion of mass percentages with other types of percentages, such as volume percentages (to be discussed later in this section).\nMass percentages are popular concentration units for consumer products. The label of a typical liquid bleach bottle (Figure 6.11) cites the concentration of its active ingredient, sodium hypochlorite ( NaOCl ), as being $7.4 \\%$. A 100.0-g sample of bleach would therefore contain 7.4 g of NaOCl .\n\n\nFIGURE 6.11 Liquid bleach is an aqueous solution of sodium hypochlorite ( NaOCl ). This brand has a concentration of $7.4 \\% \\mathrm{NaOCl}$ by mass."}
{"id": 2965, "contents": "625. Calculation of Percent by Mass - \nA $5.0-\\mathrm{g}$ sample of spinal fluid contains $3.75 \\mathrm{mg}(0.00375 \\mathrm{~g})$ of glucose. What is the percent by mass of glucose in spinal fluid?"}
{"id": 2966, "contents": "626. Solution - \nThe spinal fluid sample contains roughly 4 mg of glucose in 5000 mg of fluid, so the mass fraction of glucose should be a bit less than one part in 1000, or about $0.1 \\%$. Substituting the given masses into the equation defining mass percentage yields:\n\n$$\n\\% \\text { glucose }=\\frac{3.75 \\mathrm{mg} \\text { glucose } \\times \\frac{1 \\mathrm{~g}}{1000 \\mathrm{mg}}}{5.0 \\mathrm{~g} \\text { spinal fluid }}=0.075 \\%\n$$\n\nThe computed mass percentage agrees with our rough estimate (it's a bit less than 0.1\\%).\nNote that while any mass unit may be used to compute a mass percentage ( $\\mathrm{mg}, \\mathrm{g}, \\mathrm{kg}, \\mathrm{oz}$, and so on), the same unit must be used for both the solute and the solution so that the mass units cancel, yielding a dimensionless ratio. In this case, the solute mass unit in the numerator was converted from mg to g to match the units in the denominator. Alternatively, the spinal fluid mass unit in the denominator could have been converted from g to mg instead. As long as identical mass units are used for both solute and solution, the computed mass percentage will be correct."}
{"id": 2967, "contents": "627. Check Your Learning - \nA bottle of a tile cleanser contains 135 g of HCl and 775 g of water. What is the percent by mass of HCl in this cleanser?"}
{"id": 2968, "contents": "628. Answer: - \n14.8\\%"}
{"id": 2969, "contents": "630. Calculations using Mass Percentage - \n\"Concentrated\" hydrochloric acid is an aqueous solution of $37.2 \\% \\mathrm{HCl}$ that is commonly used as a laboratory reagent. The density of this solution is $1.19 \\mathrm{~g} / \\mathrm{mL}$. What mass of HCl is contained in 0.500 L of this solution?"}
{"id": 2970, "contents": "631. Solution - \nThe HCl concentration is near $40 \\%$, so a 100-g portion of this solution would contain about 40 g of HCl . Since the solution density isn't greatly different from that of water ( $1 \\mathrm{~g} / \\mathrm{mL}$ ), a reasonable estimate of the HCl mass in $500 \\mathrm{~g}(0.5 \\mathrm{~L})$ of the solution is about five times greater than that in a 100 g portion, or $5 \\times 40=200 \\mathrm{~g}$. In order to derive the mass of solute in a solution from its mass percentage, the mass of the solution must be known. Using the solution density given, convert the solution's volume to mass, and then use the given mass percentage to calculate the solute mass. This mathematical approach is outlined in this flowchart:\n\n\nFor proper unit cancellation, the $0.500-\\mathrm{L}$ volume is converted into 500 mL , and the mass percentage is expressed as a ratio, $37.2 \\mathrm{~g} \\mathrm{HCl} / \\mathrm{g}$ solution:\n\n$$\n500 \\mathrm{~mL} \\text { solution }\\left(\\frac{1.19 \\mathrm{~g} \\text { solution }}{\\mathrm{mL} \\text { solution }}\\right)\\left(\\frac{37.2 \\mathrm{~g} \\mathrm{HCl}}{100 \\mathrm{~g} \\text { solution }}\\right)=221 \\mathrm{~g} \\mathrm{HCl}\n$$\n\nThis mass of HCl is consistent with our rough estimate of approximately 200 g .\nCheck Your Learning\nWhat volume of concentrated HCl solution contains 125 g of HCl ?"}
{"id": 2971, "contents": "632. Answer: - \n282 mL"}
{"id": 2972, "contents": "633. Volume Percentage - \nLiquid volumes over a wide range of magnitudes are conveniently measured using common and relatively inexpensive laboratory equipment. The concentration of a solution formed by dissolving a liquid solute in a liquid solvent is therefore often expressed as a volume percentage, $\\% \\mathrm{vol}$ or $(\\mathrm{v} / \\mathrm{v}) \\%$ :\n\n$$\n\\text { volume percentage }=\\frac{\\text { volume solute }}{\\text { volume solution }} \\times 100 \\%\n$$"}
{"id": 2973, "contents": "635. Calculations using Volume Percentage - \nRubbing alcohol (isopropanol) is usually sold as a 70\\%vol aqueous solution. If the density of isopropyl alcohol is $0.785 \\mathrm{~g} / \\mathrm{mL}$, how many grams of isopropyl alcohol are present in a 355 mL bottle of rubbing alcohol?"}
{"id": 2974, "contents": "636. Solution - \nPer the definition of volume percentage, the isopropanol volume is $70 \\%$ of the total solution volume. Multiplying the isopropanol volume by its density yields the requested mass:\n$\\left(355 \\mathrm{~mL}\\right.$ solution) $\\left(\\frac{70 \\mathrm{~mL} \\text { isopropyl alcohol }}{100 \\mathrm{~mL} \\text { solution }}\\right)\\left(\\frac{0.785 \\mathrm{~g} \\text { isopropyl alcohol }}{1 \\mathrm{~mL} \\text { isopropyl alcohol }}\\right)=195 \\mathrm{~g}$ isopropyl alchol"}
{"id": 2975, "contents": "637. Check Your Learning - \nWine is approximately $12 \\%$ ethanol $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}\\right)$ by volume. Ethanol has a molar mass of $46.06 \\mathrm{~g} / \\mathrm{mol}$ and a density $0.789 \\mathrm{~g} / \\mathrm{mL}$. How many moles of ethanol are present in a $750-\\mathrm{mL}$ bottle of wine?"}
{"id": 2976, "contents": "638. Answer: - \n1.5 mol ethanol"}
{"id": 2977, "contents": "639. Mass-Volume Percentage - \n\"Mixed\" percentage units, derived from the mass of solute and the volume of solution, are popular for certain biochemical and medical applications. A mass-volume percent is a ratio of a solute's mass to the solution's volume expressed as a percentage. The specific units used for solute mass and solution volume may vary, depending on the solution. For example, physiological saline solution, used to prepare intravenous fluids, has a concentration of $0.9 \\%$ mass $/$ volume ( $\\mathrm{m} / \\mathrm{v}$ ), indicating that the composition is 0.9 g of solute per 100 mL of solution. The concentration of glucose in blood (commonly referred to as \"blood sugar\") is also typically expressed in terms of a mass-volume ratio. Though not expressed explicitly as a percentage, its concentration is usually given in milligrams of glucose per deciliter ( 100 mL ) of blood (Figure 6.12).\n\n\nFIGURE 6.12 \"Mixed\" mass-volume units are commonly encountered in medical settings. (a) The NaCl concentration of physiological saline is $0.9 \\%(\\mathrm{~m} / \\mathrm{v})$. (b) This device measures glucose levels in a sample of blood. The normal range for glucose concentration in blood (fasting) is around $70-100 \\mathrm{mg} / \\mathrm{dL}$. (credit a: modification of work by \"The National Guard\"/Flickr; credit b: modification of work by Biswarup Ganguly)"}
{"id": 2978, "contents": "640. Parts per Million and Parts per Billion - \nVery low solute concentrations are often expressed using appropriately small units such as parts per million (ppm) or parts per billion (ppb). Like percentage (\"part per hundred\") units, ppm and ppb may be defined in terms of masses, volumes, or mixed mass-volume units. There are also ppm and ppb units defined with respect to numbers of atoms and molecules.\n\nThe mass-based definitions of ppm and ppb are given here:\n\n$$\n\\begin{aligned}\n& \\mathrm{ppm}=\\frac{\\text { mass solute }}{\\text { mass solution }} \\times 10^{6} \\mathrm{ppm} \\\\\n& \\mathrm{ppb}=\\frac{\\text { mass solute }}{\\text { mass solution }} \\times 10^{9} \\mathrm{ppb}\n\\end{aligned}\n$$\n\nBoth ppm and ppb are convenient units for reporting the concentrations of pollutants and other trace contaminants in water. Concentrations of these contaminants are typically very low in treated and natural waters, and their levels cannot exceed relatively low concentration thresholds without causing adverse effects on health and wildlife. For example, the EPA has identified the maximum safe level of fluoride ion in tap water to be 4 ppm . Inline water filters are designed to reduce the concentration of fluoride and several other tracelevel contaminants in tap water (Figure 6.13).\n\n\nFIGURE 6.13 (a) In some areas, trace-level concentrations of contaminants can render unfiltered tap water unsafe for drinking and cooking. (b) Inline water filters reduce the concentration of solutes in tap water. (credit a: modification of work by Jenn Durfey; credit b: modification of work by \"vastateparkstaff\"/Wikimedia commons)"}
{"id": 2979, "contents": "642. Calculation of Parts per Million and Parts per Billion Concentrations - \nAccording to the EPA, when the concentration of lead in tap water reaches 15 ppb , certain remedial actions must be taken. What is this concentration in ppm? At this concentration, what mass of lead ( $\\mu \\mathrm{g}$ ) would be contained in a typical glass of water ( 300 mL )?"}
{"id": 2980, "contents": "643. Solution - \nThe definitions of the ppm and ppb units may be used to convert the given concentration from ppb to ppm . Comparing these two unit definitions shows that ppm is 1000 times greater than $\\mathrm{ppb}\\left(1 \\mathrm{ppm}=10^{3} \\mathrm{ppb}\\right)$. Thus:\n\n$$\n15 \\mathrm{ppb} \\times \\frac{1 \\mathrm{ppm}}{10^{3} \\mathrm{ppb}}=0.015 \\mathrm{ppm}\n$$\n\nThe definition of the ppb unit may be used to calculate the requested mass if the mass of the solution is provided. Since the volume of solution ( 300 mL ) is given, its density must be used to derive the corresponding mass. Assume the density of tap water to be roughly the same as that of pure water ( $\\sim 1.00 \\mathrm{~g} / \\mathrm{mL}$ ), since the concentrations of any dissolved substances should not be very large. Rearranging the equation defining the ppb unit and substituting the given quantities yields:\n\n$$\n\\begin{gathered}\n\\mathrm{ppb}=\\frac{\\text { mass solute }}{\\text { mass solution }} \\times 10^{9} \\mathrm{ppb} \\\\\n\\text { mass solute }=\\frac{\\mathrm{ppb} \\times \\text { mass solution }}{10^{9} \\mathrm{ppb}} \\\\\n\\text { mass solute }=\\frac{15 \\mathrm{ppb} \\times 300 \\mathrm{~mL} \\times \\frac{1.00 \\mathrm{~g}}{\\mathrm{~mL}}}{10^{9} \\mathrm{ppb}}=4.5 \\times 10^{-6} \\mathrm{~g}\n\\end{gathered}\n$$\n\nFinally, convert this mass to the requested unit of micrograms:\n\n$$\n4.5 \\times 10^{-6} \\mathrm{~g} \\times \\frac{1 \\mu \\mathrm{~g}}{10^{-6} \\mathrm{~g}}=4.5 \\mu \\mathrm{~g}\n$$"}
{"id": 2981, "contents": "644. Check Your Learning - \nA 50.0-g sample of industrial wastewater was determined to contain 0.48 mg of mercury. Express the mercury concentration of the wastewater in ppm and ppb units."}
{"id": 2982, "contents": "645. Answer: - \n9.6 ppm, 9600 ppb"}
{"id": 2983, "contents": "646. Key Terms - \naqueous solution solution for which water is the solvent\nAvogadro's number ( $\\boldsymbol{N}_{\\boldsymbol{A}}$ ) experimentally determined value of the number of entities comprising 1 mole of substance, equal to $6.022 \\times$ $10^{23} \\mathrm{~mol}^{-1}$\nconcentrated qualitative term for a solution containing solute at a relatively high concentration\nconcentration quantitative measure of the relative amounts of solute and solvent present in a solution\ndilute qualitative term for a solution containing solute at a relatively low concentration\ndilution process of adding solvent to a solution in order to lower the concentration of solutes\ndissolved describes the process by which solute components are dispersed in a solvent\nempirical formula mass sum of average atomic masses for all atoms represented in an empirical formula\nformula mass sum of the average masses for all atoms represented in a chemical formula; for covalent compounds, this is also the molecular mass\nmass percentage ratio of solute-to-solution mass\nexpressed as a percentage\nmass-volume percent ratio of solute mass to solution volume, expressed as a percentage\nmolar mass mass in grams of 1 mole of a substance\nmolarity (M) unit of concentration, defined as the number of moles of solute dissolved in 1 liter of solution\nmole amount of substance containing the same number of atoms, molecules, ions, or other entities as the number of atoms in exactly 12 grams of ${ }^{12} \\mathrm{C}$\nparts per billion (ppb) ratio of solute-to-solution mass multiplied by $10^{9}$\nparts per million (ppm) ratio of solute-to-solution mass multiplied by $10^{6}$\npercent composition percentage by mass of the various elements in a compound\nsolute solution component present in a concentration less than that of the solvent\nsolvent solution component present in a concentration that is higher relative to other components\nvolume percentage ratio of solute-to-solution volume expressed as a percentage"}
{"id": 2984, "contents": "647. Key Equations - \n$\\% \\mathrm{X}=\\frac{\\text { mass } \\mathrm{X}}{\\text { mass compound }} \\times 100 \\%$\n$\\frac{\\text { molecular or molar mass }\\left(\\mathrm{amu} \\text { or } \\frac{\\mathrm{g}}{\\mathrm{mol}}\\right)}{\\text { empirical formula mass }\\left(\\mathrm{amu} \\text { or } \\frac{\\mathrm{g}}{\\mathrm{mol}}\\right)}=n$ formula units/molecule\n$\\left(\\mathrm{A}_{\\mathrm{x}} \\mathrm{B}_{\\mathrm{y}}\\right)_{\\mathrm{n}}=\\mathrm{A}_{\\mathrm{nx}} \\mathrm{B}_{\\mathrm{ny}}$\n$M=\\frac{\\text { mol solute }}{\\mathrm{L} \\text { solution }}$\n$C_{1} V_{1}=C_{2} V_{2}$\nPercent by mass $=\\frac{\\text { mass of solute }}{\\text { mass of solution }} \\times 100$\n$\\mathrm{ppm}=\\frac{\\text { mass solute }}{\\text { mass solution }} \\times 10^{6} \\mathrm{ppm}$\n$\\mathrm{ppb}=\\frac{\\text { mass solute }}{\\text { mass solution }} \\times 10^{9} \\mathrm{ppb}$"}
{"id": 2985, "contents": "649. 1 Formula Mass - \nThe formula mass of a substance is the sum of the average atomic masses of each atom represented in the chemical formula and is expressed in atomic mass units. The formula mass of a covalent compound is also called the molecular mass. Due to the use of the same reference substance in defining the atomic mass unit and the mole, the formula\nmass (amu) and molar mass ( $\\mathrm{g} / \\mathrm{mol}$ ) for any substance are numerically equivalent (for example, one $\\mathrm{H}_{2} \\mathrm{O}$ molecule weighs approximately18 amu and 1 mole of $\\mathrm{H}_{2} \\mathrm{O}$ molecules weighs approximately 18 g)."}
{"id": 2986, "contents": "650. Formulas - \nThe chemical identity of a substance is defined by the types and relative numbers of atoms composing its fundamental entities (molecules in the case of covalent compounds, ions in the case of ionic compounds). A compound's percent composition provides the mass percentage of each element in the compound, and it is often experimentally determined and used to derive the compound's empirical formula. The empirical formula mass of a covalent compound may be compared to the compound's molecular or molar mass to derive a molecular formula."}
{"id": 2987, "contents": "650. Formulas - 650.1. Molarity\nSolutions are homogeneous mixtures. Many solutions contain one component, called the solvent, in which other components, called solutes, are dissolved. An aqueous solution is one for which the solvent is water. The concentration of a solution is a measure of the relative amount of solute in a given\namount of solution. Concentrations may be measured using various units, with one very useful unit being molarity, defined as the number of moles of solute per liter of solution. The solute concentration of a solution may be decreased by adding solvent, a process referred to as dilution. The dilution equation is a simple relation between concentrations and volumes of a solution before and after dilution."}
{"id": 2988, "contents": "650. Formulas - 650.2. Other Units for Solution Concentrations\nIn addition to molarity, a number of other solution concentration units are used in various applications. Percentage concentrations based on the solution components' masses, volumes, or both are useful for expressing relatively high concentrations, whereas lower concentrations are conveniently expressed using ppm or ppb units. These units are popular in environmental, medical, and other fields where mole-based units such as molarity are not as commonly used."}
{"id": 2989, "contents": "651. Exercises - 651.1. Formula Mass\n1. What is the total mass (amu) of carbon in each of the following molecules?\n(a) $\\mathrm{CH}_{4}$\n(b) $\\mathrm{CHCl}_{3}$\n(c) $\\mathrm{C}_{12} \\mathrm{H}_{10} \\mathrm{O}_{6}$\n(d) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$\n2. What is the total mass of hydrogen in each of the molecules?\n(a) $\\mathrm{CH}_{4}$\n(b) $\\mathrm{CHCl}_{3}$\n(c) $\\mathrm{C}_{12} \\mathrm{H}_{10} \\mathrm{O}_{6}$\n(d) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$\n3. Calculate the molecular or formula mass of each of the following:\n(a) $\\mathrm{P}_{4}$\n(b) $\\mathrm{H}_{2} \\mathrm{O}$\n(c) $\\mathrm{Ca}\\left(\\mathrm{NO}_{3}\\right)_{2}$\n(d) $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$ (acetic acid)\n(e) $\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}$ (sucrose, cane sugar)\n4. Determine the molecular mass of the following compounds:\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n5. Determine the molecular mass of the following compounds:\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n6. Which molecule has a molecular mass of 28.05 amu ?\n(a)\n\n(b)\n\n(c)"}
{"id": 2990, "contents": "651. Exercises - 651.2. Determining Empirical and Molecular Formulas\n7. What information is needed to determine the molecular formula of a compound from the empirical formula?\n8. Calculate the following to four significant figures:\n(a) the percent composition of ammonia, $\\mathrm{NH}_{3}$\n(b) the percent composition of photographic fixer solution \"hypo,\" $\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$\n(c) the percent of calcium ion in $\\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$\n9. Determine the following to four significant figures:\n(a) the percent composition of hydrazoic acid, $\\mathrm{HN}_{3}$\n(b) the percent composition of TNT, $\\mathrm{C}_{6} \\mathrm{H}_{2}\\left(\\mathrm{CH}_{3}\\right)\\left(\\mathrm{NO}_{2}\\right)_{3}$\n(c) the percent of $\\mathrm{SO}_{4}{ }^{2-}$ in $\\mathrm{Al}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}$\n10. Determine the percent ammonia, $\\mathrm{NH}_{3}$, in $\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6} \\mathrm{Cl}_{3}$, to three significant figures.\n11. Determine the percent water in $\\mathrm{CuSO}_{4} \\bullet 5 \\mathrm{H}_{2} \\mathrm{O}$ to three significant figures.\n12. Determine the empirical formulas for compounds with the following percent compositions:\n(a) $15.8 \\%$ carbon and $84.2 \\%$ sulfur\n(b) $40.0 \\%$ carbon, $6.7 \\%$ hydrogen, and $53.3 \\%$ oxygen\n13. Determine the empirical formulas for compounds with the following percent compositions:\n(a) $43.6 \\%$ phosphorus and $56.4 \\%$ oxygen\n(b) $28.7 \\% \\mathrm{~K}, 1.5 \\% \\mathrm{H}, 22.8 \\% \\mathrm{P}$, and $47.0 \\% \\mathrm{O}$"}
{"id": 2991, "contents": "651. Exercises - 651.2. Determining Empirical and Molecular Formulas\n(b) $28.7 \\% \\mathrm{~K}, 1.5 \\% \\mathrm{H}, 22.8 \\% \\mathrm{P}$, and $47.0 \\% \\mathrm{O}$\n14. A compound of carbon and hydrogen contains $92.3 \\% \\mathrm{C}$ and has a molar mass of $78.1 \\mathrm{~g} / \\mathrm{mol}$. What is its molecular formula?\n15. Dichloroethane, a compound that is often used for dry cleaning, contains carbon, hydrogen, and chlorine. It has a molar mass of $99 \\mathrm{~g} / \\mathrm{mol}$. Analysis of a sample shows that it contains $24.3 \\%$ carbon and $4.1 \\%$ hydrogen. What is its molecular formula?\n16. Determine the empirical and molecular formula for chrysotile asbestos. Chrysotile has the following percent composition: $28.03 \\% \\mathrm{Mg}, 21.60 \\% \\mathrm{Si}, 1.16 \\% \\mathrm{H}$, and $49.21 \\% \\mathrm{O}$. The molar mass for chrysotile is $520.8 \\mathrm{~g} / \\mathrm{mol}$.\n17. Polymers are large molecules composed of simple units repeated many times. Thus, they often have relatively simple empirical formulas. Calculate the empirical formulas of the following polymers:\n(a) Lucite (Plexiglas); $59.9 \\% \\mathrm{C}, 8.06 \\% \\mathrm{H}, 32.0 \\% \\mathrm{O}$\n(b) Saran; $24.8 \\%$ C, $2.0 \\%$ H, $73.1 \\%$ Cl\n(c) polyethylene; $86 \\% \\mathrm{C}, 14 \\% \\mathrm{H}$\n(d) polystyrene; $92.3 \\% \\mathrm{C}, 7.7 \\% \\mathrm{H}$\n(e) Orlon; $67.9 \\%$ C, $5.70 \\% \\mathrm{H}, 26.4 \\% \\mathrm{~N}$"}
{"id": 2992, "contents": "651. Exercises - 651.2. Determining Empirical and Molecular Formulas\n(e) Orlon; $67.9 \\%$ C, $5.70 \\% \\mathrm{H}, 26.4 \\% \\mathrm{~N}$\n18. A major textile dye manufacturer developed a new yellow dye. The dye has a percent composition of $75.95 \\% \\mathrm{C}, 17.72 \\% \\mathrm{~N}$, and $6.33 \\% \\mathrm{H}$ by mass with a molar mass of about $240 \\mathrm{~g} / \\mathrm{mol}$. Determine the molecular formula of the dye."}
{"id": 2993, "contents": "651. Exercises - 651.3. Molarity\n19. Explain what changes and what stays the same when 1.00 L of a solution of NaCl is diluted to 1.80 L .\n20. What information is needed to calculate the molarity of a sulfuric acid solution?\n21. A $200-\\mathrm{mL}$ sample and a $400-\\mathrm{mL}$ sample of a solution of salt have the same molarity. In what ways are the two samples identical? In what ways are these two samples different?\n22. Determine the molarity for each of the following solutions:\n(a) 0.444 mol of $\\mathrm{CoCl}_{2}$ in 0.654 L of solution\n(b) 98.0 g of phosphoric acid, $\\mathrm{H}_{3} \\mathrm{PO}_{4}$, in 1.00 L of solution\n(c) 0.2074 g of calcium hydroxide, $\\mathrm{Ca}(\\mathrm{OH})_{2}$, in 40.00 mL of solution\n(d) 10.5 kg of $\\mathrm{Na}_{2} \\mathrm{SO}_{4} \\cdot 10 \\mathrm{H}_{2} \\mathrm{O}$ in 18.60 L of solution\n(e) $7.0 \\times 10^{-3} \\mathrm{~mol}$ of $\\mathrm{I}_{2}$ in 100.0 mL of solution\n(f) $1.8 \\times 10^{4} \\mathrm{mg}$ of HCl in 0.075 L of solution\n23. Determine the molarity of each of the following solutions:\n(a) 1.457 mol KCl in 1.500 L of solution\n(b) 0.515 g of $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ in 1.00 L of solution\n(c) 20.54 g of $\\mathrm{Al}\\left(\\mathrm{NO}_{3}\\right)_{3}$ in 1575 mL of solution\n(d) 2.76 kg of $\\mathrm{CuSO}_{4} \\cdot 5 \\mathrm{H}_{2} \\mathrm{O}$ in 1.45 L of solution\n(e) 0.005653 mol of $\\mathrm{Br}_{2}$ in 10.00 mL of solution"}
{"id": 2994, "contents": "651. Exercises - 651.3. Molarity\n(e) 0.005653 mol of $\\mathrm{Br}_{2}$ in 10.00 mL of solution\n(f) 0.000889 g of glycine, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{NO}_{2}$, in 1.05 mL of solution\n24. Consider this question: What is the mass of the solute in 0.500 L of 0.30 M glucose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$, used for intravenous injection?\n(a) Outline the steps necessary to answer the question.\n(b) Answer the question.\n25. Consider this question: What is the mass of solute in 200.0 L of a $1.556-M$ solution of KBr ?\n(a) Outline the steps necessary to answer the question.\n(b) Answer the question.\n26. Calculate the number of moles and the mass of the solute in each of the following solutions:\n(a) 2.00 L of $18.5 \\mathrm{M} \\mathrm{H}_{2} \\mathrm{SO}_{4}$, concentrated sulfuric acid\n(b) 100.0 mL of $3.8 \\times 10^{-6} \\mathrm{M} \\mathrm{NaCN}$, the minimum lethal concentration of sodium cyanide in blood serum\n(c) 5.50 L of $13.3 \\mathrm{MH}_{2} \\mathrm{CO}$, the formaldehyde used to \"fix\" tissue samples\n(d) 325 mL of $1.8 \\times 10^{-6} \\mathrm{M} \\mathrm{FeSO}_{4}$, the minimum concentration of iron sulfate detectable by taste in drinking water\n27. Calculate the number of moles and the mass of the solute in each of the following solutions:\n(a) 325 mL of $8.23 \\times 10^{-5} \\mathrm{M} \\mathrm{KI}$, a source of iodine in the diet\n(b) 75.0 mL of $2.2 \\times 10^{-5} \\mathrm{MH}_{2} \\mathrm{SO}_{4}$, a sample of acid rain"}
{"id": 2995, "contents": "651. Exercises - 651.3. Molarity\n(b) 75.0 mL of $2.2 \\times 10^{-5} \\mathrm{MH}_{2} \\mathrm{SO}_{4}$, a sample of acid rain\n(c) 0.2500 L of $0.1135 \\mathrm{M}_{2} \\mathrm{CrO}_{4}$, an analytical reagent used in iron assays\n(d) 10.5 L of $3.716 \\mathrm{M}\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{4}$, a liquid fertilizer\n28. Consider this question: What is the molarity of $\\mathrm{KMnO}_{4}$ in a solution of 0.0908 g of $\\mathrm{KMnO}_{4}$ in 0.500 L of solution?\n(a) Outline the steps necessary to answer the question.\n(b) Answer the question.\n29. Consider this question: What is the molarity of HCl if 35.23 mL of a solution of HCl contain 0.3366 g of HCl ?\n(a) Outline the steps necessary to answer the question.\n(b) Answer the question.\n30. Calculate the molarity of each of the following solutions:\n(a) 0.195 g of cholesterol, $\\mathrm{C}_{27} \\mathrm{H}_{46} \\mathrm{O}$, in 0.100 L of serum, the average concentration of cholesterol in human serum\n(b) 4.25 g of $\\mathrm{NH}_{3}$ in 0.500 L of solution, the concentration of $\\mathrm{NH}_{3}$ in household ammonia\n(c) 1.49 kg of isopropyl alcohol, $\\mathrm{C}_{3} \\mathrm{H}_{7} \\mathrm{OH}$, in 2.50 L of solution, the concentration of isopropyl alcohol in rubbing alcohol\n(d) 0.029 g of $\\mathrm{I}_{2}$ in 0.100 L of solution, the solubility of $\\mathrm{I}_{2}$ in water at $20^{\\circ} \\mathrm{C}$\n31. Calculate the molarity of each of the following solutions:\n(a) 293 g HCl in 666 mL of solution, a concentrated HCl solution"}
{"id": 2996, "contents": "651. Exercises - 651.3. Molarity\n31. Calculate the molarity of each of the following solutions:\n(a) 293 g HCl in 666 mL of solution, a concentrated HCl solution\n(b) $2.026 \\mathrm{~g} \\mathrm{FeCl}_{3}$ in 0.1250 L of a solution used as an unknown in general chemistry laboratories\n(c) $0.001 \\mathrm{mg} \\mathrm{Cd}^{2+}$ in 0.100 L , the maximum permissible concentration of cadmium in drinking water\n(d) $0.0079 \\mathrm{~g} \\mathrm{C}_{7} \\mathrm{H}_{5} \\mathrm{SNO}_{3}$ in one ounce ( 29.6 mL ), the concentration of saccharin in a diet soft drink.\n32. There is about 1.0 g of calcium, as $\\mathrm{Ca}^{2+}$, in 1.0 L of milk. What is the molarity of $\\mathrm{Ca}^{2+}$ in milk?\n33. What volume of a $1.00-M \\mathrm{Fe}\\left(\\mathrm{NO}_{3}\\right)_{3}$ solution can be diluted to prepare 1.00 L of a solution with a concentration of 0.250 M ?\n34. If 0.1718 L of a $0.3556-M \\mathrm{C}_{3} \\mathrm{H}_{7} \\mathrm{OH}$ solution is diluted to a concentration of 0.1222 M , what is the volume of the resulting solution?\n35. If 4.12 L of a $0.850 \\mathrm{M}-\\mathrm{H}_{3} \\mathrm{PO}_{4}$ solution is be diluted to a volume of 10.00 L , what is the concentration of the resulting solution?\n36. What volume of a $0.33-M \\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}$ solution can be diluted to prepare 25 mL of a solution with a concentration of 0.025 M ?\n37. What is the concentration of the NaCl solution that results when 0.150 L of a $0.556-\\mathrm{M}$ solution is allowed to evaporate until the volume is reduced to 0.105 L ?"}
{"id": 2997, "contents": "651. Exercises - 651.3. Molarity\n37. What is the concentration of the NaCl solution that results when 0.150 L of a $0.556-\\mathrm{M}$ solution is allowed to evaporate until the volume is reduced to 0.105 L ?\n38. What is the molarity of the diluted solution when each of the following solutions is diluted to the given final volume?\n(a) 1.00 L of a $0.250-\\mathrm{M}$ solution of $\\mathrm{Fe}\\left(\\mathrm{NO}_{3}\\right)_{3}$ is diluted to a final volume of 2.00 L\n(b) 0.5000 L of a $0.1222-\\mathrm{M}$ solution of $\\mathrm{C}_{3} \\mathrm{H}_{7} \\mathrm{OH}$ is diluted to a final volume of 1.250 L\n(c) 2.35 L of a $0.350-M$ solution of $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ is diluted to a final volume of 4.00 L\n(d) 22.50 mL of a $0.025-M$ solution of $\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}$ is diluted to 100.0 mL\n39. What is the final concentration of the solution produced when 225.5 mL of a $0.09988-M$ solution of $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ is allowed to evaporate until the solution volume is reduced to 45.00 mL ?\n40. A $2.00-\\mathrm{L}$ bottle of a solution of concentrated HCl was purchased for the general chemistry laboratory. The solution contained 868.8 g of HCl . What is the molarity of the solution?\n41. An experiment in a general chemistry laboratory calls for a $2.00-M$ solution of HCl . How many mL of 11.9 $M \\mathrm{HCl}$ would be required to make 250 mL of 2.00 M HCl ?\n42. What volume of a $0.20-M \\mathrm{~K}_{2} \\mathrm{SO}_{4}$ solution contains 57 g of $\\mathrm{K}_{2} \\mathrm{SO}_{4}$ ?"}
{"id": 2998, "contents": "651. Exercises - 651.3. Molarity\n42. What volume of a $0.20-M \\mathrm{~K}_{2} \\mathrm{SO}_{4}$ solution contains 57 g of $\\mathrm{K}_{2} \\mathrm{SO}_{4}$ ?\n43. The US Environmental Protection Agency (EPA) places limits on the quantities of toxic substances that may be discharged into the sewer system. Limits have been established for a variety of substances, including hexavalent chromium, which is limited to $0.50 \\mathrm{mg} / \\mathrm{L}$. If an industry is discharging hexavalent chromium as potassium dichromate $\\left(\\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}\\right)$, what is the maximum permissible molarity of that substance?"}
{"id": 2999, "contents": "651. Exercises - 651.4. Other Units for Solution Concentrations\n44. Consider this question: What mass of a concentrated solution of nitric acid ( $68.0 \\% \\mathrm{HNO}_{3}$ by mass) is needed to prepare 400.0 g of a $10.0 \\%$ solution of $\\mathrm{HNO}_{3}$ by mass?\n(a) Outline the steps necessary to answer the question.\n(b) Answer the question.\n45. What mass of a $4.00 \\% \\mathrm{NaOH}$ solution by mass contains 15.0 g of NaOH ?\n46. What mass of solid $\\mathrm{NaOH}(97.0 \\% \\mathrm{NaOH}$ by mass) is required to prepare 1.00 L of a $10.0 \\%$ solution of NaOH by mass? The density of the $10.0 \\%$ solution is $1.109 \\mathrm{~g} / \\mathrm{mL}$.\n47. What mass of HCl is contained in 45.0 mL of an aqueous HCl solution that has a density of $1.19 \\mathrm{~g} \\mathrm{~cm}^{-3}$ and contains $37.21 \\% \\mathrm{HCl}$ by mass?\n48. The hardness of water (hardness count) is usually expressed in parts per million (by mass) of $\\mathrm{CaCO}_{3}$, which is equivalent to milligrams of $\\mathrm{CaCO}_{3}$ per liter of water. What is the molar concentration of $\\mathrm{Ca}^{2+}$ ions in a water sample with a hardness count of $175 \\mathrm{mg} \\mathrm{CaCO}_{3} / \\mathrm{L}$ ?\n49. The level of mercury in a stream was suspected to be above the minimum considered safe (1 part per billion by weight). An analysis indicated that the concentration was 0.68 parts per billion. Assume a density of $1.0 \\mathrm{~g} / \\mathrm{mL}$ and calculate the molarity of mercury in the stream."}
{"id": 3000, "contents": "651. Exercises - 651.4. Other Units for Solution Concentrations\n50. In Canada and the United Kingdom, devices that measure blood glucose levels provide a reading in millimoles per liter. If a measurement of 5.3 mM is observed, what is the concentration of glucose $\\left(\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}\\right)$ in $\\mathrm{mg} / \\mathrm{dL}$ ?\n51. A throat spray is $1.40 \\%$ by mass phenol, $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{OH}$, in water. If the solution has a density of $0.9956 \\mathrm{~g} / \\mathrm{mL}$, calculate the molarity of the solution.\n52. Copper(I) iodide (CuI) is often added to table salt as a dietary source of iodine. How many moles of CuI are contained in $1.00 \\mathrm{lb}(454 \\mathrm{~g})$ of table salt containing $0.0100 \\%$ CuI by mass?\n53. A cough syrup contains $5.0 \\%$ ethyl alcohol, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$, by mass. If the density of the solution is $0.9928 \\mathrm{~g} / \\mathrm{mL}$, determine the molarity of the alcohol in the cough syrup.\n54. D5W is a solution used as an intravenous fluid. It is a $5.0 \\%$ by mass solution of dextrose $\\left(\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}\\right)$ in water. If the density of D5W is $1.029 \\mathrm{~g} / \\mathrm{mL}$, calculate the molarity of dextrose in the solution.\n55. Find the molarity of a $40.0 \\%$ by mass aqueous solution of sulfuric acid, $\\mathrm{H}_{2} \\mathrm{SO}_{4}$, for which the density is $1.3057 \\mathrm{~g} / \\mathrm{mL}$."}
{"id": 3001, "contents": "652. CHAPTER 7 <br> Stoichiometry of Chemical Reactions - \nFigure 7.1 Many modern rocket fuels are solid mixtures of substances combined in carefully measured amounts and ignited to yield a thrust-generating chemical reaction. (credit: modification of work by NASA)"}
{"id": 3002, "contents": "653. CHAPTER OUTLINE - 653.1. Writing and Balancing Chemical Equations\n7.2 Classifying Chemical Reactions"}
{"id": 3003, "contents": "653. CHAPTER OUTLINE - 653.2. Reaction Stoichiometry\n7.4 Reaction Yields\n7.5 Quantitative Chemical Analysis\n\nINTRODUCTION Solid-fuel rockets are a central feature in the world's space exploration programs, including the new Space Launch System being developed by the National Aeronautics and Space Administration (NASA) to replace the retired Space Shuttle fleet (Figure 7.1). The engines of these rockets rely on carefully prepared solid mixtures of chemicals combined in precisely measured amounts. Igniting the mixture initiates a vigorous chemical reaction that rapidly generates large amounts of gaseous products. These gases are ejected from the rocket engine through its nozzle, providing the thrust needed to propel heavy payloads into space. Both the nature of this chemical reaction and the relationships between the amounts of the substances being consumed and produced by the reaction are critically important considerations that determine the success of the technology. This chapter will describe how to symbolize chemical reactions using chemical equations, how to classify some common chemical reactions by identifying patterns of reactivity, and how to determine the quantitative relations between the amounts of substances involved in chemical reactions-that is, the reaction stoichiometry."}
{"id": 3004, "contents": "654. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Derive chemical equations from narrative descriptions of chemical reactions.\n- Write and balance chemical equations in molecular, total ionic, and net ionic formats.\n\nAn earlier chapter of this text introduced the use of element symbols to represent individual atoms. When atoms gain or lose electrons to yield ions, or combine with other atoms to form molecules, their symbols are modified or combined to generate chemical formulas that appropriately represent these species. Extending this symbolism to represent both the identities and the relative quantities of substances undergoing a chemical (or physical) change involves writing and balancing a chemical equation. Consider as an example the reaction between one methane molecule $\\left(\\mathrm{CH}_{4}\\right)$ and two diatomic oxygen molecules $\\left(\\mathrm{O}_{2}\\right)$ to produce one carbon dioxide molecule $\\left(\\mathrm{CO}_{2}\\right)$ and two water molecules $\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)$. The chemical equation representing this process is provided in the upper half of Figure 7.2, with space-filling molecular models shown in the lower half of the figure.\n\n\nFIGURE 7.2 The reaction between methane and oxygen to yield carbon dioxide and water (shown at bottom) may be represented by a chemical equation using formulas (top).\n\nThis example illustrates the fundamental aspects of any chemical equation:\n\n1. The substances undergoing reaction are called reactants, and their formulas are placed on the left side of the equation.\n2. The substances generated by the reaction are called products, and their formulas are placed on the right side of the equation.\n3. Plus signs (+) separate individual reactant and product formulas, and an arrow $(\\longrightarrow)$ separates the reactant and product (left and right) sides of the equation.\n4. The relative numbers of reactant and product species are represented by coefficients (numbers placed immediately to the left of each formula). A coefficient of 1 is typically omitted."}
{"id": 3005, "contents": "654. LEARNING OBJECTIVES - \nIt is common practice to use the smallest possible whole-number coefficients in a chemical equation, as is done in this example. Realize, however, that these coefficients represent the relative numbers of reactants and products, and, therefore, they may be correctly interpreted as ratios. Methane and oxygen react to yield carbon dioxide and water in a 1:2:1:2 ratio. This ratio is satisfied if the numbers of these molecules are, respectively, 1-2-1-2, or 2-4-2-4, or 3-6-3-6, and so on (Figure 7.3). Likewise, these coefficients may be interpreted with regard to any amount (number) unit, and so this equation may be correctly read in many ways, including:\n\n- One methane molecule and two oxygen molecules react to yield one carbon dioxide molecule and two water molecules.\n- One dozen methane molecules and two dozen oxygen molecules react to yield one dozen carbon dioxide molecules and two dozen water molecules.\n- One mole of methane molecules and 2 moles of oxygen molecules react to yield 1 mole of carbon dioxide\nmolecules and 2 moles of water molecules.\nMixture before reaction\n\n\nMixture after reaction\n\nFIGURE 7.3 Regardless of the absolute numbers of molecules involved, the ratios between numbers of molecules of each species that react (the reactants) and molecules of each species that form (the products) are the same and are given by the chemical reaction equation."}
{"id": 3006, "contents": "655. Balancing Equations - \nThe chemical equation described in section 4.1 is balanced, meaning that equal numbers of atoms for each element involved in the reaction are represented on the reactant and product sides. This is a requirement the equation must satisfy to be consistent with the law of conservation of matter. It may be confirmed by simply summing the numbers of atoms on either side of the arrow and comparing these sums to ensure they are equal. Note that the number of atoms for a given element is calculated by multiplying the coefficient of any formula containing that element by the element's subscript in the formula. If an element appears in more than one formula on a given side of the equation, the number of atoms represented in each must be computed and then added together. For example, both product species in the example reaction, $\\mathrm{CO}_{2}$ and $\\mathrm{H}_{2} \\mathrm{O}$, contain the element oxygen, and so the number of oxygen atoms on the product side of the equation is\n\n$$\n\\left(1 \\mathrm{CO}_{2} \\text { molecule } \\times \\frac{2 \\mathrm{O} \\text { atoms }}{\\mathrm{CO}_{2} \\text { molecule }}\\right)+\\left(2 \\mathrm{H}_{2} \\mathrm{O} \\text { molecules } \\times \\frac{1 \\mathrm{O} \\text { atom }}{\\mathrm{H}_{2} \\mathrm{O} \\text { molecule }}\\right)=4 \\mathrm{O} \\text { atoms }\n$$\n\nThe equation for the reaction between methane and oxygen to yield carbon dioxide and water is confirmed to be balanced per this approach, as shown here:\n\n$$\n\\mathrm{CH}_{4}+2 \\mathrm{O}_{2} \\longrightarrow \\mathrm{CO}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}\n$$"}
{"id": 3007, "contents": "655. Balancing Equations - \n$$\n\\mathrm{CH}_{4}+2 \\mathrm{O}_{2} \\longrightarrow \\mathrm{CO}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}\n$$\n\n| Element | Reactants | Products | Balanced? |\n| :--- | :--- | :--- | :--- |\n| C | $1 \\times 1=1$ | $1 \\times 1=1$ | $1=1$, yes |\n| H | $4 \\times 1=4$ | $2 \\times 2=4$ | $4=4$, yes |\n| O | $2 \\times 2=4$ | $(1 \\times 2)+(2 \\times 1)=4$ | $4=4$, yes |\n\nA balanced chemical equation often may be derived from a qualitative description of some chemical reaction by a fairly simple approach known as balancing by inspection. Consider as an example the decomposition of water to yield molecular hydrogen and oxygen. This process is represented qualitatively by an unbalanced chemical equation:\n\n$$\n\\mathrm{H}_{2} \\mathrm{O} \\longrightarrow \\mathrm{H}_{2}+\\mathrm{O}_{2} \\quad \\text { (unbalanced) }\n$$\n\nComparing the number of H and O atoms on either side of this equation confirms its imbalance:\n\n| Element | Reactants | Products | Balanced? |\n| :--- | :--- | :--- | :--- |\n| H | $1 \\times 2=2$ | $1 \\times 2=2$ | $2=2$, yes |\n| O | $1 \\times 1=1$ | $1 \\times 2=2$ | $1 \\neq 2$, no |"}
{"id": 3008, "contents": "655. Balancing Equations - \nThe numbers of H atoms on the reactant and product sides of the equation are equal, but the numbers of O atoms are not. To achieve balance, the coefficients of the equation may be changed as needed. Keep in mind, of course, that the formula subscripts define, in part, the identity of the substance, and so these cannot be changed without altering the qualitative meaning of the equation. For example, changing the reactant formula from $\\mathrm{H}_{2} \\mathrm{O}$ to $\\mathrm{H}_{2} \\mathrm{O}_{2}$ would yield balance in the number of atoms, but doing so also changes the reactant's identity (it's now hydrogen peroxide and not water). The O atom balance may be achieved by changing the coefficient for $\\mathrm{H}_{2} \\mathrm{O}$ to 2 .\n\n| $2 \\mathrm{H}_{2} \\mathrm{O}$ |  | $\\longrightarrow \\mathrm{H}_{2}+\\mathrm{O}_{2}$ | (unbalanced) |\n| :--- | :--- | :--- | :--- |\n| Element | Reactants | Products | Balanced? |\n| H | $\\mathbf{2} \\times 2=4$ | $1 \\times 2=2$ | $4 \\neq 2$, no |\n| O | $2 \\times 1=2$ | $1 \\times 2=2$ | $2=2$, yes |\n\nThe H atom balance was upset by this change, but it is easily reestablished by changing the coefficient for the $\\mathrm{H}_{2}$ product to 2 .\n\n| $2 \\mathrm{H}_{2} \\mathrm{O}$ |  | $\\longrightarrow \\mathrm{H}_{2}+\\mathrm{O}_{2}$ | (balanced) |\n| :--- | :--- | :--- | :--- |\n| Element | Reactants | Products | Balanced? |\n| H | $2 \\times 2=4$ | $\\mathbf{2} \\times 2=4$ | $4=4$, yes |\n| O | $2 \\times 1=2$ | $1 \\times 2=2$ | $2=2$, yes |\n\nThese coefficients yield equal numbers of both H and O atoms on the reactant and product sides, and the balanced equation is, therefore:"}
{"id": 3009, "contents": "655. Balancing Equations - \nThese coefficients yield equal numbers of both H and O atoms on the reactant and product sides, and the balanced equation is, therefore:\n\n$$\n2 \\mathrm{H}_{2} \\mathrm{O} \\longrightarrow 2 \\mathrm{H}_{2}+\\mathrm{O}_{2}\n$$"}
{"id": 3010, "contents": "657. Balancing Chemical Equations - \nWrite a balanced equation for the reaction of molecular nitrogen $\\left(\\mathrm{N}_{2}\\right)$ and oxygen $\\left(\\mathrm{O}_{2}\\right)$ to form dinitrogen pentoxide."}
{"id": 3011, "contents": "658. Solution - \nFirst, write the unbalanced equation.\n\n$$\n\\mathrm{N}_{2}+\\mathrm{O}_{2} \\longrightarrow \\mathrm{~N}_{2} \\mathrm{O}_{5} \\quad \\text { (unbalanced) }\n$$\n\nNext, count the number of each type of atom present in the unbalanced equation.\n\n| Element | Reactants | Products | Balanced? |\n| :--- | :--- | :--- | :--- |\n| N | $1 \\times 2=2$ | $1 \\times 2=2$ | $2=2$, yes |\n| O | $1 \\times 2=2$ | $1 \\times 5=5$ | $2 \\neq 5$, no |\n\nThough nitrogen is balanced, changes in coefficients are needed to balance the number of oxygen atoms. To balance the number of oxygen atoms, a reasonable first attempt would be to change the coefficients for the $\\mathrm{O}_{2}$ and $\\mathrm{N}_{2} \\mathrm{O}_{5}$ to integers that will yield 10 O atoms (the least common multiple for the O atom subscripts in these two formulas).\n\n| $\\mathrm{N}_{2}+\\mathbf{5} \\mathrm{O}_{2} \\longrightarrow \\mathbf{2} \\mathrm{~N}_{2} \\mathrm{O}_{5}$ |  | (unbalanced) |  |\n| :--- | :--- | :--- | :--- |\n| Element | Reactants | Products | Balanced? |\n| N | $1 \\times 2=2$ | $\\mathbf{2} \\times 2=4$ | $2 \\neq 4$, no |\n| O | $\\mathbf{5} \\times 2=10$ | $\\mathbf{2} \\times 5=10$ | $10=10$, yes |\n\nThe N atom balance has been upset by this change; it is restored by changing the coefficient for the reactant $\\mathrm{N}_{2}$ to 2 ."}
{"id": 3012, "contents": "658. Solution - \nThe N atom balance has been upset by this change; it is restored by changing the coefficient for the reactant $\\mathrm{N}_{2}$ to 2 .\n\n| $2 \\mathrm{~N}_{2}+5 \\mathrm{O}_{2} \\longrightarrow 2 \\mathrm{~N}_{2} \\mathrm{O}_{5}$ |  |  |  |\n| :--- | :--- | :--- | :--- |\n| Element | Reactants | Products | Balanced? |\n| N | $2 \\times 2=4$ | $2 \\times 2=4$ | $4=4$, yes |\n| O | $5 \\times 2=10$ | $2 \\times 5=10$ | $10=10$, yes |\n\nThe numbers of N and O atoms on either side of the equation are now equal, and so the equation is balanced."}
{"id": 3013, "contents": "659. Check Your Learning - \nWrite a balanced equation for the decomposition of ammonium nitrate to form molecular nitrogen, molecular oxygen, and water. (Hint: Balance oxygen last, since it is present in more than one molecule on the right side of the equation.)"}
{"id": 3014, "contents": "660. Answer: - \n$2 \\mathrm{NH}_{4} \\mathrm{NO}_{3} \\longrightarrow 2 \\mathrm{~N}_{2}+\\mathrm{O}_{2}+4 \\mathrm{H}_{2} \\mathrm{O}$\n\nIt is sometimes convenient to use fractions instead of integers as intermediate coefficients in the process of balancing a chemical equation. When balance is achieved, all the equation's coefficients may then be multiplied by a whole number to convert the fractional coefficients to integers without upsetting the atom balance. For example, consider the reaction of ethane $\\left(\\mathrm{C}_{2} \\mathrm{H}_{6}\\right)$ with oxygen to yield $\\mathrm{H}_{2} \\mathrm{O}$ and $\\mathrm{CO}_{2}$, represented by the unbalanced equation:\n\n$$\n\\mathrm{C}_{2} \\mathrm{H}_{6}+\\mathrm{O}_{2} \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{CO}_{2} \\quad \\text { (unbalanced) }\n$$\n\nFollowing the usual inspection approach, one might first balance C and H atoms by changing the coefficients for the two product species, as shown:\n\n$$\n\\mathrm{C}_{2} \\mathrm{H}_{6}+\\mathrm{O}_{2} \\longrightarrow 3 \\mathrm{H}_{2} \\mathrm{O}+2 \\mathrm{CO}_{2} \\quad \\text { (unbalanced) }\n$$\n\nThis results in seven $O$ atoms on the product side of the equation, an odd number-no integer coefficient can be used with the $\\mathrm{O}_{2}$ reactant to yield an odd number, so a fractional coefficient, $\\frac{7}{2}$, is used instead to yield a\nprovisional balanced equation:\n\n$$\n\\mathrm{C}_{2} \\mathrm{H}_{6}+\\frac{7}{2} \\mathrm{O}_{2} \\longrightarrow 3 \\mathrm{H}_{2} \\mathrm{O}+2 \\mathrm{CO}_{2}\n$$\n\nA conventional balanced equation with integer-only coefficients is derived by multiplying each coefficient by 2 :"}
{"id": 3015, "contents": "660. Answer: - \nA conventional balanced equation with integer-only coefficients is derived by multiplying each coefficient by 2 :\n\n$$\n2 \\mathrm{C}_{2} \\mathrm{H}_{6}+7 \\mathrm{O}_{2} \\longrightarrow 6 \\mathrm{H}_{2} \\mathrm{O}+4 \\mathrm{CO}_{2}\n$$\n\nFinally with regard to balanced equations, recall that convention dictates use of the smallest whole-number coefficients. Although the equation for the reaction between molecular nitrogen and molecular hydrogen to produce ammonia is, indeed, balanced,\n\n$$\n3 \\mathrm{~N}_{2}+9 \\mathrm{H}_{2} \\longrightarrow 6 \\mathrm{NH}_{3}\n$$\n\nthe coefficients are not the smallest possible integers representing the relative numbers of reactant and product molecules. Dividing each coefficient by the greatest common factor, 3, gives the preferred equation:\n\n$$\n\\mathrm{N}_{2}+3 \\mathrm{H}_{2} \\longrightarrow 2 \\mathrm{NH}_{3}\n$$"}
{"id": 3016, "contents": "661. LINK TO LEARNING - \nUse this interactive tutorial (http://openstax.org/l/16BalanceEq) for additional practice balancing equations."}
{"id": 3017, "contents": "662. Additional Information in Chemical Equations - \nThe physical states of reactants and products in chemical equations very often are indicated with a parenthetical abbreviation following the formulas. Common abbreviations include $s$ for solids, $l$ for liquids, $g$ for gases, and aq for substances dissolved in water (aqueous solutions, as introduced in the preceding chapter). These notations are illustrated in the example equation here:\n\n$$\n2 \\mathrm{Na}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{NaOH}(a q)+\\mathrm{H}_{2}(g)\n$$\n\nThis equation represents the reaction that takes place when sodium metal is placed in water. The solid sodium reacts with liquid water to produce molecular hydrogen gas and the ionic compound sodium hydroxide (a solid in pure form, but readily dissolved in water).\n\nSpecial conditions necessary for a reaction are sometimes designated by writing a word or symbol above or below the equation's arrow. For example, a reaction carried out by heating may be indicated by the uppercase Greek letter delta ( $\\Delta$ ) over the arrow.\n\n$$\n\\mathrm{CaCO}_{3}(s) \\xrightarrow{\\Delta} \\mathrm{CaO}(s)+\\mathrm{CO}_{2}(g)\n$$\n\nOther examples of these special conditions will be encountered in more depth in later chapters."}
{"id": 3018, "contents": "663. Equations for Ionic Reactions - \nGiven the abundance of water on earth, it stands to reason that a great many chemical reactions take place in aqueous media. When ions are involved in these reactions, the chemical equations may be written with various levels of detail appropriate to their intended use. To illustrate this, consider a reaction between ionic compounds taking place in an aqueous solution. When aqueous solutions of $\\mathrm{CaCl}_{2}$ and $\\mathrm{AgNO}_{3}$ are mixed, a reaction takes place producing aqueous $\\mathrm{Ca}\\left(\\mathrm{NO}_{3}\\right)_{2}$ and solid AgCl :\n\n$$\n\\mathrm{CaCl}_{2}(a q)+2 \\mathrm{AgNO}_{3}(a q) \\longrightarrow \\mathrm{Ca}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+2 \\mathrm{AgCl}(s)\n$$\n\nThis balanced equation, derived in the usual fashion, is called a molecular equation because it doesn't explicitly represent the ionic species that are present in solution. When ionic compounds dissolve in water, they may dissociate into their constituent ions, which are subsequently dispersed homogenously throughout the resulting solution (a thorough discussion of this important process is provided in the chapter on solutions). Ionic compounds dissolved in water are, therefore, more realistically represented as dissociated ions, in this case:\n\n$$\n\\begin{aligned}\n& \\mathrm{CaCl}_{2}(a q) \\longrightarrow \\mathrm{Ca}^{2+}(a q)+2 \\mathrm{Cl}^{-}(a q) \\\\\n& 2 \\mathrm{AgNO}_{3}(a q) \\longrightarrow 2 \\mathrm{Ag}^{+}(a q)+2 \\mathrm{NO}_{3}^{-}(a q) \\\\\n& \\mathrm{Ca}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q) \\longrightarrow \\mathrm{Ca}^{2+}(a q)+2 \\mathrm{NO}_{3}^{-}(a q)\n\\end{aligned}\n$$\n\nUnlike these three ionic compounds, AgCl does not dissolve in water to a significant extent, as signified by its physical state notation, $s$.\n\nExplicitly representing all dissolved ions results in a complete ionic equation. In this particular case, the formulas for the dissolved ionic compounds are replaced by formulas for their dissociated ions:"}
{"id": 3019, "contents": "663. Equations for Ionic Reactions - \nExplicitly representing all dissolved ions results in a complete ionic equation. In this particular case, the formulas for the dissolved ionic compounds are replaced by formulas for their dissociated ions:\n\n$$\n\\mathrm{Ca}^{2+}(a q)+2 \\mathrm{Cl}^{-}(a q)+2 \\mathrm{Ag}^{+}(a q)+2 \\mathrm{NO}_{3}^{-}(a q) \\longrightarrow \\mathrm{Ca}^{2+}(a q)+2 \\mathrm{NO}_{3}^{-}(a q)+2 \\mathrm{Ag} \\mathrm{Cl}(s)\n$$\n\nExamining this equation shows that two chemical species are present in identical form on both sides of the arrow, $\\mathrm{Ca}^{2+}(a q)$ and $\\mathrm{NO}_{3}{ }^{-}(a q)$. These spectator ions-ions whose presence is required to maintain charge neutrality-are neither chemically nor physically changed by the process, and so they may be eliminated from the equation to yield a more succinct representation called a net ionic equation:\n\n$$\n\\begin{gathered}\n\\mathrm{Ca}^{2+}(a q)+2 \\mathrm{Cl}^{-}(a q)+2 \\mathrm{Ag}^{+}(a q)+2 \\mathrm{NO}_{3}^{-}(a q) \\longrightarrow \\mathrm{Ca}^{2+}(a q)+2 \\mathrm{NO}_{3}^{-}(a q)+2 \\mathrm{AgCl}(s) \\\\\n2 \\mathrm{Cl}^{-}(a q)+2 \\mathrm{Ag}^{+}(a q) \\longrightarrow 2 \\mathrm{AgCl}(s)\n\\end{gathered}\n$$\n\nFollowing the convention of using the smallest possible integers as coefficients, this equation is then written:\n\n$$\n\\mathrm{Cl}^{-}(a q)+\\mathrm{Ag}^{+}(a q) \\longrightarrow \\mathrm{AgCl}(s)\n$$\n\nThis net ionic equation indicates that solid silver chloride may be produced from dissolved chloride and silver(I) ions, regardless of the source of these ions. These molecular and complete ionic equations provide additional information, namely, the ionic compounds used as sources of $\\mathrm{Cl}^{-}$and $\\mathrm{Ag}^{+}$."}
{"id": 3020, "contents": "665. Ionic and Molecular Equations - \nWhen carbon dioxide is dissolved in an aqueous solution of sodium hydroxide, the mixture reacts to yield aqueous sodium carbonate and liquid water. Write balanced molecular, complete ionic, and net ionic equations for this process."}
{"id": 3021, "contents": "666. Solution - \nBegin by identifying formulas for the reactants and products and arranging them properly in chemical equation form:\n\n$$\n\\mathrm{CO}_{2}(a q)+\\mathrm{NaOH}(a q) \\longrightarrow \\mathrm{Na}_{2} \\mathrm{CO}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\quad \\text { (unbalanced) }\n$$\n\nBalance is achieved easily in this case by changing the coefficient for NaOH to 2 , resulting in the molecular equation for this reaction:\n\n$$\n\\mathrm{CO}_{2}(a q)+2 \\mathrm{NaOH}(a q) \\longrightarrow \\mathrm{Na}_{2} \\mathrm{CO}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nThe two dissolved ionic compounds, NaOH and $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$, can be represented as dissociated ions to yield the complete ionic equation:\n\n$$\n\\mathrm{CO}_{2}(a q)+2 \\mathrm{Na}^{+}(a q)+2 \\mathrm{OH}^{-}(a q) \\longrightarrow 2 \\mathrm{Na}^{+}(a q)+\\mathrm{CO}_{3}^{2-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nFinally, identify the spectator ion(s), in this case $\\mathrm{Na}^{+}(a q)$, and remove it from each side of the equation to generate the net ionic equation:\n\n$$\n\\begin{aligned}\n\\mathrm{CO}_{2}(a q)+2 \\mathrm{Na}^{+}(a q)+2 \\mathrm{OH}^{-}(a q) & \\longrightarrow 2 \\mathrm{Na}^{+}(a q)+\\mathrm{CO}_{3}^{2-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\\\\n\\mathrm{CO}_{2}(a q)+2 \\mathrm{OH}^{-}(a q) & \\longrightarrow \\mathrm{CO}_{3}^{2-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)\n\\end{aligned}\n$$"}
{"id": 3022, "contents": "667. Check Your Learning - \nDiatomic chlorine and sodium hydroxide (lye) are commodity chemicals produced in large quantities, along with diatomic hydrogen, via the electrolysis of brine, according to the following unbalanced equation:\n\n$$\n\\mathrm{NaCl}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\xrightarrow{\\text { electricity }} \\mathrm{NaOH}(a q)+\\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g)\n$$\n\nWrite balanced molecular, complete ionic, and net ionic equations for this process."}
{"id": 3023, "contents": "668. Answer: - \n$2 \\mathrm{NaCl}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{NaOH}(a q)+\\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g) \\quad$ (molecular)\n$2 \\mathrm{Na}^{+}(a q)+2 \\mathrm{Cl}^{-}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{Na}^{+}(a q)+2 \\mathrm{OH}^{-}(a q)+\\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g) \\quad$ (complete ionic)\n$2 \\mathrm{Cl}^{-}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{OH}^{-}(a q)+\\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g) \\quad$ (net ionic)"}
{"id": 3024, "contents": "669. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Define three common types of chemical reactions (precipitation, acid-base, and oxidation-reduction)\n- Classify chemical reactions as one of these three types given appropriate descriptions or chemical equations\n- Identify common acids and bases\n- Predict the solubility of common inorganic compounds by using solubility rules\n- Compute the oxidation states for elements in compounds\n\nHumans interact with one another in various and complex ways, and we classify these interactions according to common patterns of behavior. When two humans exchange information, we say they are communicating. When they exchange blows with their fists or feet, we say they are fighting. Faced with a wide range of varied interactions between chemical substances, scientists have likewise found it convenient (or even necessary) to classify chemical interactions by identifying common patterns of reactivity. This module will provide an introduction to three of the most prevalent types of chemical reactions: precipitation, acid-base, and oxidation-reduction."}
{"id": 3025, "contents": "670. Precipitation Reactions and Solubility Rules - \nA precipitation reaction is one in which dissolved substances react to form one (or more) solid products. Many reactions of this type involve the exchange of ions between ionic compounds in aqueous solution and are sometimes referred to as double displacement, double replacement, or metathesis reactions. These reactions are common in nature and are responsible for the formation of coral reefs in ocean waters and kidney stones in animals. They are used widely in industry for production of a number of commodity and specialty chemicals. Precipitation reactions also play a central role in many chemical analysis techniques, including spot tests used to identify metal ions and gravimetric methods for determining the composition of matter (see the last module of this chapter).\n\nThe extent to which a substance may be dissolved in water, or any solvent, is quantitatively expressed as its solubility, defined as the maximum concentration of a substance that can be achieved under specified conditions. Substances with relatively large solubilities are said to be soluble. A substance will precipitate when solution conditions are such that its concentration exceeds its solubility. Substances with relatively low solubilities are said to be insoluble, and these are the substances that readily precipitate from solution. More information on these important concepts is provided in a later chapter on solutions. For purposes of predicting the identities of solids formed by precipitation reactions, one may simply refer to patterns of solubility that\nhave been observed for many ionic compounds (Table 7.1)."}
{"id": 3026, "contents": "670. Precipitation Reactions and Solubility Rules - \n| Soluble Ionic Compounds | contain these ions | exceptions |\n| :---: | :---: | :---: |\n|  | $\\mathrm{NH}_{4}{ }^{+}$ <br> group I cations: <br> $\\mathrm{Li}^{+}$ <br> $\\mathrm{Na}^{+}$ <br> $\\mathrm{K}^{+}$ <br> $\\mathrm{Rb}^{+}$ <br> $\\mathrm{Cs}^{+}$ | none |\n|  | $\\begin{aligned} & \\mathrm{Cl}^{-} \\\\ & \\mathrm{Br}^{-} \\\\ & \\mathrm{I}^{-} \\end{aligned}$ | compounds with $\\mathrm{Ag}^{+}, \\mathrm{Hg}_{2}{ }^{2+}$, and $\\mathrm{Pb}^{2+}$ |\n|  | $\\mathrm{F}^{-}$ | compounds with group 2 metal cations, $\\mathrm{Pb}^{2+}$, and $\\mathrm{Fe}^{3+}$ |\n|  | $\\begin{aligned} & \\mathrm{C}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}^{-} \\\\ & \\mathrm{HCO}_{3}^{-} \\\\ & \\mathrm{NO}_{3}^{-} \\\\ & \\mathrm{ClO}_{3}^{-} \\end{aligned}$ | none |\n|  | $\\mathrm{SO}_{4}{ }^{2-}$ | compounds with $\\mathrm{Ag}^{+}, \\mathrm{Ba}^{2+}, \\mathrm{Ca}^{2+}, \\mathrm{Hg}_{2}{ }^{2+}, \\mathrm{Pb}^{2+}$ and $\\mathrm{Sr}^{2+}$ |\n| Insoluble Ionic Compounds | contain these ions | exceptions |\n|  | $\\begin{aligned} & \\mathrm{CO}_{3}{ }^{2-} \\\\ & \\mathrm{CrO}_{4}{ }^{2-} \\\\ & \\mathrm{PO}_{4}{ }^{3-} \\\\ & \\mathrm{S}^{2-} \\end{aligned}$ | compounds with group 1 cations and $\\mathrm{NH}_{4}{ }^{+}$ |"}
{"id": 3027, "contents": "670. Precipitation Reactions and Solubility Rules - \n|  | $\\mathrm{OH}^{-}$ | compounds with group 1 cations and $\\mathrm{Ba}^{2+}$ |"}
{"id": 3028, "contents": "670. Precipitation Reactions and Solubility Rules - \nTABLE 7.1\n\nA vivid example of precipitation is observed when solutions of potassium iodide and lead nitrate are mixed, resulting in the formation of solid lead iodide:\n\n$$\n2 \\mathrm{KI}(a q)+\\mathrm{Pb}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q) \\longrightarrow \\mathrm{PbI}_{2}(s)+2 \\mathrm{KNO}_{3}(a q)\n$$\n\nThis observation is consistent with the solubility guidelines: The only insoluble compound among all those involved is lead iodide, one of the exceptions to the general solubility of iodide salts.\n\nThe net ionic equation representing this reaction is:\n\n$$\n\\mathrm{Pb}^{2+}(a q)+2 \\mathrm{I}^{-}(a q) \\longrightarrow \\mathrm{PbI}_{2}(s)\n$$\n\nLead iodide is a bright yellow solid that was formerly used as an artist's pigment known as iodine yellow (Figure 7.4). The properties of pure $\\mathrm{PbI}_{2}$ crystals make them useful for fabrication of X-ray and gamma ray detectors.\n\n\nFIGURE 7.4 A precipitate of $\\mathrm{PbI}_{2}$ forms when solutions containing $\\mathrm{Pb}^{2+}$ and $\\mathrm{I}^{-}$are mixed. (credit: Der Kreole/ Wikimedia Commons)"}
{"id": 3029, "contents": "670. Precipitation Reactions and Solubility Rules - \nThe solubility guidelines in Table 7.1 may be used to predict whether a precipitation reaction will occur when solutions of soluble ionic compounds are mixed together. One merely needs to identify all the ions present in the solution and then consider if possible cation/anion pairing could result in an insoluble compound. For example, mixing solutions of silver nitrate and sodium chloride will yield a solution containing $\\mathrm{Ag}^{+}, \\mathrm{NO}_{3}{ }^{-}$, $\\mathrm{Na}^{+}$, and $\\mathrm{Cl}^{-}$ions. Aside from the two ionic compounds originally present in the solutions, $\\mathrm{AgNO}_{3}$ and NaCl , two additional ionic compounds may be derived from this collection of ions: $\\mathrm{NaNO}_{3}$ and AgCl . The solubility guidelines indicate all nitrate salts are soluble but that AgCl is one of insoluble. A precipitation reaction, therefore, is predicted to occur, as described by the following equations:\n\n$$\n\\begin{gathered}\n\\mathrm{NaCl}(a q)+\\mathrm{AgNO}_{3}(a q) \\longrightarrow \\mathrm{AgCl}(s)+\\mathrm{NaNO}_{3}(a q) \\quad \\text { (molecular) } \\\\\n\\mathrm{Ag}^{+}(a q)+\\mathrm{Cl}^{-}(a q) \\longrightarrow \\mathrm{AgCl}(s) \\quad \\text { (net ionic) }\n\\end{gathered}\n$$"}
{"id": 3030, "contents": "672. Predicting Precipitation Reactions - \nPredict the result of mixing reasonably concentrated solutions of the following ionic compounds. If precipitation is expected, write a balanced net ionic equation for the reaction.\n(a) potassium sulfate and barium nitrate\n(b) lithium chloride and silver acetate\n(c) lead nitrate and ammonium carbonate"}
{"id": 3031, "contents": "673. Solution - \n(a) The two possible products for this combination are $\\mathrm{KNO}_{3}$ and $\\mathrm{BaSO}_{4}$. The solubility guidelines indicate\n$\\mathrm{BaSO}_{4}$ is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is\n\n$$\n\\mathrm{Ba}^{2+}(a q)+\\mathrm{SO}_{4}^{2-}(a q) \\longrightarrow \\mathrm{BaSO}_{4}(s)\n$$\n\n(b) The two possible products for this combination are $\\mathrm{LiC}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}$ and AgCl . The solubility guidelines indicate AgCl is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is\n\n$$\n\\mathrm{Ag}^{+}(a q)+\\mathrm{Cl}^{-}(a q) \\longrightarrow \\mathrm{AgCl}(s)\n$$\n\n(c) The two possible products for this combination are $\\mathrm{PbCO}_{3}$ and $\\mathrm{NH}_{4} \\mathrm{NO}_{3}$. The solubility guidelines indicate $\\mathrm{PbCO}_{3}$ is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is\n\n$$\n\\mathrm{Pb}^{2+}(a q)+\\mathrm{CO}_{3}^{2-}(a q) \\longrightarrow \\mathrm{PbCO}_{3}(s)\n$$"}
{"id": 3032, "contents": "674. Check Your Learning - \nWhich solution could be used to precipitate the barium ion, $\\mathrm{Ba}^{2+}$, in a water sample: sodium chloride, sodium hydroxide, or sodium sulfate? What is the formula for the expected precipitate?"}
{"id": 3033, "contents": "675. Answer: - \nsodium sulfate, $\\mathrm{BaSO}_{4}$"}
{"id": 3034, "contents": "676. Acid-Base Reactions - \nAn acid-base reaction is one in which a hydrogen ion, $\\mathrm{H}^{+}$, is transferred from one chemical species to another. Such reactions are of central importance to numerous natural and technological processes, ranging from the chemical transformations that take place within cells and the lakes and oceans, to the industrial-scale production of fertilizers, pharmaceuticals, and other substances essential to society. The subject of acid-base chemistry, therefore, is worthy of thorough discussion, and a full chapter is devoted to this topic later in the text.\n\nFor purposes of this brief introduction, we will consider only the more common types of acid-base reactions that take place in aqueous solutions. In this context, an acid is a substance that will dissolve in water to yield hydronium ions, $\\mathrm{H}_{3} \\mathrm{O}^{+}$. As an example, consider the equation shown here:\n\n$$\n\\mathrm{HCl}(a q)+\\mathrm{H}_{2} \\mathrm{O}(a q) \\longrightarrow \\mathrm{Cl}^{-}(a q)+\\mathrm{H}_{3} \\mathrm{O}^{+}(a q)\n$$\n\nThe process represented by this equation confirms that hydrogen chloride is an acid. When dissolved in water, $\\mathrm{H}_{3} \\mathrm{O}^{+}$ions are produced by a chemical reaction in which $\\mathrm{H}^{+}$ions are transferred from HCl molecules to $\\mathrm{H}_{2} \\mathrm{O}$ molecules (Figure 7.5).\n\n\nFIGURE 7.5 When hydrogen chloride gas dissolves in water, (a) it reacts as an acid, transferring protons to water molecules to yield (b) hydronium ions (and solvated chloride ions)."}
{"id": 3035, "contents": "676. Acid-Base Reactions - \nFIGURE 7.5 When hydrogen chloride gas dissolves in water, (a) it reacts as an acid, transferring protons to water molecules to yield (b) hydronium ions (and solvated chloride ions).\n\nThe nature of HCl is such that its reaction with water as just described is essentially $100 \\%$ efficient: Virtually every HCl molecule that dissolves in water will undergo this reaction. Acids that completely react in this fashion are called strong acids, and HCl is one among just a handful of common acid compounds that are classified as strong (Table 7.2). A far greater number of compounds behave as weak acids and only partially react with water, leaving a large majority of dissolved molecules in their original form and generating a relatively small amount of hydronium ions. Weak acids are commonly encountered in nature, being the substances partly responsible for the tangy taste of citrus fruits, the stinging sensation of insect bites, and the unpleasant smells associated with body odor. A familiar example of a weak acid is acetic acid, the main ingredient in food vinegars:\n\n$$\n\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}(a q)+\\mathrm{H}_{3} \\mathrm{O}^{+}(a q)\n$$\n\nWhen dissolved in water under typical conditions, only about $1 \\%$ of acetic acid molecules are present in the ionized form, $\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}$(Figure 7.6). (The use of a double-arrow in the equation above denotes the partial reaction aspect of this process, a concept addressed fully in the chapters on chemical equilibrium.)\n\n(a)\n\n(b)\n\nFIGURE 7.6 (a) Fruits such as oranges, lemons, and grapefruit contain the weak acid citric acid. (b) Vinegars contain the weak acid acetic acid. (credit a: modification of work by Scott Bauer; credit b: modification of work by Br\u00fccke-Osteuropa/Wikimedia Commons)\n\nCommon Strong Acids"}
{"id": 3036, "contents": "676. Acid-Base Reactions - \nCommon Strong Acids\n\n| Compound Formula | Name in Aqueous Solution |\n| :--- | :--- |\n| HBr | hydrobromic acid |\n| HCl | hydrochloric acid |\n| HI | hydroiodic acid |\n| $\\mathrm{HNO}_{3}$ | nitric acid |\n| $\\mathrm{HClO}_{4}$ | perchloric acid |\n| $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ | sulfuric acid |\n\nTABLE 7.2\n\nA base is a substance that will dissolve in water to yield hydroxide ions, $\\mathrm{OH}^{-}$. The most common bases are ionic compounds composed of alkali or alkaline earth metal cations (groups 1 and 2) combined with the hydroxide ion-for example, NaOH and $\\mathrm{Ca}(\\mathrm{OH})_{2}$. Unlike the acid compounds discussed previously, these compounds do not react chemically with water; instead they dissolve and dissociate, releasing hydroxide ions directly into the solution. For example, KOH and $\\mathrm{Ba}(\\mathrm{OH})_{2}$ dissolve in water and dissociate completely to produce cations ( $\\mathrm{K}^{+}$and $\\mathrm{Ba}^{2+}$, respectively) and hydroxide ions, $\\mathrm{OH}^{-}$. These bases, along with other hydroxides that completely dissociate in water, are considered strong bases.\n\nConsider as an example the dissolution of lye (sodium hydroxide) in water:\n\n$$\n\\mathrm{NaOH}(s) \\longrightarrow \\mathrm{Na}^{+}(a q)+\\mathrm{OH}^{-}(a q)\n$$"}
{"id": 3037, "contents": "676. Acid-Base Reactions - \n$$\n\\mathrm{NaOH}(s) \\longrightarrow \\mathrm{Na}^{+}(a q)+\\mathrm{OH}^{-}(a q)\n$$\n\nThis equation confirms that sodium hydroxide is a base. When dissolved in water, NaOH dissociates to yield $\\mathrm{Na}^{+}$and $\\mathrm{OH}^{-}$ions. This is also true for any other ionic compound containing hydroxide ions. Since the dissociation process is essentially complete when ionic compounds dissolve in water under typical conditions, NaOH and other ionic hydroxides are all classified as strong bases.\nUnlike ionic hydroxides, some compounds produce hydroxide ions when dissolved by chemically reacting with water molecules. In all cases, these compounds react only partially and so are classified as weak bases. These types of compounds are also abundant in nature and important commodities in various technologies. For example, global production of the weak base ammonia is typically well over 100 metric tons annually, being widely used as an agricultural fertilizer, a raw material for chemical synthesis of other compounds, and an active ingredient in household cleaners (Figure 7.7). When dissolved in water, ammonia reacts partially to yield hydroxide ions, as shown here:\n\n$$\n\\mathrm{NH}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{NH}_{4}^{+}(a q)+\\mathrm{OH}^{-}(a q)\n$$\n\nThis is, by definition, an acid-base reaction, in this case involving the transfer of $\\mathrm{H}^{+}$ions from water molecules to ammonia molecules. Under typical conditions, only about $1 \\%$ of the dissolved ammonia is present as $\\mathrm{NH}_{4}{ }^{+}$ ions.\n\n\nFIGURE 7.7 Ammonia is a weak base used in a variety of applications. (a) Pure ammonia is commonly applied as an agricultural fertilizer. (b) Dilute solutions of ammonia are effective household cleansers. (credit a: modification of work by National Resources Conservation Service; credit b: modification of work by pat00139)\n\nA neutralization reaction is a specific type of acid-base reaction in which the reactants are an acid and a base (but not water), and the products are often a salt and water"}
{"id": 3038, "contents": "676. Acid-Base Reactions - \nA neutralization reaction is a specific type of acid-base reaction in which the reactants are an acid and a base (but not water), and the products are often a salt and water\n\n$$\n\\text { acid + base } \\longrightarrow \\text { salt + water }\n$$\n\nTo illustrate a neutralization reaction, consider what happens when a typical antacid such as milk of magnesia (an aqueous suspension of solid $\\mathrm{Mg}(\\mathrm{OH})_{2}$ ) is ingested to ease symptoms associated with excess stomach acid ( HCl ):\n\n$$\n\\mathrm{Mg}(\\mathrm{OH})_{2}(s)+2 \\mathrm{HCl}(a q) \\longrightarrow \\mathrm{MgCl}_{2}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nNote that in addition to water, this reaction produces a salt, magnesium chloride."}
{"id": 3039, "contents": "678. Writing Equations for Acid-Base Reactions - \nWrite balanced chemical equations for the acid-base reactions described here:\n(a) the weak acid hydrogen hypochlorite reacts with water\n(b) a solution of barium hydroxide is neutralized with a solution of nitric acid"}
{"id": 3040, "contents": "679. Solution - \n(a) The two reactants are provided, HOCl and $\\mathrm{H}_{2} \\mathrm{O}$. Since the substance is reported to be an acid, its reaction with water will involve the transfer of $\\mathrm{H}^{+}$from HOCl to $\\mathrm{H}_{2} \\mathrm{O}$ to generate hydronium ions, $\\mathrm{H}_{3} \\mathrm{O}^{+}$and hypochlorite ions, $\\mathrm{OCl}^{-}$.\n\n$$\n\\mathrm{HOCl}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{OCl}^{-}(a q)+\\mathrm{H}_{3} \\mathrm{O}^{+}(a q)\n$$\n\nA double-arrow is appropriate in this equation because it indicates the HOCl is a weak acid that has not reacted completely.\n(b) The two reactants are provided, $\\mathrm{Ba}(\\mathrm{OH})_{2}$ and $\\mathrm{HNO}_{3}$. Since this is a neutralization reaction, the two products will be water and a salt composed of the cation of the ionic hydroxide $\\left(\\mathrm{Ba}^{2+}\\right)$ and the anion generated when the acid transfers its hydrogen ion $\\left(\\mathrm{NO}_{3}{ }^{-}\\right)$.\n\n$$\n\\mathrm{Ba}(\\mathrm{OH})_{2}(a q)+2 \\mathrm{HNO}_{3}(a q) \\longrightarrow \\mathrm{Ba}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nCheck Your Learning\nWrite the net ionic equation representing the neutralization of any strong acid with an ionic hydroxide. (Hint: Consider the ions produced when a strong acid is dissolved in water.)"}
{"id": 3041, "contents": "680. Answer: - \n$\\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{OH}^{-}(a q) \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(l)$"}
{"id": 3042, "contents": "682. Stomach Antacids - \nOur stomachs contain a solution of roughly 0.03 M HCl , which helps us digest the food we eat. The burning sensation associated with heartburn is a result of the acid of the stomach leaking through the muscular valve at the top of the stomach into the lower reaches of the esophagus. The lining of the esophagus is not protected from the corrosive effects of stomach acid the way the lining of the stomach is, and the results can be very painful. When we have heartburn, it feels better if we reduce the excess acid in the esophagus by taking an antacid. As you may have guessed, antacids are bases. One of the most common antacids is calcium carbonate, $\\mathrm{CaCO}_{3}$. The reaction,\n\n$$\n\\mathrm{CaCO}_{3}(s)+2 \\mathrm{HCl}(a q) \\rightleftharpoons \\mathrm{CaCl}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{CO}_{2}(g)\n$$\n\nnot only neutralizes stomach acid, it also produces $\\mathrm{CO}_{2}(g)$, which may result in a satisfying belch.\nMilk of Magnesia is a suspension of the sparingly soluble base magnesium hydroxide, $\\mathrm{Mg}(\\mathrm{OH})_{2}$. It works according to the reaction:\n\n$$\n\\operatorname{Mg}(\\mathrm{OH})_{2}(s) \\rightleftharpoons \\mathrm{Mg}^{2+}(a q)+2 \\mathrm{OH}^{-}(a q)\n$$\n\nThe hydroxide ions generated in this equilibrium then go on to react with the hydronium ions from the stomach acid, so that:\n\n$$\n\\mathrm{H}_{3} \\mathrm{O}^{+}+\\mathrm{OH}^{-} \\rightleftharpoons 2 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$"}
{"id": 3043, "contents": "682. Stomach Antacids - \n$$\n\\mathrm{H}_{3} \\mathrm{O}^{+}+\\mathrm{OH}^{-} \\rightleftharpoons 2 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nThis reaction does not produce carbon dioxide, but magnesium-containing antacids can have a laxative effect. Several antacids have aluminum hydroxide, $\\mathrm{Al}(\\mathrm{OH})_{3}$, as an active ingredient. The aluminum hydroxide tends to cause constipation, and some antacids use aluminum hydroxide in concert with magnesium hydroxide to balance the side effects of the two substances."}
{"id": 3044, "contents": "684. Culinary Aspects of Chemistry - \nExamples of acid-base chemistry are abundant in the culinary world. One example is the use of baking soda, or sodium bicarbonate in baking. $\\mathrm{NaHCO}_{3}$ is a base. When it reacts with an acid such as lemon juice, buttermilk, or sour cream in a batter, bubbles of carbon dioxide gas are formed from decomposition of the resulting carbonic acid, and the batter \"rises.\" Baking powder is a combination of sodium bicarbonate, and one or more acid salts that react when the two chemicals come in contact with water in the batter.\n\nMany people like to put lemon juice or vinegar, both of which are acids, on cooked fish (Figure 7.8). It turns out that fish have volatile amines (bases) in their systems, which are neutralized by the acids to yield involatile ammonium salts. This reduces the odor of the fish, and also adds a \"sour\" taste that we seem to enjoy.\n\n\nFIGURE 7.8 A neutralization reaction takes place between citric acid in lemons or acetic acid in vinegar, and the bases in the flesh of fish.\n\nPickling is a method used to preserve vegetables using a naturally produced acidic environment. The vegetable, such as a cucumber, is placed in a sealed jar submerged in a brine solution. The brine solution favors the growth of beneficial bacteria and suppresses the growth of harmful bacteria. The beneficial bacteria feed on starches in the cucumber and produce lactic acid as a waste product in a process called fermentation. The lactic acid eventually increases the acidity of the brine to a level that kills any harmful bacteria, which require a basic environment. Without the harmful bacteria consuming the cucumbers they are able to last much longer than if they were unprotected. A byproduct of the pickling process changes the flavor of the vegetables with the acid making them taste sour."}
{"id": 3045, "contents": "685. LINK TO LEARNING - \nExplore the microscopic view (http://openstax.org/l/16AcidsBases) of strong and weak acids and bases."}
{"id": 3046, "contents": "686. Oxidation-Reduction Reactions - \nEarth's atmosphere contains about $20 \\%$ molecular oxygen, $\\mathrm{O}_{2}$, a chemically reactive gas that plays an essential role in the metabolism of aerobic organisms and in many environmental processes that shape the world. The term oxidation was originally used to describe chemical reactions involving $\\mathrm{O}_{2}$, but its meaning has evolved to refer to a broad and important reaction class known as oxidation-reduction (redox) reactions. A few examples of such reactions will be used to develop a clear picture of this classification.\n\nSome redox reactions involve the transfer of electrons between reactant species to yield ionic products, such as the reaction between sodium and chlorine to yield sodium chloride:\n\n$$\n2 \\mathrm{Na}(s)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{NaCl}(s)\n$$\n\nIt is helpful to view the process with regard to each individual reactant, that is, to represent the fate of each reactant in the form of an equation called a half-reaction:\n\n$$\n\\begin{aligned}\n& 2 \\mathrm{Na}(s) \\longrightarrow 2 \\mathrm{Na}^{+}(s)+2 \\mathrm{e}^{-} \\\\\n& \\mathrm{Cl}_{2}(g)+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Cl}^{-}(s)\n\\end{aligned}\n$$\n\nThese equations show that Na atoms lose electrons while Cl atoms (in the $\\mathrm{Cl}_{2}$ molecule) gain electrons, the \" s \" subscripts for the resulting ions signifying they are present in the form of a solid ionic compound. For redox reactions of this sort, the loss and gain of electrons define the complementary processes that occur:\n\n```\noxidation = loss of electrons\nreduction = gain of electrons\n```\n\nIn this reaction, then, sodium is oxidized and chlorine undergoes reduction. Viewed from a more active perspective, sodium functions as a reducing agent (reductant), since it provides electrons to (or reduces) chlorine. Likewise, chlorine functions as an oxidizing agent (oxidant), as it effectively removes electrons from (oxidizes) sodium."}
{"id": 3047, "contents": "686. Oxidation-Reduction Reactions - \n$$\n\\begin{aligned}\n\\text { reducing agent } & =\\text { species that is oxidized } \\\\\n\\text { oxidizing agent } & =\\text { species that is reduced }\n\\end{aligned}\n$$\n\nSome redox processes, however, do not involve the transfer of electrons. Consider, for example, a reaction similar to the one yielding NaCl :\n\n$$\n\\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{HCl}(g)\n$$\n\nThe product of this reaction is a covalent compound, so transfer of electrons in the explicit sense is not involved. To clarify the similarity of this reaction to the previous one and permit an unambiguous definition of redox reactions, a property called oxidation number has been defined. The oxidation number (or oxidation state) of an element in a compound is the charge its atoms would possess if the compound were ionic. The following guidelines are used to assign oxidation numbers to each element in a molecule or ion.\n\n1. The oxidation number of an atom in an elemental substance is zero.\n2. The oxidation number of a monatomic ion is equal to the ion's charge.\n3. Oxidation numbers for common nonmetals are usually assigned as follows:\n\n- Hydrogen: +1 when combined with nonmetals, -1 when combined with metals\n- Oxygen: - 2 in most compounds, sometimes -1 (so-called peroxides, $\\mathrm{O}_{2}{ }^{2-}$ ), very rarely $-\\frac{1}{2}$ (so-called superoxides, $\\mathrm{O}_{2}{ }^{-}$), positive values when combined with F (values vary)\n- Halogens: -1 for $F$ always, -1 for other halogens except when combined with oxygen or other halogens (positive oxidation numbers in these cases, varying values)\n\n4. The sum of oxidation numbers for all atoms in a molecule or polyatomic ion equals the charge on the molecule or ion.\n\nNote: The proper convention for reporting charge is to write the number first, followed by the sign (e.g., $2+$ ), while oxidation number is written with the reversed sequence, sign followed by number (e.g., +2 ). This convention aims to emphasize the distinction between these two related properties."}
{"id": 3048, "contents": "688. Assigning Oxidation Numbers - \nFollow the guidelines in this section of the text to assign oxidation numbers to all the elements in the following species:\n(a) $\\mathrm{H}_{2} \\mathrm{~S}$\n(b) $\\mathrm{SO}_{3}{ }^{2-}$\n(c) $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$"}
{"id": 3049, "contents": "689. Solution - \n(a) According to guideline 3 , the oxidation number for H is +1 .\n\nUsing this oxidation number and the compound's formula, guideline 4 may then be used to calculate the oxidation number for sulfur:\n\n$$\n\\begin{gathered}\n\\text { charge on } \\mathrm{H}_{2} \\mathrm{~S}=0=(2 \\times+1)+(1 \\times x) \\\\\nx=0-(2 \\times+1)=-2\n\\end{gathered}\n$$\n\n(b) Guideline 3 suggests the oxidation number for oxygen is -2 .\n\nUsing this oxidation number and the ion's formula, guideline 4 may then be used to calculate the oxidation number for sulfur:\n\n$$\n\\begin{gathered}\n\\text { charge on } \\mathrm{SO}_{3}{ }^{2-}=-2=(3 \\times-2)+(1 \\times x) \\\\\nx=-2-(3 \\times-2)=+4\n\\end{gathered}\n$$\n\n(c) For ionic compounds, it's convenient to assign oxidation numbers for the cation and anion separately.\n\nAccording to guideline 2, the oxidation number for sodium is +1 .\nAssuming the usual oxidation number for oxygen ( -2 per guideline 3 ), the oxidation number for sulfur is calculated as directed by guideline 4:\n\n$$\n\\begin{gathered}\n\\text { charge on } \\mathrm{SO}_{4}{ }^{2-}=-2=(4 \\times-2)+(1 \\times x) \\\\\nx=-2-(4 \\times-2)=+6\n\\end{gathered}\n$$"}
{"id": 3050, "contents": "690. Check Your Learning - \nAssign oxidation states to the elements whose atoms are underlined in each of the following compounds or ions:\n(a) $\\mathrm{KNO}_{3}$\n(b) $\\mathrm{AlH}_{3}$\n(c) $\\mathrm{NH}_{4}{ }^{+}$\n(d) $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}$\n\nAnswer:\n(a) N, +5; (b) Al, +3; (c) N, -3; (d) P, +5\n\nUsing the oxidation number concept, an all-inclusive definition of redox reaction has been established. Oxidation-reduction (redox) reactions are those in which one or more elements involved undergo a change in oxidation number. (While the vast majority of redox reactions involve changes in oxidation number for two or more elements, a few interesting exceptions to this rule do exist Example 7.6.) Definitions for the complementary processes of this reaction class are correspondingly revised as shown here:\n\n$$\n\\begin{aligned}\n\\text { oxidation } & =\\text { increase in oxidation number } \\\\\n\\text { reduction } & =\\text { decrease in oxidation number }\n\\end{aligned}\n$$\n\nReturning to the reactions used to introduce this topic, they may now both be identified as redox processes. In the reaction between sodium and chlorine to yield sodium chloride, sodium is oxidized (its oxidation number increases from 0 in Na to +1 in NaCl ) and chlorine is reduced (its oxidation number decreases from 0 in $\\mathrm{Cl}_{2}$ to -1 in NaCl ). In the reaction between molecular hydrogen and chlorine, hydrogen is oxidized (its oxidation number increases from 0 in $\\mathrm{H}_{2}$ to +1 in HCl ) and chlorine is reduced (its oxidation number decreases from 0 in\n$\\mathrm{Cl}_{2}$ to -1 in HCl ).\nSeveral subclasses of redox reactions are recognized, including combustion reactions in which the reductant (also called a fuel) and oxidant (often, but not necessarily, molecular oxygen) react vigorously and produce significant amounts of heat, and often light, in the form of a flame. Solid rocket-fuel reactions such as the one depicted in Figure 7.1 are combustion processes. A typical propellant reaction in which solid aluminum is oxidized by ammonium perchlorate is represented by this equation:"}
{"id": 3051, "contents": "690. Check Your Learning - \n$$\n10 \\mathrm{Al}(s)+6 \\mathrm{NH}_{4} \\mathrm{ClO}_{4}(s) \\longrightarrow 4 \\mathrm{Al}_{2} \\mathrm{O}_{3}(s)+2 \\mathrm{AlCl}_{3}(s)+12 \\mathrm{H}_{2} \\mathrm{O}(g)+3 \\mathrm{~N}_{2}(g)\n$$"}
{"id": 3052, "contents": "691. LINK TO LEARNING - \nWatch a brief video (http://openstax.org/1/16hybridrocket) showing the test firing of a small-scale, prototype, hybrid rocket engine planned for use in the new Space Launch System being developed by NASA. The first engines firing at\n3 s (green flame) use a liquid fuel/oxidant mixture, and the second, more powerful engines firing at 4 s (yellow flame) use a solid mixture.\n\nSingle-displacement (replacement) reactions are redox reactions in which an ion in solution is displaced (or replaced) via the oxidation of a metallic element. One common example of this type of reaction is the acid oxidation of certain metals:\n\n$$\n\\mathrm{Zn}(s)+2 \\mathrm{HCl}(a q) \\longrightarrow \\mathrm{ZnCl}_{2}(a q)+\\mathrm{H}_{2}(g)\n$$\n\nMetallic elements may also be oxidized by solutions of other metal salts; for example:\n\n$$\n\\mathrm{Cu}(s)+2 \\mathrm{AgNO}_{3}(a q) \\longrightarrow \\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+2 \\mathrm{Ag}(s)\n$$\n\nThis reaction may be observed by placing copper wire in a solution containing a dissolved silver salt. Silver ions in solution are reduced to elemental silver at the surface of the copper wire, and the resulting $\\mathrm{Cu}^{2+}$ ions dissolve in the solution to yield a characteristic blue color (Figure 7.9).\n\n(a)\n\n(b)\n\n(c)\n\nFIGURE 7.9 (a) A copper wire is shown next to a solution containing silver(I) ions. (b) Displacement of dissolved silver ions by copper ions results in (c) accumulation of gray-colored silver metal on the wire and development of a blue color in the solution, due to dissolved copper ions. (credit: modification of work by Mark Ott)"}
{"id": 3053, "contents": "693. Describing Redox Reactions - \nIdentify which equations represent redox reactions, providing a name for the reaction if appropriate. For those reactions identified as redox, name the oxidant and reductant.\n(a) $\\mathrm{ZnCO}_{3}(s) \\longrightarrow \\mathrm{ZnO}(s)+\\mathrm{CO}_{2}(g)$\n(b) $2 \\mathrm{Ga}(l)+3 \\mathrm{Br}_{2}(l) \\longrightarrow 2 \\mathrm{GaBr}_{3}(s)$\n(c) $2 \\mathrm{H}_{2} \\mathrm{O}_{2}(a q) \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{O}_{2}(g)$\n(d) $\\mathrm{BaCl}_{2}(a q)+\\mathrm{K}_{2} \\mathrm{SO}_{4}(a q) \\longrightarrow \\mathrm{BaSO}_{4}(s)+2 \\mathrm{KCl}(a q)$\n(e) $\\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{~g})+3 \\mathrm{O}_{2}(\\mathrm{~g}) \\longrightarrow 2 \\mathrm{CO}_{2}(\\mathrm{~g})+2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})$"}
{"id": 3054, "contents": "694. Solution - \nRedox reactions are identified per definition if one or more elements undergo a change in oxidation number.\n(a) This is not a redox reaction, since oxidation numbers remain unchanged for all elements.\n(b) This is a redox reaction. Gallium is oxidized, its oxidation number increasing from 0 in $\\mathrm{Ga}(I)$ to +3 in $\\operatorname{GaBr}_{3}(S)$. The reducing agent is $\\mathrm{Ga}(I)$. Bromine is reduced, its oxidation number decreasing from 0 in $\\mathrm{Br}_{2}(I)$ to -1 in $\\operatorname{GaBr}_{3}(S)$. The oxidizing agent is $\\mathrm{Br}_{2}(I)$.\n(c) This is a redox reaction. It is a particularly interesting process, as it involves the same element, oxygen, undergoing both oxidation and reduction (a so-called disproportionation reaction). Oxygen is oxidized, its oxidation number increasing from -1 in $\\mathrm{H}_{2} \\mathrm{O}_{2}(\\mathrm{aq})$ to 0 in $\\mathrm{O}_{2}(\\mathrm{~g})$. Oxygen is also reduced, its oxidation number decreasing from -1 in $\\mathrm{H}_{2} \\mathrm{O}_{2}(a q)$ to -2 in $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})$. For disproportionation reactions, the same substance functions as an oxidant and a reductant.\n(d) This is not a redox reaction, since oxidation numbers remain unchanged for all elements.\n(e) This is a redox reaction (combustion). Carbon is oxidized, its oxidation number increasing from -2 in $\\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{~g})$ to +4 in $\\mathrm{CO}_{2}(\\mathrm{~g})$. The reducing agent (fuel) is $\\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{~g})$. Oxygen is reduced, its oxidation number decreasing from 0 in $\\mathrm{O}_{2}(\\mathrm{~g})$ to -2 in $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})$. The oxidizing agent is $\\mathrm{O}_{2}(\\mathrm{~g})$."}
{"id": 3055, "contents": "695. Check Your Learning - \nThis equation describes the production of tin(II) chloride:\n\n$$\n\\mathrm{Sn}(s)+2 \\mathrm{HCl}(g) \\longrightarrow \\mathrm{SnCl}_{2}(s)+\\mathrm{H}_{2}(g)\n$$\n\nIs this a redox reaction? If so, provide a more specific name for the reaction if appropriate, and identify the oxidant and reductant."}
{"id": 3056, "contents": "696. Answer: - \nYes, a single-replacement reaction. $\\mathrm{Sn}(s)$ is the reductant, $\\mathrm{HCl}(g)$ is the oxidant."}
{"id": 3057, "contents": "697. Balancing Redox Reactions via the Half-Reaction Method - \nRedox reactions that take place in aqueous media often involve water, hydronium ions, and hydroxide ions as reactants or products. Although these species are not oxidized or reduced, they do participate in chemical change in other ways (e.g., by providing the elements required to form oxyanions). Equations representing these reactions are sometimes very difficult to balance by inspection, so systematic approaches have been developed to assist in the process. One very useful approach is to use the method of half-reactions, which involves the following steps:\n\n1. Write the two half-reactions representing the redox process.\n2. Balance all elements except oxygen and hydrogen.\n3. Balance oxygen atoms by adding $\\mathrm{H}_{2} \\mathrm{O}$ molecules.\n4. Balance hydrogen atoms by adding $\\mathrm{H}^{+}$ions.\n5. Balance charge by adding electrons.\n6. If necessary, multiply each half-reaction's coefficients by the smallest possible integers to yield equal numbers of electrons in each.\n7. Add the balanced half-reactions together and simplify by removing species that appear on both sides of the equation.\n8. For reactions occurring in basic media (excess hydroxide ions), carry out these additional steps:\na. Add $\\mathrm{OH}^{-}$ions to both sides of the equation in numbers equal to the number of $\\mathrm{H}^{+}$ions.\nb. On the side of the equation containing both $\\mathrm{H}^{+}$and $\\mathrm{OH}^{-}$ions, combine these ions to yield water molecules.\nc. Simplify the equation by removing any redundant water molecules.\n9. Finally, check to see that both the number of atoms and the total charges ${ }^{\\frac{1}{}}$ are balanced."}
{"id": 3058, "contents": "699. Balancing Redox Reactions in Acidic Solution - \nWrite a balanced equation for the reaction between dichromate ion and iron(II) to yield iron(III) and chromium(III) in acidic solution.\n\n$$\n\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+\\mathrm{Fe}^{2+} \\longrightarrow \\mathrm{Cr}^{3+}+\\mathrm{Fe}^{3+}\n$$"}
{"id": 3059, "contents": "700. Solution - \nStep 1.\nWrite the two half-reactions.\nEach half-reaction will contain one reactant and one product with one element in common.\n\n$$\n\\begin{gathered}\n\\mathrm{Fe}^{2+} \\longrightarrow \\mathrm{Fe}^{3+} \\\\\n\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-} \\longrightarrow \\mathrm{Cr}^{3+}\n\\end{gathered}\n$$\n\nStep 2.\nBalance all elements except oxygen and hydrogen. The iron half-reaction is already balanced, but the chromium half-reaction shows two Cr atoms on the left and one Cr atom on the right. Changing the coefficient on the right side of the equation to 2 achieves balance with regard to Cr atoms.\n\n$$\n\\begin{gathered}\n\\mathrm{Fe}^{2+} \\longrightarrow \\mathrm{Fe}^{3+} \\\\\n\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-} \\longrightarrow 2 \\mathrm{Cr}^{3+}\n\\end{gathered}\n$$\n\nStep 3.\nBalance oxygen atoms by adding $\\mathrm{H}_{2} \\mathrm{O}$ molecules. The iron half-reaction does not contain O atoms. The chromium half-reaction shows seven O atoms on the left and none on the right, so seven water molecules are added to the right side.\n\n$$\n\\begin{gathered}\n\\mathrm{Fe}^{2+} \\longrightarrow \\mathrm{Fe}^{3+} \\\\\n\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-} \\longrightarrow 2 \\mathrm{Cr}^{3+}+7 \\mathrm{H}_{2} \\mathrm{O}\n\\end{gathered}\n$$\n\nStep 4.\nBalance hydrogen atoms by adding $\\mathrm{H}^{+}$ions. The iron half-reaction does not contain H atoms. The chromium half-reaction shows 14 H atoms on the right and none on the left, so 14 hydrogen ions are added to the left side."}
{"id": 3060, "contents": "700. Solution - \n$$\n\\begin{gathered}\n\\mathrm{Fe}^{2+} \\longrightarrow \\mathrm{Fe}^{3+} \\\\\n\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+14 \\mathrm{H}^{+} \\longrightarrow 2 \\mathrm{Cr}^{3+}+7 \\mathrm{H}_{2} \\mathrm{O}\n\\end{gathered}\n$$\n\nStep 5.\nBalance charge by adding electrons. The iron half-reaction shows a total charge of $2+$ on the left side (1 $\\mathrm{Fe}^{2+}$ ion) and $3+$ on the right side ( $1 \\mathrm{Fe}^{3+}$ ion). Adding one electron to the right side brings that side's total charge to $(3+)+(1-)=2+$, and charge balance is achieved.\n\nThe chromium half-reaction shows a total charge of $(1 \\times 2-)+(14 \\times 1+)=12+$ on the left side $\\left(1 \\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}\\right.$ ion and $14 \\mathrm{H}^{+}$ions). The total charge on the right side is $(2 \\times 3+)=6+\\left(2 \\mathrm{Cr}^{3+}\\right.$ ions $)$. Adding six electrons to\nthe left side will bring that side's total charge to $(12++6-)=6+$, and charge balance is achieved.\n\n$$\n\\begin{gathered}\n\\mathrm{Fe}^{2+} \\longrightarrow \\mathrm{Fe}^{3+}+\\mathrm{e}^{-} \\\\\n\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+14 \\mathrm{H}^{+}+6 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Cr}^{3+}+7 \\mathrm{H}_{2} \\mathrm{O}\n\\end{gathered}\n$$\n\nStep 6.\nMultiply the two half-reactions so the number of electrons in one reaction equals the number of electrons in the other reaction. To be consistent with mass conservation, and the idea that redox reactions involve the transfer (not creation or destruction) of electrons, the iron half-reaction's coefficient must be multiplied by 6 ."}
{"id": 3061, "contents": "700. Solution - \n$$\n\\begin{gathered}\n6 \\mathrm{Fe}^{2+} \\longrightarrow 6 \\mathrm{Fe}^{3+}+6 \\mathrm{e}^{-} \\\\\n\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+6 \\mathrm{e}^{-}+14 \\mathrm{H}^{+} \\longrightarrow 2 \\mathrm{Cr}^{3+}+7 \\mathrm{H}_{2} \\mathrm{O}\n\\end{gathered}\n$$\n\nStep 7.\nAdd the balanced half-reactions and cancel species that appear on both sides of the equation.\n\n$$\n6 \\mathrm{Fe}^{2+}+\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+6 \\mathrm{e}^{-}+14 \\mathrm{H}^{+} \\longrightarrow 6 \\mathrm{Fe}^{3+}+6 \\mathrm{e}^{-}+2 \\mathrm{Cr}^{3+}+7 \\mathrm{H}_{2} \\mathrm{O}\n$$\n\nOnly the six electrons are redundant species. Removing them from each side of the equation yields the simplified, balanced equation here:\n\n$$\n6 \\mathrm{Fe}^{2+}+\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+14 \\mathrm{H}^{+} \\longrightarrow 6 \\mathrm{Fe}^{3+}+2 \\mathrm{Cr}^{3+}+7 \\mathrm{H}_{2} \\mathrm{O}\n$$\n\nA final check of atom and charge balance confirms the equation is balanced.\n\n|  | Reactants | Products |\n| :--- | :--- | :--- |\n| Fe | 6 | 6 |\n| Cr | 2 | 2 |\n| O | 7 | 7 |\n| H | 14 | 14 |\n| charge | $24+$ | $24+$ |"}
{"id": 3062, "contents": "701. Check Your Learning - \nIn basic solution, molecular chlorine, $\\mathrm{Cl}_{2}$, reacts with hydroxide ions, $\\mathrm{OH}^{-}$, to yield chloride ions, $\\mathrm{Cl}^{-}$. and chlorate ions, $\\mathrm{ClO}_{3}{ }^{-}$. HINT: This is a disproportionation reaction in which the element chlorine is both oxidized and reduced. Write a balanced equation for this reaction."}
{"id": 3063, "contents": "702. Answer: - \n$3 \\mathrm{Cl}_{2}(a q)+6 \\mathrm{OH}^{-}(a q) \\longrightarrow 5 \\mathrm{Cl}^{-}(a q)+\\mathrm{ClO}_{3}^{-}(a q)+3 \\mathrm{H}_{2} \\mathrm{O}(l)$"}
{"id": 3064, "contents": "703. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Explain the concept of stoichiometry as it pertains to chemical reactions\n- Use balanced chemical equations to derive stoichiometric factors relating amounts of reactants and products\n- Perform stoichiometric calculations involving mass, moles, and solution molarity\n\nA balanced chemical equation provides a great deal of information in a very succinct format. Chemical formulas provide the identities of the reactants and products involved in the chemical change, allowing\nclassification of the reaction. Coefficients provide the relative numbers of these chemical species, allowing a quantitative assessment of the relationships between the amounts of substances consumed and produced by the reaction. These quantitative relationships are known as the reaction's stoichiometry, a term derived from the Greek words stoicheion (meaning \"element\") and metron (meaning \"measure\"). In this module, the use of balanced chemical equations for various stoichiometric applications is explored.\n\nThe general approach to using stoichiometric relationships is similar in concept to the way people go about many common activities. Food preparation, for example, offers an appropriate comparison. A recipe for making eight pancakes calls for 1 cup pancake mix, $\\frac{3}{4}$ cup milk, and one egg. The \"equation\" representing the preparation of pancakes per this recipe is\n\n$$\n1 \\text { cup mix }+\\frac{3}{4} \\text { cup milk }+1 \\text { egg } \\longrightarrow 8 \\text { pancakes }\n$$\n\nIf two dozen pancakes are needed for a big family breakfast, the ingredient amounts must be increased proportionally according to the amounts given in the recipe. For example, the number of eggs required to make 24 pancakes is\n\n$$\n24 \\text { pancakes } \\times \\frac{1 \\mathrm{egg}}{8 \\text { pancakes- }}=3 \\mathrm{eggs}\n$$\n\nBalanced chemical equations are used in much the same fashion to determine the amount of one reactant required to react with a given amount of another reactant, or to yield a given amount of product, and so forth. The coefficients in the balanced equation are used to derive stoichiometric factors that permit computation of the desired quantity. To illustrate this idea, consider the production of ammonia by reaction of hydrogen and nitrogen:"}
{"id": 3065, "contents": "703. LEARNING OBJECTIVES - \n$$\n\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\longrightarrow 2 \\mathrm{NH}_{3}(g)\n$$\n\nThis equation shows ammonia molecules are produced from hydrogen molecules in a $2: 3$ ratio, and stoichiometric factors may be derived using any amount (number) unit:\n\n$$\n\\frac{2 \\mathrm{NH}_{3} \\text { molecules }}{3 \\mathrm{H}_{2} \\text { molecules }} \\text { or } \\frac{2 \\text { doz } \\mathrm{NH}_{3} \\text { molecules }}{3 \\text { doz } \\mathrm{H}_{2} \\text { molecules }} \\text { or } \\frac{2 \\mathrm{~mol} \\mathrm{NH}_{3} \\text { molecules }}{3 \\mathrm{~mol} \\mathrm{H}_{2} \\text { molecules }}\n$$\n\nThese stoichiometric factors can be used to compute the number of ammonia molecules produced from a given number of hydrogen molecules, or the number of hydrogen molecules required to produce a given number of ammonia molecules. Similar factors may be derived for any pair of substances in any chemical equation."}
{"id": 3066, "contents": "705. Moles of Reactant Required in a Reaction - \nHow many moles of $\\mathrm{I}_{2}$ are required to react with 0.429 mol of Al according to the following equation (see Figure 7.10)?\n\n$$\n2 \\mathrm{Al}+3 \\mathrm{I}_{2} \\longrightarrow 2 \\mathrm{AlI}_{3}\n$$\n\n\n\nFIGURE 7.10 Aluminum and iodine react to produce aluminum iodide. The heat of the reaction vaporizes some of the solid iodine as a purple vapor. (credit: modification of work by Mark Ott)"}
{"id": 3067, "contents": "706. Solution - \nReferring to the balanced chemical equation, the stoichiometric factor relating the two substances of interest\nis $\\frac{3 \\mathrm{~mol} \\mathrm{I}_{2}}{2 \\mathrm{~mol} \\mathrm{Al}}$. The molar amount of iodine is derived by multiplying the provided molar amount of aluminum by this factor:\n\n\n$$\n\\begin{aligned}\n\\mathrm{mol} \\mathrm{I}_{2} & =0.429 \\mathrm{~mol} \\mathrm{At} \\times \\frac{3 \\mathrm{~mol} \\mathrm{I}_{2}}{2 \\mathrm{~mol} \\mathrm{At}} \\\\\n& =0.644 \\mathrm{~mol} \\mathrm{I}_{2}\n\\end{aligned}\n$$"}
{"id": 3068, "contents": "707. Check Your Learning - \nHow many moles of $\\mathrm{Ca}(\\mathrm{OH})_{2}$ are required to react with 1.36 mol of $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ to produce $\\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$ according to the equation $3 \\mathrm{Ca}(\\mathrm{OH})_{2}+2 \\mathrm{H}_{3} \\mathrm{PO}_{4} \\longrightarrow \\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}+6 \\mathrm{H}_{2} \\mathrm{O}$ ?"}
{"id": 3069, "contents": "708. Answer: - \n2.04 mol"}
{"id": 3070, "contents": "710. Number of Product Molecules Generated by a Reaction - \nHow many carbon dioxide molecules are produced when 0.75 mol of propane is combusted according to this equation?\n\n$$\n\\mathrm{C}_{3} \\mathrm{H}_{8}+5 \\mathrm{O}_{2} \\longrightarrow 3 \\mathrm{CO}_{2}+4 \\mathrm{H}_{2} \\mathrm{O}\n$$"}
{"id": 3071, "contents": "711. Solution - \nThe approach here is the same as for Example 7.8, though the absolute number of molecules is requested, not the number of moles of molecules. This will simply require use of the moles-to-numbers conversion factor, Avogadro's number.\n\nThe balanced equation shows that carbon dioxide is produced from propane in a $3: 1$ ratio:\n\n$$\n\\frac{3 \\mathrm{~mol} \\mathrm{CO}_{2}}{1 \\mathrm{~mol} \\mathrm{C}_{3} \\mathrm{H}_{8}}\n$$\n\nUsing this stoichiometric factor, the provided molar amount of propane, and Avogadro's number,"}
{"id": 3072, "contents": "712. Check Your Learning - \nHow many $\\mathrm{NH}_{3}$ molecules are produced by the reaction of 4.0 mol of $\\mathrm{Ca}(\\mathrm{OH})_{2}$ according to the following equation:\n\n$$\n\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{4}+\\mathrm{Ca}(\\mathrm{OH})_{2} \\longrightarrow 2 \\mathrm{NH}_{3}+\\mathrm{CaSO}_{4}+2 \\mathrm{H}_{2} \\mathrm{O}\n$$"}
{"id": 3073, "contents": "713. Answer: - \n$4.8 \\times 10^{24} \\mathrm{NH}_{3}$ molecules\n\nThese examples illustrate the ease with which the amounts of substances involved in a chemical reaction of known stoichiometry may be related. Directly measuring numbers of atoms and molecules is, however, not an\neasy task, and the practical application of stoichiometry requires that we use the more readily measured property of mass."}
{"id": 3074, "contents": "715. Relating Masses of Reactants and Products - \nWhat mass of sodium hydroxide, NaOH , would be required to produce 16 g of the antacid milk of magnesia [magnesium hydroxide, $\\mathrm{Mg}(\\mathrm{OH})_{2}$ ] by the following reaction?\n\n$$\n\\mathrm{MgCl}_{2}(a q)+2 \\mathrm{NaOH}(a q) \\longrightarrow \\mathrm{Mg}(\\mathrm{OH})_{2}(s)+2 \\mathrm{NaCl}(a q)\n$$"}
{"id": 3075, "contents": "716. Solution - \nThe approach used previously in Example 7.8 and Example 7.9 is likewise used here; that is, we must derive an appropriate stoichiometric factor from the balanced chemical equation and use it to relate the amounts of the two substances of interest. In this case, however, masses (not molar amounts) are provided and requested, so additional steps of the sort learned in the previous chapter are required. The calculations required are outlined in this flowchart:"}
{"id": 3076, "contents": "717. Check Your Learning - \nWhat mass of gallium oxide, $\\mathrm{Ga}_{2} \\mathrm{O}_{3}$, can be prepared from 29.0 g of gallium metal? The equation for the reaction is $4 \\mathrm{Ga}+3 \\mathrm{O}_{2} \\longrightarrow 2 \\mathrm{Ga}_{2} \\mathrm{O}_{3}$."}
{"id": 3077, "contents": "718. Answer: - \n39.0 g"}
{"id": 3078, "contents": "720. Relating Masses of Reactants - \nWhat mass of oxygen gas, $\\mathrm{O}_{2}$, from the air is consumed in the combustion of 702 g of octane, $\\mathrm{C}_{8} \\mathrm{H}_{18}$, one of the principal components of gasoline?\n\n$$\n2 \\mathrm{C}_{8} \\mathrm{H}_{18}+25 \\mathrm{O}_{2} \\longrightarrow 16 \\mathrm{CO}_{2}+18 \\mathrm{H}_{2} \\mathrm{O}\n$$"}
{"id": 3079, "contents": "721. Solution - \nThe approach required here is the same as for the Example 7.10, differing only in that the provided and requested masses are both for reactant species."}
{"id": 3080, "contents": "722. Check Your Learning - \nWhat mass of CO is required to react with 25.13 g of $\\mathrm{Fe}_{2} \\mathrm{O}_{3}$ according to the equation\n$\\mathrm{Fe}_{2} \\mathrm{O}_{3}+3 \\mathrm{CO} \\longrightarrow 2 \\mathrm{Fe}+3 \\mathrm{CO}_{2}$ ?"}
{"id": 3081, "contents": "723. Answer: - \n13.22 g\n\nThese examples illustrate just a few instances of reaction stoichiometry calculations. Numerous variations on the beginning and ending computational steps are possible depending upon what particular quantities are provided and sought (volumes, solution concentrations, and so forth). Regardless of the details, all these calculations share a common essential component: the use of stoichiometric factors derived from balanced chemical equations. Figure 7.11 provides a general outline of the various computational steps associated with many reaction stoichiometry calculations.\n\n\nFIGURE 7.11 The flowchart depicts the various computational steps involved in most reaction stoichiometry\ncalculations."}
{"id": 3082, "contents": "725. Airbags - \nAirbags (Figure 7.12) are a safety feature provided in most automobiles since the 1990s. The effective operation of an airbag requires that it be rapidly inflated with an appropriate amount (volume) of gas when the vehicle is involved in a collision. This requirement is satisfied in many automotive airbag systems through use of explosive chemical reactions, one common choice being the decomposition of sodium azide, $\\mathrm{NaN}_{3}$. When sensors in the vehicle detect a collision, an electrical current is passed through a carefully measured amount of $\\mathrm{NaN}_{3}$ to initiate its decomposition:\n\n$$\n2 \\mathrm{NaN}_{3}(s) \\longrightarrow 3 \\mathrm{~N}_{2}(g)+2 \\mathrm{Na}(s)\n$$\n\nThis reaction is very rapid, generating gaseous nitrogen that can deploy and fully inflate a typical airbag in a fraction of a second ( $\\sim 0.03-0.1 \\mathrm{~s}$ ). Among many engineering considerations, the amount of sodium azide used must be appropriate for generating enough nitrogen gas to fully inflate the air bag and ensure its proper function. For example, a small mass $(\\sim 100 \\mathrm{~g})$ of $\\mathrm{NaN}_{3}$ will generate approximately 50 L of $\\mathrm{N}_{2}$.\n\n\nFIGURE 7.12 Airbags deploy upon impact to minimize serious injuries to passengers. (credit: Jon Seidman)"}
{"id": 3083, "contents": "726. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Explain the concepts of theoretical yield and limiting reactants/reagents.\n- Derive the theoretical yield for a reaction under specified conditions.\n- Calculate the percent yield for a reaction.\n\nThe relative amounts of reactants and products represented in a balanced chemical equation are often referred to as stoichiometric amounts. All the exercises of the preceding module involved stoichiometric amounts of reactants. For example, when calculating the amount of product generated from a given amount of reactant, it was assumed that any other reactants required were available in stoichiometric amounts (or greater). In this module, more realistic situations are considered, in which reactants are not present in stoichiometric amounts."}
{"id": 3084, "contents": "727. Limiting Reactant - \nConsider another food analogy, making grilled cheese sandwiches (Figure 7.13):\n1 slice of cheese +2 slices of bread $\\longrightarrow 1$ sandwich\nStoichiometric amounts of sandwich ingredients for this recipe are bread and cheese slices in a 2:1 ratio.\n\nProvided with 28 slices of bread and 11 slices of cheese, one may prepare 11 sandwiches per the provided recipe, using all the provided cheese and having six slices of bread left over. In this scenario, the number of sandwiches prepared has been limited by the number of cheese slices, and the bread slices have been provided in excess.\n\n\nFIGURE 7.13 Sandwich making can illustrate the concepts of limiting and excess reactants.\nConsider this concept now with regard to a chemical process, the reaction of hydrogen with chlorine to yield hydrogen chloride:\n\n$$\n\\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{HCl}(g)\n$$\n\nThe balanced equation shows the hydrogen and chlorine react in a 1:1 stoichiometric ratio. If these reactants are provided in any other amounts, one of the reactants will nearly always be entirely consumed, thus limiting the amount of product that may be generated. This substance is the limiting reactant, and the other substance is the excess reactant. Identifying the limiting and excess reactants for a given situation requires computing the molar amounts of each reactant provided and comparing them to the stoichiometric amounts represented in the balanced chemical equation. For example, imagine combining 3 moles of $\\mathrm{H}_{2}$ and 2 moles of $\\mathrm{Cl}_{2}$. This represents a $3: 2$ (or 1.5:1) ratio of hydrogen to chlorine present for reaction, which is greater than the stoichiometric ratio of 1:1. Hydrogen, therefore, is present in excess, and chlorine is the limiting reactant. Reaction of all the provided chlorine ( 2 mol ) will consume 2 mol of the 3 mol of hydrogen provided, leaving 1 mol of hydrogen unreacted."}
{"id": 3085, "contents": "727. Limiting Reactant - \nAn alternative approach to identifying the limiting reactant involves comparing the amount of product expected for the complete reaction of each reactant. Each reactant amount is used to separately calculate the amount of product that would be formed per the reaction's stoichiometry. The reactant yielding the lesser amount of product is the limiting reactant. For the example in the previous paragraph, complete reaction of the hydrogen would yield\n\n$$\n\\text { mol } \\mathrm{HCl} \\text { produced }=3 \\mathrm{~mol} \\mathrm{H}_{2} \\times \\frac{2 \\mathrm{~mol} \\mathrm{HCl}}{1 \\mathrm{~mol} \\mathrm{H}_{2}}=6 \\mathrm{~mol} \\mathrm{HCl}\n$$\n\nComplete reaction of the provided chlorine would produce\n\n$$\n\\text { mol } \\mathrm{HCl} \\text { produced }=2 \\mathrm{~mol} \\mathrm{Cl}_{2} \\times \\frac{2 \\mathrm{~mol} \\mathrm{HCl}}{1 \\mathrm{~mol} \\mathrm{Cl}_{2}}=4 \\mathrm{~mol} \\mathrm{HCl}\n$$\n\nThe chlorine will be completely consumed once 4 moles of HCl have been produced. Since enough hydrogen was provided to yield 6 moles of HCl , there will be unreacted hydrogen remaining once this reaction is complete. Chlorine, therefore, is the limiting reactant and hydrogen is the excess reactant (Figure 7.14).\n\n\nFIGURE 7.14 When $\\mathrm{H}_{2}$ and $\\mathrm{Cl}_{2}$ are combined in nonstoichiometric amounts, one of these reactants will limit the amount of HCl that can be produced. This illustration shows a reaction in which hydrogen is present in excess and chlorine is the limiting reactant."}
{"id": 3086, "contents": "728. LINK TO LEARNING - \nView this interactive simulation (http://openstax.org/l/16reactantprod) illustrating the concepts of limiting and excess reactants."}
{"id": 3087, "contents": "730. Identifying the Limiting Reactant - \nSilicon nitride is a very hard, high-temperature-resistant ceramic used as a component of turbine blades in jet engines. It is prepared according to the following equation:\n\n$$\n3 \\mathrm{Si}(s)+2 \\mathrm{~N}_{2}(g) \\longrightarrow \\mathrm{Si}_{3} \\mathrm{~N}_{4}(s)\n$$\n\nWhich is the limiting reactant when 2.00 g of Si and 1.50 g of $\\mathrm{N}_{2}$ react?"}
{"id": 3088, "contents": "731. Solution - \nCompute the provided molar amounts of reactants, and then compare these amounts to the balanced equation to identify the limiting reactant.\n\n$$\n\\begin{gathered}\n\\mathrm{mol} \\mathrm{Si}=2.00 \\mathrm{~g} \\mathrm{Si} \\times \\frac{1 \\mathrm{~mol} \\mathrm{Si}}{28.09 \\mathrm{~g} \\mathrm{Si}}=0.0712 \\mathrm{~mol} \\mathrm{Si} \\\\\n\\mathrm{~mol} \\mathrm{~N}_{2}=1.50-\\mathrm{g}_{\\mathrm{g}}^{\\mathrm{g} \\mathrm{~N}_{2}} \\times \\frac{1 \\mathrm{~mol} \\mathrm{~N}_{2}}{28.02-\\frac{\\mathrm{g} \\mathrm{~N}_{2}}{}}=0.0535 \\mathrm{~mol} \\mathrm{~N}_{2}\n\\end{gathered}\n$$\n\nThe provided $\\mathrm{Si}: \\mathrm{N}_{2}$ molar ratio is:\n\n$$\n\\frac{0.0712 \\mathrm{~mol} \\mathrm{Si}}{0.0535 \\mathrm{~mol} \\mathrm{~N}_{2}}=\\frac{1.33 \\mathrm{~mol} \\mathrm{Si}}{1 \\mathrm{~mol} \\mathrm{~N}_{2}}\n$$\n\nThe stoichiometric Si: $\\mathrm{N}_{2}$ ratio is:\n\n$$\n\\frac{3 \\mathrm{~mol} \\mathrm{Si}}{2 \\mathrm{~mol} \\mathrm{~N}_{2}}=\\frac{1.5 \\mathrm{~mol} \\mathrm{Si}}{1 \\mathrm{~mol} \\mathrm{~N}_{2}}\n$$\n\nComparing these ratios shows that Si is provided in a less-than-stoichiometric amount, and so is the limiting reactant.\n\nAlternatively, compute the amount of product expected for complete reaction of each of the provided reactants. The 0.0712 moles of silicon would yield"}
{"id": 3089, "contents": "731. Solution - \nAlternatively, compute the amount of product expected for complete reaction of each of the provided reactants. The 0.0712 moles of silicon would yield\n\n$$\n\\mathrm{mol} \\mathrm{Si}_{3} \\mathrm{~N}_{4} \\text { produced }=0.0712 \\mathrm{~mol} \\mathrm{Si} \\times \\frac{1 \\mathrm{~mol} \\mathrm{Si}_{3} \\mathrm{~N}_{4}}{3 \\mathrm{~mol} \\mathrm{Si}}=0.0237 \\mathrm{~mol} \\mathrm{Si}_{3} \\mathrm{~N}_{4}\n$$\n\nwhile the 0.0535 moles of nitrogen would produce\n\n$$\n\\mathrm{mol} \\mathrm{Si}_{3} \\mathrm{~N}_{4} \\text { produced }=0.0535 \\mathrm{~mol} \\mathrm{~N}_{2} \\times \\frac{1 \\mathrm{~mol} \\mathrm{Si}_{3} \\mathrm{~N}_{4}}{2 \\mathrm{~mol} \\mathrm{~N}_{2}}=0.0268 \\mathrm{~mol} \\mathrm{Si}_{3} \\mathrm{~N}_{4}\n$$\n\nSince silicon yields the lesser amount of product, it is the limiting reactant."}
{"id": 3090, "contents": "732. Check Your Learning - \nWhich is the limiting reactant when 5.00 g of $\\mathrm{H}_{2}$ and 10.0 g of $\\mathrm{O}_{2}$ react and form water?"}
{"id": 3091, "contents": "733. Answer: - \n$\\mathrm{O}_{2}$"}
{"id": 3092, "contents": "734. Percent Yield - \nThe amount of product that may be produced by a reaction under specified conditions, as calculated per the stoichiometry of an appropriate balanced chemical equation, is called the theoretical yield of the reaction. In practice, the amount of product obtained is called the actual yield, and it is often less than the theoretical yield for a number of reasons. Some reactions are inherently inefficient, being accompanied by side reactions that generate other products. Others are, by nature, incomplete (consider the partial reactions of weak acids and bases discussed earlier in this chapter). Some products are difficult to collect without some loss, and so less than perfect recovery will reduce the actual yield. The extent to which a reaction's theoretical yield is achieved is commonly expressed as its percent yield:\n\n$$\n\\text { percent yield }=\\frac{\\text { actual yield }}{\\text { theoretical yield }} \\times 100 \\%\n$$\n\nActual and theoretical yields may be expressed as masses or molar amounts (or any other appropriate property; e.g., volume, if the product is a gas). As long as both yields are expressed using the same units, these units will cancel when percent yield is calculated."}
{"id": 3093, "contents": "736. Calculation of Percent Yield - \nUpon reaction of 1.274 g of copper sulfate with excess zinc metal, 0.392 g copper metal was obtained according to the equation:\n\n$$\n\\mathrm{CuSO}_{4}(a q)+\\mathrm{Zn}(s) \\longrightarrow \\mathrm{Cu}(s)+\\mathrm{ZnSO}_{4}(a q)\n$$\n\nWhat is the percent yield?"}
{"id": 3094, "contents": "737. Solution - \nThe provided information identifies copper sulfate as the limiting reactant, and so the theoretical yield is found by the approach illustrated in the previous module, as shown here:\n\n$$\n1.274-\\mathrm{g} \\mathrm{CuSO}_{4} \\times \\frac{1 \\mathrm{molCuSO}_{4}}{159.62-\\mathrm{gCuSO}_{4}} \\times \\frac{1 \\mathrm{molCu}}{1 \\mathrm{molCuSO}_{4}} \\times \\frac{63.55 \\mathrm{~g} \\mathrm{Cu}}{1 \\mathrm{~mol} \\mathrm{Cu}}=0.5072 \\mathrm{~g} \\mathrm{Cu}\n$$\n\nUsing this theoretical yield and the provided value for actual yield, the percent yield is calculated to be\n\n$$\n\\begin{aligned}\n\\text { percent yield } & =\\left(\\frac{\\text { actual yield }}{\\text { theoretical yield }}\\right) \\times 100 \\\\\n\\text { percent yield } & =\\left(\\frac{0.392 \\mathrm{~g} \\mathrm{Cu}}{0.5072 \\mathrm{~g} \\mathrm{Cu}}\\right) \\times 100 \\\\\n& =77.3 \\%\n\\end{aligned}\n$$"}
{"id": 3095, "contents": "738. Check Your Learning - \nexcess HF?\n\n$$\n\\mathrm{CCl}_{4}+2 \\mathrm{HF} \\longrightarrow \\mathrm{CF}_{2} \\mathrm{Cl}_{2}+2 \\mathrm{HCl}\n$$"}
{"id": 3096, "contents": "739. Answer: - \n48.3\\%"}
{"id": 3097, "contents": "741. Green Chemistry and Atom Economy - \nThe purposeful design of chemical products and processes that minimize the use of environmentally hazardous substances and the generation of waste is known as green chemistry. Green chemistry is a philosophical approach that is being applied to many areas of science and technology, and its practice is summarized by guidelines known as the \"Twelve Principles of Green Chemistry\" (see details at this website (http://openstax.org/l/16greenchem)). One of the 12 principles is aimed specifically at maximizing the efficiency of processes for synthesizing chemical products. The atom economy of a process is a measure of this efficiency, defined as the percentage by mass of the final product of a synthesis relative to the masses of all the reactants used:\n\n$$\n\\text { atom economy }=\\frac{\\text { mass of product }}{\\text { mass of reactants }} \\times 100 \\%\n$$\n\nThough the definition of atom economy at first glance appears very similar to that for percent yield, be aware that this property represents a difference in the theoretical efficiencies of different chemical processes. The percent yield of a given chemical process, on the other hand, evaluates the efficiency of a process by comparing the yield of product actually obtained to the maximum yield predicted by stoichiometry.\n\nThe synthesis of the common nonprescription pain medication, ibuprofen, nicely illustrates the success of a green chemistry approach (Figure 7.15). First marketed in the early 1960s, ibuprofen was produced using a six-step synthesis that required 514 g of reactants to generate each mole ( 206 g ) of ibuprofen, an atom economy of $40 \\%$. In the 1990 s, an alternative process was developed by the BHC Company (now BASF Corporation) that requires only three steps and has an atom economy of $\\sim 80 \\%$, nearly twice that of the original process. The BHC process generates significantly less chemical waste; uses less-hazardous and recyclable materials; and provides significant cost-savings to the manufacturer (and, subsequently, the consumer). In recognition of the positive environmental impact of the BHC process, the company received the Environmental Protection Agency's Greener Synthetic Pathways Award in 1997.\n\n(a)\n\n\n(b)"}
{"id": 3098, "contents": "741. Green Chemistry and Atom Economy - \n(a)\n\n\n(b)\n\nFIGURE 7.15 (a) Ibuprofen is a popular nonprescription pain medication commonly sold as 200 mg tablets. (b) The BHC process for synthesizing ibuprofen requires only three steps and exhibits an impressive atom economy. (credit a: modification of work by Derrick Coetzee)"}
{"id": 3099, "contents": "742. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the fundamental aspects of titrations and gravimetric analysis.\n- Perform stoichiometric calculations using typical titration and gravimetric data.\n\nIn the 18th century, the strength (actually the concentration) of vinegar samples was determined by noting the amount of potassium carbonate, $\\mathrm{K}_{2} \\mathrm{CO}_{3}$, which had to be added, a little at a time, before bubbling ceased. The greater the weight of potassium carbonate added to reach the point where the bubbling ended, the more concentrated the vinegar.\n\nWe now know that the effervescence that occurred during this process was due to reaction with acetic acid, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$, the compound primarily responsible for the odor and taste of vinegar. Acetic acid reacts with potassium carbonate according to the following equation:\n\n$$\n2 \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(a q)+\\mathrm{K}_{2} \\mathrm{CO}_{3}(s) \\longrightarrow 2 \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{~K}(a q)+\\mathrm{CO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nThe bubbling was due to the production of $\\mathrm{CO}_{2}$.\nThe test of vinegar with potassium carbonate is one type of quantitative analysis-the determination of the\namount or concentration of a substance in a sample. In the analysis of vinegar, the concentration of the solute (acetic acid) was determined from the amount of reactant that combined with the solute present in a known volume of the solution. In other types of chemical analyses, the amount of a substance present in a sample is determined by measuring the amount of product that results."}
{"id": 3100, "contents": "743. Titration - \nThe described approach to measuring vinegar strength was an early version of the analytical technique known as titration analysis. A typical titration analysis involves the use of a buret (Figure 7.16) to make incremental additions of a solution containing a known concentration of some substance (the titrant) to a sample solution containing the substance whose concentration is to be measured (the analyte). The titrant and analyte undergo a chemical reaction of known stoichiometry, and so measuring the volume of titrant solution required for complete reaction with the analyte (the equivalence point of the titration) allows calculation of the analyte concentration. The equivalence point of a titration may be detected visually if a distinct change in the appearance of the sample solution accompanies the completion of the reaction. The halt of bubble formation in the classic vinegar analysis is one such example, though, more commonly, special dyes called indicators are added to the sample solutions to impart a change in color at or very near the equivalence point of the titration. Equivalence points may also be detected by measuring some solution property that changes in a predictable way during the course of the titration. Regardless of the approach taken to detect a titration's equivalence point, the volume of titrant actually measured is called the end point. Properly designed titration methods typically ensure that the difference between the equivalence and end points is negligible. Though any type of chemical reaction may serve as the basis for a titration analysis, the three described in this chapter (precipitation, acid-base, and redox) are most common. Additional details regarding titration analysis are provided in the chapter on acid-base equilibria.\n\n\nFIGURE 7.16 (a) A student fills a buret in preparation for a titration analysis. (b) A typical buret permits volume measurements to the nearest 0.01 mL . (credit a: modification of work by Mark Blaser and Matt Evans; credit b: modification of work by Mark Blaser and Matt Evans)"}
{"id": 3101, "contents": "745. Titration Analysis - \nThe end point in a titration of a $50.00-\\mathrm{mL}$ sample of aqueous HCl was reached by addition of 35.23 mL of 0.250 M NaOH titrant. The titration reaction is:\n\n$$\n\\mathrm{HCl}(a q)+\\mathrm{NaOH}(a q) \\longrightarrow \\mathrm{NaCl}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nWhat is the molarity of the HCl ?"}
{"id": 3102, "contents": "746. Solution - \nAs for all reaction stoichiometry calculations, the key issue is the relation between the molar amounts of the chemical species of interest as depicted in the balanced chemical equation. The approach outlined in previous modules of this chapter is followed, with additional considerations required, since the amounts of reactants provided and requested are expressed as solution concentrations.\n\nFor this exercise, the calculation will follow the following outlined steps:\n\n\nThe molar amount of HCl is calculated to be:\n\n$$\n35.23 \\mathrm{~mL} \\mathrm{NaOHI} \\times \\frac{1 \\mathrm{~L}}{1000 \\mathrm{~mL}} \\times \\frac{0.250 \\mathrm{~mol} \\mathrm{NaOH}}{1 \\mathrm{~L}} \\times \\frac{1 \\mathrm{~mol} \\mathrm{HCl}}{1 \\mathrm{~mol} \\mathrm{NaOH}}=8.81 \\times 10^{-3} \\mathrm{~mol} \\mathrm{HCl}\n$$\n\nUsing the provided volume of HCl solution and the definition of molarity, the HCl concentration is:\n\n$$\n\\begin{aligned}\nM & =\\frac{\\text { mol HCl }}{\\mathrm{L} \\text { solution }} \\\\\nM & =\\frac{8.81 \\times 10^{-3} \\mathrm{~mol} \\mathrm{HCl}}{50.00 \\mathrm{~mL} \\times \\frac{1 \\mathrm{~L}}{1000 \\mathrm{~mL}}} \\\\\nM & =0.176 M\n\\end{aligned}\n$$\n\nNote: For these types of titration calculations, it is convenient to recognize that solution molarity is also equal to the number of millimoles of solute per milliliter of solution:\n\n$$\nM=\\frac{\\text { mol solute }}{\\mathrm{L} \\text { solution }} \\times \\frac{\\frac{10^{3} \\mathrm{mmol}}{\\mathrm{~mol}}}{\\frac{10^{3} \\mathrm{~mL}}{\\mathrm{~L}}}=\\frac{\\mathrm{mmol} \\text { solute }}{\\mathrm{mL} \\text { solution }}\n$$"}
{"id": 3103, "contents": "746. Solution - \nUsing this version of the molarity unit will shorten the calculation by eliminating two conversion factors:"}
{"id": 3104, "contents": "747. Check Your Learning - \nA $20.00-\\mathrm{mL}$ sample of aqueous oxalic acid, $\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$, was titrated with a $0.09113-M$ solution of potassium permanganate, $\\mathrm{KMnO}_{4}$ (see net ionic equation below).\n\n$$\n2 \\mathrm{MnO}_{4}^{-}(a q)+5 \\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}(a q)+6 \\mathrm{H}^{+}(a q) \\longrightarrow 10 \\mathrm{CO}_{2}(g)+2 \\mathrm{Mn}^{2+}(a q)+8 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nA volume of 23.24 mL was required to reach the end point. What is the oxalic acid molarity?"}
{"id": 3105, "contents": "748. Answer: - \n0.2648 M"}
{"id": 3106, "contents": "749. Gravimetric Analysis - \nA gravimetric analysis is one in which a sample is subjected to some treatment that causes a change in the physical state of the analyte that permits its separation from the other components of the sample. Mass measurements of the sample, the isolated analyte, or some other component of the analysis system, used along with the known stoichiometry of the compounds involved, permit calculation of the analyte concentration. Gravimetric methods were the first techniques used for quantitative chemical analysis, and they remain important tools in the modern chemistry laboratory.\n\nThe required change of state in a gravimetric analysis may be achieved by various physical and chemical processes. For example, the moisture (water) content of a sample is routinely determined by measuring the mass of a sample before and after it is subjected to a controlled heating process that evaporates the water. Also common are gravimetric techniques in which the analyte is subjected to a precipitation reaction of the sort described earlier in this chapter. The precipitate is typically isolated from the reaction mixture by filtration, carefully dried, and then weighed (Figure 7.17). The mass of the precipitate may then be used, along with relevant stoichiometric relationships, to calculate analyte concentration.\n\n\nFIGURE 7.17 Precipitate may be removed from a reaction mixture by filtration."}
{"id": 3107, "contents": "751. Gravimetric Analysis - \nA $0.4550-\\mathrm{g}$ solid mixture containing $\\mathrm{MgSO}_{4}$ is dissolved in water and treated with an excess of $\\mathrm{Ba}\\left(\\mathrm{NO}_{3}\\right)_{2}$, resulting in the precipitation of 0.6168 g of $\\mathrm{BaSO}_{4}$.\n\n$$\n\\mathrm{MgSO}_{4}(a q)+\\mathrm{Ba}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q) \\longrightarrow \\mathrm{BaSO}_{4}(s)+\\mathrm{Mg}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)\n$$\n\nWhat is the concentration (mass percent) of $\\mathrm{MgSO}_{4}$ in the mixture?"}
{"id": 3108, "contents": "752. Solution - \nThe plan for this calculation is similar to others used in stoichiometric calculations, the central step being the connection between the moles of $\\mathrm{BaSO}_{4}$ and $\\mathrm{MgSO}_{4}$ through their stoichiometric factor. Once the mass of $\\mathrm{MgSO}_{4}$ is computed, it may be used along with the mass of the sample mixture to calculate the requested percentage concentration.\n\n\nThe mass of $\\mathrm{MgSO}_{4}$ that would yield the provided precipitate mass is\n\n\nThe concentration of $\\mathrm{MgSO}_{4}$ in the sample mixture is then calculated to be\n\n$$\n\\begin{aligned}\n& \\text { percent } \\mathrm{MgSO}_{4}=\\frac{\\text { mass } \\mathrm{MgSO}_{4}}{\\text { mass sample }} \\times 100 \\% \\\\\n& \\frac{0.3181 \\mathrm{~g}}{0.4550 \\mathrm{~g}} \\times 100 \\%=69.91 \\%\n\\end{aligned}\n$$"}
{"id": 3109, "contents": "753. Check Your Learning - \nWhat is the percent of chloride ion in a sample if 1.1324 g of the sample produces 1.0881 g of AgCl when treated with excess $\\mathrm{Ag}^{+}$?\n\n$$\n\\mathrm{Ag}^{+}(a q)+\\mathrm{Cl}^{-}(a q) \\longrightarrow \\mathrm{AgCl}(s)\n$$"}
{"id": 3110, "contents": "754. Answer: - \n23.76\\%\n\nThe elemental composition of hydrocarbons and related compounds may be determined via a gravimetric method known as combustion analysis. In a combustion analysis, a weighed sample of the compound is heated to a high temperature under a stream of oxygen gas, resulting in its complete combustion to yield gaseous products of known identities. The complete combustion of hydrocarbons, for example, will yield carbon dioxide and water as the only products. The gaseous combustion products are swept through separate, preweighed collection devices containing compounds that selectively absorb each product (Figure 7.18). The mass increase of each device corresponds to the mass of the absorbed product and may be used in an appropriate stoichiometric calculation to derive the mass of the relevant element.\n\n\nFIGURE 7.18 This schematic diagram illustrates the basic components of a combustion analysis device for determining the carbon and hydrogen content of a sample."}
{"id": 3111, "contents": "755. Combustion Analysis - \nPolyethylene is a hydrocarbon polymer used to produce food-storage bags and many other flexible plastic items. A combustion analysis of a $0.00126-\\mathrm{g}$ sample of polyethylene yields 0.00394 g of $\\mathrm{CO}_{2}$ and 0.00161 g of $\\mathrm{H}_{2} \\mathrm{O}$. What is the empirical formula of polyethylene?"}
{"id": 3112, "contents": "756. Solution - \nThe primary assumption in this exercise is that all the carbon in the sample combusted is converted to carbon dioxide, and all the hydrogen in the sample is converted to water:\n\n$$\n\\mathrm{C}_{\\mathrm{x}} \\mathrm{H}_{\\mathrm{y}}(s)+\\operatorname{excess} \\mathrm{O}_{2}(g) \\longrightarrow x \\mathrm{CO}_{2}(g)+\\frac{y}{2} \\mathrm{H}_{2} \\mathrm{O}(g)\n$$\n\nNote that a balanced equation is not necessary for the task at hand. To derive the empirical formula of the compound, only the subscripts $x$ and $y$ are needed.\n\nFirst, calculate the molar amounts of carbon and hydrogen in the sample, using the provided masses of the carbon dioxide and water, respectively. With these molar amounts, the empirical formula for the compound may be written as described in the previous chapter of this text. An outline of this approach is given in the following flow chart:\n\n\n$$\n\\begin{aligned}\n& \\text { mol C }=0.00394 \\mathrm{~g} \\mathrm{CO}_{2} \\times \\frac{1 \\mathrm{~mol} \\mathrm{CO}_{2}}{44.01 \\mathrm{~g}} \\times \\frac{1 \\mathrm{~mol} \\mathrm{C}_{2}}{1 \\mathrm{~mol} \\mathrm{CO}_{2}}=8.95 \\times 10^{-5} \\mathrm{~mol} \\mathrm{C} \\\\\n& \\mathrm{~mol} \\mathrm{H}\n\\end{aligned}=0.00161 \\mathrm{~g} \\mathrm{H}_{2} \\mathrm{O} \\times \\frac{1 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{O}}{18.02 \\mathrm{~g}} \\times \\frac{2 \\mathrm{~mol} \\mathrm{H}^{1 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{O}}=1.79 \\times 10^{-4} \\mathrm{~mol} \\mathrm{H}}{}\n$$"}
{"id": 3113, "contents": "756. Solution - \nThe empirical formula for the compound is then derived by identifying the smallest whole-number multiples for these molar amounts. The H -to- C molar ratio is\n\n$$\n\\frac{\\mathrm{mol} \\mathrm{H}}{\\mathrm{~mol} \\mathrm{C}}=\\frac{1.79 \\times 10^{-4} \\mathrm{~mol} \\mathrm{H}}{8.95 \\times 10^{-5} \\mathrm{~mol} \\mathrm{C}}=\\frac{2 \\mathrm{~mol} \\mathrm{H}}{1 \\mathrm{~mol} \\mathrm{C}}\n$$\n\nand the empirical formula for polyethylene is $\\mathrm{CH}_{2}$."}
{"id": 3114, "contents": "757. Check Your Learning - \nA $0.00215-\\mathrm{g}$ sample of polystyrene, a polymer composed of carbon and hydrogen, produced 0.00726 g of $\\mathrm{CO}_{2}$ and 0.00148 g of $\\mathrm{H}_{2} \\mathrm{O}$ in a combustion analysis. What is the empirical formula for polystyrene?"}
{"id": 3115, "contents": "758. Answer: - \nCH"}
{"id": 3116, "contents": "759. Key Terms - \nacid substance that produces $\\mathrm{H}_{3} \\mathrm{O}^{+}$when dissolved in water\nacid-base reaction reaction involving the transfer of a hydrogen ion between reactant species\nactual yield amount of product formed in a reaction\nanalyte chemical species of interest\nbalanced equation chemical equation with equal numbers of atoms for each element in the reactant and product\nbase substance that produces $\\mathrm{OH}^{-}$when dissolved in water\nburet device used for the precise delivery of variable liquid volumes, such as in a titration analysis\nchemical equation symbolic representation of a chemical reaction\ncoefficient number placed in front of symbols or formulas in a chemical equation to indicate their relative amount\ncombustion analysis gravimetric technique used to determine the elemental composition of a compound via the collection and weighing of its gaseous combustion products\ncombustion reaction vigorous redox reaction producing significant amounts of energy in the form of heat and, sometimes, light\ncomplete ionic equation chemical equation in which all dissolved ionic reactants and products, including spectator ions, are explicitly represented by formulas for their dissociated ions\nend point measured volume of titrant solution that yields the change in sample solution appearance or other property expected for stoichiometric equivalence (see equivalence point)\nequivalence point volume of titrant solution required to react completely with the analyte in a titration analysis; provides a stoichiometric amount of titrant for the sample's analyte according to the titration reaction\nexcess reactant reactant present in an amount greater than required by the reaction stoichiometry\ngravimetric analysis quantitative chemical analysis method involving the separation of an analyte from a sample by a physical or chemical process and subsequent mass measurements of the analyte, reaction product, and/or sample\nhalf-reaction an equation that shows whether each reactant loses or gains electrons in a reaction.\nindicator substance added to the sample in a\ntitration analysis to permit visual detection of the end point\ninsoluble of relatively low solubility; dissolving only to a slight extent\nlimiting reactant reactant present in an amount lower than required by the reaction stoichiometry, thus limiting the amount of product generated\nmolecular equation chemical equation in which all reactants and products are represented as neutral substances"}
{"id": 3117, "contents": "759. Key Terms - \nlimiting reactant reactant present in an amount lower than required by the reaction stoichiometry, thus limiting the amount of product generated\nmolecular equation chemical equation in which all reactants and products are represented as neutral substances\nnet ionic equation chemical equation in which only those dissolved ionic reactants and products that undergo a chemical or physical change are represented (excludes spectator ions)\nneutralization reaction reaction between an acid and a base to produce salt and water\noxidation process in which an element's oxidation number is increased by loss of electrons\noxidation number (also, oxidation state) the charge each atom of an element would have in a compound if the compound were ionic\noxidation-reduction reaction (also, redox reaction) reaction involving a change in oxidation number for one or more reactant elements\noxidizing agent (also, oxidant) substance that brings about the oxidation of another substance, and in the process becomes reduced\npercent yield measure of the efficiency of a reaction, expressed as a percentage of the theoretical yield\nprecipitate insoluble product that forms from reaction of soluble reactants\nprecipitation reaction reaction that produces one or more insoluble products; when reactants are ionic compounds, sometimes called doubledisplacement or metathesis\nproduct substance formed by a chemical or physical change; shown on the right side of the arrow in a chemical equation\nquantitative analysis the determination of the amount or concentration of a substance in a sample\nreactant substance undergoing a chemical or physical change; shown on the left side of the arrow in a chemical equation\nreducing agent (also, reductant) substance that brings about the reduction of another substance, and in the process becomes oxidized\nreduction process in which an element's oxidation number is decreased by gain of electrons\nsalt ionic compound that can be formed by the\nreaction of an acid with a base that contains a cation and an anion other than hydroxide or oxide\nsingle-displacement reaction (also, replacement) redox reaction involving the oxidation of an elemental substance by an ionic species\nsolubility the extent to which a substance may be dissolved in water, or any solvent\nsoluble of relatively high solubility; dissolving to a relatively large extent\nspectator ion ion that does not undergo a chemical or physical change during a reaction, but its presence is required to maintain charge neutrality"}
{"id": 3118, "contents": "759. Key Terms - \nsoluble of relatively high solubility; dissolving to a relatively large extent\nspectator ion ion that does not undergo a chemical or physical change during a reaction, but its presence is required to maintain charge neutrality\nstoichiometric factor ratio of coefficients in a balanced chemical equation, used in computations relating amounts of reactants and products\nstoichiometry relationships between the amounts"}
{"id": 3119, "contents": "760. Key Equations - \npercent yield $=\\left(\\frac{\\text { actual yield }}{\\text { theoretical yield }}\\right) \\times 100$"}
{"id": 3120, "contents": "761. Summary - 761.1. Writing and Balancing Chemical Equations\nChemical equations are symbolic representations of chemical and physical changes. Formulas for the substances undergoing the change (reactants) and substances generated by the change (products) are separated by an arrow and preceded by integer coefficients indicating their relative numbers. Balanced equations are those whose coefficients result in equal numbers of atoms for each element in the reactants and products. Chemical reactions in aqueous solution that involve ionic reactants or products may be represented more realistically by complete ionic equations and, more succinctly, by net ionic equations."}
{"id": 3121, "contents": "761. Summary - 761.2. Classifying Chemical Reactions\nChemical reactions are classified according to similar patterns of behavior. A large number of important reactions are included in three categories: precipitation, acid-base, and oxidationreduction (redox). Precipitation reactions involve the formation of one or more insoluble products. Acid-base reactions involve the transfer of hydrogen ions between reactants. Redox reactions involve a change in oxidation number for one or more reactant elements. Writing balanced equations for\nof reactants and products of a chemical reaction\nstrong acid acid that reacts completely when dissolved in water to yield hydronium ions\nstrong base base that reacts completely when dissolved in water to yield hydroxide ions\ntheoretical yield amount of product that may be produced from a given amount of reactant(s) according to the reaction stoichiometry\ntitrant solution containing a known concentration of substance that will react with the analyte in a titration analysis\ntitration analysis quantitative chemical analysis method that involves measuring the volume of a reactant solution required to completely react with the analyte in a sample\nweak acid acid that reacts only to a slight extent when dissolved in water to yield hydronium ions\nweak base base that reacts only to a slight extent when dissolved in water to yield hydroxide ions\nsome redox reactions that occur in aqueous solutions is simplified by using a systematic approach called the half-reaction method."}
{"id": 3122, "contents": "761. Summary - 761.3. Reaction Stoichiometry\nA balanced chemical equation may be used to describe a reaction's stoichiometry (the relationships between amounts of reactants and products). Coefficients from the equation are used to derive stoichiometric factors that subsequently may be used for computations relating reactant and product masses, molar amounts, and other quantitative properties."}
{"id": 3123, "contents": "761. Summary - 761.4. Reaction Yields\nWhen reactions are carried out using less-thanstoichiometric quantities of reactants, the amount of product generated will be determined by the limiting reactant. The amount of product generated by a chemical reaction is its actual yield. This yield is often less than the amount of product predicted by the stoichiometry of the balanced chemical equation representing the reaction (its theoretical yield). The extent to which a reaction generates the theoretical amount of product is expressed as its percent yield."}
{"id": 3124, "contents": "761. Summary - 761.5. Quantitative Chemical Analysis\nThe stoichiometry of chemical reactions may serve as the basis for quantitative chemical analysis methods. Titrations involve measuring the volume of a titrant solution required to completely react with a sample solution. This volume is then used to calculate the concentration of analyte in the sample using the stoichiometry of the titration reaction. Gravimetric analysis involves separating the analyte\nfrom the sample by a physical or chemical process, determining its mass, and then calculating its concentration in the sample based on the stoichiometry of the relevant process. Combustion analysis is a gravimetric method used to determine the elemental composition of a compound by collecting and weighing the gaseous products of its combustion."}
{"id": 3125, "contents": "762. Exercises - 762.1. Writing and Balancing Chemical Equations\n1. What does it mean to say an equation is balanced? Why is it important for an equation to be balanced?\n2. Consider molecular, complete ionic, and net ionic equations.\n(a) What is the difference between these types of equations?\n(b) In what circumstance would the complete and net ionic equations for a reaction be identical?\n3. Balance the following equations:\n(a) $\\mathrm{PCl}_{5}(s)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{POCl}_{3}(l)+\\mathrm{HCl}(a q)$\n(b) $\\mathrm{Cu}(s)+\\mathrm{HNO}_{3}(a q) \\longrightarrow \\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{NO}(g)$\n(c) $\\mathrm{H}_{2}(g)+\\mathrm{I}_{2}(s) \\longrightarrow \\mathrm{HI}(s)$\n(d) $\\mathrm{Fe}(s)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{Fe}_{2} \\mathrm{O}_{3}(s)$\n(e) $\\mathrm{Na}(s)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{NaOH}(a q)+\\mathrm{H}_{2}(g)$\n(f) $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}(s) \\longrightarrow \\mathrm{Cr}_{2} \\mathrm{O}_{3}(s)+\\mathrm{N}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(g)$\n(g) $\\mathrm{P}_{4}(s)+\\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{PCl}_{3}(l)$\n(h) $\\mathrm{PtCl}_{4}(s) \\longrightarrow \\mathrm{Pt}(s)+\\mathrm{Cl}_{2}(g)$\n4. Balance the following equations:"}
{"id": 3126, "contents": "762. Exercises - 762.1. Writing and Balancing Chemical Equations\n(h) $\\mathrm{PtCl}_{4}(s) \\longrightarrow \\mathrm{Pt}(s)+\\mathrm{Cl}_{2}(g)$\n4. Balance the following equations:\n(a) $\\mathrm{Ag}(s)+\\mathrm{H}_{2} \\mathrm{~S}(g)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{Ag}_{2} \\mathrm{~S}(s)+\\mathrm{H}_{2} \\mathrm{O}(l)$\n(b) $\\mathrm{P}_{4}(s)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{P}_{4} \\mathrm{O}_{10}(s)$\n(c) $\\mathrm{Pb}(s)+\\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{Pb}(\\mathrm{OH})_{2}(s)$\n(d) $\\mathrm{Fe}(s)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Fe}_{3} \\mathrm{O}_{4}(s)+\\mathrm{H}_{2}(g)$\n(e) $\\mathrm{Sc}_{2} \\mathrm{O}_{3}(s)+\\mathrm{SO}_{3}(l) \\longrightarrow \\mathrm{Sc}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}(s)$\n(f) $\\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}(a q)+\\mathrm{H}_{3} \\mathrm{PO}_{4}(a q) \\longrightarrow \\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}(a q)$\n(g) $\\mathrm{Al}(s)+\\mathrm{H}_{2} \\mathrm{SO}_{4}(a q) \\longrightarrow \\mathrm{Al}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}(s)+\\mathrm{H}_{2}(g)$\n(h) $\\mathrm{TiCl}_{4}(s)+\\mathrm{H}_{2} \\mathrm{O}(g) \\longrightarrow \\mathrm{TiO}_{2}(s)+\\mathrm{HCl}(g)$\n5. Write a balanced molecular equation describing each of the following chemical reactions."}
{"id": 3127, "contents": "762. Exercises - 762.1. Writing and Balancing Chemical Equations\n5. Write a balanced molecular equation describing each of the following chemical reactions.\n(a) Solid calcium carbonate is heated and decomposes to solid calcium oxide and carbon dioxide gas.\n(b) Gaseous butane, $\\mathrm{C}_{4} \\mathrm{H}_{10}$, reacts with diatomic oxygen gas to yield gaseous carbon dioxide and water vapor.\n(c) Aqueous solutions of magnesium chloride and sodium hydroxide react to produce solid magnesium hydroxide and aqueous sodium chloride.\n(d) Water vapor reacts with sodium metal to produce solid sodium hydroxide and hydrogen gas.\n6. Write a balanced equation describing each of the following chemical reactions.\n(a) Solid potassium chlorate, $\\mathrm{KClO}_{3}$, decomposes to form solid potassium chloride and diatomic oxygen gas.\n(b) Solid aluminum metal reacts with solid diatomic iodine to form solid $\\mathrm{Al}_{2} \\mathrm{I}_{6}$.\n(c) When solid sodium chloride is added to aqueous sulfuric acid, hydrogen chloride gas and aqueous sodium sulfate are produced.\n(d) Aqueous solutions of phosphoric acid and potassium hydroxide react to produce aqueous potassium dihydrogen phosphate and liquid water.\n7. Colorful fireworks often involve the decomposition of barium nitrate and potassium chlorate and the reaction of the metals magnesium, aluminum, and iron with oxygen.\n(a) Write the formulas of barium nitrate and potassium chlorate.\n(b) The decomposition of solid potassium chlorate leads to the formation of solid potassium chloride and diatomic oxygen gas. Write an equation for the reaction.\n(c) The decomposition of solid barium nitrate leads to the formation of solid barium oxide, diatomic nitrogen gas, and diatomic oxygen gas. Write an equation for the reaction.\n(d) Write separate equations for the reactions of the solid metals magnesium, aluminum, and iron with diatomic oxygen gas to yield the corresponding metal oxides. (Assume the iron oxide contains $\\mathrm{Fe}^{3+}$ ions.)\n8. Fill in the blank with a single chemical formula for a covalent compound that will balance the equation:"}
{"id": 3128, "contents": "762. Exercises - 762.1. Writing and Balancing Chemical Equations\n9. Aqueous hydrogen fluoride (hydrofluoric acid) is used to etch glass and to analyze minerals for their silicon content. Hydrogen fluoride will also react with sand (silicon dioxide).\n(a) Write an equation for the reaction of solid silicon dioxide with hydrofluoric acid to yield gaseous silicon tetrafluoride and liquid water.\n(b) The mineral fluorite (calcium fluoride) occurs extensively in Illinois. Solid calcium fluoride can also be prepared by the reaction of aqueous solutions of calcium chloride and sodium fluoride, yielding aqueous sodium chloride as the other product. Write complete and net ionic equations for this reaction.\n10. A novel process for obtaining magnesium from sea water involves several reactions. Write a balanced chemical equation for each step of the process.\n(a) The first step is the decomposition of solid calcium carbonate from seashells to form solid calcium oxide and gaseous carbon dioxide.\n(b) The second step is the formation of solid calcium hydroxide as the only product from the reaction of the solid calcium oxide with liquid water.\n(c) Solid calcium hydroxide is then added to the seawater, reacting with dissolved magnesium chloride to yield solid magnesium hydroxide and aqueous calcium chloride.\n(d) The solid magnesium hydroxide is added to a hydrochloric acid solution, producing dissolved magnesium chloride and liquid water.\n(e) Finally, the magnesium chloride is melted and electrolyzed to yield liquid magnesium metal and diatomic chlorine gas.\n11. From the balanced molecular equations, write the complete ionic and net ionic equations for the following:\n(a) $\\mathrm{K}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}(a q)+\\mathrm{Ba}(\\mathrm{OH})_{2}(a q) \\longrightarrow 2 \\mathrm{KOH}(a q)+\\mathrm{BaC}_{2} \\mathrm{O}_{4}(s)$\n(b) $\\mathrm{Pb}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+\\mathrm{H}_{2} \\mathrm{SO}_{4}(a q) \\longrightarrow \\mathrm{PbSO}_{4}(s)+2 \\mathrm{HNO}_{3}(a q)$"}
{"id": 3129, "contents": "762. Exercises - 762.1. Writing and Balancing Chemical Equations\n(c) $\\mathrm{CaCO}_{3}(s)+\\mathrm{H}_{2} \\mathrm{SO}_{4}(a q) \\longrightarrow \\mathrm{CaSO}_{4}(s)+\\mathrm{CO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l)$"}
{"id": 3130, "contents": "762. Exercises - 762.2. Classifying Chemical Reactions\n12. Use the following equations to answer the next four questions:\ni. $\\mathrm{H}_{2} \\mathrm{O}(s) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(l)$\nii. $\\mathrm{Na}^{+}(a q)+\\mathrm{Cl}^{-}(\\mathrm{aq})+\\mathrm{Ag}^{+}(a q)+\\mathrm{NO}_{3}{ }^{-}(a q) \\longrightarrow \\mathrm{AgCl}(s)+\\mathrm{Na}^{+}(a q)+\\mathrm{NO}_{3}{ }^{-}(a q)$\niii. $\\mathrm{CH}_{3} \\mathrm{OH}(g)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(g)$\niv. $2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{H}_{2}(g)+\\mathrm{O}_{2}(g)$\nv. $\\mathrm{H}^{+}(a q)+\\mathrm{OH}^{-}(a q) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(l)$\n(a) Which equation describes a physical change?\n(b) Which equation identifies the reactants and products of a combustion reaction?\n(c) Which equation is not balanced?\n(d) Which is a net ionic equation?\n13. Indicate what type, or types, of reaction each of the following represents:\n(a) $\\mathrm{Ca}(s)+\\mathrm{Br}_{2}(l) \\longrightarrow \\mathrm{CaBr}_{2}(s)$\n(b) $\\mathrm{Ca}(\\mathrm{OH})_{2}(a q)+2 \\mathrm{HBr}(a q) \\longrightarrow \\mathrm{CaBr}_{2}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n(c) $\\mathrm{C}_{6} \\mathrm{H}_{12}(l)+9 \\mathrm{O}_{2}(g) \\longrightarrow 6 \\mathrm{CO}_{2}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(g)$\n14. Indicate what type, or types, of reaction each of the following represents:"}
{"id": 3131, "contents": "762. Exercises - 762.2. Classifying Chemical Reactions\n14. Indicate what type, or types, of reaction each of the following represents:\n(a) $\\mathrm{H}_{2} \\mathrm{O}(g)+\\mathrm{C}(s) \\longrightarrow \\mathrm{CO}(g)+\\mathrm{H}_{2}(g)$\n(b) $2 \\mathrm{KClO}_{3}(s) \\longrightarrow 2 \\mathrm{KCl}(s)+3 \\mathrm{O}_{2}(g)$\n(c) $\\mathrm{Al}(\\mathrm{OH})_{3}(a q)+3 \\mathrm{HCl}(a q) \\longrightarrow \\mathrm{AlCl}_{3}(a q)+3 \\mathrm{H}_{2} \\mathrm{O}(l)$\n(d) $\\mathrm{Pb}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+\\mathrm{H}_{2} \\mathrm{SO}_{4}(a q) \\longrightarrow \\mathrm{PbSO}_{4}(s)+2 \\mathrm{HNO}_{3}(a q)$\n15. Silver can be separated from gold because silver dissolves in nitric acid while gold does not. Is the dissolution of silver in nitric acid an acid-base reaction or an oxidation-reduction reaction? Explain your answer.\n16. Determine the oxidation states of the elements in the following compounds:\n(a) NaI\n(b) $\\mathrm{GdCl}_{3}$\n(c) $\\mathrm{LiNO}_{3}$\n(d) $\\mathrm{H}_{2} \\mathrm{Se}$\n(e) $\\mathrm{Mg}_{2} \\mathrm{Si}$\n(f) $\\mathrm{RbO}_{2}$, rubidium superoxide\n(g) HF\n17. Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides.\n(a) $\\mathrm{H}_{3} \\mathrm{PO}_{4}$\n(b) $\\mathrm{Al}(\\mathrm{OH})_{3}$\n(c) $\\mathrm{SeO}_{2}$\n(d) $\\mathrm{KNO}_{2}$\n(e) $\\mathrm{In}_{2} \\mathrm{~S}_{3}$\n(f) $\\mathrm{P}_{4} \\mathrm{O}_{6}$"}
{"id": 3132, "contents": "762. Exercises - 762.2. Classifying Chemical Reactions\n(d) $\\mathrm{KNO}_{2}$\n(e) $\\mathrm{In}_{2} \\mathrm{~S}_{3}$\n(f) $\\mathrm{P}_{4} \\mathrm{O}_{6}$\n18. Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides.\n(a) $\\mathrm{H}_{2} \\mathrm{SO}_{4}$\n(b) $\\mathrm{Ca}(\\mathrm{OH})_{2}$\n(c) BrOH\n(d) $\\mathrm{ClNO}_{2}$\n(e) $\\mathrm{TiCl}_{4}$\n(f) NaH\n19. Classify the following as acid-base reactions or oxidation-reduction reactions:\n(a) $\\mathrm{Na}_{2} \\mathrm{~S}(a q)+2 \\mathrm{HCl}(a q) \\longrightarrow 2 \\mathrm{NaCl}(a q)+\\mathrm{H}_{2} \\mathrm{~S}(g)$\n(b) $2 \\mathrm{Na}(s)+2 \\mathrm{HCl}(a q) \\longrightarrow 2 \\mathrm{NaCl}(a q)+\\mathrm{H}_{2}(g)$\n(c) $\\mathrm{Mg}(s)+\\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{MgCl}_{2}(s)$\n(d) $\\mathrm{MgO}(s)+2 \\mathrm{HCl}(a q) \\longrightarrow \\mathrm{MgCl}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)$\n(e) $\\mathrm{K}_{3} \\mathrm{P}(s)+2 \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{K}_{3} \\mathrm{PO}_{4}(s)$\n(f) $3 \\mathrm{KOH}(a q)+\\mathrm{H}_{3} \\mathrm{PO}_{4}(a q) \\longrightarrow \\mathrm{K}_{3} \\mathrm{PO}_{4}(a q)+3 \\mathrm{H}_{2} \\mathrm{O}(l)$\n20. Identify the atoms that are oxidized and reduced, the change in oxidation state for each, and the oxidizing and reducing agents in each of the following equations:"}
{"id": 3133, "contents": "762. Exercises - 762.2. Classifying Chemical Reactions\n20. Identify the atoms that are oxidized and reduced, the change in oxidation state for each, and the oxidizing and reducing agents in each of the following equations:\n(a) $\\mathrm{Mg}(s)+\\mathrm{NiCl}_{2}(a q) \\longrightarrow \\mathrm{MgCl}_{2}(a q)+\\mathrm{Ni}(s)$\n(b) $\\mathrm{PCl}_{3}(l)+\\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{PCl}_{5}(s)$\n(c) $\\mathrm{C}_{2} \\mathrm{H}_{4}(g)+3 \\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{CO}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(g)$\n(d) $\\mathrm{Zn}(s)+\\mathrm{H}_{2} \\mathrm{SO}_{4}(a q) \\longrightarrow \\mathrm{ZnSO}_{4}(a q)+\\mathrm{H}_{2}(g)$\n(e) $2 \\mathrm{~K}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}(s)+\\mathrm{I}_{2}(s) \\longrightarrow \\mathrm{K}_{2} \\mathrm{~S}_{4} \\mathrm{O}_{6}(s)+2 \\mathrm{KI}(s)$\n(f) $3 \\mathrm{Cu}(s)+8 \\mathrm{HNO}_{3}(a q) \\longrightarrow 3 \\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+2 \\mathrm{NO}(g)+4 \\mathrm{H}_{2} \\mathrm{O}(l)$\n21. Complete and balance the following acid-base equations:\n(a) HCl gas reacts with solid $\\mathrm{Ca}(\\mathrm{OH})_{2}(s)$.\n(b) A solution of $\\mathrm{Sr}(\\mathrm{OH})_{2}$ is added to a solution of $\\mathrm{HNO}_{3}$.\n22. Complete and balance the following acid-base equations:\n(a) A solution of $\\mathrm{HClO}_{4}$ is added to a solution of LiOH .\n(b) Aqueous $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ reacts with NaOH ."}
{"id": 3134, "contents": "762. Exercises - 762.2. Classifying Chemical Reactions\n(a) A solution of $\\mathrm{HClO}_{4}$ is added to a solution of LiOH .\n(b) Aqueous $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ reacts with NaOH .\n(c) $\\mathrm{Ba}(\\mathrm{OH})_{2}$ reacts with HF gas.\n23. Complete and balance the following oxidation-reduction reactions, which give the highest possible oxidation state for the oxidized atoms.\n(a) $\\mathrm{Al}(s)+\\mathrm{F}_{2}(g) \\longrightarrow$\n(b) $\\mathrm{Al}(s)+\\mathrm{CuBr}_{2}(a q) \\longrightarrow$ (single displacement)\n(c) $\\mathrm{P}_{4}(s)+\\mathrm{O}_{2}(g) \\longrightarrow$\n(d) $\\mathrm{Ca}(s)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow$ (products are a strong base and a diatomic gas)\n24. Complete and balance the following oxidation-reduction reactions, which give the highest possible oxidation state for the oxidized atoms.\n(a) $\\mathrm{K}(s)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow$\n(b) $\\mathrm{Ba}(s)+\\mathrm{HBr}(a q) \\longrightarrow$\n(c) $\\operatorname{Sn}(s)+\\mathrm{I}_{2}(s) \\longrightarrow$\n25. Complete and balance the equations for the following acid-base neutralization reactions. If water is used as a solvent, write the reactants and products as aqueous ions. In some cases, there may be more than one correct answer, depending on the amounts of reactants used.\n(a) $\\mathrm{Mg}(\\mathrm{OH})_{2}(s)+\\mathrm{HClO}_{4}(a q) \\longrightarrow$\n(b) $\\mathrm{SO}_{3}(g)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow$ (assume an excess of water and that the product dissolves)\n(c) $\\mathrm{SrO}(s)+\\mathrm{H}_{2} \\mathrm{SO}_{4}(l) \\longrightarrow$"}
{"id": 3135, "contents": "762. Exercises - 762.2. Classifying Chemical Reactions\n(c) $\\mathrm{SrO}(s)+\\mathrm{H}_{2} \\mathrm{SO}_{4}(l) \\longrightarrow$\n26. When heated to $700-800^{\\circ} \\mathrm{C}$, diamonds, which are pure carbon, are oxidized by atmospheric oxygen. (They burn!) Write the balanced equation for this reaction.\n27. The military has experimented with lasers that produce very intense light when fluorine combines explosively with hydrogen. What is the balanced equation for this reaction?\n28. Write the molecular, total ionic, and net ionic equations for the following reactions:\n(a) $\\mathrm{Ca}(\\mathrm{OH})_{2}(a q)+\\mathrm{HC}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}(a q) \\longrightarrow$\n(b) $\\mathrm{H}_{3} \\mathrm{PO}_{4}(a q)+\\mathrm{CaCl}_{2}(a q) \\longrightarrow$\n29. Great Lakes Chemical Company produces bromine, $\\mathrm{Br}_{2}$, from bromide salts such as NaBr , in Arkansas brine by treating the brine with chlorine gas. Write a balanced equation for the reaction of NaBr with $\\mathrm{Cl}_{2}$.\n30. In a common experiment in the general chemistry laboratory, magnesium metal is heated in air to produce MgO. MgO is a white solid, but in these experiments it often looks gray, due to small amounts of $\\mathrm{Mg}_{3} \\mathrm{~N}_{2}$, a compound formed as some of the magnesium reacts with nitrogen. Write a balanced equation for each reaction.\n31. Lithium hydroxide may be used to absorb carbon dioxide in enclosed environments, such as manned spacecraft and submarines. Write an equation for the reaction that involves 2 mol of LiOH per 1 mol of $\\mathrm{CO}_{2}$. (Hint: Water is one of the products.)"}
{"id": 3136, "contents": "762. Exercises - 762.2. Classifying Chemical Reactions\n32. Calcium propionate is sometimes added to bread to retard spoilage. This compound can be prepared by the reaction of calcium carbonate, $\\mathrm{CaCO}_{3}$, with propionic acid, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{CO}_{2} \\mathrm{H}$, which has properties similar to those of acetic acid. Write the balanced equation for the formation of calcium propionate.\n33. Complete and balance the equations of the following reactions, each of which could be used to remove hydrogen sulfide from natural gas:\n(a) $\\mathrm{Ca}(\\mathrm{OH})_{2}(s)+\\mathrm{H}_{2} \\mathrm{~S}(g) \\longrightarrow$\n(b) $\\mathrm{Na}_{2} \\mathrm{CO}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{~S}(g) \\longrightarrow$\n34. Copper(II) sulfide is oxidized by molecular oxygen to produce gaseous sulfur trioxide and solid copper(II) oxide. The gaseous product then reacts with liquid water to produce liquid dihydrogen sulfate as the only product. Write the two equations which represent these reactions.\n35. Write balanced chemical equations for the reactions used to prepare each of the following compounds from the given starting material(s). In some cases, additional reactants may be required.\n(a) solid ammonium nitrate from gaseous molecular nitrogen via a two-step process (first reduce the nitrogen to ammonia, then neutralize the ammonia with an appropriate acid)\n(b) gaseous hydrogen bromide from liquid molecular bromine via a one-step redox reaction\n(c) gaseous $\\mathrm{H}_{2} \\mathrm{~S}$ from solid Zn and S via a two-step process (first a redox reaction between the starting materials, then reaction of the product with a strong acid)"}
{"id": 3137, "contents": "762. Exercises - 762.2. Classifying Chemical Reactions\n36. Calcium cyclamate $\\mathrm{Ca}\\left(\\mathrm{C}_{6} \\mathrm{H}_{11} \\mathrm{NHSO}_{3}\\right)_{2}$ is an artificial sweetener used in many countries around the world but is banned in the United States. It can be purified industrially by converting it to the barium salt through reaction of the acid $\\mathrm{C}_{6} \\mathrm{H}_{11} \\mathrm{NHSO}_{3} \\mathrm{H}$ with barium carbonate, treatment with sulfuric acid (barium sulfate is very insoluble), and then neutralization with calcium hydroxide. Write the balanced equations for these reactions.\n37. Complete and balance each of the following half-reactions (steps $2-5$ in half-reaction method):\n(a) $\\mathrm{Sn}^{4+}(a q) \\longrightarrow \\mathrm{Sn}^{2+}(a q)$\n(b) $\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}(a q) \\longrightarrow \\mathrm{Ag}(s)+\\mathrm{NH}_{3}(a q)$\n(c) $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}(s) \\longrightarrow \\mathrm{Hg}(l)+\\mathrm{Cl}^{-}(a q)$\n(d) $\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{O}_{2}(g)$ (in acidic solution)\n(e) $\\mathrm{IO}_{3}^{-}(a q) \\longrightarrow \\mathrm{I}_{2}(s)$ (in basic solution)\n(f) $\\mathrm{SO}_{3}{ }^{2-}(a q) \\longrightarrow \\mathrm{SO}_{4}{ }^{2-}(a q)$ (in acidic solution)\n(g) $\\mathrm{MnO}_{4}{ }^{-}(a q) \\longrightarrow \\mathrm{Mn}^{2+}(a q)$ (in acidic solution)\n(h) $\\mathrm{Cl}^{-}(a q) \\longrightarrow \\mathrm{ClO}_{3}{ }^{-}(a q)$ (in basic solution)\n38. Complete and balance each of the following half-reactions (steps $2-5$ in half-reaction method):"}
{"id": 3138, "contents": "762. Exercises - 762.2. Classifying Chemical Reactions\n38. Complete and balance each of the following half-reactions (steps $2-5$ in half-reaction method):\n(a) $\\mathrm{Cr}^{2+}(a q) \\longrightarrow \\mathrm{Cr}^{3+}(a q)$\n(b) $\\mathrm{Hg}(l)+\\mathrm{Br}^{-}(a q) \\longrightarrow \\mathrm{HgBr}_{4}{ }^{2-}(a q)$\n(c) $\\mathrm{ZnS}(s) \\longrightarrow \\mathrm{Zn}(s)+\\mathrm{S}^{2-}(a q)$\n(d) $\\mathrm{H}_{2}(g) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}$ (l (in basic solution)\n(e) $\\mathrm{H}_{2}(g) \\longrightarrow \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)$ (in acidic solution)\n(f) $\\mathrm{NO}_{3}{ }^{-}(a q) \\longrightarrow \\mathrm{HNO}_{2}(a q)$ (in acidic solution)\n(g) $\\mathrm{MnO}_{2}(s) \\longrightarrow \\mathrm{MnO}_{4}{ }^{-}(a q)$ (in basic solution)\n(h) $\\mathrm{Cl}^{-}(a q) \\longrightarrow \\mathrm{ClO}_{3}{ }^{-}(a q)$ (in acidic solution)\n39. Balance each of the following equations according to the half-reaction method:\n(a) $\\mathrm{Sn}^{2+}(a q)+\\mathrm{Cu}^{2+}(a q) \\longrightarrow \\mathrm{Sn}^{4+}(a q)+\\mathrm{Cu}^{+}(a q)$\n(b) $\\mathrm{H}_{2} \\mathrm{~S}(g)+\\mathrm{Hg}_{2}{ }^{2+}(a q) \\longrightarrow \\mathrm{Hg}(l)+\\mathrm{S}(s)$ (in acid)\n(c) $\\mathrm{CN}^{-}(a q)+\\mathrm{ClO}_{2}(a q) \\longrightarrow \\mathrm{CNO}^{-}(a q)+\\mathrm{Cl}^{-}(a q)$ (in acid)"}
{"id": 3139, "contents": "762. Exercises - 762.2. Classifying Chemical Reactions\n(d) $\\mathrm{Fe}^{2+}(a q)+\\mathrm{Ce}^{4+}(a q) \\longrightarrow \\mathrm{Fe}^{3+}(a q)+\\mathrm{Ce}^{3+}(a q)$\n(e) $\\mathrm{HBrO}(a q) \\longrightarrow \\mathrm{Br}^{-}(a q)+\\mathrm{O}_{2}(g)$ (in acid)\n40. Balance each of the following equations according to the half-reaction method:\n(a) $\\mathrm{Zn}(s)+\\mathrm{NO}_{3}{ }^{-}(a q) \\longrightarrow \\mathrm{Zn}^{2+}(a q)+\\mathrm{N}_{2}(g)$ (in acid)\n(b) $\\mathrm{Zn}(s)+\\mathrm{NO}_{3}{ }^{-}(a q) \\longrightarrow \\mathrm{Zn}^{2+}(a q)+\\mathrm{NH}_{3}(a q)$ (in base)\n(c) $\\mathrm{CuS}(s)+\\mathrm{NO}_{3}{ }^{-}(a q) \\longrightarrow \\mathrm{Cu}^{2+}(a q)+\\mathrm{S}(s)+\\mathrm{NO}(g)$ (in acid)\n(d) $\\mathrm{NH}_{3}(a q)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{NO}_{2}(g)$ (gas phase)\n(e) $\\mathrm{H}_{2} \\mathrm{O}_{2}(a q)+\\mathrm{MnO}_{4}^{-}(a q) \\longrightarrow \\mathrm{Mn}^{2+}(a q)+\\mathrm{O}_{2}(g)$ (in acid)\n(f) $\\mathrm{NO}_{2}(g) \\longrightarrow \\mathrm{NO}_{3}{ }^{-}(a q)+\\mathrm{NO}_{2}{ }^{-}(a q)$ (in base)\n(g) $\\mathrm{Fe}^{3+}(a q)+\\mathrm{I}^{-}(a q) \\longrightarrow \\mathrm{Fe}^{2+}(a q)+\\mathrm{I}_{2}(a q)$\n41. Balance each of the following equations according to the half-reaction method:"}
{"id": 3140, "contents": "762. Exercises - 762.2. Classifying Chemical Reactions\n41. Balance each of the following equations according to the half-reaction method:\n(a) $\\mathrm{MnO}_{4}{ }^{-}(a q)+\\mathrm{NO}_{2}{ }^{-}(a q) \\longrightarrow \\mathrm{MnO}_{2}(s)+\\mathrm{NO}_{3}{ }^{-}(a q)$ (in base)\n(b) $\\mathrm{MnO}_{4}{ }^{2-}(a q) \\longrightarrow \\mathrm{MnO}_{4}{ }^{-}(a q)+\\mathrm{MnO}_{2}(s)$ (in base)\n(c) $\\mathrm{Br}_{2}(l)+\\mathrm{SO}_{2}(g) \\longrightarrow \\mathrm{Br}^{-}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q)$ (in acid)"}
{"id": 3141, "contents": "762. Exercises - 762.3. Reaction Stoichiometry\n42. Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following:\n(a) The number of moles and the mass of chlorine, $\\mathrm{Cl}_{2}$, required to react with 10.0 g of sodium metal, Na , to produce sodium chloride, NaCl .\n(b) The number of moles and the mass of oxygen formed by the decomposition of 1.252 g of mercury(II) oxide.\n(c) The number of moles and the mass of sodium nitrate, $\\mathrm{NaNO}_{3}$, required to produce 128 g of oxygen. ( $\\mathrm{NaNO}_{2}$ is the other product.)\n(d) The number of moles and the mass of carbon dioxide formed by the combustion of 20.0 kg of carbon in an excess of oxygen.\n(e) The number of moles and the mass of copper(II) carbonate needed to produce 1.500 kg of copper(II) oxide. ( $\\mathrm{CO}_{2}$ is the other product.)\n(f)\n\nThe number of moles and the mass of\nformed by the reaction of 12.85 g of"}
{"id": 3142, "contents": "762. Exercises - 762.3. Reaction Stoichiometry\nThe number of moles and the mass of\nformed by the reaction of 12.85 g of\n\nwith an excess of $\\mathrm{Br}_{2}$.\n43. Determine the number of moles and the mass requested for each reaction in Exercise 7.42.\n44. Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following:\n(a) The number of moles and the mass of Mg required to react with 5.00 g of HCl and produce $\\mathrm{MgCl}_{2}$ and $\\mathrm{H}_{2}$.\n(b) The number of moles and the mass of oxygen formed by the decomposition of 1.252 g of $\\operatorname{silver(I)~oxide.~}$\n(c) The number of moles and the mass of magnesium carbonate, $\\mathrm{MgCO}_{3}$, required to produce 283 g of carbon dioxide. (MgO is the other product.)\n(d) The number of moles and the mass of water formed by the combustion of 20.0 kg of acetylene, $\\mathrm{C}_{2} \\mathrm{H}_{2}$, in an excess of oxygen.\n(e) The number of moles and the mass of barium peroxide, $\\mathrm{BaO}_{2}$, needed to produce 2.500 kg of barium oxide, $\\mathrm{BaO}\\left(\\mathrm{O}_{2}\\right.$ is the other product.)\n(f)\n\nThe number of moles and the mass of"}
{"id": 3143, "contents": "762. Exercises - 762.3. Reaction Stoichiometry\n45. Determine the number of moles and the mass requested for each reaction in Exercise 7.44.\n46. $\\mathrm{H}_{2}$ is produced by the reaction of 118.5 mL of a $0.8775-\\mathrm{M}$ solution of $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ according to the following equation: $2 \\mathrm{Cr}+2 \\mathrm{H}_{3} \\mathrm{PO}_{4} \\longrightarrow 3 \\mathrm{H}_{2}+2 \\mathrm{CrPO}_{4}$.\n(a) Outline the steps necessary to determine the number of moles and mass of $\\mathrm{H}_{2}$.\n(b) Perform the calculations outlined.\n47. Gallium chloride is formed by the reaction of 2.6 L of a 1.44 M solution of HCl according to the following equation: $2 \\mathrm{Ga}+6 \\mathrm{HCl} \\longrightarrow 2 \\mathrm{GaCl}_{3}+3 \\mathrm{H}_{2}$.\n(a) Outline the steps necessary to determine the number of moles and mass of gallium chloride.\n(b) Perform the calculations outlined.\n48. $\\mathrm{I}_{2}$ is produced by the reaction of 0.4235 mol of $\\mathrm{CuCl}_{2}$ according to the following equation: $2 \\mathrm{CuCl}_{2}+4 \\mathrm{KI} \\longrightarrow 2 \\mathrm{CuI}+4 \\mathrm{KCl}+\\mathrm{I}_{2}$.\n(a) How many molecules of $\\mathrm{I}_{2}$ are produced?\n(b) What mass of $\\mathrm{I}_{2}$ is produced?\n49. Silver is often extracted from ores such as $\\mathrm{K}\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]$ and then recovered by the reaction $2 \\mathrm{~K}\\left\\mathrm{Ag}(\\mathrm{CN})_{2}\\right+\\mathrm{Zn}(s) \\longrightarrow 2 \\mathrm{Ag}(s)+\\mathrm{Zn}(\\mathrm{CN})_{2}(a q)+2 \\mathrm{KCN}(a q)$"}
{"id": 3144, "contents": "762. Exercises - 762.3. Reaction Stoichiometry\n(a) How many molecules of $\\mathrm{Zn}(\\mathrm{CN})_{2}$ are produced by the reaction of 35.27 g of $\\mathrm{K}\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]$ ?\n(b) What mass of $\\mathrm{Zn}(\\mathrm{CN})_{2}$ is produced?\n50. What mass of silver oxide, $\\mathrm{Ag}_{2} \\mathrm{O}$, is required to produce 25.0 g of silver sulfadiazine, $\\mathrm{AgC}_{10} \\mathrm{H}_{9} \\mathrm{~N}_{4} \\mathrm{SO}_{2}$, from the reaction of silver oxide and sulfadiazine?\n$2 \\mathrm{C}_{10} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{SO}_{2}+\\mathrm{Ag}_{2} \\mathrm{O} \\longrightarrow 2 \\mathrm{AgC}_{10} \\mathrm{H}_{9} \\mathrm{~N}_{4} \\mathrm{SO}_{2}+\\mathrm{H}_{2} \\mathrm{O}$\n51. Carborundum is silicon carbide, SiC , a very hard material used as an abrasive on sandpaper and in other applications. It is prepared by the reaction of pure sand, $\\mathrm{SiO}_{2}$, with carbon at high temperature. Carbon monoxide, CO , is the other product of this reaction. Write the balanced equation for the reaction, and calculate how much $\\mathrm{SiO}_{2}$ is required to produce 3.00 kg of SiC .\n52. Automotive air bags inflate when a sample of sodium azide, $\\mathrm{NaN}_{3}$, is very rapidly decomposed.\n$2 \\mathrm{NaN}_{3}(s) \\longrightarrow 2 \\mathrm{Na}(s)+3 \\mathrm{~N}_{2}(g)$\nWhat mass of sodium azide is required to produce $2.6 \\mathrm{ft}^{3}(73.6 \\mathrm{~L})$ of nitrogen gas with a density of $1.25 \\mathrm{~g} /$ L?"}
{"id": 3145, "contents": "762. Exercises - 762.3. Reaction Stoichiometry\n53. Urea, $\\mathrm{CO}\\left(\\mathrm{NH}_{2}\\right)_{2}$, is manufactured on a large scale for use in producing urea-formaldehyde plastics and as a fertilizer. What is the maximum mass of urea that can be manufactured from the $\\mathrm{CO}_{2}$ produced by combustion of $1.00 \\times 10^{3} \\mathrm{~kg}$ of carbon followed by the reaction?\n$\\mathrm{CO}_{2}(g)+2 \\mathrm{NH}_{3}(g) \\longrightarrow \\mathrm{CO}\\left(\\mathrm{NH}_{2}\\right)_{2}(s)+\\mathrm{H}_{2} \\mathrm{O}(l)$\n54. In an accident, a solution containing 2.5 kg of nitric acid was spilled. Two kilograms of $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ was quickly spread on the area and $\\mathrm{CO}_{2}$ was released by the reaction. Was sufficient $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ used to neutralize all of the acid?\n55. A compact car gets 37.5 miles per gallon on the highway. If gasoline contains $84.2 \\%$ carbon by mass and has a density of $0.8205 \\mathrm{~g} / \\mathrm{mL}$, determine the mass of carbon dioxide produced during a 500-mile trip (3.785 liters per gallon).\n56. What volume of 0.750 M hydrochloric acid solution can be prepared from the HCl produced by the reaction of 25.0 g of NaCl with excess sulfuric acid?\n$\\mathrm{NaCl}(s)+\\mathrm{H}_{2} \\mathrm{SO}_{4}(l) \\longrightarrow \\mathrm{HCl}(g)+\\mathrm{NaHSO}_{4}(s)$"}
{"id": 3146, "contents": "762. Exercises - 762.3. Reaction Stoichiometry\n$\\mathrm{NaCl}(s)+\\mathrm{H}_{2} \\mathrm{SO}_{4}(l) \\longrightarrow \\mathrm{HCl}(g)+\\mathrm{NaHSO}_{4}(s)$\n57. What volume of a 0.2089 MKI solution contains enough KI to react exactly with the $\\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}$ in 43.88 mL of a 0.3842 M solution of $\\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}$ ? $2 \\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}+4 \\mathrm{KI} \\longrightarrow 2 \\mathrm{CuI}+\\mathrm{I}_{2}+4 \\mathrm{KNO}_{3}$\n58. A mordant is a substance that combines with a dye to produce a stable fixed color in a dyed fabric. Calcium acetate is used as a mordant. It is prepared by the reaction of acetic acid with calcium hydroxide.\n$2 \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}+\\mathrm{Ca}(\\mathrm{OH})_{2} \\longrightarrow \\mathrm{Ca}\\left(\\mathrm{CH}_{3} \\mathrm{CO}_{2}\\right)_{2}+2 \\mathrm{H}_{2} \\mathrm{O}$\nWhat mass of $\\mathrm{Ca}(\\mathrm{OH})_{2}$ is required to react with the acetic acid in 25.0 mL of a solution having a density of $1.065 \\mathrm{~g} / \\mathrm{mL}$ and containing $58.0 \\%$ acetic acid by mass?"}
{"id": 3147, "contents": "762. Exercises - 762.3. Reaction Stoichiometry\n59. The toxic pigment called white lead, $\\mathrm{Pb}_{3}(\\mathrm{OH})_{2}\\left(\\mathrm{CO}_{3}\\right)_{2}$, has been replaced in white paints by rutile, $\\mathrm{TiO}_{2}$. How much rutile (g) can be prepared from 379 g of an ore that contains $88.3 \\%$ ilmenite $\\left(\\mathrm{FeTiO}_{3}\\right)$ by mass? $2 \\mathrm{FeTiO}_{3}+4 \\mathrm{HCl}+\\mathrm{Cl}_{2} \\longrightarrow 2 \\mathrm{FeCl}_{3}+2 \\mathrm{TiO}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}$"}
{"id": 3148, "contents": "762. Exercises - 762.4. Reaction Yields\n60. The following quantities are placed in a container: $1.5 \\times 10^{24}$ atoms of hydrogen, 1.0 mol of sulfur, and 88.0 g of diatomic oxygen.\n(a) What is the total mass in grams for the collection of all three elements?\n(b) What is the total number of moles of atoms for the three elements?\n(c) If the mixture of the three elements formed a compound with molecules that contain two hydrogen atoms, one sulfur atom, and four oxygen atoms, which substance is consumed first?\n(d) How many atoms of each remaining element would remain unreacted in the change described in (c)?\n61. What is the limiting reactant in a reaction that produces sodium chloride from 8 g of sodium and 8 g of diatomic chlorine?\n62. Which of the postulates of Dalton's atomic theory explains why we can calculate a theoretical yield for a chemical reaction?\n63. A student isolated 25 g of a compound following a procedure that would theoretically yield 81 g . What was his percent yield?\n64. A sample of 0.53 g of carbon dioxide was obtained by heating 1.31 g of calcium carbonate. What is the percent yield for this reaction?\n$\\mathrm{CaCO}_{3}(s) \\longrightarrow \\mathrm{CaO}(s)+\\mathrm{CO}_{2}(s)$\n65. Freon-12, $\\mathrm{CCl}_{2} \\mathrm{~F}_{2}$, is prepared from $\\mathrm{CCl}_{4}$ by reaction with HF . The other product of this reaction is HCl . Outline the steps needed to determine the percent yield of a reaction that produces 12.5 g of $\\mathrm{CCl}_{2} \\mathrm{~F}_{2}$ from 32.9 g of $\\mathrm{CCl}_{4}$. Freon-12 has been banned and is no longer used as a refrigerant because it catalyzes the decomposition of ozone and has a very long lifetime in the atmosphere. Determine the percent yield."}
{"id": 3149, "contents": "762. Exercises - 762.4. Reaction Yields\n66. Citric acid, $\\mathrm{C}_{6} \\mathrm{H}_{8} \\mathrm{O}_{7}$, a component of jams, jellies, and fruity soft drinks, is prepared industrially via fermentation of sucrose by the mold Aspergillus niger. The equation representing this reaction is $\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}+\\mathrm{H}_{2} \\mathrm{O}+3 \\mathrm{O}_{2} \\longrightarrow 2 \\mathrm{C}_{6} \\mathrm{H}_{8} \\mathrm{O}_{7}+4 \\mathrm{H}_{2} \\mathrm{O}$\nWhat mass of citric acid is produced from exactly 1 metric ton $\\left(1.000 \\times 10^{3} \\mathrm{~kg}\\right)$ of sucrose if the yield is $92.30 \\%$ ?\n67. Toluene, $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CH}_{3}$, is oxidized by air under carefully controlled conditions to benzoic acid, $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CO}_{2} \\mathrm{H}$, which is used to prepare the food preservative sodium benzoate, $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CO}_{2} \\mathrm{Na}$. What is the percent yield of a reaction that converts 1.000 kg of toluene to 1.21 kg of benzoic acid?\n$2 \\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CH}_{3}+3 \\mathrm{O}_{2} \\longrightarrow 2 \\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CO}_{2} \\mathrm{H}+2 \\mathrm{H}_{2} \\mathrm{O}$\n68. In a laboratory experiment, the reaction of 3.0 mol of $\\mathrm{H}_{2}$ with 2.0 mol of $\\mathrm{I}_{2}$ produced 1.0 mol of HI . Determine the theoretical yield in grams and the percent yield for this reaction."}
{"id": 3150, "contents": "762. Exercises - 762.4. Reaction Yields\n69. Outline the steps needed to solve the following problem, then do the calculations. Ether, $\\left(\\mathrm{C}_{2} \\mathrm{H}_{5}\\right)_{2} \\mathrm{O}$, which was originally used as an anesthetic but has been replaced by safer and more effective medications, is prepared by the reaction of ethanol with sulfuric acid.\n$2 \\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}+\\mathrm{H}_{2} \\mathrm{SO}_{4} \\longrightarrow\\left(\\mathrm{C}_{2} \\mathrm{H}_{5}\\right)_{2} \\mathrm{O}+\\mathrm{H}_{2} \\mathrm{SO}_{4} \\cdot \\mathrm{H}_{2} \\mathrm{O}$\nWhat is the percent yield of ether if $1.17 \\mathrm{~L}(\\mathrm{~d}=0.7134 \\mathrm{~g} / \\mathrm{mL})$ is isolated from the reaction of 1.500 L of $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$ $(\\mathrm{d}=0.7894 \\mathrm{~g} / \\mathrm{mL})$ ?\n70. Outline the steps needed to determine the limiting reactant when 30.0 g of propane, $\\mathrm{C}_{3} \\mathrm{H}_{8}$, is burned with 75.0 g of oxygen."}
{"id": 3151, "contents": "762. Exercises - 762.4. Reaction Yields\nDetermine the limiting reactant.\n71. Outline the steps needed to determine the limiting reactant when 0.50 mol of Cr and 0.75 mol of $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ react according to the following chemical equation.\n$2 \\mathrm{Cr}+2 \\mathrm{H}_{3} \\mathrm{PO}_{4} \\longrightarrow 2 \\mathrm{CrPO}_{4}+3 \\mathrm{H}_{2}$\nDetermine the limiting reactant.\n72. What is the limiting reactant when 1.50 g of lithium and 1.50 g of nitrogen combine to form lithium nitride, a component of advanced batteries, according to the following unbalanced equation?\n$\\mathrm{Li}+\\mathrm{N}_{2} \\longrightarrow \\mathrm{Li}_{3} \\mathrm{~N}$\n73. Uranium can be isolated from its ores by dissolving it as $\\mathrm{UO}_{2}\\left(\\mathrm{NO}_{3}\\right)_{2}$, then separating it as solid $\\mathrm{UO}_{2}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}\\right) \\cdot 3 \\mathrm{H}_{2} \\mathrm{O}$. Addition of 0.4031 g of sodium oxalate, $\\mathrm{Na}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$, to a solution containing 1.481 g of uranyl nitrate, $\\mathrm{UO}_{2}\\left(\\mathrm{NO}_{3}\\right)_{2}$, yields 1.073 g of solid $\\mathrm{UO}_{2}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}\\right) \\cdot 3 \\mathrm{H}_{2} \\mathrm{O}$."}
{"id": 3152, "contents": "762. Exercises - 762.4. Reaction Yields\n$\\mathrm{Na}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}+\\mathrm{UO}_{2}\\left(\\mathrm{NO}_{3}\\right)_{2}+3 \\mathrm{H}_{2} \\mathrm{O} \\longrightarrow \\mathrm{UO}_{2}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}\\right) \\cdot 3 \\mathrm{H}_{2} \\mathrm{O}+2 \\mathrm{NaNO}_{3}$\nDetermine the limiting reactant and the percent yield of this reaction.\n74. How many molecules of $\\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{Cl}_{2}$ can be prepared from $15 \\mathrm{C}_{2} \\mathrm{H}_{4}$ molecules and $8 \\mathrm{Cl}_{2}$ molecules?\n75. How many molecules of the sweetener saccharin can be prepared from 30 C atoms, 25 H atoms, 12 O atoms, 8 S atoms, and 14 N atoms?"}
{"id": 3153, "contents": "762. Exercises - 762.4. Reaction Yields\n76. The phosphorus pentoxide used to produce phosphoric acid for cola soft drinks is prepared by burning phosphorus in oxygen.\n(a) What is the limiting reactant when 0.200 mol of $\\mathrm{P}_{4}$ and 0.200 mol of $\\mathrm{O}_{2}$ react according to\n$\\mathrm{P}_{4}+5 \\mathrm{O}_{2} \\longrightarrow \\mathrm{P}_{4} \\mathrm{O}_{10}$\n(b) Calculate the percent yield if 10.0 g of $\\mathrm{P}_{4} \\mathrm{O}_{10}$ is isolated from the reaction.\n77. Would you agree to buy 1 trillion $(1,000,000,000,000)$ gold atoms for $\\$ 5$ ? Explain why or why not. Find the current price of gold at http://money.cnn.com/data/commodities/ ( 1 troy ounce $=31.1 \\mathrm{~g}$ )"}
{"id": 3154, "contents": "762. Exercises - 762.5. Quantitative Chemical Analysis\n78. What volume of $0.0105-\\mathrm{M} \\mathrm{HBr}$ solution is required to titrate 125 mL of a $0.0100-\\mathrm{MCa}(\\mathrm{OH})_{2}$ solution? $\\mathrm{Ca}(\\mathrm{OH})_{2}(a q)+2 \\mathrm{HBr}(a q) \\longrightarrow \\mathrm{CaBr}_{2}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n79. Titration of a $20.0-\\mathrm{mL}$ sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidity of the rain is due to the presence of sulfuric acid, what was the concentration of sulfuric acid in this sample of rain?\n80. What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with $0.2503 \\mathrm{M} \\mathrm{AgNO}_{3}$ requires 20.22 mL of the $\\mathrm{AgNO}_{3}$ solution to reach the end point?\n$\\mathrm{AgNO}_{3}(a q)+\\mathrm{NaCl}(a q) \\longrightarrow \\mathrm{AgCl}(s)+\\mathrm{NaNO}_{3}(a q)$\n81. In a common medical laboratory determination of the concentration of free chloride ion in blood serum, a serum sample is titrated with a $\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}$ solution.\n$2 \\mathrm{Cl}^{-}(a q)+\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q) \\longrightarrow 2 \\mathrm{NO}_{3}{ }^{-}(a q)+\\mathrm{HgCl}_{2}(s)$\nWhat is the $\\mathrm{Cl}^{-}$concentration in a $0.25-\\mathrm{mL}$ sample of normal serum that requires 1.46 mL of $8.25 \\times 10^{-4}$ $M \\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}(\\mathrm{aq})$ to reach the end point?"}
{"id": 3155, "contents": "762. Exercises - 762.5. Quantitative Chemical Analysis\n82. Potatoes can be peeled commercially by soaking them in a $3-M$ to $6-M$ solution of sodium hydroxide, then removing the loosened skins by spraying them with water. Does a sodium hydroxide solution have a suitable concentration if titration of 12.00 mL of the solution requires 30.6 mL of 1.65 M HCI to reach the end point?\n83. A sample of gallium bromide, $\\mathrm{GaBr}_{3}$, weighing 0.165 g was dissolved in water and treated with silver nitrate, $\\mathrm{AgNO}_{3}$, resulting in the precipitation of 0.299 g AgBr . Use these data to compute the \\%Ga (by mass) $\\mathrm{GaBr}_{3}$.\n84. The principal component of mothballs is naphthalene, a compound with a molecular mass of about 130 amu, containing only carbon and hydrogen. A $3.000-\\mathrm{mg}$ sample of naphthalene burns to give 10.3 mg of $\\mathrm{CO}_{2}$. Determine its empirical and molecular formulas.\n85. A $0.025-\\mathrm{g}$ sample of a compound composed of boron and hydrogen, with a molecular mass of $\\sim 28 \\mathrm{amu}$, burns spontaneously when exposed to air, producing 0.063 g of $\\mathrm{B}_{2} \\mathrm{O}_{3}$. What are the empirical and molecular formulas of the compound?"}
{"id": 3156, "contents": "762. Exercises - 762.5. Quantitative Chemical Analysis\n86. Sodium bicarbonate (baking soda), $\\mathrm{NaHCO}_{3}$, can be purified by dissolving it in hot water $\\left(60^{\\circ} \\mathrm{C}\\right)$, filtering to remove insoluble impurities, cooling to $0^{\\circ} \\mathrm{C}$ to precipitate solid $\\mathrm{NaHCO}_{3}$, and then filtering to remove the solid, leaving soluble impurities in solution. Any $\\mathrm{NaHCO}_{3}$ that remains in solution is not recovered. The solubility of $\\mathrm{NaHCO}_{3}$ in hot water of $60^{\\circ} \\mathrm{C}$ is $164 \\mathrm{~g} / \\mathrm{L}$. Its solubility in cold water of $0{ }^{\\circ} \\mathrm{C}$ is $69 \\mathrm{~g} / \\mathrm{L}$. What is the percent yield of $\\mathrm{NaHCO}_{3}$ when it is purified by this method?\n87. What volume of 0.600 MHCl is required to react completely with 2.50 g of sodium hydrogen carbonate? $\\mathrm{NaHCO}_{3}(a q)+\\mathrm{HCl}(a q) \\longrightarrow \\mathrm{NaCl}(a q)+\\mathrm{CO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l)$\n88. What volume of $0.08892 \\mathrm{MHNO}_{3}$ is required to react completely with 0.2352 g of potassium hydrogen phosphate?\n$2 \\mathrm{HNO}_{3}(a q)+\\mathrm{K}_{2} \\mathrm{HPO}_{4}(a q) \\longrightarrow \\mathrm{H}_{3} \\mathrm{PO}_{4}(a q)+2 \\mathrm{KNO}_{3}(a q)$\n89. What volume of a $0.3300-M$ solution of sodium hydroxide would be required to titrate 15.00 mL of 0.1500 $M$ oxalic acid?"}
{"id": 3157, "contents": "762. Exercises - 762.5. Quantitative Chemical Analysis\n89. What volume of a $0.3300-M$ solution of sodium hydroxide would be required to titrate 15.00 mL of 0.1500 $M$ oxalic acid?\n$\\mathrm{C}_{2} \\mathrm{O}_{4} \\mathrm{H}_{2}(a q)+2 \\mathrm{NaOH}(a q) \\longrightarrow \\mathrm{Na}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n90. What volume of a $0.00945-M$ solution of potassium hydroxide would be required to titrate 50.00 mL of a sample of acid rain with a $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ concentration of $1.23 \\times 10^{-4} \\mathrm{M}$.\n$\\mathrm{H}_{2} \\mathrm{SO}_{4}(a q)+2 \\mathrm{KOH}(a q) \\longrightarrow \\mathrm{K}_{2} \\mathrm{SO}_{4}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n91. A sample of solid calcium hydroxide, $\\mathrm{Ca}(\\mathrm{OH})_{2}$, is allowed to stand in water until a saturated solution is formed. A titration of 75.00 mL of this solution with $5.00 \\times 10^{-2} \\mathrm{M} \\mathrm{HCl}$ requires 36.6 mL of the acid to reach the end point.\n$\\mathrm{Ca}(\\mathrm{OH})_{2}(a q)+2 \\mathrm{HCl}(a q) \\longrightarrow \\mathrm{CaCl}_{2}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$\nWhat is the molarity?\n92. What mass of $\\mathrm{Ca}(\\mathrm{OH})_{2}$ will react with 25.0 g of butanoic to form the preservative calcium butanoate according to the equation?"}
{"id": 3158, "contents": "762. Exercises - 762.5. Quantitative Chemical Analysis\n93. How many milliliters of a $0.1500-M$ solution of KOH will be required to titrate 40.00 mL of a $0.0656-M$ solution of $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ ?\n$\\mathrm{H}_{3} \\mathrm{PO}_{4}(a q)+2 \\mathrm{KOH}(a q) \\longrightarrow \\mathrm{K}_{2} \\mathrm{HPO}_{4}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n94. Potassium hydrogen phthalate, $\\mathrm{KHC}_{8} \\mathrm{H}_{4} \\mathrm{O}_{4}$, or KHP , is used in many laboratories, including general chemistry laboratories, to standardize solutions of base. KHP is one of only a few stable solid acids that can be dried by warming and weighed. A $0.3420-\\mathrm{g}$ sample of $\\mathrm{KHC}_{8} \\mathrm{H}_{4} \\mathrm{O}_{4}$ reacts with 35.73 mL of a NaOH solution in a titration. What is the molar concentration of the NaOH ?\n$\\mathrm{KHC}_{8} \\mathrm{H}_{4} \\mathrm{O}_{4}(a q)+\\mathrm{NaOH}(a q) \\longrightarrow \\mathrm{KNaC}_{8} \\mathrm{H}_{4} \\mathrm{O}_{4}(a q)+\\mathrm{H}_{2} \\mathrm{O}(a q)$\n95. The reaction of $\\mathrm{WCl}_{6}$ with Al at $\\sim 400^{\\circ} \\mathrm{C}$ gives black crystals of a compound containing only tungsten and chlorine. A sample of this compound, when reduced with hydrogen, gives 0.2232 g of tungsten metal and hydrogen chloride, which is absorbed in water. Titration of the hydrochloric acid thus produced requires 46.2 mL of 0.1051 M NaOH to reach the end point. What is the empirical formula of the black tungsten chloride?"}
{"id": 3159, "contents": "763. CHAPTER 8 <br> Gases - \nFigure 8.1 The hot air inside these balloons is less dense than the surrounding cool air. This results in a buoyant force that causes the balloons to rise when their guy lines are untied. (credit: modification of work by Anthony Quintano)"}
{"id": 3160, "contents": "764. CHAPTER OUTLINE - 764.1. Gas Pressure\n8.2 Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law\n8.3 Stoichiometry of Gaseous Substances, Mixtures, and Reactions"}
{"id": 3161, "contents": "764. CHAPTER OUTLINE - 764.2. Effusion and Diffusion of Gases\n8.5 The Kinetic-Molecular Theory\n8.6 Non-Ideal Gas Behavior\n\nINTRODUCTION We are surrounded by an ocean of gas-the atmosphere-and many of the properties of gases are familiar to us from our daily activities. Heated gases expand, which can make a hot air balloon rise (Figure 8.1) or cause a blowout in a bicycle tire left in the sun on a hot day.\n\nGases have played an important part in the development of chemistry. In the seventeenth and eighteenth centuries, many scientists investigated gas behavior, providing the first mathematical descriptions of the behavior of matter.\n\nIn this chapter, we will examine the relationships between gas temperature, pressure, amount, and volume. We will study a simple theoretical model and use it to analyze the experimental behavior of gases. The results of these analyses will show us the limitations of the theory and how to improve on it."}
{"id": 3162, "contents": "765. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Define the property of pressure\n- Define and convert among the units of pressure measurements\n- Describe the operation of common tools for measuring gas pressure\n- Calculate pressure from manometer data\n\nThe earth's atmosphere exerts a pressure, as does any other gas. Although we do not normally notice atmospheric pressure, we are sensitive to pressure changes-for example, when your ears \"pop\" during takeoff and landing while flying, or when you dive underwater. Gas pressure is caused by the force exerted by gas molecules colliding with the surfaces of objects (Figure 8.2). Although the force of each collision is very small, any surface of appreciable area experiences a large number of collisions in a short time, which can result in a high pressure. In fact, normal air pressure is strong enough to crush a metal container when not balanced by equal pressure from inside the container.\n\n\nFIGURE 8.2 The atmosphere above us exerts a large pressure on objects at the surface of the earth, roughly equal to the weight of a bowling ball pressing on an area the size of a human thumbnail."}
{"id": 3163, "contents": "766. LINK TO LEARNING - \nA dramatic illustration (http://openstax.org/l/16atmospressur1) of atmospheric pressure is provided in this brief video, which shows a railway tanker car imploding when its internal pressure is decreased.\n\nA smaller scale demonstration (http://openstax.org/l/16atmospressur2) of this phenomenon is briefly explained.\n\nAtmospheric pressure is caused by the weight of the column of air molecules in the atmosphere above an object, such as the tanker car. At sea level, this pressure is roughly the same as that exerted by a full-grown African elephant standing on a doormat, or a typical bowling ball resting on your thumbnail. These may seem like huge amounts, and they are, but life on earth has evolved under such atmospheric pressure. If you actually perch a bowling ball on your thumbnail, the pressure experienced is twice the usual pressure, and the sensation is unpleasant.\n\nIn general, pressure is defined as the force exerted on a given area: $P=\\frac{F}{A}$. Note that pressure is directly proportional to force and inversely proportional to area. Thus, pressure can be increased either by increasing the amount of force or by decreasing the area over which it is applied; pressure can be decreased by decreasing the force or increasing the area.\n\nLet's apply this concept to determine which exerts a greater pressure in Figure 8.3-the elephant or the figure skater? A large African elephant can weigh 7 tons, supported on four feet, each with a diameter of about 1.5 ft\n(footprint area of $250 \\mathrm{in}^{2}$ ), so the pressure exerted by each foot is about $14 \\mathrm{lb} / \\mathrm{in}^{2}$ :\n\n$$\n\\text { pressure per elephant foot }=14,000 \\frac{\\mathrm{lb}}{\\text { elephant }} \\times \\frac{1 \\text { elephant }}{4 \\text { feet }} \\times \\frac{1 \\text { foot }}{250 \\mathrm{in}^{2}}=14 \\mathrm{lb} / \\mathrm{in}^{2}\n$$"}
{"id": 3164, "contents": "766. LINK TO LEARNING - \nThe figure skater weighs about 120 lbs , supported on two skate blades, each with an area of about $2 \\mathrm{in}^{2}$, so the pressure exerted by each blade is about $30 \\mathrm{lb} / \\mathrm{in}^{2}$ :\n\n$$\n\\text { pressure per skate blade }=120 \\frac{\\mathrm{lb}}{\\text { skater }} \\times \\frac{1 \\text { skater }}{2 \\text { blades }} \\times \\frac{1 \\text { blade }}{2 \\mathrm{in}^{2}}=30 \\mathrm{lb} / \\mathrm{in}^{2}\n$$\n\nEven though the elephant is more than one hundred-times heavier than the skater, it exerts less than one-half of the pressure. On the other hand, if the skater removes their skates and stands with bare feet (or regular footwear) on the ice, the larger area over which their weight is applied greatly reduces the pressure exerted:\n\n$$\n\\text { pressure per human foot }=120 \\frac{\\mathrm{lb}}{\\text { skater }} \\times \\frac{1 \\text { skater }}{2 \\text { feet }} \\times \\frac{1 \\text { foot }}{30 \\mathrm{in}^{2}}=2 \\mathrm{lb} / \\mathrm{in}^{2}\n$$\n\n\n\nFIGURE 8.3 Although (a) an elephant's weight is large, creating a very large force on the ground, (b) the figure skater exerts a much higher pressure on the ice due to the small surface area of the skates. (credit a: modification of work by Guido da Rozze; credit b: modification of work by Ryosuke Yagi)"}
{"id": 3165, "contents": "766. LINK TO LEARNING - \nThe SI unit of pressure is the pascal ( $\\mathbf{P a}$ ), with $1 \\mathrm{~Pa}=1 \\mathrm{~N} / \\mathrm{m}^{2}$, where N is the newton, a unit of force defined as 1 $\\mathrm{kg} \\mathrm{m} / \\mathrm{s}^{2}$. One pascal is a small pressure; in many cases, it is more convenient to use units of kilopascal ( $1 \\mathrm{kPa}=$ 1000 Pa ) or bar ( $1 \\mathrm{bar}=100,000 \\mathrm{~Pa}$ ). In the United States, pressure is often measured in pounds of force on an area of one square inch-pounds per square inch (psi)-for example, in car tires. Pressure can also be measured using the unit atmosphere (atm), which originally represented the average sea level air pressure at the approximate latitude of Paris $\\left(45^{\\circ}\\right)$. Table 8.1 provides some information on these and a few other common units for pressure measurements\n\nPressure Units\n\n| Unit Name and Abbreviation | Definition or Relation to Other Unit |\n| :--- | :--- |\n| pascal (Pa) | $1 \\mathrm{~Pa}=1 \\mathrm{~N} / \\mathrm{m}^{2}$ <br> recommended IUPAC unit |\n| kilopascal (kPa) | $1 \\mathrm{kPa}=1000 \\mathrm{~Pa}$ |\n| pounds per square inch (psi) | air pressure at sea level is ~14.7 psi |\n| atmosphere (atm) | 1 atm $=101,325 \\mathrm{~Pa}=760$ torr <br> air pressure at sea level is $\\sim 1 \\mathrm{~atm}$ |\n\nTABLE 8.1"}
{"id": 3166, "contents": "766. LINK TO LEARNING - \nTABLE 8.1\n\n| Unit Name and Abbreviation | Definition or Relation to Other Unit |\n| :--- | :--- |\n| bar (bar, or b) | 1 bar $=100,000 \\mathrm{~Pa}$ (exactly) <br> commonly used in meteorology |\n| millibar (mbar, or mb) | $1000 \\mathrm{mbar}=1 \\mathrm{bar}$ |\n| inches of mercury (in. Hg) | $1 \\mathrm{in} . \\mathrm{Hg}=3386 \\mathrm{~Pa}$ <br> used by aviation industry, also some weather reports |\n| torr | 1 torr $=\\frac{1}{760}$ atm <br> named after Evangelista Torricelli, inventor of the barometer |\n| millimeters of mercury (mm Hg) | $1 \\mathrm{~mm} \\mathrm{Hg} \\mathrm{\\sim 1} \\mathrm{torr}$ |\n\nTABLE 8.1"}
{"id": 3167, "contents": "768. Conversion of Pressure Units - \nThe United States National Weather Service reports pressure in both inches of Hg and millibars. Convert a pressure of 29.2 in . Hg into:\n(a) torr\n(b) atm\n(c) kPa\n(d) mbar"}
{"id": 3168, "contents": "769. Solution - \nThis is a unit conversion problem. The relationships between the various pressure units are given in Table 8.1.\n(a) $29.2 \\mathrm{im} \\mathrm{Hg} \\times \\frac{25.4 \\mathrm{~mm}}{1 \\mathrm{im}} \\times \\frac{1 \\text { torr }}{1 \\mathrm{mmHg}}=742$ torr\n(b) $742 \\mathrm{tOrf} \\times \\frac{1 \\mathrm{~atm}}{760 \\text { tort }}=0.976 \\mathrm{~atm}$\n(c) 742 torf $\\times \\frac{101.325 \\mathrm{kPa}}{760 \\text { tort }}=98.9 \\mathrm{kPa}$\n(d) $98.9 \\mathrm{kPa} \\times \\frac{1000 \\mathrm{~Pa}}{1 \\mathrm{kPa}} \\times \\frac{1 \\text { bar }}{100,000 \\mathrm{~Pa}} \\times \\frac{1000 \\mathrm{mbar}}{1 \\text { bar }}=989 \\mathrm{mbar}$"}
{"id": 3169, "contents": "770. Check Your Learning - \nA typical barometric pressure in Kansas City is 740 torr. What is this pressure in atmospheres, in millimeters of mercury, in kilopascals, and in bar?"}
{"id": 3170, "contents": "771. Answer: - \n0.974 atm; $740 \\mathrm{~mm} \\mathrm{Hg} ; 98.7 \\mathrm{kPa} ; 0.987$ bar\n\nWe can measure atmospheric pressure, the force exerted by the atmosphere on the earth's surface, with a barometer (Figure 8.4). A barometer is a glass tube that is closed at one end, filled with a nonvolatile liquid such as mercury, and then inverted and immersed in a container of that liquid. The atmosphere exerts pressure on the liquid outside the tube, the column of liquid exerts pressure inside the tube, and the pressure at the liquid surface is the same inside and outside the tube. The height of the liquid in the tube is therefore\nproportional to the pressure exerted by the atmosphere.\n\n\nFIGURE 8.4 In a barometer, the height, $h$, of the column of liquid is used as a measurement of the air pressure. Using very dense liquid mercury (left) permits the construction of reasonably sized barometers, whereas using water (right) would require a barometer more than 30 feet tall.\n\nIf the liquid is water, normal atmospheric pressure will support a column of water over 10 meters high, which is rather inconvenient for making (and reading) a barometer. Because mercury ( Hg ) is about 13.6-times denser than water, a mercury barometer only needs to be $\\frac{1}{13.6}$ as tall as a water barometer-a more suitable size. Standard atmospheric pressure of 1 atm at sea level ( $101,325 \\mathrm{~Pa}$ ) corresponds to a column of mercury that is about 760 mm (29.92 in.) high. The torr was originally intended to be a unit equal to one millimeter of mercury, but it no longer corresponds exactly. The pressure exerted by a fluid due to gravity is known as hydrostatic pressure, $p$ :\n\n$$\np=h \\rho g\n$$\n\nwhere $h$ is the height of the fluid, $\\rho$ is the density of the fluid, and $g$ is acceleration due to gravity."}
{"id": 3171, "contents": "773. Calculation of Barometric Pressure - \nShow the calculation supporting the claim that atmospheric pressure near sea level corresponds to the pressure exerted by a column of mercury that is about 760 mm high. The density of mercury $=13.6 \\mathrm{~g} / \\mathrm{cm}^{3}$."}
{"id": 3172, "contents": "774. Solution - \nThe hydrostatic pressure is given by $p=h \\rho g$, with $h=760 \\mathrm{~mm}, \\rho=13.6 \\mathrm{~g} / \\mathrm{cm}^{3}$, and $g=9.81 \\mathrm{~m} / \\mathrm{s}^{2}$. Plugging these values into the equation and doing the necessary unit conversions will give us the value we seek. (Note: We are expecting to find a pressure of $\\sim 101,325 \\mathrm{~Pa}$.)\n\n$$\n\\begin{gathered}\n101,325 \\mathrm{~N} / \\mathrm{m}^{2}=101,325 \\frac{\\mathrm{~kg} \\cdot \\mathrm{~m} / \\mathrm{s}^{2}}{\\mathrm{~m}^{2}}=101,325 \\frac{\\mathrm{~kg}}{\\mathrm{~m} \\cdot \\mathrm{~s}^{2}} \\\\\np=\\left(760 \\mathrm{~mm} \\times \\frac{1 \\mathrm{~m}}{1000 \\mathrm{~mm}}\\right) \\times\\left(\\frac{13.6 \\mathrm{~g}}{1 \\mathrm{~cm}^{3}} \\times \\frac{1 \\mathrm{~kg}}{1000 \\mathrm{~g}} \\times \\frac{(100 \\mathrm{~cm})^{3}}{(1 \\mathrm{~m})^{3}}\\right) \\times\\left(\\frac{9.81 \\mathrm{~m}}{1 \\mathrm{~s}^{2}}\\right) \\\\\n=(0.760 \\mathrm{~m})\\left(13,600 \\mathrm{~kg} / \\mathrm{m}^{3}\\right)\\left(9.81 \\mathrm{~m} / \\mathrm{s}^{2}\\right)=1.01 \\times 10^{5} \\mathrm{~kg} / \\mathrm{ms}^{2}=1.01 \\times 10^{5} \\mathrm{~N} / \\mathrm{m}^{2}\n\\end{gathered}\n$$\n\n$$\n=1.01 \\times 10^{5} \\mathrm{~Pa}\n$$"}
{"id": 3173, "contents": "775. Check Your Learning - \nCalculate the height of a column of water at $25^{\\circ} \\mathrm{C}$ that corresponds to normal atmospheric pressure. The density of water at this temperature is $1.0 \\mathrm{~g} / \\mathrm{cm}^{3}$."}
{"id": 3174, "contents": "776. Answer: - \n10.3 m\n\nA manometer is a device similar to a barometer that can be used to measure the pressure of a gas trapped in a container. A closed-end manometer is a U-shaped tube with one closed arm, one arm that connects to the gas to be measured, and a nonvolatile liquid (usually mercury) in between. As with a barometer, the distance between the liquid levels in the two arms of the tube ( $h$ in the diagram) is proportional to the pressure of the gas in the container. An open-end manometer (Figure 8.5) is the same as a closed-end manometer, but one of its arms is open to the atmosphere. In this case, the distance between the liquid levels corresponds to the difference in pressure between the gas in the container and the atmosphere.\n\n\nFIGURE 8.5 A manometer can be used to measure the pressure of a gas. The (difference in) height between the liquid levels ( $h$ ) is a measure of the pressure. Mercury is usually used because of its large density."}
{"id": 3175, "contents": "778. Calculation of Pressure Using a Closed-End Manometer - \nThe pressure of a sample of gas is measured with a closed-end manometer, as shown to the right. The liquid in the manometer is mercury. Determine the pressure of the gas in:\n(a) torr\n(b) Pa\n(c) bar"}
{"id": 3176, "contents": "779. Solution - \nThe pressure of the gas is equal to a column of mercury of height 26.4 cm . (The pressure at the bottom horizontal line is equal on both sides of the tube. The pressure on the left is due to the gas and the pressure on the right is due to 26.4 cm Hg , or mercury.) We could use the equation $p=h \\rho g$ as in Example 8.2, but it is simpler to just convert between units using Table 8.1.\n(a) $26.4 \\mathrm{cmHg} \\times \\frac{10 \\mathrm{mmHz}}{1 \\mathrm{cmHg}} \\times \\frac{1 \\text { torr }}{1 \\mathrm{mmHg}}=264$ torr\n(b) 264 tori $\\times \\frac{1 \\mathrm{Am}}{760 \\text { torr }} \\times \\frac{101,325 \\mathrm{~Pa}}{1+35,200 \\mathrm{~Pa}}$\n(c) $35,200 \\mathrm{~Pa} \\times \\frac{1 \\mathrm{bar}}{100,000 \\mathrm{~Pa}}=0.352 \\mathrm{bar}$"}
{"id": 3177, "contents": "780. Check Your Learning - \nThe pressure of a sample of gas is measured with a closed-end manometer. The liquid in the manometer is mercury. Determine the pressure of the gas in:\n(a) torr\n(b) Pa\n(c) bar"}
{"id": 3178, "contents": "781. Answer: - \n(a) $\\sim 150$ torr; (b) $\\sim 20,000 \\mathrm{~Pa}$; (c) $\\sim 0.20$ bar"}
{"id": 3179, "contents": "783. Calculation of Pressure Using an Open-End Manometer - \nThe pressure of a sample of gas is measured at sea level with an open-end Hg (mercury) manometer, as shown to the right. Determine the pressure of the gas in:\n(a) mm Hg\n(b) atm\n(c) kPa"}
{"id": 3180, "contents": "784. Solution - \nThe pressure of the gas equals the hydrostatic pressure due to a column of mercury of height 13.7 cm plus the pressure of the atmosphere at sea level. (The pressure at the bottom horizontal line is equal on both sides of the tube. The pressure on the left is due to the gas and the pressure on the right is due to 13.7 cm of Hg plus atmospheric pressure.)\n(a) In mm Hg , this is: $137 \\mathrm{~mm} \\mathrm{Hg}+760 \\mathrm{~mm} \\mathrm{Hg}=897 \\mathrm{~mm} \\mathrm{Hg}$\n(b) $897 \\mathrm{mmHg} \\times \\frac{1 \\mathrm{~atm}}{760 \\mathrm{~mm} \\mathrm{Hg}}=1.18 \\mathrm{~atm}$\n(c) $1.18 \\mathrm{~atm} \\times \\frac{101.325 \\mathrm{kPa}}{1 \\mathrm{~atm}}=1.20 \\times 10^{2} \\mathrm{kPa}$"}
{"id": 3181, "contents": "785. Check Your Learning - \nThe pressure of a sample of gas is measured at sea level with an open-end Hg manometer, as shown to the right. Determine the pressure of the gas in:\n(a) mm Hg\n(b) atm\n(c) kPa\n\n\nAnswer:\n(a) 642 mm Hg ; (b) 0.845 atm ; (c) 85.6 kPa"}
{"id": 3182, "contents": "787. Measuring Blood Pressure - \nBlood pressure is measured using a device called a sphygmomanometer (Greek sphygmos = \"pulse\"). It consists of an inflatable cuff to restrict blood flow, a manometer to measure the pressure, and a method of determining when blood flow begins and when it becomes impeded (Figure 8.6). Since its invention in 1881, it has been an essential medical device. There are many types of sphygmomanometers: manual ones that require a stethoscope and are used by medical professionals; mercury ones, used when the most accuracy is required; less accurate mechanical ones; and digital ones that can be used with little training but that have limitations. When using a sphygmomanometer, the cuff is placed around the upper arm and inflated until blood flow is completely blocked, then slowly released. As the heart beats, blood forced through the arteries causes a rise in pressure. This rise in pressure at which blood flow begins is the systolic pressure-the peak pressure in the cardiac cycle. When the cuff's pressure equals the arterial systolic pressure, blood flows past the cuff, creating audible sounds that can be heard using a stethoscope. This is followed by a decrease in pressure as the heart's ventricles prepare for another beat. As cuff pressure continues to decrease, eventually sound is no longer heard; this is the diastolic pressure-the lowest pressure (resting phase) in the cardiac cycle. Blood pressure units from a sphygmomanometer are in terms of millimeters of mercury ( mm Hg ).\n\n(a)\n\n(b)\n\nFIGURE 8.6 (a) A medical technician prepares to measure a patient's blood pressure with a sphygmomanometer. (b) A typical sphygmomanometer uses a valved rubber bulb to inflate the cuff and a diaphragm gauge to measure pressure. (credit a: modification of work by Master Sgt. Jeffrey Allen)"}
{"id": 3183, "contents": "789. Meteorology, Climatology, and Atmospheric Science - \nThroughout the ages, people have observed clouds, winds, and precipitation, trying to discern patterns and make predictions: when it is best to plant and harvest; whether it is safe to set out on a sea voyage; and much more. We now face complex weather and atmosphere-related challenges that will have a major impact on our civilization and the ecosystem. Several different scientific disciplines use chemical principles to help us better understand weather, the atmosphere, and climate. These are meteorology, climatology, and atmospheric science. Meteorology is the study of the atmosphere, atmospheric phenomena, and atmospheric effects on earth's weather. Meteorologists seek to understand and predict the weather in the short term, which can save lives and benefit the economy. Weather forecasts (Figure 8.7) are the result of thousands of measurements of air pressure, temperature, and the like, which are compiled, modeled, and analyzed in weather centers worldwide.\n\n\nFIGURE 8.7 Meteorologists use weather maps to describe and predict weather. Regions of high (H) and low (L) pressure have large effects on weather conditions. The gray lines represent locations of constant pressure known as isobars. (credit: modification of work by National Oceanic and Atmospheric Administration)\n\nIn terms of weather, low-pressure systems occur when the earth's surface atmospheric pressure is lower than the surrounding environment: Moist air rises and condenses, producing clouds. Movement of moisture and air within various weather fronts instigates most weather events.\n\nThe atmosphere is the gaseous layer that surrounds a planet. Earth's atmosphere, which is roughly 100-125 km thick, consists of roughly $78.1 \\%$ nitrogen and $21.0 \\%$ oxygen, and can be subdivided further into the regions shown in Figure 8.8: the exosphere (furthest from earth, $>700 \\mathrm{~km}$ above sea level), the thermosphere ( $80-700 \\mathrm{~km}$ ), the mesosphere ( $50-80 \\mathrm{~km}$ ), the stratosphere (second lowest level of our atmosphere, $12-50 \\mathrm{~km}$ above sea level), and the troposphere (up to 12 km above sea level, roughly $80 \\%$ of the earth's atmosphere by mass and the layer where most weather events originate). As you go higher in the troposphere, air density and temperature both decrease."}
{"id": 3184, "contents": "789. Meteorology, Climatology, and Atmospheric Science - \nFIGURE 8.8 Earth's atmosphere has five layers: the troposphere, the stratosphere, the mesosphere, the thermosphere, and the exosphere.\n\nClimatology is the study of the climate, averaged weather conditions over long time periods, using atmospheric data. However, climatologists study patterns and effects that occur over decades, centuries, and millennia, rather than shorter time frames of hours, days, and weeks like meteorologists. Atmospheric science is an even broader field, combining meteorology, climatology, and other scientific disciplines that study the atmosphere."}
{"id": 3185, "contents": "790. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Identify the mathematical relationships between the various properties of gases\n- Use the ideal gas law, and related gas laws, to compute the values of various gas properties under specified conditions\n\nDuring the seventeenth and especially eighteenth centuries, driven both by a desire to understand nature and a quest to make balloons in which they could fly (Figure 8.9), a number of scientists established the relationships between the macroscopic physical properties of gases, that is, pressure, volume, temperature, and amount of gas. Although their measurements were not precise by today's standards, they were able to determine the mathematical relationships between pairs of these variables (e.g., pressure and temperature, pressure and volume) that hold for an ideal gas-a hypothetical construct that real gases approximate under certain conditions. Eventually, these individual laws were combined into a single equation-the ideal gas law-that relates gas quantities for gases and is quite accurate for low pressures and moderate temperatures. We will consider the key developments in individual relationships (for pedagogical reasons not quite in historical order), then put them together in the ideal gas law.\n\n(a)\n\n(b)\n\n(c)\n\nFIGURE 8.9 In 1783, the first (a) hydrogen-filled balloon flight, (b) manned hot air balloon flight, and (c) manned hydrogen-filled balloon flight occurred. When the hydrogen-filled balloon depicted in (a) landed, the frightened villagers of Gonesse reportedly destroyed it with pitchforks and knives. The launch of the latter was reportedly viewed by 400,000 people in Paris."}
{"id": 3186, "contents": "791. Pressure and Temperature: Amontons's Law - \nImagine filling a rigid container attached to a pressure gauge with gas and then sealing the container so that no gas may escape. If the container is cooled, the gas inside likewise gets colder and its pressure is observed to decrease. Since the container is rigid and tightly sealed, both the volume and number of moles of gas remain constant. If we heat the sphere, the gas inside gets hotter (Figure 8.10) and the pressure increases.\n\n\nFIGURE 8.10 The effect of temperature on gas pressure: When the hot plate is off, the pressure of the gas in the sphere is relatively low. As the gas is heated, the pressure of the gas in the sphere increases.\n\nThis relationship between temperature and pressure is observed for any sample of gas confined to a constant volume. An example of experimental pressure-temperature data is shown for a sample of air under these conditions in Figure 8.11. We find that temperature and pressure are linearly related, and if the temperature is on the kelvin scale, then $P$ and $T$ are directly proportional (again, when volume and moles of gas are held constant); if the temperature on the kelvin scale increases by a certain factor, the gas pressure increases by the same factor.\n\n| Temperature <br> $\\left({ }^{\\circ} \\mathbf{C}\\right)$ | Temperature <br> $\\mathbf{( K )}$ | Pressure <br> $\\mathbf{( k P a )}$ |\n| :---: | :---: | :---: |\n| -100 | 173 | 36.0 |\n| -50 | 223 | 46.4 |\n| 0 | 273 | 56.7 |\n| 50 | 323 | 67.1 |\n| 100 | 373 | 77.5 |\n| 150 | 423 | 88.0 |\n\n\n\nFIGURE 8.11 For a constant volume and amount of air, the pressure and temperature are directly proportional, provided the temperature is in kelvin. (Measurements cannot be made at lower temperatures because of the condensation of the gas.) When this line is extrapolated to lower pressures, it reaches a pressure of 0 at $-273^{\\circ} \\mathrm{C}$, which is 0 on the kelvin scale and the lowest possible temperature, called absolute zero."}
{"id": 3187, "contents": "791. Pressure and Temperature: Amontons's Law - \nGuillaume Amontons was the first to empirically establish the relationship between the pressure and the temperature of a gas ( $\\sim 1700$ ), and Joseph Louis Gay-Lussac determined the relationship more precisely (~1800). Because of this, the $P$ - $T$ relationship for gases is known as either Amontons's law or Gay-Lussac's law. Under either name, it states that the pressure of a given amount of gas is directly proportional to its temperature on the kelvin scale when the volume is held constant. Mathematically, this can be written:\n\n$$\nP \\propto T \\text { or } P=\\text { constant } \\times T \\text { or } P=k \\times T\n$$\n\nwhere $\\alpha$ means \"is proportional to,\" and $k$ is a proportionality constant that depends on the identity, amount, and volume of the gas.\n\nFor a confined, constant volume of gas, the ratio $\\frac{P}{T}$ is therefore constant (i.e., $\\frac{P}{T}=k$ ). If the gas is initially in \"Condition 1\" (with $P=P_{1}$ and $T=T_{1}$ ), and then changes to \"Condition 2\" (with $P=P_{2}$ and $T=T_{2}$ ), we have that $\\frac{P_{1}}{T_{1}}=k$ and $\\frac{P_{2}}{T_{2}}=k$, which reduces to $\\frac{P_{1}}{T_{1}}=\\frac{P_{2}}{T_{2}}$. This equation is useful for pressure-temperature calculations for a confined gas at constant volume. Note that temperatures must be on the kelvin scale for any gas law calculations ( 0 on the kelvin scale and the lowest possible temperature is called absolute zero). (Also note that there are at least three ways we can describe how the pressure of a gas changes as its temperature changes: We can use a table of values, a graph, or a mathematical equation.)"}
{"id": 3188, "contents": "793. Predicting Change in Pressure with Temperature - \nA can of hair spray is used until it is empty except for the propellant, isobutane gas.\n(a) On the can is the warning \"Store only at temperatures below $120^{\\circ} \\mathrm{F}\\left(48.8^{\\circ} \\mathrm{C}\\right)$. Do not incinerate.\" Why?\n(b) The gas in the can is initially at $24^{\\circ} \\mathrm{C}$ and 360 kPa , and the can has a volume of 350 mL . If the can is left in a car that reaches $50^{\\circ} \\mathrm{C}$ on a hot day, what is the new pressure in the can?"}
{"id": 3189, "contents": "794. Solution - \n(a) The can contains an amount of isobutane gas at a constant volume, so if the temperature is increased by heating, the pressure will increase proportionately. High temperature could lead to high pressure, causing the can to burst. (Also, isobutane is combustible, so incineration could cause the can to explode.)\n(b) We are looking for a pressure change due to a temperature change at constant volume, so we will use Amontons's/Gay-Lussac's law. Taking $P_{1}$ and $T_{1}$ as the initial values, $T_{2}$ as the temperature where the pressure is unknown and $P_{2}$ as the unknown pressure, and converting ${ }^{\\circ} \\mathrm{C}$ to K , we have:\n\n$$\n\\frac{P_{1}}{T_{1}}=\\frac{P_{2}}{T_{2}} \\text { which means that } \\frac{360 \\mathrm{kPa}}{297 \\mathrm{~K}}=\\frac{P_{2}}{323 \\mathrm{~K}}\n$$\n\nRearranging and solving gives: $P_{2}=\\frac{360 \\mathrm{kPa} \\times 323 \\mathrm{~K}}{297 \\mathrm{~K}}=390 \\mathrm{kPa}$"}
{"id": 3190, "contents": "795. Check Your Learning - \nA sample of nitrogen, $\\mathrm{N}_{2}$, occupies 45.0 mL at $27^{\\circ} \\mathrm{C}$ and 600 torr. What pressure will it have if cooled to $-73^{\\circ} \\mathrm{C}$ while the volume remains constant?"}
{"id": 3191, "contents": "796. Answer: - \n400 torr"}
{"id": 3192, "contents": "797. Volume and Temperature: Charles's Law - \nIf we fill a balloon with air and seal it, the balloon contains a specific amount of air at atmospheric pressure, let's say 1 atm. If we put the balloon in a refrigerator, the gas inside gets cold and the balloon shrinks (although both the amount of gas and its pressure remain constant). If we make the balloon very cold, it will shrink a great deal, and it expands again when it warms up."}
{"id": 3193, "contents": "798. LINK TO LEARNING - \nThis video (http://openstax.org/l/16CharlesLaw) shows how cooling and heating a gas causes its volume to decrease or increase, respectively.\n\nThese examples of the effect of temperature on the volume of a given amount of a confined gas at constant pressure are true in general: The volume increases as the temperature increases, and decreases as the temperature decreases. Volume-temperature data for a 1-mole sample of methane gas at 1 atm are listed and graphed in Figure 8.12.\n\n| Temperature $\\left({ }^{\\circ} \\mathrm{C}\\right)$ | Temperature (K) | Volume (L) |\n| :---: | :---: | :---: |\n| -3 | 270 | 22 |\n| -23 | 250 | 21 |\n| -53 | 220 | 18 |\n| -162 | 111 | 9 |\n\n\n\nFIGURE 8.12 The volume and temperature are linearly related for 1 mole of methane gas at a constant pressure of 1 atm. If the temperature is in kelvin, volume and temperature are directly proportional. The line stops at 111 K because methane liquefies at this temperature; when extrapolated, it intersects the graph's origin, representing a temperature of absolute zero.\n\nThe relationship between the volume and temperature of a given amount of gas at constant pressure is known as Charles's law in recognition of the French scientist and balloon flight pioneer Jacques Alexandre C\u00e9sar Charles. Charles's law states that the volume of a given amount of gas is directly proportional to its temperature on the kelvin scale when the pressure is held constant.\n\nMathematically, this can be written as:\n\n$$\nV \\alpha T \\text { or } V=\\text { constant } \\cdot T \\text { or } V=k \\cdot T \\text { or } V_{1} / T_{1}=V_{2} / T_{2}\n$$"}
{"id": 3194, "contents": "798. LINK TO LEARNING - \nwith $k$ being a proportionality constant that depends on the amount and pressure of the gas.\nFor a confined, constant pressure gas sample, $\\frac{V}{T}$ is constant (i.e., the ratio $=k$ ), and as seen with the $P-T$ relationship, this leads to another form of Charles's law: $\\frac{V_{1}}{T_{1}}=\\frac{V_{2}}{T_{2}}$."}
{"id": 3195, "contents": "800. Predicting Change in Volume with Temperature - \nA sample of carbon dioxide, $\\mathrm{CO}_{2}$, occupies 0.300 L at $10^{\\circ} \\mathrm{C}$ and 750 torr. What volume will the gas have at $30^{\\circ} \\mathrm{C}$ and 750 torr?"}
{"id": 3196, "contents": "801. Solution - \nBecause we are looking for the volume change caused by a temperature change at constant pressure, this is a job for Charles's law. Taking $V_{1}$ and $T_{1}$ as the initial values, $T_{2}$ as the temperature at which the volume is unknown and $V_{2}$ as the unknown volume, and converting ${ }^{\\circ} \\mathrm{C}$ into K we have:\n\n$$\n\\frac{V_{1}}{T_{1}}=\\frac{V_{2}}{T_{2}} \\text { which means that } \\frac{0.300 \\mathrm{~L}}{283 \\mathrm{~K}}=\\frac{V_{2}}{303 \\mathrm{~K}}\n$$\n\nRearranging and solving gives: $V_{2}=\\frac{0.300 \\mathrm{~L} \\times 303 \\mathrm{~K}}{283 \\mathrm{~K}}=0.321 \\mathrm{~L}$\nThis answer supports our expectation from Charles's law, namely, that raising the gas temperature (from 283 K to 303 K ) at a constant pressure will yield an increase in its volume (from 0.300 L to 0.321 L )."}
{"id": 3197, "contents": "802. Check Your Learning - \nA sample of oxygen, $\\mathrm{O}_{2}$, occupies 32.2 mL at $30^{\\circ} \\mathrm{C}$ and 452 torr. What volume will it occupy at $-70^{\\circ} \\mathrm{C}$ and the same pressure?"}
{"id": 3198, "contents": "803. Answer: - \n21.6 mL"}
{"id": 3199, "contents": "805. Measuring Temperature with a Volume Change - \nTemperature is sometimes measured with a gas thermometer by observing the change in the volume of the gas as the temperature changes at constant pressure. The hydrogen in a particular hydrogen gas thermometer has a volume of $150.0 \\mathrm{~cm}^{3}$ when immersed in a mixture of ice and water $\\left(0.00^{\\circ} \\mathrm{C}\\right)$. When immersed in boiling liquid ammonia, the volume of the hydrogen, at the same pressure, is $131.7 \\mathrm{~cm}^{3}$. Find the temperature of boiling ammonia on the kelvin and Celsius scales."}
{"id": 3200, "contents": "806. Solution - \nA volume change caused by a temperature change at constant pressure means we should use Charles's law. Taking $V_{1}$ and $T_{1}$ as the initial values, $T_{2}$ as the temperature at which the volume is unknown and $V_{2}$ as the unknown volume, and converting ${ }^{\\circ} \\mathrm{C}$ into K we have:\n\n$$\n\\frac{V_{1}}{T_{1}}=\\frac{V_{2}}{T_{2}} \\text { which means that } \\frac{150.0 \\mathrm{~cm}^{3}}{273.15 \\mathrm{~K}}=\\frac{131.7 \\mathrm{~cm}^{3}}{T_{2}}\n$$\n\nRearrangement gives $T_{2}=\\frac{131.7 \\mathrm{~cm}^{3} \\times 273.15 \\mathrm{~K}}{150.0 \\mathrm{~cm}^{3}}=239.8 \\mathrm{~K}$\nSubtracting 273.15 from 239.8 K , we find that the temperature of the boiling ammonia on the Celsius scale is $-33.4^{\\circ} \\mathrm{C}$."}
{"id": 3201, "contents": "807. Check Your Learning - \nWhat is the volume of a sample of ethane at 467 K and 1.1 atm if it occupies 405 mL at 298 K and 1.1 atm?"}
{"id": 3202, "contents": "808. Answer: - \n635 mL"}
{"id": 3203, "contents": "809. Volume and Pressure: Boyle's Law - \nIf we partially fill an airtight syringe with air, the syringe contains a specific amount of air at constant temperature, say $25^{\\circ} \\mathrm{C}$. If we slowly push in the plunger while keeping temperature constant, the gas in the syringe is compressed into a smaller volume and its pressure increases; if we pull out the plunger, the volume increases and the pressure decreases. This example of the effect of volume on the pressure of a given amount of a confined gas is true in general. Decreasing the volume of a contained gas will increase its pressure, and increasing its volume will decrease its pressure. In fact, if the volume increases by a certain factor, the pressure decreases by the same factor, and vice versa. Volume-pressure data for an air sample at room temperature are graphed in Figure 8.13.\n\n\n\n\nFIGURE 8.13 When a gas occupies a smaller volume, it exerts a higher pressure; when it occupies a larger volume, it exerts a lower pressure (assuming the amount of gas and the temperature do not change). Since $P$ and $V$ are inversely proportional, a graph of $\\frac{1}{P}$ vs. $V$ is linear.\n\nUnlike the $P-T$ and $V-T$ relationships, pressure and volume are not directly proportional to each other. Instead, $P$ and $V$ exhibit inverse proportionality: Increasing the pressure results in a decrease of the volume of the gas. Mathematically this can be written:\n\n$$\nP \\propto 1 / V \\text { or } P=k \\cdot 1 / V \\text { or } P \\cdot V=k \\text { or } P_{1} V_{1}=P_{2} V_{2}\n$$"}
{"id": 3204, "contents": "809. Volume and Pressure: Boyle's Law - \nwith $k$ being a constant. Graphically, this relationship is shown by the straight line that results when plotting the inverse of the pressure ( $\\frac{1}{P}$ ) versus the volume $(V)$, or the inverse of volume $\\left(\\frac{1}{V}\\right)$ versus the pressure $(P)$. Graphs with curved lines are difficult to read accurately at low or high values of the variables, and they are more difficult to use in fitting theoretical equations and parameters to experimental data. For those reasons, scientists often try to find a way to \"linearize\" their data. If we plot $P$ versus $V$, we obtain a hyperbola (see Figure 8.14).\n\n\nFIGURE 8.14 The relationship between pressure and volume is inversely proportional. (a) The graph of $P$ vs. $V$ is a hyperbola, whereas (b) the graph of $\\left(\\frac{1}{P}\\right)$ vs. $V$ is linear.\n\nThe relationship between the volume and pressure of a given amount of gas at constant temperature was first published by the English natural philosopher Robert Boyle over 300 years ago. It is summarized in the statement now known as Boyle's law: The volume of a given amount of gas held at constant temperature is inversely proportional to the pressure under which it is measured."}
{"id": 3205, "contents": "811. Volume of a Gas Sample - \nThe sample of gas in Figure 8.13 has a volume of 15.0 mL at a pressure of 13.0 psi. Determine the pressure of the gas at a volume of 7.5 mL , using:\n(a) the $P$ - $V$ graph in Figure 8.13\n(b) the $\\frac{1}{P}$ vs. $V$ graph in Figure 8.13\n(c) the Boyle's law equation\n\nComment on the likely accuracy of each method."}
{"id": 3206, "contents": "812. Solution - \n(a) Estimating from the $P$ - $V$ graph gives a value for $P$ somewhere around 27 psi .\n(b) Estimating from the $\\frac{1}{P}$ versus $V$ graph give a value of about 26 psi .\n(c) From Boyle's law, we know that the product of pressure and volume ( $P V$ ) for a given sample of gas at a constant temperature is always equal to the same value. Therefore we have $P_{1} V_{1}=k$ and $P_{2} V_{2}=k$ which means that $P_{1} V_{1}=P_{2} V_{2}$.\n\nUsing $P_{1}$ and $V_{1}$ as the known values 13.0 psi and $15.0 \\mathrm{~mL}, P_{2}$ as the pressure at which the volume is unknown, and $V_{2}$ as the unknown volume, we have:\n\n$$\nP_{1} V_{1}=P_{2} V_{2} \\text { or } 13.0 \\mathrm{psi} \\times 15.0 \\mathrm{~mL}=P_{2} \\times 7.5 \\mathrm{~mL}\n$$\n\nSolving:\n\n$$\nP_{2}=\\frac{13.0 \\mathrm{psi} \\times 15.0 \\mathrm{~mL}}{7.5 \\mathrm{~mL}}=26 \\mathrm{psi}\n$$\n\nIt was more difficult to estimate well from the $P$ - $V$ graph, so (a) is likely more inaccurate than (b) or (c). The calculation will be as accurate as the equation and measurements allow."}
{"id": 3207, "contents": "813. Check Your Learning - \nThe sample of gas in Figure 8.13 has a volume of 30.0 mL at a pressure of 6.5 psi . Determine the volume of the gas at a pressure of 11.0 psi , using:\n(a) the $P$ - $V$ graph in Figure 8.13\n(b) the $\\frac{1}{P}$ vs. $V$ graph in Figure 8.13\n(c) the Boyle's law equation\n\nComment on the likely accuracy of each method."}
{"id": 3208, "contents": "814. Answer: - \n(a) about $17-18 \\mathrm{~mL}$; (b) $\\sim 18 \\mathrm{~mL}$; (c) 17.7 mL ; it was more difficult to estimate well from the $P$ - $V$ graph, so (a) is likely more inaccurate than (b); the calculation will be as accurate as the equation and measurements allow"}
{"id": 3209, "contents": "816. Breathing and Boyle's Law - \nWhat do you do about 20 times per minute for your whole life, without break, and often without even being aware of it? The answer, of course, is respiration, or breathing. How does it work? It turns out that the gas laws apply here. Your lungs take in gas that your body needs (oxygen) and get rid of waste gas (carbon dioxide). Lungs are made of spongy, stretchy tissue that expands and contracts while you breathe. When you inhale, your diaphragm and intercostal muscles (the muscles between your ribs) contract, expanding your chest cavity and making your lung volume larger. The increase in volume leads to a decrease in pressure (Boyle's law). This causes air to flow into the lungs (from high pressure to low pressure). When you exhale, the process reverses: Your diaphragm and rib muscles relax, your chest cavity contracts, and your lung volume decreases, causing the pressure to increase (Boyle's law again), and air flows out of the lungs (from high pressure to low pressure). You then breathe in and out again, and again, repeating this Boyle's law cycle for the rest of your life (Figure 8.15).\n\n\nFIGURE 8.15 Breathing occurs because expanding and contracting lung volume creates small pressure differences between your lungs and your surroundings, causing air to be drawn into and forced out of your lungs."}
{"id": 3210, "contents": "817. Moles of Gas and Volume: Avogadro's Law - \nThe Italian scientist Amedeo Avogadro advanced a hypothesis in 1811 to account for the behavior of gases, stating that equal volumes of all gases, measured under the same conditions of temperature and pressure, contain the same number of molecules. Over time, this relationship was supported by many experimental observations as expressed by Avogadro's law: For a confined gas, the volume ( $V$ ) and number of moles (n) are directly proportional if the pressure and temperature both remain constant.\n\nIn equation form, this is written as:\n\n$$\nV \\propto n \\text { or } V=k \\times n \\text { or } \\frac{V_{1}}{n_{1}}=\\frac{V_{2}}{n_{2}}\n$$\n\nMathematical relationships can also be determined for the other variable pairs, such as $P$ versus $n$, and $n$ versus $T$."}
{"id": 3211, "contents": "818. LINK TO LEARNING - \nVisit this interactive PhET simulation (http://openstax.org/l/16IdealGasLaw) to investigate the relationships between pressure, volume, temperature, and amount of gas. Use the simulation to examine the effect of changing one parameter on another while holding the other parameters constant (as described in the preceding sections on the various gas laws)."}
{"id": 3212, "contents": "819. The Ideal Gas Law - \nTo this point, four separate laws have been discussed that relate pressure, volume, temperature, and the number of moles of the gas:\n\n- Boyle's law: $P V=$ constant at constant $T$ and $n$\n- Amontons's law: $\\frac{P}{T}=$ constant at constant $V$ and $n$\n- Charles's law: $\\frac{V}{T}=$ constant at constant $P$ and $n$\n- Avogadro's law: $\\frac{V}{n}=$ constant at constant $P$ and $T$\n\nCombining these four laws yields the ideal gas law, a relation between the pressure, volume, temperature, and number of moles of a gas:\n\n$$\nP V=n R T\n$$\n\nwhere $P$ is the pressure of a gas, $V$ is its volume, $n$ is the number of moles of the gas, $T$ is its temperature on the kelvin scale, and $R$ is a constant called the ideal gas constant or the universal gas constant. The units used to express pressure, volume, and temperature will determine the proper form of the gas constant as required by dimensional analysis, the most commonly encountered values being $0.08206 \\mathrm{~L} \\mathrm{~atm} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}$ and 8.314 kPa L $\\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$.\n\nGases whose properties of $P, V$, and $T$ are accurately described by the ideal gas law (or the other gas laws) are said to exhibit ideal behavior or to approximate the traits of an ideal gas. An ideal gas is a hypothetical construct that may be used along with kinetic molecular theory to effectively explain the gas laws as will be described in a later module of this chapter. Although all the calculations presented in this module assume ideal behavior, this assumption is only reasonable for gases under conditions of relatively low pressure and high temperature. In the final module of this chapter, a modified gas law will be introduced that accounts for the non-ideal behavior observed for many gases at relatively high pressures and low temperatures."}
{"id": 3213, "contents": "819. The Ideal Gas Law - \nThe ideal gas equation contains five terms, the gas constant $R$ and the variable properties $P, V, n$, and $T$. Specifying any four of these terms will permit use of the ideal gas law to calculate the fifth term as demonstrated in the following example exercises."}
{"id": 3214, "contents": "821. Using the Ideal Gas Law - \nMethane, $\\mathrm{CH}_{4}$, is being considered for use as an alternative automotive fuel to replace gasoline. One gallon of gasoline could be replaced by 655 g of $\\mathrm{CH}_{4}$. What is the volume of this much methane at $25^{\\circ} \\mathrm{C}$ and 745 torr?"}
{"id": 3215, "contents": "822. Solution - \nWe must rearrange $P V=n R T$ to solve for $V: V=\\frac{n R T}{P}$\nIf we choose to use $R=0.08206 \\mathrm{~L} \\mathrm{~atm} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}$, then the amount must be in moles, temperature must be in kelvin, and pressure must be in atm.\n\nConverting into the \"right\" units:\n\n$$\n\\begin{gathered}\nn=655 \\mathrm{gCH}_{4} \\times \\frac{1 \\mathrm{~mol}}{16.043-\\mathrm{g}_{4} \\mathrm{CH}_{4}}=40.8 \\mathrm{~mol} \\\\\nT=25^{\\circ} \\mathrm{C}+273=298 \\mathrm{~K} \\\\\nP=745 \\mathrm{tmm} \\times \\frac{1 \\mathrm{~atm}}{760 \\mathrm{tmf}}=0.980 \\mathrm{~atm} \\\\\nV=\\frac{n R T}{P}=\\frac{(40.8 \\mathrm{mor})\\left(0.08206 \\mathrm{~L} \\mathrm{atmol}^{-1} \\mathrm{~K}^{-1}\\right)(298 \\mathrm{~K})}{0.980 \\mathrm{~atm}}=1.02 \\times 10^{3} \\mathrm{~L}\n\\end{gathered}\n$$\n\nIt would require 1020 L ( 269 gal ) of gaseous methane at about 1 atm of pressure to replace 1 gal of gasoline. It requires a large container to hold enough methane at 1 atm to replace several gallons of gasoline."}
{"id": 3216, "contents": "823. Check Your Learning - \nCalculate the pressure in bar of 2520 moles of hydrogen gas stored at $27^{\\circ} \\mathrm{C}$ in the $180-\\mathrm{L}$ storage tank of a modern hydrogen-powered car."}
{"id": 3217, "contents": "824. Answer: - \n350 bar\n\nIf the number of moles of an ideal gas are kept constant under two different sets of conditions, a useful mathematical relationship called the combined gas law is obtained: $\\frac{P_{1} V_{1}}{T_{1}}=\\frac{P_{2} V_{2}}{T_{2}}$ using units of atm, L, and K. Both sets of conditions are equal to the product of $n \\times R$ (where $n=$ the number of moles of the gas and $R$ is the ideal gas law constant)."}
{"id": 3218, "contents": "826. Using the Combined Gas Law - \nWhen filled with air, a typical scuba tank with a volume of 13.2 L has a pressure of 153 atm (Figure 8.16). If the water temperature is $27^{\\circ} \\mathrm{C}$, how many liters of air will such a tank provide to a diver's lungs at a depth of approximately 70 feet in the ocean where the pressure is 3.13 atm ?\n\n\nFIGURE 8.16 Scuba divers use compressed air to breathe while underwater. (credit: modification of work by Mark Goodchild)\n\nLetting 1 represent the air in the scuba tank and 2 represent the air in the lungs, and noting that body temperature (the temperature the air will be in the lungs) is $37^{\\circ} \\mathrm{C}$, we have:\n\n$$\n\\frac{P_{1} V_{1}}{T_{1}}=\\frac{P_{2} V_{2}}{T_{2}} \\rightarrow \\frac{(153 \\mathrm{~atm})(13.2 \\mathrm{~L})}{(300 \\mathrm{~K})}=\\frac{(3.13 \\mathrm{~atm})\\left(V_{2}\\right)}{(310 \\mathrm{~K})}\n$$\n\nSolving for $V_{2}$ :\n\n$$\nV_{2}=\\frac{(153 \\mathrm{~atm})(13.2 \\mathrm{~L})(310 \\mathrm{~K})}{(300 \\mathrm{~K})(3.13 \\mathrm{~atm})}=667 \\mathrm{~L}\n$$\n\n(Note: Be advised that this particular example is one in which the assumption of ideal gas behavior is not very reasonable, since it involves gases at relatively high pressures and low temperatures. Despite this limitation, the calculated volume can be viewed as a good \"ballpark\" estimate.)"}
{"id": 3219, "contents": "827. Check Your Learning - \nA sample of ammonia is found to occupy 0.250 L under laboratory conditions of $27^{\\circ} \\mathrm{C}$ and 0.850 atm . Find the volume of this sample at $0^{\\circ} \\mathrm{C}$ and 1.00 atm ."}
{"id": 3220, "contents": "828. Answer: - \n0.193 L"}
{"id": 3221, "contents": "830. The Interdependence between Ocean Depth and Pressure in Scuba Diving - \nWhether scuba diving at the Great Barrier Reef in Australia (shown in Figure 8.17) or in the Caribbean, divers must understand how pressure affects a number of issues related to their comfort and safety.\n\n\nFIGURE 8.17 Scuba divers, whether at the Great Barrier Reef or in the Caribbean, must be aware of buoyancy, pressure equalization, and the amount of time they spend underwater, to avoid the risks associated with pressurized gases in the body. (credit: Kyle Taylor)"}
{"id": 3222, "contents": "830. The Interdependence between Ocean Depth and Pressure in Scuba Diving - \nPressure increases with ocean depth, and the pressure changes most rapidly as divers reach the surface. The pressure a diver experiences is the sum of all pressures above the diver (from the water and the air). Most pressure measurements are given in units of atmospheres, expressed as \"atmospheres absolute\" or ATA in the diving community: Every 33 feet of salt water represents 1 ATA of pressure in addition to 1 ATA of pressure from the atmosphere at sea level. As a diver descends, the increase in pressure causes the body's air pockets in the ears and lungs to compress; on the ascent, the decrease in pressure causes these air pockets to expand, potentially rupturing eardrums or bursting the lungs. Divers must therefore undergo equalization by adding air to body airspaces on the descent by breathing normally and adding air to the mask by breathing out of the nose or adding air to the ears and sinuses by equalization techniques; the corollary is also true on ascent, divers must release air from the body to maintain equalization. Buoyancy, or the ability to control whether a diver sinks or floats, is controlled by the buoyancy compensator (BCD). If a diver is ascending, the air in their BCD expands because of lower pressure according to Boyle's law (decreasing the pressure of gases increases the volume). The expanding air increases the buoyancy of the diver, and they begin to ascend. The diver must vent air from the BCD or risk an uncontrolled ascent that could rupture the lungs. In descending, the increased pressure causes the air in the BCD to compress and the diver sinks much more quickly; the diver must add air to the BCD or risk an uncontrolled descent, facing much higher pressures near the ocean floor. The pressure also impacts how long a diver can stay underwater before ascending. The deeper a diver dives, the more compressed the air that is breathed because of increased pressure: If a diver dives 33 feet, the pressure is 2 ATA and the air would be compressed to one-half of its original volume. The diver uses up available air twice as fast as at the surface."}
{"id": 3223, "contents": "831. Standard Conditions of Temperature and Pressure - \nWe have seen that the volume of a given quantity of gas and the number of molecules (moles) in a given volume of gas vary with changes in pressure and temperature. Chemists sometimes make comparisons against a standard temperature and pressure (STP) for reporting properties of gases: 273.15 K and 1 atm (101.325 $\\mathrm{kPa}) .{ }^{\\frac{1}{2}}$ At STP, one mole of an ideal gas has a volume of about $22.4 \\mathrm{~L}-$ this is referred to as the standard molar volume (Figure 8.18).\n\n[^4]\n\nFIGURE 8.18 Regardless of its chemical identity, one mole of gas behaving ideally occupies a volume of $\\sim 22.4 \\mathrm{~L}$ at STP."}
{"id": 3224, "contents": "831. Standard Conditions of Temperature and Pressure - 831.1. Stoichiometry of Gaseous Substances, Mixtures, and Reactions LEARNING OBJECTIVES\nBy the end of this section, you will be able to:\n\n- Use the ideal gas law to compute gas densities and molar masses\n- Perform stoichiometric calculations involving gaseous substances\n- State Dalton's law of partial pressures and use it in calculations involving gaseous mixtures\n\nThe study of the chemical behavior of gases was part of the basis of perhaps the most fundamental chemical revolution in history. French nobleman Antoine Lavoisier, widely regarded as the \"father of modern chemistry,\" changed chemistry from a qualitative to a quantitative science through his work with gases. He discovered the law of conservation of matter, discovered the role of oxygen in combustion reactions, determined the composition of air, explained respiration in terms of chemical reactions, and more. He was a casualty of the French Revolution, guillotined in 1794. Of his death, mathematician and astronomer JosephLouis Lagrange said, \"It took the mob only a moment to remove his head; a century will not suffice to reproduce it.\" ${ }^{2}$ Much of the knowledge we do have about Lavoisier's contributions is due to his wife, MarieAnne Paulze Lavoisier, who worked with him in his lab. A trained artist fluent in several languages, she created detailed illustrations of the equipment in his lab, and translated texts from foreign scientists to complement his knowledge. After his execution, she was instrumental in publishing Lavoisier's major treatise, which unified many concepts of chemistry and laid the groundwork for significant further study.\n\nAs described in an earlier chapter of this text, we can turn to chemical stoichiometry for answers to many of the questions that ask \"How much?\" The essential property involved in such use of stoichiometry is the amount of substance, typically measured in moles ( $n$ ). For gases, molar amount can be derived from convenient experimental measurements of pressure, temperature, and volume. Therefore, these measurements are useful in assessing the stoichiometry of pure gases, gas mixtures, and chemical reactions involving gases. This section will not introduce any new material or ideas, but will provide examples of applications and ways to integrate concepts we have already discussed.\n\n[^5]"}
{"id": 3225, "contents": "832. Gas Density and Molar Mass - \nThe ideal gas law described previously in this chapter relates the properties of pressure $P$, volume $V$, temperature $T$, and molar amount $n$. This law is universal, relating these properties in identical fashion regardless of the chemical identity of the gas:\n\n$$\nP V=n R T\n$$\n\nThe density $d$ of a gas, on the other hand, is determined by its identity. As described in another chapter of this text, the density of a substance is a characteristic property that may be used to identify the substance.\n\n$$\nd=\\frac{m}{V}\n$$\n\nRearranging the ideal gas equation to isolate $V$ and substituting into the density equation yields\n\n$$\nd=\\frac{m P}{n R T}=\\left(\\frac{m}{n}\\right) \\frac{P}{R T}\n$$\n\nThe ratio $m / n$ is the definition of molar mass, $\\mathcal{M}$ :\n\n$$\n\\mathscr{M}=\\frac{m}{n}\n$$\n\nThe density equation can then be written\n\n$$\nd=\\frac{\\mathscr{M} P}{R T}\n$$\n\nThis relation may be used for calculating the densities of gases of known identities at specified values of pressure and temperature as demonstrated in Example 8.11."}
{"id": 3226, "contents": "834. Measuring Gas Density - \nWhat is the density of molecular nitrogen gas at STP?"}
{"id": 3227, "contents": "835. Solution - \nThe molar mass of molecular nitrogen, $\\mathrm{N}_{2}$, is $28.01 \\mathrm{~g} / \\mathrm{mol}$. Substituting this value along with standard temperature and pressure into the gas density equation yields\n\n$$\nd=\\frac{\\mathscr{M} P}{R T}=\\frac{(28.01 \\mathrm{~g} / \\mathrm{mol})(1.00 \\mathrm{~atm})}{\\left(0.0821 \\mathrm{~L} \\mathrm{~atm} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}\\right)(273 \\mathrm{~K})}=1.25 \\mathrm{~g} / \\mathrm{L}\n$$"}
{"id": 3228, "contents": "836. Check Your Learning - \nWhat is the density of molecular hydrogen gas at $17.0^{\\circ} \\mathrm{C}$ and a pressure of 760 torr?"}
{"id": 3229, "contents": "837. Answer: - \n$\\mathrm{d}=0.0847 \\mathrm{~g} / \\mathrm{L}$\n\nWhen the identity of a gas is unknown, measurements of the mass, pressure, volume, and temperature of a sample can be used to calculate the molar mass of the gas (a useful property for identification purposes). Combining the ideal gas equation\n\n$$\nP V=n R T\n$$\n\nand the definition of molar mass\n\n$$\n\\mathscr{M}=\\frac{m}{n}\n$$\n\nyields the following equation:\n\n$$\n\\mathscr{M}=\\frac{m R T}{P V}\n$$\n\nDetermining the molar mass of a gas via this approach is demonstrated in Example 8.12."}
{"id": 3230, "contents": "839. Determining the Molecular Formula of a Gas from its Molar Mass and Empirical Formula - \nCyclopropane, a gas once used with oxygen as a general anesthetic, is composed of $85.7 \\%$ carbon and $14.3 \\%$ hydrogen by mass. Find the empirical formula. If 1.56 g of cyclopropane occupies a volume of 1.00 L at 0.984 atm and $50^{\\circ} \\mathrm{C}$, what is the molecular formula for cyclopropane?"}
{"id": 3231, "contents": "840. Solution - \nFirst determine the empirical formula of the gas. Assume 100 g and convert the percentage of each element into grams. Determine the number of moles of carbon and hydrogen in the $100-\\mathrm{g}$ sample of cyclopropane. Divide by the smallest number of moles to relate the number of moles of carbon to the number of moles of hydrogen. In the last step, realize that the smallest whole number ratio is the empirical formula:\n\n$$\n\\begin{array}{lll}\n85.7 \\mathrm{~g} \\mathrm{C} \\times \\frac{1 \\mathrm{~mol} \\mathrm{C}}{12.01 \\mathrm{~g} \\mathrm{C}}=7.136 \\mathrm{~mol} \\mathrm{C} & \\frac{7.136}{7.136}=1.00 \\mathrm{~mol} \\mathrm{C} \\\\\n14.3 \\mathrm{~g} \\mathrm{H} \\times \\frac{1 \\mathrm{~mol} \\mathrm{H}}{1.01 \\mathrm{~g} \\mathrm{H}}=14.158 \\mathrm{~mol} \\mathrm{H} & \\frac{14.158}{7.136}=1.98 \\mathrm{~mol} \\mathrm{H}\n\\end{array}\n$$\n\nEmpirical formula is $\\mathrm{CH}_{2}$ [empirical mass (EM) of $14.03 \\mathrm{~g} / \\mathrm{emp}$ irical unit].\nNext, use the provided values for mass, pressure, temperature and volume to compute the molar mass of the gas:\n\n$$\n\\mathscr{M}=\\frac{m R T}{P V}=\\frac{(1.56 \\mathrm{~g})\\left(0.0821 \\mathrm{~L} \\mathrm{~atm} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}\\right)(323 \\mathrm{~K})}{(0.984 \\mathrm{~atm})(1.00 \\mathrm{~L})}=42.0 \\mathrm{~g} / \\mathrm{mol}\n$$\n\nComparing the molar mass to the empirical formula mass shows how many empirical formula units make up a molecule:"}
{"id": 3232, "contents": "840. Solution - \nComparing the molar mass to the empirical formula mass shows how many empirical formula units make up a molecule:\n\n$$\n\\frac{M}{E M}=\\frac{42.0 \\mathrm{~g} / \\mathrm{mol}}{14.0 \\mathrm{~g} / \\mathrm{mol}}=3\n$$\n\nThe molecular formula is thus derived from the empirical formula by multiplying each of its subscripts by three:\n\n$$\n\\left(\\mathrm{CH}_{2}\\right)_{3}=\\mathrm{C}_{3} \\mathrm{H}_{6}\n$$"}
{"id": 3233, "contents": "841. Check Your Learning - \nAcetylene, a fuel used welding torches, is composed of $92.3 \\% \\mathrm{C}$ and $7.7 \\% \\mathrm{H}$ by mass. Find the empirical formula. If 1.10 g of acetylene occupies of volume of 1.00 L at 1.15 atm and $59.5^{\\circ} \\mathrm{C}$, what is the molecular formula for acetylene?"}
{"id": 3234, "contents": "842. Answer: - \nEmpirical formula, CH; Molecular formula, $\\mathrm{C}_{2} \\mathrm{H}_{2}$"}
{"id": 3235, "contents": "844. Determining the Molar Mass of a Volatile Liquid - \nThe approximate molar mass of a volatile liquid can be determined by:\n\n1. Heating a sample of the liquid in a flask with a tiny hole at the top, which converts the liquid into gas that\nmay escape through the hole\n2. Removing the flask from heat at the instant when the last bit of liquid becomes gas, at which time the flask will be filled with only gaseous sample at ambient pressure\n3. Sealing the flask and permitting the gaseous sample to condense to liquid, and then weighing the flask to determine the sample's mass (see Figure 8.19)\n\n\nFIGURE 8.19 When the volatile liquid in the flask is heated past its boiling point, it becomes gas and drives air out of the flask. At $t_{l \\longrightarrow g}$, the flask is filled with volatile liquid gas at the same pressure as the atmosphere. If the flask is then cooled to room temperature, the gas condenses and the mass of the gas that filled the flask, and is now liquid, can be measured. (credit: modification of work by Mark Ott)\n\nUsing this procedure, a sample of chloroform gas weighing 0.494 g is collected in a flask with a volume of 129 $\\mathrm{cm}^{3}$ at $99.6^{\\circ} \\mathrm{C}$ when the atmospheric pressure is 742.1 mm Hg . What is the approximate molar mass of chloroform?"}
{"id": 3236, "contents": "845. Solution - \nSince $\\mathcal{M}=\\frac{m}{n}$ and $n=\\frac{P V}{R T}$, substituting and rearranging gives $\\mathcal{M}=\\frac{m R T}{P V}$,\nthen\n\n$$\n\\mathcal{M}=\\frac{m R T}{P V}=\\frac{(0.494 \\mathrm{~g}) \\times 0.08206 \\mathrm{~L} \\cdot \\mathrm{~atm} / \\mathrm{mol} \\mathrm{~K} \\times 372.8 \\mathrm{~K}}{0.976 \\mathrm{~atm} \\times 0.129 \\mathrm{~L}}=120 \\mathrm{~g} / \\mathrm{mol} .\n$$"}
{"id": 3237, "contents": "846. Check Your Learning - \nA sample of phosphorus that weighs $3.243 \\times 10^{-2} \\mathrm{~g}$ exerts a pressure of 31.89 kPa in a $56.0-\\mathrm{mL}$ bulb at $550^{\\circ} \\mathrm{C}$. What are the molar mass and molecular formula of phosphorus vapor?"}
{"id": 3238, "contents": "847. Answer: - \n$124 \\mathrm{~g} / \\mathrm{mol} \\mathrm{P}_{4}$"}
{"id": 3239, "contents": "848. The Pressure of a Mixture of Gases: Dalton's Law - \nUnless they chemically react with each other, the individual gases in a mixture of gases do not affect each other's pressure. Each individual gas in a mixture exerts the same pressure that it would exert if it were present alone in the container (Figure 8.20). The pressure exerted by each individual gas in a mixture is called its partial pressure. This observation is summarized by Dalton's law of partial pressures: The total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases:\n\n$$\nP_{\\text {Total }}=P_{A}+P_{B}+P_{C}+\\ldots=\\Sigma_{\\mathrm{i}} P_{\\mathrm{i}}\n$$\n\nIn the equation $P_{\\text {Total }}$ is the total pressure of a mixture of gases, $P_{A}$ is the partial pressure of gas $A ; P_{B}$ is the partial pressure of gas $B ; P_{C}$ is the partial pressure of gas $C$; and so on.\n\n\nFIGURE 8.20 If equal-volume cylinders containing gasses at pressures of $300 \\mathrm{kPa}, 450 \\mathrm{kPa}$, and 600 kPa are all combined in the same-size cylinder, the total pressure of the gas mixture is 1350 kPa .\n\nThe partial pressure of gas A is related to the total pressure of the gas mixture via its mole fraction ( $\\boldsymbol{X}$ ), a unit of concentration defined as the number of moles of a component of a solution divided by the total number of moles of all components:\n\n$$\nP_{A}=X_{A} \\times P_{\\text {Total }} \\quad \\text { where } \\quad X_{A}=\\frac{n_{A}}{n_{\\text {Total }}}\n$$\n\nwhere $P_{A}, X_{A}$, and $n_{A}$ are the partial pressure, mole fraction, and number of moles of gas A, respectively, and $n_{\\text {Total }}$ is the number of moles of all components in the mixture."}
{"id": 3240, "contents": "850. The Pressure of a Mixture of Gases - \nA 10.0-L vessel contains $2.50 \\times 10^{-3} \\mathrm{~mol}$ of $\\mathrm{H}_{2}, 1.00 \\times 10^{-3} \\mathrm{~mol}$ of He , and $3.00 \\times 10^{-4} \\mathrm{~mol}$ of Ne at $35^{\\circ} \\mathrm{C}$.\n(a) What are the partial pressures of each of the gases?\n(b) What is the total pressure in atmospheres?"}
{"id": 3241, "contents": "851. Solution - \nThe gases behave independently, so the partial pressure of each gas can be determined from the ideal gas equation, using $P=\\frac{n R T}{V}$ :\n\n$$\n\\begin{aligned}\n& P_{\\mathrm{H}_{2}}=\\frac{\\left(2.50 \\times 10^{-3} \\mathrm{mot}\\right)\\left(0.08206 \\mathrm{~L} \\mathrm{~atm} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}\\right)(308 \\mathrm{~K})}{10.0 \\mathrm{t}}=6.32 \\times 10^{-3} \\mathrm{~atm} \\\\\n& P_{\\mathrm{He}}=\\frac{\\left(1.00 \\times 10^{-3} \\mathrm{mot}\\right)\\left(0.08206 \\mathrm{t} \\mathrm{~atm} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}\\right)(308 \\mathrm{~K})}{10.0 \\mathrm{t}}=2.53 \\times 10^{-3} \\mathrm{~atm} \\\\\n& \\left.P_{\\mathrm{Ne}}=\\frac{\\left(3.00 \\times 10^{-4} \\mathrm{mot}\\right)\\left(0.08206 \\mathrm{t} \\mathrm{~atm} \\mathrm{mmol}^{-1} \\mathrm{~K}^{-1}\\right)}{10.0 \\mathrm{t}}(308 \\mathrm{~K})\\right)=7.58 \\times 10^{-4} \\mathrm{~atm}\n\\end{aligned}\n$$\n\nThe total pressure is given by the sum of the partial pressures:\n\n$$\nP_{\\mathrm{T}}=P_{\\mathrm{H}_{2}}+P_{\\mathrm{He}}+P_{\\mathrm{Ne}}=(0.00632+0.00253+0.00076) \\mathrm{atm}=9.61 \\times 10^{-3} \\mathrm{~atm}\n$$"}
{"id": 3242, "contents": "852. Check Your Learning - \nA 5.73-L flask at $25^{\\circ} \\mathrm{C}$ contains 0.0388 mol of $\\mathrm{N}_{2}, 0.147 \\mathrm{~mol}$ of CO , and $0.0803 \\mathrm{~mol} \\mathrm{of} \\mathrm{H}_{2}$. What is the total pressure in the flask in atmospheres?"}
{"id": 3243, "contents": "853. Answer: - \n1.137 atm\n\nHere is another example of this concept, but dealing with mole fraction calculations."}
{"id": 3244, "contents": "855. The Pressure of a Mixture of Gases - \nA gas mixture used for anesthesia contains 2.83 mol oxygen, $\\mathrm{O}_{2}$, and 8.41 mol nitrous oxide, $\\mathrm{N}_{2} \\mathrm{O}$. The total pressure of the mixture is 192 kPa .\n(a) What are the mole fractions of $\\mathrm{O}_{2}$ and $\\mathrm{N}_{2} \\mathrm{O}$ ?\n(b) What are the partial pressures of $\\mathrm{O}_{2}$ and $\\mathrm{N}_{2} \\mathrm{O}$ ?"}
{"id": 3245, "contents": "856. Solution - \nThe mole fraction is given by $X_{A}=\\frac{n_{A}}{n_{\\text {Total }}}$ and the partial pressure is $P_{A}=X_{A} \\times P_{\\text {Total }}$.\nFor $\\mathrm{O}_{2}$,\n\n$$\nX_{O_{2}}=\\frac{n_{O_{2}}}{n_{\\text {Total }}}=\\frac{2.83 \\mathrm{~mol}}{(2.83+8.41) \\mathrm{mol}}=0.252\n$$\n\nand $P_{O_{2}}=X_{O_{2}} \\times P_{\\text {Total }}=0.252 \\times 192 \\mathrm{kPa}=48.4 \\mathrm{kPa}$\nFor $\\mathrm{N}_{2} \\mathrm{O}$,\n\n$$\nX_{N_{2} O}=\\frac{n_{N_{2} O}}{n_{\\text {Total }}}=\\frac{8.41 \\mathrm{~mol}}{(2.83+8.41) \\mathrm{mol}}=0.748\n$$\n\nand\n$P_{N_{2} O}=X_{N_{2} O} \\times P_{\\text {Total }}=0.748 \\times 192 \\mathrm{kPa}=144 \\mathrm{kPa}$"}
{"id": 3246, "contents": "857. Check Your Learning - \nWhat is the pressure of a mixture of 0.200 g of $\\mathrm{H}_{2}, 1.00 \\mathrm{~g}$ of $\\mathrm{N}_{2}$, and 0.820 g of Ar in a container with a volume of 2.00 L at $20^{\\circ} \\mathrm{C}$ ?"}
{"id": 3247, "contents": "858. Answer: - \n1.87 atm"}
{"id": 3248, "contents": "859. Collection of Gases over Water - \nA simple way to collect gases that do not react with water is to capture them in a bottle that has been filled with water and inverted into a dish filled with water. The pressure of the gas inside the bottle can be made equal to the air pressure outside by raising or lowering the bottle. When the water level is the same both inside and outside the bottle (Figure 8.21), the pressure of the gas is equal to the atmospheric pressure, which can be measured with a barometer.\n\n\nFIGURE 8.21 When a reaction produces a gas that is collected above water, the trapped gas is a mixture of the gas produced by the reaction and water vapor. If the collection flask is appropriately positioned to equalize the water levels both within and outside the flask, the pressure of the trapped gas mixture will equal the atmospheric pressure outside the flask (see the earlier discussion of manometers).\n\nHowever, there is another factor we must consider when we measure the pressure of the gas by this method. Water evaporates and there is always gaseous water (water vapor) above a sample of liquid water. As a gas is collected over water, it becomes saturated with water vapor and the total pressure of the mixture equals the partial pressure of the gas plus the partial pressure of the water vapor. The pressure of the pure gas is therefore equal to the total pressure minus the pressure of the water vapor-this is referred to as the \"dry\" gas pressure, that is, the pressure of the gas only, without water vapor. The vapor pressure of water, which is the pressure exerted by water vapor in equilibrium with liquid water in a closed container, depends on the temperature (Figure 8.22); more detailed information on the temperature dependence of water vapor can be found in Table 8.2, and vapor pressure will be discussed in more detail in the chapter on liquids.\n\n\nFIGURE 8.22 This graph shows the vapor pressure of water at sea level as a function of temperature.\n\nVapor Pressure of Ice and Water in Various Temperatures at Sea Level"}
{"id": 3249, "contents": "859. Collection of Gases over Water - \nFIGURE 8.22 This graph shows the vapor pressure of water at sea level as a function of temperature.\n\nVapor Pressure of Ice and Water in Various Temperatures at Sea Level\n\n| Temperature $\\left({ }^{\\circ} \\mathrm{C}\\right)$ | Pressure (torr) | Temperature ( ${ }^{\\circ} \\mathrm{C}$ ) | Pressure (torr) | Temperature ( ${ }^{\\circ} \\mathrm{C}$ ) | Pressure (torr) |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| -10 | 1.95 | 18 | 15.5 | 30 | 31.8 |\n| -5 | 3.0 | 19 | 16.5 | 35 | 42.2 |\n| -2 | 3.9 | 20 | 17.5 | 40 | 55.3 |\n| 0 | 4.6 | 21 | 18.7 | 50 | 92.5 |\n| 2 | 5.3 | 22 | 19.8 | 60 | 149.4 |\n| 4 | 6.1 | 23 | 21.1 | 70 | 233.7 |\n| 6 | 7.0 | 24 | 22.4 | 80 | 355.1 |\n| 8 | 8.0 | 25 | 23.8 | 90 | 525.8 |\n| 10 | 9.2 | 26 | 25.2 | 95 | 633.9 |\n| 12 | 10.5 | 27 | 26.7 | 99 | 733.2 |\n| 14 | 12.0 | 28 | 28.3 | 100.0 | 760.0 |\n| 16 | 13.6 | 29 | 30.0 | 101.0 | 787.6 |\n\nTABLE 8.2"}
{"id": 3250, "contents": "861. Pressure of a Gas Collected Over Water - \nIf 0.200 L of argon is collected over water at a temperature of $26^{\\circ} \\mathrm{C}$ and a pressure of 750 torr in a system like that shown in Figure 8.21, what is the partial pressure of argon?"}
{"id": 3251, "contents": "862. Solution - \nAccording to Dalton's law, the total pressure in the bottle ( 750 torr) is the sum of the partial pressure of argon and the partial pressure of gaseous water:\n\n$$\nP_{\\mathrm{T}}=P_{\\mathrm{Ar}}+P_{\\mathrm{H}_{2} \\mathrm{O}}\n$$\n\nRearranging this equation to solve for the pressure of argon gives:\n\n$$\nP_{\\mathrm{Ar}}=P_{\\mathrm{T}}-P_{\\mathrm{H}_{2} \\mathrm{O}}\n$$\n\nThe pressure of water vapor above a sample of liquid water at $26^{\\circ} \\mathrm{C}$ is 25.2 torr (Appendix E), so:\n\n$$\nP_{\\mathrm{Ar}}=750 \\text { torr }-25.2 \\text { torr }=725 \\text { torr }\n$$"}
{"id": 3252, "contents": "863. Check Your Learning - \nA sample of oxygen collected over water at a temperature of $29.0^{\\circ} \\mathrm{C}$ and a pressure of 764 torr has a volume of 0.560 L . What volume would the dry oxygen have under the same conditions of temperature and pressure?"}
{"id": 3253, "contents": "864. Answer: - \n0.583 L"}
{"id": 3254, "contents": "865. Chemical Stoichiometry and Gases - \nChemical stoichiometry describes the quantitative relationships between reactants and products in chemical reactions.\n\nWe have previously measured quantities of reactants and products using masses for solids and volumes in conjunction with the molarity for solutions; now we can also use gas volumes to indicate quantities. If we know the volume, pressure, and temperature of a gas, we can use the ideal gas equation to calculate how many moles of the gas are present. If we know how many moles of a gas are involved, we can calculate the volume of a gas at any temperature and pressure."}
{"id": 3255, "contents": "866. Avogadro's Law Revisited - \nSometimes we can take advantage of a simplifying feature of the stoichiometry of gases that solids and solutions do not exhibit: All gases that show ideal behavior contain the same number of molecules in the same volume (at the same temperature and pressure). Thus, the ratios of volumes of gases involved in a chemical reaction are given by the coefficients in the equation for the reaction, provided that the gas volumes are measured at the same temperature and pressure.\n\nWe can extend Avogadro's law (that the volume of a gas is directly proportional to the number of moles of the gas) to chemical reactions with gases: Gases combine, or react, in definite and simple proportions by volume, provided that all gas volumes are measured at the same temperature and pressure. For example, since nitrogen and hydrogen gases react to produce ammonia gas according to $\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\longrightarrow 2 \\mathrm{NH}_{3}(g)$, a given volume of nitrogen gas reacts with three times that volume of hydrogen gas to produce two times that volume of ammonia gas, if pressure and temperature remain constant.\n\nThe explanation for this is illustrated in Figure 8.23. According to Avogadro's law, equal volumes of gaseous $\\mathrm{N}_{2}$, $\\mathrm{H}_{2}$, and $\\mathrm{NH}_{3}$, at the same temperature and pressure, contain the same number of molecules. Because one molecule of $\\mathrm{N}_{2}$ reacts with three molecules of $\\mathrm{H}_{2}$ to produce two molecules of $\\mathrm{NH}_{3}$, the volume of $\\mathrm{H}_{2}$ required is three times the volume of $\\mathrm{N}_{2}$, and the volume of $\\mathrm{NH}_{3}$ produced is two times the volume of $\\mathrm{N}_{2}$.\n\n\nFIGURE 8.23 One volume of $\\mathrm{N}_{2}$ combines with three volumes of $\\mathrm{H}_{2}$ to form two volumes of $\\mathrm{NH}_{3}$."}
{"id": 3256, "contents": "868. Reaction of Gases - \nPropane, $\\mathrm{C}_{3} \\mathrm{H}_{8}(\\mathrm{~g})$, is used in gas grills to provide the heat for cooking. What volume of $\\mathrm{O}_{2}(\\mathrm{~g})$ measured at $25{ }^{\\circ} \\mathrm{C}$ and 760 torr is required to react with 2.7 L of propane measured under the same conditions of temperature and pressure? Assume that the propane undergoes complete combustion."}
{"id": 3257, "contents": "869. Solution - \nThe ratio of the volumes of $\\mathrm{C}_{3} \\mathrm{H}_{8}$ and $\\mathrm{O}_{2}$ will be equal to the ratio of their coefficients in the balanced equation for the reaction:\n\n$$\n\\begin{array}{lll}\n\\mathrm{C}_{3} \\mathrm{H}_{8}(\\mathrm{~g})+5 \\mathrm{O}_{2}(\\mathrm{~g}) & \\longrightarrow & 3 \\mathrm{CO}_{2}(\\mathrm{~g})+4 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\\\\n1 \\text { volume }+5 \\text { volumes } & & 3 \\text { volumes }+4 \\text { volumes }\n\\end{array}\n$$\n\nFrom the equation, we see that one volume of $\\mathrm{C}_{3} \\mathrm{H}_{8}$ will react with five volumes of $\\mathrm{O}_{2}$ :\n\n$$\n2.7 \\mathrm{LC}_{3} \\mathrm{H}_{8}-\\frac{5 \\mathrm{LO}_{2}}{1 \\mathrm{LC}_{3} \\mathrm{H}_{8}}=13.5 \\mathrm{~L} \\mathrm{O}_{2}\n$$\n\nA volume of 13.5 L of $\\mathrm{O}_{2}$ will be required to react with $2.7 \\mathrm{~L}^{\\text {of }} \\mathrm{C}_{3} \\mathrm{H}_{8}$."}
{"id": 3258, "contents": "870. Check Your Learning - \nAn acetylene tank for an oxyacetylene welding torch provides 9340 L of acetylene gas, $\\mathrm{C}_{2} \\mathrm{H}_{2}$, at $0^{\\circ} \\mathrm{C}$ and 1 atm. How many tanks of oxygen, each providing $7.00 \\times 10^{3} \\mathrm{~L}$ of $\\mathrm{O}_{2}$ at $0^{\\circ} \\mathrm{C}$ and 1 atm , will be required to burn the acetylene?\n\n$$\n2 \\mathrm{C}_{2} \\mathrm{H}_{2}+5 \\mathrm{O}_{2} \\longrightarrow 4 \\mathrm{CO}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}\n$$"}
{"id": 3259, "contents": "871. Answer: - \n3.34 tanks ( $2.34 \\times 10^{4} \\mathrm{~L}$ )"}
{"id": 3260, "contents": "873. Volumes of Reacting Gases - \nAmmonia is an important fertilizer and industrial chemical. Suppose that a volume of 683 billion cubic feet of gaseous ammonia, measured at $25^{\\circ} \\mathrm{C}$ and 1 atm , was manufactured. What volume of $\\mathrm{H}_{2}(\\mathrm{~g})$, measured under the same conditions, was required to prepare this amount of ammonia by reaction with $\\mathrm{N}_{2}$ ?\n\n$$\n\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\longrightarrow 2 \\mathrm{NH}_{3}(g)\n$$"}
{"id": 3261, "contents": "874. Solution - \nBecause equal volumes of $\\mathrm{H}_{2}$ and $\\mathrm{NH}_{3}$ contain equal numbers of molecules and each three molecules of $\\mathrm{H}_{2}$ that react produce two molecules of $\\mathrm{NH}_{3}$, the ratio of the volumes of $\\mathrm{H}_{2}$ and $\\mathrm{NH}_{3}$ will be equal to $3: 2$. Two volumes of $\\mathrm{NH}_{3}$, in this case in units of billion $\\mathrm{ft}^{3}$, will be formed from three volumes of $\\mathrm{H}_{2}$ :\n\n$$\n683 \\text { billion } \\mathrm{ft}^{3} \\mathrm{NH}_{3} \\times \\frac{3 \\text { billion } \\mathrm{ft}^{3} \\mathrm{H}_{2}}{2 \\text { bitlion } \\mathrm{ft}^{3} \\mathrm{NH}_{3}}=1.02 \\times 10^{3} \\text { billion } \\mathrm{ft}^{3} \\mathrm{H}_{2}\n$$\n\nThe manufacture of 683 billion $\\mathrm{ft}^{3}$ of $\\mathrm{NH}_{3}$ required 1020 billion $\\mathrm{ft}^{3}$ of $\\mathrm{H}_{2}$. (At $25^{\\circ} \\mathrm{C}$ and 1 atm, this is the volume of a cube with an edge length of approximately 1.9 miles.)"}
{"id": 3262, "contents": "875. Check Your Learning - \nWhat volume of $\\mathrm{O}_{2}(\\mathrm{~g})$ measured at $25{ }^{\\circ} \\mathrm{C}$ and 760 torr is required to react with 17.0 L of ethylene, $\\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{~g})$, measured under the same conditions of temperature and pressure? The products are $\\mathrm{CO}_{2}$ and water vapor."}
{"id": 3263, "contents": "876. Answer: - \n51.0 L"}
{"id": 3264, "contents": "878. Volume of Gaseous Product - \nWhat volume of hydrogen at $27^{\\circ} \\mathrm{C}$ and 723 torr may be prepared by the reaction of 8.88 g of gallium with an excess of hydrochloric acid?\n\n$$\n2 \\mathrm{Ga}(s)+6 \\mathrm{HCl}(a q) \\longrightarrow 2 \\mathrm{GaCl}_{3}(a q)+3 \\mathrm{H}_{2}(g)\n$$"}
{"id": 3265, "contents": "879. Solution - \nConvert the provided mass of the limiting reactant, Ga , to moles of hydrogen produced:\n\n$$\n8.88 \\mathrm{~g} \\mathrm{Ga} \\times \\frac{1 \\mathrm{molGa}}{69.723 \\mathrm{~g} \\mathrm{ga}} \\times \\frac{3 \\mathrm{~mol} \\mathrm{H}_{2}}{2 \\mathrm{molGa}}=0.191 \\mathrm{~mol} \\mathrm{H}_{2}\n$$\n\nConvert the provided temperature and pressure values to appropriate units ( K and atm, respectively), and then use the molar amount of hydrogen gas and the ideal gas equation to calculate the volume of gas:\n\n$$\nV=\\left(\\frac{n R T}{P}\\right)=\\frac{0.191 \\mathrm{mot} \\times 0.08206 \\mathrm{~L} \\mathrm{atmmol}^{-1} \\mathrm{~K}^{-1} \\times 300 \\mathrm{~K}}{0.951 \\mathrm{~atm}}=4.94 \\mathrm{~L}\n$$"}
{"id": 3266, "contents": "880. Check Your Learning - \nSulfur dioxide is an intermediate in the preparation of sulfuric acid. What volume of $\\mathrm{SO}_{2}$ at $343{ }^{\\circ} \\mathrm{C}$ and 1.21 atm is produced by burning 1.00 kg of sulfur in excess oxygen?"}
{"id": 3267, "contents": "881. Answer: - \n$1.30 \\times 10^{3} \\mathrm{~L}$"}
{"id": 3268, "contents": "883. Greenhouse Gases and Climate Change - \nThe thin skin of our atmosphere keeps the earth from being an ice planet and makes it habitable. In fact, this is due to less than $0.5 \\%$ of the air molecules. Of the energy from the sun that reaches the earth, almost $\\frac{1}{3}$ is reflected back into space, with the rest absorbed by the atmosphere and the surface of the earth. Some of the energy that the earth absorbs is re-emitted as infrared (IR) radiation, a portion of which passes back out through the atmosphere into space. Most of this IR radiation, however, is absorbed by certain atmospheric gases, effectively trapping heat within the atmosphere in a phenomenon known as the greenhouse effect. This effect maintains global temperatures within the range needed to sustain life on earth. Without our atmosphere, the earth's average temperature would be lower by more than $30^{\\circ} \\mathrm{C}$ (nearly $60^{\\circ} \\mathrm{F}$ ). The major greenhouse gases (GHGs) are water vapor, carbon dioxide, methane, and ozone. Since the Industrial Revolution, human activity has been increasing the concentrations of GHGs, which have changed the energy balance and are significantly altering the earth's climate (Figure 8.24).\n\n\nFIGURE 8.24 Greenhouse gases trap enough of the sun's energy to make the planet habitable-this is known as the greenhouse effect. Human activities are increasing greenhouse gas levels, warming the planet and causing more extreme weather events.\n\nThere is strong evidence from multiple sources that higher atmospheric levels of $\\mathrm{CO}_{2}$ are caused by human activity, with fossil fuel burning accounting for about $\\frac{3}{4}$ of the recent increase in $\\mathrm{CO}_{2}$. Reliable data from ice cores reveals that $\\mathrm{CO}_{2}$ concentration in the atmosphere is at the highest level in the past 800,000 years; other evidence indicates that it may be at its highest level in 20 million years. In recent years, the $\\mathrm{CO}_{2}$ concentration has increased from preindustrial levels of $\\sim 280 \\mathrm{ppm}$ to more than 400 ppm today (Figure 8.25).\n\nCarbon Dioxide in the Atmosphere"}
{"id": 3269, "contents": "883. Greenhouse Gases and Climate Change - \nCarbon Dioxide in the Atmosphere\n\n\nFIGURE $8.25 \\mathrm{CO}_{2}$ levels over the past 700,000 years were typically from 200-300 ppm, with a steep, unprecedented increase over the past 50 years."}
{"id": 3270, "contents": "884. LINK TO LEARNING - \nClick here (http://openstax.org/l/16GlobalWarming) to see a 2-minute video explaining greenhouse gases and global warming."}
{"id": 3271, "contents": "886. Susan Solomon - \nAtmospheric and climate scientist Susan Solomon (Figure 8.26) is the author of one of The New York Times books of the year (The Coldest March, 2001), one of Time magazine's 100 most influential people in the world (2008), and a working group leader of the Intergovernmental Panel on Climate Change (IPCC), which was the recipient of the 2007 Nobel Peace Prize. She helped determine and explain the cause of the formation of the ozone hole over Antarctica, and has authored many important papers on climate change.\n\nShe has been awarded the top scientific honors in the US and France (the National Medal of Science and the Grande Medaille, respectively), and is a member of the National Academy of Sciences, the Royal Society, the French Academy of Sciences, and the European Academy of Sciences. Formerly a professor at the University of Colorado, she is now at MIT, and continues to work at NOAA.\n\nFor more information, watch this video (http://openstax.org/l/16SusanSolomon) about Susan Solomon.\n\n\nFIGURE 8.26 Susan Solomon's research focuses on climate change and has been instrumental in determining the cause of the ozone hole over Antarctica. (credit: National Oceanic and Atmospheric Administration)"}
{"id": 3272, "contents": "887. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Define and explain effusion and diffusion\n- State Graham's law and use it to compute relevant gas properties\n\nIf you have ever been in a room when a piping hot pizza was delivered, you have been made aware of the fact that gaseous molecules can quickly spread throughout a room, as evidenced by the pleasant aroma that soon reaches your nose. Although gaseous molecules travel at tremendous speeds (hundreds of meters per second), they collide with other gaseous molecules and travel in many different directions before reaching the desired target. At room temperature, a gaseous molecule will experience billions of collisions per second. The mean free path is the average distance a molecule travels between collisions. The mean free path increases with decreasing pressure; in general, the mean free path for a gaseous molecule will be hundreds of times the diameter of the molecule\n\nIn general, we know that when a sample of gas is introduced to one part of a closed container, its molecules very quickly disperse throughout the container; this process by which molecules disperse in space in response to differences in concentration is called diffusion (shown in Figure 8.27). The gaseous atoms or molecules are, of course, unaware of any concentration gradient, they simply move randomly-regions of higher concentration have more particles than regions of lower concentrations, and so a net movement of species from high to low concentration areas takes place. In a closed environment, diffusion will ultimately result in equal concentrations of gas throughout, as depicted in Figure 8.27. The gaseous atoms and molecules continue to move, but since their concentrations are the same in both bulbs, the rates of transfer between the bulbs are equal (no net transfer of molecules occurs).\n\n\nFIGURE 8.27 (a) Two gases, $\\mathrm{H}_{2}$ and $\\mathrm{O}_{2}$, are initially separated. (b) When the stopcock is opened, they mix together."}
{"id": 3273, "contents": "887. LEARNING OBJECTIVES - \nFIGURE 8.27 (a) Two gases, $\\mathrm{H}_{2}$ and $\\mathrm{O}_{2}$, are initially separated. (b) When the stopcock is opened, they mix together.\n\nThe lighter gas, $\\mathrm{H}_{2}$, passes through the opening faster than $\\mathrm{O}_{2}$, so just after the stopcock is opened, more $\\mathrm{H}_{2}$ molecules move to the $\\mathrm{O}_{2}$ side than $\\mathrm{O}_{2}$ molecules move to the $\\mathrm{H}_{2}$ side. (c) After a short time, both the slowermoving $\\mathrm{O}_{2}$ molecules and the faster-moving $\\mathrm{H}_{2}$ molecules have distributed themselves evenly on both sides of the vessel.\n\nWe are often interested in the rate of diffusion, the amount of gas passing through some area per unit time:\n\n$$\n\\text { rate of diffusion }=\\frac{\\text { amount of gas passing through an area }}{\\text { unit of time }}\n$$\n\nThe diffusion rate depends on several factors: the concentration gradient (the increase or decrease in concentration from one point to another); the amount of surface area available for diffusion; and the distance the gas particles must travel. Note also that the time required for diffusion to occur is inversely proportional to the rate of diffusion, as shown in the rate of diffusion equation.\n\nA process involving movement of gaseous species similar to diffusion is effusion, the escape of gas molecules through a tiny hole such as a pinhole in a balloon into a vacuum (Figure 8.28). Although diffusion and effusion rates both depend on the molar mass of the gas involved, their rates are not equal; however, the ratios of their rates are the same.\n\n\nDiffusion\n\n\nEffusion\n\nFIGURE 8.28 Diffusion involves the unrestricted dispersal of molecules throughout space due to their random motion. When this process is restricted to passage of molecules through very small openings in a physical barrier, the process is called effusion."}
{"id": 3274, "contents": "887. LEARNING OBJECTIVES - \nFIGURE 8.28 Diffusion involves the unrestricted dispersal of molecules throughout space due to their random motion. When this process is restricted to passage of molecules through very small openings in a physical barrier, the process is called effusion.\n\nIf a mixture of gases is placed in a container with porous walls, the gases effuse through the small openings in the walls. The lighter gases pass through the small openings more rapidly (at a higher rate) than the heavier ones (Figure 8.29). In 1832, Thomas Graham studied the rates of effusion of different gases and formulated Graham's law of effusion: The rate of effusion of a gas is inversely proportional to the square root of the mass of its particles:\n\n$$\n\\text { rate of effusion } \\propto \\frac{1}{\\sqrt{\\mathcal{M}}}\n$$\n\nThis means that if two gases A and B are at the same temperature and pressure, the ratio of their effusion rates is inversely proportional to the ratio of the square roots of the masses of their particles:\n\n$$\n\\frac{\\text { rate of effusion of } A}{\\text { rate of effusion of } B}=\\frac{\\sqrt{\\mathcal{N}_{B}}}{\\sqrt{\\mathcal{N}_{A}}}\n$$\n\n\n\nFIGURE 8.29 The left photograph shows two balloons inflated with different gases, helium (orange) and argon (blue).The right-side photograph shows the balloons approximately 12 hours after being filled, at which time the helium balloon has become noticeably more deflated than the argon balloon, due to the greater effusion rate of the lighter helium gas. (credit: modification of work by Paul Flowers)"}
{"id": 3275, "contents": "889. Applying Graham's Law to Rates of Effusion - \nCalculate the ratio of the rate of effusion of hydrogen to the rate of effusion of oxygen."}
{"id": 3276, "contents": "890. Solution - \nFrom Graham's law, we have:\n\n$$\n\\frac{\\text { rate of effusion of hydrogen }}{\\text { rate of effusion of oxygen }}=\\frac{\\sqrt{32 \\mathrm{gmol}^{-1}}}{\\sqrt{2 \\mathrm{gmol}^{-1}}}=\\frac{\\sqrt{16}}{\\sqrt{1}}=\\frac{4}{1}\n$$\n\nHydrogen effuses four times as rapidly as oxygen."}
{"id": 3277, "contents": "891. Check Your Learning - \nAt a particular pressure and temperature, nitrogen gas effuses at the rate of $79 \\mathrm{~mL} / \\mathrm{s}$. Under the same conditions, at what rate will sulfur dioxide effuse?"}
{"id": 3278, "contents": "892. Answer: - \n$52 \\mathrm{~mL} / \\mathrm{s}$"}
{"id": 3279, "contents": "894. Effusion Time Calculations - \nIt takes 243 s for $4.46 \\times 10^{-5} \\mathrm{~mol}$ Xe to effuse through a tiny hole. Under the same conditions, how long will it take $4.46 \\times 10^{-5} \\mathrm{~mol} \\mathrm{Ne}$ to effuse?"}
{"id": 3280, "contents": "895. Solution - \nIt is important to resist the temptation to use the times directly, and to remember how rate relates to time as well as how it relates to mass. Recall the definition of rate of effusion:\n\n$$\n\\text { rate of effusion }=\\frac{\\text { amount of gas transferred }}{\\text { time }}\n$$\n\nand combine it with Graham's law:\n\n$$\n\\frac{\\text { rate of effusion of gas } \\mathrm{Xe}}{\\text { rate of effusion of gas Ne }}=\\frac{\\sqrt{\\mathcal{M}_{\\mathrm{Ne}}}}{\\sqrt{\\mathcal{M}_{\\mathrm{Xe}}}}\n$$\n\nTo get:\n\n$$\n\\frac{\\frac{\\text { amount of Xe transferred }}{\\text { time for Xe }}}{\\frac{\\text { amount of Ne transferred }}{\\text { time for Ne }}}=\\frac{\\sqrt{\\mathcal{M}_{\\mathrm{Ne}}}}{\\sqrt{\\mathcal{M}_{\\mathrm{Xe}}}}\n$$\n\nNoting that amount of $A=$ amount of $B$, and solving for time for $N e$ :\n\n$$\n\\frac{\\frac{\\text { amount of } \\mathrm{Xe}}{\\text { time for } \\mathrm{Xe}}}{\\frac{\\text { amount of Ne }}{\\text { time for } \\mathrm{Ne}}}=\\frac{\\text { time for } \\mathrm{Ne}}{\\text { time for } \\mathrm{Xe}}=\\frac{\\sqrt{\\mathcal{M}_{\\mathrm{Ne}}}}{\\sqrt{\\mathcal{M}_{\\mathrm{Xe}}}}=\\frac{\\sqrt{\\mathcal{M}_{\\mathrm{Ne}}}}{\\sqrt{\\mathcal{M}_{\\mathrm{Xe}}}}\n$$\n\nand substitute values:\n\n$$\n\\frac{\\text { time for } \\mathrm{Ne}}{243 \\mathrm{~s}}=\\sqrt{\\frac{20.2 \\mathrm{~g} \\mathrm{~mol}}{131.3 \\mathrm{~m} \\mathrm{~mol}}}=0.392\n$$\n\nFinally, solve for the desired quantity:\n\n$$\n\\text { time for } \\mathrm{Ne}=0.392 \\times 243 \\mathrm{~s}=95.3 \\mathrm{~s}\n$$"}
{"id": 3281, "contents": "895. Solution - \nFinally, solve for the desired quantity:\n\n$$\n\\text { time for } \\mathrm{Ne}=0.392 \\times 243 \\mathrm{~s}=95.3 \\mathrm{~s}\n$$\n\nNote that this answer is reasonable: Since Ne is lighter than Xe , the effusion rate for Ne will be larger than that for Xe , which means the time of effusion for Ne will be smaller than that for Xe ."}
{"id": 3282, "contents": "896. Check Your Learning - \nA party balloon filled with helium deflates to $\\frac{2}{3}$ of its original volume in 8.0 hours. How long will it take an identical balloon filled with the same number of moles of air ( $\\mathcal{M}=28.2 \\mathrm{~g} / \\mathrm{mol}$ ) to deflate to $\\frac{1}{2}$ of its original volume?"}
{"id": 3283, "contents": "897. Answer: - \n32 h"}
{"id": 3284, "contents": "899. Determining Molar Mass Using Graham's Law - \nAn unknown gas effuses 1.66 times more rapidly than $\\mathrm{CO}_{2}$. What is the molar mass of the unknown gas? Can you make a reasonable guess as to its identity?"}
{"id": 3285, "contents": "900. Solution - \nFrom Graham's law, we have:\n\n$$\n\\frac{\\text { rate of effusion of Unknown }}{\\text { rate of effusion of } \\mathrm{CO}_{2}}=\\frac{\\sqrt{\\mathcal{M}_{\\mathrm{CO}_{2}}}}{\\sqrt{\\mathcal{M}_{\\text {Unknown }}}}\n$$\n\nPlug in known data:\n\n$$\n\\frac{1.66}{1}=\\frac{\\sqrt{44.0 \\mathrm{~g} / \\mathrm{mol}}}{\\sqrt{\\mathcal{M}_{\\text {Unknown }}}}\n$$\n\nSolve:\n\n$$\n\\mathcal{M}_{\\text {Unknown }}=\\frac{44.0 \\mathrm{~g} / \\mathrm{mol}}{(1.66)^{2}}=16.0 \\mathrm{~g} / \\mathrm{mol}\n$$\n\nThe gas could well be $\\mathrm{CH}_{4}$, the only gas with this molar mass."}
{"id": 3286, "contents": "901. Check Your Learning - \nHydrogen gas effuses through a porous container 8.97-times faster than an unknown gas. Estimate the molar mass of the unknown gas."}
{"id": 3287, "contents": "902. Answer: - \n$163 \\mathrm{~g} / \\mathrm{mol}$"}
{"id": 3288, "contents": "904. Use of Diffusion for Nuclear Energy Applications: Uranium Enrichment - \nGaseous diffusion has been used to produce enriched uranium for use in nuclear power plants and weapons. Naturally occurring uranium contains only $0.72 \\%$ of ${ }^{235} \\mathrm{U}$, the kind of uranium that is \"fissile,\" that is, capable of sustaining a nuclear fission chain reaction. Nuclear reactors require fuel that is $2-5 \\%{ }^{235} \\mathrm{U}$, and nuclear bombs need even higher concentrations. One way to enrich uranium to the desired levels is to take advantage of Graham's law. In a gaseous diffusion enrichment plant, uranium hexafluoride ( $\\mathrm{UF}_{6}$, the only uranium compound that is volatile enough to work) is slowly pumped through large cylindrical vessels called diffusers, which contain porous barriers with microscopic openings. The process is one of diffusion because the other side of the barrier is not evacuated. The ${ }^{235} \\mathrm{UF}_{6}$ molecules have a higher average speed and diffuse through the barrier a little faster than the heavier ${ }^{238} \\mathrm{UF}_{6}$ molecules. The gas that has passed through the barrier is slightly enriched in ${ }^{235} \\mathrm{UF}_{6}$ and the residual gas is slightly depleted. The small difference in molecular weights between ${ }^{235} \\mathrm{UF}_{6}$ and ${ }^{238} \\mathrm{UF}_{6}$ only about $0.4 \\%$ enrichment, is achieved in one diffuser (Figure 8.30). But by connecting many diffusers in a sequence of stages (called a cascade), the desired level of enrichment can be attained.\n\n\nFIGURE 8.30 In a diffuser, gaseous $U F_{6}$ is pumped through a porous barrier, which partially separates ${ }^{235} \\mathrm{UF}_{6}$ from ${ }^{238} \\mathrm{UF}_{6}$ The $\\mathrm{UF}_{6}$ must pass through many large diffuser units to achieve sufficient enrichment in ${ }^{235} \\mathrm{U}$."}
{"id": 3289, "contents": "904. Use of Diffusion for Nuclear Energy Applications: Uranium Enrichment - \nThe large scale separation of gaseous ${ }^{235} \\mathrm{UF}_{6}$ from ${ }^{238} \\mathrm{UF}_{6}$ was first done during the World War II, at the atomic energy installation in Oak Ridge, Tennessee, as part of the Manhattan Project (the development of the first atomic bomb). Although the theory is simple, this required surmounting many daunting technical challenges to make it work in practice. The barrier must have tiny, uniform holes (about $10^{-6} \\mathrm{~cm}$ in diameter) and be porous enough to produce high flow rates. All materials (the barrier, tubing, surface coatings, lubricants, and gaskets) need to be able to contain, but not react with, the highly reactive and corrosive $\\mathrm{UF}_{6}$.\n\nBecause gaseous diffusion plants require very large amounts of energy (to compress the gas to the high\npressures required and drive it through the diffuser cascade, to remove the heat produced during compression, and so on), it is now being replaced by gas centrifuge technology, which requires far less energy. A current hot political issue is how to deny this technology to Iran, to prevent it from producing enough enriched uranium for them to use to make nuclear weapons."}
{"id": 3290, "contents": "904. Use of Diffusion for Nuclear Energy Applications: Uranium Enrichment - 904.1. The Kinetic-Molecular Theory\nLEARNING OBJECTIVES\n\n- State the postulates of the kinetic-molecular theory\n- Use this theory's postulates to explain the gas laws\n\nThe gas laws that we have seen to this point, as well as the ideal gas equation, are empirical, that is, they have been derived from experimental observations. The mathematical forms of these laws closely describe the macroscopic behavior of most gases at pressures less than about 1 or 2 atm . Although the gas laws describe relationships that have been verified by many experiments, they do not tell us why gases follow these relationships.\n\nThe kinetic molecular theory (KMT) is a simple microscopic model that effectively explains the gas laws described in previous modules of this chapter. This theory is based on the following five postulates described here. (Note: The term \"molecule\" will be used to refer to the individual chemical species that compose the gas, although some gases are composed of atomic species, for example, the noble gases.)\n\n1. Gases are composed of molecules that are in continuous motion, travelling in straight lines and changing direction only when they collide with other molecules or with the walls of a container.\n2. The molecules composing the gas are negligibly small compared to the distances between them.\n3. The pressure exerted by a gas in a container results from collisions between the gas molecules and the container walls.\n4. Gas molecules exert no attractive or repulsive forces on each other or the container walls; therefore, their collisions are elastic (do not involve a loss of energy).\n5. The average kinetic energy of the gas molecules is proportional to the kelvin temperature of the gas.\n\nThe test of the KMT and its postulates is its ability to explain and describe the behavior of a gas. The various gas laws can be derived from the assumptions of the KMT, which have led chemists to believe that the assumptions of the theory accurately represent the properties of gas molecules. We will first look at the individual gas laws (Boyle's, Charles's, Amontons's, Avogadro's, and Dalton's laws) conceptually to see how the KMT explains them. Then, we will more carefully consider the relationships between molecular masses, speeds, and kinetic energies with temperature, and explain Graham's law."}
{"id": 3291, "contents": "905. The Kinetic-Molecular Theory Explains the Behavior of Gases, Part I - \nRecalling that gas pressure is exerted by rapidly moving gas molecules and depends directly on the number of molecules hitting a unit area of the wall per unit of time, we see that the KMT conceptually explains the behavior of a gas as follows:\n\n- Amontons's law. If the temperature is increased, the average speed and kinetic energy of the gas molecules increase. If the volume is held constant, the increased speed of the gas molecules results in more frequent and more forceful collisions with the walls of the container, therefore increasing the pressure (Figure 8.31).\n- Charles's law. If the temperature of a gas is increased, a constant pressure may be maintained only if the volume occupied by the gas increases. This will result in greater average distances traveled by the molecules to reach the container walls, as well as increased wall surface area. These conditions will decrease the both the frequency of molecule-wall collisions and the number of collisions per unit area, the combined effects of which balance the effect of increased collision forces due to the greater kinetic energy at the higher temperature.\n- Boyle's law. If the gas volume volume of a given amount of gas at a given temperature is decreased (that is, if the gas is compressed), the molecules will be exposed to a decreased container wall area. Collisions with\nthe container wall will therefore occur more frequently and the pressure exerted by the gas will increase (Figure 8.31).\n- Avogadro's law. At constant pressure and temperature, the frequency and force of molecule-wall collisions are constant. Under such conditions, increasing the number of gaseous molecules will require a proportional increase in the container volume in order to yield a decrease in the number of collisions per unit area to compensate for the increased frequency of collisions (Figure 8.31).\n- Dalton's Law. Because of the large distances between them, the molecules of one gas in a mixture bombard the container walls with the same frequency whether other gases are present or not, and the total pressure of a gas mixture equals the sum of the (partial) pressures of the individual gases.\n\n\nFIGURE 8.31 (a) When gas temperature increases, gas pressure increases due to increased force and frequency of molecular collisions. (b) When volume decreases, gas pressure increases due to increased frequency of molecular collisions. (c) When the amount of gas increases at a constant pressure, volume increases to yield a constant number of collisions per unit wall area per unit time."}
{"id": 3292, "contents": "906. molecular speeds and Kinetic Energy - \nThe previous discussion showed that the KMT qualitatively explains the behaviors described by the various gas laws. The postulates of this theory may be applied in a more quantitative fashion to derive these individual laws. To do this, we must first look at speeds and kinetic energies of gas molecules, and the temperature of a gas sample.\n\nIn a gas sample, individual molecules have widely varying speeds; however, because of the vast number of molecules and collisions involved, the molecular speed distribution and average speed are constant. This molecular speed distribution is known as a Maxwell-Boltzmann distribution, and it depicts the relative numbers of molecules in a bulk sample of gas that possesses a given speed (Figure 8.32).\n\n\nFIGURE 8.32 The molecular speed distribution for oxygen gas at 300 K is shown here. Very few molecules move at either very low or very high speeds. The number of molecules with intermediate speeds increases rapidly up to a maximum, which is the most probable speed, then drops off rapidly. Note that the most probable speed, $v_{p}$, is a little less than $400 \\mathrm{~m} / \\mathrm{s}$, while the root mean square speed, $u_{\\mathrm{rms}}$, is closer to $500 \\mathrm{~m} / \\mathrm{s}$.\n\nThe kinetic energy (KE) of a particle of mass ( $m$ ) and speed $(u)$ is given by:\n\n$$\n\\mathrm{KE}=\\frac{1}{2} m u^{2}\n$$\n\nExpressing mass in kilograms and speed in meters per second will yield energy values in units of joules ( $\\mathrm{J}=\\mathrm{kg}$ $\\mathrm{m}^{2} \\mathrm{~s}^{-2}$ ). To deal with a large number of gas molecules, we use averages for both speed and kinetic energy. In the KMT, the root mean square speed of a particle, $\\mathbf{u}_{\\mathbf{r m s}}$, is defined as the square root of the average of the squares of the speeds with $n=$ the number of particles:"}
{"id": 3293, "contents": "906. molecular speeds and Kinetic Energy - \n$$\nu_{\\mathrm{rms}}=\\sqrt{\\overline{u^{2}}}=\\sqrt{\\frac{u_{1}^{2}+u_{2}^{2}+u_{3}^{2}+u_{4}^{2}+\\ldots}{n}}\n$$\n\nThe average kinetic energy for a mole of particles, $\\mathrm{KE}_{\\text {avg }}$, is then equal to:\n\n$$\n\\mathrm{KE}_{\\mathrm{avg}}=\\frac{1}{2} M u_{\\mathrm{rms}}^{2}\n$$\n\nwhere $M$ is the molar mass expressed in units of $\\mathrm{kg} / \\mathrm{mol}$. The $\\mathrm{KE}_{\\text {avg }}$ of a mole of gas molecules is also directly proportional to the temperature of the gas and may be described by the equation:\n\n$$\n\\mathrm{KE}_{\\mathrm{avg}}=\\frac{3}{2} R T\n$$\n\nwhere $R$ is the gas constant and T is the kelvin temperature. When used in this equation, the appropriate form of the gas constant is $8.314 \\mathrm{~J} / \\mathrm{mol} \\cdot \\mathrm{K}\\left(8.314 \\mathrm{~kg} \\mathrm{~m}^{2} \\mathrm{~s}^{-2} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}\\right)$. These two separate equations for $\\mathrm{KE}_{\\text {avg }}$ may be combined and rearranged to yield a relation between molecular speed and temperature:\n\n$$\n\\begin{gathered}\n\\frac{1}{2} M u_{\\mathrm{rms}}^{2}=\\frac{3}{2} R T \\\\\nu_{\\mathrm{rms}}=\\sqrt{\\frac{3 R T}{M}}\n\\end{gathered}\n$$"}
{"id": 3294, "contents": "908. Calculation of $u_{\\text {rms }}$ - \nCalculate the root-mean-square speed for a nitrogen molecule at $30^{\\circ} \\mathrm{C}$."}
{"id": 3295, "contents": "909. Solution - \nConvert the temperature into Kelvin:\n\n$$\n30^{\\circ} \\mathrm{C}+273=303 \\mathrm{~K}\n$$\n\nDetermine the molar mass of nitrogen in kilograms:\n\n$$\n\\frac{28.0 \\frac{\\mathrm{~g}}{\\mathrm{o}}}{1 \\mathrm{~mol}} \\times \\frac{1 \\mathrm{~kg}}{1000 \\frac{\\mathrm{~g}}{\\mathrm{o}}}=0.028 \\mathrm{~kg} / \\mathrm{mol}\n$$\n\nReplace the variables and constants in the root-mean-square speed equation, replacing Joules with the equivalent $\\mathrm{kg} \\mathrm{m}{ }^{2} \\mathrm{~s}^{-2}$ :\n\n$$\n\\begin{gathered}\nu_{\\mathrm{rms}}=\\sqrt{\\frac{3 R T}{M}} \\\\\nu_{\\mathrm{rms}}=\\sqrt{\\frac{3(8.314 \\mathrm{~J} / \\mathrm{mol} \\mathrm{~K})(303 \\mathrm{~K})}{(0.028 \\mathrm{~kg} / \\mathrm{mol})}}=\\sqrt{2.70 \\times 10^{5} \\mathrm{~m}^{2} \\mathrm{~s}^{-2}}=519 \\mathrm{~m} / \\mathrm{s}\n\\end{gathered}\n$$"}
{"id": 3296, "contents": "910. Check Your Learning - \nCalculate the root-mean-square speed for a mole of oxygen molecules at $-23^{\\circ} \\mathrm{C}$."}
{"id": 3297, "contents": "911. Answer: - \n441 m/s\n\nIf the temperature of a gas increases, its $\\mathrm{KE}_{\\text {avg }}$ increases, more molecules have higher speeds and fewer molecules have lower speeds, and the distribution shifts toward higher speeds overall, that is, to the right. If temperature decreases, $\\mathrm{KE}_{\\text {avg }}$ decreases, more molecules have lower speeds and fewer molecules have higher speeds, and the distribution shifts toward lower speeds overall, that is, to the left. This behavior is illustrated for nitrogen gas in Figure 8.33.\n\n\nFIGURE 8.33 The molecular speed distribution for nitrogen gas $\\left(N_{2}\\right)$ shifts to the right and flattens as the temperature increases; it shifts to the left and heightens as the temperature decreases.\n\nAt a given temperature, all gases have the same $\\mathrm{KE}_{\\text {avg }}$ for their molecules. Gases composed of lighter molecules have more high-speed particles and a higher $u_{r m s}$, with a speed distribution that peaks at relatively higher speeds. Gases consisting of heavier molecules have more low-speed particles, a lower $u_{r m s}$, and a speed distribution that peaks at relatively lower speeds. This trend is demonstrated by the data for a series of noble gases shown in Figure 8.34.\n\n\nFIGURE 8.34 molecular speed is directly related to molecular mass. At a given temperature, lighter molecules move faster on average than heavier molecules."}
{"id": 3298, "contents": "912. LINK TO LEARNING - \nThe gas simulator (http://openstax.org/l/16MolecVelocity) may be used to examine the effect of temperature on molecular speeds. Examine the simulator's \"energy histograms\" (molecular speed distributions) and \"species information\" (which gives average speed values) for molecules of different masses at various temperatures.\n\nThe Kinetic-Molecular Theory Explains the Behavior of Gases, Part II\nAccording to Graham's law, the molecules of a gas are in rapid motion and the molecules themselves are small. The average distance between the molecules of a gas is large compared to the size of the molecules. As a consequence, gas molecules can move past each other easily and diffuse at relatively fast rates.\n\nThe rate of effusion of a gas depends directly on the (average) speed of its molecules:\neffusion rate $\\propto u_{\\text {rms }}$\nUsing this relation, and the equation relating molecular speed to mass, Graham's law may be easily derived as shown here:\n\n$$\n\\begin{gathered}\nu_{\\mathrm{rms}}=\\sqrt{\\frac{3 R T}{M}} \\\\\nM=\\frac{3 R T}{u_{\\mathrm{rms}}^{2}}=\\frac{3 R T}{\\bar{u}^{2}} \\\\\n\\frac{\\text { effusion rate } \\mathrm{A}}{\\text { effusion rate } \\mathrm{B}}=\\frac{u_{\\mathrm{rms} \\mathrm{~A}}}{u_{\\mathrm{rms} \\mathrm{~B}}}=\\frac{\\sqrt{\\frac{3 R T}{M_{\\mathrm{A}}}}}{\\sqrt{\\frac{3 R T}{M_{\\mathrm{B}}}}}=\\sqrt{\\frac{M_{\\mathrm{B}}}{M_{\\mathrm{A}}}}\n\\end{gathered}\n$$\n\nThe ratio of the rates of effusion is thus derived to be inversely proportional to the ratio of the square roots of their masses. This is the same relation observed experimentally and expressed as Graham's law."}
{"id": 3299, "contents": "913. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the physical factors that lead to deviations from ideal gas behavior\n- Explain how these factors are represented in the van der Waals equation\n- Define compressibility (Z) and describe how its variation with pressure reflects non-ideal behavior\n- Quantify non-ideal behavior by comparing computations of gas properties using the ideal gas law and the van der Waals equation\n\nThus far, the ideal gas law, $P V=n R T$, has been applied to a variety of different types of problems, ranging from\nreaction stoichiometry and empirical and molecular formula problems to determining the density and molar mass of a gas. As mentioned in the previous modules of this chapter, however, the behavior of a gas is often non-ideal, meaning that the observed relationships between its pressure, volume, and temperature are not accurately described by the gas laws. In this section, the reasons for these deviations from ideal gas behavior are considered.\n\nOne way in which the accuracy of $P V=n R T$ can be judged is by comparing the actual volume of 1 mole of gas (its molar volume, $V_{\\mathrm{m}}$ ) to the molar volume of an ideal gas at the same temperature and pressure. This ratio is called the compressibility factor ( $\\mathbf{Z}$ ) with:\n\n$$\n\\mathrm{Z}=\\frac{\\text { molar volume of gas at same } T \\text { and } P}{\\text { molar volume of ideal gas at same } T \\text { and } P}=\\left(\\frac{P V_{m}}{R T}\\right)_{\\text {measured }}\n$$\n\nIdeal gas behavior is therefore indicated when this ratio is equal to 1 , and any deviation from 1 is an indication of non-ideal behavior. Figure 8.35 shows plots of Z over a large pressure range for several common gases.\n\n\nFIGURE 8.35 A graph of the compressibility factor $(Z)$ vs. pressure shows that gases can exhibit significant deviations from the behavior predicted by the ideal gas law.\n\nAs is apparent from Figure 8.35, the ideal gas law does not describe gas behavior well at relatively high pressures. To determine why this is, consider the differences between real gas properties and what is expected of a hypothetical ideal gas."}
{"id": 3300, "contents": "913. LEARNING OBJECTIVES - \nAs is apparent from Figure 8.35, the ideal gas law does not describe gas behavior well at relatively high pressures. To determine why this is, consider the differences between real gas properties and what is expected of a hypothetical ideal gas.\n\nParticles of a hypothetical ideal gas have no significant volume and do not attract or repel each other. In general, real gases approximate this behavior at relatively low pressures and high temperatures. However, at high pressures, the molecules of a gas are crowded closer together, and the amount of empty space between the molecules is reduced. At these higher pressures, the volume of the gas molecules themselves becomes appreciable relative to the total volume occupied by the gas. The gas therefore becomes less compressible at these high pressures, and although its volume continues to decrease with increasing pressure, this decrease is not proportional as predicted by Boyle's law.\n\nAt relatively low pressures, gas molecules have practically no attraction for one another because they are (on average) so far apart, and they behave almost like particles of an ideal gas. At higher pressures, however, the force of attraction is also no longer insignificant. This force pulls the molecules a little closer together, slightly decreasing the pressure (if the volume is constant) or decreasing the volume (at constant pressure) (Figure 8.36). This change is more pronounced at low temperatures because the molecules have lower KE relative to the attractive forces, and so they are less effective in overcoming these attractions after colliding with one another.\n\n\nFIGURE 8.36 (a) Attractions between gas molecules serve to decrease the gas volume at constant pressure compared to an ideal gas whose molecules experience no attractive forces. (b) These attractive forces will decrease the force of collisions between the molecules and container walls, therefore reducing the pressure exerted at constant volume compared to an ideal gas.\n\nThere are several different equations that better approximate gas behavior than does the ideal gas law. The first, and simplest, of these was developed by the Dutch scientist Johannes van der Waals in 1879. The van der Waals equation improves upon the ideal gas law by adding two terms: one to account for the volume of the gas molecules and another for the attractive forces between them.\n\n$$\nP V=n R T \\longrightarrow\\left(P+\\frac{a n^{2}}{V^{2}}\\right)(V-n b)=n R T\n$$"}
{"id": 3301, "contents": "913. LEARNING OBJECTIVES - \n$$\nP V=n R T \\longrightarrow\\left(P+\\frac{a n^{2}}{V^{2}}\\right)(V-n b)=n R T\n$$\n\nThe constant a corresponds to the strength of the attraction between molecules of a particular gas, and the constant $b$ corresponds to the size of the molecules of a particular gas. The \"correction\" to the pressure term in the ideal gas law is $\\frac{n^{2} a}{V^{2}}$, and the \"correction\" to the volume is $n b$. Note that when $V$ is relatively large and $n$ is relatively small, both of these correction terms become negligible, and the van der Waals equation reduces to the ideal gas law, $P V=n R T$. Such a condition corresponds to a gas in which a relatively low number of molecules is occupying a relatively large volume, that is, a gas at a relatively low pressure. Experimental values for the van der Waals constants of some common gases are given in Table 8.3.\n\nValues of van der Waals Constants for Some Common Gases\n\n| Gas | $\\boldsymbol{a}\\left(\\mathrm{L}^{2} \\mathrm{~atm} / \\mathrm{mol}^{2}\\right)$ | $\\boldsymbol{b}(\\mathrm{L} / \\mathrm{mol})$ |\n| :--- | :--- | :--- |\n| $\\mathrm{N}_{2}$ | 1.39 | 0.0391 |\n| $\\mathrm{O}_{2}$ | 1.36 | 0.0318 |\n| $\\mathrm{CO}_{2}$ | 3.59 | 0.0427 |\n| $\\mathrm{H}_{2} \\mathrm{O}$ | 5.46 | 0.0305 |\n| He | 0.0342 | 0.0237 |\n| $\\mathrm{CCl}_{4}$ | 20.4 | 0.1383 |\n\nTABLE 8.3"}
{"id": 3302, "contents": "913. LEARNING OBJECTIVES - \nTABLE 8.3\n\nAt low pressures, the correction for intermolecular attraction, $a$, is more important than the one for molecular volume, $b$. At high pressures and small volumes, the correction for the volume of the molecules becomes important because the molecules themselves are incompressible and constitute an appreciable fraction of the total volume. At some intermediate pressure, the two corrections have opposing influences and the gas appears to follow the relationship given by $P V=n R T$ over a small range of pressures. This behavior is reflected by the \"dips\" in several of the compressibility curves shown in Figure 8.35. The attractive force between molecules initially makes the gas more compressible than an ideal gas, as pressure is raised ( Z decreases with increasing $P$ ). At very high pressures, the gas becomes less compressible ( Z increases with $P$ ), as the gas molecules begin to occupy an increasingly significant fraction of the total gas volume.\n\nStrictly speaking, the ideal gas equation functions well when intermolecular attractions between gas molecules are negligible and the gas molecules themselves do not occupy an appreciable part of the whole volume. These criteria are satisfied under conditions of low pressure and high temperature. Under such conditions, the gas is said to behave ideally, and deviations from the gas laws are small enough that they may be disregarded-this is, however, very often not the case."}
{"id": 3303, "contents": "915. Comparison of Ideal Gas Law and van der Waals Equation - \nA 4.25-L flask contains $3.46 \\mathrm{~mol} \\mathrm{CO}_{2}$ at $229^{\\circ} \\mathrm{C}$. Calculate the pressure of this sample of $\\mathrm{CO}_{2}$ :\n(a) from the ideal gas law\n(b) from the van der Waals equation\n(c) Explain the reason(s) for the difference."}
{"id": 3304, "contents": "916. Solution - \n(a) From the ideal gas law:\n\n$$\nP=\\frac{n R T}{V}=\\frac{3.46 \\mathrm{mot} \\times 0.08206 \\mathrm{t} \\mathrm{~atm} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1} \\times 502 \\mathrm{~K}}{4.25 \\mathrm{t}}=33.5 \\mathrm{~atm}\n$$\n\n(b) From the van der Waals equation:\n\n$$\n\\begin{gathered}\n\\left(P+\\frac{n^{2} a}{V^{2}}\\right) \\times(V-n b)=n R T \\rightarrow P=\\frac{n R T}{(V-n b)}-\\frac{n^{2} a}{V^{2}} \\\\\nP=\\frac{3.46 \\mathrm{~mol} \\times 0.08206 \\mathrm{~L} \\mathrm{~atm} \\mathrm{~mol}}{}=1 \\mathrm{~K}^{-1} \\times 502 \\mathrm{~K} \\\\\n\\left(4.25 \\mathrm{~L}-3.46 \\mathrm{~mol} \\times 0.0427 \\mathrm{~L} \\mathrm{~mol}^{-1}\\right)\n\\end{gathered} \\frac{(3.46 \\mathrm{~mol})^{2} \\times 3.59 \\mathrm{~L}^{2} \\mathrm{~atm} \\mathrm{~mol}^{2}}{(4.25 \\mathrm{~L})^{2}} .\n$$\n\nThis finally yields $P=32.4 \\mathrm{~atm}$.\n(c) This is not very different from the value from the ideal gas law because the pressure is not very high and the temperature is not very low. The value is somewhat different because $\\mathrm{CO}_{2}$ molecules do have some volume and attractions between molecules, and the ideal gas law assumes they do not have volume or attractions."}
{"id": 3305, "contents": "917. Check your Learning - \nA $560-\\mathrm{mL}$ flask contains $21.3 \\mathrm{~g} \\mathrm{~N} \\mathrm{~N}_{2}$ at $145^{\\circ} \\mathrm{C}$. Calculate the pressure of $\\mathrm{N}_{2}$ :\n(a) from the ideal gas law\n(b) from the van der Waals equation\n(c) Explain the reason(s) for the difference."}
{"id": 3306, "contents": "918. Answer: - \n(a) 46.562 atm ; (b) 46.594 atm ; (c) The van der Waals equation takes into account the volume of the gas molecules themselves as well as intermolecular attractions."}
{"id": 3307, "contents": "919. Key Terms - \nabsolute zero temperature at which the volume of a gas would be zero according to Charles's law.\nAmontons's law (also, Gay-Lussac's law) pressure of a given number of moles of gas is directly proportional to its kelvin temperature when the volume is held constant\natmosphere (atm) unit of pressure; 1 atm = $101,325 \\mathrm{~Pa}$\nAvogadro's law volume of a gas at constant temperature and pressure is proportional to the number of gas molecules\nbar (bar or b) unit of pressure; $1 \\mathrm{bar}=100,000 \\mathrm{~Pa}$\nbarometer device used to measure atmospheric pressure\nBoyle's law volume of a given number of moles of gas held at constant temperature is inversely proportional to the pressure under which it is measured\nCharles's law volume of a given number of moles of gas is directly proportional to its kelvin temperature when the pressure is held constant\ncompressibility factor ( $Z$ ) ratio of the experimentally measured molar volume for a gas to its molar volume as computed from the ideal gas equation\nDalton's law of partial pressures total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases.\ndiffusion movement of an atom or molecule from a region of relatively high concentration to one of relatively low concentration (discussed in this chapter with regard to gaseous species, but applicable to species in any phase)\neffusion transfer of gaseous atoms or molecules from a container to a vacuum through very small openings\nGraham's law of effusion rates of diffusion and effusion of gases are inversely proportional to the square roots of their molecular masses\nhydrostatic pressure pressure exerted by a fluid due to gravity\nideal gas hypothetical gas whose physical properties are perfectly described by the gas laws\nideal gas constant ( $\\boldsymbol{R}$ ) constant derived from the\n$8.314 \\mathrm{~L} \\mathrm{kPa} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$\nideal gas law relation between the pressure, volume, amount, and temperature of a gas under conditions derived by combination of the simple gas laws\nkinetic molecular theory theory based on simple principles and assumptions that effectively explains ideal gas behavior\nmanometer device used to measure the pressure of a gas trapped in a container"}
{"id": 3308, "contents": "919. Key Terms - \nkinetic molecular theory theory based on simple principles and assumptions that effectively explains ideal gas behavior\nmanometer device used to measure the pressure of a gas trapped in a container\nmean free path average distance a molecule travels between collisions\nmole fraction ( $\\boldsymbol{X}$ ) concentration unit defined as the ratio of the molar amount of a mixture component to the total number of moles of all mixture components\npartial pressure pressure exerted by an individual gas in a mixture\npascal (Pa) SI unit of pressure; $1 \\mathrm{~Pa}=1 \\mathrm{~N} / \\mathrm{m}^{2}$\npounds per square inch (psi) unit of pressure common in the US\npressure force exerted per unit area\nrate of diffusion amount of gas diffusing through a given area over a given time\nroot mean square speed ( $\\boldsymbol{u}_{\\text {rms }}$ ) measure of average speed for a group of particles calculated as the square root of the average squared speed\nstandard conditions of temperature and pressure (STP) $273.15 \\mathrm{~K}\\left(0^{\\circ} \\mathrm{C}\\right)$ and $1 \\mathrm{~atm}(101.325 \\mathrm{kPa})$\nstandard molar volume volume of 1 mole of gas at STP, approximately 22.4 L for gases behaving ideally\ntorr unit of pressure; 1 torr $=\\frac{1}{760} \\mathrm{~atm}$\nvan der Waals equation modified version of the ideal gas equation containing additional terms to account for non-ideal gas behavior\nvapor pressure of water pressure exerted by water vapor in equilibrium with liquid water in a closed container at a specific temperature"}
{"id": 3309, "contents": "920. Key Equations - \n$$\nP=\\frac{F}{A}\n$$\n\n$p=h \\rho g$\n$P V=n R T$\n$P_{\\text {Total }}=P_{A}+P_{B}+P_{C}+\\ldots=\\Sigma_{\\mathrm{i}} P_{\\mathrm{i}}$\n$P_{A}=X_{A} P_{\\text {Total }}$\n$X_{A}=\\frac{n_{A}}{n_{\\text {Total }}}$\nrate of diffusion $=\\frac{\\text { amount of gas passing through an area }}{\\text { unit of time }}$\n$\\frac{\\text { rate of effusion of gas } \\mathrm{A}}{\\text { rate of effusion of gas } \\mathrm{B}}=\\frac{\\sqrt{m_{B}}}{\\sqrt{m_{A}}}=\\frac{\\sqrt{\\mathcal{N}_{B}}}{\\sqrt{\\mathcal{N}_{A}}}$\n$u_{\\mathrm{rms}}=\\sqrt{\\overline{u^{2}}}=\\sqrt{\\frac{u_{1}^{2}+u_{2}^{2}+u_{3}^{2}+u_{4}^{2}+\\ldots}{n}}$\n$\\mathrm{KE}_{\\mathrm{avg}}=\\frac{3}{2} R T$\n$u_{\\mathrm{rms}}=\\sqrt{\\frac{3 R T}{M}}$\n$\\mathrm{Z}=\\frac{\\text { molar volume of gas at same } T \\text { and } P}{\\text { molar volume of ideal gas at same } T \\text { and } P}=\\left(\\frac{P \\times V_{m}}{R \\times T}\\right)_{\\text {measured }}$\n$\\left(P+\\frac{n^{2} a}{V^{2}}\\right) \\times(V-n b)=n R T$"}
{"id": 3310, "contents": "921. Summary - 921.1. Gas Pressure\nGases exert pressure, which is force per unit area. The pressure of a gas may be expressed in the SI unit of pascal or kilopascal, as well as in many other units including torr, atmosphere, and bar. Atmospheric pressure is measured using a barometer; other gas pressures can be measured using one of several types of manometers."}
{"id": 3311, "contents": "921. Summary - 921.2. Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law\nThe behavior of gases can be described by several laws based on experimental observations of their properties. The pressure of a given amount of gas is directly proportional to its absolute temperature, provided that the volume does not change (Amontons's law). The volume of a given gas sample is directly proportional to its absolute temperature at constant pressure (Charles's law). The volume of a given amount of gas is inversely proportional to its pressure when temperature is held constant (Boyle's law). Under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules (Avogadro's law).\n\nThe equations describing these laws are special cases of the ideal gas law, $P V=n R T$, where $P$ is the pressure of the gas, $V$ is its volume, $n$ is the number of moles of the gas, $T$ is its kelvin temperature, and $R$ is the ideal (universal) gas constant."}
{"id": 3312, "contents": "921. Summary - 921.3. Stoichiometry of Gaseous Substances, Mixtures, and Reactions\nThe ideal gas law can be used to derive a number of convenient equations relating directly measured\nquantities to properties of interest for gaseous substances and mixtures. Appropriate rearrangement of the ideal gas equation may be made to permit the calculation of gas densities and molar masses. Dalton's law of partial pressures may be used to relate measured gas pressures for gaseous mixtures to their compositions. Avogadro's law may be used in stoichiometric computations for chemical reactions involving gaseous reactants or products."}
{"id": 3313, "contents": "921. Summary - 921.4. Effusion and Diffusion of Gases\nGaseous atoms and molecules move freely and randomly through space. Diffusion is the process whereby gaseous atoms and molecules are transferred from regions of relatively high concentration to regions of relatively low concentration. Effusion is a similar process in which gaseous species pass from a container to a vacuum through very small orifices. The rates of effusion of gases are inversely proportional to the square roots of their densities or to the square roots of their atoms/molecules' masses (Graham's law)."}
{"id": 3314, "contents": "921. Summary - 921.5. The Kinetic-Molecular Theory\nThe kinetic molecular theory is a simple but very effective model that effectively explains ideal gas behavior. The theory assumes that gases consist of widely separated molecules of negligible volume that are in constant motion, colliding elastically with one another and the walls of their container with average speeds determined by their absolute temperatures. The individual molecules of a gas exhibit a range of speeds, the distribution of these speeds being dependent on the temperature of the\ngas and the mass of its molecules."}
{"id": 3315, "contents": "921. Summary - 921.6. Non-Ideal Gas Behavior\nGas molecules possess a finite volume and experience forces of attraction for one another. Consequently, gas behavior is not necessarily described well by the ideal gas law. Under conditions of low pressure and high temperature, these factors are negligible, the ideal gas equation is an accurate\ndescription of gas behavior, and the gas is said to exhibit ideal behavior. However, at lower temperatures and higher pressures, corrections for molecular volume and molecular attractions are required to account for finite molecular size and attractive forces. The van der Waals equation is a modified version of the ideal gas law that can be used to account for the non-ideal behavior of gases under these conditions."}
{"id": 3316, "contents": "922. Exercises - 922.1. Gas Pressure\n1. Why are sharp knives more effective than dull knives? (Hint: Think about the definition of pressure.)\n2. Why do some small bridges have weight limits that depend on how many wheels or axles the crossing vehicle has?\n3. Why should you roll or belly-crawl rather than walk across a thinly-frozen pond?\n4. A typical barometric pressure in Redding, California, is about 750 mm Hg . Calculate this pressure in atm and kPa.\n5. A typical barometric pressure in Denver, Colorado, is 615 mm Hg . What is this pressure in atmospheres and kilopascals?\n6. A typical barometric pressure in Kansas City is 740 torr. What is this pressure in atmospheres, in millimeters of mercury, and in kilopascals?\n7. Canadian tire pressure gauges are marked in units of kilopascals. What reading on such a gauge corresponds to 32 psi?\n8. During the Viking landings on Mars, the atmospheric pressure was determined to be on the average about 6.50 millibars ( $1 \\mathrm{bar}=0.987 \\mathrm{~atm}$ ). What is that pressure in torr and kPa ?\n9. The pressure of the atmosphere on the surface of the planet Venus is about 88.8 atm. Compare that pressure in psi to the normal pressure on earth at sea level in psi.\n10. A medical laboratory catalog describes the pressure in a cylinder of a gas as 14.82 MPa . What is the pressure of this gas in atmospheres and torr?\n11. Consider this scenario and answer the following questions: On a mid-August day in the northeastern United States, the following information appeared in the local newspaper: atmospheric pressure at sea level 29.97 in. Hg, 1013.9 mbar.\n(a) What was the pressure in kPa ?\n(b) The pressure near the seacoast in the northeastern United States is usually reported near $30.0 \\mathrm{in} . \\mathrm{Hg}$. During a hurricane, the pressure may fall to near 28.0 in . Hg. Calculate the drop in pressure in torr.\n12. Why is it necessary to use a nonvolatile liquid in a barometer or manometer?"}
{"id": 3317, "contents": "922. Exercises - 922.1. Gas Pressure\n12. Why is it necessary to use a nonvolatile liquid in a barometer or manometer?\n13. The pressure of a sample of gas is measured at sea level with a closed-end manometer. The liquid in the manometer is mercury. Determine the pressure of the gas in:\n(a) torr\n(b) Pa\n(c) bar"}
{"id": 3318, "contents": "922. Exercises - 922.1. Gas Pressure\n14. The pressure of a sample of gas is measured with an open-end manometer, partially shown to the right. The liquid in the manometer is mercury. Assuming atmospheric pressure is 29.92 in . Hg, determine the pressure of the gas in:\n(a) torr\n(b) Pa\n(c) bar\n\n15. The pressure of a sample of gas is measured at sea level with an open-end mercury manometer. Assuming atmospheric pressure is 760.0 mm Hg , determine the pressure of the gas in:\n(a) mm Hg\n(b) atm\n(c) kPa\n\n16. The pressure of a sample of gas is measured at sea level with an open-end mercury manometer. Assuming atmospheric pressure is 760 mm Hg , determine the pressure of the gas in:\n(a) mm Hg\n(b) atm\n(c) kPa\n\n17. How would the use of a volatile liquid affect the measurement of a gas using open-ended manometers vs. closed-end manometers?"}
{"id": 3319, "contents": "922. Exercises - 922.2. Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law\n18. Sometimes leaving a bicycle in the sun on a hot day will cause a blowout. Why?\n19. Explain how the volume of the bubbles exhausted by a scuba diver (Figure 8.16) change as they rise to the surface, assuming that they remain intact.\n20. One way to state Boyle's law is \"All other things being equal, the pressure of a gas is inversely proportional to its volume.\" (a) What is the meaning of the term \"inversely proportional?\" (b) What are the \"other things\" that must be equal?\n21. An alternate way to state Avogadro's law is \"All other things being equal, the number of molecules in a gas is directly proportional to the volume of the gas.\" (a) What is the meaning of the term \"directly proportional?\" (b) What are the \"other things\" that must be equal?\n22. How would the graph in Figure 8.12 change if the number of moles of gas in the sample used to determine the curve were doubled?\n23. How would the graph in Figure 8.13 change if the number of moles of gas in the sample used to determine the curve were doubled?\n24. In addition to the data found in Figure 8.13, what other information do we need to find the mass of the sample of air used to determine the graph?\n25. Determine the volume of 1 mol of $\\mathrm{CH}_{4}$ gas at 150 K and 1 atm, using Figure 8.12.\n26. Determine the pressure of the gas in the syringe shown in Figure 8.13 when its volume is 12.5 mL , using:\n(a) the appropriate graph\n(b) Boyle's law\n27. A spray can is used until it is empty except for the propellant gas, which has a pressure of 1344 torr at 23 ${ }^{\\circ} \\mathrm{C}$. If the can is thrown into a fire $\\left(\\mathrm{T}=475^{\\circ} \\mathrm{C}\\right)$, what will be the pressure in the hot can?"}
{"id": 3320, "contents": "922. Exercises - 922.2. Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law\n28. What is the temperature of an 11.2 - L sample of carbon monoxide, CO , at 744 torr if it occupies 13.3 L at 55 ${ }^{\\circ} \\mathrm{C}$ and 744 torr?\n29. A $2.50-\\mathrm{L}$ volume of hydrogen measured at $-196^{\\circ} \\mathrm{C}$ is warmed to $100^{\\circ} \\mathrm{C}$. Calculate the volume of the gas at the higher temperature, assuming no change in pressure.\n30. A balloon inflated with three breaths of air has a volume of 1.7 L . At the same temperature and pressure, what is the volume of the balloon if five more same-sized breaths are added to the balloon?\n31. A weather balloon contains 8.80 moles of helium at a pressure of 0.992 atm and a temperature of $25^{\\circ} \\mathrm{C}$ at ground level. What is the volume of the balloon under these conditions?"}
{"id": 3321, "contents": "922. Exercises - 922.2. Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law\n32. The volume of an automobile air bag was 66.8 L when inflated at $25^{\\circ} \\mathrm{C}$ with 77.8 g of nitrogen gas. What was the pressure in the bag in kPa ?\n33. How many moles of gaseous boron trifluoride, $\\mathrm{BF}_{3}$, are contained in a $4.3410-\\mathrm{L}$ bulb at 788.0 K if the pressure is 1.220 atm ? How many grams of $\\mathrm{BF}_{3}$ ?\n34. Iodine, $\\mathrm{I}_{2}$, is a solid at room temperature but sublimes (converts from a solid into a gas) when warmed. What is the temperature in a $73.3-\\mathrm{mL}$ bulb that contains 0.292 g of $\\mathrm{I}_{2}$ vapor at a pressure of 0.462 atm ?\n35. How many grams of gas are present in each of the following cases?\n(a) 0.100 L of $\\mathrm{CO}_{2}$ at 307 torr and $26^{\\circ} \\mathrm{C}$\n(b) $8.75 \\mathrm{~L}^{\\text {of }} \\mathrm{C}_{2} \\mathrm{H}_{4}$, at 378.3 kPa and 483 K\n(c) 221 mL of Ar at 0.23 torr and $-54^{\\circ} \\mathrm{C}$\n36. A high altitude balloon is filled with $1.41 \\times 10^{4} \\mathrm{~L}$ of hydrogen at a temperature of $21^{\\circ} \\mathrm{C}$ and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km , where the temperature is $-48^{\\circ} \\mathrm{C}$ and the pressure is 63.1 torr?"}
{"id": 3322, "contents": "922. Exercises - 922.2. Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law\n37. A cylinder of medical oxygen has a volume of 35.4 L , and contains $\\mathrm{O}_{2}$ at a pressure of 151 atm and a temperature of $25^{\\circ} \\mathrm{C}$. What volume of $\\mathrm{O}_{2}$ does this correspond to at normal body conditions, that is, 1 atm and $37^{\\circ} \\mathrm{C}$ ?\n38. A large scuba tank (Figure 8.16) with a volume of 18 L is rated for a pressure of 220 bar. The tank is filled at $20^{\\circ} \\mathrm{C}$ and contains enough air to supply 1860 L of air to a diver at a pressure of 2.37 atm (a depth of 45 feet). Was the tank filled to capacity at $20^{\\circ} \\mathrm{C}$ ?\n39. A 20.0-L cylinder containing 11.34 kg of butane, $\\mathrm{C}_{4} \\mathrm{H}_{10}$, was opened to the atmosphere. Calculate the mass of the gas remaining in the cylinder if it were opened and the gas escaped until the pressure in the cylinder was equal to the atmospheric pressure, 0.983 atm , and a temperature of $27^{\\circ} \\mathrm{C}$.\n40. While resting, the average $70-\\mathrm{kg}$ human male consumes 14 L of pure $\\mathrm{O}_{2}$ per hour at $25^{\\circ} \\mathrm{C}$ and 100 kPa . How many moles of $\\mathrm{O}_{2}$ are consumed by a 70 kg man while resting for 1.0 h ?\n41. For a given amount of gas showing ideal behavior, draw labeled graphs of:\n(a) the variation of $P$ with $V$\n(b) the variation of $V$ with $T$\n(c) the variation of $P$ with $T$\n(d) the variation of $\\frac{1}{P}$ with $V$\n42. A liter of methane gas, $\\mathrm{CH}_{4}$, at STP contains more atoms of hydrogen than does a liter of pure hydrogen gas, $\\mathrm{H}_{2}$, at STP. Using Avogadro's law as a starting point, explain why."}
{"id": 3323, "contents": "922. Exercises - 922.2. Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law\n43. The effect of chlorofluorocarbons (such as $\\mathrm{CCl}_{2} \\mathrm{~F}_{2}$ ) on the depletion of the ozone layer is well known. The use of substitutes, such as $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{~F}(g)$, for the chlorofluorocarbons, has largely corrected the problem. Calculate the volume occupied by 10.0 g of each of these compounds at STP:\n(a) $\\mathrm{CCl}_{2} \\mathrm{~F}_{2}(g)$\n(b) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{~F}(g)$\n44. As 1 g of the radioactive element radium decays over 1 year, it produces $1.16 \\times 10^{18}$ alpha particles (helium nuclei). Each alpha particle becomes an atom of helium gas. What is the pressure in pascal of the helium gas produced if it occupies a volume of 125 mL at a temperature of $25^{\\circ} \\mathrm{C}$ ?\n45. A balloon with a volume of 100.21 L at $21^{\\circ} \\mathrm{C}$ and 0.981 atm is released and just barely clears the top of Mount Crumpit in British Columbia. If the final volume of the balloon is 144.53 L at a temperature of 5.24 ${ }^{\\circ} \\mathrm{C}$, what is the pressure experienced by the balloon as it clears Mount Crumpet?\n46. If the temperature of a fixed amount of a gas is doubled at constant volume, what happens to the pressure?\n47. If the volume of a fixed amount of a gas is tripled at constant temperature, what happens to the pressure?"}
{"id": 3324, "contents": "922. Exercises - 922.3. Stoichiometry of Gaseous Substances, Mixtures, and Reactions\n48. What is the density of laughing gas, dinitrogen monoxide, $\\mathrm{N}_{2} \\mathrm{O}$, at a temperature of 325 K and a pressure of 113.0 kPa ?\n49. Calculate the density of Freon $12, \\mathrm{CF}_{2} \\mathrm{Cl}_{2}$, at $30.0^{\\circ} \\mathrm{C}$ and 0.954 atm.\n50. Which is denser at the same temperature and pressure, dry air or air saturated with water vapor? Explain.\n51. A cylinder of $\\mathrm{O}_{2}(\\mathrm{~g})$ used in breathing by patients with emphysema has a volume of 3.00 L at a pressure of 10.0 atm . If the temperature of the cylinder is $28.0^{\\circ} \\mathrm{C}$, what mass of oxygen is in the cylinder?\n52. What is the molar mass of a gas if 0.0494 g of the gas occupies a volume of 0.100 L at a temperature $26^{\\circ} \\mathrm{C}$ and a pressure of 307 torr?\n53. What is the molar mass of a gas if 0.281 g of the gas occupies a volume of 125 mL at a temperature $126^{\\circ} \\mathrm{C}$ and a pressure of 777 torr?\n54. How could you show experimentally that the molecular formula of propene is $\\mathrm{C}_{3} \\mathrm{H}_{6}$, not $\\mathrm{CH}_{2}$ ?\n55. The density of a certain gaseous fluoride of phosphorus is $3.93 \\mathrm{~g} / \\mathrm{L}$ at STP. Calculate the molar mass of this fluoride and determine its molecular formula.\n56. Consider this question: What is the molecular formula of a compound that contains $39 \\% \\mathrm{C}, 45 \\% \\mathrm{~N}$, and $16 \\% \\mathrm{H}$ if 0.157 g of the compound occupies 125 mL with a pressure of 99.5 kPa at $22^{\\circ} \\mathrm{C}$ ?\n(a) Outline the steps necessary to answer the question.\n(b) Answer the question."}
{"id": 3325, "contents": "922. Exercises - 922.3. Stoichiometry of Gaseous Substances, Mixtures, and Reactions\n(a) Outline the steps necessary to answer the question.\n(b) Answer the question.\n57. A 36.0-L cylinder of a gas used for calibration of blood gas analyzers in medical laboratories contains 350 $\\mathrm{g} \\mathrm{CO}_{2}, 805 \\mathrm{~g} \\mathrm{O}_{2}$, and $4,880 \\mathrm{~g} \\mathrm{~N}_{2}$. At 25 degrees C , what is the pressure in the cylinder in atmospheres?\n58. A cylinder of a gas mixture used for calibration of blood gas analyzers in medical laboratories contains $5.0 \\% \\mathrm{CO}_{2}, 12.0 \\% \\mathrm{O}_{2}$, and the remainder $\\mathrm{N}_{2}$ at a total pressure of 146 atm . What is the partial pressure of each component of this gas? (The percentages given indicate the percent of the total pressure that is due to each component.)\n59. A sample of gas isolated from unrefined petroleum contains $90.0 \\% \\mathrm{CH}_{4}, 8.9 \\% \\mathrm{C}_{2} \\mathrm{H}_{6}$, and $1.1 \\% \\mathrm{C}_{3} \\mathrm{H}_{8}$ at a total pressure of 307.2 kPa . What is the partial pressure of each component of this gas? (The percentages given indicate the percent of the total pressure that is due to each component.)\n60. A mixture of 0.200 g of $\\mathrm{H}_{2}, 1.00 \\mathrm{~g}$ of $\\mathrm{N}_{2}$, and 0.820 g of Ar is stored in a closed container at STP. Find the volume of the container, assuming that the gases exhibit ideal behavior.\n61. Most mixtures of hydrogen gas with oxygen gas are explosive. However, a mixture that contains less than $3.0 \\% \\mathrm{O}_{2}$ is not. If enough $\\mathrm{O}_{2}$ is added to a cylinder of $\\mathrm{H}_{2}$ at 33.2 atm to bring the total pressure to 34.5 atm , is the mixture explosive?"}
{"id": 3326, "contents": "922. Exercises - 922.3. Stoichiometry of Gaseous Substances, Mixtures, and Reactions\n62. A commercial mercury vapor analyzer can detect, in air, concentrations of gaseous Hg atoms (which are poisonous) as low as $2 \\times 10^{-6} \\mathrm{mg} / \\mathrm{L}$ of air. At this concentration, what is the partial pressure of gaseous mercury if the atmospheric pressure is 733 torr at $26^{\\circ} \\mathrm{C}$ ?\n63. A sample of carbon monoxide was collected over water at a total pressure of 756 torr and a temperature of $18{ }^{\\circ} \\mathrm{C}$. What is the pressure of the carbon monoxide? (See Table 8.2 for the vapor pressure of water.)\n64. In an experiment in a general chemistry laboratory, a student collected a sample of a gas over water. The volume of the gas was 265 mL at a pressure of 753 torr and a temperature of $27^{\\circ} \\mathrm{C}$. The mass of the gas was 0.472 g . What was the molar mass of the gas?\n65. Joseph Priestley first prepared pure oxygen by heating mercuric oxide, HgO :\n$2 \\mathrm{HgO}(s) \\longrightarrow 2 \\mathrm{Hg}(l)+\\mathrm{O}_{2}(g)$\n(a) Outline the steps necessary to answer the following question: What volume of $\\mathrm{O}_{2}$ at $23{ }^{\\circ} \\mathrm{C}$ and 0.975 atm is produced by the decomposition of 5.36 g of HgO ?\n(b) Answer the question.\n66. Cavendish prepared hydrogen in 1766 by the novel method of passing steam through a red-hot gun barrel:\n$4 \\mathrm{H}_{2} \\mathrm{O}(g)+3 \\mathrm{Fe}(s) \\longrightarrow \\mathrm{Fe}_{3} \\mathrm{O}_{4}(s)+4 \\mathrm{H}_{2}(g)$"}
{"id": 3327, "contents": "922. Exercises - 922.3. Stoichiometry of Gaseous Substances, Mixtures, and Reactions\n(a) Outline the steps necessary to answer the following question: What volume of $\\mathrm{H}_{2}$ at a pressure of 745 torr and a temperature of $20^{\\circ} \\mathrm{C}$ can be prepared from the reaction of 15.0 g of $\\mathrm{H}_{2} \\mathrm{O}$ ?\n(b) Answer the question.\n67. The chlorofluorocarbon $\\mathrm{CCl}_{2} \\mathrm{~F}_{2}$ can be recycled into a different compound by reaction with hydrogen to produce $\\mathrm{CH}_{2} \\mathrm{~F}_{2}(\\mathrm{~g})$, a compound useful in chemical manufacturing:\n$\\mathrm{CCl}_{2} \\mathrm{~F}_{2}(g)+4 \\mathrm{H}_{2}(g) \\longrightarrow \\mathrm{CH}_{2} \\mathrm{~F}_{2}(g)+2 \\mathrm{HCl}(g)$\n(a) Outline the steps necessary to answer the following question: What volume of hydrogen at 225 atm and $35.5^{\\circ} \\mathrm{C}$ would be required to react with 1 ton $\\left(1.000 \\times 10^{3} \\mathrm{~kg}\\right)$ of $\\mathrm{CCl}_{2} \\mathrm{~F}_{2}$ ?\n(b) Answer the question.\n68. Automobile air bags are inflated with nitrogen gas, which is formed by the decomposition of solid sodium azide $\\left(\\mathrm{NaN}_{3}\\right)$. The other product is sodium metal. Calculate the volume of nitrogen gas at $27{ }^{\\circ} \\mathrm{C}$ and 756 torr formed by the decomposition of 125 g of sodium azide.\n69. Lime, CaO , is produced by heating calcium carbonate, $\\mathrm{CaCO}_{3}$; carbon dioxide is the other product.\n(a) Outline the steps necessary to answer the following question: What volume of carbon dioxide at 875 K and 0.966 atm is produced by the decomposition of 1 ton $\\left(1.000 \\times 10^{3} \\mathrm{~kg}\\right)$ of calcium carbonate?\n(b) Answer the question."}
{"id": 3328, "contents": "922. Exercises - 922.3. Stoichiometry of Gaseous Substances, Mixtures, and Reactions\n(b) Answer the question.\n70. Before small batteries were available, carbide lamps were used for bicycle lights. Acetylene gas, $\\mathrm{C}_{2} \\mathrm{H}_{2}$, and solid calcium hydroxide were formed by the reaction of calcium carbide, $\\mathrm{CaC}_{2}$, with water. The ignition of the acetylene gas provided the light. Currently, the same lamps are used by some cavers, and calcium carbide is used to produce acetylene for carbide cannons.\n(a) Outline the steps necessary to answer the following question: What volume of $\\mathrm{C}_{2} \\mathrm{H}_{2}$ at 1.005 atm and $12.2{ }^{\\circ} \\mathrm{C}$ is formed by the reaction of 15.48 g of $\\mathrm{CaC}_{2}$ with water?\n(b) Answer the question.\n71. Calculate the volume of oxygen required to burn 12.00 L of ethane gas, $\\mathrm{C}_{2} \\mathrm{H}_{6}$, to produce carbon dioxide and water, if the volumes of $\\mathrm{C}_{2} \\mathrm{H}_{6}$ and $\\mathrm{O}_{2}$ are measured under the same conditions of temperature and pressure.\n72. What volume of $\\mathrm{O}_{2}$ at STP is required to oxidize 8.0 L of NO at STP to $\\mathrm{NO}_{2}$ ? What volume of $\\mathrm{NO}_{2}$ is produced at STP?\n73. Consider the following questions:\n(a) What is the total volume of the $\\mathrm{CO}_{2}(\\mathrm{~g})$ and $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ at $600^{\\circ} \\mathrm{C}$ and 0.888 atm produced by the combustion of 1.00 L of $\\mathrm{C}_{2} \\mathrm{H}_{6}(\\mathrm{~g})$ measured at STP?\n(b) What is the partial pressure of $\\mathrm{H}_{2} \\mathrm{O}$ in the product gases?\n74. Methanol, $\\mathrm{CH}_{3} \\mathrm{OH}$, is produced industrially by the following reaction:"}
{"id": 3329, "contents": "922. Exercises - 922.3. Stoichiometry of Gaseous Substances, Mixtures, and Reactions\n74. Methanol, $\\mathrm{CH}_{3} \\mathrm{OH}$, is produced industrially by the following reaction:\n$\\mathrm{CO}(g)+2 \\mathrm{H}_{2}(g) \\xrightarrow{\\text { copper catalyst } 300^{\\circ} \\mathrm{C}, 300 \\mathrm{~atm}} \\mathrm{CH}_{3} \\mathrm{OH}(g)$\nAssuming that the gases behave as ideal gases, find the ratio of the total volume of the reactants to the final volume.\n75. What volume of oxygen at 423.0 K and a pressure of 127.4 kPa is produced by the decomposition of 129.7 g of $\\mathrm{BaO}_{2}$ to BaO and $\\mathrm{O}_{2}$ ?\n76. A 2.50-L sample of a colorless gas at STP decomposed to give 2.50 L of $\\mathrm{N}_{2}$ and 1.25 L of $\\mathrm{O}_{2}$ at STP. What is the colorless gas?\n77. Ethanol, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$, is produced industrially from ethylene, $\\mathrm{C}_{2} \\mathrm{H}_{4}$, by the following sequence of reactions:\n$3 \\mathrm{C}_{2} \\mathrm{H}_{4}+2 \\mathrm{H}_{2} \\mathrm{SO}_{4} \\longrightarrow \\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{HSO}_{4}+\\left(\\mathrm{C}_{2} \\mathrm{H}_{5}\\right)_{2} \\mathrm{SO}_{4}$\n$\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{HSO}_{4}+\\left(\\mathrm{C}_{2} \\mathrm{H}_{5}\\right)_{2} \\mathrm{SO}_{4}+3 \\mathrm{H}_{2} \\mathrm{O} \\longrightarrow 3 \\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}+2 \\mathrm{H}_{2} \\mathrm{SO}_{4}$"}
{"id": 3330, "contents": "922. Exercises - 922.3. Stoichiometry of Gaseous Substances, Mixtures, and Reactions\nWhat volume of ethylene at STP is required to produce 1.000 metric ton ( 1000 kg ) of ethanol if the overall yield of ethanol is $90.1 \\%$ ?\n78. One molecule of hemoglobin will combine with four molecules of oxygen. If 1.0 g of hemoglobin combines with 1.53 mL of oxygen at body temperature $\\left(37^{\\circ} \\mathrm{C}\\right)$ and a pressure of 743 torr, what is the molar mass of hemoglobin?\n79. A sample of a compound of xenon and fluorine was confined in a bulb with a pressure of 18 torr. Hydrogen was added to the bulb until the pressure was 72 torr. Passage of an electric spark through the mixture produced Xe and HF. After the HF was removed by reaction with solid KOH , the final pressure of xenon and unreacted hydrogen in the bulb was 36 torr. What is the empirical formula of the xenon fluoride in the original sample? (Note: Xenon fluorides contain only one xenon atom per molecule.)\n80. One method of analyzing amino acids is the van Slyke method. The characteristic amino groups $\\left(-\\mathrm{NH}_{2}\\right)$ in protein material are allowed to react with nitrous acid, $\\mathrm{HNO}_{2}$, to form $\\mathrm{N}_{2}$ gas. From the volume of the gas, the amount of amino acid can be determined. A $0.0604-\\mathrm{g}$ sample of a biological sample containing glycine, $\\mathrm{CH}_{2}\\left(\\mathrm{NH}_{2}\\right) \\mathrm{COOH}$, was analyzed by the van Slyke method and yielded 3.70 mL of $\\mathrm{N}_{2}$ collected over water at a pressure of 735 torr and $29^{\\circ} \\mathrm{C}$. What was the percentage of glycine in the sample?\n$\\mathrm{CH}_{2}\\left(\\mathrm{NH}_{2}\\right) \\mathrm{CO}_{2} \\mathrm{H}+\\mathrm{HNO}_{2} \\longrightarrow \\mathrm{CH}_{2}(\\mathrm{OH}) \\mathrm{CO}_{2} \\mathrm{H}+\\mathrm{H}_{2} \\mathrm{O}+\\mathrm{N}_{2}$"}
{"id": 3331, "contents": "922. Exercises - 922.4. Effusion and Diffusion of Gases\n81. A balloon filled with helium gas takes 6 hours to deflate to $50 \\%$ of its original volume. How long will it take for an identical balloon filled with the same volume of hydrogen gas (instead of helium) to decrease its volume by $50 \\%$ ?\n82. Explain why the numbers of molecules are not identical in the left- and right-hand bulbs shown in the center illustration of Figure 8.27.\n83. Starting with the definition of rate of effusion and Graham's finding relating rate and molar mass, show how to derive the Graham's law equation, relating the relative rates of effusion for two gases to their molecular masses.\n84. Heavy water, $\\mathrm{D}_{2} \\mathrm{O}$ (molar mass $=20.03 \\mathrm{~g} \\mathrm{~mol}^{-1}$ ), can be separated from ordinary water, $\\mathrm{H}_{2} \\mathrm{O}$ (molar mass $=$ 18.01), as a result of the difference in the relative rates of diffusion of the molecules in the gas phase. Calculate the relative rates of diffusion of $\\mathrm{H}_{2} \\mathrm{O}$ and $\\mathrm{D}_{2} \\mathrm{O}$.\n85. Which of the following gases diffuse more slowly than oxygen? $\\mathrm{F}_{2}, \\mathrm{Ne}, \\mathrm{N}_{2} \\mathrm{O}, \\mathrm{C}_{2} \\mathrm{H}_{2}, \\mathrm{NO}, \\mathrm{Cl}_{2}, \\mathrm{H}_{2} \\mathrm{~S}$"}
{"id": 3332, "contents": "922. Exercises - 922.4. Effusion and Diffusion of Gases\n86. During the discussion of gaseous diffusion for enriching uranium, it was claimed that ${ }^{235} \\mathrm{UF}_{6}$ diffuses $0.4 \\%$ faster than ${ }^{238} \\mathrm{UF}_{6}$. Show the calculation that supports this value. The molar mass of ${ }^{235} \\mathrm{UF}_{6}=$ $235.043930+6 \\times 18.998403=349.034348 \\mathrm{~g} / \\mathrm{mol}$, and the molar mass of ${ }^{238} \\mathrm{UF}_{6}=238.050788+6 \\times$ $18.998403=352.041206 \\mathrm{~g} / \\mathrm{mol}$.\n87. Calculate the relative rate of diffusion of ${ }^{1} \\mathrm{H}_{2}$ (molar mass $2.0 \\mathrm{~g} / \\mathrm{mol}$ ) compared with ${ }^{2} \\mathrm{H}_{2}$ (molar mass 4.0 $\\mathrm{g} / \\mathrm{mol}$ ) and the relative rate of diffusion of $\\mathrm{O}_{2}$ (molar mass $32 \\mathrm{~g} / \\mathrm{mol}$ ) compared with $\\mathrm{O}_{3}$ (molar mass $48 \\mathrm{~g} /$ mol ).\n88. A gas of unknown identity diffuses at a rate of $83.3 \\mathrm{~mL} / \\mathrm{s}$ in a diffusion apparatus in which carbon dioxide diffuses at the rate of $102 \\mathrm{~mL} / \\mathrm{s}$. Calculate the molecular mass of the unknown gas."}
{"id": 3333, "contents": "922. Exercises - 922.4. Effusion and Diffusion of Gases\n89. When two cotton plugs, one moistened with ammonia and the other with hydrochloric acid, are simultaneously inserted into opposite ends of a glass tube that is 87.0 cm long, a white ring of $\\mathrm{NH}_{4} \\mathrm{Cl}$ forms where gaseous $\\mathrm{NH}_{3}$ and gaseous HCl first come into contact. $\\mathrm{NH}_{3}(g)+\\mathrm{HCl}(g) \\longrightarrow \\mathrm{NH}_{4} \\mathrm{Cl}(s)$ At approximately what distance from the ammonia moistened plug does this occur? (Hint: Calculate the rates of diffusion for both $\\mathrm{NH}_{3}$ and HCl , and find out how much faster $\\mathrm{NH}_{3}$ diffuses than HCl .)"}
{"id": 3334, "contents": "922. Exercises - 922.5. The Kinetic-Molecular Theory\n90. Using the postulates of the kinetic molecular theory, explain why a gas uniformly fills a container of any shape.\n91. Can the speed of a given molecule in a gas double at constant temperature? Explain your answer.\n92. Describe what happens to the average kinetic energy of ideal gas molecules when the conditions are changed as follows:\n(a) The pressure of the gas is increased by reducing the volume at constant temperature.\n(b) The pressure of the gas is increased by increasing the temperature at constant volume.\n(c) The average speed of the molecules is increased by a factor of 2.\n93. The distribution of molecular speeds in a sample of helium is shown in Figure 8.34. If the sample is cooled, will the distribution of speeds look more like that of $\\mathrm{H}_{2}$ or of $\\mathrm{H}_{2} \\mathrm{O}$ ? Explain your answer.\n94. What is the ratio of the average kinetic energy of a $\\mathrm{SO}_{2}$ molecule to that of an $\\mathrm{O}_{2}$ molecule in a mixture of two gases? What is the ratio of the root mean square speeds, $u_{\\text {rms }}$, of the two gases?\n95. A 1-L sample of CO initially at STP is heated to 546 K , and its volume is increased to 2 L .\n(a) What effect do these changes have on the number of collisions of the molecules of the gas per unit area of the container wall?\n(b) What is the effect on the average kinetic energy of the molecules?\n(c) What is the effect on the root mean square speed of the molecules?\n96. The root mean square speed of $\\mathrm{H}_{2}$ molecules at $25^{\\circ} \\mathrm{C}$ is about $1.6 \\mathrm{~km} / \\mathrm{s}$. What is the root mean square speed of a $\\mathrm{N}_{2}$ molecule at $25^{\\circ} \\mathrm{C}$ ?\n97. Answer the following questions:\n(a) Is the pressure of the gas in the hot-air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?"}
{"id": 3335, "contents": "922. Exercises - 922.5. The Kinetic-Molecular Theory\n97. Answer the following questions:\n(a) Is the pressure of the gas in the hot-air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?\n(b) Is the density of the gas in the hot-air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?\n(c) At a pressure of 1 atm and a temperature of $20^{\\circ} \\mathrm{C}$, dry air has a density of $1.2256 \\mathrm{~g} / \\mathrm{L}$. What is the (average) molar mass of dry air?\n(d) The average temperature of the gas in a hot-air balloon is $1.30 \\times 10^{2}{ }^{\\circ} \\mathrm{F}$. Calculate its density, assuming the molar mass equals that of dry air.\n(e) The lifting capacity of a hot-air balloon is equal to the difference in the mass of the cool air displaced by the balloon and the mass of the gas in the balloon. What is the difference in the mass of 1.00 L of the cool air in part (c) and the hot air in part (d)?\n(f) An average balloon has a diameter of 60 feet and a volume of $1.1 \\times 10^{5} \\mathrm{ft}^{3}$. What is the lifting power of such a balloon? If the weight of the balloon and its rigging is 500 pounds, what is its capacity for carrying passengers and cargo?\n(g) A balloon carries 40.0 gallons of liquid propane (density $0.5005 \\mathrm{~g} / \\mathrm{L}$ ). What volume of $\\mathrm{CO}_{2}$ and $\\mathrm{H}_{2} \\mathrm{O}$ gas is produced by the combustion of this propane?\n(h) A balloon flight can last about 90 minutes. If all of the fuel is burned during this time, what is the approximate rate of heat loss (in $\\mathrm{kJ} / \\mathrm{min}$ ) from the hot air in the bag during the flight?"}
{"id": 3336, "contents": "922. Exercises - 922.5. The Kinetic-Molecular Theory\n98. Show that the ratio of the rate of diffusion of Gas 1 to the rate of diffusion of Gas $2, \\frac{R_{1}}{R_{2}}$, is the same at $0^{\\circ} \\mathrm{C}$ and $100^{\\circ} \\mathrm{C}$."}
{"id": 3337, "contents": "922. Exercises - 922.6. Non-Ideal Gas Behavior\n99. Graphs showing the behavior of several different gases follow. Which of these gases exhibit behavior significantly different from that expected for ideal gases?\n\n\nTemperature\n\n\nPV/RT\n\n\nTemperature\n\n\n\n\nTemperature\n100. Explain why the plot of $P V$ for $\\mathrm{CO}_{2}$ differs from that of an ideal gas.\n\n101. Under which of the following sets of conditions does a real gas behave most like an ideal gas, and for which conditions is a real gas expected to deviate from ideal behavior? Explain.\n(a) high pressure, small volume\n(b) high temperature, low pressure\n(c) low temperature, high pressure\n102. Describe the factors responsible for the deviation of the behavior of real gases from that of an ideal gas.\n103. For which of the following gases should the correction for the molecular volume be largest: $\\mathrm{CO}, \\mathrm{CO}_{2}, \\mathrm{H}_{2}, \\mathrm{He}, \\mathrm{NH}_{3}, \\mathrm{SF}_{6}$ ?\n104. A 0.245 -L flask contains $0.467 \\mathrm{~mol} \\mathrm{CO}_{2}$ at $159{ }^{\\circ} \\mathrm{C}$. Calculate the pressure:\n(a) using the ideal gas law\n(b) using the van der Waals equation\n(c) Explain the reason for the difference.\n(d) Identify which correction (that for P or V ) is dominant and why.\n105. Answer the following questions:\n(a) If XX behaved as an ideal gas, what would its graph of Z vs. P look like?\n(b) For most of this chapter, we performed calculations treating gases as ideal. Was this justified?\n(c) What is the effect of the volume of gas molecules on Z? Under what conditions is this effect small? When is it large? Explain using an appropriate diagram.\n(d) What is the effect of intermolecular attractions on the value of Z? Under what conditions is this effect small? When is it large? Explain using an appropriate diagram.\n(e) In general, under what temperature conditions would you expect Z to have the largest deviations from the Z for an ideal gas?\n\n418 8\u2022Exercises"}
{"id": 3338, "contents": "923. CHAPTER 9 Thermochemistry - \nFigure 9.1 Sliding a match head along a rough surface initiates a combustion reaction that produces energy in the form of heat and light. (credit: modification of work by Laszlo Ilyes)"}
{"id": 3339, "contents": "924. CHAPTER OUTLINE - 924.3. Enthalpy\n9.4 Strengths of Ionic and Covalent Bonds\n\nINTRODUCTION Chemical reactions, such as those that occur when you light a match, involve changes in energy as well as matter. Societies at all levels of development could not function without the energy released by chemical reactions. In 2012, about $85 \\%$ of US energy consumption came from the combustion of petroleum products, coal, wood, and garbage. We use this energy to produce electricity ( $38 \\%$ ); to transport food, raw materials, manufactured goods, and people (27\\%); for industrial production (21\\%); and to heat and power our homes and businesses (10\\%). ${ }^{\\frac{1}{2}}$ While these combustion reactions help us meet our essential energy needs, they are also recognized by the majority of the scientific community as a major contributor to global climate change.\n\nUseful forms of energy are also available from a variety of chemical reactions other than combustion. For example, the energy produced by the batteries in a cell phone, car, or flashlight results from chemical reactions. This chapter introduces many of the basic ideas necessary to explore the relationships between chemical changes and energy, with a focus on thermal energy."}
{"id": 3340, "contents": "924. CHAPTER OUTLINE - 924.3. Enthalpy\n[^6]The chemical bond is simply another form of energy, and its strength is indicated in exactly the same units as any other process involving energy. That is, in joules (or kilojoules) - or, when taken on a molar basis, in J/mol or $\\mathrm{kJ} / \\mathrm{mol}$. All of the principles involving energy we have examined up to now apply to bond energies exactly as they do to all other forms of energy: that is, energy may change form or be absorbed or released, but it cannot be created or destroyed. The energy of a chemical bond is indicated by the bond enthalpy, taken by convention to be the energy required to break a chemical bond. The same energy is released when a chemical bond is formed. Given that a chemical reaction involves breaking some bonds and making others, the enthalpy change for a chemical reaction can be estimated by analyzing the bonds broken and formed during the reaction. While this procedure gives an estimate of the overall enthalpy, it should be noted that a far more precise method involves using enthalpies of formation; the fact that there can be significant differences between, for example, an $\\mathrm{O}-\\mathrm{H}$ bond in water and that in acetic acid accounts for the discrepancy in calculating $\\Delta H$ using bond enthalpies versus the use of enthalpies of formation."}
{"id": 3341, "contents": "925. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Define energy, distinguish types of energy, and describe the nature of energy changes that accompany chemical and physical changes\n- Distinguish the related properties of heat, thermal energy, and temperature\n- Define and distinguish specific heat and heat capacity, and describe the physical implications of both\n- Perform calculations involving heat, specific heat, and temperature change\n\nChemical changes and their accompanying changes in energy are important parts of our everyday world (Figure 9.2). The macronutrients in food (proteins, fats, and carbohydrates) undergo metabolic reactions that provide the energy to keep our bodies functioning. We burn a variety of fuels (gasoline, natural gas, coal) to produce energy for transportation, heating, and the generation of electricity. Industrial chemical reactions use enormous amounts of energy to produce raw materials (such as iron and aluminum). Energy is then used to manufacture those raw materials into useful products, such as cars, skyscrapers, and bridges.\n\n\nFIGURE 9.2 The energy involved in chemical changes is important to our daily lives: (a) A cheeseburger for lunch provides the energy you need to get through the rest of the day; (b) the combustion of gasoline provides the energy that moves your car (and you) between home, work, and school; and (c) coke, a processed form of coal, provides the energy needed to convert iron ore into iron, which is essential for making many of the products we use daily. (credit a: modification of work by \"Pink Sherbet Photography\"/Flickr; credit b: modification of work by Jeffery Turner)"}
{"id": 3342, "contents": "925. LEARNING OBJECTIVES - \nOver $90 \\%$ of the energy we use comes originally from the sun. Every day, the sun provides the earth with almost 10,000 times the amount of energy necessary to meet all of the world's energy needs for that day. Our challenge is to find ways to convert and store incoming solar energy so that it can be used in reactions or chemical processes that are both convenient and nonpolluting. Plants and many bacteria capture solar energy through photosynthesis. We release the energy stored in plants when we burn wood or plant products such as ethanol. We also use this energy to fuel our bodies by eating food that comes directly from plants or from animals that got their energy by eating plants. Burning coal and petroleum also releases stored solar energy: These fuels are fossilized plant and animal matter.\n\nThis chapter will introduce the basic ideas of an important area of science concerned with the amount of heat absorbed or released during chemical and physical changes-an area called thermochemistry. The concepts introduced in this chapter are widely used in almost all scientific and technical fields. Food scientists use them to determine the energy content of foods. Biologists study the energetics of living organisms, such as the metabolic combustion of sugar into carbon dioxide and water. The oil, gas, and transportation industries, renewable energy providers, and many others endeavor to find better methods to produce energy for our commercial and personal needs. Engineers strive to improve energy efficiency, find better ways to heat and cool our homes, refrigerate our food and drinks, and meet the energy and cooling needs of computers and electronics, among other applications. Understanding thermochemical principles is essential for chemists, physicists, biologists, geologists, every type of engineer, and just about anyone who studies or does any kind of science."}
{"id": 3343, "contents": "926. Energy - \nEnergy can be defined as the capacity to supply heat or do work. One type of work ( $\\boldsymbol{w}$ ) is the process of causing matter to move against an opposing force. For example, we do work when we inflate a bicycle tire-we move matter (the air in the pump) against the opposing force of the air already in the tire.\n\nLike matter, energy comes in different types. One scheme classifies energy into two types: potential energy, the energy an object has because of its relative position, composition, or condition, and kinetic energy, the energy that an object possesses because of its motion. Water at the top of a waterfall or dam has potential energy because of its position; when it flows downward through generators, it has kinetic energy that can be used to do work and produce electricity in a hydroelectric plant (Figure 9.3). A battery has potential energy because the chemicals within it can produce electricity that can do work.\n\n(a)\n\n(b)\n\nFIGURE 9.3 (a) Water at a higher elevation, for example, at the top of Victoria Falls, has a higher potential energy than water at a lower elevation. As the water falls, some of its potential energy is converted into kinetic energy. (b) If the water flows through generators at the bottom of a dam, such as the Hoover Dam shown here, its kinetic energy is converted into electrical energy. (credit a: modification of work by Steve Jurvetson; credit b: modification of work by \"curimedia\"/Wikimedia commons)\n\nEnergy can be converted from one form into another, but all of the energy present before a change occurs always exists in some form after the change is completed. This observation is expressed in the law of conservation of energy: during a chemical or physical change, energy can be neither created nor destroyed, although it can be changed in form. (This is also one version of the first law of thermodynamics, as you will learn later.)"}
{"id": 3344, "contents": "926. Energy - \nWhen one substance is converted into another, there is always an associated conversion of one form of energy into another. Heat is usually released or absorbed, but sometimes the conversion involves light, electrical energy, or some other form of energy. For example, chemical energy (a type of potential energy) is stored in the molecules that compose gasoline. When gasoline is combusted within the cylinders of a car's engine, the rapidly expanding gaseous products of this chemical reaction generate mechanical energy (a type of kinetic\nenergy) when they move the cylinders' pistons.\nAccording to the law of conservation of matter (seen in an earlier chapter), there is no detectable change in the total amount of matter during a chemical change. When chemical reactions occur, the energy changes are relatively modest and the mass changes are too small to measure, so the laws of conservation of matter and energy hold well. However, in nuclear reactions, the energy changes are much larger (by factors of a million or so), the mass changes are measurable, and matter-energy conversions are significant. This will be examined in more detail in a later chapter on nuclear chemistry."}
{"id": 3345, "contents": "927. Thermal Energy, Temperature, and Heat - \nThermal energy is kinetic energy associated with the random motion of atoms and molecules. Temperature is a quantitative measure of \"hot\" or \"cold.\" When the atoms and molecules in an object are moving or vibrating quickly, they have a higher average kinetic energy (KE), and we say that the object is \"hot.\" When the atoms and molecules are moving slowly, they have lower average KE, and we say that the object is \"cold\" (Figure 9.4). Assuming that no chemical reaction or phase change (such as melting or vaporizing) occurs, increasing the amount of thermal energy in a sample of matter will cause its temperature to increase. And, assuming that no chemical reaction or phase change (such as condensation or freezing) occurs, decreasing the amount of thermal energy in a sample of matter will cause its temperature to decrease.\n\n\nHot water\n(a)\n\n\nCold water\n(b)\n\nFIGURE 9.4 (a) The molecules in a sample of hot water move more rapidly than (b) those in a sample of cold water."}
{"id": 3346, "contents": "928. LINK TO LEARNING - \nClick on this interactive simulation (http://openstax.org/l/16PHETtempFX) to view the effects of temperature on molecular motion.\n\nMost substances expand as their temperature increases and contract as their temperature decreases. This property can be used to measure temperature changes, as shown in Figure 9.5. The operation of many thermometers depends on the expansion and contraction of substances in response to temperature changes.\n\n\nFIGURE 9.5 (a) In an alcohol or mercury thermometer, the liquid (dyed red for visibility) expands when heated and contracts when cooled, much more so than the glass tube that contains the liquid. (b) In a bimetallic thermometer, two different metals (such as brass and steel) form a two-layered strip. When heated or cooled, one of the metals (brass) expands or contracts more than the other metal (steel), causing the strip to coil or uncoil. Both types of thermometers have a calibrated scale that indicates the temperature. (credit a: modification of work by \"dwstucke\"/Flickr)"}
{"id": 3347, "contents": "929. LINK TO LEARNING - \nThe following demonstration (http://openstax.org/l/16Bimetallic) allows one to view the effects of heating and cooling a coiled bimetallic strip.\n\nHeat ( $\\boldsymbol{q}$ ) is the transfer of thermal energy between two bodies at different temperatures. Heat flow (a redundant term, but one commonly used) increases the thermal energy of one body and decreases the thermal energy of the other. Suppose we initially have a high temperature (and high thermal energy) substance (H) and a low temperature (and low thermal energy) substance ( L ). The atoms and molecules in H have a higher average KE than those in L . If we place substance H in contact with substance L , the thermal energy will flow spontaneously from substance H to substance L . The temperature of substance H will decrease, as will the average KE of its molecules; the temperature of substance $L$ will increase, along with the average KE of its molecules. Heat flow will continue until the two substances are at the same temperature (Figure 9.6).\n\n\nFIGURE 9.6 (a) Substances H and L are initially at different temperatures, and their atoms have different average kinetic energies. (b) When they contact each other, collisions between the molecules result in the transfer of kinetic (thermal) energy from the hotter to the cooler matter. (c) The two objects reach \"thermal equilibrium\" when both substances are at the same temperature and their molecules have the same average kinetic energy."}
{"id": 3348, "contents": "930. LINK TO LEARNING - \nClick on the PhET simulation (http://openstax.org/l/16PHETenergy) to explore energy forms and changes. Visit\nthe Energy Systems tab to create combinations of energy sources, transformation methods, and outputs. Click on Energy Symbols to visualize the transfer of energy.\n\nMatter undergoing chemical reactions and physical changes can release or absorb heat. A change that releases heat is called an exothermic process. For example, the combustion reaction that occurs when using an oxyacetylene torch is an exothermic process-this process also releases energy in the form of light as evidenced by the torch's flame (Figure 9.7). A reaction or change that absorbs heat is an endothermic process. A cold pack used to treat muscle strains provides an example of an endothermic process. When the substances in the cold pack (water and a salt like ammonium nitrate) are brought together, the resulting process absorbs heat, leading to the sensation of cold.\n\n(a)\n\n(b)\n\nFIGURE 9.7 (a) An oxyacetylene torch produces heat by the combustion of acetylene in oxygen. The energy released by this exothermic reaction heats and then melts the metal being cut. The sparks are tiny bits of the molten metal flying away. (b) A cold pack uses an endothermic process to create the sensation of cold. (credit a: modification of work by \"Skatebiker\"/Wikimedia commons)"}
{"id": 3349, "contents": "930. LINK TO LEARNING - \nHistorically, energy was measured in units of calories (cal). A calorie is the amount of energy required to raise one gram of water by 1 degree C (1 kelvin). However, this quantity depends on the atmospheric pressure and the starting temperature of the water. The ease of measurement of energy changes in calories has meant that the calorie is still frequently used. The Calorie (with a capital C), or large calorie, commonly used in quantifying food energy content, is a kilocalorie. The SI unit of heat, work, and energy is the joule. A joule (J) is defined as the amount of energy used when a force of 1 newton moves an object 1 meter. It is named in honor of the English physicist James Prescott Joule. One joule is equivalent to $1 \\mathrm{~kg} \\mathrm{~m}{ }^{2} / \\mathrm{s}^{2}$, which is also called 1 newton-meter. A kilojoule (kJ) is 1000 joules. To standardize its definition, 1 calorie has been set to equal 4.184 joules.\n\nWe now introduce two concepts useful in describing heat flow and temperature change. The heat capacity (C) of a body of matter is the quantity of heat ( $q$ ) it absorbs or releases when it experiences a temperature change $(\\Delta T)$ of 1 degree Celsius (or equivalently, 1 kelvin):\n\n$$\nC=\\frac{q}{\\Delta T}\n$$\n\nHeat capacity is determined by both the type and amount of substance that absorbs or releases heat. It is therefore an extensive property-its value is proportional to the amount of the substance.\n\nFor example, consider the heat capacities of two cast iron frying pans. The heat capacity of the large pan is five times greater than that of the small pan because, although both are made of the same material, the mass of the large pan is five times greater than the mass of the small pan. More mass means more atoms are present in the larger pan, so it takes more energy to make all of those atoms vibrate faster. The heat capacity of the small cast\niron frying pan is found by observing that it takes $18,150 \\mathrm{~J}$ of energy to raise the temperature of the pan by $50.0^{\\circ} \\mathrm{C}$ :"}
{"id": 3350, "contents": "930. LINK TO LEARNING - \n$$\nC_{\\text {small pan }}=\\frac{18,140 \\mathrm{~J}}{50.0^{\\circ} \\mathrm{C}}=363 \\mathrm{~J} /{ }^{\\circ} \\mathrm{C}\n$$\n\nThe larger cast iron frying pan, while made of the same substance, requires $90,700 \\mathrm{~J}$ of energy to raise its temperature by $50.0^{\\circ} \\mathrm{C}$. The larger pan has a (proportionally) larger heat capacity because the larger amount of material requires a (proportionally) larger amount of energy to yield the same temperature change:\n\n$$\nC_{\\text {large pan }}=\\frac{90,700 \\mathrm{~J}}{50.0^{\\circ} \\mathrm{C}}=1814 \\mathrm{~J} /{ }^{\\circ} \\mathrm{C}\n$$\n\nThe specific heat capacity (c) of a substance, commonly called its \"specific heat,\" is the quantity of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius (or 1 kelvin):\n\n$$\nc=\\frac{q}{\\mathrm{~m} \\Delta \\mathrm{~T}}\n$$\n\nSpecific heat capacity depends only on the kind of substance absorbing or releasing heat. It is an intensive property-the type, but not the amount, of the substance is all that matters. For example, the small cast iron frying pan has a mass of 808 g . The specific heat of iron (the material used to make the pan) is therefore:\n\n$$\nc_{\\text {iron }}=\\frac{18,140 \\mathrm{~J}}{(808 \\mathrm{~g})\\left(50.0^{\\circ} \\mathrm{C}\\right)}=0.449 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\n$$\n\nThe large frying pan has a mass of 4040 g . Using the data for this pan, we can also calculate the specific heat of iron:"}
{"id": 3351, "contents": "930. LINK TO LEARNING - \nThe large frying pan has a mass of 4040 g . Using the data for this pan, we can also calculate the specific heat of iron:\n\n$$\nc_{\\text {iron }}=\\frac{90,700 \\mathrm{~J}}{(4040 \\mathrm{~g})\\left(50.0^{\\circ} \\mathrm{C}\\right)}=0.449 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\n$$\n\nAlthough the large pan is more massive than the small pan, since both are made of the same material, they both yield the same value for specific heat (for the material of construction, iron). Note that specific heat is measured in units of energy per temperature per mass and is an intensive property, being derived from a ratio of two extensive properties (heat and mass). The molar heat capacity, also an intensive property, is the heat capacity per mole of a particular substance and has units of $\\mathrm{J} / \\mathrm{mol}{ }^{\\circ} \\mathrm{C}$ (Figure 9.8).\n\n\nFIGURE 9.8 Because of its larger mass, a large frying pan has a larger heat capacity than a small frying pan. Because they are made of the same material, both frying pans have the same specific heat. (credit: Mark Blaser)\nWater has a relatively high specific heat (about $4.2 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$ for the liquid and $2.09 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$ for the solid); most metals have much lower specific heats (usually less than $1 \\mathrm{~J} / \\mathrm{g}^{\\circ} \\mathrm{C}$ ). The specific heat of a substance varies somewhat with temperature. However, this variation is usually small enough that we will treat specific heat as constant over the range of temperatures that will be considered in this chapter. Specific heats of some common substances are listed in Table 9.1.\n\nSpecific Heats of Common Substances at $25^{\\circ} \\mathrm{C}$ and 1 bar"}
{"id": 3352, "contents": "930. LINK TO LEARNING - \nSpecific Heats of Common Substances at $25^{\\circ} \\mathrm{C}$ and 1 bar\n\n| Substance | Symbol (state) | Specific Heat ( $\\mathrm{J} / \\mathrm{g}^{\\circ} \\mathrm{C}$ ) |\n| :---: | :---: | :---: |\n| helium | $\\mathrm{He}(\\mathrm{g})$ | 5.193 |\n| water | $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{I})$ | 4.184 |\n| ethanol | $\\mathrm{C}_{2} \\mathrm{H}_{6} \\mathrm{O}(\\mathrm{I})$ | 2.376 |\n| ice | $\\mathrm{H}_{2} \\mathrm{O}(s)$ | 2.093 (at $\\left.-10^{\\circ} \\mathrm{C}\\right)$ |\n| water vapor | $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ | 1.864 |\n| nitrogen | $\\mathrm{N}_{2}(\\mathrm{~g})$ | 1.040 |\n| air |  | 1.007 |\n| oxygen | $\\mathrm{O}_{2}(\\mathrm{~g})$ | 0.918 |\n| aluminum | $\\mathrm{Al}(\\mathrm{s})$ | 0.897 |\n| carbon dioxide | $\\mathrm{CO}_{2}(\\mathrm{~g})$ | 0.853 |\n| argon | $\\operatorname{Ar}(\\mathrm{g})$ | 0.522 |\n| iron | $\\mathrm{Fe}(s)$ | 0.449 |\n| copper | $\\mathrm{Cu}(s)$ | 0.385 |\n| lead | $\\mathrm{Pb}(s)$ | 0.130 |\n| gold | $\\mathrm{Au}(s)$ | 0.129 |\n| silicon | Si(s) | 0.712 |\n\nTABLE 9.1\n\nIf we know the mass of a substance and its specific heat, we can determine the amount of heat, $q$, entering or leaving the substance by measuring the temperature change before and after the heat is gained or lost:"}
{"id": 3353, "contents": "930. LINK TO LEARNING - \nIf we know the mass of a substance and its specific heat, we can determine the amount of heat, $q$, entering or leaving the substance by measuring the temperature change before and after the heat is gained or lost:\n\n$$\n\\begin{aligned}\n& q=(\\text { specific heat }) \\times(\\text { mass of substance }) \\times(\\text { temperature change }) \\\\\n& q=c \\times m \\times \\Delta T=c \\times m \\times\\left(T_{\\text {final }}-T_{\\text {initial }}\\right)\n\\end{aligned}\n$$\n\nIn this equation, $c$ is the specific heat of the substance, $m$ is its mass, and $\\Delta T$ (which is read \"delta $T$ \") is the temperature change, $T_{\\text {final }}-T_{\\text {initial }}$. If a substance gains thermal energy, its temperature increases, its final temperature is higher than its initial temperature, $T_{\\text {final }}-T_{\\text {initial }}$ has a positive value, and the value of $q$ is positive. If a substance loses thermal energy, its temperature decreases, the final temperature is lower than the initial temperature, $T_{\\text {final }}-T_{\\text {initial }}$ has a negative value, and the value of $q$ is negative."}
{"id": 3354, "contents": "932. Measuring Heat - \nA flask containing $8.0 \\times 10^{2} \\mathrm{~g}$ of water is heated, and the temperature of the water increases from $21^{\\circ} \\mathrm{C}$ to 85 ${ }^{\\circ} \\mathrm{C}$. How much heat did the water absorb?"}
{"id": 3355, "contents": "933. Solution - \nTo answer this question, consider these factors:\n\n- the specific heat of the substance being heated (in this case, water)\n- the amount of substance being heated (in this case, $8.0 \\times 10^{2} \\mathrm{~g}$ )\n- the magnitude of the temperature change (in this case, from $21^{\\circ} \\mathrm{C}$ to $85^{\\circ} \\mathrm{C}$ ).\n\nThe specific heat of water is $4.184 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$, so to heat 1 g of water by $1^{\\circ} \\mathrm{C}$ requires 4.184 J . We note that since 4.184 J is required to heat 1 g of water by $1^{\\circ} \\mathrm{C}$, we will need 800 times as much to heat $8.0 \\times 10^{2} \\mathrm{~g}$ of water by 1 ${ }^{\\circ} \\mathrm{C}$. Finally, we observe that since 4.184 J are required to heat 1 g of water by $1^{\\circ} \\mathrm{C}$, we will need 64 times as much to heat it by $64^{\\circ} \\mathrm{C}$ (that is, from $21^{\\circ} \\mathrm{C}$ to $85^{\\circ} \\mathrm{C}$ ).\n\nThis can be summarized using the equation:"}
{"id": 3356, "contents": "933. Solution - \nThis can be summarized using the equation:\n\n$$\n\\begin{aligned}\n& q=c \\times m \\times \\Delta T=c \\times m \\times\\left(T_{\\text {final }}-T_{\\text {initial }}\\right) \\\\\n= & \\left(4.184 \\mathrm{~J} / \\frac{\\mathrm{g}}{}{ }^{\\circ} \\mathrm{C}\\right) \\times\\left(8.0 \\times 10^{2} \\frac{\\mathrm{~g}}{\\mathrm{~g}}\\right) \\times(85-21)^{\\circ} \\mathrm{C} \\\\\n= & \\left(4.184 \\mathrm{~J} / \\frac{\\mathrm{g}}{}{ }^{\\circ} \\mathrm{C}\\right) \\times\\left(8.0 \\times 10^{2} \\frac{\\mathrm{~g}}{\\mathrm{C}}\\right) \\times(64)^{\\circ} \\mathrm{C} \\\\\n= & 210,000 \\mathrm{~J}\\left(=2.1 \\times 10^{2} \\mathrm{~kJ}\\right)\n\\end{aligned}\n$$\n\nBecause the temperature increased, the water absorbed heat and $q$ is positive."}
{"id": 3357, "contents": "934. Check Your Learning - \nHow much heat, in joules, must be added to a 502 g iron skillet to increase its temperature from $25^{\\circ} \\mathrm{C}$ to 250 ${ }^{\\circ} \\mathrm{C}$ ? The specific heat of iron is $0.449 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$."}
{"id": 3358, "contents": "935. Answer: - \n$5.07 \\times 10^{4} \\mathrm{~J}$\n\nNote that the relationship between heat, specific heat, mass, and temperature change can be used to determine any of these quantities (not just heat) if the other three are known or can be deduced."}
{"id": 3359, "contents": "937. Determining Other Quantities - \nA piece of unknown metal weighs 348 g . When the metal piece absorbs 6.64 kJ of heat, its temperature increases from $22.4^{\\circ} \\mathrm{C}$ to $43.6^{\\circ} \\mathrm{C}$. Determine the specific heat of this metal (which might provide a clue to its identity)."}
{"id": 3360, "contents": "938. Solution - \nSince mass, heat, and temperature change are known for this metal, we can determine its specific heat using the relationship:\n\n$$\nq=c \\times m \\times \\Delta T=c \\times m \\times\\left(T_{\\text {final }}-T_{\\text {initial }}\\right)\n$$\n\nSubstituting the known values:\n\n$$\n6640 \\mathrm{~J}=c \\times(348 \\mathrm{~g}) \\times(43.6-22.4)^{\\circ} \\mathrm{C}\n$$\n\nSolving:\n\n$$\nc=\\frac{6640 \\mathrm{~J}}{(348 \\mathrm{~g}) \\times\\left(21.2^{\\circ} \\mathrm{C}\\right)}=0.900 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\n$$\n\nComparing this value with the values in Table 9.1, this value matches the specific heat of aluminum, which suggests that the unknown metal may be aluminum."}
{"id": 3361, "contents": "939. Check Your Learning - \nA piece of unknown metal weighs 217 g . When the metal piece absorbs 1.43 kJ of heat, its temperature increases from $24.5^{\\circ} \\mathrm{C}$ to $39.1^{\\circ} \\mathrm{C}$. Determine the specific heat of this metal, and predict its identity."}
{"id": 3362, "contents": "940. Answer: - \n$c=0.451 \\mathrm{~J} / \\mathrm{g}^{\\circ} \\mathrm{C}$; the metal is likely to be iron"}
{"id": 3363, "contents": "942. Solar Thermal Energy Power Plants - \nThe sunlight that reaches the earth contains thousands of times more energy than we presently capture. Solar thermal systems provide one possible solution to the problem of converting energy from the sun into energy we can use. Large-scale solar thermal plants have different design specifics, but all concentrate sunlight to heat some substance; the heat \"stored\" in that substance is then converted into electricity.\n\nThe Solana Generating Station in Arizona's Sonora Desert produces 280 megawatts of electrical power. It uses parabolic mirrors that focus sunlight on pipes filled with a heat transfer fluid (HTF) (Figure 9.9). The HTF then does two things: It turns water into steam, which spins turbines, which in turn produces electricity, and it melts and heats a mixture of salts, which functions as a thermal energy storage system. After the sun goes down, the molten salt mixture can then release enough of its stored heat to produce steam to run the turbines for 6 hours. Molten salts are used because they possess a number of beneficial properties, including high heat capacities and thermal conductivities.\n\n\nFIGURE 9.9 This solar thermal plant uses parabolic trough mirrors to concentrate sunlight. (credit a: modification of work by Bureau of Land Management)\n\nThe 377-megawatt Ivanpah Solar Generating System, located in the Mojave Desert in California, is the largest solar thermal power plant in the world (Figure 9.10). Its 170,000 mirrors focus huge amounts of sunlight on three water-filled towers, producing steam at over $538^{\\circ} \\mathrm{C}$ that drives electricity-producing turbines. It produces enough energy to power 140,000 homes. Water is used as the working fluid because of its large heat capacity and heat of vaporization.\n\n(a)\n\n(b)\n\nFIGURE 9.10 (a) The Ivanpah solar thermal plant uses 170,000 mirrors to concentrate sunlight on water-filled towers. (b) It covers 4000 acres of public land near the Mojave Desert and the California-Nevada border. (credit a: modification of work by Craig Dietrich; credit b: modification of work by \"USFWS Pacific Southwest Region\"/Flickr)"}
{"id": 3364, "contents": "943. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Explain the technique of calorimetry\n- Calculate and interpret heat and related properties using typical calorimetry data\n\nOne technique we can use to measure the amount of heat involved in a chemical or physical process is known as calorimetry. Calorimetry is used to measure amounts of heat transferred to or from a substance. To do so, the heat is exchanged with a calibrated object (calorimeter). The temperature change measured by the calorimeter is used to derive the amount of heat transferred by the process under study. The measurement of heat transfer using this approach requires the definition of a system (the substance or substances undergoing the chemical or physical change) and its surroundings (all other matter, including components of the measurement apparatus, that serve to either provide heat to the system or absorb heat from the system).\n\nA calorimeter is a device used to measure the amount of heat involved in a chemical or physical process. For example, when an exothermic reaction occurs in solution in a calorimeter, the heat produced by the reaction is absorbed by the solution, which increases its temperature. When an endothermic reaction occurs, the heat required is absorbed from the thermal energy of the solution, which decreases its temperature (Figure 9.11). The temperature change, along with the specific heat and mass of the solution, can then be used to calculate the amount of heat involved in either case.\n\n\nFIGURE 9.11 In a calorimetric determination, either (a) an exothermic process occurs and heat, $q$, is negative, indicating that thermal energy is transferred from the system to its surroundings, or (b) an endothermic process occurs and heat, $q$, is positive, indicating that thermal energy is transferred from the surroundings to the system.\n\nCalorimetry measurements are important in understanding the heat transferred in reactions involving everything from microscopic proteins to massive machines. During her time at the National Bureau of Standards, research chemist Reatha Clark King performed calorimetric experiments to understand the precise heats of various flourine compounds. Her work was important to NASA in their quest for better rocket fuels."}
{"id": 3365, "contents": "943. LEARNING OBJECTIVES - \nScientists use well-insulated calorimeters that all but prevent the transfer of heat between the calorimeter and its environment, which effectively limits the \"surroundings\" to the nonsystem components with the calorimeter (and the calorimeter itself). This enables the accurate determination of the heat involved in chemical processes, the energy content of foods, and so on. General chemistry students often use simple calorimeters constructed from polystyrene cups (Figure 9.12). These easy-to-use \"coffee cup\" calorimeters allow more heat exchange with the outside environment, and therefore produce less accurate energy values.\n\n\nFIGURE 9.12 A simple calorimeter can be constructed from two polystyrene cups. A thermometer and stirrer extend through the cover into the reaction mixture.\n\nCommercial solution calorimeters are also available. Relatively inexpensive calorimeters often consist of two thin-walled cups that are nested in a way that minimizes thermal contact during use, along with an insulated cover, handheld stirrer, and simple thermometer. More expensive calorimeters used for industry and research typically have a well-insulated, fully enclosed reaction vessel, motorized stirring mechanism, and a more accurate temperature sensor (Figure 9.13).\n\n\nFIGURE 9.13 Commercial solution calorimeters range from (a) simple, inexpensive models for student use to (b) expensive, more accurate models for industry and research.\n\nBefore discussing the calorimetry of chemical reactions, consider a simpler example that illustrates the core idea behind calorimetry. Suppose we initially have a high-temperature substance, such as a hot piece of metal $(M)$, and a low-temperature substance, such as cool water (W). If we place the metal in the water, heat will flow from M to W . The temperature of M will decrease, and the temperature of W will increase, until the two substances have the same temperature-that is, when they reach thermal equilibrium (Figure 9.14). If this occurs in a calorimeter, ideally all of this heat transfer occurs between the two substances, with no heat gained or lost by either its external environment. Under these ideal circumstances, the net heat change is zero:\n\n$$\nq_{\\text {substance } \\mathrm{M}}+q_{\\text {substance } \\mathrm{W}}=0\n$$"}
{"id": 3366, "contents": "943. LEARNING OBJECTIVES - \n$$\nq_{\\text {substance } \\mathrm{M}}+q_{\\text {substance } \\mathrm{W}}=0\n$$\n\nThis relationship can be rearranged to show that the heat gained by substance M is equal to the heat lost by substance W:\n\n$$\nq_{\\text {substance } \\mathrm{M}}=-q_{\\text {substance } \\mathrm{W}}\n$$\n\nThe magnitude of the heat (change) is therefore the same for both substances, and the negative sign merely shows that $q_{\\text {substance } \\mathrm{M}}$ and $q_{\\text {substance W }}$ are opposite in direction of heat flow (gain or loss) but does not indicate the arithmetic sign of either $q$ value (that is determined by whether the matter in question gains or loses heat, per definition). In the specific situation described, $q_{\\text {substance } M}$ is a negative value and $q_{\\text {substance } \\mathrm{W}}$ is positive, since heat is transferred from M to W .\n\n\nFIGURE 9.14 In a simple calorimetry process, (a) heat, $q$, is transferred from the hot metal, $M$, to the cool water, W, until (b) both are at the same temperature."}
{"id": 3367, "contents": "945. Heat Transfer between Substances at Different Temperatures - \nA 360.0-g piece of rebar (a steel rod used for reinforcing concrete) is dropped into 425 mL of water at $24.0^{\\circ} \\mathrm{C}$. The final temperature of the water was measured as $42.7^{\\circ} \\mathrm{C}$. Calculate the initial temperature of the piece of rebar. Assume the specific heat of steel is approximately the same as that for iron (Table 9.1), and that all heat transfer occurs between the rebar and the water (there is no heat exchange with the surroundings)."}
{"id": 3368, "contents": "946. Solution - \nThe temperature of the water increases from $24.0^{\\circ} \\mathrm{C}$ to $42.7^{\\circ} \\mathrm{C}$, so the water absorbs heat. That heat came from the piece of rebar, which initially was at a higher temperature. Assuming that all heat transfer was between the rebar and the water, with no heat \"lost\" to the outside environment, then heat given off by rebar = -heat taken in by water, or:\n\n$$\nq_{\\text {rebar }}=-q_{\\text {water }}\n$$\n\nSince we know how heat is related to other measurable quantities, we have:\n\n$$\n(c \\times m \\times \\Delta T)_{\\mathrm{rebar}}=-(c \\times m \\times \\Delta T)_{\\mathrm{water}}\n$$\n\nLetting $\\mathrm{f}=$ final and $\\mathrm{i}=$ initial, in expanded form, this becomes:\n\n$$\nc_{\\text {rebar }} \\times m_{\\text {rebar }} \\times\\left(T_{\\mathrm{f}, \\text { rebar }}-T_{\\mathrm{i}, \\text { rebar }}\\right)=-c_{\\text {water }} \\times m_{\\text {water }} \\times\\left(T_{\\mathrm{f}, \\text { water }}-T_{\\mathrm{i}, \\text { water }}\\right)\n$$\n\nThe density of water is $1.0 \\mathrm{~g} / \\mathrm{mL}$, so 425 mL of water $=425 \\mathrm{~g}$. Noting that the final temperature of both the rebar and water is $42.7^{\\circ} \\mathrm{C}$, substituting known values yields:"}
{"id": 3369, "contents": "946. Solution - \n$$\n\\begin{gathered}\n\\left(0.449 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\\right)(360.0 \\mathrm{~g})\\left(42.7{ }^{\\circ} \\mathrm{C}-T_{\\mathrm{i}, \\text { rebar }}\\right)=-\\left(4.184 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\\right)(425 \\mathrm{~g})\\left(42.7^{\\circ} \\mathrm{C}-24.0^{\\circ} \\mathrm{C}\\right) \\\\\nT_{i, r e b a r}=\\frac{\\left(4.184 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\\right)(425 \\mathrm{~g})\\left(42.7^{\\circ} \\mathrm{C}-24.0^{\\circ} \\mathrm{C}\\right)}{\\left(0.449 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\\right)(360.0 \\mathrm{~g})}+42.7^{\\circ} \\mathrm{C}\n\\end{gathered}\n$$\n\nSolving this gives $T_{i, \\text { rebar }}=248^{\\circ} \\mathrm{C}$, so the initial temperature of the rebar was $248{ }^{\\circ} \\mathrm{C}$."}
{"id": 3370, "contents": "947. Check Your Learning - \nA $248-\\mathrm{g}$ piece of copper is dropped into 390 mL of water at $22.6^{\\circ} \\mathrm{C}$. The final temperature of the water was\nmeasured as $39.9^{\\circ} \\mathrm{C}$. Calculate the initial temperature of the piece of copper. Assume that all heat transfer occurs between the copper and the water."}
{"id": 3371, "contents": "948. Answer: - \nThe initial temperature of the copper was $335.6^{\\circ} \\mathrm{C}$."}
{"id": 3372, "contents": "949. Check Your Learning - \nA 248 -g piece of copper initially at $314^{\\circ} \\mathrm{C}$ is dropped into 390 mL of water initially at $22.6^{\\circ} \\mathrm{C}$. Assuming that all heat transfer occurs between the copper and the water, calculate the final temperature."}
{"id": 3373, "contents": "950. Answer: - \nThe final temperature (reached by both copper and water) is $38.7^{\\circ} \\mathrm{C}$.\n\nThis method can also be used to determine other quantities, such as the specific heat of an unknown metal."}
{"id": 3374, "contents": "952. Identifying a Metal by Measuring Specific Heat - \nA 59.7 g piece of metal that had been submerged in boiling water was quickly transferred into 60.0 mL of water initially at $22.0^{\\circ} \\mathrm{C}$. The final temperature is $28.5^{\\circ} \\mathrm{C}$. Use these data to determine the specific heat of the metal. Use this result to identify the metal."}
{"id": 3375, "contents": "953. Solution - \nAssuming perfect heat transfer, heat given off by metal $=-$ heat taken in by water, or:\n\n$$\nq_{\\text {metal }}=-q_{\\text {water }}\n$$\n\nIn expanded form, this is:\n\n$$\nc_{\\text {metal }} \\times m_{\\text {metal }} \\times\\left(T_{\\mathrm{f}, \\text { metal }}-T_{\\mathrm{i}, \\text { metal }}\\right)=-c_{\\text {water }} \\times m_{\\text {water }} \\times\\left(T_{\\mathrm{f}, \\text { water }}-T_{\\mathrm{i}, \\text { water }}\\right)\n$$\n\nNoting that since the metal was submerged in boiling water, its initial temperature was $100.0^{\\circ} \\mathrm{C}$; and that for water, $60.0 \\mathrm{~mL}=60.0 \\mathrm{~g}$; we have:\n\n$$\n\\left(c_{\\text {metal }}\\right)(59.7 \\mathrm{~g})\\left(28.5^{\\circ} \\mathrm{C}-100.0^{\\circ} \\mathrm{C}\\right)=-\\left(4.18 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\\right)(60.0 \\mathrm{~g})\\left(28.5^{\\circ} \\mathrm{C}-22.0^{\\circ} \\mathrm{C}\\right)\n$$\n\nSolving this:\n\n$$\nc_{\\text {metal }}=\\frac{-\\left(4.184 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\\right)(60.0 \\mathrm{~g})\\left(6.5^{\\circ} \\mathrm{C}\\right)}{(59.7 \\mathrm{~g})\\left(-71.5^{\\circ} \\mathrm{C}\\right)}=0.38 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\n$$"}
{"id": 3376, "contents": "953. Solution - \nComparing this with values in Table 9.1, our experimental specific heat is closest to the value for copper (0.39 $\\mathrm{J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$ ), so we identify the metal as copper."}
{"id": 3377, "contents": "954. Check Your Learning - \nA 92.9-g piece of a silver/gray metal is heated to $178.0^{\\circ} \\mathrm{C}$, and then quickly transferred into 75.0 mL of water initially at $24.0^{\\circ} \\mathrm{C}$. After 5 minutes, both the metal and the water have reached the same temperature: $29.7^{\\circ} \\mathrm{C}$. Determine the specific heat and the identity of the metal. (Note: You should find that the specific heat is close to that of two different metals. Explain how you can confidently determine the identity of the metal)."}
{"id": 3378, "contents": "955. Answer: - \n$c_{\\text {metal }}=0.13 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$\nThis specific heat is close to that of either gold or lead. It would be difficult to determine which metal this was based solely on the numerical values. However, the observation that the metal is silver/gray in addition to the value for the specific heat indicates that the metal is lead.\n\nWhen we use calorimetry to determine the heat involved in a chemical reaction, the same principles we have\nbeen discussing apply. The amount of heat absorbed by the calorimeter is often small enough that we can neglect it (though not for highly accurate measurements, as discussed later), and the calorimeter minimizes energy exchange with the outside environment. Because energy is neither created nor destroyed during a chemical reaction, the heat produced or consumed in the reaction (the \"system\"), $q_{\\text {reaction }}$, plus the heat absorbed or lost by the solution (the \"surroundings\"), $q_{\\text {solution }}$, must add up to zero:\n\n$$\nq_{\\text {reaction }}+q_{\\text {solution }}=0\n$$\n\nThis means that the amount of heat produced or consumed in the reaction equals the amount of heat absorbed or lost by the solution:\n\n$$\nq_{\\text {reaction }}=-q_{\\text {solution }}\n$$\n\nThis concept lies at the heart of all calorimetry problems and calculations."}
{"id": 3379, "contents": "957. Heat Produced by an Exothermic Reaction - \nWhen 50.0 mL of $1.00 \\mathrm{M} \\mathrm{HCl}(\\mathrm{aq})$ and 50.0 mL of $1.00 \\mathrm{M} \\mathrm{NaOH}(\\mathrm{aq})$, both at $22.0^{\\circ} \\mathrm{C}$, are added to a coffee cup calorimeter, the temperature of the mixture reaches a maximum of $28.9^{\\circ} \\mathrm{C}$. What is the approximate amount of heat produced by this reaction?\n\n$$\n\\mathrm{HCl}(a q)+\\mathrm{NaOH}(a q) \\longrightarrow \\mathrm{NaCl}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)\n$$"}
{"id": 3380, "contents": "958. Solution - \nTo visualize what is going on, imagine that you could combine the two solutions so quickly that no reaction took place while they mixed; then after mixing, the reaction took place. At the instant of mixing, you have 100.0 mL of a mixture of HCl and NaOH at $22.0^{\\circ} \\mathrm{C}$. The HCl and NaOH then react until the solution temperature reaches $28.9^{\\circ} \\mathrm{C}$.\n\nThe heat given off by the reaction is equal to that taken in by the solution. Therefore:\n\n$$\nq_{\\text {reaction }}=-q_{\\text {solution }}\n$$\n\n(It is important to remember that this relationship only holds if the calorimeter does not absorb any heat from the reaction, and there is no heat exchange between the calorimeter and the outside environment.)\n\nNext, we know that the heat absorbed by the solution depends on its specific heat, mass, and temperature change:\n\n$$\nq_{\\text {solution }}=(c \\times m \\times \\Delta T)_{\\text {solution }}\n$$\n\nTo proceed with this calculation, we need to make a few more reasonable assumptions or approximations. Since the solution is aqueous, we can proceed as if it were water in terms of its specific heat and mass values. The density of water is approximately $1.0 \\mathrm{~g} / \\mathrm{mL}$, so 100.0 mL has a mass of about $1.0 \\times 10^{2} \\mathrm{~g}$ (two significant figures). The specific heat of water is approximately $4.184 \\mathrm{~J} / \\mathrm{g}^{\\circ} \\mathrm{C}$, so we use that for the specific heat of the solution. Substituting these values gives:\n\n$$\nq_{\\text {solution }}=\\left(4.184 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\\right)\\left(1.0 \\times 10^{2} \\mathrm{~g}\\right)\\left(28.9^{\\circ} \\mathrm{C}-22.0^{\\circ} \\mathrm{C}\\right)=2.9 \\times 10^{3} \\mathrm{~J}\n$$"}
{"id": 3381, "contents": "958. Solution - \nFinally, since we are trying to find the heat of the reaction, we have:\n\n$$\nq_{\\text {reaction }}=-q_{\\text {solution }}=-2.9 \\times 10^{3} \\mathrm{~J}\n$$\n\nThe negative sign indicates that the reaction is exothermic. It produces 2.9 kJ of heat."}
{"id": 3382, "contents": "959. Check Your Learning - \nWhen 100 mL of $0.200 \\mathrm{M} \\mathrm{NaCl}(\\mathrm{aq})$ and 100 mL of $0.200 \\mathrm{M} \\mathrm{AgNO}_{3}(\\mathrm{aq})$, both at $21.9^{\\circ} \\mathrm{C}$, are mixed in a coffee cup calorimeter, the temperature increases to $23.5^{\\circ} \\mathrm{C}$ as solid AgCl forms. How much heat is produced by this precipitation reaction? What assumptions did you make to determine your value?"}
{"id": 3383, "contents": "960. Answer: - \n$1.34 \\times 1.3 \\mathrm{~kJ}$; assume no heat is absorbed by the calorimeter, no heat is exchanged between the calorimeter and its surroundings, and that the specific heat and mass of the solution are the same as those for water"}
{"id": 3384, "contents": "962. Thermochemistry of Hand Warmers - \nWhen working or playing outdoors on a cold day, you might use a hand warmer to warm your hands (Figure 9.15). A common reusable hand warmer contains a supersaturated solution of $\\mathrm{NaC}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}$ (sodium acetate) and a metal disc. Bending the disk creates nucleation sites around which the metastable $\\mathrm{NaC}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}$ quickly crystallizes (a later chapter on solutions will investigate saturation and supersaturation in more detail).\n\nThe process $\\mathrm{NaC}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}(a q) \\longrightarrow \\mathrm{NaC}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}(s)$ is exothermic, and the heat produced by this process is absorbed by your hands, thereby warming them (at least for a while). If the hand warmer is reheated, the $\\mathrm{NaC}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}$ redissolves and can be reused.\n\n\nFIGURE 9.15 Chemical hand warmers produce heat that warms your hand on a cold day. In this one, you can see the metal disc that initiates the exothermic precipitation reaction. (credit: modification of work by Science Buddies TV/YouTube)\n\nAnother common hand warmer produces heat when it is ripped open, exposing iron and water in the hand warmer to oxygen in the air. One simplified version of this exothermic reaction is\n$2 \\mathrm{Fe}(s)+\\frac{3}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{Fe}_{2} \\mathrm{O}_{3}(s)$. Salt in the hand warmer catalyzes the reaction, so it produces heat more rapidly; cellulose, vermiculite, and activated carbon help distribute the heat evenly. Other types of hand warmers use lighter fluid (a platinum catalyst helps lighter fluid oxidize exothermically), charcoal (charcoal oxidizes in a special case), or electrical units that produce heat by passing an electrical current from a battery through resistive wires."}
{"id": 3385, "contents": "963. LINK TO LEARNING - \nThis link (http://openstax.org/l/16Handwarmer) shows the precipitation reaction that occurs when the disk in a chemical hand warmer is flexed."}
{"id": 3386, "contents": "965. Heat Flow in an Instant Ice Pack - \nWhen solid ammonium nitrate dissolves in water, the solution becomes cold. This is the basis for an \"instant ice pack\" (Figure 9.16). When 3.21 g of solid $\\mathrm{NH}_{4} \\mathrm{NO}_{3}$ dissolves in 50.0 g of water at $24.9^{\\circ} \\mathrm{C}$ in a calorimeter, the temperature decreases to $20.3^{\\circ} \\mathrm{C}$.\n\nCalculate the value of $q$ for this reaction and explain the meaning of its arithmetic sign. State any assumptions that you made.\n\n\nFIGURE 9.16 An instant cold pack consists of a bag containing solid ammonium nitrate and a second bag of water. When the bag of water is broken, the pack becomes cold because the dissolution of ammonium nitrate is an endothermic process that removes thermal energy from the water. The cold pack then removes thermal energy from your body."}
{"id": 3387, "contents": "966. Solution - \nWe assume that the calorimeter prevents heat transfer between the solution and its external environment (including the calorimeter itself), in which case:\n\n$$\nq_{\\mathrm{rxn}}=-q_{\\mathrm{soln}}\n$$\n\nwith \"rxn\" and \"soln\" used as shorthand for \"reaction\" and \"solution,\" respectively.\nAssuming also that the specific heat of the solution is the same as that for water, we have:\n\n$$\n\\begin{aligned}\n& q_{\\mathrm{rxn}}=-q_{\\mathrm{soln}}=-(c \\times m \\times \\Delta T)_{\\text {soln }} \\\\\n& =-\\left[\\left(4.184 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\\right) \\times(53.2 \\mathrm{~g}) \\times\\left(20.3^{\\circ} \\mathrm{C}-24.9^{\\circ} \\mathrm{C}\\right)\\right] \\\\\n& =-\\left[\\left(4.184 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\\right) \\times(53.2 \\mathrm{~g}) \\times\\left(-4.6^{\\circ} \\mathrm{C}\\right)\\right] \\\\\n& +1.0 \\times 10^{3} \\mathrm{~J}=+1.0 \\mathrm{~kJ}\n\\end{aligned}\n$$\n\nThe positive sign for $q$ indicates that the dissolution is an endothermic process."}
{"id": 3388, "contents": "967. Check Your Learning - \nWhen a $3.00-\\mathrm{g}$ sample of KCl was added to $3.00 \\times 10^{2} \\mathrm{~g}$ of water in a coffee cup calorimeter, the temperature decreased by $1.05^{\\circ} \\mathrm{C}$. How much heat is involved in the dissolution of the KCl ? What assumptions did you make?"}
{"id": 3389, "contents": "968. Answer: - \n1.33 kJ ; assume that the calorimeter prevents heat transfer between the solution and its external environment (including the calorimeter itself) and that the specific heat of the solution is the same as that for water\n\nIf the amount of heat absorbed by a calorimeter is too large to neglect or if we require more accurate results, then we must take into account the heat absorbed both by the solution and by the calorimeter.\n\nThe calorimeters described are designed to operate at constant (atmospheric) pressure and are convenient to measure heat flow accompanying processes that occur in solution. A different type of calorimeter that operates at constant volume, colloquially known as a bomb calorimeter, is used to measure the energy produced by reactions that yield large amounts of heat and gaseous products, such as combustion reactions. (The term \"bomb\" comes from the observation that these reactions can be vigorous enough to resemble explosions that would damage other calorimeters.) This type of calorimeter consists of a robust steel container (the \"bomb\") that contains the reactants and is itself submerged in water (Figure 9.17). The sample is placed in the bomb, which is then filled with oxygen at high pressure. A small electrical spark is used to ignite the sample. The energy produced by the reaction is absorbed by the steel bomb and the surrounding water. The temperature increase is measured and, along with the known heat capacity of the calorimeter, is used to calculate the energy produced by the reaction. Bomb calorimeters require calibration to determine the heat capacity of the calorimeter and ensure accurate results. The calibration is accomplished using a reaction with a known $q$, such as a measured quantity of benzoic acid ignited by a spark from a nickel fuse wire that is weighed before and after the reaction. The temperature change produced by the known reaction is used to determine the heat capacity of the calorimeter. The calibration is generally performed each time before the calorimeter is used to gather research data.\n\n\nFIGURE 9.17 (a) A bomb calorimeter is used to measure heat produced by reactions involving gaseous reactants or products, such as combustion. (b) The reactants are contained in the gas-tight \"bomb,\" which is submerged in water and surrounded by insulating materials. (credit a: modification of work by \"Harbor1\"/Wikimedia commons)"}
{"id": 3390, "contents": "969. LINK TO LEARNING - \nClick on this link (http://openstax.org/l/16BombCal) to view how a bomb calorimeter is prepared for action. This site (http://openstax.org/l/16Calorcalcs) shows calorimetric calculations using sample data."}
{"id": 3391, "contents": "971. Bomb Calorimetry - \nWhen 3.12 g of glucose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$, is burned in a bomb calorimeter, the temperature of the calorimeter increases from $23.8^{\\circ} \\mathrm{C}$ to $35.6^{\\circ} \\mathrm{C}$. The calorimeter contains 775 g of water, and the bomb itself has a heat\ncapacity of $893 \\mathrm{~J} /{ }^{\\circ} \\mathrm{C}$. How much heat was produced by the combustion of the glucose sample?"}
{"id": 3392, "contents": "972. Solution - \nThe combustion produces heat that is primarily absorbed by the water and the bomb. (The amounts of heat absorbed by the reaction products and the unreacted excess oxygen are relatively small and dealing with them is beyond the scope of this text. We will neglect them in our calculations.)\n\nThe heat produced by the reaction is absorbed by the water and the bomb:\n\n$$\n\\begin{aligned}\n& q_{\\mathrm{rxn}}=-\\left(q_{\\text {water }}+q_{\\text {bomb }}\\right) \\\\\n& =-\\left[\\left(4.184 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}\\right) \\times(775 \\mathrm{~g}) \\times\\left(35.6^{\\circ} \\mathrm{C}-23.8{ }^{\\circ} \\mathrm{C}\\right)+893 \\mathrm{~J} /{ }^{\\circ} \\mathrm{C} \\times\\left(35.6^{\\circ} \\mathrm{C}-23.8{ }^{\\circ} \\mathrm{C}\\right)\\right] \\\\\n& =-(38,300 \\mathrm{~J}+10,500 \\mathrm{~J}) \\\\\n& =-48,800 \\mathrm{~J}=-48.8 \\mathrm{~kJ}\n\\end{aligned}\n$$\n\nThis reaction released 48.7 kJ of heat when 3.12 g of glucose was burned."}
{"id": 3393, "contents": "973. Check Your Learning - \nWhen 0.963 g of benzene, $\\mathrm{C}_{6} \\mathrm{H}_{6}$, is burned in a bomb calorimeter, the temperature of the calorimeter increases by $8.39^{\\circ} \\mathrm{C}$. The bomb has a heat capacity of $784 \\mathrm{~J} /{ }^{\\circ} \\mathrm{C}$ and is submerged in 925 mL of water. How much heat was produced by the combustion of the benzene sample?"}
{"id": 3394, "contents": "974. Answer: - \n$\\mathrm{q}_{\\mathrm{rx}}=-39.0 \\mathrm{~kJ}$ (the reaction produced 39.0 kJ of heat)\n\nSince the first one was constructed in 1899, 35 calorimeters have been built to measure the heat produced by a living person. ${ }^{2}$ These whole-body calorimeters of various designs are large enough to hold an individual human being. More recently, whole-room calorimeters allow for relatively normal activities to be performed, and these calorimeters generate data that more closely reflect the real world. These calorimeters are used to measure the metabolism of individuals under different environmental conditions, different dietary regimes, and with different health conditions, such as diabetes.\n\nFor example Carla Prado's team at University of Alberta undertook whole-body calorimetry to understand the energy expenditures of women who had recently given birth. Studies like this help develop better recommendations and regimens for nutrition, exercise, and general wellbeing during this period of significant physiological change. In humans, metabolism is typically measured in Calories per day. A nutritional calorie (Calorie) is the energy unit used to quantify the amount of energy derived from the metabolism of foods; one Calorie is equal to 1000 calories ( 1 kcal ), the amount of energy needed to heat 1 kg of water by $1^{\\circ} \\mathrm{C}$."}
{"id": 3395, "contents": "976. Measuring Nutritional Calories - \nIn your day-to-day life, you may be more familiar with energy being given in Calories, or nutritional calories, which are used to quantify the amount of energy in foods. One calorie (cal) = exactly 4.184 joules, and one Calorie (note the capitalization) $=1000 \\mathrm{cal}$, or 1 kcal . (This is approximately the amount of energy needed to heat 1 kg of water by $1^{\\circ} \\mathrm{C}$.)\n\nThe macronutrients in food are proteins, carbohydrates, and fats or oils. Proteins provide about 4 Calories per gram, carbohydrates also provide about 4 Calories per gram, and fats and oils provide about 9 Calories/ g. Nutritional labels on food packages show the caloric content of one serving of the food, as well as the breakdown into Calories from each of the three macronutrients (Figure 9.18).\n\n[^7]\n\nFIGURE 9.18 (a) Macaroni and cheese contain energy in the form of the macronutrients in the food. (b) The food's nutritional information is shown on the package label. In the US, the energy content is given in Calories (per serving); the rest of the world usually uses kilojoules. (credit a: modification of work by \"Rex Roof\"/Flickr)\n\nFor the example shown in (b), the total energy per 228-g portion is calculated by:\n$(5 \\mathrm{~g}$ protein $\\times 4$ Calories $/ \\mathrm{g})+(31 \\mathrm{~g}$ carb $\\times 4$ Calories $/ \\mathrm{g})+(12 \\mathrm{~g}$ fat $\\times 9$ Calories $/ \\mathrm{g})=252$ Calories\nSo, you can use food labels to count your Calories. But where do the values come from? And how accurate are they? The caloric content of foods can be determined by using bomb calorimetry; that is, by burning the food and measuring the energy it contains. A sample of food is weighed, mixed in a blender, freezedried, ground into powder, and formed into a pellet. The pellet is burned inside a bomb calorimeter, and the measured temperature change is converted into energy per gram of food."}
{"id": 3396, "contents": "976. Measuring Nutritional Calories - \nToday, the caloric content on food labels is derived using a method called the Atwater system that uses the average caloric content of the different chemical constituents of food, protein, carbohydrate, and fats. The average amounts are those given in the equation and are derived from the various results given by bomb calorimetry of whole foods. The carbohydrate amount is discounted a certain amount for the fiber content, which is indigestible carbohydrate. To determine the energy content of a food, the quantities of carbohydrate, protein, and fat are each multiplied by the average Calories per gram for each and the products summed to obtain the total energy."}
{"id": 3397, "contents": "977. LINK TO LEARNING - \nClick on this link (http://openstax.org/l/16USDA) to access the US Department of Agriculture (USDA) National Nutrient Database, containing nutritional information on over 8000 foods."}
{"id": 3398, "contents": "978. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- State the first law of thermodynamics\n- Define enthalpy and explain its classification as a state function\n- Write and balance thermochemical equations\n- Calculate enthalpy changes for various chemical reactions\n- Explain Hess's law and use it to compute reaction enthalpies\n\nThermochemistry is a branch of chemical thermodynamics, the science that deals with the relationships between heat, work, and other forms of energy in the context of chemical and physical processes. As we concentrate on thermochemistry in this chapter, we need to consider some widely used concepts of thermodynamics.\n\nSubstances act as reservoirs of energy, meaning that energy can be added to them or removed from them. Energy is stored in a substance when the kinetic energy of its atoms or molecules is raised. The greater kinetic energy may be in the form of increased translations (travel or straight-line motions), vibrations, or rotations of the atoms or molecules. When thermal energy is lost, the intensities of these motions decrease and the kinetic energy falls. The total of all possible kinds of energy present in a substance is called the internal energy ( $\\boldsymbol{U}$ ), sometimes symbolized as $E$.\n\nAs a system undergoes a change, its internal energy can change, and energy can be transferred from the system to the surroundings, or from the surroundings to the system. Energy is transferred into a system when it absorbs heat $(q)$ from the surroundings or when the surroundings do work ( $w$ ) on the system. For example, energy is transferred into room-temperature metal wire if it is immersed in hot water (the wire absorbs heat from the water), or if you rapidly bend the wire back and forth (the wire becomes warmer because of the work done on it). Both processes increase the internal energy of the wire, which is reflected in an increase in the wire's temperature. Conversely, energy is transferred out of a system when heat is lost from the system, or when the system does work on the surroundings.\n\nThe relationship between internal energy, heat, and work can be represented by the equation:\n\n$$\n\\Delta U=q+w\n$$"}
{"id": 3399, "contents": "978. LEARNING OBJECTIVES - \nThe relationship between internal energy, heat, and work can be represented by the equation:\n\n$$\n\\Delta U=q+w\n$$\n\nas shown in Figure 9.19. This is one version of the first law of thermodynamics, and it shows that the internal energy of a system changes through heat flow into or out of the system (positive $q$ is heat flow in; negative $q$ is heat flow out) or work done on or by the system. The work, $w$, is positive if it is done on the system and negative if it is done by the system.\n\n\nFIGURE 9.19 The internal energy, $U$, of a system can be changed by heat flow and work. If heat flows into the system, $q_{i n}$, or work is done on the system, $w_{\\text {on }}$, its internal energy increases, $\\Delta U>0$. If heat flows out of the system, $q_{\\text {out }}$, or work is done by the system, $w_{\\text {by }}$, its internal energy decreases, $\\Delta U<0$.\nA type of work called expansion work (or pressure-volume work) occurs when a system pushes back the\nsurroundings against a restraining pressure, or when the surroundings compress the system. An example of this occurs during the operation of an internal combustion engine. The reaction of gasoline and oxygen is exothermic. Some of this energy is given off as heat, and some does work pushing the piston in the cylinder. The substances involved in the reaction are the system, and the engine and the rest of the universe are the surroundings. The system loses energy by both heating and doing work on the surroundings, and its internal energy decreases. (The engine is able to keep the car moving because this process is repeated many times per second while the engine is running.) We will consider how to determine the amount of work involved in a chemical or physical change in the chapter on thermodynamics."}
{"id": 3400, "contents": "979. LINK TO LEARNING - \nThis view of an internal combustion engine (http://openstax.org/l/16combustion) illustrates the conversion of energy produced by the exothermic combustion reaction of a fuel such as gasoline into energy of motion.\n\nAs discussed, the relationship between internal energy, heat, and work can be represented as $\\Delta U=q+w$. Internal energy is an example of a state function (or state variable), whereas heat and work are not state functions. The value of a state function depends only on the state that a system is in, and not on how that state is reached. If a quantity is not a state function, then its value does depend on how the state is reached. An example of a state function is altitude or elevation. If you stand on the summit of Mt. Kilimanjaro, you are at an altitude of 5895 m , and it does not matter whether you hiked there or parachuted there. The distance you traveled to the top of Kilimanjaro, however, is not a state function. You could climb to the summit by a direct route or by a more roundabout, circuitous path (Figure 9.20). The distances traveled would differ (distance is not a state function) but the elevation reached would be the same (altitude is a state function).\n\n\nFIGURE 9.20 Paths $X$ and $Y$ represent two different routes to the summit of Mt. Kilimanjaro. Both have the same change in elevation (altitude or elevation on a mountain is a state function; it does not depend on path), but they have very different distances traveled (distance walked is not a state function; it depends on the path). (credit: modification of work by Paul Shaffner)\n\nChemists ordinarily use a property known as enthalpy (H) to describe the thermodynamics of chemical and physical processes. Enthalpy is defined as the sum of a system's internal energy $(U)$ and the mathematical product of its pressure ( $P$ ) and volume ( $V$ ):\n\n$$\nH=U+P V\n$$"}
{"id": 3401, "contents": "979. LINK TO LEARNING - \n$$\nH=U+P V\n$$\n\nEnthalpy is also a state function. Enthalpy values for specific substances cannot be measured directly; only enthalpy changes for chemical or physical processes can be determined. For processes that take place at constant pressure (a common condition for many chemical and physical changes), the enthalpy change ( $\\boldsymbol{\\Delta H} \\boldsymbol{H})$ is:\n\n$$\n\\Delta H=\\Delta U+P \\Delta V\n$$\n\nThe mathematical product $P \\Delta V$ represents work ( $w$ ), namely, expansion or pressure-volume work as noted. By their definitions, the arithmetic signs of $\\Delta V$ and $w$ will always be opposite:\n\n$$\nP \\Delta V=-w\n$$\n\nSubstituting this equation and the definition of internal energy into the enthalpy-change equation yields:\n\n$$\n\\begin{aligned}\n& \\Delta H=\\Delta U+P \\Delta V \\\\\n& =q_{\\mathrm{p}}+w-w \\\\\n& =q_{\\mathrm{p}}\n\\end{aligned}\n$$\n\nwhere $q_{p}$ is the heat of reaction under conditions of constant pressure.\nAnd so, if a chemical or physical process is carried out at constant pressure with the only work done caused by expansion or contraction, then the heat flow $\\left(q_{p}\\right)$ and enthalpy change $(\\Delta H)$ for the process are equal.\n\nThe heat given off when you operate a Bunsen burner is equal to the enthalpy change of the methane combustion reaction that takes place, since it occurs at the essentially constant pressure of the atmosphere. On the other hand, the heat produced by a reaction measured in a bomb calorimeter (Figure 9.17) is not equal to $\\Delta H$ because the closed, constant-volume metal container prevents the pressure from remaining constant (it may increase or decrease if the reaction yields increased or decreased amounts of gaseous species). Chemists usually perform experiments under normal atmospheric conditions, at constant external pressure with $q=$ $\\Delta H$, which makes enthalpy the most convenient choice for determining heat changes for chemical reactions.\n\nThe following conventions apply when using $\\Delta H$ :"}
{"id": 3402, "contents": "979. LINK TO LEARNING - \nThe following conventions apply when using $\\Delta H$ :\n\n- A negative value of an enthalpy change, $\\Delta H<0$, indicates an exothermic reaction; a positive value, $\\Delta H>0$, indicates an endothermic reaction. If the direction of a chemical equation is reversed, the arithmetic sign of its $\\Delta H$ is changed (a process that is endothermic in one direction is exothermic in the opposite direction).\n- Chemists use a thermochemical equation to represent the changes in both matter and energy. In a thermochemical equation, the enthalpy change of a reaction is shown as a $\\Delta H$ value following the equation for the reaction. This $\\Delta H$ value indicates the amount of heat associated with the reaction involving the number of moles of reactants and products as shown in the chemical equation. For example, consider this equation:\n\n$$\n\\mathrm{H}_{2}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(l) \\quad \\Delta H=-286 \\mathrm{~kJ}\n$$\n\nThis equation indicates that when 1 mole of hydrogen gas and $\\frac{1}{2}$ mole of oxygen gas at some temperature and pressure change to 1 mole of liquid water at the same temperature and pressure, 286 kJ of heat are released to the surroundings. If the coefficients of the chemical equation are multiplied by some factor, the enthalpy change must be multiplied by that same factor ( $\\Delta H$ is an extensive property):\n(two-fold increase in amounts)"}
{"id": 3403, "contents": "979. LINK TO LEARNING - \n$$\n\\begin{array}{lr}\n2 \\mathrm{H}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(l) & \\Delta H=2 \\times(-286 \\mathrm{~kJ})=-572 \\mathrm{~kJ} \\\\\n\\text { (two-fold decrease in amounts) } & \\\\\n\\frac{1}{2} \\mathrm{H}_{2}(g)+\\frac{1}{4} \\mathrm{O}_{2}(g) \\longrightarrow \\frac{1}{2} \\mathrm{H}_{2} \\mathrm{O}(l) & \\Delta H=\\frac{1}{2} \\times(-286 \\mathrm{~kJ})=-143 \\mathrm{~kJ}\n\\end{array}\n$$\n\n- The enthalpy change of a reaction depends on the physical states of the reactants and products, so these must be shown. For example, when 1 mole of hydrogen gas and $\\frac{1}{2}$ mole of oxygen gas change to 1 mole of liquid water at the same temperature and pressure, 286 kJ of heat are released. If gaseous water forms, only 242 kJ of heat are released.\n\n$$\n\\mathrm{H}_{2}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(g) \\quad \\Delta H=-242 \\mathrm{~kJ}\n$$"}
{"id": 3404, "contents": "981. Writing Thermochemical Equations - \nWhen 0.0500 mol of $\\mathrm{HCl}(\\mathrm{aq})$ reacts with 0.0500 mol of $\\mathrm{NaOH}(\\mathrm{aq})$ to form 0.0500 mol of $\\mathrm{NaCl}(\\mathrm{aq}), 2.9 \\mathrm{~kJ}$ of heat are produced. Write a balanced thermochemical equation for the reaction of one mole of HCl .\n\n$$\n\\mathrm{HCl}(a q)+\\mathrm{NaOH}(a q) \\longrightarrow \\mathrm{NaCl}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)\n$$"}
{"id": 3405, "contents": "982. Solution - \nFor the reaction of 0.0500 mol acid $(\\mathrm{HCl}), q=-2.9 \\mathrm{~kJ}$. The reactants are provided in stoichiometric amounts (same molar ratio as in the balanced equation), and so the amount of acid may be used to calculate a molar enthalpy change. Since $\\Delta H$ is an extensive property, it is proportional to the amount of acid neutralized:\n\n$$\n\\Delta H=1 \\mathrm{molHCt} \\times \\frac{-2.9 \\mathrm{~kJ}}{0.0500 \\mathrm{molHCt}}=-58 \\mathrm{~kJ}\n$$\n\nThe thermochemical equation is then\n\n$$\n\\mathrm{HCl}(a q)+\\mathrm{NaOH}(a q) \\longrightarrow \\mathrm{NaCl}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\quad \\Delta H=-58 \\mathrm{~kJ}\n$$"}
{"id": 3406, "contents": "983. Check Your Learning - \nWhen $1.34 \\mathrm{~g} \\mathrm{Zn}(s)$ reacts with 60.0 mL of $0.750 \\mathrm{M} \\mathrm{HCl}(\\mathrm{aq}), 3.14 \\mathrm{~kJ}$ of heat are produced. Determine the enthalpy change per mole of zinc reacting for the reaction:\n\n$$\n\\mathrm{Zn}(s)+2 \\mathrm{HCl}(a q) \\longrightarrow \\mathrm{ZnCl}_{2}(a q)+\\mathrm{H}_{2}(g)\n$$"}
{"id": 3407, "contents": "984. Answer: - \n$\\Delta H=-153 \\mathrm{~kJ}$\n\nBe sure to take both stoichiometry and limiting reactants into account when determining the $\\Delta H$ for a chemical reaction."}
{"id": 3408, "contents": "986. Writing Thermochemical Equations - \nA gummy bear contains 2.67 g sucrose, $\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}$. When it reacts with 7.19 g potassium chlorate, $\\mathrm{KClO}_{3}, 43.7$ kJ of heat are produced. Write a thermochemical equation for the reaction of one mole of sucrose:\n\n$$\n\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}(a q)+8 \\mathrm{KClO}_{3}(a q) \\longrightarrow 12 \\mathrm{CO}_{2}(g)+11 \\mathrm{H}_{2} \\mathrm{O}(l)+8 \\mathrm{KCl}(a q) .\n$$"}
{"id": 3409, "contents": "987. Solution - \nUnlike the previous example exercise, this one does not involve the reaction of stoichiometric amounts of reactants, and so the limiting reactant must be identified (it limits the yield of the reaction and the amount of thermal energy produced or consumed).\n\nThe provided amounts of the two reactants are\n\n$$\n\\begin{aligned}\n& (2.67 \\mathrm{~g})(1 \\mathrm{~mol} / 342.3 \\mathrm{~g})=0.00780 \\mathrm{~mol} \\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11} \\\\\n& (7.19 \\mathrm{~g})(1 \\mathrm{~mol} / 122.5 \\mathrm{~g})=0.0587 \\mathrm{~mol} \\mathrm{KCIO}\n\\end{aligned}\n$$\n\nThe provided molar ratio of perchlorate-to-sucrose is then\n\n$$\n0.0587 \\mathrm{~mol} \\mathrm{KCIO}_{3} / 0.00780 \\mathrm{~mol} \\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}=7.52\n$$\n\nThe balanced equation indicates $8 \\mathrm{~mol}_{\\mathrm{KClO}}^{3}$ are required for reaction with $1 \\mathrm{~mol} \\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}$. Since the provided amount of $\\mathrm{KClO}_{3}$ is less than the stoichiometric amount, it is the limiting reactant and may be used to compute the enthalpy change:\n\n$$\n\\triangle \\mathrm{H}=-43.7 \\mathrm{~kJ} / 0.0587 \\mathrm{~mol} \\mathrm{KCIO}_{3}=744 \\mathrm{~kJ} / \\mathrm{mol} \\mathrm{KCIO}_{3}\n$$\n\nBecause the equation, as written, represents the reaction of $8 \\mathrm{~mol}_{\\mathrm{KClO}}^{3}$, the enthalpy change is"}
{"id": 3410, "contents": "987. Solution - \nBecause the equation, as written, represents the reaction of $8 \\mathrm{~mol}_{\\mathrm{KClO}}^{3}$, the enthalpy change is\n\n$$\n(744 \\mathrm{~kJ} / \\mathrm{mol} \\mathrm{KCIO} 3)\\left(8 \\mathrm{~mol} \\mathrm{KCIO}_{3}\\right)=5960 \\mathrm{~kJ}\n$$\n\nThe enthalpy change for this reaction is -5960 kJ , and the thermochemical equation is:\n\n$$\n\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}+8 \\mathrm{KClO}_{3} \\longrightarrow 12 \\mathrm{CO}_{2}+11 \\mathrm{H}_{2} \\mathrm{O}+8 \\mathrm{KCl} \\quad \\Delta H=-5960 \\mathrm{~kJ}\n$$"}
{"id": 3411, "contents": "988. Check Your Learning - \nWhen 1.42 g of iron reacts with 1.80 g of chlorine, 3.22 g of $\\mathrm{FeCl}_{2}(s)$ and 8.60 kJ of heat is produced. What is the enthalpy change for the reaction when 1 mole of $\\mathrm{FeCl}_{2}(s)$ is produced?"}
{"id": 3412, "contents": "989. Answer: - \n$\\Delta H=-338 \\mathrm{~kJ}$\n\nEnthalpy changes are typically tabulated for reactions in which both the reactants and products are at the same conditions. A standard state is a commonly accepted set of conditions used as a reference point for the determination of properties under other different conditions. For chemists, the IUPAC standard state refers to materials under a pressure of 1 bar and solutions at 1 M , and does not specify a temperature. Many thermochemical tables list values with a standard state of 1 atm . Because the $\\Delta H$ of a reaction changes very little with such small changes in pressure ( $1 \\mathrm{bar}=0.987 \\mathrm{~atm}$ ), $\\Delta H$ values (except for the most precisely measured values) are essentially the same under both sets of standard conditions. We will include a superscripted \" o \" in the enthalpy change symbol to designate standard state. Since the usual (but not technically standard) temperature is 298.15 K , this temperature will be assumed unless some other temperature is specified. Thus, the symbol $\\left(\\Delta H^{\\circ}\\right)$ is used to indicate an enthalpy change for a process occurring under these conditions. (The symbol $\\Delta H$ is used to indicate an enthalpy change for a reaction occurring under nonstandard conditions.)\n\nThe enthalpy changes for many types of chemical and physical processes are available in the reference literature, including those for combustion reactions, phase transitions, and formation reactions. As we discuss these quantities, it is important to pay attention to the extensive nature of enthalpy and enthalpy changes. Since the enthalpy change for a given reaction is proportional to the amounts of substances involved, it may be reported on that basis (i.e., as the $\\Delta H$ for specific amounts of reactants). However, we often find it more useful to divide one extensive property $(\\Delta H)$ by another (amount of substance), and report a per-amount intensive value of $\\Delta H$, often \"normalized\" to a per-mole basis. (Note that this is similar to determining the intensive property specific heat from the extensive property heat capacity, as seen previously.)"}
{"id": 3413, "contents": "990. Standard Enthalpy of Combustion - \nStandard enthalpy of combustion ( $\\Delta \\boldsymbol{H}_{C}^{\\circ}$ ) is the enthalpy change when 1 mole of a substance burns (combines vigorously with oxygen) under standard state conditions; it is sometimes called \"heat of combustion.\" For example, the enthalpy of combustion of ethanol, $-1366.8 \\mathrm{~kJ} / \\mathrm{mol}$, is the amount of heat produced when one mole of ethanol undergoes complete combustion at $25^{\\circ} \\mathrm{C}$ and 1 atmosphere pressure, yielding products also at $25^{\\circ} \\mathrm{C}$ and 1 atm .\n\n$$\n\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(l)+3 \\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{CO}_{2}+3 \\mathrm{H}_{2} \\mathrm{O}(l) \\quad \\Delta H^{\\circ}=-1366.8 \\mathrm{~kJ}\n$$\n\nEnthalpies of combustion for many substances have been measured; a few of these are listed in Table 9.2. Many readily available substances with large enthalpies of combustion are used as fuels, including hydrogen, carbon (as coal or charcoal), and hydrocarbons (compounds containing only hydrogen and carbon), such as methane, propane, and the major components of gasoline."}
{"id": 3414, "contents": "990. Standard Enthalpy of Combustion - \n| Substance | Combustion Reaction | Enthalpy of Combustion, $\\Delta H_{c}^{\\circ}$ <br> $\\left(\\frac{\\mathrm{kJ}}{\\mathrm{mol}}\\right.$ at $\\left.25^{\\circ} \\mathrm{C}\\right)$ |\n| :--- | :--- | :--- |\n| carbon | $\\mathrm{C}(s)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)$ | -393.5 |\n| hydrogen | $\\mathrm{H}_{2}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(l)$ | -285.8 |\n| magnesium | $\\mathrm{Mg}(s)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{MgO}(s)$ | -601.6 |\n| sulfur | $\\mathrm{S}(s)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{SO}_{2}(g)$ | -296.8 |\n| carbon <br> monoxide | $\\mathrm{CO}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)$ | -283.0 |\n| methane | $\\mathrm{CH}_{4}(g)+2 \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$ | -890.8 |\n| acetylene | $\\mathrm{C}_{2} \\mathrm{H}_{2}(g)+\\frac{5}{2} \\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{CO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l)$ | -1301.1 |\n| ethanol | $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(l)+3 \\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{CO}_{2}(g)+3 \\mathrm{H}_{2} \\mathrm{O}(l)$ | -1366.8 |"}
{"id": 3415, "contents": "990. Standard Enthalpy of Combustion - \n| methanol | $\\mathrm{CH}_{3} \\mathrm{OH}(l)+\\frac{3}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$ | -726.1 |\n| isooctane | $\\mathrm{C}_{8} \\mathrm{H}_{18}(l)+\\frac{25}{2} \\mathrm{O}_{2}(g) \\longrightarrow 8 \\mathrm{CO}_{2}(g)+9 \\mathrm{H}_{2} \\mathrm{O}(l)$ | -5461 |"}
{"id": 3416, "contents": "990. Standard Enthalpy of Combustion - \nTABLE 9.2"}
{"id": 3417, "contents": "992. Using Enthalpy of Combustion - \nAs Figure 9.21 suggests, the combustion of gasoline is a highly exothermic process. Let us determine the approximate amount of heat produced by burning 1.00 L of gasoline, assuming the enthalpy of combustion of gasoline is the same as that of isooctane, a common component of gasoline. The density of isooctane is 0.692 $\\mathrm{g} / \\mathrm{mL}$.\n\n\nFIGURE 9.21 The combustion of gasoline is very exothermic. (credit: modification of work by \"AlexEagle\"/Flickr)"}
{"id": 3418, "contents": "993. Solution - \nStarting with a known amount (1.00 L of isooctane), we can perform conversions between units until we arrive at the desired amount of heat or energy. The enthalpy of combustion of isooctane provides one of the\nnecessary conversions. Table 9.2 gives this value as -5460 kJ per 1 mole of isooctane $\\left(\\mathrm{C}_{8} \\mathrm{H}_{18}\\right)$.\nUsing these data,\n\n\nThe combustion of 1.00 L of isooctane produces $33,100 \\mathrm{~kJ}$ of heat. (This amount of energy is enough to melt 99.2 kg , or about 218 lbs , of ice.)\n\nNote: If you do this calculation one step at a time, you would find:\n\n$$\n\\begin{aligned}\n& 1.00 \\mathrm{LC}_{8} \\mathrm{H}_{18} \\longrightarrow 1.00 \\times 10^{3} \\mathrm{~mL} \\mathrm{C}_{8} \\mathrm{H}_{18} \\\\\n& 1.00 \\times 10^{3} \\mathrm{mLC}_{8} \\mathrm{H}_{18} \\longrightarrow 692 \\mathrm{~g} \\mathrm{C}_{8} \\mathrm{H}_{18} \\\\\n& 692 \\mathrm{~g} \\mathrm{C}_{8} \\mathrm{H}_{18} \\longrightarrow 6.07 \\mathrm{~mol} \\mathrm{C}_{8} \\mathrm{H}_{18} \\\\\n& 6.07 \\mathrm{~mol} \\mathrm{C}_{8} \\mathrm{H}_{18} \\longrightarrow-3.31 \\times 10^{4} \\mathrm{~kJ}\n\\end{aligned}\n$$"}
{"id": 3419, "contents": "994. Check Your Learning - \nHow much heat is produced by the combustion of 125 g of acetylene?"}
{"id": 3420, "contents": "995. Answer: - \n$6.25 \\times 10^{3} \\mathrm{~kJ}$"}
{"id": 3421, "contents": "997. Emerging Algae-Based Energy Technologies (Biofuels) - \nAs reserves of fossil fuels diminish and become more costly to extract, the search is ongoing for replacement fuel sources for the future. Among the most promising biofuels are those derived from algae (Figure 9.22). The species of algae used are nontoxic, biodegradable, and among the world's fastest growing organisms. About $50 \\%$ of algal weight is oil, which can be readily converted into fuel such as biodiesel. Algae can yield 26,000 gallons of biofuel per hectare-much more energy per acre than other crops. Some strains of algae can flourish in brackish water that is not usable for growing other crops. Algae can produce biodiesel, biogasoline, ethanol, butanol, methane, and even jet fuel.\n\n\nFIGURE 9.22 (a) Tiny algal organisms can be (b) grown in large quantities and eventually (c) turned into a useful fuel such as biodiesel. (credit a: modification of work by Micah Sittig; credit b: modification of work by Robert Kerton; credit c: modification of work by John F. Williams)\n\nAccording to the US Department of Energy, only 39,000 square kilometers (about 0.4\\% of the land mass of the US or less than $\\frac{1}{7}$ of the area used to grow corn) can produce enough algal fuel to replace all the petroleum-based fuel used in the US. The cost of algal fuels is becoming more competitive-for instance, the US Air Force is producing jet fuel from algae at a total cost of under $\\$ 5$ per gallon. ${ }^{\\frac{3}{3}}$ The process used to produce algal fuel is as follows: grow the algae (which use sunlight as their energy source and $\\mathrm{CO}_{2}$ as a raw material); harvest the algae; extract the fuel compounds (or precursor compounds); process as necessary (e.g., perform a transesterification reaction to make biodiesel); purify; and distribute (Figure 9.23).\n\n\nFIGURE 9.23 Algae convert sunlight and carbon dioxide into oil that is harvested, extracted, purified, and transformed into a variety of renewable fuels."}
{"id": 3422, "contents": "998. LINK TO LEARNING - \nClick here (http://openstax.org/l/16biofuel) to learn more about the process of creating algae biofuel."}
{"id": 3423, "contents": "999. Standard Enthalpy of Formation - \nA standard enthalpy of formation $\\Delta H_{\\mathrm{f}}^{\\circ}$ is an enthalpy change for a reaction in which exactly 1 mole of a pure substance is formed from free elements in their most stable states under standard state conditions. These values are especially useful for computing or predicting enthalpy changes for chemical reactions that are impractical or dangerous to carry out, or for processes for which it is difficult to make measurements. If we have values for the appropriate standard enthalpies of formation, we can determine the enthalpy change for any reaction, which we will practice in the next section on Hess's law.\n\nThe standard enthalpy of formation of $\\mathrm{CO}_{2}(\\mathrm{~g})$ is $-393.5 \\mathrm{~kJ} / \\mathrm{mol}$. This is the enthalpy change for the exothermic reaction:\n\n$$\n\\mathrm{C}(s)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g) \\quad \\Delta H_{\\mathrm{f}}^{\\circ}=\\Delta H^{\\circ}=-393.5 \\mathrm{~kJ}\n$$\n\nstarting with the reactants at a pressure of 1 atm and $25^{\\circ} \\mathrm{C}$ (with the carbon present as graphite, the most stable form of carbon under these conditions) and ending with one mole of $\\mathrm{CO}_{2}$, also at 1 atm and $25^{\\circ} \\mathrm{C}$. For nitrogen dioxide, $\\mathrm{NO}_{2}(\\mathrm{~g}), \\Delta \\boldsymbol{H}_{\\mathrm{f}}^{\\circ}$ is $33.2 \\mathrm{~kJ} / \\mathrm{mol}$. This is the enthalpy change for the reaction:\n\n$$\n\\frac{1}{2} \\mathrm{~N}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{NO}_{2}(g) \\quad \\Delta H_{\\mathrm{f}}^{\\circ}=\\Delta H^{\\circ}=+33.2 \\mathrm{~kJ}\n$$"}
{"id": 3424, "contents": "999. Standard Enthalpy of Formation - \nA reaction equation with $\\frac{1}{2}$ mole of $\\mathrm{N}_{2}$ and 1 mole of $\\mathrm{O}_{2}$ is correct in this case because the standard enthalpy of formation always refers to 1 mole of product, $\\mathrm{NO}_{2}(g)$.\n\nYou will find a table of standard enthalpies of formation of many common substances in Appendix G. These values indicate that formation reactions range from highly exothermic (such as $-2984 \\mathrm{~kJ} / \\mathrm{mol}$ for the formation of $\\mathrm{P}_{4} \\mathrm{O}_{10}$ ) to strongly endothermic (such as $+226.7 \\mathrm{~kJ} / \\mathrm{mol}$ for the formation of acetylene, $\\mathrm{C}_{2} \\mathrm{H}_{2}$ ). By definition, the standard enthalpy of formation of an element in its most stable form is equal to zero under\n\n3 For more on algal fuel, see http://www.theguardian.com/environment/2010/feb/13/algae-solve-pentagon-fuel-problem.\nstandard conditions, which is 1 atm for gases and 1 M for solutions."}
{"id": 3425, "contents": "1001. Evaluating an Enthalpy of Formation - \nOzone, $\\mathrm{O}_{3}(\\mathrm{~g})$, forms from oxygen, $\\mathrm{O}_{2}(\\mathrm{~g})$, by an endothermic process. Ultraviolet radiation is the source of the energy that drives this reaction in the upper atmosphere. Assuming that both the reactants and products of the reaction are in their standard states, determine the standard enthalpy of formation, $\\Delta \\boldsymbol{H}_{\\mathrm{f}}^{\\circ}$ of ozone from the following information:\n\n$$\n3 \\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{O}_{3}(g) \\quad \\Delta H^{\\circ}=+286 \\mathrm{~kJ}\n$$"}
{"id": 3426, "contents": "1002. Solution - \n$\\Delta H_{\\mathrm{f}}^{\\circ}$ is the enthalpy change for the formation of one mole of a substance in its standard state from the elements in their standard states. Thus, $\\Delta H_{\\mathrm{f}}^{\\circ}$ for $\\mathrm{O}_{3}(g)$ is the enthalpy change for the reaction:\n\n$$\n\\frac{3}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{O}_{3}(g)\n$$\n\nFor the formation of 2 mol of $\\mathrm{O}_{3}(\\mathrm{~g}), \\Delta H^{\\circ}=+286 \\mathrm{~kJ}$. This ratio, $\\left(\\frac{286 \\mathrm{~kJ}}{2 \\mathrm{~mol} \\mathrm{O}_{3}}\\right)$, can be used as a conversion factor to find the heat produced when 1 mole of $\\mathrm{O}_{3}(\\mathrm{~g})$ is formed, which is the enthalpy of formation for $\\mathrm{O}_{3}(\\mathrm{~g})$ :\n\n$$\n\\Delta H^{\\circ} \\text { for } 1 \\text { mole of } \\mathrm{O}_{3}(g)=1 \\mathrm{mel}_{3} \\times \\frac{286 \\mathrm{~kJ}}{2 \\mathrm{~mol} \\Theta_{3}}=143 \\mathrm{~kJ}\n$$\n\nTherefore, $\\Delta H_{\\mathrm{f}}^{\\circ}\\left[\\mathrm{O}_{3}(g)\\right]=+143 \\mathrm{~kJ} / \\mathrm{mol}$."}
{"id": 3427, "contents": "1003. Check Your Learning - \nHydrogen gas, $\\mathrm{H}_{2}$, reacts explosively with gaseous chlorine, $\\mathrm{Cl}_{2}$, to form hydrogen chloride, $\\mathrm{HCl}(g)$. What is the enthalpy change for the reaction of 1 mole of $\\mathrm{H}_{2}(\\mathrm{~g})$ with 1 mole of $\\mathrm{Cl}_{2}(\\mathrm{~g})$ if both the reactants and products are at standard state conditions? The standard enthalpy of formation of $\\mathrm{HCl}(\\mathrm{g})$ is $-92.3 \\mathrm{~kJ} / \\mathrm{mol}$."}
{"id": 3428, "contents": "1004. Answer: - \nFor the reaction $\\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{HCl}(g) \\quad \\Delta H^{\\circ}=-184.6 \\mathrm{~kJ}$"}
{"id": 3429, "contents": "1006. Writing Reaction Equations for $\\Delta H_{\\mathrm{f}}^{\\circ}$ - \nWrite the heat of formation reaction equations for:\n(a) $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(\\mathrm{I})$\n(b) $\\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}(s)$"}
{"id": 3430, "contents": "1007. Solution - \nRemembering that $\\Delta H_{\\mathrm{f}}^{\\circ}$ reaction equations are for forming 1 mole of the compound from its constituent elements under standard conditions, we have:\n(a) $2 \\mathrm{C}(s$, graphite $)+3 \\mathrm{H}_{2}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(l)$\n(b) $3 \\mathrm{Ca}(s)+\\frac{1}{2} \\mathrm{P}_{4}(s)+4 \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}(s)$\n\nNote: The standard state of carbon is graphite, and phosphorus exists as $\\mathrm{P}_{4}$."}
{"id": 3431, "contents": "1008. Check Your Learning - \nWrite the heat of formation reaction equations for:\n(a) $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OC}_{2} \\mathrm{H}_{5}($ I $)$\n(b) $\\mathrm{Na}_{2} \\mathrm{CO}_{3}(s)$"}
{"id": 3432, "contents": "1009. Answer: - \n(a) $4 \\mathrm{C}\\left(s\\right.$, graphite) $+5 \\mathrm{H}_{2}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OC}_{2} \\mathrm{H}_{5}(l)$; (b)\n$2 \\mathrm{Na}(s)+\\mathrm{C}(s$, graphite $)+\\frac{3}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{Na}_{2} \\mathrm{CO}_{3}(s)$"}
{"id": 3433, "contents": "1010. Hess's Law - \nThere are two ways to determine the amount of heat involved in a chemical change: measure it experimentally, or calculate it from other experimentally determined enthalpy changes. Some reactions are difficult, if not impossible, to investigate and make accurate measurements for experimentally. And even when a reaction is not hard to perform or measure, it is convenient to be able to determine the heat involved in a reaction without having to perform an experiment.\n\nThis type of calculation usually involves the use of Hess's law, which states: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps. Hess's law is valid because enthalpy is a state function: Enthalpy changes depend only on where a chemical process starts and ends, but not on the path it takes from start to finish. For example, we can think of the reaction of carbon with oxygen to form carbon dioxide as occurring either directly or by a two-step process. The direct process is written:\n\n$$\n\\mathrm{C}(s)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g) \\quad \\Delta H^{\\circ}=-394 \\mathrm{~kJ}\n$$\n\nIn the two-step process, first carbon monoxide is formed:\n\n$$\n\\mathrm{C}(s)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}(g) \\quad \\Delta H^{\\circ}=-111 \\mathrm{~kJ}\n$$\n\nThen, carbon monoxide reacts further to form carbon dioxide:\n\n$$\n\\mathrm{CO}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g) \\quad \\Delta H^{\\circ}=-283 \\mathrm{~kJ}\n$$\n\nThe equation describing the overall reaction is the sum of these two chemical changes:"}
{"id": 3434, "contents": "1010. Hess's Law - \nThe equation describing the overall reaction is the sum of these two chemical changes:\n\n$$\n\\begin{aligned}\n& \\text { Step 1: } \\mathrm{C}(s)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}(g) \\\\\n& \\text { Step 2: } \\mathrm{CO}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g) \\\\\n& \\text { Sum: } \\mathrm{C}(s)+\\frac{1}{2} \\mathrm{O}_{2}(g)+\\mathrm{CO}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}(g)+\\mathrm{CO}_{2}(g)\n\\end{aligned}\n$$\n\nBecause the CO produced in Step 1 is consumed in Step 2, the net change is:\n\n$$\n\\mathrm{C}(s)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)\n$$\n\nAccording to Hess's law, the enthalpy change of the reaction will equal the sum of the enthalpy changes of the steps.\n\n$$\n\\begin{array}{ll}\n\\mathrm{C}(s)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}(g) & \\Delta H^{\\circ}=-111 \\mathrm{~kJ} \\\\\n\\frac{\\mathrm{CO}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)}{\\mathrm{C}(s)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)} & \\frac{\\Delta H^{\\circ}=-283 \\mathrm{~kJ}}{\\Delta H^{\\circ}=-394 \\mathrm{~kJ}}\n\\end{array}\n$$"}
{"id": 3435, "contents": "1010. Hess's Law - \nThe result is shown in Figure 9.24. We see that $\\Delta H$ of the overall reaction is the same whether it occurs in one step or two. This finding (overall $\\Delta H$ for the reaction = sum of $\\Delta H$ values for reaction \"steps\" in the overall reaction) is true in general for chemical and physical processes.\n\n\nFIGURE 9.24 The formation of $\\mathrm{CO}_{2}(g)$ from its elements can be thought of as occurring in two steps, which sum to the overall reaction, as described by Hess's law. The horizontal blue lines represent enthalpies. For an exothermic process, the products are at lower enthalpy than are the reactants.\n\nBefore we further practice using Hess's law, let us recall two important features of $\\Delta H$.\n\n1. $\\Delta H$ is directly proportional to the quantities of reactants or products. For example, the enthalpy change for the reaction forming 1 mole of $\\mathrm{NO}_{2}(g)$ is +33.2 kJ :\n\n$$\n\\frac{1}{2} \\mathrm{~N}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{NO}_{2}(g) \\quad \\Delta H=+33.2 \\mathrm{~kJ}\n$$\n\nWhen 2 moles of $\\mathrm{NO}_{2}$ (twice as much) are formed, the $\\Delta H$ will be twice as large:\n\n$$\n\\mathrm{N}_{2}(g)+2 \\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{NO}_{2}(g) \\quad \\Delta H=+66.4 \\mathrm{~kJ}\n$$\n\nIn general, if we multiply or divide an equation by a number, then the enthalpy change should also be multiplied or divided by the same number.\n2. $\\Delta H$ for a reaction in one direction is equal in magnitude and opposite in sign to $\\Delta H$ for the reaction in the reverse direction. For example, given that:\n\n$$\n\\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{HCl}(g) \\quad \\Delta H=-184.6 \\mathrm{~kJ}\n$$"}
{"id": 3436, "contents": "1010. Hess's Law - \n$$\n\\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{HCl}(g) \\quad \\Delta H=-184.6 \\mathrm{~kJ}\n$$\n\nThen, for the \"reverse\" reaction, the enthalpy change is also \"reversed\":\n\n$$\n2 \\mathrm{HCl}(g) \\longrightarrow \\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g) \\quad \\Delta H=+184.6 \\mathrm{~kJ}\n$$"}
{"id": 3437, "contents": "1012. Stepwise Calculation of $\\Delta H_{\\mathrm{f}}^{\\circ}$ Using Hess's Law - \nDetermine the enthalpy of formation, $\\Delta H_{f}^{\\circ}$, of $\\mathrm{FeCl}_{3}(s)$ from the enthalpy changes of the following two-step process that occurs under standard state conditions:\n\n$$\n\\begin{array}{cr}\n\\mathrm{Fe}(s)+\\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{FeCl}_{2}(s) & \\Delta H^{\\circ}=-341.8 \\mathrm{~kJ} \\\\\n\\mathrm{FeCl}_{2}(s)+\\frac{1}{2} \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{FeCl}_{3}(s) & \\Delta H^{\\circ}=-57.7 \\mathrm{~kJ}\n\\end{array}\n$$"}
{"id": 3438, "contents": "1013. Solution - \nWe are trying to find the standard enthalpy of formation of $\\mathrm{FeCl}_{3}(s)$, which is equal to $\\Delta H^{\\circ}$ for the reaction:\n\n$$\n\\mathrm{Fe}(s)+\\frac{3}{2} \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{FeCl}_{3}(s) \\quad \\Delta H_{\\mathrm{f}}^{\\circ}=?\n$$\n\nLooking at the reactions, we see that the reaction for which we want to find $\\Delta H^{\\circ}$ is the sum of the two reactions with known $\\Delta H$ values, so we must sum their $\\Delta H \\mathrm{~s}$ :\n\n$$\n\\begin{array}{ll}\n\\mathrm{Fe}(s)+\\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{FeCl}_{2}(s) & \\Delta H^{\\circ}=-341.8 \\mathrm{~kJ} \\\\\n\\frac{\\mathrm{FeCl}_{2}(s)+\\frac{1}{2} \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{FeCl}_{3}(s)}{\\mathrm{Fe}(s)+\\frac{3}{2} \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{FeCl}_{3}(s)} & \\frac{\\Delta H^{\\circ}=-57.7 \\mathrm{~kJ}}{\\Delta H^{\\circ}=-399.5 \\mathrm{~kJ}}\n\\end{array}\n$$\n\nThe enthalpy of formation, $\\Delta \\boldsymbol{H}_{\\mathrm{f}}^{\\circ}$, of $\\mathrm{FeCl}_{3}(s)$ is $-399.5 \\mathrm{~kJ} / \\mathrm{mol}$."}
{"id": 3439, "contents": "1014. Check Your Learning - \nCalculate $\\Delta H$ for the process:\n\n$$\n\\mathrm{N}_{2}(g)+2 \\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{NO}_{2}(g)\n$$\n\nfrom the following information:\n\n$$\n\\begin{aligned}\n\\mathrm{N}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{NO}(g) & \\Delta H=180.5 \\mathrm{~kJ} \\\\\n\\mathrm{NO}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{NO}_{2}(g) & \\Delta H=-57.06 \\mathrm{~kJ}\n\\end{aligned}\n$$"}
{"id": 3440, "contents": "1015. Answer: - \n66.4 kJ\n\nHere is a less straightforward example that illustrates the thought process involved in solving many Hess's law problems. It shows how we can find many standard enthalpies of formation (and other values of $\\Delta H$ ) if they are difficult to determine experimentally."}
{"id": 3441, "contents": "1017. A More Challenging Problem Using Hess's Law - \nChlorine monofluoride can react with fluorine to form chlorine trifluoride:\n(i) $\\mathrm{ClF}(g)+\\mathrm{F}_{2}(g) \\longrightarrow \\mathrm{ClF}_{3}(g)$\n$\\Delta H^{\\circ}=$ ?\n\nUse the reactions here to determine the $\\Delta H^{\\circ}$ for reaction (i):\n\n$$\n\\text { (ii) } 2 \\mathrm{OF}_{2}(g) \\longrightarrow \\mathrm{O}_{2}(g)+2 \\mathrm{~F}_{2}(g) \\quad \\Delta H_{(i i)}^{\\circ}=-49.4 \\mathrm{~kJ}\n$$\n\n$$\n\\text { (iii) } 2 \\mathrm{ClF}(g)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{Cl}_{2} \\mathrm{O}(g)+\\mathrm{OF}_{2}(g) \\quad \\Delta H_{(i i i)}^{\\circ}=+214.0 \\mathrm{~kJ}\n$$\n\n$$\n\\text { (iv) } \\mathrm{ClF}_{3}(g)+\\mathrm{O}_{2}(g) \\longrightarrow \\frac{1}{2} \\mathrm{Cl}_{2} \\mathrm{O}(g)+\\frac{3}{2} \\mathrm{OF}_{2}(g) \\quad \\Delta H_{(i v)}^{\\circ}=+236.2 \\mathrm{~kJ}\n$$"}
{"id": 3442, "contents": "1018. Solution - \nOur goal is to manipulate and combine reactions (ii), (iii), and (iv) such that they add up to reaction (i). Going from left to right in (i), we first see that $\\operatorname{ClF}(g)$ is needed as a reactant. This can be obtained by multiplying reaction (iii) by $\\frac{1}{2}$, which means that the $\\Delta H^{\\circ}$ change is also multiplied by $\\frac{1}{2}$ :\n\n$$\n\\mathrm{ClF}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\frac{1}{2} \\mathrm{Cl}_{2} \\mathrm{O}(g)+\\frac{1}{2} \\mathrm{OF}_{2}(g) \\quad \\Delta H^{\\circ}=\\frac{1}{2}(214.0)=+107.0 \\mathrm{~kJ}\n$$\n\nNext, we see that $\\mathrm{F}_{2}$ is also needed as a reactant. To get this, reverse and halve reaction (ii), which means that the $\\Delta H^{\\circ}$ changes sign and is halved:\n\n$$\n\\frac{1}{2} \\mathrm{O}_{2}(g)+\\mathrm{F}_{2}(g) \\longrightarrow \\mathrm{OF}_{2}(g) \\quad \\Delta H^{\\circ}=+24.7 \\mathrm{~kJ}\n$$\n\nTo get $\\mathrm{ClF}_{3}$ as a product, reverse (iv), changing the sign of $\\Delta H^{\\circ}$ :\n\n$$\n\\frac{1}{2} \\mathrm{Cl}_{2} \\mathrm{O}(g)+\\frac{3}{2} \\mathrm{OF}_{2}(g) \\longrightarrow \\mathrm{ClF}_{3}(g)+\\mathrm{O}_{2}(g) \\quad \\Delta H^{\\circ}=-236.2 \\mathrm{~kJ}\n$$\n\nNow check to make sure that these reactions add up to the reaction we want:"}
{"id": 3443, "contents": "1018. Solution - \nNow check to make sure that these reactions add up to the reaction we want:\n\n$$\n\\begin{array}{ll}\n\\mathrm{ClF}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\frac{1}{2} \\mathrm{Cl}_{2} \\mathrm{O}(g)+\\frac{1}{2} \\mathrm{OF}_{2}(g) & \\Delta H^{\\circ}=+107.0 \\mathrm{~kJ} \\\\\n\\frac{1}{2} \\mathrm{O}_{2}(g)+\\mathrm{F}_{2}(g) \\longrightarrow \\mathrm{OF}_{2}(g) & \\Delta H^{\\circ}=+24.7 \\mathrm{~kJ} \\\\\n\\frac{1}{2} \\mathrm{Cl}_{2} \\mathrm{O}(g)+\\frac{3}{2} \\mathrm{OF}_{2}(g) \\longrightarrow \\mathrm{ClF}_{3}(g)+\\mathrm{O}_{2}(g) & \\frac{\\Delta H^{\\circ}=-236.2 \\mathrm{~kJ}}{\\Delta H^{\\circ}=-104.5 \\mathrm{~kJ}}\n\\end{array}\n$$\n\nReactants $\\frac{1}{2} \\mathrm{O}_{2}$ and $\\frac{1}{2} \\mathrm{O}_{2}$ cancel out product $\\mathrm{O}_{2}$; product $\\frac{1}{2} \\mathrm{Cl}_{2} \\mathrm{O}$ cancels reactant $\\frac{1}{2} \\mathrm{Cl}_{2} \\mathrm{O}$; and reactant $\\frac{3}{2} \\mathrm{OF}_{2}$ is cancelled by products $\\frac{1}{2} \\mathrm{OF}_{2}$ and $\\mathrm{OF}_{2}$. This leaves only reactants $\\mathrm{ClF}(g)$ and $\\mathrm{F}_{2}(g)$ and product $\\mathrm{ClF}_{3}(g)$, which are what we want. Since summing these three modified reactions yields the reaction of interest, summing the three modified $\\Delta H^{\\circ}$ values will give the desired $\\Delta H^{\\circ}$ :"}
{"id": 3444, "contents": "1018. Solution - \n$$\n\\Delta H^{\\circ}=(+107.0 \\mathrm{~kJ})+(24.7 \\mathrm{~kJ})+(-236.2 \\mathrm{~kJ})=-104.5 \\mathrm{~kJ}\n$$"}
{"id": 3445, "contents": "1019. Check Your Learning - \nAluminum chloride can be formed from its elements:\n(i) $2 \\mathrm{Al}(s)+3 \\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{AlCl}_{3}(s) \\quad \\Delta H^{\\circ}=$ ?\n\nUse the reactions here to determine the $\\Delta H^{\\circ}$ for reaction (i):\n(ii) $\\mathrm{HCl}(g) \\longrightarrow \\mathrm{HCl}(a q) \\quad \\Delta H_{(i i)}^{\\circ}=-74.8 \\mathrm{~kJ}$\n(iii) $\\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{HCl}(g) \\quad \\Delta H_{(i i i)}^{\\circ}=-185 \\mathrm{~kJ}$\n(iv) $\\mathrm{AlCl}_{3}(a q) \\longrightarrow \\mathrm{AlCl}_{3}(s) \\quad \\Delta H_{\\text {(iv) }}^{\\circ}=+323 \\mathrm{~kJ} / \\mathrm{mol}$\n(v) $2 \\mathrm{Al}(s)+6 \\mathrm{HCl}(a q) \\longrightarrow 2 \\mathrm{AlCl}_{3}(a q)+3 \\mathrm{H}_{2}(g) \\quad \\Delta H_{(v)}^{\\circ}=-1049 \\mathrm{~kJ}$"}
{"id": 3446, "contents": "1020. Answer: - \n-1407 kJ\n\nWe also can use Hess's law to determine the enthalpy change of any reaction if the corresponding enthalpies of formation of the reactants and products are available. The stepwise reactions we consider are: (i) decompositions of the reactants into their component elements (for which the enthalpy changes are proportional to the negative of the enthalpies of formation of the reactants), followed by (ii) re-combinations of the elements to give the products (with the enthalpy changes proportional to the enthalpies of formation of the products). The standard enthalpy change of the overall reaction is therefore equal to: (ii) the sum of the standard enthalpies of formation of all the products plus (i) the sum of the negatives of the standard enthalpies of formation of the reactants. This is usually rearranged slightly to be written as follows, with $\\sum$ representing \"the sum of\" and $n$ standing for the stoichiometric coefficients:\n\n$$\n\\Delta H_{\\text {reaction }}^{\\circ}=\\sum n \\times \\Delta H_{\\mathrm{f}}^{\\circ}(\\text { products })-\\sum n \\times \\Delta H_{\\mathrm{f}}^{\\circ}(\\text { reactants })\n$$\n\nThe following example shows in detail why this equation is valid, and how to use it to calculate the enthalpy change for a reaction of interest.\n\nEXAMPLE 9.15"}
{"id": 3447, "contents": "1021. Using Hess's Law - \nWhat is the standard enthalpy change for the reaction:\n\n$$\n3 \\mathrm{NO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{HNO}_{3}(a q)+\\mathrm{NO}(g) \\quad \\Delta H^{\\circ}=?\n$$"}
{"id": 3448, "contents": "1022. Solution: Using the Equation - \nUse the special form of Hess's law given previously, and values from Appendix G:\n\n$$\n\\left.\\left.\\begin{array}{rl}\n& \\Delta H_{\\text {reaction }}^{\\circ}=\\sum n \\times \\Delta H_{\\mathrm{f}}^{\\circ}(\\text { products })-\\sum n \\times \\Delta H_{\\mathrm{f}}^{\\circ}(\\text { reactants }) \\\\\n= & {[2 \\mathrm{molHNO}(\\mathrm{qq})}\n\\end{array} \\times \\frac{-207.4 \\mathrm{~kJ}}{m+1 \\mathrm{mNO}(\\mathrm{~mol})}+1 \\mathrm{NO}(\\mathrm{~g})-\\times \\frac{+90.2 \\mathrm{~kJ}}{\\mathrm{molNO}(\\mathrm{~g})}\\right]\\right] .\n$$"}
{"id": 3449, "contents": "1023. Solution: Supporting Why the General Equation Is Valid - \nAlternatively, we can write this reaction as the sum of the decompositions of $3 \\mathrm{NO}_{2}(g)$ and $1 \\mathrm{H}_{2} \\mathrm{O}(I)$ into their constituent elements, and the formation of $2 \\mathrm{HNO}_{3}(\\mathrm{aq})$ and $1 \\mathrm{NO}(g)$ from their constituent elements. Writing out these reactions, and noting their relationships to the $\\Delta H_{\\mathrm{f}}^{\\circ}$ values for these compounds (from Appendix G), we have:\n\n$$\n\\begin{gathered}\n3 \\mathrm{NO}_{2}(g) \\longrightarrow 3 / 2 \\mathrm{~N}_{2}(g)+3 \\mathrm{O}_{2}(g) \\quad \\Delta H_{1}^{\\circ}=-99.6 \\mathrm{~kJ} \\\\\n\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{H}_{2}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\quad \\Delta H_{2}^{\\circ}=+285.8 \\mathrm{~kJ}\\left[-1 \\times \\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)\\right] \\\\\n\\mathrm{H}_{2}(g)+\\mathrm{N}_{2}(g)+3 \\mathrm{O}_{2}(\\mathrm{~g}) \\longrightarrow 2 \\mathrm{HNO}_{3}(a q) \\quad \\Delta H_{3}^{\\circ}=-414.8 \\mathrm{~kJ}\\left[2 \\times \\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{HNO}_{3}\\right)\\right] \\\\\n\\frac{1}{2} \\mathrm{~N}_{2}(g)+\\frac{1}{2} \\mathrm{O}_{2}(\\mathrm{~g}) \\longrightarrow \\mathrm{NO}(\\mathrm{~g})\n\\end{gathered} \\Delta H_{4}^{\\circ}=+90.2 \\mathrm{~kJ}[1 \\times(\\mathrm{NO})] \\mathrm{l}\n$$\n\nSumming these reaction equations gives the reaction we are interested in:"}
{"id": 3450, "contents": "1023. Solution: Supporting Why the General Equation Is Valid - \nSumming these reaction equations gives the reaction we are interested in:\n\n$$\n3 \\mathrm{NO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{HNO}_{3}(a q)+\\mathrm{NO}(g)\n$$\n\nSumming their enthalpy changes gives the value we want to determine:\n\n$$\n\\begin{aligned}\n\\Delta H_{\\mathrm{rxn}}^{\\circ} & =\\Delta H_{1}^{\\circ}+\\Delta H_{2}^{\\circ}+\\Delta H_{3}^{\\circ}+\\Delta H_{4}^{\\circ}=(-99.6 \\mathrm{~kJ})+(+285.8 \\mathrm{~kJ})+(-414.8 \\mathrm{~kJ})+(+90.2 \\mathrm{~kJ}) \\\\\n& =-138.4 \\mathrm{~kJ}\n\\end{aligned}\n$$\n\nSo the standard enthalpy change for this reaction is $\\Delta H^{\\circ}=-138.4 \\mathrm{~kJ}$.\nNote that this result was obtained by (1) multiplying the $\\Delta H_{\\mathrm{f}}^{\\circ}$ of each product by its stoichiometric coefficient and summing those values, (2) multiplying the $\\Delta H_{\\mathrm{f}}^{\\circ}$ of each reactant by its stoichiometric coefficient and summing those values, and then (3) subtracting the result found in (2) from the result found in (1). This is also the procedure in using the general equation, as shown."}
{"id": 3451, "contents": "1024. Check Your Learning - \nCalculate the heat of combustion of 1 mole of ethanol, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(I)$, when $\\mathrm{H}_{2} \\mathrm{O}(I)$ and $\\mathrm{CO}_{2}(g)$ are formed. Use the following enthalpies of formation: $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(\\mathrm{I}),-278 \\mathrm{~kJ} / \\mathrm{mol} ; \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{I}),-286 \\mathrm{~kJ} / \\mathrm{mol}$; and $\\mathrm{CO}_{2}(\\mathrm{~g}),-394 \\mathrm{~kJ} / \\mathrm{mol}$."}
{"id": 3452, "contents": "1025. Answer: - \n-1368 kJ/mol"}
{"id": 3453, "contents": "1026. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the energetics of covalent and ionic bond formation and breakage\n- Use the Born-Haber cycle to compute lattice energies for ionic compounds\n- Use average covalent bond energies to estimate enthalpies of reaction\n\nA bond's strength describes how strongly each atom is joined to another atom, and therefore how much energy is required to break the bond between the two atoms.\n\nIt is essential to remember that energy must be added to break chemical bonds (an endothermic process), whereas forming chemical bonds releases energy (an exothermic process). In the case of $\\mathrm{H}_{2}$, the covalent bond is very strong; a large amount of energy, 436 kJ , must be added to break the bonds in one mole of hydrogen molecules and cause the atoms to separate:\n\n$$\n\\mathrm{H}_{2}(\\mathrm{~g}) \\longrightarrow 2 \\mathrm{H}(\\mathrm{~g}) \\quad \\text { bond energy }=436 \\mathrm{~kJ}\n$$\n\nConversely, the same amount of energy is released when one mole of $\\mathrm{H}_{2}$ molecules forms from two moles of H atoms:\n\n$$\n2 \\mathrm{H}(g) \\longrightarrow \\mathrm{H}_{2}(g) \\quad \\text { bond energy }=-436 \\mathrm{~kJ}\n$$"}
{"id": 3454, "contents": "1027. Bond Strength: Covalent Bonds - \nStable molecules exist because covalent bonds hold the atoms together. We measure the strength of a covalent bond by the energy required to break it, that is, the energy necessary to separate the bonded atoms. Separating any pair of bonded atoms requires energy (see Figure 4.4). The stronger a bond, the greater the energy required to break it.\n\nThe energy required to break a specific covalent bond in one mole of gaseous molecules is called the bond energy or the bond dissociation energy. The bond energy for a diatomic molecule, $\\mathrm{D}_{\\mathrm{X}-\\mathrm{Y}}$, is defined as the standard enthalpy change for the endothermic reaction:\n\n$$\n\\mathrm{XY}(g) \\longrightarrow \\mathrm{X}(g)+\\mathrm{Y}(g) \\quad \\mathrm{D}_{\\mathrm{X}-\\mathrm{Y}}=\\Delta H^{\\circ}\n$$\n\nFor example, the bond energy of the pure covalent $\\mathrm{H}-\\mathrm{H}$ bond, $\\mathrm{D}_{\\mathrm{H}-\\mathrm{H}}$, is 436 kJ per mole of $\\mathrm{H}-\\mathrm{H}$ bonds broken:\n\n$$\n\\mathrm{H}_{2}(g) \\longrightarrow 2 \\mathrm{H}(g) \\quad \\mathrm{D}_{\\mathrm{H}-\\mathrm{H}}=\\Delta H^{\\circ}=436 \\mathrm{~kJ}\n$$\n\nMolecules with three or more atoms have two or more bonds. The sum of all bond energies in such a molecule is equal to the standard enthalpy change for the endothermic reaction that breaks all the bonds in the molecule. For example, the sum of the four $\\mathrm{C}-\\mathrm{H}$ bond energies in $\\mathrm{CH}_{4}, 1660 \\mathrm{~kJ}$, is equal to the standard enthalpy change of the reaction:"}
{"id": 3455, "contents": "1027. Bond Strength: Covalent Bonds - \nThe average $\\mathrm{C}-\\mathrm{H}$ bond energy, $\\mathrm{D}_{\\mathrm{C}-\\mathrm{H}}$, is $1660 / 4=415 \\mathrm{~kJ} / \\mathrm{mol}$ because there are four moles of $\\mathrm{C}-\\mathrm{H}$ bonds broken per mole of the reaction. Although the four $\\mathrm{C}-\\mathrm{H}$ bonds are equivalent in the original molecule, they do not each require the same energy to break; once the first bond is broken (which requires $439 \\mathrm{~kJ} / \\mathrm{mol}$ ), the remaining bonds are easier to break. The $415 \\mathrm{~kJ} / \\mathrm{mol}$ value is the average, not the exact value required to break any one bond.\n\nThe strength of a bond between two atoms increases as the number of electron pairs in the bond increases. Generally, as the bond strength increases, the bond length decreases. Thus, we find that triple bonds are stronger and shorter than double bonds between the same two atoms; likewise, double bonds are stronger and shorter than single bonds between the same two atoms. Average bond energies for some common bonds appear in Table 9.3, and a comparison of bond lengths and bond strengths for some common bonds appears in\n\nTable 9.4. When one atom bonds to various atoms in a group, the bond strength typically decreases as we move down the group. For example, $\\mathrm{C}-\\mathrm{F}$ is $439 \\mathrm{~kJ} / \\mathrm{mol}, \\mathrm{C}-\\mathrm{Cl}$ is $330 \\mathrm{~kJ} / \\mathrm{mol}$, and $\\mathrm{C}-\\mathrm{Br}$ is $275 \\mathrm{~kJ} / \\mathrm{mol}$.\n\nBond Energies (kJ/mol)"}
{"id": 3456, "contents": "1027. Bond Strength: Covalent Bonds - \n| Bond | Bond Energy | Bond | Bond Energy | Bond | Bond Energy |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| $\\mathrm{H}-\\mathrm{H}$ | 436 | C-S | 260 | $\\mathrm{F}-\\mathrm{Cl}$ | 255 |\n| H-C | 415 | $\\mathrm{C}-\\mathrm{Cl}$ | 330 | $\\mathrm{F}-\\mathrm{Br}$ | 235 |\n| $\\mathrm{H}-\\mathrm{N}$ | 390 | $\\mathrm{C}-\\mathrm{Br}$ | 275 | Si-Si | 230 |\n| H-O | 464 | C-I | 240 | Si-P | 215 |\n| H-F | 569 | $\\mathrm{N}-\\mathrm{N}$ | 160 | Si-S | 225 |\n| H-Si | 395 | $\\mathrm{N}=\\mathrm{N}$ | 418 | $\\mathrm{Si}-\\mathrm{Cl}$ | 359 |\n| $\\mathrm{H}-\\mathrm{P}$ | 320 | $\\mathrm{N} \\equiv \\mathrm{N}$ | 946 | $\\mathrm{Si}-\\mathrm{Br}$ | 290 |\n| H-S | 340 | $\\mathrm{N}-\\mathrm{O}$ | 200 | Si-I | 215 |\n| $\\mathrm{H}-\\mathrm{Cl}$ | 432 | N-F | 270 | P-P | 215 |\n| $\\mathrm{H}-\\mathrm{Br}$ | 370 | N-P | 210 | P-S | 230 |\n| H-I | 295 | $\\mathrm{N}-\\mathrm{Cl}$ | 200 | $\\mathrm{P}-\\mathrm{Cl}$ | 330 |\n| C-C | 345 | $\\mathrm{N}-\\mathrm{Br}$ | 245 | $\\mathrm{P}-\\mathrm{Br}$ | 270 |\n| $\\mathrm{C}=\\mathrm{C}$ | 611 | O-O | 140 | P-I | 215 |"}
{"id": 3457, "contents": "1027. Bond Strength: Covalent Bonds - \n| $\\mathrm{C}=\\mathrm{C}$ | 611 | O-O | 140 | P-I | 215 |\n| $\\mathrm{C} \\equiv \\mathrm{C}$ | 837 | $\\mathrm{O}=\\mathrm{O}$ | 498 | S-S | 215 |\n| $\\mathrm{C}-\\mathrm{N}$ | 290 | O-F | 160 | $\\mathrm{S}-\\mathrm{Cl}$ | 250 |\n| $\\mathrm{C}=\\mathrm{N}$ | 615 | O-Si | 370 | $\\mathrm{S}-\\mathrm{Br}$ | 215 |\n| $\\mathrm{C} \\equiv \\mathrm{N}$ | 891 | O-P | 350 | $\\mathrm{Cl}-\\mathrm{Cl}$ | 243 |\n| $\\mathrm{C}-\\mathrm{O}$ | 350 | $\\mathrm{O}-\\mathrm{Cl}$ | 205 | $\\mathrm{Cl}-\\mathrm{Br}$ | 220 |\n| $\\mathrm{C}=\\mathrm{O}$ | 741 | O-I | 200 | Cl-I | 210 |\n| $\\mathrm{C} \\equiv \\mathrm{O}$ | 1080 | F-F | 160 | $\\mathrm{Br}-\\mathrm{Br}$ | 190 |\n| C-F | 439 | F-Si | 540 | $\\mathrm{Br}-\\mathrm{I}$ | 180 |\n| $\\mathrm{C}-\\mathrm{Si}$ | 360 | F-P | 489 | I-I | 150 |"}
{"id": 3458, "contents": "1027. Bond Strength: Covalent Bonds - \nTABLE 9.3\n\n| Bond | Bond Energy | Bond | Bond Energy | Bond | Bond Energy |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| C-P | 265 | F-S | 285 |  |  |\n\nTABLE 9.3\n\nAverage Bond Lengths and Bond Energies for Some Common Bonds\n\n| Bond | Bond Length (\u00c5) | Bond Energy (kJ/mol) |\n| :--- | :--- | :--- |\n| $\\mathrm{C}-\\mathrm{C}$ | 1.54 | 345 |\n| $\\mathrm{C}=\\mathrm{C}$ | 1.34 | 611 |\n| $\\mathrm{C} \\equiv \\mathrm{C}$ | 1.20 | 837 |\n| $\\mathrm{C}-\\mathrm{N}$ | 1.43 | 290 |\n| $\\mathrm{C}=\\mathrm{N}$ | 1.38 | 615 |\n| $\\mathrm{C} \\equiv \\mathrm{N}$ | 1.16 | 891 |\n| $\\mathrm{C}-\\mathrm{O}$ | 1.43 | 741 |\n| $\\mathrm{C}=\\mathrm{O}$ | 1.23 | 1.13 |\n\nTABLE 9.4"}
{"id": 3459, "contents": "1027. Bond Strength: Covalent Bonds - \nTABLE 9.4\n\nThe bond energy is the difference between the energy minimum (which occurs at the bond distance) and the energy of the two separated atoms. This is the quantity of energy released when the bond is formed. Conversely, the same amount of energy is required to break the bond. For the $\\mathrm{H}_{2}$ molecule shown in Figure 5.2, at the bond distance of 74 pm the system is $7.24 \\times 10^{-19} \\mathrm{~J}$ lower in energy than the two separated hydrogen atoms. This may seem like a small number. However, as we will learn in more detail later, bond energies are often discussed on a per-mole basis. For example, it requires $7.24 \\times 10^{-19} \\mathrm{~J}$ to break one $\\mathrm{H}-\\mathrm{H}$ bond, but it takes $4.36 \\times 10^{5} \\mathrm{~J}$ to break 1 mole of $\\mathrm{H}-\\mathrm{H}$ bonds. A comparison of some bond lengths and energies is shown in Figure 5.2 and Table 9.3. We can find many of these bonds in a variety of molecules, and this table provides average values. For example, breaking the first $\\mathrm{C}-\\mathrm{H}$ bond in $\\mathrm{CH}_{4}$ requires $439.3 \\mathrm{~kJ} / \\mathrm{mol}$, while breaking the first $\\mathrm{C}-\\mathrm{H}$ bond in $\\mathrm{H}-\\mathrm{CH}_{2} \\mathrm{C}_{6} \\mathrm{H}_{5}$ (a common paint thinner) requires $375.5 \\mathrm{~kJ} / \\mathrm{mol}$."}
{"id": 3460, "contents": "1027. Bond Strength: Covalent Bonds - \nAs seen in Table 9.3 and Table 9.4, an average carbon-carbon single bond is $347 \\mathrm{~kJ} / \\mathrm{mol}$, while in a carboncarbon double bond, the $\\pi$ bond increases the bond strength by $267 \\mathrm{~kJ} / \\mathrm{mol}$. Adding an additional $\\pi$ bond causes a further increase of $225 \\mathrm{~kJ} / \\mathrm{mol}$. We can see a similar pattern when we compare other $\\sigma$ and $\\pi$ bonds. Thus, each individual $\\pi$ bond is generally weaker than a corresponding $\\sigma$ bond between the same two atoms. In a $\\sigma$ bond, there is a greater degree of orbital overlap than in a $\\pi$ bond.\n\nWe can use bond energies to calculate approximate enthalpy changes for reactions where enthalpies of formation are not available. Calculations of this type will also tell us whether a reaction is exothermic or endothermic. An exothermic reaction ( $\\Delta H$ negative, heat produced) results when the bonds in the products are stronger than the bonds in the reactants. An endothermic reaction ( $\\Delta H$ positive, heat absorbed) results when the bonds in the products are weaker than those in the reactants.\n\nThe enthalpy change, $\\Delta H$, for a chemical reaction is approximately equal to the sum of the energy required to break all bonds in the reactants (energy \"in\", positive sign) plus the energy released when all bonds are formed in the products (energy \"out,\" negative sign). This can be expressed mathematically in the following way:\n\n$$\n\\Delta H=\\Sigma \\mathrm{D}_{\\text {bonds broken }}-\\Sigma \\mathrm{D}_{\\text {bonds formed }}\n$$"}
{"id": 3461, "contents": "1027. Bond Strength: Covalent Bonds - \n$$\n\\Delta H=\\Sigma \\mathrm{D}_{\\text {bonds broken }}-\\Sigma \\mathrm{D}_{\\text {bonds formed }}\n$$\n\nIn this expression, the symbol $\\Sigma$ means \"the sum of\" and D represents the bond energy in kilojoules per mole, which is always a positive number. The bond energy is obtained from a table (like Table 9.4) and will depend on whether the particular bond is a single, double, or triple bond. Thus, in calculating enthalpies in this manner, it is important that we consider the bonding in all reactants and products. Because D values are typically averages for one type of bond in many different molecules, this calculation provides a rough estimate, not an exact value, for the enthalpy of reaction.\n\nConsider the following reaction:\n\n$$\n\\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{HCl}(g)\n$$\n\nor\n\n$$\n\\mathrm{H}-\\mathrm{H}(g)+\\mathrm{Cl}-\\mathrm{Cl}(g) \\longrightarrow 2 \\mathrm{H}-\\mathrm{Cl}(g)\n$$\n\nTo form two moles of HCl , one mole of $\\mathrm{H}-\\mathrm{H}$ bonds and one mole of $\\mathrm{Cl}-\\mathrm{Cl}$ bonds must be broken. The energy required to break these bonds is the sum of the bond energy of the $\\mathrm{H}-\\mathrm{H}$ bond $(436 \\mathrm{~kJ} / \\mathrm{mol})$ and the $\\mathrm{Cl}-\\mathrm{Cl}$ bond ( $243 \\mathrm{~kJ} / \\mathrm{mol}$ ). During the reaction, two moles of $\\mathrm{H}-\\mathrm{Cl}$ bonds are formed (bond energy $=432 \\mathrm{~kJ} / \\mathrm{mol}$ ), releasing 2 $\\times 432 \\mathrm{~kJ}$; or 864 kJ . Because the bonds in the products are stronger than those in the reactants, the reaction releases more energy than it consumes:"}
{"id": 3462, "contents": "1027. Bond Strength: Covalent Bonds - \n$$\n\\begin{aligned}\n\\Delta H & =\\sum \\mathrm{D}_{\\text {bonds broken }}-\\sum \\mathrm{D}_{\\text {bonds formed }} \\\\\n\\Delta H & =\\left[\\mathrm{D}_{\\mathrm{H}-\\mathrm{H}}+\\mathrm{D}_{\\mathrm{Cl}-\\mathrm{Cl}}\\right]-2 \\mathrm{D}_{\\mathrm{H}-\\mathrm{Cl}} \\\\\n& =[436+243]-2(432)=-185 \\mathrm{~kJ}\n\\end{aligned}\n$$\n\nThis excess energy is released as heat, so the reaction is exothermic. Appendix $G$ gives a value for the standard molar enthalpy of formation of $\\mathrm{HCl}(\\mathrm{g}), \\Delta H_{\\mathrm{f}}^{\\circ}$, of $-92.307 \\mathrm{~kJ} / \\mathrm{mol}$. Twice that value is -184.6 kJ , which agrees well with the answer obtained earlier for the formation of two moles of HCl ."}
{"id": 3463, "contents": "1029. Using Bond Energies to Calculate Approximate Enthalpy Changes - \nMethanol, $\\mathrm{CH}_{3} \\mathrm{OH}$, may be an excellent alternative fuel. The high-temperature reaction of steam and carbon produces a mixture of the gases carbon monoxide, CO , and hydrogen, $\\mathrm{H}_{2}$, from which methanol can be produced. Using the bond energies in Table 9.4, calculate the approximate enthalpy change, $\\Delta H$, for the reaction here:\n\n$$\n\\mathrm{CO}(\\mathrm{~g})+2 \\mathrm{H}_{2}(\\mathrm{~g}) \\longrightarrow \\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{~g})\n$$"}
{"id": 3464, "contents": "1030. Solution - \nFirst, we need to write the Lewis structures of the reactants and the products:\n\n\nFrom this, we see that $\\Delta H$ for this reaction involves the energy required to break a $\\mathrm{C}-\\mathrm{O}$ triple bond and two $\\mathrm{H}-\\mathrm{H}$ single bonds, as well as the energy produced by the formation of three $\\mathrm{C}-\\mathrm{H}$ single bonds, a $\\mathrm{C}-\\mathrm{O}$ single bond, and an $\\mathrm{O}-\\mathrm{H}$ single bond. We can express this as follows:\n\n$$\n\\begin{aligned}\n\\Delta H & =\\Sigma \\mathrm{D}_{\\text {bonds broken }}-\\Sigma \\mathrm{D}_{\\text {bonds formed }} \\\\\n\\Delta H & =\\left[\\mathrm{D}_{\\mathrm{C} \\equiv \\mathrm{O}}+2\\left(\\mathrm{D}_{\\mathrm{H}-\\mathrm{H}}\\right)\\right]-\\left[3\\left(\\mathrm{D}_{\\mathrm{C}-\\mathrm{H}}\\right)+\\mathrm{D}_{\\mathrm{C}-\\mathrm{O}}+\\mathrm{D}_{\\mathrm{O}-\\mathrm{H}}\\right]\n\\end{aligned}\n$$\n\nUsing the bond energy values in Table 9.4, we obtain:\n\n$$\n\\begin{aligned}\n\\Delta H & =[1080+2(436)]-[3(415)+350+464] \\\\\n& =-107 \\mathrm{~kJ}\n\\end{aligned}\n$$\n\nWe can compare this value to the value calculated based on $\\Delta H_{\\mathrm{f}}^{\\circ}$ data from Appendix G :\n\n$$\n\\begin{aligned}\n\\Delta H & =\\left[\\Delta H_{\\mathrm{f}}^{\\circ} \\mathrm{CH}_{3} \\mathrm{OH}(g)\\right]-\\left[\\Delta H_{\\mathrm{f}}^{\\circ} \\mathrm{CO}(g)+2 \\times \\Delta H_{\\mathrm{f}}^{\\circ} \\mathrm{H}_{2}\\right] \\\\\n& =[-201.0]-[-110.52+2 \\times 0] \\\\\n& =-90.5 \\mathrm{~kJ}\n\\end{aligned}\n$$"}
{"id": 3465, "contents": "1030. Solution - \nNote that there is a fairly significant gap between the values calculated using the two different methods. This occurs because D values are the average of different bond strengths; therefore, they often give only rough agreement with other data."}
{"id": 3466, "contents": "1031. Check Your Learning - \nEthyl alcohol, $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$, was one of the first organic chemicals deliberately synthesized by humans. It has many uses in industry, and it is the alcohol contained in alcoholic beverages. It can be obtained by the fermentation of sugar or synthesized by the hydration of ethylene in the following reaction:\n\n\nUsing the bond energies in Table 9.4, calculate an approximate enthalpy change, $\\Delta H$, for this reaction."}
{"id": 3467, "contents": "1032. Answer: - \n-35 kJ"}
{"id": 3468, "contents": "1033. Ionic Bond Strength and Lattice Energy - \nAn ionic compound is stable because of the electrostatic attraction between its positive and negative ions. The lattice energy of a compound is a measure of the strength of this attraction. The lattice energy ( $\\boldsymbol{\\Delta} \\boldsymbol{H}_{\\text {lattice }}$ ) of an ionic compound is defined as the energy required to separate one mole of the solid into its component gaseous ions. For the ionic solid MX, the lattice energy is the enthalpy change of the process:\n\n$$\n\\mathrm{MX}(s) \\longrightarrow \\mathrm{M}^{n+}(g)+\\mathrm{X}^{n-}(g) \\quad \\Delta H_{\\text {lattice }}\n$$\n\nNote that we are using the convention where the ionic solid is separated into ions, so our lattice energies will be endothermic (positive values). Some texts use the equivalent but opposite convention, defining lattice energy as the energy released when separate ions combine to form a lattice and giving negative (exothermic) values. Thus, if you are looking up lattice energies in another reference, be certain to check which definition is being used. In both cases, a larger magnitude for lattice energy indicates a more stable ionic compound. For sodium chloride, $\\Delta H_{\\text {lattice }}=769 \\mathrm{~kJ}$. Thus, it requires 769 kJ to separate one mole of solid NaCl into gaseous $\\mathrm{Na}^{+}$ and $\\mathrm{Cl}^{-}$ions. When one mole each of gaseous $\\mathrm{Na}^{+}$and $\\mathrm{Cl}^{-}$ions form solid $\\mathrm{NaCl}, 769 \\mathrm{~kJ}$ of heat is released.\n\nThe lattice energy $\\Delta H_{\\text {lattice }}$ of an ionic crystal can be expressed by the following equation (derived from Coulomb's law, governing the forces between electric charges):\n\n$$\n\\Delta H_{\\text {lattice }}=\\frac{\\mathrm{C}\\left(\\mathrm{Z}^{+}\\right)\\left(\\mathrm{Z}^{-}\\right)}{\\mathrm{R}_{\\mathrm{o}}}\n$$"}
{"id": 3469, "contents": "1033. Ionic Bond Strength and Lattice Energy - \nin which C is a constant that depends on the type of crystal structure; $\\mathrm{Z}^{+}$and $\\mathrm{Z}^{-}$are the charges on the ions; and $R_{0}$ is the interionic distance (the sum of the radii of the positive and negative ions). Thus, the lattice energy\nof an ionic crystal increases rapidly as the charges of the ions increase and the sizes of the ions decrease. When all other parameters are kept constant, doubling the charge of both the cation and anion quadruples the lattice energy. For example, the lattice energy of $\\mathrm{LiF}\\left(\\mathrm{Z}^{+}\\right.$and $\\mathrm{Z}^{-}=1$ ) is $1023 \\mathrm{~kJ} / \\mathrm{mol}$, whereas that of $\\mathrm{MgO}\\left(\\mathrm{Z}^{+}\\right.$and $\\mathrm{Z}^{-}=2$ ) is $3900 \\mathrm{~kJ} / \\mathrm{mol}\\left(\\mathrm{R}_{\\mathrm{o}}\\right.$ is nearly the same-about 200 pm for both compounds).\n\nDifferent interatomic distances produce different lattice energies. For example, we can compare the lattice energy of $\\mathrm{MgF}_{2}(2957 \\mathrm{~kJ} / \\mathrm{mol})$ to that of $\\mathrm{MgI}_{2}(2327 \\mathrm{~kJ} / \\mathrm{mol})$ to observe the effect on lattice energy of the smaller ionic size of $\\mathrm{F}^{-}$as compared to $\\mathrm{I}^{-}$."}
{"id": 3470, "contents": "1035. Lattice Energy Comparisons - \nThe precious gem ruby is aluminum oxide, $\\mathrm{Al}_{2} \\mathrm{O}_{3}$, containing traces of $\\mathrm{Cr}^{3+}$. The compound $\\mathrm{Al}_{2} \\mathrm{Se}_{3}$ is used in the fabrication of some semiconductor devices. Which has the larger lattice energy, $\\mathrm{Al}_{2} \\mathrm{O}_{3}$ or $\\mathrm{Al}_{2} \\mathrm{Se}_{3}$ ?"}
{"id": 3471, "contents": "1036. Solution - \nIn these two ionic compounds, the charges $\\mathrm{Z}^{+}$and $\\mathrm{Z}^{-}$are the same, so the difference in lattice energy will depend upon $\\mathrm{R}_{0}$. The $\\mathrm{O}^{2-}$ ion is smaller than the $\\mathrm{Se}^{2-}$ ion. Thus, $\\mathrm{Al}_{2} \\mathrm{O}_{3}$ would have a shorter interionic distance than $\\mathrm{Al}_{2} \\mathrm{Se}_{3}$, and $\\mathrm{Al}_{2} \\mathrm{O}_{3}$ would have the larger lattice energy."}
{"id": 3472, "contents": "1037. Check Your Learning - \nZinc oxide, ZnO , is a very effective sunscreen. How would the lattice energy of ZnO compare to that of NaCl ?"}
{"id": 3473, "contents": "1038. Answer: - \nZnO would have the larger lattice energy because the Z values of both the cation and the anion in ZnO are greater, and the interionic distance of ZnO is smaller than that of NaCl ."}
{"id": 3474, "contents": "1039. The Born-Haber Cycle - \nIt is not possible to measure lattice energies directly. However, the lattice energy can be calculated using the equation given in the previous section or by using a thermochemical cycle. The Born-Haber cycle is an application of Hess's law that breaks down the formation of an ionic solid into a series of individual steps:\n\n- $\\Delta H_{\\mathrm{f}}^{\\circ}$, the standard enthalpy of formation of the compound\n- IE, the ionization energy of the metal\n- $E A$, the electron affinity of the nonmetal\n- $\\Delta H_{s}^{\\circ}$, the enthalpy of sublimation of the metal\n- $D$, the bond dissociation energy of the nonmetal\n- $\\Delta H_{\\text {lattice }}$, the lattice energy of the compound\n\nFigure 9.25 diagrams the Born-Haber cycle for the formation of solid cesium fluoride.\n\n\nFIGURE 9.25 The Born-Haber cycle shows the relative energies of each step involved in the formation of an ionic solid from the necessary elements in their reference states.\n\nWe begin with the elements in their most common states, $\\operatorname{Cs}(s)$ and $\\mathrm{F}_{2}(g)$. The $\\Delta H_{s}^{\\circ}$ represents the conversion of solid cesium into a gas, and then the ionization energy converts the gaseous cesium atoms into cations. In the next step, we account for the energy required to break the $F-F$ bond to produce fluorine atoms. Converting one mole of fluorine atoms into fluoride ions is an exothermic process, so this step gives off energy (the electron affinity) and is shown as decreasing along the $y$-axis. We now have one mole of Cs cations and one mole of F anions. These ions combine to produce solid cesium fluoride. The enthalpy change in this step is the negative of the lattice energy, so it is also an exothermic quantity. The total energy involved in this conversion is equal to the experimentally determined enthalpy of formation, $\\Delta \\boldsymbol{H}_{\\mathrm{f}}^{\\circ}$, of the compound from its elements. In this case, the overall change is exothermic."}
{"id": 3475, "contents": "1039. The Born-Haber Cycle - \nHess's law can also be used to show the relationship between the enthalpies of the individual steps and the enthalpy of formation. Table 9.5 shows this for cesium fluoride, CsF."}
{"id": 3476, "contents": "1039. The Born-Haber Cycle - \n| Enthalpy of <br> sublimation of <br> $\\mathrm{Cs}(s)$ | $\\mathrm{Cs}(s) \\longrightarrow \\mathrm{Cs}(g)$ | $\\Delta H=\\Delta H_{s}^{\\circ}=76.5 \\mathrm{~kJ} / \\mathrm{mol}$ |\n| :--- | :--- | :--- |\n| One-half of the <br> bond energy of <br> $\\mathrm{F}_{2}$ | $\\frac{1}{2} \\mathrm{~F}_{2}(\\mathrm{~g}) \\longrightarrow \\mathrm{F}(\\mathrm{g})$ | $\\Delta H=\\frac{1}{2} D=79.4 \\mathrm{~kJ} / \\mathrm{mol}$ |\n| Ionization <br> energy of $\\mathrm{Cs}(g)$ | $\\mathrm{Cs}(g) \\longrightarrow \\mathrm{Cs}^{+}(g)+\\mathrm{e}^{-}$ | $\\Delta H=I E=375.7 \\mathrm{~kJ} / \\mathrm{mol}$ |\n| Electron <br> affinity of F | $\\mathrm{F}(g)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{F}^{-}(g)$ | $\\Delta H=E A=-328.2 \\mathrm{~kJ} / \\mathrm{mol}$ |\n| Negative of the <br> lattice energy <br> of CsF $(s)$ | $\\mathrm{Cs}{ }^{+}(g)+\\mathrm{F}^{-}(g) \\longrightarrow \\mathrm{CsF}(s)$ | $\\Delta H=-\\Delta H_{\\text {lattice }}=?$ |\n| Enthalpy of <br> formation of <br> CsF $(s)$, add <br> steps $1-5$ | $\\Delta H=\\Delta H_{f}^{\\circ}=\\Delta H_{s}^{\\circ}+\\frac{1}{2} D+I E+(E A)+\\left(-\\Delta H_{\\mathrm{lattice}}\\right)$ | $\\Delta H=-553.5 \\mathrm{~kJ} / \\mathrm{mol}$ |"}
{"id": 3477, "contents": "1040. TABLE 9.5 - \nThus, the lattice energy can be calculated from other values. For cesium fluoride, using this data, the lattice energy is:\n\n$$\n\\Delta H_{\\text {lattice }}=76.5+79.4+375.7+(-328.2)-(-553.5)=756.9 \\mathrm{~kJ} / \\mathrm{mol}\n$$\n\nThe Born-Haber cycle may also be used to calculate any one of the other quantities in the equation for lattice energy, provided that the remainder is known. For example, if the relevant enthalpy of sublimation $\\Delta H_{s}^{\\circ}$, ionization energy (IE), bond dissociation enthalpy (D), lattice energy $\\Delta H_{\\text {lattice, }}$, and standard enthalpy of formation $\\Delta H_{\\mathrm{f}}^{\\circ}$ are known, the Born-Haber cycle can be used to determine the electron affinity of an atom.\n\nLattice energies calculated for ionic compounds are typically much higher than bond dissociation energies measured for covalent bonds. Whereas lattice energies typically fall in the range of 600-4000 kJ/mol (some even higher), covalent bond dissociation energies are typically between $150-400 \\mathrm{~kJ} / \\mathrm{mol}$ for single bonds. Keep in mind, however, that these are not directly comparable values. For ionic compounds, lattice energies are associated with many interactions, as cations and anions pack together in an extended lattice. For covalent bonds, the bond dissociation energy is associated with the interaction of just two atoms."}
{"id": 3478, "contents": "1041. Key Terms - \nbomb calorimeter device designed to measure the energy change for processes occurring under conditions of constant volume; commonly used for reactions involving solid and gaseous reactants or products\nbond energy (also, bond dissociation energy) energy required to break a covalent bond in a gaseous substance\nBorn-Haber cycle thermochemical cycle relating the various energetic steps involved in the formation of an ionic solid from the relevant elements\ncalorie (cal) unit of heat or other energy; the amount of energy required to raise 1 gram of water by 1 degree Celsius; 1 cal is defined as 4.184 J\ncalorimeter device used to measure the amount of heat absorbed or released in a chemical or physical process\ncalorimetry process of measuring the amount of heat involved in a chemical or physical process\nchemical thermodynamics area of science that deals with the relationships between heat, work, and all forms of energy associated with chemical and physical processes\nendothermic process chemical reaction or physical change that absorbs heat\nenergy capacity to supply heat or do work\nenthalpy ( $\\boldsymbol{H}$ ) sum of a system's internal energy and the mathematical product of its pressure and volume\nenthalpy change ( $\\mathbf{\\Delta H}$ ) heat released or absorbed by a system under constant pressure during a chemical or physical process\nexothermic process chemical reaction or physical change that releases heat\nexpansion work (pressure-volume work) work done as a system expands or contracts against external pressure\nfirst law of thermodynamics internal energy of a system changes due to heat flow in or out of the system or work done on or by the system\nheat (q) transfer of thermal energy between two bodies\nheat capacity (C) extensive property of a body of matter that represents the quantity of heat required to increase its temperature by 1 degree Celsius (or 1 kelvin)\nHess's law if a process can be represented as the sum of several steps, the enthalpy change of the process equals the sum of the enthalpy changes of the steps\nhydrocarbon compound composed only of hydrogen and carbon; the major component of fossil fuels\ninternal energy ( $\\boldsymbol{U}$ ) total of all possible kinds of energy present in a substance or substances"}
{"id": 3479, "contents": "1041. Key Terms - \nhydrocarbon compound composed only of hydrogen and carbon; the major component of fossil fuels\ninternal energy ( $\\boldsymbol{U}$ ) total of all possible kinds of energy present in a substance or substances\njoule (J) SI unit of energy; 1 joule is the kinetic energy of an object with a mass of 2 kilograms moving with a velocity of 1 meter per second, 1 J $=1 \\mathrm{~kg} \\mathrm{~m}^{2} / \\mathrm{s}$ and $4.184 \\mathrm{~J}=1 \\mathrm{cal}$\nkinetic energy energy of a moving body, in joules, equal to $\\frac{1}{2} m v^{2}$ (where $m=$ mass and $v=$ velocity)\nlattice energy ( $\\boldsymbol{\\Delta} \\boldsymbol{H}_{\\text {lattice }}$ ) energy required to separate one mole of an ionic solid into its component gaseous ions\nnutritional calorie (Calorie) unit used for quantifying energy provided by digestion of foods, defined as 1000 cal or 1 kcal\npotential energy energy of a particle or system of particles derived from relative position, composition, or condition\nspecific heat capacity (c) intensive property of a substance that represents the quantity of heat required to raise the temperature of 1 gram of the substance by 1 degree Celsius (or 1 kelvin)\nstandard enthalpy of combustion $\\left(\\Delta H_{c}^{\\circ}\\right)$ heat released when one mole of a compound undergoes complete combustion under standard conditions\nstandard enthalpy of formation $\\left(\\Delta H_{\\mathrm{f}}^{\\circ}\\right)$ enthalpy change of a chemical reaction in which 1 mole of a pure substance is formed from its elements in their most stable states under standard state conditions\nstandard state set of physical conditions as accepted as common reference conditions for reporting thermodynamic properties; 1 bar of pressure, and solutions at 1 molar concentrations, usually at a temperature of 298.15 K\nstate function property depending only on the state of a system, and not the path taken to reach that state\nsurroundings all matter other than the system being studied\nsystem portion of matter undergoing a chemical or physical change being studied"}
{"id": 3480, "contents": "1041. Key Terms - \nstate function property depending only on the state of a system, and not the path taken to reach that state\nsurroundings all matter other than the system being studied\nsystem portion of matter undergoing a chemical or physical change being studied\ntemperature intensive property of matter that is a quantitative measure of \"hotness\" and \"coldness\"\nthermal energy kinetic energy associated with the random motion of atoms and molecules\nthermochemistry study of measuring the amount\nof heat absorbed or released during a chemical reaction or a physical change\nwork (w) energy transfer due to changes in\nexternal, macroscopic variables such as pressure and volume; or causing matter to move against an opposing force"}
{"id": 3481, "contents": "1042. Key Equations - \n$q=c \\times m \\times \\Delta T=c \\times m \\times\\left(T_{\\text {final }}-T_{\\text {initial }}\\right)$\n$\\Delta U=q+w$\n$\\Delta H_{\\text {reaction }}^{\\circ}=\\sum n \\times \\Delta H_{\\mathrm{f}}^{\\circ}($ products $)-\\sum n \\times \\Delta H_{\\mathrm{f}}^{\\circ}($ reactants $)$\nBond energy for a diatomic molecule: XY $(g) \\longrightarrow \\mathrm{X}(g)+\\mathrm{Y}(g)$\n\n$$\n\\mathrm{D}_{\\mathrm{X}-\\mathrm{Y}}=\\Delta H^{\\circ}\n$$\n\nEnthalpy change: $\\Delta H=\\Sigma \\mathrm{D}_{\\text {bonds broken }}-\\Sigma \\mathrm{D}_{\\text {bonds formed }}$\nLattice energy for a solid MX: MX $(s) \\longrightarrow \\mathrm{M}^{n+}(g)+\\mathrm{X}^{n-}(g) \\quad \\Delta H_{\\text {lattice }}$\nLattice energy for an ionic crystal: $\\Delta H_{\\text {lattice }}=\\frac{\\mathrm{C}\\left(\\mathrm{Z}^{+}\\right)\\left(\\mathrm{Z}^{-}\\right)}{\\mathrm{R}_{\\mathrm{o}}}$"}
{"id": 3482, "contents": "1043. Summary - 1043.1. Energy Basics\nEnergy is the capacity to supply heat or do work (applying a force to move matter). Kinetic energy (KE) is the energy of motion; potential energy is energy due to relative position, composition, or condition. When energy is converted from one form into another, energy is neither created nor destroyed (law of conservation of energy or first law of thermodynamics).\n\nThe thermal energy of matter is due to the kinetic energies of its constituent atoms or molecules. Temperature is an intensive property of matter reflecting hotness or coldness that increases as the average kinetic energy increases. Heat is the transfer of thermal energy between objects at different temperatures. Chemical and physical processes can absorb heat (endothermic) or release heat (exothermic). The SI unit of energy, heat, and work is the joule ( J ).\n\nSpecific heat and heat capacity are measures of the energy needed to change the temperature of a substance or object. The amount of heat absorbed or released by a substance depends directly on the type of substance, its mass, and the temperature change it undergoes."}
{"id": 3483, "contents": "1043. Summary - 1043.2. Calorimetry\nCalorimetry is used to measure the amount of thermal energy transferred in a chemical or physical process. This requires careful measurement of the temperature change that occurs during the process and the masses of the system and surroundings. These measured quantities are then used to compute the amount of heat produced or consumed\nin the process using known mathematical relations.\nCalorimeters are designed to minimize energy exchange between their contents and the external environment. They range from simple coffee cup calorimeters used by introductory chemistry students to sophisticated bomb calorimeters used to determine the energy content of food."}
{"id": 3484, "contents": "1043. Summary - 1043.3. Enthalpy\nIf a chemical change is carried out at constant pressure and the only work done is caused by expansion or contraction, $q$ for the change is called the enthalpy change with the symbol $\\Delta H$, or $\\Delta H^{\\circ}$ for reactions occurring under standard state conditions at 298 K . The value of $\\Delta H$ for a reaction in one direction is equal in magnitude, but opposite in sign, to $\\Delta H$ for the reaction in the opposite direction, and $\\Delta H$ is directly proportional to the quantity of reactants and products. The standard enthalpy of formation, $\\Delta \\boldsymbol{H}_{\\mathrm{f}}^{\\circ}$, is the enthalpy change accompanying the formation of 1 mole of a substance from the elements in their most stable states at 1 bar and 298.15 K. If the enthalpies of formation are available for the reactants and products of a reaction, the enthalpy change can be calculated using Hess's law: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps."}
{"id": 3485, "contents": "1043. Summary - 1043.4. Strengths of Ionic and Covalent Bonds\nThe strength of a covalent bond is measured by its bond dissociation energy, that is, the amount of energy required to break that particular bond in a mole of molecules. Multiple bonds are stronger than\nsingle bonds between the same atoms. The enthalpy of a reaction can be estimated based on the energy input required to break bonds and the energy released when new bonds are formed. For ionic bonds, the lattice energy is the energy required to separate one mole of a compound into its gas phase\nions. Lattice energy increases for ions with higher charges and shorter distances between ions. Lattice energies are often calculated using the Born-Haber cycle, a thermochemical cycle including all of the energetic steps involved in converting elements into an ionic compound."}
{"id": 3486, "contents": "1044. Exercises - 1044.1. Energy Basics\n1. A burning match and a bonfire may have the same temperature, yet you would not sit around a burning match on a fall evening to stay warm. Why not?\n2. Prepare a table identifying several energy transitions that take place during the typical operation of an automobile.\n3. Explain the difference between heat capacity and specific heat of a substance.\n4. Calculate the heat capacity, in joules and in calories per degree, of the following:\n(a) 28.4 g of water\n(b) 1.00 oz of lead\n5. Calculate the heat capacity, in joules and in calories per degree, of the following:\n(a) 45.8 g of nitrogen gas\n(b) 1.00 pound of aluminum metal\n6. How much heat, in joules and in calories, must be added to a $75.0-\\mathrm{g}$ iron block with a specific heat of $0.449 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$ to increase its temperature from $25^{\\circ} \\mathrm{C}$ to its melting temperature of $1535{ }^{\\circ} \\mathrm{C}$ ?\n7. How much heat, in joules and in calories, is required to heat a $28.4-\\mathrm{g}(1-\\mathrm{oz})$ ice cube from $-23.0^{\\circ} \\mathrm{C}$ to -1.0 ${ }^{\\circ} \\mathrm{C}$ ?\n8. How much would the temperature of 275 g of water increase if 36.5 kJ of heat were added?\n9. If 14.5 kJ of heat were added to 485 g of liquid water, how much would its temperature increase?\n10. A piece of unknown substance weighs 44.7 g and requires 2110 J to increase its temperature from $23.2^{\\circ} \\mathrm{C}$ to $89.6^{\\circ} \\mathrm{C}$.\n(a) What is the specific heat of the substance?\n(b) If it is one of the substances found in Table 9.1, what is its likely identity?"}
{"id": 3487, "contents": "1044. Exercises - 1044.1. Energy Basics\n(a) What is the specific heat of the substance?\n(b) If it is one of the substances found in Table 9.1, what is its likely identity?\n11. A piece of unknown solid substance weighs 437.2 g , and requires 8460 J to increase its temperature from $19.3^{\\circ} \\mathrm{C}$ to $68.9^{\\circ} \\mathrm{C}$.\n(a) What is the specific heat of the substance?\n(b) If it is one of the substances found in Table 9.1, what is its likely identity?\n12. An aluminum kettle weighs 1.05 kg .\n(a) What is the heat capacity of the kettle?\n(b) How much heat is required to increase the temperature of this kettle from $23.0^{\\circ} \\mathrm{C}$ to $99.0^{\\circ} \\mathrm{C}$ ?\n(c) How much heat is required to heat this kettle from $23.0^{\\circ} \\mathrm{C}$ to $99.0^{\\circ} \\mathrm{C}$ if it contains 1.25 L of water (density of $0.997 \\mathrm{~g} / \\mathrm{mL}$ and a specific heat of $4.184 \\mathrm{~J} / \\mathrm{g}^{\\circ} \\mathrm{C}$ )?\n13. Most people find waterbeds uncomfortable unless the water temperature is maintained at about $85^{\\circ} \\mathrm{F}$. Unless it is heated, a waterbed that contains 892 L of water cools from $85^{\\circ} \\mathrm{F}$ to $72^{\\circ} \\mathrm{F}$ in 24 hours. Estimate the amount of electrical energy required over 24 hours, in kWh , to keep the bed from cooling. Note that 1 kilowatt-hour $(\\mathrm{kWh})=3.6 \\times 10^{6} \\mathrm{~J}$, and assume that the density of water is $1.0 \\mathrm{~g} / \\mathrm{mL}$ (independent of temperature). What other assumptions did you make? How did they affect your calculated result (i.e., were they likely to yield \"positive\" or \"negative\" errors)?"}
{"id": 3488, "contents": "1044. Exercises - 1044.2. Calorimetry\n14. A $500-\\mathrm{mL}$ bottle of water at room temperature and a $2-\\mathrm{L}$ bottle of water at the same temperature were placed in a refrigerator. After 30 minutes, the $500-\\mathrm{mL}$ bottle of water had cooled to the temperature of the refrigerator. An hour later, the 2-L of water had cooled to the same temperature. When asked which sample of water lost the most heat, one student replied that both bottles lost the same amount of heat because they started at the same temperature and finished at the same temperature. A second student thought that the 2-L bottle of water lost more heat because there was more water. A third student believed that the $500-\\mathrm{mL}$ bottle of water lost more heat because it cooled more quickly. A fourth student thought that it was not possible to tell because we do not know the initial temperature and the final temperature of the water. Indicate which of these answers is correct and describe the error in each of the other answers.\n15. Would the amount of heat measured for the reaction in Example 9.5 be greater, lesser, or remain the same if we used a calorimeter that was a poorer insulator than a coffee cup calorimeter? Explain your answer.\n16. Would the amount of heat absorbed by the dissolution in Example 9.6 appear greater, lesser, or remain the same if the experimenter used a calorimeter that was a poorer insulator than a coffee cup calorimeter? Explain your answer.\n17. Would the amount of heat absorbed by the dissolution in Example 9.6 appear greater, lesser, or remain the same if the heat capacity of the calorimeter were taken into account? Explain your answer.\n18. How many milliliters of water at $23^{\\circ} \\mathrm{C}$ with a density of $1.00 \\mathrm{~g} / \\mathrm{mL}$ must be mixed with 180 mL (about 6 oz ) of coffee at $95^{\\circ} \\mathrm{C}$ so that the resulting combination will have a temperature of $60^{\\circ} \\mathrm{C}$ ? Assume that coffee and water have the same density and the same specific heat."}
{"id": 3489, "contents": "1044. Exercises - 1044.2. Calorimetry\n19. How much will the temperature of a cup ( 180 g ) of coffee at $95^{\\circ} \\mathrm{C}$ be reduced when a 45 g silver spoon (specific heat $0.24 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$ ) at $25^{\\circ} \\mathrm{C}$ is placed in the coffee and the two are allowed to reach the same temperature? Assume that the coffee has the same density and specific heat as water.\n20. A $45-\\mathrm{g}$ aluminum spoon (specific heat $0.88 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$ ) at $24^{\\circ} \\mathrm{C}$ is placed in $180 \\mathrm{~mL}(180 \\mathrm{~g})$ of coffee at $85^{\\circ} \\mathrm{C}$ and the temperature of the two become equal.\n(a) What is the final temperature when the two become equal? Assume that coffee has the same specific heat as water.\n(b) The first time a student solved this problem she got an answer of $88^{\\circ} \\mathrm{C}$. Explain why this is clearly an incorrect answer.\n21. The temperature of the cooling water as it leaves the hot engine of an automobile is $240^{\\circ} \\mathrm{F}$. After it passes through the radiator it has a temperature of $175^{\\circ} \\mathrm{F}$. Calculate the amount of heat transferred from the engine to the surroundings by one gallon of water with a specific heat of $4.184 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$.\n22. A $70.0-\\mathrm{g}$ piece of metal at $80.0^{\\circ} \\mathrm{C}$ is placed in 100 g of water at $22.0^{\\circ} \\mathrm{C}$ contained in a calorimeter like that shown in Figure 9.12. The metal and water come to the same temperature at $24.6^{\\circ} \\mathrm{C}$. How much heat did the metal give up to the water? What is the specific heat of the metal?"}
{"id": 3490, "contents": "1044. Exercises - 1044.2. Calorimetry\n23. If a reaction produces 1.506 kJ of heat, which is trapped in 30.0 g of water initially at $26.5^{\\circ} \\mathrm{C}$ in a calorimeter like that in Figure 9.12, what is the resulting temperature of the water?\n24. A $0.500-\\mathrm{g}$ sample of KCl is added to 50.0 g of water in a calorimeter (Figure 9.12). If the temperature decreases by $1.05^{\\circ} \\mathrm{C}$, what is the approximate amount of heat involved in the dissolution of the KCl , assuming the specific heat of the resulting solution is $4.18 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$ ? Is the reaction exothermic or endothermic?\n25. Dissolving 3.0 g of $\\mathrm{CaCl}_{2}(s)$ in 150.0 g of water in a calorimeter (Figure 9.12) at $22.4^{\\circ} \\mathrm{C}$ causes the temperature to rise to $25.8^{\\circ} \\mathrm{C}$. What is the approximate amount of heat involved in the dissolution, assuming the specific heat of the resulting solution is $4.18 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$ ? Is the reaction exothermic or endothermic?\n26. When 50.0 g of $0.200 \\mathrm{M} \\mathrm{NaCl}(\\mathrm{aq})$ at $24.1^{\\circ} \\mathrm{C}$ is added to 100.0 g of $0.100 \\mathrm{M} \\mathrm{AgNO}_{3}(\\mathrm{aq})$ at $24.1^{\\circ} \\mathrm{C}$ in a calorimeter, the temperature increases to $25.2^{\\circ} \\mathrm{C}$ as $\\mathrm{AgCl}(s)$ forms. Assuming the specific heat of the solution and products is $4.20 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$, calculate the approximate amount of heat in joules produced."}
{"id": 3491, "contents": "1044. Exercises - 1044.2. Calorimetry\n27. The addition of 3.15 g of $\\mathrm{Ba}(\\mathrm{OH})_{2} \\cdot 8 \\mathrm{H}_{2} \\mathrm{O}$ to a solution of 1.52 g of $\\mathrm{NH}_{4} \\mathrm{SCN}$ in 100 g of water in a calorimeter caused the temperature to fall by $3.1^{\\circ} \\mathrm{C}$. Assuming the specific heat of the solution and products is 4.20 $\\mathrm{J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$, calculate the approximate amount of heat absorbed by the reaction, which can be represented by the following equation:\n$\\mathrm{Ba}(\\mathrm{OH})_{2} \\cdot 8 \\mathrm{H}_{2} \\mathrm{O}(s)+2 \\mathrm{NH}_{4} \\mathrm{SCN}(\\mathrm{aq}) \\longrightarrow \\mathrm{Ba}(\\mathrm{SCN})_{2}(\\mathrm{aq})+2 \\mathrm{NH}_{3}(\\mathrm{aq})+10 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{I})$\n28. The reaction of 50 mL of acid and 50 mL of base described in Example 9.5 increased the temperature of the solution by $6.9^{\\circ} \\mathrm{C}$. How much would the temperature have increased if 100 mL of acid and 100 mL of base had been used in the same calorimeter starting at the same temperature of $22.0^{\\circ} \\mathrm{C}$ ? Explain your answer.\n29. If the 3.21 g of $\\mathrm{NH}_{4} \\mathrm{NO}_{3}$ in Example 9.6 were dissolved in 100.0 g of water under the same conditions, how much would the temperature change? Explain your answer."}
{"id": 3492, "contents": "1044. Exercises - 1044.2. Calorimetry\n30. When 1.0 g of fructose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}(s)$, a sugar commonly found in fruits, is burned in oxygen in a bomb calorimeter, the temperature of the calorimeter increases by $1.58^{\\circ} \\mathrm{C}$. If the heat capacity of the calorimeter and its contents is $9.90 \\mathrm{~kJ} /{ }^{\\circ} \\mathrm{C}$, what is $q$ for this combustion?\n31. When a $0.740-\\mathrm{g}$ sample of trinitrotoluene (TNT), $\\mathrm{C}_{7} \\mathrm{H}_{5} \\mathrm{~N}_{2} \\mathrm{O}_{6}$, is burned in a bomb calorimeter, the temperature increases from $23.4^{\\circ} \\mathrm{C}$ to $26.9^{\\circ} \\mathrm{C}$. The heat capacity of the calorimeter is $534 \\mathrm{~J} /{ }^{\\circ} \\mathrm{C}$, and it contains 675 mL of water. How much heat was produced by the combustion of the TNT sample?\n32. One method of generating electricity is by burning coal to heat water, which produces steam that drives an electric generator. To determine the rate at which coal is to be fed into the burner in this type of plant, the heat of combustion per ton of coal must be determined using a bomb calorimeter. When 1.00 g of coal is burned in a bomb calorimeter (Figure 9.17), the temperature increases by $1.48{ }^{\\circ} \\mathrm{C}$. If the heat capacity of the calorimeter is $21.6 \\mathrm{~kJ} /{ }^{\\circ} \\mathrm{C}$, determine the heat produced by combustion of a ton of coal $\\left(2.000 \\times 10^{3}\\right.$ pounds).\n33. The amount of fat recommended for someone with a daily diet of 2000 Calories is 65 g . What percent of the calories in this diet would be supplied by this amount of fat if the average number of Calories for fat is 9.1 Calories/g?"}
{"id": 3493, "contents": "1044. Exercises - 1044.2. Calorimetry\n34. A teaspoon of the carbohydrate sucrose (common sugar) contains 16 Calories ( 16 kcal ). What is the mass of one teaspoon of sucrose if the average number of Calories for carbohydrates is 4.1 Calories/g?\n35. What is the maximum mass of carbohydrate in a $6-$ oz serving of diet soda that contains less than 1 Calorie per can if the average number of Calories for carbohydrates is 4.1 Calories/g?\n36. A pint of premium ice cream can contain 1100 Calories. What mass of fat, in grams and pounds, must be produced in the body to store an extra $1.1 \\times 10^{3}$ Calories if the average number of Calories for fat is 9.1 Calories/g?\n37. A serving of a breakfast cereal contains 3 g of protein, 18 g of carbohydrates, and 6 g of fat. What is the Calorie content of a serving of this cereal if the average number of Calories for fat is 9.1 Calories $/ \\mathrm{g}$, for carbohydrates is 4.1 Calories/g, and for protein is 4.1 Calories/g?\n38. Which is the least expensive source of energy in kilojoules per dollar: a box of breakfast cereal that weighs 32 ounces and costs $\\$ 4.23$, or a liter of isooctane (density, $0.6919 \\mathrm{~g} / \\mathrm{mL}$ ) that costs $\\$ 0.45$ ? Compare the nutritional value of the cereal with the heat produced by combustion of the isooctane under standard conditions. A 1.0-ounce serving of the cereal provides 130 Calories."}
{"id": 3494, "contents": "1044. Exercises - 1044.3. Enthalpy\n39. Explain how the heat measured in Example 9.5 differs from the enthalpy change for the exothermic reaction described by the following equation:\n$\\mathrm{HCl}(a q)+\\mathrm{NaOH}(a q) \\longrightarrow \\mathrm{NaCl}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)$\n40. Using the data in the check your learning section of Example 9.5, calculate $\\Delta H$ in $\\mathrm{kJ} / \\mathrm{mol}$ of $\\mathrm{AgNO}_{3}(\\mathrm{aq})$ for the reaction: $\\mathrm{NaCl}(a q)+\\mathrm{AgNO}_{3}(a q) \\longrightarrow \\mathrm{AgCl}(s)+\\mathrm{NaNO}_{3}(a q)$\n41. Calculate the enthalpy of solution ( $\\Delta H$ for the dissolution) per mole of $\\mathrm{NH}_{4} \\mathrm{NO}_{3}$ under the conditions described in Example 9.6.\n42. Calculate $\\Delta H$ for the reaction described by the equation. (Hint: Use the value for the approximate amount of heat absorbed by the reaction that you calculated in a previous exercise.)\n$\\mathrm{Ba}(\\mathrm{OH})_{2} \\cdot 8 \\mathrm{H}_{2} \\mathrm{O}(s)+2 \\mathrm{NH}_{4} \\mathrm{SCN}(a q) \\longrightarrow \\mathrm{Ba}(\\mathrm{SCN})_{2}(a q)+2 \\mathrm{NH}_{3}(a q)+10 \\mathrm{H}_{2} \\mathrm{O}(l)$\n43. Calculate the enthalpy of solution ( $\\Delta H$ for the dissolution) per mole of $\\mathrm{CaCl}_{2}$ (refer to Exercise 9.25).\n44. Although the gas used in an oxyacetylene torch (Figure 9.7) is essentially pure acetylene, the heat produced by combustion of one mole of acetylene in such a torch is likely not equal to the enthalpy of combustion of acetylene listed in Table 9.2. Considering the conditions for which the tabulated data are reported, suggest an explanation."}
{"id": 3495, "contents": "1044. Exercises - 1044.3. Enthalpy\n45. How much heat is produced by burning 4.00 moles of acetylene under standard state conditions?\n46. How much heat is produced by combustion of 125 g of methanol under standard state conditions?\n47. How many moles of isooctane must be burned to produce 100 kJ of heat under standard state conditions?\n48. What mass of carbon monoxide must be burned to produce 175 kJ of heat under standard state conditions?\n49. When 2.50 g of methane burns in oxygen, 125 kJ of heat is produced. What is the enthalpy of combustion per mole of methane under these conditions?\n50. How much heat is produced when 100 mL of 0.250 M HCl (density, $1.00 \\mathrm{~g} / \\mathrm{mL}$ ) and 200 mL of 0.150 M NaOH (density, $1.00 \\mathrm{~g} / \\mathrm{mL}$ ) are mixed?\n$\\mathrm{HCl}(a q)+\\mathrm{NaOH}(a q) \\longrightarrow \\mathrm{NaCl}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\quad \\Delta H^{\\circ}=-58 \\mathrm{~kJ}$\nIf both solutions are at the same temperature and the specific heat of the products is $4.19 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$, how much will the temperature increase? What assumption did you make in your calculation?\n51. A sample of 0.562 g of carbon is burned in oxygen in a bomb calorimeter, producing carbon dioxide. Assume both the reactants and products are under standard state conditions, and that the heat released is directly proportional to the enthalpy of combustion of graphite. The temperature of the calorimeter increases from $26.74^{\\circ} \\mathrm{C}$ to $27.93^{\\circ} \\mathrm{C}$. What is the heat capacity of the calorimeter and its contents?"}
{"id": 3496, "contents": "1044. Exercises - 1044.3. Enthalpy\n52. Before the introduction of chlorofluorocarbons, sulfur dioxide (enthalpy of vaporization, $6.00 \\mathrm{kcal} / \\mathrm{mol}$ ) was used in household refrigerators. What mass of $\\mathrm{SO}_{2}$ must be evaporated to remove as much heat as evaporation of 1.00 kg of $\\mathrm{CCl}_{2} \\mathrm{~F}_{2}$ (enthalpy of vaporization is $17.4 \\mathrm{~kJ} / \\mathrm{mol}$ )?\nThe vaporization reactions for $\\mathrm{SO}_{2}$ and $\\mathrm{CCl}_{2} \\mathrm{~F}_{2}$ are $\\mathrm{SO}_{2}(l) \\longrightarrow \\mathrm{SO}_{2}(g)$ and $\\mathrm{CCl}_{2} \\mathrm{~F}(l) \\longrightarrow \\mathrm{CCl}_{2} \\mathrm{~F}_{2}(g)$, respectively.\n53. Homes may be heated by pumping hot water through radiators. What mass of water will provide the same amount of heat when cooled from 95.0 to $35.0^{\\circ} \\mathrm{C}$, as the heat provided when 100 g of steam is cooled from $110^{\\circ} \\mathrm{C}$ to $100^{\\circ} \\mathrm{C}$.\n54. Which of the enthalpies of combustion in Table 9.2 the table are also standard enthalpies of formation?\n55. Does the standard enthalpy of formation of $\\mathrm{H}_{2} \\mathrm{O}(g)$ differ from $\\Delta H^{\\circ}$ for the reaction\n$2 \\mathrm{H}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(g) ?$\n56. Joseph Priestly prepared oxygen in 1774 by heating red mercury(II) oxide with sunlight focused through a lens. How much heat is required to decompose exactly 1 mole of red $\\mathrm{HgO}(s)$ to $\\mathrm{Hg}(\\mathrm{I})$ and $\\mathrm{O}_{2}(g)$ under standard conditions?"}
{"id": 3497, "contents": "1044. Exercises - 1044.3. Enthalpy\n57. How many kilojoules of heat will be released when exactly 1 mole of manganese, Mn , is burned to form $\\mathrm{Mn}_{3} \\mathrm{O}_{4}(s)$ at standard state conditions?\n58. How many kilojoules of heat will be released when exactly 1 mole of iron, Fe , is burned to form $\\mathrm{Fe}_{2} \\mathrm{O}_{3}(s)$ at standard state conditions?\n59. The following sequence of reactions occurs in the commercial production of aqueous nitric acid:\n$4 \\mathrm{NH}_{3}(g)+5 \\mathrm{O}_{2}(g) \\longrightarrow 4 \\mathrm{NO}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(l) \\quad \\Delta H=-907 \\mathrm{~kJ}$\n$2 \\mathrm{NO}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{NO}_{2}(g) \\quad \\Delta H=-113 \\mathrm{~kJ}$\n$3 \\mathrm{NO}_{2}+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{HNO}_{3}(a q)+\\mathrm{NO}(g) \\quad \\Delta H=-139 \\mathrm{~kJ}$\nDetermine the total energy change for the production of one mole of aqueous nitric acid by this process.\n60. Both graphite and diamond burn.\n$\\mathrm{C}(s$, diamond $)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)$\nFor the conversion of graphite to diamond:\n$\\mathrm{C}(s$, graphite $) \\longrightarrow \\mathrm{C}(s$, diamond $) \\quad \\Delta H^{\\circ}=1.90 \\mathrm{~kJ}$\nWhich produces more heat, the combustion of graphite or the combustion of diamond?\n61. From the molar heats of formation in Appendix G, determine how much heat is required to evaporate one mole of water: $\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(g)$\n62. Which produces more heat?"}
{"id": 3498, "contents": "1044. Exercises - 1044.3. Enthalpy\n62. Which produces more heat?\n$\\mathrm{Os}(s) \\longrightarrow 2 \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{OsO}_{4}(s)$\nor\n$\\mathrm{Os}(s) \\longrightarrow 2 \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{OsO}_{4}(g)$\nfor the phase change $\\mathrm{OsO}_{4}(s) \\longrightarrow \\mathrm{OsO}_{4}(g) \\quad \\Delta H=56.4 \\mathrm{~kJ}$\n63. Calculate $\\Delta \\boldsymbol{H}^{\\circ}$ for the process\n$\\mathrm{Sb}(s)+\\frac{5}{2} \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{SbCl}_{5}(s)$\nfrom the following information:\n$\\mathrm{Sb}(s)+\\frac{3}{2} \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{SbCl}_{3}(s) \\quad \\Delta H^{\\circ}=-314 \\mathrm{~kJ}$\n$\\mathrm{SbCl}_{3}(s)+\\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{SbCl}_{5}(s) \\quad \\Delta H^{\\circ}=-80 \\mathrm{~kJ}$\n64. Calculate $\\Delta H^{\\circ}$ for the process $\\mathrm{Zn}(s)+\\mathrm{S}(s)+2 \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{ZnSO}_{4}(s)$\nfrom the following information:\n$\\mathrm{Zn}(s)+\\mathrm{S}(s) \\longrightarrow \\mathrm{ZnS}(s) \\quad \\Delta H^{\\circ}=-206.0 \\mathrm{~kJ}$\n$\\mathrm{ZnS}(s)+2 \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{ZnSO}_{4}(s) \\quad \\Delta H^{\\circ}=-776.8 \\mathrm{~kJ}$\n65. Calculate $\\Delta H$ for the process $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}(s) \\longrightarrow 2 \\mathrm{Hg}(l)+\\mathrm{Cl}_{2}(g)$ from the following information:"}
{"id": 3499, "contents": "1044. Exercises - 1044.3. Enthalpy\n$$\n\\begin{array}{lr}\n\\mathrm{Hg}(l)+\\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{HgCl}_{2}(s) & \\Delta H=-224 \\mathrm{~kJ} \\\\\n\\mathrm{Hg}(l)+\\mathrm{HgCl}_{2}(s) \\longrightarrow \\mathrm{Hg}_{2} \\mathrm{Cl}_{2}(s) & \\Delta H=-41.2 \\mathrm{~kJ}\n\\end{array}\n$$\n\n66. Calculate $\\Delta H^{\\circ}$ for the process $\\mathrm{Co}_{3} \\mathrm{O}_{4}(s) \\longrightarrow 3 \\mathrm{Co}(s)+2 \\mathrm{O}_{2}(g)$ from the following information:\n\n$$\n\\begin{array}{lr}\n\\mathrm{Co}(s)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CoO}(s) & \\Delta H^{\\circ}=-237.9 \\mathrm{~kJ} \\\\\n3 \\mathrm{CoO}(s)+\\frac{1}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{Co}_{3} \\mathrm{O}_{4}(s) & \\Delta H^{\\circ}=-177.5 \\mathrm{~kJ}\n\\end{array}\n$$"}
{"id": 3500, "contents": "1044. Exercises - 1044.3. Enthalpy\n67. Calculate the standard molar enthalpy of formation of $\\mathrm{NO}(g)$ from the following data:\n$\\mathrm{N}_{2}(g)+2 \\mathrm{O}_{2} \\longrightarrow 2 \\mathrm{NO}_{2}(g) \\quad \\Delta H^{\\circ}=66.4 \\mathrm{~kJ}$\n$2 \\mathrm{NO}(g)+\\mathrm{O}_{2} \\longrightarrow 2 \\mathrm{NO}_{2}(g) \\quad \\Delta H^{\\circ}=-114.1 \\mathrm{~kJ}$\n68. Using the data in Appendix G, calculate the standard enthalpy change for each of the following reactions:\n(a) $\\mathrm{N}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{NO}(g)$\n(b) $\\mathrm{Si}(s)+2 \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{SiCl}_{4}(g)$\n(c) $\\mathrm{Fe}_{2} \\mathrm{O}_{3}(s)+3 \\mathrm{H}_{2}(g) \\longrightarrow 2 \\mathrm{Fe}(s)+3 \\mathrm{H}_{2} \\mathrm{O}(l)$\n(d) $2 \\mathrm{LiOH}(s)+\\mathrm{CO}_{2}(g) \\longrightarrow \\mathrm{Li}_{2} \\mathrm{CO}_{3}(s)+\\mathrm{H}_{2} \\mathrm{O}(g)$\n69. Using the data in Appendix G, calculate the standard enthalpy change for each of the following reactions:\n(a) $\\mathrm{Si}(s)+2 \\mathrm{~F}_{2}(g) \\longrightarrow \\mathrm{SiF}_{4}(g)$\n(b) $2 \\mathrm{C}(s)+2 \\mathrm{H}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(l)$\n(c) $\\mathrm{CH}_{4}(g)+\\mathrm{N}_{2}(g) \\longrightarrow \\mathrm{HCN}(g)+\\mathrm{NH}_{3}(g)$;"}
{"id": 3501, "contents": "1044. Exercises - 1044.3. Enthalpy\n(c) $\\mathrm{CH}_{4}(g)+\\mathrm{N}_{2}(g) \\longrightarrow \\mathrm{HCN}(g)+\\mathrm{NH}_{3}(g)$;\n(d) $\\mathrm{CS}_{2}(g)+3 \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{CCl}_{4}(g)+\\mathrm{S}_{2} \\mathrm{Cl}_{2}(g)$\n70. The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions for each.\n(a) $2 \\mathrm{Ag}_{2} \\mathrm{O}(s) \\longrightarrow 4 \\mathrm{Ag}(s)+\\mathrm{O}_{2}(g)$\n(b) $\\mathrm{SnO}(s)+\\mathrm{CO}(g) \\longrightarrow \\mathrm{Sn}(s)+\\mathrm{CO}_{2}(g)$\n(c) $\\mathrm{Cr}_{2} \\mathrm{O}_{3}(s)+3 \\mathrm{H}_{2}(g) \\longrightarrow 2 \\mathrm{Cr}(s)+3 \\mathrm{H}_{2} \\mathrm{O}(l)$\n(d) $2 \\mathrm{Al}(s)+\\mathrm{Fe}_{2} \\mathrm{O}_{3}(s) \\longrightarrow \\mathrm{Al}_{2} \\mathrm{O}_{3}(s)+2 \\mathrm{Fe}(s)$\n71. The decomposition of hydrogen peroxide, $\\mathrm{H}_{2} \\mathrm{O}_{2}$, has been used to provide thrust in the control jets of various space vehicles. Using the data in Appendix G, determine how much heat is produced by the decomposition of exactly 1 mole of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ under standard conditions.\n$2 \\mathrm{H}_{2} \\mathrm{O}_{2}(\\mathrm{l}) \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(g)+\\mathrm{O}_{2}(g)$"}
{"id": 3502, "contents": "1044. Exercises - 1044.3. Enthalpy\n$2 \\mathrm{H}_{2} \\mathrm{O}_{2}(\\mathrm{l}) \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(g)+\\mathrm{O}_{2}(g)$\n72. Calculate the enthalpy of combustion of propane, $\\mathrm{C}_{3} \\mathrm{H}_{8}(\\mathrm{~g})$, for the formation of $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ and $\\mathrm{CO}_{2}(\\mathrm{~g})$. The enthalpy of formation of propane is $-104 \\mathrm{~kJ} / \\mathrm{mol}$.\n73. Calculate the enthalpy of combustion of butane, $\\mathrm{C}_{4} \\mathrm{H}_{10}(g)$ for the formation of $\\mathrm{H}_{2} \\mathrm{O}(g)$ and $\\mathrm{CO}_{2}(g)$. The enthalpy of formation of butane is $-126 \\mathrm{~kJ} / \\mathrm{mol}$.\n74. Both propane and butane are used as gaseous fuels. Which compound produces more heat per gram when burned?\n75. The white pigment $\\mathrm{TiO}_{2}$ is prepared by the reaction of titanium tetrachloride, $\\mathrm{TiCl}_{4}$, with water vapor in the gas phase: $\\mathrm{TiCl}_{4}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(g) \\longrightarrow \\mathrm{TiO}_{2}(s)+4 \\mathrm{HCl}(g)$.\nHow much heat is evolved in the production of exactly 1 mole of $\\mathrm{TiO}_{2}(s)$ under standard state conditions?\n76. Water gas, a mixture of $\\mathrm{H}_{2}$ and CO , is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon: $\\mathrm{C}(s)+\\mathrm{H}_{2} \\mathrm{O}(g) \\longrightarrow \\mathrm{CO}(g)+\\mathrm{H}_{2}(g)$.\n(a) Assuming that coke has the same enthalpy of formation as graphite, calculate $\\Delta H^{\\circ}$ for this reaction."}
{"id": 3503, "contents": "1044. Exercises - 1044.3. Enthalpy\n(a) Assuming that coke has the same enthalpy of formation as graphite, calculate $\\Delta H^{\\circ}$ for this reaction.\n(b) Methanol, a liquid fuel that could possibly replace gasoline, can be prepared from water gas and additional hydrogen at high temperature and pressure in the presence of a suitable catalyst:\n$2 \\mathrm{H}_{2}(g)+\\mathrm{CO}(g) \\longrightarrow \\mathrm{CH}_{3} \\mathrm{OH}(g)$.\nUnder the conditions of the reaction, methanol forms as a gas. Calculate $\\Delta H^{\\circ}$ for this reaction and for the condensation of gaseous methanol to liquid methanol.\n(c) Calculate the heat of combustion of 1 mole of liquid methanol to $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ and $\\mathrm{CO}_{2}(\\mathrm{~g})$.\n77. In the early days of automobiles, illumination at night was provided by burning acetylene, $\\mathrm{C}_{2} \\mathrm{H}_{2}$. Though no longer used as auto headlamps, acetylene is still used as a source of light by some cave explorers. The acetylene is (was) prepared in the lamp by the reaction of water with calcium carbide, $\\mathrm{CaC}_{2}$ :\n$\\mathrm{CaC}_{2}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Ca}(\\mathrm{OH})_{2}(s)+\\mathrm{C}_{2} \\mathrm{H}_{2}(g)$.\nCalculate the standard enthalpy of the reaction. The $\\Delta H_{\\mathrm{f}}^{\\circ}$ of $\\mathrm{CaC}_{2}$ is $-15.14 \\mathrm{kcal} / \\mathrm{mol}$.\n78. From the data in Table 9.2, determine which of the following fuels produces the greatest amount of heat per gram when burned under standard conditions: $\\mathrm{CO}(g), \\mathrm{CH}_{4}(g)$, or $\\mathrm{C}_{2} \\mathrm{H}_{2}(g)$."}
{"id": 3504, "contents": "1044. Exercises - 1044.3. Enthalpy\n79. The enthalpy of combustion of hard coal averages $-35 \\mathrm{~kJ} / \\mathrm{g}$, that of gasoline, $1.28 \\times 10^{5} \\mathrm{~kJ} / \\mathrm{gal}$. How many kilograms of hard coal provide the same amount of heat as is available from 1.0 gallon of gasoline? Assume that the density of gasoline is $0.692 \\mathrm{~g} / \\mathrm{mL}$ (the same as the density of isooctane).\n80. Ethanol, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$, is used as a fuel for motor vehicles, particularly in Brazil.\n(a) Write the balanced equation for the combustion of ethanol to $\\mathrm{CO}_{2}(\\mathrm{~g})$ and $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$, and, using the data in Appendix G, calculate the enthalpy of combustion of 1 mole of ethanol.\n(b) The density of ethanol is $0.7893 \\mathrm{~g} / \\mathrm{mL}$. Calculate the enthalpy of combustion of exactly 1 L of ethanol.\n(c) Assuming that an automobile's mileage is directly proportional to the heat of combustion of the fuel, calculate how much farther an automobile could be expected to travel on 1 L of gasoline than on 1 L of ethanol. Assume that gasoline has the heat of combustion and the density of n -octane, $\\mathrm{C}_{8} \\mathrm{H}_{18}$\n$\\left(\\Delta H_{\\mathrm{f}}^{\\circ}=-208.4 \\mathrm{~kJ} / \\mathrm{mol}\\right.$; density $\\left.=0.7025 \\mathrm{~g} / \\mathrm{mL}\\right)$."}
{"id": 3505, "contents": "1044. Exercises - 1044.3. Enthalpy\n81. Among the substances that react with oxygen and that have been considered as potential rocket fuels are diborane $\\left[\\mathrm{B}_{2} \\mathrm{H}_{6}\\right.$, produces $\\mathrm{B}_{2} \\mathrm{O}_{3}(\\mathrm{~s})$ and $\\left.\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})\\right]$, methane $\\left[\\mathrm{CH}_{4}\\right.$, produces $\\mathrm{CO}_{2}(\\mathrm{~g})$ and $\\left.\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})\\right]$, and hydrazine [ $\\mathrm{N}_{2} \\mathrm{H}_{4}$, produces $\\mathrm{N}_{2}(\\mathrm{~g})$ and $\\left.\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})\\right]$. On the basis of the heat released by 1.00 g of each substance in its reaction with oxygen, which of these compounds offers the best possibility as a rocket fuel? The $\\Delta H_{\\mathrm{f}}^{\\circ}$ of $\\mathrm{B}_{2} \\mathrm{H}_{6}(\\mathrm{~g}), \\mathrm{CH}_{4}(\\mathrm{~g})$, and $\\mathrm{N}_{2} \\mathrm{H}_{4}(I)$ may be found in Appendix G .\n82. How much heat is produced when 1.25 g of chromium metal reacts with oxygen gas under standard conditions?\n83. Ethylene, $\\mathrm{C}_{2} \\mathrm{H}_{4}$, a byproduct from the fractional distillation of petroleum, is fourth among the 50 chemical compounds produced commercially in the largest quantities. About $80 \\%$ of synthetic ethanol is manufactured from ethylene by its reaction with water in the presence of a suitable catalyst.\n$\\mathrm{C}_{2} \\mathrm{H}_{4}(g)+\\mathrm{H}_{2} \\mathrm{O}(g) \\longrightarrow \\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(l)$\nUsing the data in the table in Appendix G, calculate $\\Delta H^{\\circ}$ for the reaction."}
{"id": 3506, "contents": "1044. Exercises - 1044.3. Enthalpy\nUsing the data in the table in Appendix G, calculate $\\Delta H^{\\circ}$ for the reaction.\n84. The oxidation of the sugar glucose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$, is described by the following equation:\n$\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}(s)+6 \\mathrm{O}_{2}(g) \\longrightarrow 6 \\mathrm{CO}_{2}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(l) \\quad \\Delta H=-2816 \\mathrm{~kJ}$\nThe metabolism of glucose gives the same products, although the glucose reacts with oxygen in a series of steps in the body.\n(a) How much heat in kilojoules can be produced by the metabolism of 1.0 g of glucose?\n(b) How many Calories can be produced by the metabolism of 1.0 g of glucose?\n85. Propane, $\\mathrm{C}_{3} \\mathrm{H}_{8}$, is a hydrocarbon that is commonly used as a fuel.\n(a) Write a balanced equation for the complete combustion of propane gas.\n(b) Calculate the volume of air at $25^{\\circ} \\mathrm{C}$ and 1.00 atmosphere that is needed to completely combust 25.0 grams of propane. Assume that air is 21.0 percent $\\mathrm{O}_{2}$ by volume. (Hint: We will see how to do this calculation in a later chapter on gases-for now use the information that 1.00 L of air at $25^{\\circ} \\mathrm{C}$ and 1.00 atm contains 0.275 g of $\\mathrm{O}_{2}$.)"}
{"id": 3507, "contents": "1044. Exercises - 1044.3. Enthalpy\n(c) The heat of combustion of propane is $-2,219.2 \\mathrm{~kJ} / \\mathrm{mol}$. Calculate the heat of formation, $\\Delta \\boldsymbol{H}_{\\mathrm{f}}^{\\circ}$ of propane given that $\\Delta H_{\\mathrm{f}}^{\\circ}$ of $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{I})=-285.8 \\mathrm{~kJ} / \\mathrm{mol}$ and $\\Delta H_{\\mathrm{f}}^{\\circ}$ of $\\mathrm{CO}_{2}(\\mathrm{~g})=-393.5 \\mathrm{~kJ} / \\mathrm{mol}$.\n(d) Assuming that all of the heat released in burning 25.0 grams of propane is transferred to 4.00\nkilograms of water, calculate the increase in temperature of the water.\n86. During a recent winter month in Sheboygan, Wisconsin, it was necessary to obtain 3500 kWh of heat provided by a natural gas furnace with $89 \\%$ efficiency to keep a small house warm (the efficiency of a gas furnace is the percent of the heat produced by combustion that is transferred into the house).\n(a) Assume that natural gas is pure methane and determine the volume of natural gas in cubic feet that was required to heat the house. The average temperature of the natural gas was $56^{\\circ} \\mathrm{F}$; at this temperature and a pressure of 1 atm, natural gas has a density of $0.681 \\mathrm{~g} / \\mathrm{L}$.\n(b) How many gallons of LPG (liquefied petroleum gas) would be required to replace the natural gas used? Assume the LPG is liquid propane [ $\\mathrm{C}_{3} \\mathrm{H}_{8}$ : density, $0.5318 \\mathrm{~g} / \\mathrm{mL}$; enthalpy of combustion, $2219 \\mathrm{~kJ} / \\mathrm{mol}$ for the formation of $\\mathrm{CO}_{2}(g)$ and $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{I}]$ and the furnace used to burn the LPG has the same efficiency as the gas furnace."}
{"id": 3508, "contents": "1044. Exercises - 1044.3. Enthalpy\n(c) What mass of carbon dioxide is produced by combustion of the methane used to heat the house?\n(d) What mass of water is produced by combustion of the methane used to heat the house?\n(e) What volume of air is required to provide the oxygen for the combustion of the methane used to heat the house? Air contains $23 \\%$ oxygen by mass. The average density of air during the month was $1.22 \\mathrm{~g} / \\mathrm{L}$. (f) How many kilowatt-hours ( $1 \\mathrm{kWh}=3.6 \\times 10^{6} \\mathrm{~J}$ ) of electricity would be required to provide the heat necessary to heat the house? Note electricity is $100 \\%$ efficient in producing heat inside a house.\n(g) Although electricity is $100 \\%$ efficient in producing heat inside a house, production and distribution of electricity is not $100 \\%$ efficient. The efficiency of production and distribution of electricity produced in a coal-fired power plant is about $40 \\%$. A certain type of coal provides 2.26 kWh per pound upon combustion. What mass of this coal in kilograms will be required to produce the electrical energy necessary to heat the house if the efficiency of generation and distribution is $40 \\%$ ?"}
{"id": 3509, "contents": "1044. Exercises - 1044.4. Strengths of Ionic and Covalent Bonds\n87. Which bond in each of the following pairs of bonds is the strongest?\n(a) $\\mathrm{C}-\\mathrm{C}$ or $\\mathrm{C}=\\mathrm{C}$\n(b) $\\mathrm{C}-\\mathrm{N}$ or $\\mathrm{C} \\equiv \\mathrm{N}$\n(c) $\\mathrm{C} \\equiv \\mathrm{O}$ or $\\mathrm{C}=\\mathrm{O}$\n(d) $\\mathrm{H}-\\mathrm{F}$ or $\\mathrm{H}-\\mathrm{Cl}$\n(e) $\\mathrm{C}-\\mathrm{H}$ or $\\mathrm{O}-\\mathrm{H}$\n(f) $\\mathrm{C}-\\mathrm{N}$ or $\\mathrm{C}-\\mathrm{O}$\n88. Using the bond energies in Table 9.3, determine the approximate enthalpy change for each of the following reactions:\n(a) $\\mathrm{H}_{2}(g)+\\mathrm{Br}_{2}(g) \\longrightarrow 2 \\mathrm{HBr}(g)$\n(b) $\\mathrm{CH}_{4}(g)+\\mathrm{I}_{2}(g) \\longrightarrow \\mathrm{CH}_{3} \\mathrm{I}(g)+\\mathrm{HI}(g)$\n(c) $\\mathrm{C}_{2} \\mathrm{H}_{4}(g)+3 \\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{CO}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(g)$\n89. Using the bond energies in Table 9.3, determine the approximate enthalpy change for each of the following reactions:\n(a) $\\mathrm{Cl}_{2}(g)+3 \\mathrm{~F}_{2}(g) \\longrightarrow 2 \\mathrm{ClF}_{3}(g)$\n(b) $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}_{2}(g)+\\mathrm{H}_{2}(g) \\longrightarrow \\mathrm{H}_{3} \\mathrm{CCH}_{3}(g)$"}
{"id": 3510, "contents": "1044. Exercises - 1044.4. Strengths of Ionic and Covalent Bonds\n(c) $2 \\mathrm{C}_{2} \\mathrm{H}_{6}(g)+7 \\mathrm{O}_{2}(g) \\longrightarrow 4 \\mathrm{CO}_{2}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(g)$\n90. Draw a curve that describes the energy of a system with H and Cl atoms at varying distances. Then, find the minimum energy of this curve two ways.\n(a) Use the bond energy found in Table 9.3 and Table 9.4 to calculate the energy for one single HCl bond (Hint: How many bonds are in a mole?)\n(b) Use the enthalpy of reaction and the bond energies for $\\mathrm{H}_{2}$ and $\\mathrm{Cl}_{2}$ to solve for the energy of one mole of HCl bonds."}
{"id": 3511, "contents": "1044. Exercises - 1044.4. Strengths of Ionic and Covalent Bonds\n$$\n\\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g) \\rightleftharpoons 2 \\mathrm{HCl}(g) \\quad \\Delta H_{\\mathrm{rxn}}^{\\circ}=-184.7 \\mathrm{~kJ} / \\mathrm{mol}\n$$\n\n91. Explain why bonds occur at specific average bond distances instead of the atoms approaching each other infinitely close.\n92. When a molecule can form two different structures, the structure with the stronger bonds is usually the more stable form. Use bond energies to predict the correct structure of the hydroxylamine molecule:\n\n93. How does the bond energy of $\\mathrm{HCl}(\\mathrm{g})$ differ from the standard enthalpy of formation of $\\mathrm{HCl}(\\mathrm{g})$ ?\n94. Using the standard enthalpy of formation data in Appendix G, show how the standard enthalpy of formation of $\\mathrm{HCl}(\\mathrm{g})$ can be used to determine the bond energy.\n95. Using the standard enthalpy of formation data in Appendix G, calculate the bond energy of the carbonsulfur double bond in $\\mathrm{CS}_{2}$.\n96. Using the standard enthalpy of formation data in Appendix $G$, determine which bond is stronger: the $S-F$ bond in $\\mathrm{SF}_{4}(g)$ or in $\\mathrm{SF}_{6}(g)$ ?\n97. Using the standard enthalpy of formation data in Appendix G , determine which bond is stronger: the $\\mathrm{P}-\\mathrm{Cl}$ bond in $\\mathrm{PCl}_{3}(g)$ or in $\\mathrm{PCl}_{5}(g)$ ?\n98. Complete the following Lewis structure by adding bonds (not atoms), and then indicate the longest bond:\n\n$$\n\\begin{array}{llll}\n\\mathrm{H} \\quad \\mathrm{H} & \\mathrm{H} & \\mathrm{H}\n\\end{array}\n$$"}
{"id": 3512, "contents": "1044. Exercises - 1044.4. Strengths of Ionic and Covalent Bonds\n$$\n\\begin{array}{llll}\n\\mathrm{H} \\quad \\mathrm{H} & \\mathrm{H} & \\mathrm{H}\n\\end{array}\n$$\n\n$\\mathrm{H} C \\quad \\mathrm{C} \\quad \\mathrm{C} \\quad \\mathrm{C} \\quad \\mathrm{C} \\quad \\mathrm{H}$\nH H\n99. Use the bond energy to calculate an approximate value of $\\Delta H$ for the following reaction. Which is the more stable form of $\\mathrm{FNO}_{2}$ ?\n\n100. Use principles of atomic structure to answer each of the following: ${ }^{4}$\n(a) The radius of the Ca atom is 197 pm ; the radius of the $\\mathrm{Ca}^{2+}$ ion is 99 pm . Account for the difference.\n(b) The lattice energy of $\\mathrm{CaO}(s)$ is $-3460 \\mathrm{~kJ} / \\mathrm{mol}$; the lattice energy of $\\mathrm{K}_{2} \\mathrm{O}$ is $-2240 \\mathrm{~kJ} / \\mathrm{mol}$. Account for the difference.\n(c) Given these ionization values, explain the difference between Ca and K with regard to their first and second ionization energies.\n\n| Element | First lonization Energy (kJ/mol) | Second lonization Energy (kJ/ <br> mol) |\n| :---: | :---: | :---: |\n| K | 419 | 3050 |\n| Ca | 590 | 1140 |\n\n(d) The first ionization energy of Mg is $738 \\mathrm{~kJ} / \\mathrm{mol}$ and that of Al is $578 \\mathrm{~kJ} / \\mathrm{mol}$. Account for this difference."}
{"id": 3513, "contents": "1044. Exercises - 1044.4. Strengths of Ionic and Covalent Bonds\n[^8]101. The lattice energy of LiF is $1023 \\mathrm{~kJ} / \\mathrm{mol}$, and the Li-F distance is 200.8 pm . NaF crystallizes in the same structure as LiF but with a $\\mathrm{Na}-\\mathrm{F}$ distance of 231 pm . Which of the following values most closely approximates the lattice energy of NaF: 510, 890, 1023, 1175 , or $4090 \\mathrm{~kJ} / \\mathrm{mol}$ ? Explain your choice.\n102. For which of the following substances is the least energy required to convert one mole of the solid into separate ions?\n(a) MgO\n(b) SrO\n(c) KF\n(d) CsF\n(e) $\\mathrm{MgF}_{2}$\n103. The reaction of a metal, $M$, with a halogen, $X_{2}$, proceeds by an exothermic reaction as indicated by this equation: $\\mathrm{M}(s)+\\mathrm{X}_{2}(g) \\longrightarrow \\mathrm{MX}_{2}(s)$. For each of the following, indicate which option will make the reaction more exothermic. Explain your answers.\n(a) a large radius vs. a small radius for $\\mathrm{M}^{+2}$\n(b) a high ionization energy vs. a low ionization energy for $M$\n(c) an increasing bond energy for the halogen\n(d) a decreasing electron affinity for the halogen\n(e) an increasing size of the anion formed by the halogen"}
{"id": 3514, "contents": "1044. Exercises - 1044.4. Strengths of Ionic and Covalent Bonds\n(c) an increasing bond energy for the halogen\n(d) a decreasing electron affinity for the halogen\n(e) an increasing size of the anion formed by the halogen\n104. The lattice energy of LiF is $1023 \\mathrm{~kJ} / \\mathrm{mol}$, and the Li-F distance is 201 pm . MgO crystallizes in the same structure as LiF but with a $\\mathrm{Mg}-\\mathrm{O}$ distance of 205 pm . Which of the following values most closely approximates the lattice energy of MgO: $256 \\mathrm{~kJ} / \\mathrm{mol}, 512 \\mathrm{~kJ} / \\mathrm{mol}, 1023 \\mathrm{~kJ} / \\mathrm{mol}, 2046 \\mathrm{~kJ} / \\mathrm{mol}$, or $4008 \\mathrm{~kJ} /$ mol? Explain your choice.\n105. Which compound in each of the following pairs has the larger lattice energy? Note: $\\mathrm{Mg}^{2+}$ and $\\mathrm{Li}^{+}$have similar radii; $\\mathrm{O}^{2-}$ and $\\mathrm{F}^{-}$have similar radii. Explain your choices.\n(a) MgO or MgSe\n(b) LiF or MgO\n(c) $\\mathrm{Li}_{2} \\mathrm{O}$ or LiCl\n(d) $\\mathrm{Li}_{2} \\mathrm{Se}$ or MgO\n106. Which compound in each of the following pairs has the larger lattice energy? Note: $\\mathrm{Ba}^{2+}$ and $\\mathrm{K}^{+}$have similar radii; $\\mathrm{S}^{2-}$ and $\\mathrm{Cl}^{-}$have similar radii. Explain your choices.\n(a) $\\mathrm{K}_{2} \\mathrm{O}$ or $\\mathrm{Na}_{2} \\mathrm{O}$\n(b) $\\mathrm{K}_{2} \\mathrm{~S}$ or BaS\n(c) KCl or BaS\n(d) BaS or $\\mathrm{BaCl}_{2}$\n107. Which of the following compounds requires the most energy to convert one mole of the solid into separate ions?\n(a) MgO\n(b) SrO\n(c) KF\n(d) CsF"}
{"id": 3515, "contents": "1044. Exercises - 1044.4. Strengths of Ionic and Covalent Bonds\n107. Which of the following compounds requires the most energy to convert one mole of the solid into separate ions?\n(a) MgO\n(b) SrO\n(c) KF\n(d) CsF\n(e) $\\mathrm{MgF}_{2}$\n108. Which of the following compounds requires the most energy to convert one mole of the solid into separate ions?\n(a) $\\mathrm{K}_{2} \\mathrm{~S}$\n(b) $\\mathrm{K}_{2} \\mathrm{O}$\n(c) CaS\n(d) $\\mathrm{Cs}_{2} \\mathrm{~S}$\n(e) CaO\n109. The lattice energy of $K F$ is $794 \\mathrm{~kJ} / \\mathrm{mol}$, and the interionic distance is 269 pm . The $\\mathrm{Na}-\\mathrm{F}$ distance in NaF, which has the same structure as KF, is 231 pm . Which of the following values is the closest approximation of the lattice energy of NaF: $682 \\mathrm{~kJ} / \\mathrm{mol}, 794 \\mathrm{~kJ} / \\mathrm{mol}, 924 \\mathrm{~kJ} / \\mathrm{mol}, 1588 \\mathrm{~kJ} / \\mathrm{mol}$, or 3175 kJ/mol? Explain your answer."}
{"id": 3516, "contents": "1044. Exercises - 1044.4. Strengths of Ionic and Covalent Bonds\n474 9\u2022Exercises"}
{"id": 3517, "contents": "1045. CHAPTER 10 <br> Liquids and Solids - \nFigure 10.1 Solid carbon dioxide (\"dry ice\", left) sublimes vigorously when placed in a liquid (right), cooling the liquid and generating a dense mist of water above the cylinder. (credit: modification of work by Paul Flowers)"}
{"id": 3518, "contents": "1046. CHAPTER OUTLINE - 1046.1. Intermolecular Forces\n10.2 Properties of Liquids\n10.3 Phase Transitions\n10.4 Phase Diagrams\n10.5 The Solid State of Matter\n10.6 Lattice Structures in Crystalline Solids\n\nINTRODUCTION Leprosy has been a devastating disease throughout much of human history. Aside from the symptoms and complications of the illness, its social stigma led sufferers to be cast out of communities and isolated in colonies; in some regions this practice lasted well into the twentieth century. At that time, the best potential treatment for leprosy was oil from the chaulmoogra tree, but the oil was extremely thick, causing blisters and making usage painful and ineffective. Healthcare professionals seeking a better application contacted Alice Ball, a young chemist at the University of Hawaii, who had focused her masters thesis on a similar plant. Ball initiated a sequence of procedures (repeated acidification and purification to change the characteristics of the oil and isolate the active substances (esters, discussed later in this text). The \"Ball Method\" as it later came to be called, became the standard treatment for leprosy for decades. In the liquid and solid states, atomic and molecular interactions are of considerable strength and play an important role in determining a number of physical properties of the substance. For example, the thickness, or viscosity, of the chaulmoogra oil was due to its intermolecular forces. In this chapter, the nature of these interactions and their effects on various physical properties of liquid and solid phases will be examined."}
{"id": 3519, "contents": "1047. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the types of intermolecular forces possible between atoms or molecules in condensed phases (dispersion forces, dipole-dipole attractions, and hydrogen bonding)\n- Identify the types of intermolecular forces experienced by specific molecules based on their structures\n- Explain the relation between the intermolecular forces present within a substance and the temperatures associated with changes in its physical state\n\nAs was the case for gaseous substances, the kinetic molecular theory may be used to explain the behavior of solids and liquids. In the following description, the term particle will be used to refer to an atom, molecule, or ion. Note that we will use the popular phrase \"intermolecular attraction\" to refer to attractive forces between the particles of a substance, regardless of whether these particles are molecules, atoms, or ions.\n\nConsider these two aspects of the molecular-level environments in solid, liquid, and gaseous matter:\n\n- Particles in a solid are tightly packed together and often arranged in a regular pattern; in a liquid, they are close together with no regular arrangement; in a gas, they are far apart with no regular arrangement.\n- Particles in a solid vibrate about fixed positions and do not generally move in relation to one another; in a liquid, they move past each other but remain in essentially constant contact; in a gas, they move independently of one another except when they collide.\n\nThe differences in the properties of a solid, liquid, or gas reflect the strengths of the attractive forces between the atoms, molecules, or ions that make up each phase. The phase in which a substance exists depends on the relative extents of its intermolecular forces (IMFs) and the kinetic energies (KE) of its molecules. IMFs are the various forces of attraction that may exist between the atoms and molecules of a substance due to electrostatic phenomena, as will be detailed in this module. These forces serve to hold particles close together, whereas the particles' KE provides the energy required to overcome the attractive forces and thus increase the distance between particles. Figure 10.2 illustrates how changes in physical state may be induced by changing the temperature, hence, the average KE , of a given substance."}
{"id": 3520, "contents": "1047. LEARNING OBJECTIVES - \nFIGURE 10.2 Transitions between solid, liquid, and gaseous states of a substance occur when conditions of temperature or pressure favor the associated changes in intermolecular forces. (Note: The space between particles in the gas phase is much greater than shown.)\n\nAs an example of the processes depicted in this figure, consider a sample of water. When gaseous water is cooled sufficiently, the attractions between $\\mathrm{H}_{2} \\mathrm{O}$ molecules will be capable of holding them together when they come into contact with each other; the gas condenses, forming liquid $\\mathrm{H}_{2} \\mathrm{O}$. For example, liquid water forms on the outside of a cold glass as the water vapor in the air is cooled by the cold glass, as seen in Figure 10.3.\n\n\nFIGURE 10.3 Condensation forms when water vapor in the air is cooled enough to form liquid water, such as (a) on the outside of a cold beverage glass or (b) in the form of fog. (credit a: modification of work by Jenny Downing; credit b: modification of work by Cory Zanker)\n\nWe can also liquefy many gases by compressing them, if the temperature is not too high. The increased pressure brings the molecules of a gas closer together, such that the attractions between the molecules become strong relative to their KE. Consequently, they form liquids. Butane, $\\mathrm{C}_{4} \\mathrm{H}_{10}$, is the fuel used in disposable lighters and is a gas at standard temperature and pressure. Inside the lighter's fuel compartment, the butane is compressed to a pressure that results in its condensation to the liquid state, as shown in Figure 10.4.\n\n\nFIGURE 10.4 Gaseous butane is compressed within the storage compartment of a disposable lighter, resulting in its condensation to the liquid state. (credit: modification of work by \"Sam-Cat\"/Flickr)\n\nFinally, if the temperature of a liquid becomes sufficiently low, or the pressure on the liquid becomes sufficiently high, the molecules of the liquid no longer have enough KE to overcome the IMF between them, and a solid forms. A more thorough discussion of these and other changes of state, or phase transitions, is provided in a later module of this chapter."}
{"id": 3521, "contents": "1048. LINK TO LEARNING - \nAccess this interactive simulation (http://openstax.org/l/16phetvisual) on states of matter, phase transitions, and intermolecular forces. This simulation is useful for visualizing concepts introduced throughout this chapter."}
{"id": 3522, "contents": "1049. Forces between Molecules - \nUnder appropriate conditions, the attractions between all gas molecules will cause them to form liquids or solids. This is due to intermolecular forces, not intramolecular forces. Intramolecular forces are those within the molecule that keep the molecule together, for example, the bonds between the atoms. Intermolecular forces are the attractions between molecules, which determine many of the physical properties of a substance. Figure 10.5 illustrates these different molecular forces. The strengths of these attractive forces vary widely, though usually the IMFs between small molecules are weak compared to the intramolecular forces that bond atoms together within a molecule. For example, to overcome the IMFs in one mole of liquid HCl and convert it into gaseous HCl requires only about 17 kilojoules. However, to break the covalent bonds between the hydrogen and chlorine atoms in one mole of HCl requires about 25 times more energy-430 kilojoules.\n\n\nFIGURE 10.5 Intramolecular forces keep a molecule intact. Intermolecular forces hold multiple molecules together and determine many of a substance's properties.\n\nAll of the attractive forces between neutral atoms and molecules are known as van der Waals forces, although they are usually referred to more informally as intermolecular attraction. We will consider the various types of IMFs in the next three sections of this module."}
{"id": 3523, "contents": "1050. Dispersion Forces - \nOne of the three van der Waals forces is present in all condensed phases, regardless of the nature of the atoms or molecules composing the substance. This attractive force is called the London dispersion force in honor of German-born American physicist Fritz London who, in 1928, first explained it. This force is often referred to as simply the dispersion force. Because the electrons of an atom or molecule are in constant motion (or, alternatively, the electron's location is subject to quantum-mechanical variability), at any moment in time, an atom or molecule can develop a temporary, instantaneous dipole if its electrons are distributed asymmetrically. The presence of this dipole can, in turn, distort the electrons of a neighboring atom or molecule, producing an induced dipole. These two rapidly fluctuating, temporary dipoles thus result in a relatively weak electrostatic attraction between the species-a so-called dispersion force like that illustrated in Figure 10.6.\n\n\nFIGURE 10.6 Dispersion forces result from the formation of temporary dipoles, as illustrated here for two nonpolar diatomic molecules.\n\nDispersion forces that develop between atoms in different molecules can attract the two molecules to each other. The forces are relatively weak, however, and become significant only when the molecules are very close.\n\nLarger and heavier atoms and molecules exhibit stronger dispersion forces than do smaller and lighter atoms and molecules. $\\mathrm{F}_{2}$ and $\\mathrm{Cl}_{2}$ are gases at room temperature (reflecting weaker attractive forces); $\\mathrm{Br}_{2}$ is a liquid, and $\\mathrm{I}_{2}$ is a solid (reflecting stronger attractive forces). Trends in observed melting and boiling points for the halogens clearly demonstrate this effect, as seen in Table 10.1.\n\nMelting and Boiling Points of the Halogens"}
{"id": 3524, "contents": "1050. Dispersion Forces - \nMelting and Boiling Points of the Halogens\n\n| Halogen | Molar Mass | Atomic Radius | Melting Point | Boiling Point |\n| :---: | :---: | :---: | :---: | :---: |\n| fluorine, $\\mathrm{F}_{2}$ | $38 \\mathrm{~g} / \\mathrm{mol}$ | 72 pm | 53 K | 85 K |\n| chlorine, $\\mathrm{Cl}_{2}$ | $71 \\mathrm{~g} / \\mathrm{mol}$ | 99 pm | 172 K | 238 K |\n| bromine, $\\mathrm{Br}_{2}$ | $160 \\mathrm{~g} / \\mathrm{mol}$ | 114 pm | 266 K | 332 K |\n| iodine, $\\mathrm{I}_{2}$ | $254 \\mathrm{~g} / \\mathrm{mol}$ | 133 pm | 387 K | 457 K |\n| astatine, $\\mathrm{At}_{2}$ | $420 \\mathrm{~g} / \\mathrm{mol}$ | 150 pm | 575 K | 610 K |\n\nTABLE 10.1\n\nThe increase in melting and boiling points with increasing atomic/molecular size may be rationalized by considering how the strength of dispersion forces is affected by the electronic structure of the atoms or molecules in the substance. In a larger atom, the valence electrons are, on average, farther from the nuclei than in a smaller atom. Thus, they are less tightly held and can more easily form the temporary dipoles that produce the attraction. The measure of how easy or difficult it is for another electrostatic charge (for example, a nearby ion or polar molecule) to distort a molecule's charge distribution (its electron cloud) is known as polarizability. A molecule that has a charge cloud that is easily distorted is said to be very polarizable and will have large dispersion forces; one with a charge cloud that is difficult to distort is not very polarizable and will have small dispersion forces."}
{"id": 3525, "contents": "1052. London Forces and Their Effects - \nOrder the following compounds of a group 14 element and hydrogen from lowest to highest boiling point: $\\mathrm{CH}_{4}$, $\\mathrm{SiH}_{4}, \\mathrm{GeH}_{4}$, and $\\mathrm{SnH}_{4}$. Explain your reasoning."}
{"id": 3526, "contents": "1053. Solution - \nApplying the skills acquired in the chapter on chemical bonding and molecular geometry, all of these compounds are predicted to be nonpolar, so they may experience only dispersion forces: the smaller the molecule, the less polarizable and the weaker the dispersion forces; the larger the molecule, the larger the dispersion forces. The molar masses of $\\mathrm{CH}_{4}, \\mathrm{SiH}_{4}, \\mathrm{GeH}_{4}$, and $\\mathrm{SnH}_{4}$ are approximately $16 \\mathrm{~g} / \\mathrm{mol}, 32 \\mathrm{~g} / \\mathrm{mol}, 77 \\mathrm{~g} /$ mol, and $123 \\mathrm{~g} / \\mathrm{mol}$, respectively. Therefore, $\\mathrm{CH}_{4}$ is expected to have the lowest boiling point and $\\mathrm{SnH}_{4}$ the highest boiling point. The ordering from lowest to highest boiling point is expected to be $\\mathrm{CH}_{4}<\\mathrm{SiH}_{4}<\\mathrm{GeH}_{4}<$ $\\mathrm{SnH}_{4}$.\n\nA graph of the actual boiling points of these compounds versus the period of the group 14 element shows this prediction to be correct:"}
{"id": 3527, "contents": "1054. Check Your Learning - \nOrder the following hydrocarbons from lowest to highest boiling point: $\\mathrm{C}_{2} \\mathrm{H}_{6}, \\mathrm{C}_{3} \\mathrm{H}_{8}$, and $\\mathrm{C}_{4} \\mathrm{H}_{10}$."}
{"id": 3528, "contents": "1055. Answer: - \n$\\mathrm{C}_{2} \\mathrm{H}_{6}<\\mathrm{C}_{3} \\mathrm{H}_{8}<\\mathrm{C}_{4} \\mathrm{H}_{10}$. All of these compounds are nonpolar and only have London dispersion forces: the larger the molecule, the larger the dispersion forces and the higher the boiling point. The ordering from lowest to highest boiling point is therefore $\\mathrm{C}_{2} \\mathrm{H}_{6}<\\mathrm{C}_{3} \\mathrm{H}_{8}<\\mathrm{C}_{4} \\mathrm{H}_{10}$.\n\nThe shapes of molecules also affect the magnitudes of the dispersion forces between them. For example, boiling points for the isomers $n$-pentane, isopentane, and neopentane (shown in Figure 10.7) are $36^{\\circ} \\mathrm{C}, 27^{\\circ} \\mathrm{C}$, and $9.5^{\\circ} \\mathrm{C}$, respectively. Even though these compounds are composed of molecules with the same chemical formula, $\\mathrm{C}_{5} \\mathrm{H}_{12}$, the difference in boiling points suggests that dispersion forces in the liquid phase are different, being greatest for $n$-pentane and least for neopentane. The elongated shape of $n$-pentane provides a greater surface area available for contact between molecules, resulting in correspondingly stronger dispersion forces. The more compact shape of isopentane offers a smaller surface area available for intermolecular contact and, therefore, weaker dispersion forces. Neopentane molecules are the most compact of the three, offering the least available surface area for intermolecular contact and, hence, the weakest dispersion forces. This behavior is analogous to the connections that may be formed between strips of VELCRO brand fasteners: the greater the area of the strip's contact, the stronger the connection.\n\n\nFIGURE 10.7 The strength of the dispersion forces increases with the contact area between molecules, as demonstrated by the boiling points of these pentane isomers."}
{"id": 3529, "contents": "1057. Geckos and Intermolecular Forces - \nGeckos have an amazing ability to adhere to most surfaces. They can quickly run up smooth walls and across ceilings that have no toe-holds, and they do this without having suction cups or a sticky substance on their toes. And while a gecko can lift its feet easily as it walks along a surface, if you attempt to pick it up, it sticks to the surface. How are geckos (as well as spiders and some other insects) able to do this? Although this phenomenon has been investigated for hundreds of years, scientists only recently uncovered the details of the process that allows geckos' feet to behave this way.\n\nGeckos' toes are covered with hundreds of thousands of tiny hairs known as setae, with each seta, in turn, branching into hundreds of tiny, flat, triangular tips called spatulae. The huge numbers of spatulae on its setae provide a gecko, shown in Figure 10.8, with a large total surface area for sticking to a surface. In 2000, Kellar Autumn, who leads a multi-institutional gecko research team, found that geckos adhered equally well to both polar silicon dioxide and nonpolar gallium arsenide. This proved that geckos stick to surfaces because of dispersion forces-weak intermolecular attractions arising from temporary, synchronized charge distributions between adjacent molecules. Although dispersion forces are very weak, the total attraction over millions of spatulae is large enough to support many times the gecko's weight."}
{"id": 3530, "contents": "1057. Geckos and Intermolecular Forces - \nIn 2014, two scientists developed a model to explain how geckos can rapidly transition from \"sticky\" to \"non-sticky.\" Alex Greaney and Congcong Hu at Oregon State University described how geckos can achieve this by changing the angle between their spatulae and the surface. Geckos' feet, which are normally nonsticky, become sticky when a small shear force is applied. By curling and uncurling their toes, geckos can alternate between sticking and unsticking from a surface, and thus easily move across it. Later research led by Alyssa Stark at University of Akron showed that geckos can maintain their hold on hydrophobic surfaces (similar to the leaves in their habitats) equally well whether the surfaces were wet or dry. Stark's experiment used a ribbon to gently pull the geckos until they slipped, so that the researchers could determine the geckos' ability to hold various surfaces under wet and dry conditions. Further investigations may eventually lead to the development of better adhesives and other applications.\n\n\nFIGURE 10.8 Geckos' toes contain large numbers of tiny hairs (setae), which branch into many triangular tips (spatulae). Geckos adhere to surfaces because of van der Waals attractions between the surface and a gecko's millions of spatulae. By changing how the spatulae contact the surface, geckos can turn their stickiness \"on\" and \"off.\" (credit photo: modification of work by \"JC*+A!\"/Flickr)"}
{"id": 3531, "contents": "1058. LINK TO LEARNING - \nWatch this video (http://openstax.org/l/16kellaraut) to learn more about Kellar Autumn's research that determined that van der Waals forces are responsible for a gecko's ability to cling and climb."}
{"id": 3532, "contents": "1059. Dipole-Dipole Attractions - \nRecall from the chapter on chemical bonding and molecular geometry that polar molecules have a partial positive charge on one side and a partial negative charge on the other side of the molecule-a separation of charge called a dipole. Consider a polar molecule such as hydrogen chloride, HCl . In the HCl molecule, the more electronegative Cl atom bears the partial negative charge, whereas the less electronegative H atom bears the partial positive charge. An attractive force between HCl molecules results from the attraction between the positive end of one HCl molecule and the negative end of another. This attractive force is called a dipole-dipole attraction-the electrostatic force between the partially positive end of one polar molecule and the partially negative end of another, as illustrated in Figure 10.9.\n\n\nFIGURE 10.9 This image shows two arrangements of polar molecules, such as HCl , that allow an attraction between the partial negative end of one molecule and the partial positive end of another.\n\nThe effect of a dipole-dipole attraction is apparent when we compare the properties of HCl molecules to nonpolar $\\mathrm{F}_{2}$ molecules. Both HCl and $\\mathrm{F}_{2}$ consist of the same number of atoms and have approximately the same molecular mass. At a temperature of 150 K , molecules of both substances would have the same average KE. However, the dipole-dipole attractions between HCl molecules are sufficient to cause them to \"stick together\" to form a liquid, whereas the relatively weaker dispersion forces between nonpolar $\\mathrm{F}_{2}$ molecules are not, and so this substance is gaseous at this temperature. The higher normal boiling point of HCl ( 188 K ) compared to $\\mathrm{F}_{2}(85 \\mathrm{~K})$ is a reflection of the greater strength of dipole-dipole attractions between HCl molecules, compared to the attractions between nonpolar $\\mathrm{F}_{2}$ molecules. We will often use values such as boiling or freezing points, or enthalpies of vaporization or fusion, as indicators of the relative strengths of IMFs of attraction present within different substances."}
{"id": 3533, "contents": "1060. Dipole-Dipole Forces and Their Effects - \nPredict which will have the higher boiling point: $\\mathrm{N}_{2}$ or CO. Explain your reasoning."}
{"id": 3534, "contents": "1061. Solution - \nCO and $\\mathrm{N}_{2}$ are both diatomic molecules with masses of about 28 amu , so they experience similar London dispersion forces. Because CO is a polar molecule, it experiences dipole-dipole attractions. Because $\\mathrm{N}_{2}$ is nonpolar, its molecules cannot exhibit dipole-dipole attractions. The dipole-dipole attractions between CO molecules are comparably stronger than the dispersion forces between nonpolar $\\mathrm{N}_{2}$ molecules, so CO is expected to have the higher boiling point."}
{"id": 3535, "contents": "1062. Check Your Learning - \nPredict which will have the higher boiling point: ICl or $\\mathrm{Br}_{2}$. Explain your reasoning."}
{"id": 3536, "contents": "1063. Answer: - \nICl. ICl and $\\mathrm{Br}_{2}$ have similar masses ( $\\sim 160 \\mathrm{amu}$ ) and therefore experience similar London dispersion forces. ICl is polar and thus also exhibits dipole-dipole attractions; $\\mathrm{Br}_{2}$ is nonpolar and does not. The relatively stronger dipole-dipole attractions require more energy to overcome, so ICl will have the higher boiling point."}
{"id": 3537, "contents": "1064. Hydrogen Bonding - \nNitrosyl fluoride (ONF, molecular mass 49 amu ) is a gas at room temperature. Water ( $\\mathrm{H}_{2} \\mathrm{O}$, molecular mass 18 amu ) is a liquid, even though it has a lower molecular mass. We clearly cannot attribute this difference between the two compounds to dispersion forces. Both molecules have about the same shape and ONF is the heavier and larger molecule. It is, therefore, expected to experience more significant dispersion forces. Additionally, we cannot attribute this difference in boiling points to differences in the dipole moments of the molecules. Both molecules are polar and exhibit comparable dipole moments. The large difference between the boiling points is due to a particularly strong dipole-dipole attraction that may occur when a molecule contains a hydrogen atom bonded to a fluorine, oxygen, or nitrogen atom (the three most electronegative elements). The very large difference in electronegativity between the H atom (2.1) and the atom to which it is bonded ( 4.0 for an F atom, 3.5 for an O atom, or 3.0 for a N atom), combined with the very small size of a H atom and the relatively small sizes of $\\mathrm{F}, \\mathrm{O}$, or N atoms, leads to highly concentrated partial charges with these atoms. Molecules with $\\mathrm{F}-\\mathrm{H}, \\mathrm{O}-\\mathrm{H}$, or $\\mathrm{N}-\\mathrm{H}$ moieties are very strongly attracted to similar moieties in nearby molecules, a particularly strong type of dipole-dipole attraction called hydrogen bonding. Examples of hydrogen bonds include HF $\\cdots \\mathrm{HF}, \\mathrm{H}_{2} \\mathrm{O} \\cdots \\mathrm{HOH}$, and $\\mathrm{H}_{3} \\mathrm{~N} \\cdots \\mathrm{HNH}_{2}$, in which the hydrogen bonds are denoted by dots. Figure 10.10 illustrates hydrogen bonding between water molecules.\n\n\nFIGURE 10.10 Water molecules participate in multiple hydrogen-bonding interactions with nearby water molecules."}
{"id": 3538, "contents": "1064. Hydrogen Bonding - \nFIGURE 10.10 Water molecules participate in multiple hydrogen-bonding interactions with nearby water molecules.\n\nDespite use of the word \"bond,\" keep in mind that hydrogen bonds are intermolecular attractive forces, not intramolecular attractive forces (covalent bonds). Hydrogen bonds are much weaker than covalent bonds, only about 5 to $10 \\%$ as strong, but are generally much stronger than other dipole-dipole attractions and dispersion forces.\n\nHydrogen bonds have a pronounced effect on the properties of condensed phases (liquids and solids). For example, consider the trends in boiling points for the binary hydrides of group $15\\left(\\mathrm{NH}_{3}, \\mathrm{PH}_{3}, \\mathrm{AsH}_{3}\\right.$, and $\\left.\\mathrm{SbH}_{3}\\right)$, group 16 hydrides $\\left(\\mathrm{H}_{2} \\mathrm{O}, \\mathrm{H}_{2} \\mathrm{~S}, \\mathrm{H}_{2} \\mathrm{Se}\\right.$, and $\\mathrm{H}_{2} \\mathrm{Te}$ ), and group 17 hydrides ( $\\mathrm{HF}, \\mathrm{HCl}, \\mathrm{HBr}$, and HI ). The boiling points of the heaviest three hydrides for each group are plotted in Figure 10.11. As we progress down any of these groups, the polarities of the molecules decrease slightly, whereas the sizes of the molecules increase substantially. The effect of increasingly stronger dispersion forces dominates that of increasingly weaker dipole-dipole attractions, and the boiling points are observed to increase steadily.\n\n\nFIGURE 10.11 For the group 15, 16, and 17 hydrides, the boiling points for each class of compounds increase with increasing molecular mass for elements in periods 3,4 , and 5 ."}
{"id": 3539, "contents": "1064. Hydrogen Bonding - \nFIGURE 10.11 For the group 15, 16, and 17 hydrides, the boiling points for each class of compounds increase with increasing molecular mass for elements in periods 3,4 , and 5 .\n\nIf we use this trend to predict the boiling points for the lightest hydride for each group, we would expect $\\mathrm{NH}_{3}$ to boil at about $-120^{\\circ} \\mathrm{C}, \\mathrm{H}_{2} \\mathrm{O}$ to boil at about $-80^{\\circ} \\mathrm{C}$, and HF to boil at about $-110^{\\circ} \\mathrm{C}$. However, when we measure the boiling points for these compounds, we find that they are dramatically higher than the trends would predict, as shown in Figure 10.12. The stark contrast between our na\u00efve predictions and reality provides compelling evidence for the strength of hydrogen bonding.\n\n\nFIGURE 10.12 In comparison to periods 3-5, the binary hydrides of period 2 elements in groups 17, 16 and 15 (F, O and N , respectively) exhibit anomalously high boiling points due to hydrogen bonding."}
{"id": 3540, "contents": "1066. Effect of Hydrogen Bonding on Boiling Points - \nConsider the compounds dimethylether $\\left(\\mathrm{CH}_{3} \\mathrm{OCH}_{3}\\right)$, ethanol $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}\\right)$, and propane $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{3}\\right)$. Their boiling points, not necessarily in order, are $-42.1^{\\circ} \\mathrm{C},-24.8^{\\circ} \\mathrm{C}$, and $78.4^{\\circ} \\mathrm{C}$. Match each compound with its boiling point. Explain your reasoning."}
{"id": 3541, "contents": "1067. Solution - \nThe VSEPR-predicted shapes of $\\mathrm{CH}_{3} \\mathrm{OCH}_{3}, \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$, and $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$ are similar, as are their molar masses ( $46 \\mathrm{~g} / \\mathrm{mol}, 46 \\mathrm{~g} / \\mathrm{mol}$, and $44 \\mathrm{~g} / \\mathrm{mol}$, respectively), so they will exhibit similar dispersion forces. Since $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$ is nonpolar, it may exhibit only dispersion forces. Because $\\mathrm{CH}_{3} \\mathrm{OCH}_{3}$ is polar, it will also experience dipole-dipole attractions. Finally, $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$ has an - OH group, and so it will experience the uniquely strong dipole-dipole attraction known as hydrogen bonding. So the ordering in terms of strength of IMFs, and thus boiling points, is $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{3}<\\mathrm{CH}_{3} \\mathrm{OCH}_{3}<\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$. The boiling point of propane is -42.1 ${ }^{\\circ} \\mathrm{C}$, the boiling point of dimethylether is $-24.8^{\\circ} \\mathrm{C}$, and the boiling point of ethanol is $78.5^{\\circ} \\mathrm{C}$."}
{"id": 3542, "contents": "1068. Check Your Learning - \nEthane $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{3}\\right)$ has a melting point of $-183^{\\circ} \\mathrm{C}$ and a boiling point of $-89^{\\circ} \\mathrm{C}$. Predict the melting and boiling points for methylamine $\\left(\\mathrm{CH}_{3} \\mathrm{NH}_{2}\\right)$. Explain your reasoning."}
{"id": 3543, "contents": "1069. Answer: - \nThe melting point and boiling point for methylamine are predicted to be significantly greater than those of ethane. $\\mathrm{CH}_{3} \\mathrm{CH}_{3}$ and $\\mathrm{CH}_{3} \\mathrm{NH}_{2}$ are similar in size and mass, but methylamine possesses an - NH group and therefore may exhibit hydrogen bonding. This greatly increases its IMFs, and therefore its melting and boiling points. It is difficult to predict values, but the known values are a melting point of $-93^{\\circ} \\mathrm{C}$ and a boiling point of $-6^{\\circ} \\mathrm{C}$."}
{"id": 3544, "contents": "1071. Hydrogen Bonding and DNA - \nDeoxyribonucleic acid (DNA) is found in every living organism and contains the genetic information that determines the organism's characteristics, provides the blueprint for making the proteins necessary for life, and serves as a template to pass this information on to the organism's offspring. A DNA molecule consists of two (anti-)parallel chains of repeating nucleotides, which form its well-known double helical structure, as shown in Figure 10.13.\n\n\nFIGURE 10.13 Two separate DNA molecules form a double-stranded helix in which the molecules are held together via hydrogen bonding. (credit: modification of work by Jerome Walker, Dennis Myts)\n\nEach nucleotide contains a (deoxyribose) sugar bound to a phosphate group on one side, and one of four nitrogenous bases on the other. Two of the bases, cytosine (C) and thymine (T), are single-ringed structures known as pyrimidines. The other two, adenine (A) and guanine (G), are double-ringed structures called purines. These bases form complementary base pairs consisting of one purine and one pyrimidine, with adenine pairing with thymine, and cytosine with guanine. Each base pair is held together by hydrogen bonding. A and $T$ share two hydrogen bonds, C and G share three, and both pairings have a similar shape and structure Figure 10.14.\n\n\nFIGURE 10.14 The geometries of the base molecules result in maximum hydrogen bonding between adenine and thymine (AT) and between guanine and cytosine (GC), so-called \"complementary base pairs.\"\n\nThe cumulative effect of millions of hydrogen bonds effectively holds the two strands of DNA together. Importantly, the two strands of DNA can relatively easily \"unzip\" down the middle since hydrogen bonds are relatively weak compared to the covalent bonds that hold the atoms of the individual DNA molecules together. This allows both strands to function as a template for replication."}
{"id": 3545, "contents": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids\nLEARNING OBJECTIVES\nBy the end of this section, you will be able to:\n\n- Distinguish between adhesive and cohesive forces\n- Define viscosity, surface tension, and capillary rise\n- Describe the roles of intermolecular attractive forces in each of these properties/phenomena\n\nWhen you pour a glass of water, or fill a car with gasoline, you observe that water and gasoline flow freely. But when you pour syrup on pancakes or add oil to a car engine, you note that syrup and motor oil do not flow as readily. The viscosity of a liquid is a measure of its resistance to flow. Water, gasoline, and other liquids that flow freely have a low viscosity. Honey, syrup, motor oil, and other liquids that do not flow freely, like those shown in Figure 10.15, have higher viscosities. We can measure viscosity by measuring the rate at which a metal ball falls through a liquid (the ball falls more slowly through a more viscous liquid) or by measuring the rate at which a liquid flows through a narrow tube (more viscous liquids flow more slowly).\n\n\nFIGURE 10.15 (a) Honey and (b) motor oil are examples of liquids with high viscosities; they flow slowly. (credit a: modification of work by Scott Bauer; credit b: modification of work by David Nagy)\n\nThe IMFs between the molecules of a liquid, the size and shape of the molecules, and the temperature determine how easily a liquid flows. As Table 10.2 shows, the more structurally complex are the molecules in a liquid and the stronger the IMFs between them, the more difficult it is for them to move past each other and the greater is the viscosity of the liquid. As the temperature increases, the molecules move more rapidly and their kinetic energies are better able to overcome the forces that hold them together; thus, the viscosity of the liquid decreases.\n\nViscosities of Common Substances at $25^{\\circ} \\mathrm{C}$"}
{"id": 3546, "contents": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids\nViscosities of Common Substances at $25^{\\circ} \\mathrm{C}$\n\n| Substance | Formula | Viscosity (mPa\u2022s) |\n| :---: | :---: | :---: |\n| water | $\\mathrm{H}_{2} \\mathrm{O}$ | 0.890 |\n| mercury | Hg | 1.526 |\n| ethanol | $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$ | 1.074 |\n| octane | $\\mathrm{C}_{8} \\mathrm{H}_{18}$ | 0.508 |\n| ethylene glycol | $\\mathrm{CH}_{2}(\\mathrm{OH}) \\mathrm{CH}_{2}(\\mathrm{OH})$ | 16.1 |\n| honey | variable | $\\sim 2,000-10,000$ |\n| motor oil | variable | $\\sim 50-500$ |\n\nTABLE 10.2\n\nThe various IMFs between identical molecules of a substance are examples of cohesive forces. The molecules within a liquid are surrounded by other molecules and are attracted equally in all directions by the cohesive forces within the liquid. However, the molecules on the surface of a liquid are attracted only by about one-half as many molecules. Because of the unbalanced molecular attractions on the surface molecules, liquids contract to form a shape that minimizes the number of molecules on the surface-that is, the shape with the minimum surface area. A small drop of liquid tends to assume a spherical shape, as shown in Figure 10.16, because in a sphere, the ratio of surface area to volume is at a minimum. Larger drops are more greatly affected by gravity, air resistance, surface interactions, and so on, and as a result, are less spherical.\n\n\nFIGURE 10.16 Attractive forces result in a spherical water drop that minimizes surface area; cohesive forces hold the sphere together; adhesive forces keep the drop attached to the web. (credit photo: modification of work by \"OliBac\"/Flickr)"}
{"id": 3547, "contents": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids\nSurface tension is defined as the energy required to increase the surface area of a liquid, or the force required to increase the length of a liquid surface by a given amount. This property results from the cohesive forces between molecules at the surface of a liquid, and it causes the surface of a liquid to behave like a stretched rubber membrane. Surface tensions of several liquids are presented in Table 10.3. Among common liquids, water exhibits a distinctly high surface tension due to strong hydrogen bonding between its molecules. As a result of this high surface tension, the surface of water represents a relatively \"tough skin\" that can withstand considerable force without breaking. A steel needle carefully placed on water will float. Some insects, like the one shown in Figure 10.17, even though they are denser than water, move on its surface because they are supported by the surface tension.\n\nSurface Tensions of Common Substances at $25^{\\circ} \\mathrm{C}$\n\n| Substance | Formula | Surface Tension (mN/m) |\n| :---: | :---: | :---: |\n| water | $\\mathrm{H}_{2} \\mathrm{O}$ | 71.99 |\n| mercury | Hg | 458.48 |\n| ethanol | $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$ | 21.97 |\n| octane | $\\mathrm{C}_{8} \\mathrm{H}_{18}$ | 21.14 |\n| ethylene glycol | $\\mathrm{CH}_{2}(\\mathrm{OH}) \\mathrm{CH}_{2}(\\mathrm{OH})$ | 47.99 |\n\nTABLE 10.3"}
{"id": 3548, "contents": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids\nTABLE 10.3\n\n\nFIGURE 10.17 Surface tension (right) prevents this insect, a \"water strider,\" from sinking into the water.\nSurface tension is affected by a variety of variables, including the introduction of additional substances on the surface. In the late 1800s, Agnes Pockels, who was initially blocked from pursuing a scientific career but\nstudied on her own, began investigating the impact and characteristics of soapy and greasy films in water. Using homemade materials, she developed an instrument known as a trough for measuring surface contaminants and their effects. With the support of renowned scientist Lord Rayleigh, her 1891 paper showed that surface contamination significantly reduces surface tension, and also that changing the characteristics of the surface (compressing or expanding it) also affects surface tension. Decades later, Irving Langmuir and Katharine Blodgett built on Pockels' work in their own trough and important advances in surface chemistry. Langmuir pioneered methods for producing single-molecule layers of film; Blodgett applied these to the development of non-reflective glass (critical for film-making and other applications), and also studied methods related to cleaning surfaces, which are important in semiconductor fabrication.\n\nThe IMFs of attraction between two different molecules are called adhesive forces. Consider what happens when water comes into contact with some surface. If the adhesive forces between water molecules and the molecules of the surface are weak compared to the cohesive forces between the water molecules, the water does not \"wet\" the surface. For example, water does not wet waxed surfaces or many plastics such as polyethylene. Water forms drops on these surfaces because the cohesive forces within the drops are greater than the adhesive forces between the water and the plastic. Water spreads out on glass because the adhesive force between water and glass is greater than the cohesive forces within the water. When water is confined in a glass tube, its meniscus (surface) has a concave shape because the water wets the glass and creeps up the side of the tube. On the other hand, the cohesive forces between mercury atoms are much greater than the adhesive forces between mercury and glass. Mercury therefore does not wet glass, and it forms a convex meniscus when confined in a tube because the cohesive forces within the mercury tend to draw it into a drop (Figure 10.18)."}
{"id": 3549, "contents": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids\nFIGURE 10.18 Differences in the relative strengths of cohesive and adhesive forces result in different meniscus shapes for mercury (left) and water (right) in glass tubes. (credit: Mark Ott)\n\nIf you place one end of a paper towel in spilled wine, as shown in Figure 10.19, the liquid wicks up the paper towel. A similar process occurs in a cloth towel when you use it to dry off after a shower. These are examples of capillary action-when a liquid flows within a porous material due to the attraction of the liquid molecules to the surface of the material and to other liquid molecules. The adhesive forces between the liquid and the porous material, combined with the cohesive forces within the liquid, may be strong enough to move the liquid upward against gravity.\n\n\nFIGURE 10.19 Wine wicks up a paper towel (left) because of the strong attractions of water (and ethanol) molecules to the -OH groups on the towel's cellulose fibers and the strong attractions of water molecules to other water (and ethanol) molecules (right). (credit photo: modification of work by Mark Blaser)\n\nTowels soak up liquids like water because the fibers of a towel are made of molecules that are attracted to water molecules. Most cloth towels are made of cotton, and paper towels are generally made from paper pulp. Both consist of long molecules of cellulose that contain many -OH groups. Water molecules are attracted to these -OH groups and form hydrogen bonds with them, which draws the $\\mathrm{H}_{2} \\mathrm{O}$ molecules up the cellulose molecules. The water molecules are also attracted to each other, so large amounts of water are drawn up the cellulose fibers."}
{"id": 3550, "contents": "1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids\nCapillary action can also occur when one end of a small diameter tube is immersed in a liquid, as illustrated in Figure 10.20. If the liquid molecules are strongly attracted to the tube molecules, the liquid creeps up the inside of the tube until the weight of the liquid and the adhesive forces are in balance. The smaller the diameter of the tube is, the higher the liquid climbs. It is partly by capillary action occurring in plant cells called xylem that water and dissolved nutrients are brought from the soil up through the roots and into a plant. Capillary action is the basis for thin layer chromatography, a laboratory technique commonly used to separate small quantities of mixtures. You depend on a constant supply of tears to keep your eyes lubricated and on capillary action to pump tear fluid away.\n\n\nFIGURE 10.20 Depending upon the relative strengths of adhesive and cohesive forces, a liquid may rise (such as water) or fall (such as mercury) in a glass capillary tube. The extent of the rise (or fall) is directly proportional to the surface tension of the liquid and inversely proportional to the density of the liquid and the radius of the tube.\n\nThe height to which a liquid will rise in a capillary tube is determined by several factors as shown in the\nfollowing equation:\n\n$$\nh=\\frac{2 T \\cos \\theta}{r \\rho g}\n$$\n\nIn this equation, $h$ is the height of the liquid inside the capillary tube relative to the surface of the liquid outside the tube, $T$ is the surface tension of the liquid, $\\theta$ is the contact angle between the liquid and the tube, $r$ is the radius of the tube, $\\rho$ is the density of the liquid, and $g$ is the acceleration due to gravity, $9.8 \\mathrm{~m} / \\mathrm{s}^{2}$. When the tube is made of a material to which the liquid molecules are strongly attracted, they will spread out completely on the surface, which corresponds to a contact angle of $0^{\\circ}$. This is the situation for water rising in a glass tube."}
{"id": 3551, "contents": "1073. Capillary Rise - \nAt $25^{\\circ} \\mathrm{C}$, how high will water rise in a glass capillary tube with an inner diameter of 0.25 mm ?\nFor water, $T=71.99 \\mathrm{mN} / \\mathrm{m}$ and $\\rho=1.0 \\mathrm{~g} / \\mathrm{cm}^{3}$."}
{"id": 3552, "contents": "1074. Solution - \nThe liquid will rise to a height $h$ given by: $h=\\frac{2 T \\cos \\theta}{r \\rho g}$\nThe Newton is defined as a $\\mathrm{kg} \\mathrm{m} / \\mathrm{s}^{2}$, and so the provided surface tension is equivalent to $0.07199 \\mathrm{~kg} / \\mathrm{s}^{2}$. The provided density must be converted into units that will cancel appropriately: $\\rho=1000 \\mathrm{~kg} / \\mathrm{m}^{3}$. The diameter of the tube in meters is 0.00025 m , so the radius is 0.000125 m . For a glass tube immersed in water, the contact angle is $\\theta=0^{\\circ}$, so $\\cos \\theta=1$. Finally, acceleration due to gravity on the earth is $g=9.8 \\mathrm{~m} / \\mathrm{s}^{2}$. Substituting these values into the equation, and cancelling units, we have:\n\n$$\nh=\\frac{2\\left(0.07199 \\mathrm{~kg} / \\mathrm{s}^{2}\\right)}{(0.000125 \\mathrm{~m})\\left(1000 \\mathrm{~kg} / \\mathrm{m}^{3}\\right)\\left(9.8 \\mathrm{~m} / \\mathrm{s}^{2}\\right)}=0.12 \\mathrm{~m}=12 \\mathrm{~cm}\n$$"}
{"id": 3553, "contents": "1075. Check Your Learning - \nWater rises in a glass capillary tube to a height of 8.4 cm . What is the diameter of the capillary tube?"}
{"id": 3554, "contents": "1076. Answer: - \ndiameter $=0.36 \\mathrm{~mm}$"}
{"id": 3555, "contents": "1078. Biomedical Applications of Capillary Action - \nMany medical tests require drawing a small amount of blood, for example to determine the amount of glucose in someone with diabetes or the hematocrit level in an athlete. This procedure can be easily done because of capillary action, the ability of a liquid to flow up a small tube against gravity, as shown in Figure 10.21. When your finger is pricked, a drop of blood forms and holds together due to surface tension-the unbalanced intermolecular attractions at the surface of the drop. Then, when the open end of a narrowdiameter glass tube touches the drop of blood, the adhesive forces between the molecules in the blood and those at the glass surface draw the blood up the tube. How far the blood goes up the tube depends on the diameter of the tube (and the type of fluid). A small tube has a relatively large surface area for a given volume of blood, which results in larger (relative) attractive forces, allowing the blood to be drawn farther up the tube. The liquid itself is held together by its own cohesive forces. When the weight of the liquid in the tube generates a downward force equal to the upward force associated with capillary action, the liquid stops rising.\n\n\nFIGURE 10.21 Blood is collected for medical analysis by capillary action, which draws blood into a small diameter glass tube. (credit: modification of work by Centers for Disease Control and Prevention)"}
{"id": 3556, "contents": "1079. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Define phase transitions and phase transition temperatures\n- Explain the relation between phase transition temperatures and intermolecular attractive forces\n- Describe the processes represented by typical heating and cooling curves, and compute heat flows and enthalpy changes accompanying these processes\n\nWe witness and utilize changes of physical state, or phase transitions, in a great number of ways. As one example of global significance, consider the evaporation, condensation, freezing, and melting of water. These changes of state are essential aspects of our earth's water cycle as well as many other natural phenomena and technological processes of central importance to our lives. In this module, the essential aspects of phase transitions are explored."}
{"id": 3557, "contents": "1080. Vaporization and Condensation - \nWhen a liquid vaporizes in a closed container, gas molecules cannot escape. As these gas phase molecules move randomly about, they will occasionally collide with the surface of the condensed phase, and in some cases, these collisions will result in the molecules re-entering the condensed phase. The change from the gas phase to the liquid is called condensation. When the rate of condensation becomes equal to the rate of vaporization, neither the amount of the liquid nor the amount of the vapor in the container changes. The vapor in the container is then said to be in equilibrium with the liquid. Keep in mind that this is not a static situation, as molecules are continually exchanged between the condensed and gaseous phases. Such is an example of a dynamic equilibrium, the status of a system in which reciprocal processes (for example, vaporization and condensation) occur at equal rates. The pressure exerted by the vapor in equilibrium with a liquid in a closed container at a given temperature is called the liquid's vapor pressure (or equilibrium vapor pressure). The area of the surface of the liquid in contact with a vapor and the size of the vessel have no effect on the vapor pressure, although they do affect the time required for the equilibrium to be reached. We can measure the vapor pressure of a liquid by placing a sample in a closed container, like that illustrated in Figure 10.22 , and using a manometer to measure the increase in pressure that is due to the vapor in equilibrium with the condensed phase.\n\n(a)\n\n(b)\n\n(c)\n\nFIGURE 10.22 In a closed container, dynamic equilibrium is reached when (a) the rate of molecules escaping from\nthe liquid to become the gas (b) increases and eventually (c) equals the rate of gas molecules entering the liquid. When this equilibrium is reached, the vapor pressure of the gas is constant, although the vaporization and condensation processes continue."}
{"id": 3558, "contents": "1080. Vaporization and Condensation - \nThe chemical identities of the molecules in a liquid determine the types (and strengths) of intermolecular attractions possible; consequently, different substances will exhibit different equilibrium vapor pressures. Relatively strong intermolecular attractive forces will serve to impede vaporization as well as favoring \"recapture\" of gas-phase molecules when they collide with the liquid surface, resulting in a relatively low vapor pressure. Weak intermolecular attractions present less of a barrier to vaporization, and a reduced likelihood of gas recapture, yielding relatively high vapor pressures. The following example illustrates this dependence of vapor pressure on intermolecular attractive forces."}
{"id": 3559, "contents": "1082. Explaining Vapor Pressure in Terms of IMFs - \nGiven the shown structural formulas for these four compounds, explain their relative vapor pressures in terms of types and extents of IMFs:"}
{"id": 3560, "contents": "1083. Solution - \nDiethyl ether has a very small dipole and most of its intermolecular attractions are London forces. Although this molecule is the largest of the four under consideration, its IMFs are the weakest and, as a result, its molecules most readily escape from the liquid. It also has the highest vapor pressure. Due to its smaller size, ethanol exhibits weaker dispersion forces than diethyl ether. However, ethanol is capable of hydrogen bonding and, therefore, exhibits stronger overall IMFs, which means that fewer molecules escape from the liquid at any given temperature, and so ethanol has a lower vapor pressure than diethyl ether. Water is much smaller than either of the previous substances and exhibits weaker dispersion forces, but its extensive hydrogen bonding provides stronger intermolecular attractions, fewer molecules escaping the liquid, and a lower vapor pressure than for either diethyl ether or ethanol. Ethylene glycol has two -OH groups, so, like water, it exhibits extensive hydrogen bonding. It is much larger than water and thus experiences larger London forces. Its overall IMFs are the largest of these four substances, which means its vaporization rate will be the slowest and, consequently, its vapor pressure the lowest."}
{"id": 3561, "contents": "1084. Check Your Learning - \nAt $20^{\\circ} \\mathrm{C}$, the vapor pressures of several alcohols are given in this table. Explain these vapor pressures in terms of types and extents of IMFs for these alcohols:\n\n| Compound | methanol $\\mathrm{CH}_{3} \\mathrm{OH}$ | ethanol $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$ | propanol $\\mathrm{C}_{3} \\mathrm{H}_{7} \\mathrm{OH}$ | butanol $\\mathrm{C}_{4} \\mathrm{H}_{9} \\mathrm{OH}$ |\n| :--- | :--- | :--- | :--- | :--- |\n| Vapor Pressure at $20^{\\circ} \\mathrm{C}$ | 11.9 kPa | 5.95 kPa | 2.67 kPa | 0.56 kPa |"}
{"id": 3562, "contents": "1085. Answer: - \nAll these compounds exhibit hydrogen bonding; these strong IMFs are difficult for the molecules to overcome, so the vapor pressures are relatively low. As the size of molecule increases from methanol to butanol, dispersion forces increase, which means that the vapor pressures decrease as observed:\n$\\mathrm{P}_{\\text {methanol }}>\\mathrm{P}_{\\text {ethanol }}>\\mathrm{P}_{\\text {propanol }}>\\mathrm{P}_{\\text {butanol }}$.\n\nAs temperature increases, the vapor pressure of a liquid also increases due to the increased average KE of its molecules. Recall that at any given temperature, the molecules of a substance experience a range of kinetic\nenergies, with a certain fraction of molecules having a sufficient energy to overcome IMF and escape the liquid (vaporize). At a higher temperature, a greater fraction of molecules have enough energy to escape from the liquid, as shown in Figure 10.23. The escape of more molecules per unit of time and the greater average speed of the molecules that escape both contribute to the higher vapor pressure.\n\n\nFIGURE 10.23 Temperature affects the distribution of kinetic energies for the molecules in a liquid. At the higher temperature, more molecules have the necessary kinetic energy, $K E$, to escape from the liquid into the gas phase."}
{"id": 3563, "contents": "1086. Boiling Points - \nWhen the vapor pressure increases enough to equal the external atmospheric pressure, the liquid reaches its boiling point. The boiling point of a liquid is the temperature at which its equilibrium vapor pressure is equal to the pressure exerted on the liquid by its gaseous surroundings. For liquids in open containers, this pressure is that due to the earth's atmosphere. The normal boiling point of a liquid is defined as its boiling point when surrounding pressure is equal to $1 \\mathrm{~atm}(101.3 \\mathrm{kPa})$. Figure 10.24 shows the variation in vapor pressure with temperature for several different substances. Considering the definition of boiling point, these curves may be seen as depicting the dependence of a liquid's boiling point on surrounding pressure.\n\n\nFIGURE 10.24 The boiling points of liquids are the temperatures at which their equilibrium vapor pressures equal the pressure of the surrounding atmosphere. Normal boiling points are those corresponding to a pressure of 1 atm (101.3 kPa.)"}
{"id": 3564, "contents": "1088. A Boiling Point at Reduced Pressure - \nA typical atmospheric pressure in Leadville, Colorado (elevation 10,200 feet) is 68 kPa . Use the graph in Figure 10.24 to determine the boiling point of water at this elevation."}
{"id": 3565, "contents": "1089. Solution - \nThe graph of the vapor pressure of water versus temperature in Figure 10.24 indicates that the vapor pressure of water is 68 kPa at about $90^{\\circ} \\mathrm{C}$. Thus, at about $90^{\\circ} \\mathrm{C}$, the vapor pressure of water will equal the atmospheric pressure in Leadville, and water will boil."}
{"id": 3566, "contents": "1090. Check Your Learning - \nThe boiling point of ethyl ether was measured to be $10^{\\circ} \\mathrm{C}$ at a base camp on the slopes of Mount Everest. Use Figure 10.24 to determine the approximate atmospheric pressure at the camp."}
{"id": 3567, "contents": "1091. Answer: - \nApproximately 40 kPa (0.4 atm)\n\nThe quantitative relation between a substance's vapor pressure and its temperature is described by the Clausius-Clapeyron equation:\n\n$$\nP=A e^{-\\Delta H_{\\mathrm{vap}} / R T}\n$$\n\nwhere $\\Delta H_{\\text {vap }}$ is the enthalpy of vaporization for the liquid, $R$ is the gas constant, and $A$ is a constant whose value depends on the chemical identity of the substance. Temperature T must be in Kelvin in this equation. This equation is often rearranged into logarithmic form to yield the linear equation:\n\n$$\n\\ln P=-\\frac{\\Delta H_{\\mathrm{vap}}}{R T}+\\ln A\n$$\n\nThis linear equation may be expressed in a two-point format that is convenient for use in various computations, as demonstrated in the example exercises that follow. If at temperature $T_{1}$, the vapor pressure is $\\mathrm{P}_{1}$, and at temperature $\\mathrm{T}_{2}$, the vapor pressure is $\\mathrm{P}_{2}$, the corresponding linear equations are:\n\n$$\n\\ln P_{1}=-\\frac{\\Delta H_{\\mathrm{vap}}}{R T_{1}}+\\ln A \\quad \\text { and } \\quad \\ln P_{2}=-\\frac{\\Delta H_{\\mathrm{vap}}}{R T_{2}}+\\ln A\n$$\n\nSince the constant, $A$, is the same, these two equations may be rearranged to isolate $\\ln A$ and then set them equal to one another:\n\n$$\n\\ln P_{1}+\\frac{\\Delta H_{\\mathrm{vap}}}{R T_{1}}=\\ln P_{2}+\\frac{\\Delta H_{\\mathrm{vap}}}{R T_{2}}\n$$\n\nwhich can be combined into:\n\n$$\n\\ln \\left(\\frac{P_{2}}{P_{1}}\\right)=\\frac{\\Delta H_{\\mathrm{vap}}}{R}\\left(\\frac{1}{T_{1}}-\\frac{1}{T_{2}}\\right)\n$$"}
{"id": 3568, "contents": "1093. Estimating Enthalpy of Vaporization - \nIsooctane (2,2,4-trimethylpentane) has an octane rating of 100. It is used as one of the standards for the octane-rating system for gasoline. At $34.0^{\\circ} \\mathrm{C}$, the vapor pressure of isooctane is 10.0 kPa , and at $98.8^{\\circ} \\mathrm{C}$, its vapor pressure is 100.0 kPa . Use this information to estimate the enthalpy of vaporization for isooctane."}
{"id": 3569, "contents": "1094. Solution - \nThe enthalpy of vaporization, $\\Delta H_{\\text {vap }}$, can be determined by using the Clausius-Clapeyron equation:\n\n$$\n\\ln \\left(\\frac{P_{2}}{P_{1}}\\right)=\\frac{\\Delta H_{\\mathrm{vap}}}{R}\\left(\\frac{1}{T_{1}}-\\frac{1}{T_{2}}\\right)\n$$\n\nSince we have two vapor pressure-temperature values ( $T_{1}=34.0^{\\circ} \\mathrm{C}=307.2 \\mathrm{~K}, P_{1}=10.0 \\mathrm{kPa}$ and $T_{2}=98.8^{\\circ} \\mathrm{C}=$ $372.0 \\mathrm{~K}, P_{2}=100 \\mathrm{kPa}$ ), we can substitute them into this equation and solve for $\\Delta H_{\\text {vap }}$. Rearranging the Clausius-Clapeyron equation and solving for $\\Delta H_{\\text {vap }}$ yields:\n\n$$\n\\Delta H_{\\mathrm{vap}}=\\frac{R \\cdot \\ln \\left(\\frac{P_{2}}{P_{1}}\\right)}{\\left(\\frac{1}{T_{1}}-\\frac{1}{T_{2}}\\right)}=\\frac{(8.3145 \\mathrm{~J} / \\mathrm{mol} \\cdot \\mathrm{~K}) \\cdot \\ln \\left(\\frac{100 \\mathrm{kPa}}{10.0 \\mathrm{kPa}}\\right)}{\\left(\\frac{1}{307.2 \\mathrm{~K}}-\\frac{1}{372.0 \\mathrm{~K}}\\right)}=33,800 \\mathrm{~J} / \\mathrm{mol}=33.8 \\mathrm{~kJ} / \\mathrm{mol}\n$$\n\nNote that the pressure can be in any units, so long as they agree for both $P$ values, but the temperature must be in kelvin for the Clausius-Clapeyron equation to be valid."}
{"id": 3570, "contents": "1095. Check Your Learning - \nAt $20.0^{\\circ} \\mathrm{C}$, the vapor pressure of ethanol is 5.95 kPa , and at $63.5^{\\circ} \\mathrm{C}$, its vapor pressure is 53.3 kPa . Use this information to estimate the enthalpy of vaporization for ethanol."}
{"id": 3571, "contents": "1096. Answer: - \n$41,360 \\mathrm{~J} / \\mathrm{mol}$ or $41.4 \\mathrm{~kJ} / \\mathrm{mol}$"}
{"id": 3572, "contents": "1098. Estimating Temperature (or Vapor Pressure) - \nFor benzene $\\left(\\mathrm{C}_{6} \\mathrm{H}_{6}\\right)$, the normal boiling point is $80.1^{\\circ} \\mathrm{C}$ and the enthalpy of vaporization is $30.8 \\mathrm{~kJ} / \\mathrm{mol}$. What is the boiling point of benzene in Denver, where atmospheric pressure $=83.4 \\mathrm{kPa}$ ?"}
{"id": 3573, "contents": "1099. Solution - \nIf the temperature and vapor pressure are known at one point, along with the enthalpy of vaporization, $\\Delta H_{\\text {vap, }}$, then the temperature that corresponds to a different vapor pressure (or the vapor pressure that corresponds to a different temperature) can be determined by using the Clausius-Clapeyron equation:\n\n$$\n\\ln \\left(\\frac{P_{2}}{P_{1}}\\right)=\\frac{\\Delta H_{\\mathrm{vap}}}{R}\\left(\\frac{1}{T_{1}}-\\frac{1}{T_{2}}\\right)\n$$\n\nSince the normal boiling point is the temperature at which the vapor pressure equals atmospheric pressure at sea level, we know one vapor pressure-temperature value ( $T_{1}=80.1^{\\circ} \\mathrm{C}=353.3 \\mathrm{~K}, P_{1}=101.3 \\mathrm{kPa}, \\Delta H_{\\text {vap }}=30.8$ $\\mathrm{kJ} / \\mathrm{mol}$ ) and want to find the temperature ( $T_{2}$ ) that corresponds to vapor pressure $P_{2}=83.4 \\mathrm{kPa}$. We can substitute these values into the Clausius-Clapeyron equation and then solve for $T_{2}$. Rearranging the ClausiusClapeyron equation and solving for $T_{2}$ yields:\n$T_{2}=\\left(\\frac{-R \\cdot \\ln \\left(\\frac{P_{2}}{P_{1}}\\right)}{\\Delta H_{\\text {vap }}}+\\frac{1}{T_{1}}\\right)^{-1}=\\left(\\frac{-(8.3145 \\mathrm{~J} / \\mathrm{mol} \\cdot \\mathrm{K}) \\cdot \\ln \\left(\\frac{83.4 \\mathrm{kPa}}{101.3 \\mathrm{kPa}}\\right)}{30,800 \\mathrm{~J} / \\mathrm{mol}}+\\frac{1}{353.3 \\mathrm{~K}}\\right)^{-1}=346.9 \\mathrm{~K}$ or $73.8^{\\circ} \\mathrm{C}$"}
{"id": 3574, "contents": "1100. Check Your Learning - \nFor acetone $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CO}$, the normal boiling point is $56.5^{\\circ} \\mathrm{C}$ and the enthalpy of vaporization is $31.3 \\mathrm{~kJ} / \\mathrm{mol}$. What is the vapor pressure of acetone at $25.0^{\\circ} \\mathrm{C}$ ?"}
{"id": 3575, "contents": "1101. Answer: - \n30.1 kPa"}
{"id": 3576, "contents": "1102. Enthalpy of Vaporization - \nVaporization is an endothermic process. The cooling effect can be evident when you leave a swimming pool or\na shower. When the water on your skin evaporates, it removes heat from your skin and causes you to feel cold. The energy change associated with the vaporization process is the enthalpy of vaporization, $\\Delta H_{\\text {vap }}$. For example, the vaporization of water at standard temperature is represented by:\n\n$$\n\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(g) \\quad \\Delta H_{\\mathrm{vap}}=44.01 \\mathrm{~kJ} / \\mathrm{mol}\n$$\n\nAs described in the chapter on thermochemistry, the reverse of an endothermic process is exothermic. And so, the condensation of a gas releases heat:\n\n$$\n\\mathrm{H}_{2} \\mathrm{O}(g) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(l) \\quad \\Delta H_{\\mathrm{con}}=-\\Delta H_{\\mathrm{vap}}=-44.01 \\mathrm{~kJ} / \\mathrm{mol}\n$$"}
{"id": 3577, "contents": "1104. Using Enthalpy of Vaporization - \nOne way our body is cooled is by evaporation of the water in sweat (Figure 10.25). In very hot climates, we can lose as much as 1.5 L of sweat per day. Although sweat is not pure water, we can get an approximate value of the amount of heat removed by evaporation by assuming that it is. How much heat is required to evaporate 1.5 L of water ( 1.5 kg ) at $T=37{ }^{\\circ} \\mathrm{C}$ (normal body temperature); $\\Delta H_{\\text {vap }}=43.46 \\mathrm{~kJ} / \\mathrm{mol}$ at $37^{\\circ} \\mathrm{C}$.\n\n\nFIGURE 10.25 Evaporation of sweat helps cool the body. (credit: \"Kullez\"/Flickr)"}
{"id": 3578, "contents": "1105. Solution - \nWe start with the known volume of sweat (approximated as just water) and use the given information to convert to the amount of heat needed:\n\n$$\n1.5 \\mathrm{~L} \\times \\frac{1000 \\mathrm{~g}}{1 \\mathrm{~g}} \\times \\frac{1 \\mathrm{mot}}{18 \\frac{\\mathrm{~g}}{\\mathrm{o}}} \\times \\frac{43.46 \\mathrm{~kJ}}{1 \\mathrm{mot}}=3.6 \\times 10^{3} \\mathrm{~kJ}\n$$\n\nThus, 3600 kJ of heat are removed by the evaporation of 1.5 L of water."}
{"id": 3579, "contents": "1106. Check Your Learning - \nHow much heat is required to evaporate 100.0 g of liquid ammonia, $\\mathrm{NH}_{3}$, at its boiling point if its enthalpy of vaporization is $4.8 \\mathrm{~kJ} / \\mathrm{mol}$ ?"}
{"id": 3580, "contents": "1107. Answer: - \n28 kJ"}
{"id": 3581, "contents": "1108. Melting and Freezing - \nWhen we heat a crystalline solid, we increase the average energy of its atoms, molecules, or ions and the solid gets hotter. At some point, the added energy becomes large enough to partially overcome the forces holding the molecules or ions of the solid in their fixed positions, and the solid begins the process of transitioning to the liquid state, or melting. At this point, the temperature of the solid stops rising, despite the continual input of heat, and it remains constant until all of the solid is melted. Only after all of the solid has melted will continued\nheating increase the temperature of the liquid (Figure 10.26).\n\n\nFIGURE 10.26 (a) This beaker of ice has a temperature of $-12.0^{\\circ} \\mathrm{C}$. (b) After 10 minutes the ice has absorbed enough heat from the air to warm to $0^{\\circ} \\mathrm{C}$. A small amount has melted. (c) Thirty minutes later, the ice has absorbed more heat, but its temperature is still $0^{\\circ} \\mathrm{C}$. The ice melts without changing its temperature. (d) Only after all the ice has melted does the heat absorbed cause the temperature to increase to $22.2^{\\circ} \\mathrm{C}$. (credit: modification of work by Mark Ott)\n\nIf we stop heating during melting and place the mixture of solid and liquid in a perfectly insulated container so no heat can enter or escape, the solid and liquid phases remain in equilibrium. This is almost the situation with a mixture of ice and water in a very good thermos bottle; almost no heat gets in or out, and the mixture of solid ice and liquid water remains for hours. In a mixture of solid and liquid at equilibrium, the reciprocal processes of melting and freezing occur at equal rates, and the quantities of solid and liquid therefore remain constant. The temperature at which the solid and liquid phases of a given substance are in equilibrium is called the melting point of the solid or the freezing point of the liquid. Use of one term or the other is normally dictated by the direction of the phase transition being considered, for example, solid to liquid (melting) or liquid to solid (freezing)."}
{"id": 3582, "contents": "1108. Melting and Freezing - \nThe enthalpy of fusion and the melting point of a crystalline solid depend on the strength of the attractive forces between the units present in the crystal. Molecules with weak attractive forces form crystals with low melting points. Crystals consisting of particles with stronger attractive forces melt at higher temperatures.\n\nThe amount of heat required to change one mole of a substance from the solid state to the liquid state is the enthalpy of fusion, $\\Delta \\mathrm{H}_{\\text {fus }}$ of the substance. The enthalpy of fusion of ice is $6.0 \\mathrm{~kJ} / \\mathrm{mol}$ at $0^{\\circ} \\mathrm{C}$. Fusion (melting) is an endothermic process:\n\n$$\n\\mathrm{H}_{2} \\mathrm{O}(s) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(l) \\quad \\Delta H_{\\text {fus }}=6.01 \\mathrm{~kJ} / \\mathrm{mol}\n$$\n\nThe reciprocal process, freezing, is an exothermic process whose enthalpy change is $-6.0 \\mathrm{~kJ} / \\mathrm{mol}$ at $0^{\\circ} \\mathrm{C}$ :\n\n$$\n\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(s)\n$$\n\n$$\n\\Delta H_{\\mathrm{frz}}=-\\Delta H_{\\mathrm{fus}}=-6.01 \\mathrm{~kJ} / \\mathrm{mol}\n$$"}
{"id": 3583, "contents": "1109. Sublimation and Deposition - \nSome solids can transition directly into the gaseous state, bypassing the liquid state, via a process known as sublimation. At room temperature and standard pressure, a piece of dry ice (solid $\\mathrm{CO}_{2}$ ) sublimes, appearing to gradually disappear without ever forming any liquid. Snow and ice sublime at temperatures below the melting point of water, a slow process that may be accelerated by winds and the reduced atmospheric pressures at high altitudes. When solid iodine is warmed, the solid sublimes and a vivid purple vapor forms (Figure 10.27). The reverse of sublimation is called deposition, a process in which gaseous substances condense directly into the solid state, bypassing the liquid state. The formation of frost is an example of deposition.\n\n\nFIGURE 10.27 Sublimation of solid iodine in the bottom of the tube produces a purple gas that subsequently deposits as solid iodine on the colder part of the tube above. (credit: modification of work by Mark Ott)\n\nLike vaporization, the process of sublimation requires an input of energy to overcome intermolecular attractions. The enthalpy of sublimation, $\\Delta \\mathrm{H}_{\\text {sub }}$, is the energy required to convert one mole of a substance from the solid to the gaseous state. For example, the sublimation of carbon dioxide is represented by:\n\n$$\n\\mathrm{CO}_{2}(s) \\longrightarrow \\mathrm{CO}_{2}(g) \\quad \\Delta H_{\\text {sub }}=26.1 \\mathrm{~kJ} / \\mathrm{mol}\n$$\n\nLikewise, the enthalpy change for the reverse process of deposition is equal in magnitude but opposite in sign to that for sublimation:\n\n$$\n\\mathrm{CO}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(s) \\quad \\Delta H_{\\mathrm{dep}}=-\\Delta H_{\\mathrm{sub}}=-26.1 \\mathrm{~kJ} / \\mathrm{mol}\n$$"}
{"id": 3584, "contents": "1109. Sublimation and Deposition - \nConsider the extent to which intermolecular attractions must be overcome to achieve a given phase transition. Converting a solid into a liquid requires that these attractions be only partially overcome; transition to the gaseous state requires that they be completely overcome. As a result, the enthalpy of fusion for a substance is less than its enthalpy of vaporization. This same logic can be used to derive an approximate relation between the enthalpies of all phase changes for a given substance. Though not an entirely accurate description, sublimation may be conveniently modeled as a sequential two-step process of melting followed by vaporization in order to apply Hess's Law. Viewed in this manner, the enthalpy of sublimation for a substance may be estimated as the sum of its enthalpies of fusion and vaporization, as illustrated in Figure 10.28. For example:\n\n\nFIGURE 10.28 For a given substance, the sum of its enthalpy of fusion and enthalpy of vaporization is approximately equal to its enthalpy of sublimation."}
{"id": 3585, "contents": "1110. Heating and Cooling Curves - \nIn the chapter on thermochemistry, the relation between the amount of heat absorbed or released by a substance, $q$, and its accompanying temperature change, $\\Delta T$, was introduced:\n\n$$\nq=m c \\Delta T\n$$\n\nwhere $m$ is the mass of the substance and $c$ is its specific heat. The relation applies to matter being heated or cooled, but not undergoing a change in state. When a substance being heated or cooled reaches a temperature corresponding to one of its phase transitions, further gain or loss of heat is a result of diminishing or enhancing intermolecular attractions, instead of increasing or decreasing molecular kinetic energies. While a substance is undergoing a change in state, its temperature remains constant. Figure 10.29 shows a typical heating curve.\n\nConsider the example of heating a pot of water to boiling. A stove burner will supply heat at a roughly constant rate; initially, this heat serves to increase the water's temperature. When the water reaches its boiling point, the temperature remains constant despite the continued input of heat from the stove burner. This same temperature is maintained by the water as long as it is boiling. If the burner setting is increased to provide heat at a greater rate, the water temperature does not rise, but instead the boiling becomes more vigorous (rapid). This behavior is observed for other phase transitions as well: For example, temperature remains constant while the change of state is in progress.\n\n\nFIGURE 10.29 A typical heating curve for a substance depicts changes in temperature that result as the substance absorbs increasing amounts of heat. Plateaus in the curve (regions of constant temperature) are exhibited when the substance undergoes phase transitions."}
{"id": 3586, "contents": "1111. EXAMPLE 10.10 - \nTotal Heat Needed to Change Temperature and Phase for a Substance\nHow much heat is required to convert 135 g of ice at $-15^{\\circ} \\mathrm{C}$ into water vapor at $120^{\\circ} \\mathrm{C}$ ?"}
{"id": 3587, "contents": "1112. Solution - \nThe transition described involves the following steps:\n\n1. Heat ice from $-15^{\\circ} \\mathrm{C}$ to $0^{\\circ} \\mathrm{C}$\n2. Melt ice\n3. Heat water from $0^{\\circ} \\mathrm{C}$ to $100^{\\circ} \\mathrm{C}$\n4. Boil water\n5. Heat steam from $100^{\\circ} \\mathrm{C}$ to $120^{\\circ} \\mathrm{C}$\n\nThe heat needed to change the temperature of a given substance (with no change in phase) is: $q=m \\times c \\times \\Delta T$ (see previous chapter on thermochemistry). The heat needed to induce a given change in phase is given by $q=$ $n \\times \\Delta H$.\n\nUsing these equations with the appropriate values for specific heat of ice, water, and steam, and enthalpies of fusion and vaporization, we have:\n\n$$\nq_{\\text {total }}=(m \\cdot c \\cdot \\Delta T)_{\\text {ice }}+n \\cdot \\Delta H_{\\text {fus }}+(m \\cdot c \\cdot \\Delta T)_{\\text {water }}+n \\cdot \\Delta H_{\\mathrm{vap}}+(m \\cdot c \\cdot \\Delta T)_{\\text {steam }}\n$$"}
{"id": 3588, "contents": "1112. Solution - \n$$\n\\begin{aligned}\n& =\\left(135 \\mathrm{~g} \\cdot 2.09 \\mathrm{~J} / \\mathrm{g} \\cdot{ }^{\\circ} \\mathrm{C} \\cdot 15^{\\circ} \\mathrm{C}\\right)+\\left(135 \\cdot \\frac{1 \\mathrm{~mol}}{18.02 \\mathrm{~g}} \\cdot 6.01 \\mathrm{~kJ} / \\mathrm{mol}\\right) \\\\\n& +\\left(135 \\mathrm{~g} \\cdot 4.18 \\mathrm{~J} / \\mathrm{g} \\cdot{ }^{\\circ} \\mathrm{C} \\cdot 100^{\\circ} \\mathrm{C}\\right)+\\left(135 \\mathrm{~g} \\cdot \\frac{1 \\mathrm{~mol}}{18.02 \\mathrm{~g}} \\cdot 40.67 \\mathrm{~kJ} / \\mathrm{mol}\\right) \\\\\n& +\\left(135 \\mathrm{~g} \\cdot 1.84 \\mathrm{~J} / \\mathrm{g} \\cdot{ }^{\\circ} \\mathrm{C} \\cdot 20^{\\circ} \\mathrm{C}\\right) \\\\\n& =4230 \\mathrm{~J}+45.0 \\mathrm{~kJ}+56,500 \\mathrm{~J}+305 \\mathrm{~kJ}+4970 \\mathrm{~J}\n\\end{aligned}\n$$\n\nConverting the quantities in J to kJ permits them to be summed, yielding the total heat required:\n\n$$\n=4.23 \\mathrm{~kJ}+45.0 \\mathrm{~kJ}+56.5 \\mathrm{~kJ}+305 \\mathrm{~kJ}+4.97 \\mathrm{~kJ}=416 \\mathrm{~kJ}\n$$"}
{"id": 3589, "contents": "1113. Check Your Learning - \nWhat is the total amount of heat released when 94.0 g water at $80.0^{\\circ} \\mathrm{C}$ cools to form ice at $-30.0^{\\circ} \\mathrm{C}$ ?"}
{"id": 3590, "contents": "1114. Answer: - \n68.7 kJ"}
{"id": 3591, "contents": "1115. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Explain the construction and use of a typical phase diagram\n- Use phase diagrams to identify stable phases at given temperatures and pressures, and to describe phase transitions resulting from changes in these properties\n- Describe the supercritical fluid phase of matter\n\nIn the previous module, the variation of a liquid's equilibrium vapor pressure with temperature was described. Considering the definition of boiling point, plots of vapor pressure versus temperature represent how the boiling point of the liquid varies with pressure. Also described was the use of heating and cooling curves to determine a substance's melting (or freezing) point. Making such measurements over a wide range of pressures yields data that may be presented graphically as a phase diagram. A phase diagram combines plots of pressure versus temperature for the liquid-gas, solid-liquid, and solid-gas phase-transition equilibria of a substance. These diagrams indicate the physical states that exist under specific conditions of pressure and temperature, and also provide the pressure dependence of the phase-transition temperatures (melting points, sublimation points, boiling points). A typical phase diagram for a pure substance is shown in Figure 10.30.\n\n\nFIGURE 10.30 The physical state of a substance and its phase-transition temperatures are represented graphically in a phase diagram.\n\nTo illustrate the utility of these plots, consider the phase diagram for water shown in Figure 10.31.\n\n\nFIGURE 10.31 The pressure and temperature axes on this phase diagram of water are not drawn to constant scale in order to illustrate several important properties."}
{"id": 3592, "contents": "1115. LEARNING OBJECTIVES - \nFIGURE 10.31 The pressure and temperature axes on this phase diagram of water are not drawn to constant scale in order to illustrate several important properties.\n\nWe can use the phase diagram to identify the physical state of a sample of water under specified conditions of pressure and temperature. For example, a pressure of 50 kPa and a temperature of $-10^{\\circ} \\mathrm{C}$ correspond to the region of the diagram labeled \"ice.\" Under these conditions, water exists only as a solid (ice). A pressure of 50 kPa and a temperature of $50^{\\circ} \\mathrm{C}$ correspond to the \"water\" region-here, water exists only as a liquid. At 25 kPa and $200^{\\circ} \\mathrm{C}$, water exists only in the gaseous state. Note that on the $\\mathrm{H}_{2} \\mathrm{O}$ phase diagram, the pressure and temperature axes are not drawn to a constant scale in order to permit the illustration of several important features as described here.\n\nThe curve BC in Figure 10.31 is the plot of vapor pressure versus temperature as described in the previous module of this chapter. This \"liquid-vapor\" curve separates the liquid and gaseous regions of the phase diagram and provides the boiling point for water at any pressure. For example, at 1 atm , the boiling point is $100^{\\circ} \\mathrm{C}$. Notice that the liquid-vapor curve terminates at a temperature of $374^{\\circ} \\mathrm{C}$ and a pressure of 218 atm ,\nindicating that water cannot exist as a liquid above this temperature, regardless of the pressure. The physical properties of water under these conditions are intermediate between those of its liquid and gaseous phases. This unique state of matter is called a supercritical fluid, a topic that will be described in the next section of this module."}
{"id": 3593, "contents": "1115. LEARNING OBJECTIVES - \nThe solid-vapor curve, labeled AB in Figure 10.31, indicates the temperatures and pressures at which ice and water vapor are in equilibrium. These temperature-pressure data pairs correspond to the sublimation, or deposition, points for water. If we could zoom in on the solid-gas line in Figure 10.31, we would see that ice has a vapor pressure of about 0.20 kPa at $-10^{\\circ} \\mathrm{C}$. Thus, if we place a frozen sample in a vacuum with a pressure less than 0.20 kPa , ice will sublime. This is the basis for the \"freeze-drying\" process often used to preserve foods, such as the ice cream shown in Figure 10.32.\n\n\nFIGURE 10.32 Freeze-dried foods, like this ice cream, are dehydrated by sublimation at pressures below the triple point for water. (credit: \"lwao\"/Flickr)\n\nThe solid-liquid curve labeled BD shows the temperatures and pressures at which ice and liquid water are in equilibrium, representing the melting/freezing points for water. Note that this curve exhibits a slight negative slope (greatly exaggerated for clarity), indicating that the melting point for water decreases slightly as pressure increases. Water is an unusual substance in this regard, as most substances exhibit an increase in melting point with increasing pressure. This behavior is partly responsible for the movement of glaciers, like the one shown in Figure 10.33. The bottom of a glacier experiences an immense pressure due to its weight that can melt some of the ice, forming a layer of liquid water on which the glacier may more easily slide.\n\n\nFIGURE 10.33 The immense pressures beneath glaciers result in partial melting to produce a layer of water that provides lubrication to assist glacial movement. This satellite photograph shows the advancing edge of the Perito Moreno glacier in Argentina. (credit: NASA)\n\nThe point of intersection of all three curves is labeled B in Figure 10.31. At the pressure and temperature represented by this point, all three phases of water coexist in equilibrium. This temperature-pressure data\npair is called the triple point. At pressures lower than the triple point, water cannot exist as a liquid, regardless of the temperature."}
{"id": 3594, "contents": "1117. Determining the State of Water - \nUsing the phase diagram for water given in Figure 10.31, determine the state of water at the following temperatures and pressures:\n(a) $-10^{\\circ} \\mathrm{C}$ and 50 kPa\n(b) $25^{\\circ} \\mathrm{C}$ and 90 kPa\n(c) $50^{\\circ} \\mathrm{C}$ and 40 kPa\n(d) $80^{\\circ} \\mathrm{C}$ and 5 kPa\n(e) $-10^{\\circ} \\mathrm{C}$ and 0.3 kPa\n(f) $50^{\\circ} \\mathrm{C}$ and 0.3 kPa"}
{"id": 3595, "contents": "1118. Solution - \nUsing the phase diagram for water, we can determine that the state of water at each temperature and pressure given are as follows: (a) solid; (b) liquid; (c) liquid; (d) gas; (e) solid; (f) gas."}
{"id": 3596, "contents": "1119. Check Your Learning - \nWhat phase changes can water undergo as the temperature changes if the pressure is held at 0.3 kPa ? If the pressure is held at 50 kPa ?"}
{"id": 3597, "contents": "1120. Answer: - \nAt $0.3 \\mathrm{kPa}: \\mathrm{s} \\longrightarrow \\mathrm{g}$ at $-58^{\\circ} \\mathrm{C}$. At $50 \\mathrm{kPa}: \\mathrm{s} \\longrightarrow 1$ at $0^{\\circ} \\mathrm{C}, \\mathrm{l} \\longrightarrow \\mathrm{g}$ at $78^{\\circ} \\mathrm{C}$\n\nConsider the phase diagram for carbon dioxide shown in Figure 10.34 as another example. The solid-liquid curve exhibits a positive slope, indicating that the melting point for $\\mathrm{CO}_{2}$ increases with pressure as it does for most substances (water being a notable exception as described previously). Notice that the triple point is well above 1 atm, indicating that carbon dioxide cannot exist as a liquid under ambient pressure conditions. Instead, cooling gaseous carbon dioxide at 1 atm results in its deposition into the solid state. Likewise, solid carbon dioxide does not melt at 1 atm pressure but instead sublimes to yield gaseous $\\mathrm{CO}_{2}$. Finally, notice that the critical point for carbon dioxide is observed at a relatively modest temperature and pressure in comparison to water.\n\n\nFIGURE 10.34 A phase diagram for carbon dioxide is shown. The pressure axis is plotted on a logarithmic scale to accommodate the large range of values."}
{"id": 3598, "contents": "1121. Determining the State of Carbon Dioxide - \nUsing the phase diagram for carbon dioxide shown in Figure 10.34, determine the state of $\\mathrm{CO}_{2}$ at the following temperatures and pressures:\n(a) $-30^{\\circ} \\mathrm{C}$ and 2000 kPa\n(b) $-90^{\\circ} \\mathrm{C}$ and 1000 kPa\n(c) $-60^{\\circ} \\mathrm{C}$ and 100 kPa\n(d) $-40^{\\circ} \\mathrm{C}$ and 1500 kPa\n(e) $0^{\\circ} \\mathrm{C}$ and 100 kPa\n(f) $20^{\\circ} \\mathrm{C}$ and 100 kPa"}
{"id": 3599, "contents": "1122. Solution - \nUsing the phase diagram for carbon dioxide provided, we can determine that the state of $\\mathrm{CO}_{2}$ at each temperature and pressure given are as follows: (a) liquid; (b) solid; (c) gas; (d) liquid; (e) gas; (f) gas."}
{"id": 3600, "contents": "1123. Check Your Learning - \nIdentify the phase changes that carbon dioxide will undergo as its temperature is increased from $-100^{\\circ} \\mathrm{C}$ while holding its pressure constant at 1500 kPa . At 50 kPa . At what approximate temperatures do these phase changes occur?"}
{"id": 3601, "contents": "1124. Answer: - \nat $1500 \\mathrm{kPa}: \\mathrm{s} \\longrightarrow 1$ at $-55^{\\circ} \\mathrm{C}, 1 \\longrightarrow \\mathrm{~g}$ at $-10^{\\circ} \\mathrm{C}$;\nat $50 \\mathrm{kPa}: \\mathrm{s} \\longrightarrow \\mathrm{g}$ at $-60^{\\circ} \\mathrm{C}$"}
{"id": 3602, "contents": "1125. Supercritical Fluids - \nIf we place a sample of water in a sealed container at $25^{\\circ} \\mathrm{C}$, remove the air, and let the vaporizationcondensation equilibrium establish itself, we are left with a mixture of liquid water and water vapor at a pressure of 0.03 atm . A distinct boundary between the more dense liquid and the less dense gas is clearly\nobserved. As we increase the temperature, the pressure of the water vapor increases, as described by the liquid-gas curve in the phase diagram for water (Figure 10.31), and a two-phase equilibrium of liquid and gaseous phases remains. At a temperature of $374^{\\circ} \\mathrm{C}$, the vapor pressure has risen to 218 atm , and any further increase in temperature results in the disappearance of the boundary between liquid and vapor phases. All of the water in the container is now present in a single phase whose physical properties are intermediate between those of the gaseous and liquid states. This phase of matter is called a supercritical fluid, and the temperature and pressure above which this phase exists is the critical point (Figure 10.35). Above its critical temperature, a gas cannot be liquefied no matter how much pressure is applied. The pressure required to liquefy a gas at its critical temperature is called the critical pressure. The critical temperatures and critical pressures of some common substances are given in the following table.\n\n| Substance | Critical Temperature $\\left({ }^{\\circ} \\mathrm{C}\\right)$ | Critical Pressure (kPa) |\n| :--- | :--- | :--- |\n| hydrogen | -240.0 | 1300 |\n| nitrogen | -147.2 | 3400 |\n| oxygen | -118.9 | 5000 |\n| carbon dioxide | 31.1 | 7400 |\n| ammonia | 132.4 | 11,300 |\n| sulfur dioxide | 157.2 | 7800 |\n| water | 374.0 | 22,000 |"}
{"id": 3603, "contents": "1125. Supercritical Fluids - \nFIGURE 10.35 (a) A sealed container of liquid carbon dioxide slightly below its critical point is heated, resulting in (b) the formation of the supercritical fluid phase. Cooling the supercritical fluid lowers its temperature and pressure below the critical point, resulting in the reestablishment of separate liquid and gaseous phases (c and d). Colored floats illustrate differences in density between the liquid, gaseous, and supercritical fluid states. (credit: modification of work by \"mrmrobin\"/YouTube)"}
{"id": 3604, "contents": "1126. LINK TO LEARNING - \nObserve the liquid-to-supercritical fluid transition (http://openstax.org/l/16supercrit) for carbon dioxide.\n\nLike a gas, a supercritical fluid will expand and fill a container, but its density is much greater than typical gas densities, typically being close to those for liquids. Similar to liquids, these fluids are capable of dissolving nonvolatile solutes. They exhibit essentially no surface tension and very low viscosities, however, so they can more effectively penetrate very small openings in a solid mixture and remove soluble components. These properties make supercritical fluids extremely useful solvents for a wide range of applications. For example, supercritical carbon dioxide has become a very popular solvent in the food industry, being used to decaffeinate coffee, remove fats from potato chips, and extract flavor and fragrance compounds from citrus\noils. It is nontoxic, relatively inexpensive, and not considered to be a pollutant. After use, the $\\mathrm{CO}_{2}$ can be easily recovered by reducing the pressure and collecting the resulting gas."}
{"id": 3605, "contents": "1128. The Critical Temperature of Carbon Dioxide - \nIf we shake a carbon dioxide fire extinguisher on a cool day ( $18{ }^{\\circ} \\mathrm{C}$ ), we can hear liquid $\\mathrm{CO}_{2}$ sloshing around inside the cylinder. However, the same cylinder appears to contain no liquid on a hot summer day ( $35^{\\circ} \\mathrm{C}$ ). Explain these observations."}
{"id": 3606, "contents": "1129. Solution - \nOn the cool day, the temperature of the $\\mathrm{CO}_{2}$ is below the critical temperature of $\\mathrm{CO}_{2}, 304 \\mathrm{~K}$ or $31^{\\circ} \\mathrm{C}$, so liquid $\\mathrm{CO}_{2}$ is present in the cylinder. On the hot day, the temperature of the $\\mathrm{CO}_{2}$ is greater than its critical temperature of $31^{\\circ} \\mathrm{C}$. Above this temperature no amount of pressure can liquefy $\\mathrm{CO}_{2}$ so no liquid $\\mathrm{CO}_{2}$ exists in the fire extinguisher."}
{"id": 3607, "contents": "1130. Check Your Learning - \nAmmonia can be liquefied by compression at room temperature; oxygen cannot be liquefied under these conditions. Why do the two gases exhibit different behavior?"}
{"id": 3608, "contents": "1131. Answer: - \nThe critical temperature of ammonia is 405.5 K , which is higher than room temperature. The critical temperature of oxygen is below room temperature; thus oxygen cannot be liquefied at room temperature."}
{"id": 3609, "contents": "1133. Decaffeinating Coffee Using Supercritical $\\mathbf{C O}_{\\mathbf{2}}$ - \nCoffee is the world's second most widely traded commodity, following only petroleum. Across the globe, people love coffee's aroma and taste. Many of us also depend on one component of coffee-caffeine-to help us get going in the morning or stay alert in the afternoon. But late in the day, coffee's stimulant effect can keep you from sleeping, so you may choose to drink decaffeinated coffee in the evening.\n\nSince the early 1900s, many methods have been used to decaffeinate coffee. All have advantages and disadvantages, and all depend on the physical and chemical properties of caffeine. Because caffeine is a somewhat polar molecule, it dissolves well in water, a polar liquid. However, since many of the other 400-plus compounds that contribute to coffee's taste and aroma also dissolve in $\\mathrm{H}_{2} \\mathrm{O}$, hot water decaffeination processes can also remove some of these compounds, adversely affecting the smell and taste of the decaffeinated coffee. Dichloromethane $\\left(\\mathrm{CH}_{2} \\mathrm{Cl}_{2}\\right)$ and ethyl acetate $\\left(\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{C}_{2} \\mathrm{H}_{5}\\right)$ have similar polarity to caffeine, and are therefore very effective solvents for caffeine extraction, but both also remove some flavor and aroma components, and their use requires long extraction and cleanup times. Because both of these solvents are toxic, health concerns have been raised regarding the effect of residual solvent remaining in the decaffeinated coffee."}
{"id": 3610, "contents": "1133. Decaffeinating Coffee Using Supercritical $\\mathbf{C O}_{\\mathbf{2}}$ - \nSupercritical fluid extraction using carbon dioxide is now being widely used as a more effective and environmentally friendly decaffeination method (Figure 10.36). At temperatures above 304.2 K and pressures above $7376 \\mathrm{kPa}, \\mathrm{CO}_{2}$ is a supercritical fluid, with properties of both gas and liquid. Like a gas, it penetrates deep into the coffee beans; like a liquid, it effectively dissolves certain substances. Supercritical carbon dioxide extraction of steamed coffee beans removes $97-99 \\%$ of the caffeine, leaving coffee's flavor and aroma compounds intact. Because $\\mathrm{CO}_{2}$ is a gas under standard conditions, its removal from the extracted coffee beans is easily accomplished, as is the recovery of the caffeine from the extract. The caffeine recovered from coffee beans via this process is a valuable product that can be used subsequently as an additive to other foods or drugs.\n\n\nFIGURE 10.36 (a) Caffeine molecules have both polar and nonpolar regions, making it soluble in solvents of varying polarities. (b) The schematic shows a typical decaffeination process involving supercritical carbon dioxide."}
{"id": 3611, "contents": "1134. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Define and describe the bonding and properties of ionic, molecular, metallic, and covalent network crystalline solids\n- Describe the main types of crystalline solids: ionic solids, metallic solids, covalent network solids, and molecular solids\n- Explain the ways in which crystal defects can occur in a solid\n\nWhen most liquids are cooled, they eventually freeze and form crystalline solids, solids in which the atoms, ions, or molecules are arranged in a definite repeating pattern. It is also possible for a liquid to freeze before its molecules become arranged in an orderly pattern. The resulting materials are called amorphous solids or noncrystalline solids (or, sometimes, glasses). The particles of such solids lack an ordered internal structure and are randomly arranged (Figure 10.37).\n\n\nFIGURE 10.37 The entities of a solid phase may be arranged in a regular, repeating pattern (crystalline solids) or\nrandomly (amorphous).\nMetals and ionic compounds typically form ordered, crystalline solids. Substances that consist of large molecules, or a mixture of molecules whose movements are more restricted, often form amorphous solids. For examples, candle waxes are amorphous solids composed of large hydrocarbon molecules. Some substances, such as silicon dioxide (shown in Figure 10.38), can form either crystalline or amorphous solids, depending on the conditions under which it is produced. Also, amorphous solids may undergo a transition to the crystalline state under appropriate conditions.\n\n(a)\n\n(b)\n\nFIGURE 10.38 (a) Silicon dioxide, $\\mathrm{SiO}_{2}$, is abundant in nature as one of several crystalline forms of the mineral quartz. (b) Rapid cooling of molten $\\mathrm{SiO}_{2}$ yields an amorphous solid known as \"fused silica\".\n\nCrystalline solids are generally classified according to the nature of the forces that hold its particles together. These forces are primarily responsible for the physical properties exhibited by the bulk solids. The following sections provide descriptions of the major types of crystalline solids: ionic, metallic, covalent network, and molecular."}
{"id": 3612, "contents": "1135. Ionic Solids - \nIonic solids, such as sodium chloride and nickel oxide, are composed of positive and negative ions that are held together by electrostatic attractions, which can be quite strong (Figure 10.39). Many ionic crystals also have high melting points. This is due to the very strong attractions between the ions-in ionic compounds, the attractions between full charges are (much) larger than those between the partial charges in polar molecular compounds. This will be looked at in more detail in a later discussion of lattice energies. Although they are hard, they also tend to be brittle, and they shatter rather than bend. Ionic solids do not conduct electricity; however, they do conduct when molten or dissolved because their ions are free to move. Many simple compounds formed by the reaction of a metallic element with a nonmetallic element are ionic.\n\n\nFIGURE 10.39 Sodium chloride is an ionic solid."}
{"id": 3613, "contents": "1136. Metallic Solids - \nMetallic solids such as crystals of copper, aluminum, and iron are formed by metal atoms Figure 10.40. The structure of metallic crystals is often described as a uniform distribution of atomic nuclei within a \"sea\" of delocalized electrons. The atoms within such a metallic solid are held together by a unique force known as metallic bonding that gives rise to many useful and varied bulk properties. All exhibit high thermal and electrical conductivity, metallic luster, and malleability. Many are very hard and quite strong. Because of their\nmalleability (the ability to deform under pressure or hammering), they do not shatter and, therefore, make useful construction materials. The melting points of the metals vary widely. Mercury is a liquid at room temperature, and the alkali metals melt below $200^{\\circ} \\mathrm{C}$. Several post-transition metals also have low melting points, whereas the transition metals melt at temperatures above $1000^{\\circ} \\mathrm{C}$. These differences reflect differences in strengths of metallic bonding among the metals.\n\n\nFIGURE 10.40 Copper is a metallic solid."}
{"id": 3614, "contents": "1137. Covalent Network Solid - \nCovalent network solids include crystals of diamond, silicon, some other nonmetals, and some covalent compounds such as silicon dioxide (sand) and silicon carbide (carborundum, the abrasive on sandpaper). Many minerals have networks of covalent bonds. The atoms in these solids are held together by a network of covalent bonds, as shown in Figure 10.41. To break or to melt a covalent network solid, covalent bonds must be broken. Because covalent bonds are relatively strong, covalent network solids are typically characterized by hardness, strength, and high melting points. For example, diamond is one of the hardest substances known and melts above $3500^{\\circ} \\mathrm{C}$.\n\n\nFIGURE 10.41 A covalent crystal contains a three-dimensional network of covalent bonds, as illustrated by the structures of diamond, silicon dioxide, silicon carbide, and graphite. Graphite is an exceptional example, composed of planar sheets of covalent crystals that are held together in layers by noncovalent forces. Unlike typical covalent solids, graphite is very soft and electrically conductive."}
{"id": 3615, "contents": "1138. Molecular Solid - \nMolecular solids, such as ice, sucrose (table sugar), and iodine, as shown in Figure 10.42, are composed of neutral molecules. The strengths of the attractive forces between the units present in different crystals vary widely, as indicated by the melting points of the crystals. Small symmetrical molecules (nonpolar molecules), such as $\\mathrm{H}_{2}, \\mathrm{~N}_{2}, \\mathrm{O}_{2}$, and $\\mathrm{F}_{2}$, have weak attractive forces and form molecular solids with very low melting points (below $-200^{\\circ} \\mathrm{C}$ ). Substances consisting of larger, nonpolar molecules have larger attractive forces and melt at higher temperatures. Molecular solids composed of molecules with permanent dipole moments (polar\nmolecules) melt at still higher temperatures. Examples include ice (melting point, $0^{\\circ} \\mathrm{C}$ ) and table sugar (melting point, $185^{\\circ} \\mathrm{C}$ ).\n\ncarbon dioxide\n\niodine\n\nFIGURE 10.42 Carbon dioxide $\\left(\\mathrm{CO}_{2}\\right)$ consists of small, nonpolar molecules and forms a molecular solid with a melting point of $-78^{\\circ} \\mathrm{C}$. Iodine $\\left(\\mathrm{I}_{2}\\right)$ consists of larger, nonpolar molecules and forms a molecular solid that melts at $114^{\\circ} \\mathrm{C}$."}
{"id": 3616, "contents": "1139. Properties of Solids - \nA crystalline solid, like those listed in Table 10.4, has a precise melting temperature because each atom or molecule of the same type is held in place with the same forces or energy. Thus, the attractions between the units that make up the crystal all have the same strength and all require the same amount of energy to be broken. The gradual softening of an amorphous material differs dramatically from the distinct melting of a crystalline solid. This results from the structural nonequivalence of the molecules in the amorphous solid. Some forces are weaker than others, and when an amorphous material is heated, the weakest intermolecular attractions break first. As the temperature is increased further, the stronger attractions are broken. Thus amorphous materials soften over a range of temperatures.\n\nTypes of Crystalline Solids and Their Properties"}
{"id": 3617, "contents": "1139. Properties of Solids - \nTypes of Crystalline Solids and Their Properties\n\n| Type of Solid | Type of Particles | Type of Attractions | Properties | Examples |\n| :---: | :---: | :---: | :---: | :---: |\n| ionic | ions | ionic <br> bonds | hard, brittle, conducts electricity as a liquid but not as a solid, high to very high melting points | $\\begin{aligned} & \\mathrm{NaCl}, \\\\ & \\mathrm{Al}_{2} \\mathrm{O}_{3} \\end{aligned}$ |\n| metallic | atoms of electropositive elements | metallic <br> bonds | shiny, malleable, ductile, conducts heat and electricity well, variable hardness and melting temperature | $\\begin{aligned} & \\mathrm{Cu}, \\mathrm{Fe}, \\mathrm{Ti}, \\\\ & \\mathrm{~Pb}, \\mathrm{U} \\end{aligned}$ |\n| covalent network | atoms of electronegative elements | covalent bonds | very hard, not conductive, very high melting points | C <br> (diamond), $\\mathrm{SiO}_{2}, \\mathrm{SiC}$ |\n| molecular | molecules (or atoms) | IMFs | variable hardness, variable brittleness, not conductive, low melting points | $\\begin{aligned} & \\mathrm{H}_{2} \\mathrm{O}, \\mathrm{CO}_{2}, \\\\ & \\mathrm{I}_{2}, \\\\ & \\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11} \\end{aligned}$ |\n\nTABLE 10.4"}
{"id": 3618, "contents": "1141. Graphene: Material of the Future - \nCarbon is an essential element in our world. The unique properties of carbon atoms allow the existence of carbon-based life forms such as ourselves. Carbon forms a huge variety of substances that we use on a daily basis, including those shown in Figure 10.43. You may be familiar with diamond and graphite, the two most common allotropes of carbon. (Allotropes are different structural forms of the same element.) Diamond is one of the hardest-known substances, whereas graphite is soft enough to be used as pencil lead. These very different properties stem from the different arrangements of the carbon atoms in the different allotropes.\n\n\nFIGURE 10.43 Diamond is extremely hard because of the strong bonding between carbon atoms in all directions. Graphite (in pencil lead) rubs off onto paper due to the weak attractions between the carbon layers. An image of a graphite surface shows the distance between the centers of adjacent carbon atoms. (credit left photo: modification of work by Steve Jurvetson; credit middle photo: modification of work by United States Geological Survey)\n\nYou may be less familiar with a recently discovered form of carbon: graphene. Graphene was first isolated in 2004 by using tape to peel off thinner and thinner layers from graphite. It is essentially a single sheet (one atom thick) of graphite. Graphene, illustrated in Figure 10.44, is not only strong and lightweight, but it is also an excellent conductor of electricity and heat. These properties may prove very useful in a wide range of applications, such as vastly improved computer chips and circuits, better batteries and solar cells, and stronger and lighter structural materials. The 2010 Nobel Prize in Physics was awarded to Andre Geim and Konstantin Novoselov for their pioneering work with graphene.\nGraphene sheet\n\n\n\nBuckyball\n\n\nNanotube\n\n\nStacked sheets\n\nFIGURE 10.44 Graphene sheets can be formed into buckyballs, nanotubes, and stacked layers."}
{"id": 3619, "contents": "1142. Crystal Defects - \nIn a crystalline solid, the atoms, ions, or molecules are arranged in a definite repeating pattern, but occasional defects may occur in the pattern. Several types of defects are known, as illustrated in Figure 10.45. Vacancies are defects that occur when positions that should contain atoms or ions are vacant. Less commonly, some atoms or ions in a crystal may occupy positions, called interstitial sites, located between the regular positions for atoms. Other distortions are found in impure crystals, as, for example, when the cations, anions, or molecules of the impurity are too large to fit into the regular positions without distorting the structure. Trace amounts of impurities are sometimes added to a crystal (a process known as doping) in order to create defects in the structure that yield desirable changes in its properties. For example, silicon crystals are doped with varying amounts of different elements to yield suitable electrical properties for their use in the manufacture of semiconductors and computer chips.\n\n\nFIGURE 10.45 Types of crystal defects include vacancies, interstitial atoms, and substitutions impurities."}
{"id": 3620, "contents": "1143. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the arrangement of atoms and ions in crystalline structures\n- Compute ionic radii using unit cell dimensions\n- Explain the use of X-ray diffraction measurements in determining crystalline structures\n\nOver $90 \\%$ of naturally occurring and man-made solids are crystalline. Most solids form with a regular arrangement of their particles because the overall attractive interactions between particles are maximized, and the total intermolecular energy is minimized, when the particles pack in the most efficient manner. The regular arrangement at an atomic level is often reflected at a macroscopic level. In this module, we will explore some of the details about the structures of metallic and ionic crystalline solids, and learn how these structures are determined experimentally."}
{"id": 3621, "contents": "1144. The Structures of Metals - \nWe will begin our discussion of crystalline solids by considering elemental metals, which are relatively simple because each contains only one type of atom. A pure metal is a crystalline solid with metal atoms packed closely together in a repeating pattern. Some of the properties of metals in general, such as their malleability and ductility, are largely due to having identical atoms arranged in a regular pattern. The different properties of one metal compared to another partially depend on the sizes of their atoms and the specifics of their spatial arrangements. We will explore the similarities and differences of four of the most common metal crystal geometries in the sections that follow."}
{"id": 3622, "contents": "1145. Unit Cells of Metals - \nThe structure of a crystalline solid, whether a metal or not, is best described by considering its simplest repeating unit, which is referred to as its unit cell. The unit cell consists of lattice points that represent the locations of atoms or ions. The entire structure then consists of this unit cell repeating in three dimensions, as illustrated in Figure 10.46.\n\n\nFIGURE 10.46 A unit cell shows the locations of lattice points repeating in all directions.\nLet us begin our investigation of crystal lattice structure and unit cells with the most straightforward structure and the most basic unit cell. To visualize this, imagine taking a large number of identical spheres, such as tennis balls, and arranging them uniformly in a container. The simplest way to do this would be to make layers in which the spheres in one layer are directly above those in the layer below, as illustrated in Figure 10.47. This arrangement is called simple cubic structure, and the unit cell is called the simple cubic unit cell or primitive cubic unit cell.\n\n\nFIGURE 10.47 When metal atoms are arranged with spheres in one layer directly above or below spheres in another layer, the lattice structure is called simple cubic. Note that the spheres are in contact.\n\nIn a simple cubic structure, the spheres are not packed as closely as they could be, and they only \"fill\" about $52 \\%$ of the volume of the container. This is a relatively inefficient arrangement, and only one metal (polonium, Po) crystallizes in a simple cubic structure. As shown in Figure 10.48, a solid with this type of arrangement consists of planes (or layers) in which each atom contacts only the four nearest neighbors in its layer; one atom directly above it in the layer above; and one atom directly below it in the layer below. The number of other particles that each particle in a crystalline solid contacts is known as its coordination number. For a polonium atom in a simple cubic array, the coordination number is, therefore, six.\n\n\nFIGURE 10.48 An atom in a simple cubic lattice structure contacts six other atoms, so it has a coordination number of six."}
{"id": 3623, "contents": "1145. Unit Cells of Metals - \nFIGURE 10.48 An atom in a simple cubic lattice structure contacts six other atoms, so it has a coordination number of six.\n\nIn a simple cubic lattice, the unit cell that repeats in all directions is a cube defined by the centers of eight atoms, as shown in Figure 10.49. Atoms at adjacent corners of this unit cell contact each other, so the edge length of this cell is equal to two atomic radii, or one atomic diameter. A cubic unit cell contains only the parts of these atoms that are within it. Since an atom at a corner of a simple cubic unit cell is contained by a total of eight unit cells, only one-eighth of that atom is within a specific unit cell. And since each simple cubic unit cell has one atom at each of its eight \"corners,\" there is $8 \\times \\frac{1}{8}=1$ atom within one simple cubic unit cell.\n\n\n\n8 corners\n\nSimple cubic lattice cell\nFIGURE 10.49 A simple cubic lattice unit cell contains one-eighth of an atom at each of its eight corners, so it contains one atom total."}
{"id": 3624, "contents": "1147. Calculation of Atomic Radius and Density for Metals, Part 1 - \nThe edge length of the unit cell of alpha polonium is 336 pm .\n(a) Determine the radius of a polonium atom.\n(b) Determine the density of alpha polonium."}
{"id": 3625, "contents": "1148. Solution - \nAlpha polonium crystallizes in a simple cubic unit cell:\n\n(a) Two adjacent Po atoms contact each other, so the edge length of this cell is equal to two Po atomic radii: $1=$ $2 r$. Therefore, the radius of Po is $r=\\frac{1}{2}=\\frac{336 \\mathrm{pm}}{2}=168 \\mathrm{pm}$.\n(b) Density is given by density $=\\frac{\\text { mass }}{\\text { volume }}$. The density of polonium can be found by determining the density of its unit cell (the mass contained within a unit cell divided by the volume of the unit cell). Since a Po unit cell contains one-eighth of a Po atom at each of its eight corners, a unit cell contains one Po atom.\n\nThe mass of a Po unit cell can be found by:\n\n$$\n1 \\text { Po unit cell } \\times \\frac{1 \\text { Po atom }}{1 \\text { Po unit cell }} \\times \\frac{1 \\mathrm{~mol} \\mathrm{Po}}{6.022 \\times 10^{23} \\text { Po atoms }} \\times \\frac{208.998 \\mathrm{~g}}{1 \\mathrm{~mol} \\mathrm{Po}}=3.47 \\times 10^{-22} \\mathrm{~g}\n$$\n\nThe volume of a Po unit cell can be found by:\n\n$$\nV=l^{3}=\\left(336 \\times 10^{-10} \\mathrm{~cm}\\right)^{3}=3.79 \\times 10^{-23} \\mathrm{~cm}^{3}\n$$\n\n(Note that the edge length was converted from pm to cm to get the usual volume units for density.)\nTherefore, the density of $\\mathrm{Po}=\\frac{3.471 \\times 10^{-22} \\mathrm{~g}}{3.79 \\times 10^{-23} \\mathrm{~cm}^{3}}=9.16 \\mathrm{~g} / \\mathrm{cm}^{3}$"}
{"id": 3626, "contents": "1149. Check Your Learning - \nThe edge length of the unit cell for nickel is 0.3524 nm . The density of Ni is $8.90 \\mathrm{~g} / \\mathrm{cm}^{3}$. Does nickel crystallize in a simple cubic structure? Explain."}
{"id": 3627, "contents": "1150. Answer: - \nNo. If Ni was simple cubic, its density would be given by:\n1 Ni atom $\\times \\frac{1 \\mathrm{~mol} \\mathrm{Ni}}{6.022 \\times 10^{23} \\mathrm{Ni} \\text { atoms }} \\times \\frac{58.693 \\mathrm{~g}}{1 \\mathrm{~mol} \\mathrm{Ni}}=9.746 \\times 10^{-23} \\mathrm{~g}$\n$V=l^{3}=\\left(3.524 \\times 10^{-8} \\mathrm{~cm}\\right)^{3}=4.376 \\times 10^{-23} \\mathrm{~cm}^{3}$\nThen the density of Ni would be $=\\frac{9.746 \\times 10^{-23} \\mathrm{~g}}{4.376 \\times 10^{-23} \\mathrm{~cm}^{3}}=2.23 \\mathrm{~g} / \\mathrm{cm}^{3}$\nSince the actual density of Ni is not close to this, Ni does not form a simple cubic structure.\n\nMost metal crystals are one of the four major types of unit cells. For now, we will focus on the three cubic unit cells: simple cubic (which we have already seen), body-centered cubic unit cell, and face-centered cubic unit cell-all of which are illustrated in Figure 10.50. (Note that there are actually seven different lattice systems, some of which have more than one type of lattice, for a total of 14 different types of unit cells. We leave the more complicated geometries for later in this module.)"}
{"id": 3628, "contents": "1151. Lattice point locations - \nFIGURE 10.50 Cubic unit cells of metals show (in the upper figures) the locations of lattice points and (in the lower figures) metal atoms located in the unit cell.\n\nSome metals crystallize in an arrangement that has a cubic unit cell with atoms at all of the corners and an atom in the center, as shown in Figure 10.51. This is called a body-centered cubic (BCC) solid. Atoms in the corners of a BCC unit cell do not contact each other but contact the atom in the center. A BCC unit cell contains two atoms: one-eighth of an atom at each of the eight corners ( $8 \\times \\frac{1}{8}=1$ atom from the corners) plus one atom from the center. Any atom in this structure touches four atoms in the layer above it and four atoms in the layer below it. Thus, an atom in a BCC structure has a coordination number of eight.\n\n\nBody-centered cubic structure\nFIGURE 10.51 In a body-centered cubic structure, atoms in a specific layer do not touch each other. Each atom touches four atoms in the layer above it and four atoms in the layer below it.\n\nAtoms in BCC arrangements are much more efficiently packed than in a simple cubic structure, occupying about $68 \\%$ of the total volume. Isomorphous metals with a BCC structure include $\\mathrm{K}, \\mathrm{Ba}, \\mathrm{Cr}, \\mathrm{Mo}, \\mathrm{W}$, and Fe at room temperature. (Elements or compounds that crystallize with the same structure are said to be isomorphous.)"}
{"id": 3629, "contents": "1151. Lattice point locations - \nMany other metals, such as aluminum, copper, and lead, crystallize in an arrangement that has a cubic unit cell with atoms at all of the corners and at the centers of each face, as illustrated in Figure 10.52. This arrangement is called a face-centered cubic (FCC) solid. A FCC unit cell contains four atoms: one-eighth of an atom at each of the eight corners ( $8 \\times \\frac{1}{8}=1$ atom from the corners) and one-half of an atom on each of the six faces ( $6 \\times \\frac{1}{2}=3$ atoms from the faces). The atoms at the corners touch the atoms in the centers of the adjacent faces along the face diagonals of the cube. Because the atoms are on identical lattice points, they have identical environments.\n\n\nFace-centered cubic structure\nFIGURE 10.52 A face-centered cubic solid has atoms at the corners and, as the name implies, at the centers of the faces of its unit cells.\n\nAtoms in an FCC arrangement are packed as closely together as possible, with atoms occupying $74 \\%$ of the volume. This structure is also called cubic closest packing (CCP). In CCP, there are three repeating layers of hexagonally arranged atoms. Each atom contacts six atoms in its own layer, three in the layer above, and three in the layer below. In this arrangement, each atom touches 12 near neighbors, and therefore has a coordination number of 12 . The fact that FCC and CCP arrangements are equivalent may not be immediately obvious, but why they are actually the same structure is illustrated in Figure 10.53.\n\n\nSide view\n\n\nTop view\n\n\nRotated view\n\nCubic closest packed structure\nFIGURE 10.53 A CCP arrangement consists of three repeating layers (ABCABC...) of hexagonally arranged atoms. Atoms in a CCP structure have a coordination number of 12 because they contact six atoms in their layer, plus three atoms in the layer above and three atoms in the layer below. By rotating our perspective, we can see that a CCP structure has a unit cell with a face containing an atom from layer A at one corner, atoms from layer B across a diagonal (at two corners and in the middle of the face), and an atom from layer C at the remaining corner. This is the same as a face-centered cubic arrangement."}
{"id": 3630, "contents": "1151. Lattice point locations - \nBecause closer packing maximizes the overall attractions between atoms and minimizes the total intermolecular energy, the atoms in most metals pack in this manner. We find two types of closest packing in simple metallic crystalline structures: CCP, which we have already encountered, and hexagonal closest packing (HCP) shown in Figure 10.54. Both consist of repeating layers of hexagonally arranged atoms. In both types, a second layer (B) is placed on the first layer (A) so that each atom in the second layer is in contact with three atoms in the first layer. The third layer is positioned in one of two ways. In HCP, atoms in the third layer are directly above atoms in the first layer (i.e., the third layer is also type A), and the stacking consists of alternating type A and type B close-packed layers (i.e., ABABAB $\\cdot \\cdot$ ). In CCP, atoms in the third layer are not above atoms in either of the first two layers (i.e., the third layer is type C), and the stacking consists of alternating type $A$, type $B$, and type $C$ close-packed layers (i.e., ABCABCABC\u2026). About two-thirds of all metals crystallize in closest-packed arrays with coordination numbers of 12. Metals that crystallize in an HCP structure include $\\mathrm{Cd}, \\mathrm{Co}, \\mathrm{Li}, \\mathrm{Mg}, \\mathrm{Na}$, and Zn , and metals that crystallize in a CCP structure include $\\mathrm{Ag}, \\mathrm{Al}, \\mathrm{Ca}$, $\\mathrm{Cu}, \\mathrm{Ni}, \\mathrm{Pb}$, and Pt .\n\n\nFIGURE 10.54 In both types of closest packing, atoms are packed as compactly as possible. Hexagonal closest packing consists of two alternating layers (ABABAB...). Cubic closest packing consists of three alternating layers (ABCABCABC...)."}
{"id": 3631, "contents": "1153. Calculation of Atomic Radius and Density for Metals, Part 2 - \nCalcium crystallizes in a face-centered cubic structure. The edge length of its unit cell is 558.8 pm .\n(a) What is the atomic radius of Ca in this structure?\n(b) Calculate the density of Ca."}
{"id": 3632, "contents": "1154. Solution - \n(a) In an FCC structure, Ca atoms contact each other across the diagonal of the face, so the length of the diagonal is equal to four Ca atomic radii $(\\mathrm{d}=4 r)$. Two adjacent edges and the diagonal of the face form a right triangle, with the length of each side equal to 558.8 pm and the length of the hypotenuse equal to four Ca atomic radii:\n\n$$\na^{2}+a^{2}=d^{2} \\longrightarrow(558.8 \\mathrm{pm})^{2}+(558.5 \\mathrm{pm})^{2}=(4 r)^{2}\n$$\n\nSolving this gives $r=\\sqrt{\\frac{(558.8 \\mathrm{pm})^{2}+(558.5 \\mathrm{pm})^{2}}{16}}=197.6 \\mathrm{pm}$ for a Ca radius.\n(b) Density is given by density $=\\frac{\\text { mass }}{\\text { volume }}$. The density of calcium can be found by determining the density of its unit cell: for example, the mass contained within a unit cell divided by the volume of the unit cell. A facecentered Ca unit cell has one-eighth of an atom at each of the eight corners ( $8 \\times \\frac{1}{8}=1$ atom $)$ and one-half of an atom on each of the six faces $6 \\times \\frac{1}{2}=3$ atoms), for a total of four atoms in the unit cell.\n\nThe mass of the unit cell can be found by:\n\n$$\n1 \\mathrm{Ca} \\text { unit cell } \\times \\frac{4 \\mathrm{Ca} \\text { atoms }}{1 \\mathrm{Ca} \\text { unit cell }} \\times \\frac{1 \\mathrm{~mol} \\mathrm{Ca}}{6.022 \\times 10^{23} \\mathrm{Ca} \\text { atoms }} \\times \\frac{40.078 \\mathrm{~g}}{1 \\mathrm{~mol} \\mathrm{Ca}}=2.662 \\times 10^{-22} \\mathrm{~g}\n$$\n\nThe volume of a Ca unit cell can be found by:"}
{"id": 3633, "contents": "1154. Solution - \nThe volume of a Ca unit cell can be found by:\n\n$$\nV=a^{3}=\\left(558.8 \\times 10^{-10} \\mathrm{~cm}\\right)^{3}=1.745 \\times 10^{-22} \\mathrm{~cm}^{3}\n$$\n\n(Note that the edge length was converted from pm to cm to get the usual volume units for density.)\nThen, the density of $\\mathrm{Ca}=\\frac{2.662 \\times 10^{-22} \\mathrm{~g}}{1.745 \\times 10^{-22} \\mathrm{~cm}^{3}}=1.53 \\mathrm{~g} / \\mathrm{cm}^{3}$"}
{"id": 3634, "contents": "1155. Check Your Learning - \nSilver crystallizes in an FCC structure. The edge length of its unit cell is 409 pm .\n(a) What is the atomic radius of Ag in this structure?\n(b) Calculate the density of Ag."}
{"id": 3635, "contents": "1156. Answer: - \n(a) 144 pm ; (b) $10.5 \\mathrm{~g} / \\mathrm{cm}^{3}$\n\nIn general, a unit cell is defined by the lengths of three axes ( $a, b$, and $c$ ) and the angles ( $\\alpha, \\beta$, and $\\gamma$ ) between them, as illustrated in Figure 10.55. The axes are defined as being the lengths between points in the space lattice. Consequently, unit cell axes join points with identical environments.\n\n\nFIGURE 10.55 A unit cell is defined by the lengths of its three axes ( $a, b$, and $c$ ) and the angles ( $\\alpha, \\beta$, and $\\gamma$ ) between the axes.\n\nThere are seven different lattice systems, some of which have more than one type of lattice, for a total of fourteen different unit cells, which have the shapes shown in Figure 10.56."}
{"id": 3636, "contents": "1156. Answer: - \nThere are seven different lattice systems, some of which have more than one type of lattice, for a total of fourteen different unit cells, which have the shapes shown in Figure 10.56.\n\n| System/Axes/Angles | Unit Cells |\n| :---: | :---: |\n| Cubic $\\begin{gathered} a=b=c \\\\ \\alpha=\\beta=\\gamma=90^{\\circ} \\end{gathered}$ | Simple <br> Face-centered <br> Body-centered |\n| Tetragonal $\\begin{gathered} a=b \\neq c \\\\ \\alpha=\\beta=\\gamma=90^{\\circ} \\end{gathered}$ | Simple <br> Body-centered |\n| Orthorhombic $\\begin{gathered} a \\neq b \\neq c \\\\ \\alpha=\\beta=\\gamma=90^{\\circ} \\end{gathered}$ | Simple <br> Body-centered <br> Base-centered <br> Face-centered |\n| Monoclinic $\\begin{gathered} a \\neq b \\neq c \\\\ \\alpha=\\gamma=90^{\\circ} ; \\beta \\neq 90^{\\circ} \\end{gathered}$ | Simple <br> Base-centered |\n| Triclinic $\\begin{gathered} a \\neq b \\neq c \\\\ \\alpha \\neq \\beta \\neq \\gamma \\neq 90^{\\circ} \\end{gathered}$ |  |\n| Hexagonal $\\begin{gathered} a=b \\neq c \\\\ \\alpha=\\beta=90^{\\circ} ; \\gamma=120^{\\circ} \\end{gathered}$ |  |\n| Rhombohedral $\\begin{gathered} a=b=c \\\\ \\alpha=\\beta=\\gamma \\neq 90^{\\circ} \\end{gathered}$ |  |\n\nFIGURE 10.56 There are seven different lattice systems and 14 different unit cells."}
{"id": 3637, "contents": "1157. The Structures of Ionic Crystals - \nIonic crystals consist of two or more different kinds of ions that usually have different sizes. The packing of these ions into a crystal structure is more complex than the packing of metal atoms that are the same size.\n\nMost monatomic ions behave as charged spheres, and their attraction for ions of opposite charge is the same in every direction. Consequently, stable structures for ionic compounds result (1) when ions of one charge are\nsurrounded by as many ions as possible of the opposite charge and (2) when the cations and anions are in contact with each other. Structures are determined by two principal factors: the relative sizes of the ions and the ratio of the numbers of positive and negative ions in the compound.\n\nIn simple ionic structures, we usually find the anions, which are normally larger than the cations, arranged in a closest-packed array. (As seen previously, additional electrons attracted to the same nucleus make anions larger and fewer electrons attracted to the same nucleus make cations smaller when compared to the atoms from which they are formed.) The smaller cations commonly occupy one of two types of holes (or interstices) remaining between the anions. The smaller of the holes is found between three anions in one plane and one anion in an adjacent plane. The four anions surrounding this hole are arranged at the corners of a tetrahedron, so the hole is called a tetrahedral hole. The larger type of hole is found at the center of six anions (three in one layer and three in an adjacent layer) located at the corners of an octahedron; this is called an octahedral hole. Figure 10.57 illustrates both of these types of holes.\n\n\nFIGURE 10.57 Cations may occupy two types of holes between anions: octahedral holes or tetrahedral holes.\nDepending on the relative sizes of the cations and anions, the cations of an ionic compound may occupy tetrahedral or octahedral holes, as illustrated in Figure 10.58. Relatively small cations occupy tetrahedral holes, and larger cations occupy octahedral holes. If the cations are too large to fit into the octahedral holes, the anions may adopt a more open structure, such as a simple cubic array. The larger cations can then occupy the larger cubic holes made possible by the more open spacing."}
{"id": 3638, "contents": "1157. The Structures of Ionic Crystals - \nFIGURE 10.58 A cation's size and the shape of the hole occupied by the compound are directly related.\nThere are two tetrahedral holes for each anion in either an HCP or CCP array of anions. A compound that crystallizes in a closest-packed array of anions with cations in the tetrahedral holes can have a maximum cation:anion ratio of $2: 1$; all of the tetrahedral holes are filled at this ratio. Examples include $\\mathrm{Li}_{2} \\mathrm{O}, \\mathrm{Na}_{2} \\mathrm{O}, \\mathrm{Li}_{2} \\mathrm{~S}$,\nand $\\mathrm{Na}_{2} \\mathrm{~S}$. Compounds with a ratio of less than $2: 1$ may also crystallize in a closest-packed array of anions with cations in the tetrahedral holes, if the ionic sizes fit. In these compounds, however, some of the tetrahedral holes remain vacant."}
{"id": 3639, "contents": "1159. Occupancy of Tetrahedral Holes - \nZinc sulfide is an important industrial source of zinc and is also used as a white pigment in paint. Zinc sulfide crystallizes with zinc ions occupying one-half of the tetrahedral holes in a closest-packed array of sulfide ions. What is the formula of zinc sulfide?"}
{"id": 3640, "contents": "1160. Solution - \nBecause there are two tetrahedral holes per anion (sulfide ion) and one-half of these holes are occupied by zinc ions, there must be $\\frac{1}{2} \\times 2$, or 1 , zinc ion per sulfide ion. Thus, the formula is ZnS ."}
{"id": 3641, "contents": "1161. Check Your Learning - \nLithium selenide can be described as a closest-packed array of selenide ions with lithium ions in all of the tetrahedral holes. What it the formula of lithium selenide?"}
{"id": 3642, "contents": "1162. Answer: - \n$\\mathrm{Li}_{2} \\mathrm{Se}$\n\nThe ratio of octahedral holes to anions in either an HCP or CCP structure is 1:1. Thus, compounds with cations in octahedral holes in a closest-packed array of anions can have a maximum cation:anion ratio of 1:1. In NiO , $\\mathrm{MnS}, \\mathrm{NaCl}$, and KH, for example, all of the octahedral holes are filled. Ratios of less than 1:1 are observed when some of the octahedral holes remain empty."}
{"id": 3643, "contents": "1164. Stoichiometry of Ionic Compounds - \nSapphire is aluminum oxide. Aluminum oxide crystallizes with aluminum ions in two-thirds of the octahedral holes in a closest-packed array of oxide ions. What is the formula of aluminum oxide?"}
{"id": 3644, "contents": "1165. Solution - \nBecause there is one octahedral hole per anion (oxide ion) and only two-thirds of these holes are occupied, the ratio of aluminum to oxygen must be $\\frac{2}{3}: 1$, which would give $\\mathrm{Al}_{2 / 3} \\mathrm{O}$. The simplest whole number ratio is $2: 3$, so the formula is $\\mathrm{Al}_{2} \\mathrm{O}_{3}$."}
{"id": 3645, "contents": "1166. Check Your Learning - \nThe white pigment titanium oxide crystallizes with titanium ions in one-half of the octahedral holes in a closest-packed array of oxide ions. What is the formula of titanium oxide?"}
{"id": 3646, "contents": "1167. Answer: - \n$\\mathrm{TiO}_{2}$\n\nIn a simple cubic array of anions, there is one cubic hole that can be occupied by a cation for each anion in the array. In CsCl , and in other compounds with the same structure, all of the cubic holes are occupied. Half of the cubic holes are occupied in $\\mathrm{SrH}_{2}, \\mathrm{UO}_{2}, \\mathrm{SrCl}_{2}$, and $\\mathrm{CaF}_{2}$.\n\nDifferent types of ionic compounds often crystallize in the same structure when the relative sizes of their ions and their stoichiometries (the two principal features that determine structure) are similar."}
{"id": 3647, "contents": "1168. Unit Cells of Ionic Compounds - \nMany ionic compounds crystallize with cubic unit cells, and we will use these compounds to describe the general features of ionic structures.\n\nWhen an ionic compound is composed of cations and anions of similar size in a 1:1 ratio, it typically forms a simple cubic structure. Cesium chloride, CsCl , (illustrated in Figure 10.59) is an example of this, with $\\mathrm{Cs}^{+}$and $\\mathrm{Cl}^{-}$having radii of 174 pm and 181 pm , respectively. We can think of this as chloride ions forming a simple cubic unit cell, with a cesium ion in the center; or as cesium ions forming a unit cell with a chloride ion in the center; or as simple cubic unit cells formed by $\\mathrm{Cs}^{+}$ions overlapping unit cells formed by $\\mathrm{Cl}^{-}$ions. Cesium ions and chloride ions touch along the body diagonals of the unit cells. One cesium ion and one chloride ion are present per unit cell, giving the l:l stoichiometry required by the formula for cesium chloride. Note that there is no lattice point in the center of the cell, and CsCl is not a BCC structure because a cesium ion is not identical to a chloride ion.\n\n\nSimple cubic structure\nFIGURE 10.59 Ionic compounds with similar-sized cations and anions, such as CsCl , usually form a simple cubic structure. They can be described by unit cells with either cations at the corners or anions at the corners.\nWe have said that the location of lattice points is arbitrary. This is illustrated by an alternate description of the CsCl structure in which the lattice points are located in the centers of the cesium ions. In this description, the cesium ions are located on the lattice points at the corners of the cell, and the chloride ion is located at the center of the cell. The two unit cells are different, but they describe identical structures."}
{"id": 3648, "contents": "1168. Unit Cells of Ionic Compounds - \nWhen an ionic compound is composed of a 1:1 ratio of cations and anions that differ significantly in size, it typically crystallizes with an FCC unit cell, like that shown in Figure 10.60. Sodium chloride, NaCl , is an example of this, with $\\mathrm{Na}^{+}$and $\\mathrm{Cl}^{-}$having radii of 102 pm and 181 pm , respectively. We can think of this as chloride ions forming an FCC cell, with sodium ions located in the octahedral holes in the middle of the cell edges and in the center of the cell. The sodium and chloride ions touch each other along the cell edges. The unit cell contains four sodium ions and four chloride ions, giving the 1:1 stoichiometry required by the formula, NaCl .\n\n\nFace-centered simple cubic structure\nFIGURE 10.60 Ionic compounds with anions that are much larger than cations, such as NaCl , usually form an FCC structure. They can be described by FCC unit cells with cations in the octahedral holes.\n\nThe cubic form of zinc sulfide, zinc blende, also crystallizes in an FCC unit cell, as illustrated in Figure 10.61. This structure contains sulfide ions on the lattice points of an FCC lattice. (The arrangement of sulfide ions is identical to the arrangement of chloride ions in sodium chloride.) The radius of a zinc ion is only about $40 \\%$ of the radius of a sulfide ion, so these small $\\mathrm{Zn}^{2+}$ ions are located in alternating tetrahedral holes, that is, in one half of the tetrahedral holes. There are four zinc ions and four sulfide ions in the unit cell, giving the empirical formula ZnS .\n\n\nZnS face-centered unit cell\nFIGURE 10.61 ZnS , zinc sulfide (or zinc blende) forms an FCC unit cell with sulfide ions at the lattice points and much smaller zinc ions occupying half of the tetrahedral holes in the structure."}
{"id": 3649, "contents": "1168. Unit Cells of Ionic Compounds - \nA calcium fluoride unit cell, like that shown in Figure 10.62, is also an FCC unit cell, but in this case, the cations are located on the lattice points; equivalent calcium ions are located on the lattice points of an FCC lattice. All of the tetrahedral sites in the FCC array of calcium ions are occupied by fluoride ions. There are four calcium ions and eight fluoride ions in a unit cell, giving a calcium:fluorine ratio of 1:2, as required by the chemical formula, $\\mathrm{CaF}_{2}$. Close examination of Figure 10.62 will reveal a simple cubic array of fluoride ions with calcium ions in one half of the cubic holes. The structure cannot be described in terms of a space lattice of points on the fluoride ions because the fluoride ions do not all have identical environments. The orientation of the four calcium ions about the fluoride ions differs.\n\n$\\mathrm{CaF}_{2}$ face-centered unit cell\nFIGURE 10.62 Calcium fluoride, $\\mathrm{CaF}_{2}$, forms an FCC unit cell with calcium ions (green) at the lattice points and fluoride ions (red) occupying all of the tetrahedral sites between them."}
{"id": 3650, "contents": "1169. Calculation of Ionic Radii - \nIf we know the edge length of a unit cell of an ionic compound and the position of the ions in the cell, we can calculate ionic radii for the ions in the compound if we make assumptions about individual ionic shapes and contacts."}
{"id": 3651, "contents": "1171. Calculation of Ionic Radii - \nThe edge length of the unit cell of LiCl ( NaCl -like structure, FCC ) is 0.514 nm or $5.14 \\AA$. Assuming that the lithium ion is small enough so that the chloride ions are in contact, as in Figure 10.60, calculate the ionic radius for the chloride ion.\n\nNote: The length unit angstrom, $\\AA$, is often used to represent atomic-scale dimensions and is equivalent to $10^{-10} \\mathrm{~m}$."}
{"id": 3652, "contents": "1172. Solution - \nOn the face of a LiCl unit cell, chloride ions contact each other across the diagonal of the face:\n\n\nDrawing a right triangle on the face of the unit cell, we see that the length of the diagonal is equal to four chloride radii (one radius from each corner chloride and one diameter-which equals two radii-from the chloride ion in the center of the face), so $d=4 r$. From the Pythagorean theorem, we have:\n\n$$\na^{2}+a^{2}=d^{2}\n$$\n\nwhich yields:\n\n$$\n(0.514 \\mathrm{~nm})^{2}+(0.514 \\mathrm{~nm})^{2}=(4 r)^{2}=16 r^{2}\n$$\n\nSolving this gives:\n\n$$\nr=\\sqrt{\\frac{(0.514 \\mathrm{~nm})^{2}+(0.514 \\mathrm{~nm})^{2}}{16}}=0.182 \\mathrm{~nm}(1.82 \\AA) \\text { for a } \\mathrm{Cl}^{-} \\text {radius. }\n$$"}
{"id": 3653, "contents": "1173. Check Your Learning - \nThe edge length of the unit cell of KCl ( NaCl -like structure, FCC ) is 6.28 \u00c5. Assuming anion-cation contact along the cell edge, calculate the radius of the potassium ion. The radius of the chloride ion is $1.82 \\AA$."}
{"id": 3654, "contents": "1174. Answer: - \nThe radius of the potassium ion is $1.33 \\AA$.\n\nIt is important to realize that values for ionic radii calculated from the edge lengths of unit cells depend on numerous assumptions, such as a perfect spherical shape for ions, which are approximations at best. Hence, such calculated values are themselves approximate and comparisons cannot be pushed too far. Nevertheless, this method has proved useful for calculating ionic radii from experimental measurements such as X-ray crystallographic determinations."}
{"id": 3655, "contents": "1175. X-Ray Crystallography - \nThe size of the unit cell and the arrangement of atoms in a crystal may be determined from measurements of the diffraction of X-rays by the crystal, termed X-ray crystallography. Diffraction is the change in the direction of travel experienced by an electromagnetic wave when it encounters a physical barrier whose dimensions are comparable to those of the wavelength of the light. X-rays are electromagnetic radiation with wavelengths about as long as the distance between neighboring atoms in crystals (on the order of a few $\\AA$ ).\n\nWhen a beam of monochromatic X-rays strikes a crystal, its rays are scattered in all directions by the atoms within the crystal. When scattered waves traveling in the same direction encounter one another, they undergo interference, a process by which the waves combine to yield either an increase or a decrease in amplitude (intensity) depending upon the extent to which the combining waves' maxima are separated (see Figure 10.63).\n\n\nConstructive interface\n(a)\n\n\nDestructive interface\n(b)\n\nFIGURE 10.63 Light waves occupying the same space experience interference, combining to yield waves of greater (a) or lesser (b) intensity, depending upon the separation of their maxima and minima.\n\nWhen X-rays of a certain wavelength, $\\lambda$, are scattered by atoms in adjacent crystal planes separated by a distance, $d$, they may undergo constructive interference when the difference between the distances traveled by the two waves prior to their combination is an integer factor, $n$, of the wavelength. This condition is satisfied when the angle of the diffracted beam, $\\theta$, is related to the wavelength and interatomic distance by the equation:\n\n$$\nn \\lambda=2 d \\sin \\theta\n$$\n\nThis relation is known as the Bragg equation in honor of W. H. Bragg, the English physicist who first explained this phenomenon. Figure 10.64 illustrates two examples of diffracted waves from the same two crystal planes. The figure on the left depicts waves diffracted at the Bragg angle, resulting in constructive interference, while that on the right shows diffraction and a different angle that does not satisfy the Bragg condition, resulting in destructive interference."}
{"id": 3656, "contents": "1175. X-Ray Crystallography - \nFIGURE 10.64 The diffraction of X-rays scattered by the atoms within a crystal permits the determination of the distance between the atoms. The top image depicts constructive interference between two scattered waves and a resultant diffracted wave of high intensity. The bottom image depicts destructive interference and a low intensity diffracted wave."}
{"id": 3657, "contents": "1176. LINK TO LEARNING - \nVisit this site (http://openstax.org/l/16bragg) for more details on the Bragg equation and a simulator that allows you to explore the effect of each variable on the intensity of the diffracted wave.\n\nAn X-ray diffractometer, such as the one illustrated in Figure 10.65, may be used to measure the angles at which X-rays are diffracted when interacting with a crystal as described earlier. From such measurements, the Bragg equation may be used to compute distances between atoms as demonstrated in the following example exercise.\n\n\nFIGURE 10.65 (a) In a diffractometer, a beam of X-rays strikes a crystalline material, producing (b) an X-ray diffraction pattern that can be analyzed to determine the crystal structure."}
{"id": 3658, "contents": "1178. Using the Bragg Equation - \nIn a diffractometer, X-rays with a wavelength of 0.1315 nm were used to produce a diffraction pattern for copper. The first order diffraction $(n=1)$ occurred at an angle $\\theta=25.25^{\\circ}$. Determine the spacing between the diffracting planes in copper."}
{"id": 3659, "contents": "1179. Solution - \nThe distance between the planes is found by solving the Bragg equation, $n \\lambda=2 d \\sin \\theta$, for $d$.\nThis gives: $d=\\frac{n \\lambda}{2 \\sin \\theta}=\\frac{1(0.1315 \\mathrm{~nm})}{2 \\sin \\left(25.25^{\\circ}\\right)}=0.154 \\mathrm{~nm}$"}
{"id": 3660, "contents": "1180. Check Your Learning - \nA crystal with spacing between planes equal to 0.394 nm diffracts X-rays with a wavelength of 0.147 nm . What is the angle for the first order diffraction?"}
{"id": 3661, "contents": "1181. Answer: - \n$10.8^{\\circ}$"}
{"id": 3662, "contents": "1183. X-ray Crystallographer Rosalind Franklin - \nThe discovery of the structure of DNA in 1953 by Francis Crick and James Watson is one of the great achievements in the history of science. They were awarded the 1962 Nobel Prize in Physiology or Medicine, along with Maurice Wilkins, who provided experimental proof of DNA's structure. British chemist Rosalind Franklin made invaluable contributions to this monumental achievement through her work in measuring X-ray diffraction images of DNA. Early in her career, Franklin's research on the structure of coals proved helpful to the British war effort. After shifting her focus to biological systems in the early 1950s, Franklin and doctoral student Raymond Gosling discovered that DNA consists of two forms: a long, thin fiber formed when wet (type \"B\") and a short, wide fiber formed when dried (type \"A\"). Her X-ray diffraction images of DNA (Figure 10.66) provided the crucial information that allowed Watson and Crick to confirm that DNA forms a double helix, and to determine details of its size and structure. Franklin also conducted pioneering research on viruses and the RNA that contains their genetic information, uncovering new information that radically changed the body of knowledge in the field. After developing ovarian cancer, Franklin continued\nto work until her death in 1958 at age 37. Among many posthumous recognitions of her work, the Chicago Medical School of Finch University of Health Sciences changed its name to the Rosalind Franklin University of Medicine and Science in 2004, and adopted an image of her famous X-ray diffraction image of DNA as its official university logo.\n\n\nFIGURE 10.66 This illustration shows an X-ray diffraction image similar to the one Franklin found in her research. (credit: National Institutes of Health)"}
{"id": 3663, "contents": "1184. Key Terms - \nadhesive force force of attraction between molecules of different chemical identities\namorphous solid (also, noncrystalline solid) solid in which the particles lack an ordered internal structure\nbody-centered cubic (BCC) solid crystalline structure that has a cubic unit cell with lattice points at the corners and in the center of the cell\nbody-centered cubic unit cell simplest repeating unit of a body-centered cubic crystal; it is a cube containing lattice points at each corner and in the center of the cube\nboiling point temperature at which the vapor pressure of a liquid equals the pressure of the gas above it\nBragg equation equation that relates the angles at which X-rays are diffracted by the atoms within a crystal\ncapillary action flow of liquid within a porous material due to the attraction of the liquid molecules to the surface of the material and to other liquid molecules\nClausius-Clapeyron equation mathematical relationship between the temperature, vapor pressure, and enthalpy of vaporization for a substance\ncohesive force force of attraction between identical molecules\ncondensation change from a gaseous to a liquid state\ncoordination number number of atoms closest to any given atom in a crystal or to the central metal atom in a complex\ncovalent network solid solid whose particles are held together by covalent bonds\ncritical point temperature and pressure above which a gas cannot be condensed into a liquid\ncrystalline solid solid in which the particles are arranged in a definite repeating pattern\ncubic closest packing (CCP) crystalline structure in which planes of closely packed atoms or ions are stacked as a series of three alternating layers of different relative orientations ( ABC )\ndeposition change from a gaseous state directly to a solid state\ndiffraction redirection of electromagnetic radiation that occurs when it encounters a physical barrier of appropriate dimensions\ndipole-dipole attraction intermolecular attraction between two permanent dipoles\ndispersion force (also, London dispersion force) attraction between two rapidly fluctuating,\ntemporary dipoles; significant only when particles are very close together\ndynamic equilibrium state of a system in which reciprocal processes are occurring at equal rates\nface-centered cubic (FCC) solid crystalline structure consisting of a cubic unit cell with lattice points on the corners and in the center of each face"}
{"id": 3664, "contents": "1184. Key Terms - \ndynamic equilibrium state of a system in which reciprocal processes are occurring at equal rates\nface-centered cubic (FCC) solid crystalline structure consisting of a cubic unit cell with lattice points on the corners and in the center of each face\nface-centered cubic unit cell simplest repeating unit of a face-centered cubic crystal; it is a cube containing lattice points at each corner and in the center of each face\nfreezing change from a liquid state to a solid state\nfreezing point temperature at which the solid and liquid phases of a substance are in equilibrium; see also melting point\nhexagonal closest packing (HCP) crystalline structure in which close packed layers of atoms or ions are stacked as a series of two alternating layers of different relative orientations (AB)\nhole (also, interstice) space between atoms within a crystal\nhydrogen bonding occurs when exceptionally strong dipoles attract; bonding that exists when hydrogen is bonded to one of the three most electronegative elements: F, O, or N\ninduced dipole temporary dipole formed when the electrons of an atom or molecule are distorted by the instantaneous dipole of a neighboring atom or molecule\ninstantaneous dipole temporary dipole that occurs for a brief moment in time when the electrons of an atom or molecule are distributed asymmetrically\nintermolecular force noncovalent attractive force between atoms, molecules, and/or ions\ninterstitial sites spaces between the regular particle positions in any array of atoms or ions\nionic solid solid composed of positive and negative ions held together by strong electrostatic attractions\nisomorphous possessing the same crystalline structure\nmelting change from a solid state to a liquid state\nmelting point temperature at which the solid and liquid phases of a substance are in equilibrium; see also freezing point\nmetallic solid solid composed of metal atoms\nmolecular solid solid composed of neutral molecules held together by intermolecular forces of attraction\nnormal boiling point temperature at which a\nliquid's vapor pressure equals 1 atm ( 760 torr)\noctahedral hole open space in a crystal at the center of six particles located at the corners of an octahedron\nphase diagram pressure-temperature graph summarizing conditions under which the phases of a substance can exist\npolarizability measure of the ability of a charge to distort a molecule's charge distribution (electron cloud)\nsimple cubic structure crystalline structure with a cubic unit cell with lattice points only at the corners"}
{"id": 3665, "contents": "1184. Key Terms - \npolarizability measure of the ability of a charge to distort a molecule's charge distribution (electron cloud)\nsimple cubic structure crystalline structure with a cubic unit cell with lattice points only at the corners\nsimple cubic unit cell (also, primitive cubic unit cell) unit cell in the simple cubic structure\nspace lattice all points within a crystal that have identical environments\nsublimation change from solid state directly to gaseous state\nsupercritical fluid substance at a temperature and pressure higher than its critical point; exhibits properties intermediate between those of gaseous and liquid states\nsurface tension energy required to increase the area, or length, of a liquid surface by a given amount"}
{"id": 3666, "contents": "1185. Key Equations - \n$h=\\frac{2 T \\cos \\theta}{r \\rho g}$\n$P=A e^{-\\Delta H_{\\mathrm{vap}} / R T}$\n$\\ln P=-\\frac{\\Delta H_{\\mathrm{vap}}}{R T}+\\ln A$\n$\\ln \\left(\\frac{P_{2}}{P_{1}}\\right)=\\frac{\\Delta H_{\\mathrm{vap}}}{R}\\left(\\frac{1}{T_{1}}-\\frac{1}{T_{2}}\\right)$\n$n \\lambda=2 d \\sin \\theta$\n\nSummary"}
{"id": 3667, "contents": "1185. Key Equations - 1185.1. Intermolecular Forces\nThe physical properties of condensed matter (liquids and solids) can be explained in terms of the kinetic molecular theory. In a liquid, intermolecular attractive forces hold the molecules in contact, although they still have sufficient KE to move past each other.\n\nIntermolecular attractive forces, collectively referred to as van der Waals forces, are responsible for the behavior of liquids and solids and are electrostatic in nature. Dipole-dipole attractions result from the electrostatic attraction of the partial negative end of one polar molecule for the partial positive end of another. The temporary dipole that results from the\ntetrahedral hole tetrahedral space formed by four atoms or ions in a crystal\ntriple point temperature and pressure at which the vapor, liquid, and solid phases of a substance are in equilibrium\nunit cell smallest portion of a space lattice that is repeated in three dimensions to form the entire lattice\nvacancy defect that occurs when a position that should contain an atom or ion is vacant\nvan der Waals force attractive or repulsive force between molecules, including dipole-dipole, dipole-induced dipole, and London dispersion forces; does not include forces due to covalent or ionic bonding, or the attraction between ions and molecules\nvapor pressure (also, equilibrium vapor pressure) pressure exerted by a vapor in equilibrium with a solid or a liquid at a given temperature\nvaporization change from liquid state to gaseous state\nviscosity measure of a liquid's resistance to flow\nX-ray crystallography experimental technique for determining distances between atoms in a crystal by measuring the angles at which X-rays are diffracted when passing through the crystal\nmotion of the electrons in an atom can induce a dipole in an adjacent atom and give rise to the London dispersion force. London forces increase with increasing molecular size. Hydrogen bonds are a special type of dipole-dipole attraction that results when hydrogen is bonded to one of the three most electronegative elements: $\\mathrm{F}, \\mathrm{O}$, or N ."}
{"id": 3668, "contents": "1185. Key Equations - 1185.2. Properties of Liquids\nThe intermolecular forces between molecules in the liquid state vary depending upon their chemical identities and result in corresponding variations in various physical properties. Cohesive forces between like molecules are responsible for a liquid's viscosity (resistance to flow) and surface tension\n(elasticity of a liquid surface). Adhesive forces between the molecules of a liquid and different molecules composing a surface in contact with the liquid are responsible for phenomena such as surface wetting and capillary rise."}
{"id": 3669, "contents": "1185. Key Equations - 1185.3. Phase Transitions\nPhase transitions are processes that convert matter from one physical state into another. There are six phase transitions between the three phases of matter. Melting, vaporization, and sublimation are all endothermic processes, requiring an input of heat to overcome intermolecular attractions. The reciprocal transitions of freezing, condensation, and deposition are all exothermic processes, involving heat as intermolecular attractive forces are established or strengthened. The temperatures at which phase transitions occur are determined by the relative strengths of intermolecular attractions and are, therefore, dependent on the chemical identity of the substance."}
{"id": 3670, "contents": "1185. Key Equations - 1185.4. Phase Diagrams\nThe temperature and pressure conditions at which a substance exists in solid, liquid, and gaseous states are summarized in a phase diagram for that substance. Phase diagrams are combined plots of three pressure-temperature equilibrium curves: solid-liquid, liquid-gas, and solid-gas. These curves represent the relationships between phasetransition temperatures and pressures. The point of intersection of all three curves represents the substance's triple point-the temperature and pressure at which all three phases are in equilibrium. At pressures below the triple point, a substance cannot exist in the liquid state, regardless of its temperature. The terminus of the liquid-gas curve represents the substance's critical point, the pressure and temperature above which a liquid phase cannot exist."}
{"id": 3671, "contents": "1185. Key Equations - 1185.5. The Solid State of Matter\nSome substances form crystalline solids consisting\nof particles in a very organized structure; others form amorphous (noncrystalline) solids with an internal structure that is not ordered. The main types of crystalline solids are ionic solids, metallic solids, covalent network solids, and molecular solids. The properties of the different kinds of crystalline solids are due to the types of particles of which they consist, the arrangements of the particles, and the strengths of the attractions between them. Because their particles experience identical attractions, crystalline solids have distinct melting temperatures; the particles in amorphous solids experience a range of interactions, so they soften gradually and melt over a range of temperatures. Some crystalline solids have defects in the definite repeating pattern of their particles. These defects (which include vacancies, atoms or ions not in the regular positions, and impurities) change physical properties such as electrical conductivity, which is exploited in the silicon crystals used to manufacture computer chips."}
{"id": 3672, "contents": "1185. Key Equations - 1185.6. Lattice Structures in Crystalline Solids\nThe structures of crystalline metals and simple ionic compounds can be described in terms of packing of spheres. Metal atoms can pack in hexagonal closestpacked structures, cubic closest-packed structures, body-centered structures, and simple cubic structures. The anions in simple ionic structures commonly adopt one of these structures, and the cations occupy the spaces remaining between the anions. Small cations usually occupy tetrahedral holes in a closest-packed array of anions. Larger cations usually occupy octahedral holes. Still larger cations can occupy cubic holes in a simple cubic array of anions. The structure of a solid can be described by indicating the size and shape of a unit cell and the contents of the cell. The type of structure and dimensions of the unit cell can be determined by X-ray diffraction measurements."}
{"id": 3673, "contents": "1186. Exercises - 1186.1. Intermolecular Forces\n1. In terms of their bulk properties, how do liquids and solids differ? How are they similar?\n2. In terms of the kinetic molecular theory, in what ways are liquids similar to solids? In what ways are liquids different from solids?\n3. In terms of the kinetic molecular theory, in what ways are liquids similar to gases? In what ways are liquids different from gases?\n4. Explain why liquids assume the shape of any container into which they are poured, whereas solids are rigid and retain their shape.\n5. What is the evidence that all neutral atoms and molecules exert attractive forces on each other?\n6. Open the PhET States of Matter Simulation (http://openstax.org/l/16phetvisual) to answer the following questions:\n(a) Select the Solid, Liquid, Gas tab. Explore by selecting different substances, heating and cooling the systems, and changing the state. What similarities do you notice between the four substances for each phase (solid, liquid, gas)? What differences do you notice?\n(b) For each substance, select each of the states and record the given temperatures. How do the given temperatures for each state correlate with the strengths of their intermolecular attractions? Explain.\n(c) Select the Interaction Potential tab, and use the default neon atoms. Move the Ne atom on the right and observe how the potential energy changes. Select the Total Force button, and move the Ne atom as before. When is the total force on each atom attractive and large enough to matter? Then select the Component Forces button, and move the Ne atom. When do the attractive (van der Waals) and repulsive (electron overlap) forces balance? How does this relate to the potential energy versus the distance between atoms graph? Explain.\n7. Define the following and give an example of each:\n(a) dispersion force\n(b) dipole-dipole attraction\n(c) hydrogen bond\n8. The types of intermolecular forces in a substance are identical whether it is a solid, a liquid, or a gas. Why then does a substance change phase from a gas to a liquid or to a solid?\n9. Why do the boiling points of the noble gases increase in the order $\\mathrm{He}<\\mathrm{Ne}<\\mathrm{Ar}<\\mathrm{Kr}<\\mathrm{Xe}$ ?\n10. Neon and HF have approximately the same molecular masses."}
{"id": 3674, "contents": "1186. Exercises - 1186.1. Intermolecular Forces\n10. Neon and HF have approximately the same molecular masses.\n(a) Explain why the boiling points of Neon and HF differ.\n(b) Compare the change in the boiling points of $\\mathrm{Ne}, \\mathrm{Ar}, \\mathrm{Kr}$, and Xe with the change of the boiling points of $\\mathrm{HF}, \\mathrm{HCl}, \\mathrm{HBr}$, and HI , and explain the difference between the changes with increasing atomic or molecular mass.\n11. Arrange each of the following sets of compounds in order of increasing boiling point temperature:\n(a) $\\mathrm{HCl}, \\mathrm{H}_{2} \\mathrm{O}, \\mathrm{SiH}_{4}$\n(b) $\\mathrm{F}_{2}, \\mathrm{Cl}_{2}, \\mathrm{Br}_{2}$\n(c) $\\mathrm{CH}_{4}, \\mathrm{C}_{2} \\mathrm{H}_{6}, \\mathrm{C}_{3} \\mathrm{H}_{8}$\n(d) $\\mathrm{O}_{2}, \\mathrm{NO}, \\mathrm{N}_{2}$\n12. The molecular mass of butanol, $\\mathrm{C}_{4} \\mathrm{H}_{9} \\mathrm{OH}$, is 74.14 ; that of ethylene glycol, $\\mathrm{CH}_{2}(\\mathrm{OH}) \\mathrm{CH}_{2} \\mathrm{OH}$, is 62.08 , yet their boiling points are $117.2^{\\circ} \\mathrm{C}$ and $174^{\\circ} \\mathrm{C}$, respectively. Explain the reason for the difference.\n13. On the basis of intermolecular attractions, explain the differences in the boiling points of $n$-butane $\\left(-1^{\\circ} \\mathrm{C}\\right)$ and chloroethane $\\left(12^{\\circ} \\mathrm{C}\\right)$, which have similar molar masses.\n14. On the basis of dipole moments and/or hydrogen bonding, explain in a qualitative way the differences in the boiling points of acetone $\\left(56.2^{\\circ} \\mathrm{C}\\right)$ and 1-propanol $\\left(97.4^{\\circ} \\mathrm{C}\\right)$, which have similar molar masses."}
{"id": 3675, "contents": "1186. Exercises - 1186.1. Intermolecular Forces\n15. The melting point of $\\mathrm{H}_{2} \\mathrm{O}(s)$ is $0^{\\circ} \\mathrm{C}$. Would you expect the melting point of $\\mathrm{H}_{2} \\mathrm{~S}(s)$ to be $-85^{\\circ} \\mathrm{C}, 0^{\\circ} \\mathrm{C}$, or 185 ${ }^{\\circ} \\mathrm{C}$ ? Explain your answer.\n16. Silane $\\left(\\mathrm{SiH}_{4}\\right)$, phosphine $\\left(\\mathrm{PH}_{3}\\right)$, and hydrogen sulfide $\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)$ melt at $-185^{\\circ} \\mathrm{C},-133^{\\circ} \\mathrm{C}$, and $-85^{\\circ} \\mathrm{C}$, respectively. What does this suggest about the polar character and intermolecular attractions of the three compounds?\n17. Explain why a hydrogen bond between two water molecules is weaker than a hydrogen bond between two hydrogen fluoride molecules.\n18. Under certain conditions, molecules of acetic acid, $\\mathrm{CH}_{3} \\mathrm{COOH}$, form \"dimers,\" pairs of acetic acid molecules held together by strong intermolecular attractions:"}
{"id": 3676, "contents": "1186. Exercises - 1186.1. Intermolecular Forces\nDraw a dimer of acetic acid, showing how two $\\mathrm{CH}_{3} \\mathrm{COOH}$ molecules are held together, and stating the type of IMF that is responsible.\n19. Proteins are chains of amino acids that can form in a variety of arrangements, one of which is a helix. What kind of IMF is responsible for holding the protein strand in this shape? On the protein image, show the locations of the IMFs that hold the protein together:\n\n20. The density of liquid $\\mathrm{NH}_{3}$ is $0.64 \\mathrm{~g} / \\mathrm{mL}$; the density of gaseous $\\mathrm{NH}_{3}$ at STP is $0.0007 \\mathrm{~g} / \\mathrm{mL}$. Explain the difference between the densities of these two phases.\n21. Identify the intermolecular forces present in the following solids:\n(a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$\n(b) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$\n(c) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{Cl}$"}
{"id": 3677, "contents": "1186. Exercises - 1186.2. Properties of Liquids\n22. The test tubes shown here contain equal amounts of the specified motor oils. Identical metal spheres were dropped at the same time into each of the tubes, and a brief moment later, the spheres had fallen to the heights indicated in the illustration.\nRank the motor oils in order of increasing viscosity, and explain your reasoning:\n\n23. Although steel is denser than water, a steel needle or paper clip placed carefully lengthwise on the surface of still water can be made to float. Explain at a molecular level how this is possible.\n\n\nFIGURE 10.67 (credit: Cory Zanker)\n24. The surface tension and viscosity values for diethyl ether, acetone, ethanol, and ethylene glycol are shown here.\n\n| Compound | Molecule | Surface Tension <br> $(\\mathbf{m N / m})$ | Viscosity <br> $(\\mathbf{m P a} \\mathbf{s})$ |\n| :---: | :---: | :---: | :---: |\n| diethyl ether <br> $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OC}_{2} \\mathrm{H}_{5}$ |  | 17 | 0.22 |\n| acetone <br> $\\mathrm{CH}_{3} \\mathrm{COCH}_{3}$ |  | 23 | 0.31 |\n| ethanol <br> $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$ |  | 22 | 1.07 |\n| ethylene glycol <br> $\\mathrm{CH}_{2}(\\mathrm{OH}) \\mathrm{CH}(\\mathrm{OH})$ |  | 48 | 16.1 |\n\n(a) Explain their differences in viscosity in terms of the size and shape of their molecules and their IMFs.\n(b) Explain their differences in surface tension in terms of the size and shape of their molecules and their"}
{"id": 3678, "contents": "1186. Exercises - 1186.2. Properties of Liquids\n(a) Explain their differences in viscosity in terms of the size and shape of their molecules and their IMFs.\n(b) Explain their differences in surface tension in terms of the size and shape of their molecules and their\n\nIMFs:\n25. You may have heard someone use the figure of speech \"slower than molasses in winter\" to describe a process that occurs slowly. Explain why this is an apt idiom, using concepts of molecular size and shape, molecular interactions, and the effect of changing temperature.\n26. It is often recommended that you let your car engine run idle to warm up before driving, especially on cold winter days. While the benefit of prolonged idling is dubious, it is certainly true that a warm engine is more fuel efficient than a cold one. Explain the reason for this.\n27. The surface tension and viscosity of water at several different temperatures are given in this table.\n\n| Water | Surface Tension (mN/m) | Viscosity (mPa s) |\n| :---: | :---: | :---: |\n| $0{ }^{\\circ} \\mathrm{C}$ | 75.6 | 1.79 |\n| $20^{\\circ} \\mathrm{C}$ | 72.8 | 1.00 |\n| $60^{\\circ} \\mathrm{C}$ | 66.2 | 0.47 |\n| $100^{\\circ} \\mathrm{C}$ | 58.9 | 0.28 |\n\n(a) As temperature increases, what happens to the surface tension of water? Explain why this occurs, in terms of molecular interactions and the effect of changing temperature.\n(b) As temperature increases, what happens to the viscosity of water? Explain why this occurs, in terms of molecular interactions and the effect of changing temperature.\n28. At $25^{\\circ} \\mathrm{C}$, how high will water rise in a glass capillary tube with an inner diameter of 0.63 mm ? Refer to Example 10.4 for the required information.\n29. Water rises in a glass capillary tube to a height of 17 cm . What is the diameter of the capillary tube?"}
{"id": 3679, "contents": "1186. Exercises - 1186.3. Phase Transitions\n30. Heat is added to boiling water. Explain why the temperature of the boiling water does not change. What does change?\n31. Heat is added to ice at $0^{\\circ} \\mathrm{C}$. Explain why the temperature of the ice does not change. What does change?\n32. What feature characterizes the dynamic equilibrium between a liquid and its vapor in a closed container?\n33. Identify two common observations indicating some liquids have sufficient vapor pressures to noticeably evaporate?\n34. Identify two common observations indicating some solids, such as dry ice and mothballs, have vapor pressures sufficient to sublime?\n35. What is the relationship between the intermolecular forces in a liquid and its vapor pressure?\n36. What is the relationship between the intermolecular forces in a solid and its melting temperature?\n37. Why does spilled gasoline evaporate more rapidly on a hot day than on a cold day?\n38. Carbon tetrachloride, $\\mathrm{CCl}_{4}$, was once used as a dry cleaning solvent, but is no longer used because it is carcinogenic. At $57.8^{\\circ} \\mathrm{C}$, the vapor pressure of $\\mathrm{CCl}_{4}$ is 54.0 kPa , and its enthalpy of vaporization is 33.05 $\\mathrm{kJ} / \\mathrm{mol}$. Use this information to estimate the normal boiling point for $\\mathrm{CCl}_{4}$.\n39. When is the boiling point of a liquid equal to its normal boiling point?\n40. How does the boiling of a liquid differ from its evaporation?\n41. Use the information in Figure 10.24 to estimate the boiling point of water in Denver when the atmospheric pressure is 83.3 kPa .\n42. A syringe at a temperature of $20^{\\circ} \\mathrm{C}$ is filled with liquid ether in such a way that there is no space for any vapor. If the temperature is kept constant and the plunger is withdrawn to create a volume that can be occupied by vapor, what would be the approximate pressure of the vapor produced?\n43. Explain the following observations:\n(a) It takes longer to cook an egg in Ft. Davis, Texas (altitude, 5000 feet above sea level) than it does in Boston (at sea level)."}
{"id": 3680, "contents": "1186. Exercises - 1186.3. Phase Transitions\n43. Explain the following observations:\n(a) It takes longer to cook an egg in Ft. Davis, Texas (altitude, 5000 feet above sea level) than it does in Boston (at sea level).\n(b) Perspiring is a mechanism for cooling the body.\n44. The enthalpy of vaporization of water is larger than its enthalpy of fusion. Explain why.\n45. Explain why the molar enthalpies of vaporization of the following substances increase in the order $\\mathrm{CH}_{4}<$ $\\mathrm{C}_{2} \\mathrm{H}_{6}<\\mathrm{C}_{3} \\mathrm{H}_{8}$, even though the type of IMF (dispersion) is the same.\n46. Explain why the enthalpies of vaporization of the following substances increase in the order $\\mathrm{CH}_{4}<\\mathrm{NH}_{3}<$ $\\mathrm{H}_{2} \\mathrm{O}$, even though all three substances have approximately the same molar mass.\n47. The enthalpy of vaporization of $\\mathrm{CO}_{2}(I)$ is $9.8 \\mathrm{~kJ} / \\mathrm{mol}$. Would you expect the enthalpy of vaporization of $\\mathrm{CS}_{2}(\\mathrm{I})$ to be $28 \\mathrm{~kJ} / \\mathrm{mol}, 9.8 \\mathrm{~kJ} / \\mathrm{mol}$, or $-8.4 \\mathrm{~kJ} / \\mathrm{mol}$ ? Discuss the plausibility of each of these answers.\n48. The hydrogen fluoride molecule, HF , is more polar than a water molecule, $\\mathrm{H}_{2} \\mathrm{O}$ (for example, has a greater dipole moment), yet the molar enthalpy of vaporization for liquid hydrogen fluoride is lesser than that for water. Explain.\n49. Ethyl chloride (boiling point, $13^{\\circ} \\mathrm{C}$ ) is used as a local anesthetic. When the liquid is sprayed on the skin, it cools the skin enough to freeze and numb it. Explain the cooling effect of liquid ethyl chloride.\n50. Which contains the compounds listed correctly in order of increasing boiling points?"}
{"id": 3681, "contents": "1186. Exercises - 1186.3. Phase Transitions\n50. Which contains the compounds listed correctly in order of increasing boiling points?\n(a) $\\mathrm{N}_{2}<\\mathrm{CS}_{2}<\\mathrm{H}_{2} \\mathrm{O}<\\mathrm{KCl}$\n(b) $\\mathrm{H}_{2} \\mathrm{O}<\\mathrm{N}_{2}<\\mathrm{CS}_{2}<\\mathrm{KCl}$\n(c) $\\mathrm{N}_{2}<\\mathrm{KCl}<\\mathrm{CS}_{2}<\\mathrm{H}_{2} \\mathrm{O}$\n(d) $\\mathrm{CS}_{2}<\\mathrm{N}_{2}<\\mathrm{KCl}<\\mathrm{H}_{2} \\mathrm{O}$\n(e) $\\mathrm{KCl}<\\mathrm{H}_{2} \\mathrm{O}<\\mathrm{CS}_{2}<\\mathrm{N}_{2}$\n51. How much heat is required to convert 422 g of liquid $\\mathrm{H}_{2} \\mathrm{O}$ at $23.5^{\\circ} \\mathrm{C}$ into steam at $150^{\\circ} \\mathrm{C}$ ?\n52. Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventually be converted into sweat and evaporate. If you drink a 20 -ounce bottle of water that had been in the refrigerator at $3.8^{\\circ} \\mathrm{C}$, how much heat is needed to convert all of that water into sweat and then to vapor? (Note: Your body temperature is $36.6^{\\circ} \\mathrm{C}$. For the purpose of solving this problem, assume that the thermal properties of sweat are the same as for water.)\n53. Titanium tetrachloride, $\\mathrm{TiCl}_{4}$, has a melting point of $-23.2^{\\circ} \\mathrm{C}$ and has a $\\Delta H_{\\text {fusion }}=9.37 \\mathrm{~kJ} / \\mathrm{mol}$.\n(a) How much energy is required to melt 263.1 g TiCl 4 ?"}
{"id": 3682, "contents": "1186. Exercises - 1186.3. Phase Transitions\n(a) How much energy is required to melt 263.1 g TiCl 4 ?\n(b) For $\\mathrm{TiCl}_{4}$, which will likely have the larger magnitude: $\\Delta H_{\\text {fusion }}$ or $\\Delta H_{\\text {vaporization }}$ ? Explain your reasoning."}
{"id": 3683, "contents": "1186. Exercises - 1186.4. Phase Diagrams\n54. From the phase diagram for water (Figure 10.31), determine the state of water at:\n(a) $35^{\\circ} \\mathrm{C}$ and 85 kPa\n(b) $-15^{\\circ} \\mathrm{C}$ and 40 kPa\n(c) $-15^{\\circ} \\mathrm{C}$ and 0.1 kPa\n(d) $75^{\\circ} \\mathrm{C}$ and 3 kPa\n(e) $40^{\\circ} \\mathrm{C}$ and 0.1 kPa\n(f) $60^{\\circ} \\mathrm{C}$ and 50 kPa\n55. What phase changes will take place when water is subjected to varying pressure at a constant temperature of $0.005^{\\circ} \\mathrm{C}$ ? At $40^{\\circ} \\mathrm{C}$ ? At $-40^{\\circ} \\mathrm{C}$ ?\n56. Pressure cookers allow food to cook faster because the higher pressure inside the pressure cooker increases the boiling temperature of water. A particular pressure cooker has a safety valve that is set to vent steam if the pressure exceeds 3.4 atm . What is the approximate maximum temperature that can be reached inside this pressure cooker? Explain your reasoning.\n57. From the phase diagram for carbon dioxide in Figure 10.34, determine the state of $\\mathrm{CO}_{2}$ at:\n(a) $20^{\\circ} \\mathrm{C}$ and 1000 kPa\n(b) $10^{\\circ} \\mathrm{C}$ and 2000 kPa\n(c) $10^{\\circ} \\mathrm{C}$ and 100 kPa\n(d) $-40^{\\circ} \\mathrm{C}$ and 500 kPa\n(e) $-80^{\\circ} \\mathrm{C}$ and 1500 kPa\n(f) $-80^{\\circ} \\mathrm{C}$ and 10 kPa"}
{"id": 3684, "contents": "1186. Exercises - 1186.4. Phase Diagrams\n(e) $-80^{\\circ} \\mathrm{C}$ and 1500 kPa\n(f) $-80^{\\circ} \\mathrm{C}$ and 10 kPa\n58. Determine the phase changes that carbon dioxide undergoes as pressure is increased at a constant temperature of (a) $-50^{\\circ} \\mathrm{C}$ and (b) $50^{\\circ} \\mathrm{C}$. If the temperature is held at $-40^{\\circ} \\mathrm{C}$ ? At $20^{\\circ} \\mathrm{C}$ ? (See the phase diagram in Figure 10.34.)\n59. Consider a cylinder containing a mixture of liquid carbon dioxide in equilibrium with gaseous carbon dioxide at an initial pressure of 65 atm and a temperature of $20^{\\circ} \\mathrm{C}$. Sketch a plot depicting the change in the cylinder pressure with time as gaseous carbon dioxide is released at constant temperature.\n60. Dry ice, $\\mathrm{CO}_{2}(s)$, does not melt at atmospheric pressure. It sublimes at a temperature of $-78{ }^{\\circ} \\mathrm{C}$. What is the lowest pressure at which $\\mathrm{CO}_{2}(s)$ will melt to give $\\mathrm{CO}_{2}(I)$ ? At approximately what temperature will this occur? (See Figure 10.34 for the phase diagram.)\n61. If a severe storm results in the loss of electricity, it may be necessary to use a clothesline to dry laundry. In many parts of the country in the dead of winter, the clothes will quickly freeze when they are hung on the line. If it does not snow, will they dry anyway? Explain your answer.\n62. Is it possible to liquefy nitrogen at room temperature (about $25^{\\circ} \\mathrm{C}$ )? Is it possible to liquefy sulfur dioxide at room temperature? Explain your answers.\n63. Elemental carbon has one gas phase, one liquid phase, and two different solid phases, as shown in the phase diagram:"}
{"id": 3685, "contents": "1186. Exercises - 1186.4. Phase Diagrams\n(a) On the phase diagram, label the gas and liquid regions.\n(b) Graphite is the most stable phase of carbon at normal conditions. On the phase diagram, label the graphite phase.\n(c) If graphite at normal conditions is heated to 2500 K while the pressure is increased to $10^{10} \\mathrm{~Pa}$, it is converted into diamond. Label the diamond phase.\n(d) Circle each triple point on the phase diagram.\n(e) In what phase does carbon exist at 5000 K and $10^{8} \\mathrm{~Pa}$ ?\n(f) If the temperature of a sample of carbon increases from 3000 K to 5000 K at a constant pressure of $10^{6}$ Pa , which phase transition occurs, if any?"}
{"id": 3686, "contents": "1186. Exercises - 1186.5. The Solid State of Matter\n64. What types of liquids typically form amorphous solids?\n65. At very low temperatures oxygen, $\\mathrm{O}_{2}$, freezes and forms a crystalline solid. Which best describes these crystals?\n(a) ionic\n(b) covalent network\n(c) metallic\n(d) amorphous\n(e) molecular crystals\n66. As it cools, olive oil slowly solidifies and forms a solid over a range of temperatures. Which best describes the solid?\n(a) ionic\n(b) covalent network\n(c) metallic\n(d) amorphous\n(e) molecular crystals\n67. Explain why ice, which is a crystalline solid, has a melting temperature of $0^{\\circ} \\mathrm{C}$, whereas butter, which is an amorphous solid, softens over a range of temperatures.\n68. Identify the type of crystalline solid (metallic, network covalent, ionic, or molecular) formed by each of the following substances:\n(a) $\\mathrm{SiO}_{2}$\n(b) KCl\n(c) Cu\n(d) $\\mathrm{CO}_{2}$\n(e) C (diamond)\n(f) $\\mathrm{BaSO}_{4}$\n(g) $\\mathrm{NH}_{3}$\n(h) $\\mathrm{NH}_{4} \\mathrm{~F}$\n(i) $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$\n69. Identify the type of crystalline solid (metallic, network covalent, ionic, or molecular) formed by each of the following substances:\n(a) $\\mathrm{CaCl}_{2}$\n(b) SiC\n(c) $\\mathrm{N}_{2}$\n(d) Fe\n(e) C (graphite)\n(f) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$\n(g) HCl\n(h) $\\mathrm{NH}_{4} \\mathrm{NO}_{3}$\n(i) $\\mathrm{K}_{3} \\mathrm{PO}_{4}$\n70. Classify each substance in the table as either a metallic, ionic, molecular, or covalent network solid:"}
{"id": 3687, "contents": "1186. Exercises - 1186.5. The Solid State of Matter\n| Substance | Appearance | Melting Point | Electrical <br> Conductivity | Solubility in Water |\n| :---: | :---: | :---: | :---: | :---: |\n| X | lustrous, malleable | $1500^{\\circ} \\mathrm{C}$ | high | insoluble |\n| Y | soft, yellow | $113^{\\circ} \\mathrm{C}$ | none | insoluble |\n| Z | hard, white | $800^{\\circ} \\mathrm{C}$ | only if melted/ <br> dissolved | soluble |\n\n71. Classify each substance in the table as either a metallic, ionic, molecular, or covalent network solid:\n\n| Appearance |  | Melting Point | Electrical <br> Conductivity | Solubility in Water |\n| :---: | :---: | :---: | :---: | :---: |\n| X | brittle, white | $800^{\\circ} \\mathrm{C}$ | only if melted/ <br> dissolved | soluble |\n| Y | shiny, malleable | $1100^{\\circ} \\mathrm{C}$ | high | insoluble |\n| Z | hard, colorless | $3550^{\\circ} \\mathrm{C}$ | none | insoluble |\n\n72. Identify the following substances as ionic, metallic, covalent network, or molecular solids:"}
{"id": 3688, "contents": "1186. Exercises - 1186.5. The Solid State of Matter\n72. Identify the following substances as ionic, metallic, covalent network, or molecular solids:\n\nSubstance A is malleable, ductile, conducts electricity well, and has a melting point of $1135^{\\circ} \\mathrm{C}$. Substance $B$ is brittle, does not conduct electricity as a solid but does when molten, and has a melting point of 2072\n${ }^{\\circ} \\mathrm{C}$. Substance C is very hard, does not conduct electricity, and has a melting point of $3440^{\\circ} \\mathrm{C}$. Substance D is soft, does not conduct electricity, and has a melting point of $185^{\\circ} \\mathrm{C}$.\n73. Substance A is shiny, conducts electricity well, and melts at $975^{\\circ} \\mathrm{C}$. Substance A is likely a(n):\n(a) ionic solid\n(b) metallic solid\n(c) molecular solid\n(d) covalent network solid\n74. Substance B is hard, does not conduct electricity, and melts at $1200^{\\circ} \\mathrm{C}$. Substance $B$ is likely a(n):\n(a) ionic solid\n(b) metallic solid\n(c) molecular solid\n(d) covalent network solid"}
{"id": 3689, "contents": "1186. Exercises - 1186.6. Lattice Structures in Crystalline Solids\n75. Describe the crystal structure of iron, which crystallizes with two equivalent metal atoms in a cubic unit cell.\n76. Describe the crystal structure of Pt, which crystallizes with four equivalent metal atoms in a cubic unit cell.\n77. What is the coordination number of a chromium atom in the body-centered cubic structure of chromium?\n78. What is the coordination number of an aluminum atom in the face-centered cubic structure of aluminum?\n79. Cobalt metal crystallizes in a hexagonal closest packed structure. What is the coordination number of a cobalt atom?\n80. Nickel metal crystallizes in a cubic closest packed structure. What is the coordination number of a nickel atom?\n81. Tungsten crystallizes in a body-centered cubic unit cell with an edge length of 3.165 \u00c5.\n(a) What is the atomic radius of tungsten in this structure?\n(b) Calculate the density of tungsten.\n82. Platinum (atomic radius $=1.38 \\AA$ \u00c5) crystallizes in a cubic closely packed structure. Calculate the edge length of the face-centered cubic unit cell and the density of platinum.\n83. Barium crystallizes in a body-centered cubic unit cell with an edge length of $5.025 \\AA$\n(a) What is the atomic radius of barium in this structure?\n(b) Calculate the density of barium.\n84. Aluminum (atomic radius $=1.43 \\AA$ ) crystallizes in a cubic closely packed structure. Calculate the edge length of the face-centered cubic unit cell and the density of aluminum.\n85. The density of aluminum is $2.7 \\mathrm{~g} / \\mathrm{cm}^{3}$; that of silicon is $2.3 \\mathrm{~g} / \\mathrm{cm}^{3}$. Explain why Si has the lower density even though it has heavier atoms.\n86. The free space in a metal may be found by subtracting the volume of the atoms in a unit cell from the volume of the cell. Calculate the percentage of free space in each of the three cubic lattices if all atoms in each are of equal size and touch their nearest neighbors. Which of these structures represents the most efficient packing? That is, which packs with the least amount of unused space?"}
{"id": 3690, "contents": "1186. Exercises - 1186.6. Lattice Structures in Crystalline Solids\n87. Cadmium sulfide, sometimes used as a yellow pigment by artists, crystallizes with cadmium, occupying one-half of the tetrahedral holes in a closest packed array of sulfide ions. What is the formula of cadmium sulfide? Explain your answer.\n88. A compound of cadmium, tin, and phosphorus is used in the fabrication of some semiconductors. It crystallizes with cadmium occupying one-fourth of the tetrahedral holes and tin occupying one-fourth of the tetrahedral holes in a closest packed array of phosphide ions. What is the formula of the compound? Explain your answer.\n89. What is the formula of the magnetic oxide of cobalt, used in recording tapes, that crystallizes with cobalt atoms occupying one-eighth of the tetrahedral holes and one-half of the octahedral holes in a closely packed array of oxide ions?\n90. A compound containing zinc, aluminum, and sulfur crystallizes with a closest-packed array of sulfide ions. Zinc ions are found in one-eighth of the tetrahedral holes and aluminum ions in one-half of the octahedral holes. What is the empirical formula of the compound?\n91. A compound of thallium and iodine crystallizes in a simple cubic array of iodide ions with thallium ions in all of the cubic holes. What is the formula of this iodide? Explain your answer.\n92. Which of the following elements reacts with sulfur to form a solid in which the sulfur atoms form a closestpacked array with all of the octahedral holes occupied: $\\mathrm{Li}, \\mathrm{Na}, \\mathrm{Be}, \\mathrm{Ca}$, or Al ?\n93. What is the percent by mass of titanium in rutile, a mineral that contains titanium and oxygen, if structure can be described as a closest packed array of oxide ions with titanium ions in one-half of the octahedral holes? What is the oxidation number of titanium?\n94. Explain why the chemically similar alkali metal chlorides NaCl and CsCl have different structures, whereas the chemically different NaCl and MnS have the same structure.\n95. As minerals were formed from the molten magma, different ions occupied the same cites in the crystals. Lithium often occurs along with magnesium in minerals despite the difference in the charge on their ions. Suggest an explanation."}
{"id": 3691, "contents": "1186. Exercises - 1186.6. Lattice Structures in Crystalline Solids\n95. As minerals were formed from the molten magma, different ions occupied the same cites in the crystals. Lithium often occurs along with magnesium in minerals despite the difference in the charge on their ions. Suggest an explanation.\n96. Rubidium iodide crystallizes with a cubic unit cell that contains iodide ions at the corners and a rubidium ion in the center. What is the formula of the compound?\n97. One of the various manganese oxides crystallizes with a cubic unit cell that contains manganese ions at the corners and in the center. Oxide ions are located at the center of each edge of the unit cell. What is the formula of the compound?\n98. NaH crystallizes with the same crystal structure as NaCl . The edge length of the cubic unit cell of NaH is 4.880 \u00c5.\n(a) Calculate the ionic radius of $\\mathrm{H}^{-}$. (The ionic radius of $\\mathrm{Li}^{+}$is 0.0.95 $\\AA$.)\n(b) Calculate the density of NaH .\n99. Thallium(I) iodide crystallizes with the same structure as CsCl . The edge length of the unit cell of TlI is $4.20 \\AA$. Calculate the ionic radius of $\\mathrm{TI}^{+}$. (The ionic radius of $\\mathrm{I}^{-}$is $2.16 \\AA$.)\n100. A cubic unit cell contains manganese ions at the corners and fluoride ions at the center of each edge.\n(a) What is the empirical formula of this compound? Explain your answer.\n(b) What is the coordination number of the $\\mathrm{Mn}^{3+}$ ion?\n(c) Calculate the edge length of the unit cell if the radius of a $\\mathrm{Mn}^{3+}$ ion is 0.65 A .\n(d) Calculate the density of the compound.\n101. What is the spacing between crystal planes that diffract $X$-rays with a wavelength of 1.541 nm at an angle $\\theta$ of $15.55^{\\circ}$ (first order reflection)?"}
{"id": 3692, "contents": "1186. Exercises - 1186.6. Lattice Structures in Crystalline Solids\n101. What is the spacing between crystal planes that diffract $X$-rays with a wavelength of 1.541 nm at an angle $\\theta$ of $15.55^{\\circ}$ (first order reflection)?\n102. A diffractometer using $X$-rays with a wavelength of 0.2287 nm produced first order diffraction peak for a crystal angle $\\theta=16.21^{\\circ}$. Determine the spacing between the diffracting planes in this crystal.\n103. A metal with spacing between planes equal to 0.4164 nm diffracts $X$-rays with a wavelength of 0.2879 nm . What is the diffraction angle for the first order diffraction peak?\n104. Gold crystallizes in a face-centered cubic unit cell. The second-order reflection ( $n=2$ ) of $X$-rays for the planes that make up the tops and bottoms of the unit cells is at $\\theta=22.20^{\\circ}$. The wavelength of the X -rays is $1.54 \\AA$. What is the density of metallic gold?\n105. When an electron in an excited molybdenum atom falls from the $L$ to the $K$ shell, an $X$-ray is emitted. These X-rays are diffracted at an angle of $7.75^{\\circ}$ by planes with a separation of $2.64 \\AA$. What is the difference in energy between the $K$ shell and the $L$ shell in molybdenum assuming a first order diffraction?"}
{"id": 3693, "contents": "1187. CHAPTER 11 - \nSolutions and Colloids\n\n\nFigure 11.1 Coral reefs, such as this one at the Palmyra Atoll National Wildlife Refuge, are vital to the ecosystem of earth's oceans. The health of coral reefs and all marine life depends on the specific chemical composition of the complex mixture known as seawater. (credit: modification of work by \"USFWS - Pacific Region\"/Wikimedia Commons)"}
{"id": 3694, "contents": "1188. CHAPTER OUTLINE - 1188.1. The Dissolution Process\n11.2 Electrolytes\n11.3 Solubility\n11.4 Colligative Properties\n11.5 Colloids\n\nINTRODUCTION Coral reefs are home to about $25 \\%$ of all marine species. They are being threatened by climate change, oceanic acidification, and water pollution, all of which change the composition of the solution known as seawater. Dissolved oxygen in seawater is critical for sea creatures, but as the oceans warm, oxygen becomes less soluble. As the concentration of carbon dioxide in the atmosphere increases, the concentration of carbon dioxide in the oceans increases, contributing to oceanic acidification. Coral reefs are particularly sensitive to the acidification of the ocean, since the exoskeletons of the coral polyps are soluble in acidic solutions. Humans contribute to the changing of seawater composition by allowing agricultural runoff and other forms of pollution to affect our oceans.\n\nSolutions are crucial to the processes that sustain life and to many other processes involving chemical reactions. This chapter considers the nature of solutions and examines factors that determine whether a solution will form and what properties it may have. The properties of colloids-mixtures containing dispersed particles larger than the molecules and ions of typical solutions-are also discussed."}
{"id": 3695, "contents": "1189. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the basic properties of solutions and how they form\n- Predict whether a given mixture will yield a solution based on molecular properties of its components\n- Explain why some solutions either produce or absorb heat when they form\n\nAn earlier chapter of this text introduced solutions, defined as homogeneous mixtures of two or more substances. Often, one component of a solution is present at a significantly greater concentration, in which case it is called the solvent. The other components of the solution present in relatively lesser concentrations are called solutes. Sugar is a covalent solid composed of sucrose molecules, $\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}$. When this compound dissolves in water, its molecules become uniformly distributed among the molecules of water:\n\n$$\n\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}(s) \\longrightarrow \\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}(a q)\n$$\n\nThe subscript \"aq\" in the equation signifies that the sucrose molecules are solutes and are therefore individually dispersed throughout the aqueous solution (water is the solvent). Although sucrose molecules are heavier than water molecules, they remain dispersed throughout the solution; gravity does not cause them to \"settle out\" over time.\n\nPotassium dichromate, $\\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$, is an ionic compound composed of colorless potassium ions, $\\mathrm{K}^{+}$, and orange dichromate ions, $\\mathrm{Cr}_{2} \\mathrm{O}_{7}{ }^{2-}$. When a small amount of solid potassium dichromate is added to water, the compound dissolves and dissociates to yield potassium ions and dichromate ions uniformly distributed throughout the mixture (Figure 11.2), as indicated in this equation:\n\n$$\n\\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}(s) \\longrightarrow 2 \\mathrm{~K}^{+}(a q)+\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}(a q)\n$$"}
{"id": 3696, "contents": "1189. LEARNING OBJECTIVES - \nAs with the mixture of sugar and water, this mixture is also an aqueous solution. Its solutes, potassium and dichromate ions, remain individually dispersed among the solvent (water) molecules.\n\n\nFIGURE 11.2 When potassium dichromate $\\left(\\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}\\right)$ is mixed with water, it forms a homogeneous orange solution. (credit: modification of work by Mark Ott)"}
{"id": 3697, "contents": "1190. LINK TO LEARNING - \nVisit this virtual lab (http://openstax.org/l/16Phetsugar) to view simulations of the dissolution of common covalent and ionic substances (sugar and salt) in water.\n\nWater is used so often as a solvent that the word solution has come to imply an aqueous solution to many people. However, almost any gas, liquid, or solid can act as a solvent. Many alloys are solid solutions of one metal dissolved in another; for example, US five-cent coins contain nickel dissolved in copper. Air is a gaseous solution, a homogeneous mixture of nitrogen, oxygen, and several other gases. Oxygen (a gas), alcohol (a liquid), and sugar (a solid) all dissolve in water (a liquid) to form liquid solutions. Table 11.1 gives examples of several different solutions and the phases of the solutes and solvents.\n\n| Solution | Solute | Solvent |\n| :--- | :--- | :--- |\n| air | $\\mathrm{O}_{2}(g)$ | $\\mathrm{N}_{2}(g)$ |\n| soft drinks ${ }^{\\underline{1}}$ | $\\mathrm{CO}_{2}(g)$ | $\\mathrm{H}_{2} \\mathrm{O}(I)$ |\n| hydrogen in palladium | $\\mathrm{H}_{2}(g)$ | $\\mathrm{Pd}(s)$ |\n| rubbing alcohol | $\\mathrm{H}_{2} \\mathrm{O}(I)$ | $\\mathrm{C}_{3} \\mathrm{H}_{8} \\mathrm{O}(I)(2$-propanol $)$ |\n| saltwater | $\\mathrm{NaCl}^{(s)}$ | $\\mathrm{H}_{2} \\mathrm{O}(I)$ |\n| brass | $\\mathrm{Zn}(s)$ | $\\mathrm{Cu}(s)$ |\n\nTABLE 11.1\nSolutions exhibit these defining traits:"}
{"id": 3698, "contents": "1190. LINK TO LEARNING - \nTABLE 11.1\nSolutions exhibit these defining traits:\n\n- They are homogeneous; after a solution is mixed, it has the same composition at all points throughout (its composition is uniform).\n- The physical state of a solution-solid, liquid, or gas-is typically the same as that of the solvent, as demonstrated by the examples in Table 11.1.\n- The components of a solution are dispersed on a molecular scale; they consist of a mixture of separated solute particles (molecules, atoms, and/or ions) each closely surrounded by solvent species.\n- The dissolved solute in a solution will not settle out or separate from the solvent.\n- The composition of a solution, or the concentrations of its components, can be varied continuously (within limits determined by the solubility of the components, discussed in detail later in this chapter)."}
{"id": 3699, "contents": "1191. The Formation of Solutions - \nThe formation of a solution is an example of a spontaneous process, a process that occurs under specified conditions without the requirement of energy from some external source. Sometimes a mixture is stirred to speed up the dissolution process, but this is not necessary; a homogeneous solution will form eventually. The topic of spontaneity is critically important to the study of chemical thermodynamics and is treated more thoroughly in a later chapter of this text. For purposes of this chapter's discussion, it will suffice to consider two criteria that favor, but do not guarantee, the spontaneous formation of a solution:\n\n1. a decrease in the internal energy of the system (an exothermic change, as discussed in the previous chapter on thermochemistry)\n2. an increased dispersal of matter in the system (which indicates an increase in the entropy of the system, as you will learn about in the later chapter on thermodynamics)\n\nIn the process of dissolution, an internal energy change often, but not always, occurs as heat is absorbed or evolved. An increase in matter dispersal always results when a solution forms from the uniform distribution of solute molecules throughout a solvent.\n\nWhen the strengths of the intermolecular forces of attraction between solute and solvent species in a solution are no different than those present in the separated components, the solution is formed with no accompanying energy change. Such a solution is called an ideal solution. A mixture of ideal gases (or gases such as helium and argon, which closely approach ideal behavior) is an example of an ideal solution, since the entities comprising these gases experience no significant intermolecular attractions.\n\n1 If bubbles of gas are observed within the liquid, the mixture is not homogeneous and, thus, not a solution.\n\nWhen containers of helium and argon are connected, the gases spontaneously mix due to diffusion and form a solution (Figure 11.3). The formation of this solution clearly involves an increase in matter dispersal, since the helium and argon atoms occupy a volume twice as large as that which each occupied before mixing.\n\n\nValve closed\n\n\nValve open"}
{"id": 3700, "contents": "1191. The Formation of Solutions - \nValve closed\n\n\nValve open\n\nFIGURE 11.3 Samples of helium and argon spontaneously mix to give a solution.\nIdeal solutions may also form when structurally similar liquids are mixed. For example, mixtures of the alcohols methanol $\\left(\\mathrm{CH}_{3} \\mathrm{OH}\\right)$ and ethanol $\\left(\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}\\right)$ form ideal solutions, as do mixtures of the hydrocarbons pentane, $\\mathrm{C}_{5} \\mathrm{H}_{12}$, and hexane, $\\mathrm{C}_{6} \\mathrm{H}_{14}$. Placing methanol and ethanol, or pentane and hexane, in the bulbs shown in Figure 11.3 will result in the same diffusion and subsequent mixing of these liquids as is observed for the He and Ar gases (although at a much slower rate), yielding solutions with no significant change in energy. Unlike a mixture of gases, however, the components of these liquid-liquid solutions do, indeed, experience intermolecular attractive forces. But since the molecules of the two substances being mixed are structurally very similar, the intermolecular attractive forces between like and unlike molecules are essentially the same, and the dissolution process, therefore, does not entail any appreciable increase or decrease in energy. These examples illustrate how increased matter dispersal alone can provide the driving force required to cause the spontaneous formation of a solution. In some cases, however, the relative magnitudes of intermolecular forces of attraction between solute and solvent species may prevent dissolution.\n\nThree types of intermolecular attractive forces are relevant to the dissolution process: solute-solute, solventsolvent, and solute-solvent. As illustrated in Figure 11.4, the formation of a solution may be viewed as a stepwise process in which energy is consumed to overcome solute-solute and solvent-solvent attractions (endothermic processes) and released when solute-solvent attractions are established (an exothermic process referred to as solvation). The relative magnitudes of the energy changes associated with these stepwise processes determine whether the dissolution process overall will release or absorb energy. In some cases, solutions do not form because the energy required to separate solute and solvent species is so much greater than the energy released by solvation."}
{"id": 3701, "contents": "1191. The Formation of Solutions - \nFIGURE 11.4 This schematic representation of dissolution shows a stepwise process involving the endothermic separation of solute and solvent species (Steps 1 and 2) and exothermic solvation (Step 3).\n\nConsider the example of an ionic compound dissolving in water. Formation of the solution requires the electrostatic forces between the cations and anions of the compound (solute-solute) be overcome completely as attractive forces are established between these ions and water molecules (solute-solvent). Hydrogen bonding between a relatively small fraction of the water molecules must also be overcome to accommodate any dissolved solute. If the solute's electrostatic forces are significantly greater than the solvation forces, the dissolution process is significantly endothermic and the compound may not dissolve to an appreciable extent. Calcium carbonate, the major component of coral reefs, is one example of such an \"insoluble\" ionic compound (see Figure 11.1). On the other hand, if the solvation forces are much stronger than the compound's electrostatic forces, the dissolution is significantly exothermic and the compound may be highly soluble. A common example of this type of ionic compound is sodium chloride, commonly known as table salt.\n\nAs noted at the beginning of this module, spontaneous solution formation is favored, but not guaranteed, by exothermic dissolution processes. While many soluble compounds do, indeed, dissolve with the release of heat, some dissolve endothermically. Ammonium nitrate $\\left(\\mathrm{NH}_{4} \\mathrm{NO}_{3}\\right)$ is one such example and is used to make instant cold packs, like the one pictured in Figure 11.5, which are used for treating injuries. A thin-walled plastic bag of water is sealed inside a larger bag with solid $\\mathrm{NH}_{4} \\mathrm{NO}_{3}$. When the smaller bag is broken, a solution of $\\mathrm{NH}_{4} \\mathrm{NO}_{3}$ forms, absorbing heat from the surroundings (the injured area to which the pack is applied) and providing a cold compress that decreases swelling. Endothermic dissolutions such as this one require a greater energy input to separate the solute species than is recovered when the solutes are solvated, but they are spontaneous nonetheless due to the increase in disorder that accompanies formation of the solution."}
{"id": 3702, "contents": "1191. The Formation of Solutions - \nFIGURE 11.5 An instant cold pack gets cold when certain salts, such as ammonium nitrate, dissolve in water-an endothermic process."}
{"id": 3703, "contents": "1192. LINK TO LEARNING - \nWatch this brief video (http://openstax.org/l/16endoexo) illustrating endothermic and exothermic dissolution processes."}
{"id": 3704, "contents": "1193. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Define and give examples of electrolytes\n- Distinguish between the physical and chemical changes that accompany dissolution of ionic and covalent electrolytes\n- Relate electrolyte strength to solute-solvent attractive forces\n\nWhen some substances are dissolved in water, they undergo either a physical or a chemical change that yields ions in solution. These substances constitute an important class of compounds called electrolytes. Substances that do not yield ions when dissolved are called nonelectrolytes. If the physical or chemical process that generates the ions is essentially 100\\% efficient (all of the dissolved compound yields ions), then the substance is known as a strong electrolyte. If only a relatively small fraction of the dissolved substance undergoes the ion-producing process, it is called a weak electrolyte.\n\nSubstances may be identified as strong, weak, or nonelectrolytes by measuring the electrical conductance of an aqueous solution containing the substance. To conduct electricity, a substance must contain freely mobile, charged species. Most familiar is the conduction of electricity through metallic wires, in which case the mobile, charged entities are electrons. Solutions may also conduct electricity if they contain dissolved ions, with conductivity increasing as ion concentration increases. Applying a voltage to electrodes immersed in a solution permits assessment of the relative concentration of dissolved ions, either quantitatively, by measuring the electrical current flow, or qualitatively, by observing the brightness of a light bulb included in the circuit\n(Figure 11.6).\n\n\nacetic acid solution Low conductivity\n\nFIGURE 11.6 Solutions of nonelectrolytes such as ethanol do not contain dissolved ions and cannot conduct electricity. Solutions of electrolytes contain ions that permit the passage of electricity. The conductivity of an electrolyte solution is related to the strength of the electrolyte."}
{"id": 3705, "contents": "1194. Ionic Electrolytes - \nWater and other polar molecules are attracted to ions, as shown in Figure 11.7. The electrostatic attraction between an ion and a molecule with a dipole is called an ion-dipole attraction. These attractions play an important role in the dissolution of ionic compounds in water.\n\n\nFIGURE 11.7 As potassium chloride ( KCl ) dissolves in water, the ions are hydrated. The polar water molecules are attracted by the charges on the $\\mathrm{K}^{+}$and $\\mathrm{Cl}^{-}$ions. Water molecules in front of and behind the ions are not shown.\n\nWhen ionic compounds dissolve in water, the ions in the solid separate and disperse uniformly throughout the solution because water molecules surround and solvate the ions, reducing the strong electrostatic forces between them. This process represents a physical change known as dissociation. Under most conditions, ionic compounds will dissociate nearly completely when dissolved, and so they are classified as strong electrolytes. Even sparingly, soluble ionic compounds are strong electrolytes, since the small amount that does dissolve will dissociate completely.\n\nConsider what happens at the microscopic level when solid KCl is added to water. Ion-dipole forces attract the positive (hydrogen) end of the polar water molecules to the negative chloride ions at the surface of the solid, and they attract the negative (oxygen) ends to the positive potassium ions. The water molecules surround individual $\\mathrm{K}^{+}$and $\\mathrm{Cl}^{-}$ions, reducing the strong interionic forces that bind the ions together and letting them move off into solution as solvated ions, as Figure 11.7 shows. Overcoming the electrostatic attraction permits the independent motion of each hydrated ion in a dilute solution as the ions transition from fixed positions in the undissolved compound to widely dispersed, solvated ions in solution."}
{"id": 3706, "contents": "1195. Covalent Electrolytes - \nPure water is an extremely poor conductor of electricity because it is only very slightly ionized-only about two out of every 1 billion molecules ionize at $25^{\\circ} \\mathrm{C}$. Water ionizes when one molecule of water gives up a proton $\\left(\\mathrm{H}^{+}\\right.$ ion) to another molecule of water, yielding hydronium and hydroxide ions.\n\n$$\n\\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{OH}^{-}(a q)\n$$\n\nIn some cases, solutions prepared from covalent compounds conduct electricity because the solute molecules react chemically with the solvent to produce ions. For example, pure hydrogen chloride is a gas consisting of covalent HCl molecules. This gas contains no ions. However, an aqueous solution of HCl is a very good conductor, indicating that an appreciable concentration of ions exists within the solution.\n\nBecause HCl is an acid, its molecules react with water, transferring $\\mathrm{H}^{+}$ions to form hydronium ions $\\left(\\mathrm{H}_{3} \\mathrm{O}^{+}\\right)$and\nchloride ions $\\left(\\mathrm{Cl}^{-}\\right)$:\n\n\nThis reaction is essentially $100 \\%$ complete for HCl (i.e., it is a strong acid and, consequently, a strong electrolyte). Likewise, weak acids and bases that only react partially generate relatively low concentrations of ions when dissolved in water and are classified as weak electrolytes. The reader may wish to review the discussion of strong and weak acids provided in the earlier chapter of this text on reaction classes and stoichiometry."}
{"id": 3707, "contents": "1195. Covalent Electrolytes - 1195.1. Solubility\nLEARNING OBJECTIVES\nBy the end of this section, you will be able to:\n\n- Describe the effects of temperature and pressure on solubility\n- State Henry's law and use it in calculations involving the solubility of a gas in a liquid\n- Explain the degrees of solubility possible for liquid-liquid solutions\n\nImagine adding a small amount of sugar to a glass of water, stirring until all the sugar has dissolved, and then adding a bit more. You can repeat this process until the sugar concentration of the solution reaches its natural limit, a limit determined primarily by the relative strengths of the solute-solute, solute-solvent, and solventsolvent attractive forces discussed in the previous two modules of this chapter. You can be certain that you have reached this limit because, no matter how long you stir the solution, undissolved sugar remains. The concentration of sugar in the solution at this point is known as its solubility.\nThe solubility of a solute in a particular solvent is the maximum concentration that may be achieved under given conditions when the dissolution process is at equilibrium.\nWhen a solute's concentration is equal to its solubility, the solution is said to be saturated with that solute. If the solute's concentration is less than its solubility, the solution is said to be unsaturated. A solution that contains a relatively low concentration of solute is called dilute, and one with a relatively high concentration is called concentrated."}
{"id": 3708, "contents": "1196. LINK TO LEARNING - \nUse this interactive simulation (http://openstax.org/l/16Phetsoluble) to prepare various saturated solutions.\n\nSolutions may be prepared in which a solute concentration exceeds its solubility. Such solutions are said to be supersaturated, and they are interesting examples of nonequilibrium states (a detailed treatment of this important concept is provided in the text chapters on equilibrium). For example, the carbonated beverage in an open container that has not yet \"gone flat\" is supersaturated with carbon dioxide gas; given time, the $\\mathrm{CO}_{2}$ concentration will decrease until it reaches its solubility."}
{"id": 3709, "contents": "1197. LINK TO LEARNING - \nWatch this impressive video (http://openstax.org/1/16NaAcetate) showing the precipitation of sodium acetate from a supersaturated solution."}
{"id": 3710, "contents": "1198. Solutions of Gases in Liquids - \nAs for any solution, the solubility of a gas in a liquid is affected by the intermolecular attractive forces between solute and solvent species. Unlike solid and liquid solutes, however, there is no solute-solute intermolecular attraction to overcome when a gaseous solute dissolves in a liquid solvent (see Figure 11.4) since the atoms or molecules comprising a gas are far separated and experience negligible interactions. Consequently, solutesolvent interactions are the sole energetic factor affecting solubility. For example, the water solubility of\noxygen is approximately three times greater than that of helium (there are greater dispersion forces between water and the larger oxygen molecules) but 100 times less than the solubility of chloromethane, $\\mathrm{CHCl}_{3}$ (polar chloromethane molecules experience dipole-dipole attraction to polar water molecules). Likewise note the solubility of oxygen in hexane, $\\mathrm{C}_{6} \\mathrm{H}_{14}$, is approximately 20 times greater than it is in water because greater dispersion forces exist between oxygen and the larger hexane molecules.\n\nTemperature is another factor affecting solubility, with gas solubility typically decreasing as temperature increases (Figure 11.8). This inverse relation between temperature and dissolved gas concentration is responsible for one of the major impacts of thermal pollution in natural waters.\n\n\nFIGURE 11.8 The solubilities of these gases in water decrease as the temperature increases. All solubilities were measured with a constant pressure of 101.3 kPa ( 1 atm ) of gas above the solutions.\n\nWhen the temperature of a river, lake, or stream is raised, the solubility of oxygen in the water is decreased. Decreased levels of dissolved oxygen may have serious consequences for the health of the water's ecosystems and, in severe cases, can result in large-scale fish kills (Figure 11.9).\n\n(a)\n\n(b)"}
{"id": 3711, "contents": "1198. Solutions of Gases in Liquids - \n(a)\n\n(b)\n\nFIGURE 11.9 (a) The small bubbles of air in this glass of chilled water formed when the water warmed to room temperature and the solubility of its dissolved air decreased. (b) The decreased solubility of oxygen in natural waters subjected to thermal pollution can result in large-scale fish kills. (credit a: modification of work by Liz West; credit b: modification of work by U.S. Fish and Wildlife Service)\n\nThe solubility of a gaseous solute is also affected by the partial pressure of solute in the gas to which the\nsolution is exposed. Gas solubility increases as the pressure of the gas increases. Carbonated beverages provide a nice illustration of this relationship. The carbonation process involves exposing the beverage to a relatively high pressure of carbon dioxide gas and then sealing the beverage container, thus saturating the beverage with $\\mathrm{CO}_{2}$ at this pressure. When the beverage container is opened, a familiar hiss is heard as the carbon dioxide gas pressure is released, and some of the dissolved carbon dioxide is typically seen leaving solution in the form of small bubbles (Figure 11.10). At this point, the beverage is supersaturated with carbon dioxide and, with time, the dissolved carbon dioxide concentration will decrease to its equilibrium value and the beverage will become \"flat.\"\n\n\nFIGURE 11.10 Opening the bottle of carbonated beverage reduces the pressure of the gaseous carbon dioxide above the beverage. The solubility of $\\mathrm{CO}_{2}$ is thus lowered, and some dissolved carbon dioxide may be seen leaving the solution as small gas bubbles. (credit: modification of work by Derrick Coetzee)\n\nFor many gaseous solutes, the relation between solubility, $C_{\\mathrm{g}}$, and partial pressure, $P_{\\mathrm{g}}$, is a proportional one:\n\n$$\nC_{\\mathrm{g}}=k P_{\\mathrm{g}}\n$$\n\nwhere $k$ is a proportionality constant that depends on the identities of the gaseous solute and solvent, and on the solution temperature. This is a mathematical statement of Henry's law: The quantity of an ideal gas that dissolves in a definite volume of liquid is directly proportional to the pressure of the gas."}
{"id": 3712, "contents": "1200. Application of Henry's Law - \nAt $20^{\\circ} \\mathrm{C}$, the concentration of dissolved oxygen in water exposed to gaseous oxygen at a partial pressure of 101.3 kPa is $1.38 \\times 10^{-3} \\mathrm{~mol} \\mathrm{~L}^{-1}$. Use Henry's law to determine the solubility of oxygen when its partial pressure is 20.7 kPa , the approximate pressure of oxygen in earth's atmosphere."}
{"id": 3713, "contents": "1201. Solution - \nAccording to Henry's law, for an ideal solution the solubility, $C_{\\mathrm{g}}$, of a gas ( $1.38 \\times 10^{-3} \\mathrm{~mol} \\mathrm{~L}^{-1}$, in this case) is directly proportional to the pressure, $P_{\\mathrm{g}}$, of the undissolved gas above the solution ( 101.3 kPa in this case). Because both $C_{\\mathrm{g}}$ and $P_{\\mathrm{g}}$ are known, this relation can be rearragned and used to solve for $k$.\n\n$$\n\\begin{aligned}\nC_{\\mathrm{g}} & =k P_{\\mathrm{g}} \\\\\nk & =\\frac{C_{\\mathrm{g}}}{P_{\\mathrm{g}}} \\\\\n& =\\frac{1.38 \\times 10^{-3} \\mathrm{~mol} \\mathrm{~L}^{-1}}{101.3 \\mathrm{kPa}} \\\\\n& =1.36 \\times 10^{-5} \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{kPa}^{-1}\n\\end{aligned}\n$$\n\nNow, use $k$ to find the solubility at the lower pressure.\n\n$$\n\\begin{aligned}\nC_{\\mathrm{g}}= & k P_{\\mathrm{g}} \\\\\n& 1.36 \\times 10^{-5} \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{kPa}^{-1} \\times 20.7 \\mathrm{kPa} \\\\\n= & 2.82 \\times 10^{-4} \\mathrm{~mol} \\mathrm{~L}^{-1}\n\\end{aligned}\n$$\n\nNote that various units may be used to express the quantities involved in these sorts of computations. Any combination of units that yield to the constraints of dimensional analysis are acceptable."}
{"id": 3714, "contents": "1202. Check Your Learning - \nExposing a 100.0 mL sample of water at $0^{\\circ} \\mathrm{C}$ to an atmosphere containing a gaseous solute at 152 torr resulted in the dissolution of $1.45 \\times 10^{-3} \\mathrm{~g}$ of the solute. Use Henry's law to determine the solubility of this gaseous solute when its pressure is 760 torr."}
{"id": 3715, "contents": "1203. Answer: - \n$7.25 \\times 10^{-3}$ in 100.0 mL or $0.0725 \\mathrm{~g} / \\mathrm{L}$"}
{"id": 3716, "contents": "1205. Thermal Pollution and Oxygen Solubility - \nA certain species of freshwater trout requires a dissolved oxygen concentration of $7.5 \\mathrm{mg} / \\mathrm{L}$. Could these fish thrive in a thermally polluted mountain stream (water temperature is $30.0^{\\circ} \\mathrm{C}$, partial pressure of atmospheric oxygen is 0.17 atm )? Use the data in Figure 11.8 to estimate a value for the Henry's law constant at this temperature."}
{"id": 3717, "contents": "1206. Solution - \nFirst, estimate the Henry's law constant for oxygen in water at the specified temperature of $30.0^{\\circ} \\mathrm{C}$ (Figure 11.8 indicates the solubility at this temperature is approximately $\\sim 1.2 \\mathrm{~mol} / \\mathrm{L})$.\n\n$$\nk=\\frac{C_{\\mathrm{g}}}{P_{\\mathrm{g}}}=1.2 \\times 10^{-3} \\mathrm{~mol} / \\mathrm{L} / 1.00 \\mathrm{~atm}=1.2 \\times 10^{-3} \\mathrm{~mol} / \\mathrm{L} \\mathrm{~atm}\n$$\n\nThen, use this $k$ value to compute the oxygen solubility at the specified oxygen partial pressure, 0.17 atm .\n\n$$\nC_{g}=k P_{g}=\\left(1.2 \\times 10^{-3} \\mathrm{~mol} / \\mathrm{L} \\mathrm{~atm}\\right)(0.17 \\mathrm{~atm})=2.0 \\times 10^{-4} \\mathrm{~mol} / \\mathrm{L}\n$$\n\nFinally, convert this dissolved oxygen concentration from mol/L to mg/L.\n\n$$\n\\left(2.0 \\times 10^{-4} \\mathrm{~mol} / \\mathrm{L}\\right)(32.0 \\mathrm{~g} / 1 \\mathrm{~mol})(1000 \\mathrm{mg} / \\mathrm{g})=6.4 \\mathrm{mg} / \\mathrm{L}\n$$\n\nThis concentration is lesser than the required minimum value of $7.5 \\mathrm{mg} / \\mathrm{L}$, and so these trout would likely not thrive in the polluted stream."}
{"id": 3718, "contents": "1207. Check Your Learning - \nWhat dissolved oxygen concentration is expected for the stream above when it returns to a normal summer time temperature of $15^{\\circ} \\mathrm{C}$ ?"}
{"id": 3719, "contents": "1208. Answer: - \n$8.2 \\mathrm{mg} / \\mathrm{L}$"}
{"id": 3720, "contents": "1210. Decompression Sickness or \"The Bends\" - \nDecompression sickness (DCS), or \"the bends,\" is an effect of the increased pressure of the air inhaled by scuba divers when swimming underwater at considerable depths. In addition to the pressure exerted by the atmosphere, divers are subjected to additional pressure due to the water above them, experiencing an increase of approximately 1 atm for each 10 m of depth. Therefore, the air inhaled by a diver while submerged contains gases at the corresponding higher ambient pressure, and the concentrations of the gases dissolved in the diver's blood are proportionally higher per Henry's law.\n\nAs the diver ascends to the surface of the water, the ambient pressure decreases and the dissolved gases becomes less soluble. If the ascent is too rapid, the gases escaping from the diver's blood may form bubbles that can cause a variety of symptoms ranging from rashes and joint pain to paralysis and death. To avoid DCS, divers must ascend from depths at relatively slow speeds ( 10 or $20 \\mathrm{~m} / \\mathrm{min}$ ) or otherwise make several decompression stops, pausing for several minutes at given depths during the ascent. When these preventive measures are unsuccessful, divers with DCS are often provided hyperbaric oxygen therapy in pressurized vessels called decompression (or recompression) chambers (Figure 11.11). Researchers are also investigating related body reactions and defenses in order to develop better testing and treatment for decompression sicknetss. For example, Ingrid Eftedal, a barophysiologist specializing in bodily reactions to diving, has shown that white blood cells undergo chemical and genetic changes as a result of the condition; these can potentially be used to create biomarker tests and other methods to manage decompression sickness.\n\n\nFIGURE 11.11 (a) US Navy divers undergo training in a recompression chamber. (b) Divers receive hyperbaric oxygen therapy.\n\nDeviations from Henry's law are observed when a chemical reaction takes place between the gaseous solute and the solvent. Thus, for example, the solubility of ammonia in water increases more rapidly with increasing pressure than predicted by the law because ammonia, being a base, reacts to some extent with water to form ammonium ions and hydroxide ions."}
{"id": 3721, "contents": "1210. Decompression Sickness or \"The Bends\" - \nGases can form supersaturated solutions. If a solution of a gas in a liquid is prepared either at low temperature or under pressure (or both), then as the solution warms or as the gas pressure is reduced, the solution may become supersaturated. In 1986, more than 1700 people in Cameroon were killed when a cloud of gas, almost certainly carbon dioxide, bubbled from Lake Nyos (Figure 11.12), a deep lake in a volcanic crater. The water at the bottom of Lake Nyos is saturated with carbon dioxide by volcanic activity beneath the lake. It is believed\nthat the lake underwent a turnover due to gradual heating from below the lake, and the warmer, less-dense water saturated with carbon dioxide reached the surface. Consequently, tremendous quantities of dissolved $\\mathrm{CO}_{2}$ were released, and the colorless gas, which is denser than air, flowed down the valley below the lake and suffocated humans and animals living in the valley.\n\n\nFIGURE 11.12 (a) It is believed that the 1986 disaster that killed more than 1700 people near Lake Nyos in Cameroon resulted when a large volume of carbon dioxide gas was released from the lake. (b) $\\mathrm{A} \\mathrm{CO}_{2}$ vent has since been installed to help outgas the lake in a slow, controlled fashion and prevent a similar catastrophe from happening in the future. (credit a: modification of work by Jack Lockwood; credit b: modification of work by Bill Evans)"}
{"id": 3722, "contents": "1211. Solutions of Liquids in Liquids - \nSome liquids may be mixed in any proportions to yield solutions; in other words, they have infinite mutual solubility and are said to be miscible. Ethanol, sulfuric acid, and ethylene glycol (popular for use as antifreeze, pictured in Figure 11.13) are examples of liquids that are completely miscible with water. Two-cycle motor oil is miscible with gasoline, mixtures of which are used as lubricating fuels for various types of outdoor power equipment (chainsaws, leaf blowers, and so on).\n\n\nFIGURE 11.13 Water and antifreeze are miscible; mixtures of the two are homogeneous in all proportions. (credit: \"dno1967\"/Wikimedia commons)\n\nMiscible liquids are typically those with very similar polarities. Consider, for example, liquids that are polar or capable of hydrogen bonding. For such liquids, the dipole-dipole attractions (or hydrogen bonding) of the solute molecules with the solvent molecules are at least as strong as those between molecules in the pure solute or in the pure solvent. Hence, the two kinds of molecules mix easily. Likewise, nonpolar liquids are miscible with each other because there is no appreciable difference in the strengths of solute-solute, solventsolvent, and solute-solvent intermolecular attractions. The solubility of polar molecules in polar solvents and of nonpolar molecules in nonpolar solvents is, again, an illustration of the chemical axiom \"like dissolves like.\"\n\nTwo liquids that do not mix to an appreciable extent are called immiscible. Separate layers are formed when immiscible liquids are poured into the same container. Gasoline, oil (Figure 11.14), benzene, carbon tetrachloride, some paints, and many other nonpolar liquids are immiscible with water. Relatively weak attractive forces between the polar water molecules and the nonpolar liquid molecules are not adequate to overcome much stronger hydrogen bonding between water molecules. The distinction between immiscibility and miscibility is really one of extent, so that miscible liquids are of infinite mutual solubility, while liquids said to be immiscible are of very low (though not zero) mutual solubility."}
{"id": 3723, "contents": "1211. Solutions of Liquids in Liquids - \nFIGURE 11.14 Water and oil are immiscible. Mixtures of these two substances will form two separate layers with the less dense oil floating on top of the water. (credit: \"Yortw\"/Flickr)\n\nTwo liquids, such as bromine and water, that are of moderate mutual solubility are said to be partially miscible. Two partially miscible liquids usually form two layers when mixed. In the case of the bromine and water mixture, the upper layer is water, saturated with bromine, and the lower layer is bromine saturated with water. Since bromine is nonpolar, and, thus, not very soluble in water, the water layer is only slightly discolored by the bright orange bromine dissolved in it. Since the solubility of water in bromine is very low, there is no noticeable effect on the dark color of the bromine layer (Figure 11.15).\n\n\nFIGURE 11.15 Bromine (the deep orange liquid on the left) and water (the clear liquid in the middle) are partially miscible. The top layer in the mixture on the right is a saturated solution of bromine in water; the bottom layer is a saturated solution of water in bromine. (credit: Paul Flowers)"}
{"id": 3724, "contents": "1212. Solutions of Solids in Liquids - \nThe dependence of solubility on temperature for a number of solids in water is shown by the solubility curves in Figure 11.16. Reviewing these data indicates a general trend of increasing solubility with temperature, although there are exceptions, as illustrated by the ionic compound cerium sulfate.\n\n\nFIGURE 11.16 This graph shows how the solubility of several solids changes with temperature.\n\nThe temperature dependence of solubility can be exploited to prepare supersaturated solutions of certain compounds. A solution may be saturated with the compound at an elevated temperature (where the solute is more soluble) and subsequently cooled to a lower temperature without precipitating the solute. The resultant solution contains solute at a concentration greater than its equilibrium solubility at the lower temperature (i.e., it is supersaturated) and is relatively stable. Precipitation of the excess solute can be initiated by adding a seed crystal (see the video in the Link to Learning earlier in this module) or by mechanically agitating the solution. Some hand warmers, such as the one pictured in Figure 11.17, take advantage of this behavior.\n\n\nFIGURE 11.17 This hand warmer produces heat when the sodium acetate in a supersaturated solution precipitates. Precipitation of the solute is initiated by a mechanical shockwave generated when the flexible metal disk within the solution is \"clicked.\" (credit: modification of work by \"Velela\"/Wikimedia Commons)"}
{"id": 3725, "contents": "1213. LINK TO LEARNING - \nThis video (http://openstax.org/l/16handwarmer) shows the crystallization process occurring in a hand warmer."}
{"id": 3726, "contents": "1214. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Express concentrations of solution components using mole fraction and molality\n- Describe the effect of solute concentration on various solution properties (vapor pressure, boiling point, freezing point, and osmotic pressure)\n- Perform calculations using the mathematical equations that describe these various colligative effects\n- Describe the process of distillation and its practical applications\n- Explain the process of osmosis and describe how it is applied industrially and in nature\n\nThe properties of a solution are different from those of either the pure solute(s) or solvent. Many solution properties are dependent upon the chemical identity of the solute. Compared to pure water, a solution of hydrogen chloride is more acidic, a solution of ammonia is more basic, a solution of sodium chloride is more dense, and a solution of sucrose is more viscous. There are a few solution properties, however, that depend only upon the total concentration of solute species, regardless of their identities. These colligative properties include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure. This small set of properties is of central importance to many natural phenomena and technological applications, as will be described in this module."}
{"id": 3727, "contents": "1215. Mole Fraction and Molality - \nSeveral units commonly used to express the concentrations of solution components were introduced in an earlier chapter of this text, each providing certain benefits for use in different applications. For example, molarity $(M)$ is a convenient unit for use in stoichiometric calculations, since it is defined in terms of the molar amounts of solute species:\n\n$$\nM=\\frac{\\text { mol solute }}{\\mathrm{L} \\text { solution }}\n$$\n\nBecause solution volumes vary with temperature, molar concentrations will likewise vary. When expressed as molarity, the concentration of a solution with identical numbers of solute and solvent species will be different\nat different temperatures, due to the contraction/expansion of the solution. More appropriate for calculations involving many colligative properties are mole-based concentration units whose values are not dependent on temperature. Two such units are mole fraction (introduced in the previous chapter on gases) and molality.\n\nThe mole fraction, $X$, of a component is the ratio of its molar amount to the total number of moles of all solution components:\n\n$$\nX_{\\mathrm{A}}=\\frac{\\mathrm{mol} \\mathrm{~A}}{\\text { total mol of all components }}\n$$\n\nBy this definition, the sum of mole fractions for all solution components (the solvent and all solutes) is equal to one.\n\nMolality is a concentration unit defined as the ratio of the numbers of moles of solute to the mass of the solvent in kilograms:\n\n$$\nm=\\frac{\\mathrm{mol} \\text { solute }}{\\mathrm{kg} \\text { solvent }}\n$$\n\nSince these units are computed using only masses and molar amounts, they do not vary with temperature and, thus, are better suited for applications requiring temperature-independent concentrations, including several colligative properties, as will be described in this chapter module."}
{"id": 3728, "contents": "1217. Calculating Mole Fraction and Molality - \nThe antifreeze in most automobile radiators is a mixture of equal volumes of ethylene glycol and water, with minor amounts of other additives that prevent corrosion. What are the (a) mole fraction and (b) molality of ethylene glycol, $\\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{OH})_{2}$, in a solution prepared from $2.22 \\times 10^{3} \\mathrm{~g}$ of ethylene glycol and $2.00 \\times 10^{3} \\mathrm{~g}$ of water (approximately 2 L of glycol and 2 L of water)?"}
{"id": 3729, "contents": "1218. Solution - \n(a) The mole fraction of ethylene glycol may be computed by first deriving molar amounts of both solution components and then substituting these amounts into the definition of mole fraction.\n\n$$\n\\begin{gathered}\n\\mathrm{mol} \\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{OH})_{2}=2.22 \\times 10^{3} \\mathrm{~g} \\times \\frac{1 \\mathrm{~mol} \\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{OH})_{2}}{62.07 \\mathrm{~g} \\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{OH})_{2}}=35.8 \\mathrm{~mol} \\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{OH})_{2} \\\\\n\\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{O}=2.00 \\times 10^{3} \\mathrm{~g} \\times \\frac{1 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{O}}{18.02 \\mathrm{~g} \\mathrm{H}_{2} \\mathrm{O}}=111 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{O} \\\\\nX_{\\text {ethylene glycol }}=\\frac{35.8 \\mathrm{~mol} \\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{OH})_{2}}{(35.8+111) \\mathrm{mol} \\text { total }}=0.244\n\\end{gathered}\n$$\n\nNotice that mole fraction is a dimensionless property, being the ratio of properties with identical units (moles).\n(b) Derive moles of solute and mass of solvent (in kg).\n\nFirst, use the given mass of ethylene glycol and its molar mass to find the moles of solute:"}
{"id": 3730, "contents": "1218. Solution - \nFirst, use the given mass of ethylene glycol and its molar mass to find the moles of solute:\n\n$$\n2.22 \\times 10^{3} \\mathrm{~g} \\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{OH})_{2}\\left(\\frac{\\mathrm{~mol} \\mathrm{C}_{2} \\mathrm{H}_{2}(\\mathrm{OH})_{2}}{62.07 \\mathrm{~g}}\\right)=35.8 \\mathrm{~mol} \\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{OH})_{2}\n$$\n\nThen, convert the mass of the water from grams to kilograms:\n\n$$\n2.00 \\times 10^{3} \\mathrm{~g} \\mathrm{H}_{2} \\mathrm{O}\\left(\\frac{1 \\mathrm{~kg}}{1000 \\mathrm{~g}}\\right)=2.00 \\mathrm{~kg} \\mathrm{H}_{2} \\mathrm{O}\n$$\n\nFinally, calculate molality per its definition:\n\n$$\n\\begin{aligned}\n& \\text { molality }=\\frac{\\text { mol solute }}{\\mathrm{kg} \\text { solvent }} \\\\\n& \\text { molality }=\\frac{35.8 \\mathrm{~mol} \\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{OH})_{2}}{2 \\mathrm{~kg} \\mathrm{H}_{2} \\mathrm{O}} \\\\\n& \\text { molality }=17.9 \\mathrm{~m}\n\\end{aligned}\n$$"}
{"id": 3731, "contents": "1219. Check Your Learning - \nWhat are the mole fraction and molality of a solution that contains 0.850 g of ammonia, $\\mathrm{NH}_{3}$, dissolved in 125 g of water?"}
{"id": 3732, "contents": "1220. Answer: - \n$7.14 \\times 10^{-3} ; 0.399 \\mathrm{~m}$"}
{"id": 3733, "contents": "1222. Converting Mole Fraction and Molal Concentrations - \nCalculate the mole fraction of solute and solvent in a 3.0 m solution of sodium chloride."}
{"id": 3734, "contents": "1223. Solution - \nConverting from one concentration unit to another is accomplished by first comparing the two unit definitions. In this case, both units have the same numerator (moles of solute) but different denominators. The provided molal concentration may be written as:\n\n$$\n\\frac{3.0 \\mathrm{~mol} \\mathrm{NaCl}}{1 \\mathrm{~kg} \\mathrm{H}_{2} \\mathrm{O}}\n$$\n\nThe numerator for this solution's mole fraction is, therefore, 3.0 mol NaCl . The denominator may be computed by deriving the molar amount of water corresponding to 1.0 kg\n\n$$\n1.0 \\mathrm{~kg} \\mathrm{H}_{2} \\mathrm{O}\\left(\\frac{1000 \\mathrm{~g}}{1 \\mathrm{~kg}}\\right)\\left(\\frac{\\mathrm{mol} \\mathrm{H}_{2} \\mathrm{O}}{18.02 \\mathrm{~g}}\\right)=55 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{O}\n$$\n\nand then substituting these molar amounts into the definition for mole fraction."}
{"id": 3735, "contents": "1223. Solution - \nand then substituting these molar amounts into the definition for mole fraction.\n\n$$\n\\begin{aligned}\n& X_{\\mathrm{H}_{2} \\mathrm{O}}=\\frac{\\mathrm{mol} \\mathrm{H}_{2} \\mathrm{O}}{\\mathrm{~mol} \\mathrm{NaCl}+\\mathrm{mol} \\mathrm{H}_{2} \\mathrm{O}} \\\\\n& X_{\\mathrm{H}_{2} \\mathrm{O}}=\\frac{55 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{O}}{3.0 \\mathrm{~mol} \\mathrm{NaCl}+55 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{O}} \\\\\n& X_{\\mathrm{H}_{2} \\mathrm{O}}=0.95 \\\\\n& X_{\\mathrm{NaCl}}=\\frac{\\mathrm{mol} \\mathrm{NaCl}}{\\mathrm{~mol} \\mathrm{NaCl}+\\mathrm{mol} \\mathrm{H} \\mathrm{O}} \\\\\n& X_{\\mathrm{NaCl}}\n\\end{aligned}=\\frac{3.0 \\mathrm{~mol} \\mathrm{NaCl}}{3.0 \\mathrm{~mol} \\mathrm{NaCl}+55 \\mathrm{~mol} \\mathrm{H}}{ }_{2} \\mathrm{O} .0 .052 .\n$$"}
{"id": 3736, "contents": "1224. Check Your Learning - \nThe mole fraction of iodine, $\\mathrm{I}_{2}$, dissolved in dichloromethane, $\\mathrm{CH}_{2} \\mathrm{Cl}_{2}$, is 0.115 . What is the molal concentration, $m$, of iodine in this solution?"}
{"id": 3737, "contents": "1225. Answer: - \n1.50 m"}
{"id": 3738, "contents": "1227. Molality and Molarity Conversions - \nIntravenous infusion of a 0.556 M aqueous solution of glucose (density of $1.04 \\mathrm{~g} / \\mathrm{mL}$ ) is part of some postoperative recovery therapies. What is the molal concentration of glucose in this solution?"}
{"id": 3739, "contents": "1228. Solution - \nThe provided molal concentration may be explicitly written as:\n\n$$\nM=0.556 \\mathrm{~mol} \\text { glucose } / 1 \\mathrm{~L} \\text { solution }\n$$\n\nConsider the definition of molality:\n\n$$\nm=\\text { mol solute } / \\mathrm{kg} \\text { solvent }\n$$\n\nThe amount of glucose in 1-L of this solution is 0.556 mol , so the mass of water in this volume of solution is needed.\n\nFirst, compute the mass of 1.00 L of the solution:\n\n$$\n(1.0 \\mathrm{~L} \\text { soln })(1.04 \\mathrm{~g} / \\mathrm{mL})(1000 \\mathrm{~mL} / 1 \\mathrm{~L})(1 \\mathrm{~kg} / 1000 \\mathrm{~g})=1.04 \\mathrm{~kg} \\text { soln }\n$$\n\nThis is the mass of both the water and its solute, glucose, and so the mass of glucose must be subtracted. Compute the mass of glucose from its molar amount:\n\n$$\n(0.556 \\mathrm{~mol} \\text { glucose })(180.2 \\mathrm{~g} / 1 \\mathrm{~mol})=100.2 \\mathrm{~g} \\text { or } 0.1002 \\mathrm{~kg}\n$$\n\nSubtracting the mass of glucose yields the mass of water in the solution:\n\n$$\n1.04 \\mathrm{~kg} \\text { solution }-0.1002 \\mathrm{~kg} \\text { glucose }=0.94 \\mathrm{~kg} \\text { water }\n$$\n\nFinally, the molality of glucose in this solution is computed as:\n\n$$\nm=0.556 \\mathrm{~mol} \\text { glucose } / 0.94 \\mathrm{~kg} \\text { water }=0.59 \\mathrm{~m}\n$$"}
{"id": 3740, "contents": "1229. Check Your Learning - \nNitric acid, $\\mathrm{HNO}_{3}(\\mathrm{aq})$, is commercially available as a 33.7 m aqueous solution (density $=1.35 \\mathrm{~g} / \\mathrm{mL}$ ). What is the molarity of this solution?"}
{"id": 3741, "contents": "1230. Answer: - \n14.6 M"}
{"id": 3742, "contents": "1231. Vapor Pressure Lowering - \nAs described in the chapter on liquids and solids, the equilibrium vapor pressure of a liquid is the pressure exerted by its gaseous phase when vaporization and condensation are occurring at equal rates:\n\n$$\n\\text { liquid } \\rightleftharpoons \\text { gas }\n$$\n\nDissolving a nonvolatile substance in a volatile liquid results in a lowering of the liquid's vapor pressure. This phenomenon can be rationalized by considering the effect of added solute molecules on the liquid's vaporization and condensation processes. To vaporize, solvent molecules must be present at the surface of the solution. The presence of solute decreases the surface area available to solvent molecules and thereby reduces the rate of solvent vaporization. Since the rate of condensation is unaffected by the presence of solute, the net result is that the vaporization-condensation equilibrium is achieved with fewer solvent molecules in the vapor phase (i.e., at a lower vapor pressure) (Figure 11.18). While this interpretation is useful, it does not account for several important aspects of the colligative nature of vapor pressure lowering. A more rigorous explanation involves the property of entropy, a topic of discussion in a later text chapter on thermodynamics. For purposes of understanding the lowering of a liquid's vapor pressure, it is adequate to note that the more dispersed nature of matter in a solution, compared to separate solvent and solute phases, serves to effectively stabilize the solvent molecules and hinder their vaporization. A lower vapor pressure results, and a correspondingly\nhigher boiling point as described in the next section of this module.\n\n\nFIGURE 11.18 The presence of nonvolatile solutes lowers the vapor pressure of a solution by impeding the evaporation of solvent molecules.\n\nThe relationship between the vapor pressures of solution components and the concentrations of those components is described by Raoult's law: The partial pressure exerted by any component of an ideal solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.\n\n$$\nP_{\\mathrm{A}}=X_{\\mathrm{A}} P_{\\mathrm{A}}^{*}\n$$"}
{"id": 3743, "contents": "1231. Vapor Pressure Lowering - \n$$\nP_{\\mathrm{A}}=X_{\\mathrm{A}} P_{\\mathrm{A}}^{*}\n$$\n\nwhere $P_{\\mathrm{A}}$ is the partial pressure exerted by component A in the solution, $P_{\\mathrm{A}}^{*}$ is the vapor pressure of pure A, and $X_{\\mathrm{A}}$ is the mole fraction of A in the solution.\n\nRecalling that the total pressure of a gaseous mixture is equal to the sum of partial pressures for all its components (Dalton's law of partial pressures), the total vapor pressure exerted by a solution containing $i$ components is\n\n$$\nP_{\\text {solution }}=\\sum_{i} P_{i}=\\sum_{i} X_{i} P_{i}^{*}\n$$\n\nA nonvolatile substance is one whose vapor pressure is negligible ( $P^{*} \\approx 0$ ), and so the vapor pressure above a solution containing only nonvolatile solutes is due only to the solvent:\n\n$$\nP_{\\text {solution }}=X_{\\text {solvent }} P_{\\text {solvent }}^{*}\n$$\n\nethanol, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$, at $40^{\\circ} \\mathrm{C}$. The vapor pressure of pure ethanol is 0.178 atm at $40^{\\circ} \\mathrm{C}$. Glycerin is essentially nonvolatile at this temperature."}
{"id": 3744, "contents": "1232. Solution - \nSince the solvent is the only volatile component of this solution, its vapor pressure may be computed per Raoult's law as:\n\n$$\nP_{\\text {solution }}=X_{\\text {solvent }} P_{\\text {solvent }}^{*}\n$$\n\nFirst, calculate the molar amounts of each solution component using the provided mass data.\n$92.1-\\mathrm{g}_{3} \\mathrm{H}_{5}(\\mathrm{OH})_{3} \\times \\frac{1 \\mathrm{~mol} \\mathrm{C}_{3} \\mathrm{H}_{5}(\\mathrm{OH})_{3}}{92.094-\\frac{\\mathrm{C}_{3} \\mathrm{H}_{5}(\\mathrm{OH})_{3}}{}}=1.00 \\mathrm{~mol} \\mathrm{C}_{3} \\mathrm{H}_{5}(\\mathrm{OH})_{3}$\n$184.4-\\mathrm{EC}_{2} \\mathrm{H}_{5} \\mathrm{OH} \\times \\frac{1 \\mathrm{~mol} \\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}}{46.069-\\mathrm{gC}_{2} \\mathrm{H}_{5} \\mathrm{OH}}=4.000 \\mathrm{~mol} \\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$\nNext, calculate the mole fraction of the solvent (ethanol) and use Raoult's law to compute the solution's vapor pressure.\n\n$$\n\\begin{aligned}\n& X_{\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}}=\\frac{4.000 \\mathrm{~mol}}{(1.00 \\mathrm{~mol}+4.000 \\mathrm{~mol})}=0.800 \\\\\n& P_{\\text {solv }}=X_{\\text {solv }} P_{\\text {solv }}^{*}=0.800 \\times 0.178 \\mathrm{~atm}=0.142 \\mathrm{~atm}\n\\end{aligned}\n$$"}
{"id": 3745, "contents": "1233. Check Your Learning - \nA solution contains 5.00 g of urea, $\\mathrm{CO}\\left(\\mathrm{NH}_{2}\\right)_{2}$ (a nonvolatile solute) and 0.100 kg of water. If the vapor pressure of pure water at $25^{\\circ} \\mathrm{C}$ is 23.7 torr, what is the vapor pressure of the solution assuming ideal behavior?"}
{"id": 3746, "contents": "1234. Answer: - \n23.4 torr"}
{"id": 3747, "contents": "1235. Distillation of Solutions - \nSolutions whose components have significantly different vapor pressures may be separated by a selective vaporization process known as distillation. Consider the simple case of a mixture of two volatile liquids, A and B, with A being the more volatile liquid. Raoult's law can be used to show that the vapor above the solution is enriched in component A, that is, the mole fraction of A in the vapor is greater than the mole fraction of A in the liquid (see end-of-chapter Exercise 65). By appropriately heating the mixture, component A may be vaporized, condensed, and collected-effectively separating it from component B.\n\nDistillation is widely applied in both laboratory and industrial settings, being used to refine petroleum, to isolate fermentation products, and to purify water. A typical apparatus for laboratory-scale distillations is shown in Figure 11.19.\n\n\nFIGURE 11.19 A typical laboratory distillation unit is shown in (a) a photograph and (b) a schematic diagram of the components. (credit a: modification of work by \"Rifleman82\"/Wikimedia commons; credit b: modification of work by \"Slashme\"/Wikimedia Commons)\n\nOil refineries use large-scale fractional distillation to separate the components of crude oil. The crude oil is heated to high temperatures at the base of a tall fractionating column, vaporizing many of the components that rise within the column. As vaporized components reach adequately cool zones during their ascent, they condense and are collected. The collected liquids are simpler mixtures of hydrocarbons and other petroleum compounds that are of appropriate composition for various applications (e.g., diesel fuel, kerosene, gasoline), as depicted in Figure 11.20.\n\n\nSmall molecules:\n\n- Low boiling point\n- Very volatile\n- Flows easily\n- Ignites easily\n\n\nLarge molecules:\n\n- High boiling point\n- Not very volatile\n- Does not flow easily\n- Does not ignite easily\n\nFIGURE 11.20 Crude oil is a complex mixture that is separated by large-scale fractional distillation to isolate various simpler mixtures."}
{"id": 3748, "contents": "1236. Boiling Point Elevation - \nAs described in the chapter on liquids and solids, the boiling point of a liquid is the temperature at which its vapor pressure is equal to ambient atmospheric pressure. Since the vapor pressure of a solution is lowered due to the presence of nonvolatile solutes, it stands to reason that the solution's boiling point will subsequently be increased. Vapor pressure increases with temperature, and so a solution will require a higher temperature than will pure solvent to achieve any given vapor pressure, including one equivalent to that of the surrounding atmosphere. The increase in boiling point observed when nonvolatile solute is dissolved in a solvent, $\\Delta T_{\\mathrm{b}}$, is called boiling point elevation and is directly proportional to the molal concentration of solute species:\n\n$$\n\\Delta T_{\\mathrm{b}}=K_{\\mathrm{b}} m\n$$\n\nwhere $K_{\\mathrm{b}}$ is the boiling point elevation constant, or the ebullioscopic constant and $m$ is the molal concentration (molality) of all solute species.\n\nBoiling point elevation constants are characteristic properties that depend on the identity of the solvent. Values of $K_{\\mathrm{b}}$ for several solvents are listed in Table 11.2.\n\nBoiling Point Elevation and Freezing Point Depression Constants for Several Solvents\n\n| Solvent | Boiling Point ( ${ }^{\\circ} \\mathrm{C}$ at 1 atm$)$ | $K_{\\mathrm{b}}\\left({ }^{\\circ} \\mathrm{Cm}^{-1}\\right)$ | Freezing Point $\\left({ }^{\\circ} \\mathrm{C}\\right.$ at 1 atm$)$ | $K_{\\mathrm{f}}\\left({ }^{\\circ} \\mathrm{Cm}^{-1}\\right)$ |\n| :--- | :--- | :--- | :--- | :--- |\n| water | 100.0 | 0.512 | 0.0 | 1.86 |\n\nTABLE 11.2"}
{"id": 3749, "contents": "1236. Boiling Point Elevation - \nTABLE 11.2\n\n| Solvent | Boiling Point ( ${ }^{\\circ} \\mathrm{C}$ at $\\left.\\mathbf{1} \\mathbf{~ a t m}\\right)$ | $K_{\\mathrm{b}}\\left({ }^{\\circ} \\mathrm{Cm}^{-1}\\right)$ | Freezing Point ( ${ }^{\\circ} \\mathrm{C}$ at 1 atm$)$ | $K_{\\mathrm{f}}\\left({ }^{\\circ} \\mathrm{Cm}^{-1}\\right)$ |\n| :--- | :--- | :--- | :--- | :--- |\n| hydrogen acetate | 118.1 | 3.07 | 16.6 | 3.9 |\n| benzene | 80.1 | 2.53 | 5.5 | 5.12 |\n| chloroform | 61.26 | 3.63 | -63.5 | 4.68 |\n| nitrobenzene | 210.9 | 5.24 | 5.67 | 8.1 |\n\nTABLE 11.2\n\nThe extent to which the vapor pressure of a solvent is lowered and the boiling point is elevated depends on the total number of solute particles present in a given amount of solvent, not on the mass or size or chemical identities of the particles. A 1 m aqueous solution of sucrose ( $342 \\mathrm{~g} / \\mathrm{mol}$ ) and a 1 m aqueous solution of ethylene glycol ( $62 \\mathrm{~g} / \\mathrm{mol}$ ) will exhibit the same boiling point because each solution has one mole of solute particles (molecules) per kilogram of solvent."}
{"id": 3750, "contents": "1238. Calculating the Boiling Point of a Solution - \nAssuming ideal solution behavior, what is the boiling point of a 0.33 m solution of a nonvolatile solute in benzene?"}
{"id": 3751, "contents": "1239. Solution - \nUse the equation relating boiling point elevation to solute molality to solve this problem in two steps.\n\n\nStep 1. Calculate the change in boiling point.\n\n$$\n\\Delta T_{\\mathrm{b}}=K_{\\mathrm{b}} m=2.53{ }^{\\circ} \\mathrm{Cm}^{-1} \\times 0.33 m=0.83^{\\circ} \\mathrm{C}\n$$\n\nStep 2. Add the boiling point elevation to the pure solvent's boiling point.\nBoiling temperature $=80.1^{\\circ} \\mathrm{C}+0.83{ }^{\\circ} \\mathrm{C}=80.9^{\\circ} \\mathrm{C}$"}
{"id": 3752, "contents": "1240. Check Your Learning - \nAssuming ideal solution behavior, what is the boiling point of the antifreeze described in Example 11.3?"}
{"id": 3753, "contents": "1241. Answer: - \n$109.2^{\\circ} \\mathrm{C}$"}
{"id": 3754, "contents": "1243. The Boiling Point of an Iodine Solution - \nFind the boiling point of a solution of 92.1 g of iodine, $\\mathrm{I}_{2}$, in 800.0 g of chloroform, $\\mathrm{CHCl}_{3}$, assuming that the iodine is nonvolatile and that the solution is ideal."}
{"id": 3755, "contents": "1244. Solution - \nA four-step approach to solving this problem is outlined below.\n\n\nStep 1. Convert from grams to moles of $\\mathrm{I}_{2}$ using the molar mass of $\\mathrm{I}_{2}$ in the unit conversion factor. Result: 0.363 mol\nStep 2. Determine the molality of the solution from the number of moles of solute and the mass of solvent, in kilograms.\nResult: 0.454 m\nStep 3. Use the direct proportionality between the change in boiling point and molal concentration to determine how much the boiling point changes.\nResult: $1.65^{\\circ} \\mathrm{C}$\nStep 4. Determine the new boiling point from the boiling point of the pure solvent and the change.\nResult: $62.91{ }^{\\circ} \\mathrm{C}$\nCheck each result as a self-assessment."}
{"id": 3756, "contents": "1245. Check Your Learning - \nWhat is the boiling point of a solution of 1.0 g of glycerin, $\\mathrm{C}_{3} \\mathrm{H}_{5}(\\mathrm{OH})_{3}$, in 47.8 g of water? Assume an ideal solution."}
{"id": 3757, "contents": "1246. Answer: - \n$100.12{ }^{\\circ} \\mathrm{C}$"}
{"id": 3758, "contents": "1247. Freezing Point Depression - \nSolutions freeze at lower temperatures than pure liquids. This phenomenon is exploited in \"de-icing\" schemes that use salt (Figure 11.21), calcium chloride, or urea to melt ice on roads and sidewalks, and in the use of ethylene glycol as an \"antifreeze\" in automobile radiators. Seawater freezes at a lower temperature than fresh water, and so the Arctic and Antarctic oceans remain unfrozen even at temperatures below $0^{\\circ} \\mathrm{C}$ (as do the body fluids of fish and other cold-blooded sea animals that live in these oceans).\n\n\nFIGURE 11.21 Rock salt $(\\mathrm{NaCl})$, calcium chloride $\\left(\\mathrm{CaCl}_{2}\\right)$, or a mixture of the two are used to melt ice. (credit: modification of work by Eddie Welker)\n\nThe decrease in freezing point of a dilute solution compared to that of the pure solvent, $\\Delta T_{\\mathrm{f}}$, is called the freezing point depression and is directly proportional to the molal concentration of the solute\n\n$$\n\\Delta T_{\\mathrm{f}}=K_{\\mathrm{f}} m\n$$\n\nwhere $m$ is the molal concentration of the solute and $K_{\\mathrm{f}}$ is called the freezing point depression constant (or cryoscopic constant). Just as for boiling point elevation constants, these are characteristic properties whose\nvalues depend on the chemical identity of the solvent. Values of $K_{\\mathrm{f}}$ for several solvents are listed in Table 11.2."}
{"id": 3759, "contents": "1249. Calculation of the Freezing Point of a Solution - \nAssuming ideal solution behavior, what is the freezing point of the 0.33 m solution of a nonvolatile nonelectrolyte solute in benzene described in Example 11.4?"}
{"id": 3760, "contents": "1250. Solution - \nUse the equation relating freezing point depression to solute molality to solve this problem in two steps.\n\n\nStep 1. Calculate the change in freezing point.\n\n$$\n\\Delta T_{\\mathrm{f}}=K_{\\mathrm{f}} m=5.12^{\\circ} \\mathrm{C} \\mathrm{~m}^{-1} \\times 0.33 \\mathrm{~m}=1.7^{\\circ} \\mathrm{C}\n$$\n\nStep 2. Subtract the freezing point change observed from the pure solvent's freezing point.\nFreezing Temperature $=5.5^{\\circ} \\mathrm{C}-1.7^{\\circ} \\mathrm{C}=3.8^{\\circ} \\mathrm{C}$"}
{"id": 3761, "contents": "1251. Check Your Learning - \nAssuming ideal solution behavior, what is the freezing point of a 1.85 m solution of a nonvolatile nonelectrolyte solute in nitrobenzene?"}
{"id": 3762, "contents": "1252. Answer: - \n$-9.3^{\\circ} \\mathrm{C}$"}
{"id": 3763, "contents": "1254. Colligative Properties and De-Icing - \nSodium chloride and its group 2 analogs calcium and magnesium chloride are often used to de-ice roadways and sidewalks, due to the fact that a solution of any one of these salts will have a freezing point lower than $0{ }^{\\circ} \\mathrm{C}$, the freezing point of pure water. The group 2 metal salts are frequently mixed with the cheaper and more readily available sodium chloride (\"rock salt\") for use on roads, since they tend to be somewhat less corrosive than the NaCl , and they provide a larger depression of the freezing point, since they dissociate to yield three particles per formula unit, rather than two particles like the sodium chloride.\n\nBecause these ionic compounds tend to hasten the corrosion of metal, they would not be a wise choice to use in antifreeze for the radiator in your car or to de-ice a plane prior to takeoff. For these applications, covalent compounds, such as ethylene or propylene glycol, are often used. The glycols used in radiator fluid not only lower the freezing point of the liquid, but they elevate the boiling point, making the fluid useful in both winter and summer. Heated glycols are often sprayed onto the surface of airplanes prior to takeoff in inclement weather in the winter to remove ice that has already formed and prevent the formation of more ice, which would be particularly dangerous if formed on the control surfaces of the aircraft (Figure 11.22).\n\n(a)\n\n(b)\n\nFIGURE 11.22 Freezing point depression is exploited to remove ice from (a) roadways and (b) the control surfaces of aircraft."}
{"id": 3764, "contents": "1255. Phase Diagram for a Solution - \nThe colligative effects on vapor pressure, boiling point, and freezing point described in the previous section are conveniently summarized by comparing the phase diagrams for a pure liquid and a solution derived from that liquid (Figure 11.23).\n\n\nFIGURE 11.23 Phase diagrams for a pure solvent (solid curves) and a solution formed by dissolving nonvolatile solute in the solvent (dashed curves).\n\nThe liquid-vapor curve for the solution is located beneath the corresponding curve for the solvent, depicting the vapor pressure lowering, $\\Delta P$, that results from the dissolution of nonvolatile solute. Consequently, at any given pressure, the solution's boiling point is observed at a higher temperature than that for the pure solvent, reflecting the boiling point elevation, $\\Delta T_{\\mathrm{b}}$, associated with the presence of nonvolatile solute. The solid-liquid curve for the solution is displaced left of that for the pure solvent, representing the freezing point depression, $\\Delta T_{\\mathrm{f}}$, that accompanies solution formation. Finally, notice that the solid-gas curves for the solvent and its solution are identical. This is the case for many solutions comprising liquid solvents and nonvolatile solutes. Just as for vaporization, when a solution of this sort is frozen, it is actually just the solvent molecules that\nundergo the liquid-to-solid transition, forming pure solid solvent that excludes solute species. The solid and gaseous phases, therefore, are composed of solvent only, and so transitions between these phases are not subject to colligative effects."}
{"id": 3765, "contents": "1256. Osmosis and Osmotic Pressure of Solutions - \nA number of natural and synthetic materials exhibit selective permeation, meaning that only molecules or ions of a certain size, shape, polarity, charge, and so forth, are capable of passing through (permeating) the material. Biological cell membranes provide elegant examples of selective permeation in nature, while dialysis tubing used to remove metabolic wastes from blood is a more simplistic technological example. Regardless of how they may be fabricated, these materials are generally referred to as semipermeable membranes.\n\nConsider the apparatus illustrated in Figure 11.24, in which samples of pure solvent and a solution are separated by a membrane that only solvent molecules may permeate. Solvent molecules will diffuse across the membrane in both directions. Since the concentration of solvent is greater in the pure solvent than the solution, these molecules will diffuse from the solvent side of the membrane to the solution side at a faster rate than they will in the reverse direction. The result is a net transfer of solvent molecules from the pure solvent to the solution. Diffusion-driven transfer of solvent molecules through a semipermeable membrane is a process known as osmosis.\n\n\nFIGURE 11.24 (a) A solution and pure solvent are initially separated by an osmotic membrane. (b) Net transfer of solvent molecules to the solution occurs until its osmotic pressure yields equal rates of transfer in both directions.\n\nWhen osmosis is carried out in an apparatus like that shown in Figure 11.24, the volume of the solution increases as it becomes diluted by accumulation of solvent. This causes the level of the solution to rise, increasing its hydrostatic pressure (due to the weight of the column of solution in the tube) and resulting in a faster transfer of solvent molecules back to the pure solvent side. When the pressure reaches a value that yields a reverse solvent transfer rate equal to the osmosis rate, bulk transfer of solvent ceases. This pressure is called the osmotic pressure ( $\\boldsymbol{\\Pi}$ ) of the solution. The osmotic pressure of a dilute solution is related to its solute molarity, $M$, and absolute temperature, $T$, according to the equation\n\n$$\n\\Pi=M R T\n$$\n\nwhere $R$ is the universal gas constant."}
{"id": 3766, "contents": "1258. Calculation of Osmotic Pressure - \nAssuming ideal solution behavior, what is the osmotic pressure (atm) of a 0.30 M solution of glucose in water that is used for intravenous infusion at body temperature, $37^{\\circ} \\mathrm{C}$ ?"}
{"id": 3767, "contents": "1259. Solution - \nFind the osmotic pressure, $\\Pi$, using the formula $\\Pi=M R T$, where $T$ is on the Kelvin scale ( 310 K ) and the value of $R$ is expressed in appropriate units ( $0.08206 \\mathrm{~L} \\mathrm{~atm} / \\mathrm{mol} \\mathrm{K}$ ).\n\n$$\n\\begin{aligned}\n\\Pi & =M R T \\\\\n& =0.30 \\mathrm{~mol} / \\mathrm{L} \\times 0.08206 \\mathrm{~L} \\mathrm{~atm} / \\mathrm{mol} \\mathrm{~K} \\times 310 \\mathrm{~K} \\\\\n& =7.6 \\mathrm{~atm}\n\\end{aligned}\n$$"}
{"id": 3768, "contents": "1260. Check Your Learning - \nAssuming ideal solution behavior, what is the osmotic pressure (atm) a solution with a volume of 0.750 L that contains 5.0 g of methanol, $\\mathrm{CH}_{3} \\mathrm{OH}$, in water at $37^{\\circ} \\mathrm{C}$ ?"}
{"id": 3769, "contents": "1261. Answer: - \n5.3 atm\n\nIf a solution is placed in an apparatus like the one shown in Figure 11.25, applying pressure greater than the osmotic pressure of the solution reverses the osmosis and pushes solvent molecules from the solution into the pure solvent. This technique of reverse osmosis is used for large-scale desalination of seawater and on smaller scales to produce high-purity tap water for drinking.\n\n\nFIGURE 11.25 Applying a pressure greater than the osmotic pressure of a solution will reverse osmosis. Solvent molecules from the solution are pushed into the pure solvent."}
{"id": 3770, "contents": "1263. Reverse Osmosis Water Purification - \nIn the process of osmosis, diffusion serves to move water through a semipermeable membrane from a less concentrated solution to a more concentrated solution. Osmotic pressure is the amount of pressure that must be applied to the more concentrated solution to cause osmosis to stop. If greater pressure is applied, the water will go from the more concentrated solution to a less concentrated (more pure) solution. This is called reverse osmosis. Reverse osmosis (RO) is used to purify water in many applications, from desalination plants in coastal cities, to water-purifying machines in grocery stores (Figure 11.26), and smaller reverse-osmosis household units. With a hand-operated pump, small RO units can be used in third-world countries, disaster areas, and in lifeboats. Our military forces have a variety of generatoroperated RO units that can be transported in vehicles to remote locations.\n\n\nFIGURE 11.26 Reverse osmosis systems for purifying drinking water are shown here on (a) small and (b) large scales. (credit a: modification of work by Jerry Kirkhart; credit b: modification of work by Willard J. Lathrop)"}
{"id": 3771, "contents": "1263. Reverse Osmosis Water Purification - \nExamples of osmosis are evident in many biological systems because cells are surrounded by semipermeable membranes. Carrots and celery that have become limp because they have lost water can be made crisp again by placing them in water. Water moves into the carrot or celery cells by osmosis. A cucumber placed in a concentrated salt solution loses water by osmosis and absorbs some salt to become a pickle. Osmosis can also affect animal cells. Solute concentrations are particularly important when solutions are injected into the body. Solutes in body cell fluids and blood serum give these solutions an osmotic pressure of approximately 7.7 atm . Solutions injected into the body must have the same osmotic pressure as blood serum; that is, they should be isotonic with blood serum. If a less concentrated solution, a hypotonic solution, is injected in sufficient quantity to dilute the blood serum, water from the diluted serum passes into the blood cells by osmosis, causing the cells to expand and rupture. This process is called hemolysis. When a more concentrated solution, a hypertonic solution, is injected, the cells lose water to the more concentrated solution, shrivel, and possibly die in a process called crenation. These effects are illustrated in Figure 11.27.\n\n\nFIGURE 11.27 Red blood cell membranes are water permeable and will (a) swell and possibly rupture in a hypotonic solution; (b) maintain normal volume and shape in an isotonic solution; and (c) shrivel and possibly die in a hypertonic solution. (credit a/b/c: modifications of work by \"LadyofHats\"/Wikimedia commons)"}
{"id": 3772, "contents": "1264. Determination of Molar Masses - \nOsmotic pressure and changes in freezing point, boiling point, and vapor pressure are directly proportional to the number of solute species present in a given amount of solution. Consequently, measuring one of these properties for a solution prepared using a known mass of solute permits determination of the solute's molar mass."}
{"id": 3773, "contents": "1266. Determination of a Molar Mass from a Freezing Point Depression - \nA solution of 4.00 g of a nonelectrolyte dissolved in 55.0 g of benzene is found to freeze at $2.32^{\\circ} \\mathrm{C}$. Assuming ideal solution behavior, what is the molar mass of this compound?"}
{"id": 3774, "contents": "1267. Solution - \nSolve this problem using the following steps.\n\n\nStep 1. Determine the change in freezing point from the observed freezing point and the freezing point of pure benzene (Table 11.2).\n\n$$\n\\Delta T_{\\mathrm{f}}=5.5^{\\circ} \\mathrm{C}-2.32^{\\circ} \\mathrm{C}=3.2^{\\circ} \\mathrm{C}\n$$\n\nStep 2. Determine the molal concentration from $K_{\\mathrm{f}}$, the freezing point depression constant for benzene (Table 11.2), and $\\Delta T_{\\mathrm{f}}$.\n\n$$\n\\begin{gathered}\n\\Delta T_{\\mathrm{f}}=K_{\\mathrm{f}} m \\\\\nm=\\frac{\\Delta T_{\\mathrm{f}}}{K_{\\mathrm{f}}}=\\frac{3.2^{\\circ} \\mathrm{C}}{5.12^{\\circ} \\mathrm{C} m^{-1}}=0.63 \\mathrm{~m}\n\\end{gathered}\n$$\n\nStep 3. Determine the number of moles of compound in the solution from the molal concentration and the mass of solvent used to make the solution.\n\n$$\n\\text { Moles of solute }=\\frac{0.63 \\mathrm{~mol} \\text { solute }}{1.00 \\mathrm{~kg} \\text { solvent }} \\times 0.0550 \\mathrm{~kg} \\text { solvent }=0.035 \\mathrm{~mol}\n$$\n\nStep 4. Determine the molar mass from the mass of the solute and the number of moles in that mass.\n\n$$\n\\text { Molar mass }=\\frac{4.00 \\mathrm{~g}}{0.035 \\mathrm{~mol}}=1.1 \\times 10^{2} \\mathrm{~g} / \\mathrm{mol}\n$$"}
{"id": 3775, "contents": "1268. Check Your Learning - \nA solution of 35.7 g of a nonelectrolyte in 220.0 g of chloroform has a boiling point of $64.5^{\\circ} \\mathrm{C}$. Assuming ideal solution behavior, what is the molar mass of this compound?"}
{"id": 3776, "contents": "1269. Answer: - \n$1.8 \\times 10^{2} \\mathrm{~g} / \\mathrm{mol}$"}
{"id": 3777, "contents": "1271. Determination of a Molar Mass from Osmotic Pressure - \nA 0.500 L sample of an aqueous solution containing 10.0 g of hemoglobin has an osmotic pressure of 5.9 torr at $22{ }^{\\circ} \\mathrm{C}$. Assuming ideal solution behavior, what is the molar mass of hemoglobin?"}
{"id": 3778, "contents": "1272. Solution - \nHere is one set of steps that can be used to solve the problem:\n\n\nStep 1. Convert the osmotic pressure to atmospheres, then determine the molar concentration from the osmotic pressure.\n\n$$\n\\begin{gathered}\n\\Pi=\\frac{5.9 \\mathrm{torr} \\times 1 \\mathrm{~atm}}{760 \\mathrm{torr}}=7.8 \\times 10^{-3} \\mathrm{~atm} \\\\\n\\Pi=M R T \\\\\nM=\\frac{\\Pi}{R T}=\\frac{7.8 \\times 10^{-3} \\mathrm{~atm}}{(0.08206 \\mathrm{~L} \\mathrm{~atm} / \\mathrm{mol} \\mathrm{~K})(295 \\mathrm{~K})}=3.2 \\times 10^{-4} \\mathrm{M}\n\\end{gathered}\n$$\n\nStep 2. Determine the number of moles of hemoglobin in the solution from the concentration and the volume of the solution.\n\n$$\n\\text { moles of hemoglobin }=\\frac{3.2 \\times 10^{-4} \\mathrm{~mol}}{1 \\mathrm{Lsolttion}} \\times 0.500 \\mathrm{Lsolution}=1.6 \\times 10^{-4} \\mathrm{~mol}\n$$\n\nStep 3. Determine the molar mass from the mass of hemoglobin and the number of moles in that mass.\n\n$$\n\\text { molar mass }=\\frac{10.0 \\mathrm{~g}}{1.6 \\times 10^{-4} \\mathrm{~mol}}=6.2 \\times 10^{4} \\mathrm{~g} / \\mathrm{mol}\n$$"}
{"id": 3779, "contents": "1273. Check Your Learning - \nAssuming ideal solution behavior, what is the molar mass of a protein if a solution of 0.02 g of the protein in 25.0 mL of solution has an osmotic pressure of 0.56 torr at $25^{\\circ} \\mathrm{C}$ ?"}
{"id": 3780, "contents": "1274. Answer: - \n$3 \\times 10^{4} \\mathrm{~g} / \\mathrm{mol}$"}
{"id": 3781, "contents": "1275. Colligative Properties of Electrolytes - \nAs noted previously in this module, the colligative properties of a solution depend only on the number, not on the identity, of solute species dissolved. The concentration terms in the equations for various colligative properties (freezing point depression, boiling point elevation, osmotic pressure) pertain to all solute species present in the solution. For the solutions considered thus far in this chapter, the solutes have been nonelectrolytes that dissolve physically without dissociation or any other accompanying process. Each molecule that dissolves yields one dissolved solute molecule. The dissolution of an electroyte, however, is not this simple, as illustrated by the two common examples below:\n\n$$\n\\begin{array}{ll}\n\\text { dissociation } & \\mathrm{NaCl}(\\mathrm{~s}) \\longrightarrow \\mathrm{Na}^{+}(\\mathrm{aq})+\\mathrm{Cl}^{-}(\\mathrm{aq}) \\\\\n\\text { ionization } & \\mathrm{HCl}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\longrightarrow \\mathrm{Cl}^{-}(\\mathrm{aq})+\\mathrm{H}_{3} \\mathrm{O}^{+}(\\mathrm{aq})\n\\end{array}\n$$\n\nConsidering the first of these examples, and assuming complete dissociation, a 1.0 m aqueous solution of NaCl contains 2.0 mole of ions ( 1.0 mol Na and $1.0 \\mathrm{~mol} \\mathrm{Cl}^{-}$) per each kilogram of water, and its freezing point depression is expected to be\n\n$$\n\\Delta T_{\\mathrm{f}}=2.0 \\mathrm{~mol} \\text { ions } / \\mathrm{kg} \\text { water } \\times 1.86^{\\circ} \\mathrm{C} \\mathrm{~kg} \\text { water } / \\mathrm{mol} \\text { ion }=3.7^{\\circ} \\mathrm{C} .\n$$"}
{"id": 3782, "contents": "1275. Colligative Properties of Electrolytes - \nWhen this solution is actually prepared and its freezing point depression measured, however, a value of $3.4^{\\circ} \\mathrm{C}$ is obtained. Similar discrepancies are observed for other ionic compounds, and the differences between the measured and expected colligative property values typically become more significant as solute concentrations increase. These observations suggest that the ions of sodium chloride (and other strong electrolytes) are not completely dissociated in solution.\n\nTo account for this and avoid the errors accompanying the assumption of total dissociation, an experimentally measured parameter named in honor of Nobel Prize-winning German chemist Jacobus Henricus van't Hoff is used. The van't Hoff factor (i) is defined as the ratio of solute particles in solution to the number of formula units dissolved:\n\n$$\ni=\\frac{\\text { moles of particles in solution }}{\\text { moles of formula units dissolved }}\n$$\n\nValues for measured van't Hoff factors for several solutes, along with predicted values assuming complete dissociation, are shown in Table 11.3.\n\nPredicated and Measured van't Hoff Factors for Several 0.050 m Aqueous Solutions\n\n| Formula unit | Classification | Dissolution products | $\\boldsymbol{i}$ (predicted) | $\\boldsymbol{i}$ (measured) |\n| :---: | :---: | :---: | :---: | :---: |\n| $\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}$ (glucose) | Nonelectrolyte | $\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}$ | 1 | 1.0 |\n| NaCl | Strong electrolyte | $\\mathrm{Na}^{+}, \\mathrm{Cl}^{-}$ | 2 | 1.9 |\n| HCl | Strong electrolyte (acid) | $\\mathrm{H}_{3} \\mathrm{O}^{+}, \\mathrm{Cl}^{-}$ | 2 | 1.9 |\n| $\\mathrm{MgSO}_{4}$ | Strong electrolyte | $\\mathrm{Mg}^{2+}, \\mathrm{SO}_{4}{ }^{2-}$, | 2 | 1.3 |\n\nTABLE 11.3"}
{"id": 3783, "contents": "1275. Colligative Properties of Electrolytes - \nTABLE 11.3\n\n| Formula unit | Classification | Dissolution products | $\\boldsymbol{i}$ (predicted) | $\\boldsymbol{i}$ (measured) |\n| :---: | :---: | :---: | :---: | :---: |\n| $\\mathrm{MgCl}_{2}$ | Strong electrolyte | $\\mathrm{Mg}^{2+}, 2 \\mathrm{Cl}^{-}$ | 3 | 2.7 |\n| $\\mathrm{FeCl}_{3}$ | Strong electrolyte | $\\mathrm{Fe}^{3+}, 3 \\mathrm{Cl}^{-}$ | 4 | 3.4 |\n\nTABLE 11.3\n\nIn 1923, the chemists Peter Debye and Erich H\u00fcckel proposed a theory to explain the apparent incomplete ionization of strong electrolytes. They suggested that although interionic attraction in an aqueous solution is very greatly reduced by solvation of the ions and the insulating action of the polar solvent, it is not completely nullified. The residual attractions prevent the ions from behaving as totally independent particles (Figure 11.28). In some cases, a positive and negative ion may actually touch, giving a solvated unit called an ion pair. Thus, the activity, or the effective concentration, of any particular kind of ion is less than that indicated by the actual concentration. Ions become more and more widely separated the more dilute the solution, and the residual interionic attractions become less and less. Thus, in extremely dilute solutions, the effective concentrations of the ions (their activities) are essentially equal to the actual concentrations. Note that the van't Hoff factors for the electrolytes in Table 11.3 are for 0.05 m solutions, at which concentration the value of $i$ for NaCl is 1.9 , as opposed to an ideal value of 2 .\n\n\nFIGURE 11.28 Dissociation of ionic compounds in water is not always complete due to the formation of ion pairs."}
{"id": 3784, "contents": "1277. The Freezing Point of a Solution of an Electrolyte - \nThe concentration of ions in seawater is approximately the same as that in a solution containing 4.2 g of NaCl dissolved in 125 g of water. Use this information and a predicted value for the van't Hoff factor (Table 11.3) to\ndetermine the freezing temperature the solution (assume ideal solution behavior)."}
{"id": 3785, "contents": "1278. Solution - \nSolve this problem using the following series of steps.\n\n\nStep 1. Convert from grams to moles of NaCl using the molar mass of NaCl in the unit conversion factor. Result: 0.072 mol NaCl\nStep 2. Determine the number of moles of ions present in the solution using the number of moles of ions in 1 mole of NaCl as the conversion factor ( 2 mol ions $/ 1 \\mathrm{~mol} \\mathrm{NaCl}$ ).\nResult: 0.14 mol ions\nStep 3. Determine the molality of the ions in the solution from the number of moles of ions and the mass of solvent, in kilograms.\nResult: 1.2 m\nStep 4. Use the direct proportionality between the change in freezing point and molal concentration to determine how much the freezing point changes.\nResult: $2.1^{\\circ} \\mathrm{C}$\nStep 5. Determine the new freezing point from the freezing point of the pure solvent and the change.\nResult: $-2.1^{\\circ} \\mathrm{C}$\nCheck each result as a self-assessment, taking care to avoid rounding errors by retaining guard digits in each step's result for computing the next step's result."}
{"id": 3786, "contents": "1279. Check Your Learning - \nAssuming complete dissociation and ideal solution behavior, calculate the freezing point of a solution of 0.724 g of $\\mathrm{CaCl}_{2}$ in 175 g of water."}
{"id": 3787, "contents": "1280. Answer: - \n$-0.208{ }^{\\circ} \\mathrm{C}$"}
{"id": 3788, "contents": "1281. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the composition and properties of colloidal dispersions\n- List and explain several technological applications of colloids\n\nAs a child, you may have made suspensions such as mixtures of mud and water, flour and water, or a suspension of solid pigments in water, known as tempera paint. These suspensions are heterogeneous mixtures composed of relatively large particles that are visible (or that can be seen with a magnifying glass). They are cloudy, and the suspended particles settle out after mixing. On the other hand, a solution is a homogeneous mixture in which no settling occurs and in which the dissolved species are molecules or ions. Solutions exhibit completely different behavior from suspensions. A solution may be colored, but it is transparent, the molecules or ions are invisible, and they do not settle out on standing. Another class of mixtures called colloids (or colloidal dispersions) exhibit properties intermediate between those of suspensions and solutions (Figure 11.29). The particles in a colloid are larger than most simple molecules;\nhowever, colloidal particles are small enough that they do not settle out upon standing.\n\n\nFIGURE 11.29 (a) A solution is a homogeneous mixture that appears clear, such as the saltwater in this aquarium. (b) In a colloid, such as milk, the particles are much larger but remain dispersed and do not settle. (c) A suspension, such as mud, is a heterogeneous mixture of suspended particles that appears cloudy and in which the particles can settle. (credit a photo: modification of work by Adam Wimsatt; credit b photo: modification of work by Melissa Wiese; credit c photo: modification of work by Peter Burgess)\n\nThe particles in a colloid are large enough to scatter light, a phenomenon called the Tyndall effect. This can make colloidal mixtures appear cloudy or opaque, such as the searchlight beams shown in Figure 11.30. Clouds are colloidal mixtures. They are composed of water droplets that are much larger than molecules, but that are small enough that they do not settle out.\n\n\nFIGURE 11.30 The paths of searchlight beams are made visible when light is scattered by colloidal-size particles in the air (fog, smoke, etc.). (credit: \"Bahman\"/Wikimedia Commons)"}
{"id": 3789, "contents": "1281. LEARNING OBJECTIVES - \nFIGURE 11.30 The paths of searchlight beams are made visible when light is scattered by colloidal-size particles in the air (fog, smoke, etc.). (credit: \"Bahman\"/Wikimedia Commons)\n\nThe term \"colloid\"-from the Greek words kolla, meaning \"glue,\" and eidos, meaning \"like\"-was first used in 1861 by Thomas Graham to classify mixtures such as starch in water and gelatin. Many colloidal particles are aggregates of hundreds or thousands of molecules, but others (such as proteins and polymer molecules) consist of a single extremely large molecule. The protein and synthetic polymer molecules that form colloids may have molecular masses ranging from a few thousand to many million atomic mass units.\n\nAnalogous to the identification of solution components as \"solute\" and \"solvent,\" the components of a colloid are likewise classified according to their relative amounts. The particulate component typically present in a relatively minor amount is called the dispersed phase and the substance or solution throughout which the particulate is dispersed is called the dispersion medium. Colloids may involve virtually any combination of physical states (gas in liquid, liquid in solid, solid in gas, etc.), as illustrated by the examples of colloidal systems given in Table 11.4.\n\nExamples of Colloidal Systems\n\n| Dispersed Phase | Dispersion Medium | Common Examples | Name |\n| :--- | :--- | :--- | :--- |\n| solid | gas | smoke, dust | - |\n| solid | liquid | starch in water, some inks, paints, milk of magnesia | sol |\n| solid | solid | some colored gems, some alloys | - |\n| liquid | liquid | clouds, fogs, mists, sprays | aerosol |\n| liquid | solid | milk, mayonnaise, butter | emulsion |\n| liquid | jellies, gels, pearl, opal $\\left(\\mathrm{H}_{2} \\mathrm{O}\\right.$ in $\\left.\\mathrm{SiO}_{2}\\right)$ | gel |  |\n\nTABLE 11.4\n\n| Dispersed Phase | Dispersion Medium | Common Examples | Name |\n| :--- | :--- | :--- | :--- |\n| gas | liquid | foams, whipped cream, beaten egg whites | foam |\n| gas | solid | pumice, floating soaps | - |"}
{"id": 3790, "contents": "1281. LEARNING OBJECTIVES - \nTABLE 11.4"}
{"id": 3791, "contents": "1282. Preparation of Colloidal Systems - \nColloids are prepared by producing particles of colloidal dimensions and distributing these particles throughout a dispersion medium. Particles of colloidal size are formed by two methods:\n\n1. Dispersion methods: breaking down larger particles. For example, paint pigments are produced by dispersing large particles by grinding in special mills.\n2. Condensation methods: growth from smaller units, such as molecules or ions. For example, clouds form when water molecules condense and form very small droplets.\n\nA few solid substances, when brought into contact with water, disperse spontaneously and form colloidal systems. Gelatin, glue, starch, and dehydrated milk powder behave in this manner. The particles are already of colloidal size; the water simply disperses them. Powdered milk particles of colloidal size are produced by dehydrating milk spray. Some atomizers produce colloidal dispersions of a liquid in air.\n\nAn emulsion may be prepared by shaking together or blending two immiscible liquids. This breaks one liquid into droplets of colloidal size, which then disperse throughout the other liquid. Oil spills in the ocean may be difficult to clean up, partly because wave action can cause the oil and water to form an emulsion. In many emulsions, however, the dispersed phase tends to coalesce, form large drops, and separate. Therefore, emulsions are usually stabilized by an emulsifying agent, a substance that inhibits the coalescence of the dispersed liquid. For example, a little soap will stabilize an emulsion of kerosene in water. Milk is an emulsion of butterfat in water, with the protein casein serving as the emulsifying agent. Mayonnaise is an emulsion of oil in vinegar, with egg yolk components as the emulsifying agents.\n\nCondensation methods form colloidal particles by aggregation of molecules or ions. If the particles grow beyond the colloidal size range, drops or precipitates form, and no colloidal system results. Clouds form when water molecules aggregate and form colloid-sized particles. If these water particles coalesce to form adequately large water drops of liquid water or crystals of solid water, they settle from the sky as rain, sleet, or snow. Many condensation methods involve chemical reactions. A red colloidal suspension of iron(III) hydroxide may be prepared by mixing a concentrated solution of iron(III) chloride with hot water:"}
{"id": 3792, "contents": "1282. Preparation of Colloidal Systems - \n$$\n\\mathrm{Fe}^{3+}(\\mathrm{aq})+3 \\mathrm{Cl}^{-}(\\mathrm{aq})+6 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Fe}(\\mathrm{OH})_{3}(s)+\\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+3 \\mathrm{Cl}^{-}(a q) .\n$$\n\nA colloidal gold sol results from the reduction of a very dilute solution of gold(III) chloride by a reducing agent such as formaldehyde, tin(II) chloride, or iron(II) sulfate:\n\n$$\n\\mathrm{Au}^{3+}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Au}\n$$\n\nSome gold sols prepared in 1857 are still intact (the particles have not coalesced and settled), illustrating the long-term stability of many colloids."}
{"id": 3793, "contents": "1283. Soaps and Detergents - \nPioneers made soap by boiling fats with a strongly basic solution made by leaching potassium carbonate, $\\mathrm{K}_{2} \\mathrm{CO}_{3}$, from wood ashes with hot water. Animal fats contain polyesters of fatty acids (long-chain carboxylic acids). When animal fats are treated with a base like potassium carbonate or sodium hydroxide, glycerol and salts of fatty acids such as palmitic, oleic, and stearic acid are formed. The salts of fatty acids are called soaps. The sodium salt of stearic acid, sodium stearate, has the formula $\\mathrm{C}_{17} \\mathrm{H}_{35} \\mathrm{CO}_{2} \\mathrm{Na}$ and contains an uncharged nonpolar hydrocarbon chain, the $\\mathrm{C}_{17} \\mathrm{H}_{35}$ - unit, and an ionic carboxylate group, the $-\\mathrm{CO}_{2}{ }^{-}$unit (Figure 11.31).\n\n\nsodium stearate (soap)\nFIGURE 11.31 Soaps contain a nonpolar hydrocarbon end (blue) and an ionic end (red). The ionic end is a carboxylate group. The length of the hydrocarbon end can vary from soap to soap.\n\nDetergents (soap substitutes) also contain nonpolar hydrocarbon chains, such as $\\mathrm{C}_{12} \\mathrm{H}_{25}$-, and an ionic group, such as a sulfate- $\\mathrm{OSO}_{3}{ }^{-}$, or a sulfonate $-\\mathrm{SO}_{3}{ }^{-}$(Figure 11.32). Soaps form insoluble calcium and magnesium compounds in hard water; detergents form water-soluble products-a definite advantage for detergents.\n\n\nFIGURE 11.32 Detergents contain a nonpolar hydrocarbon end (blue) and an ionic end (red). The ionic end can be either a sulfate or a sulfonate. The length of the hydrocarbon end can vary from detergent to detergent."}
{"id": 3794, "contents": "1283. Soaps and Detergents - \nThe cleaning action of soaps and detergents can be explained in terms of the structures of the molecules involved. The hydrocarbon (nonpolar) end of a soap or detergent molecule dissolves in, or is attracted to, nonpolar substances such as oil, grease, or dirt particles. The ionic end is attracted by water (polar), illustrated in Figure 11.33. As a result, the soap or detergent molecules become oriented at the interface between the dirt particles and the water so they act as a kind of bridge between two different kinds of matter, nonpolar and polar. Molecules such as this are termed amphiphilic since they have both a hydrophobic (\"water-fearing\") part and a hydrophilic (\"water-loving\") part. As a consequence, dirt particles become suspended as colloidal particles and are readily washed away.\n\n\nFIGURE 11.33 This diagrammatic cross section of an emulsified drop of oil in water shows how soap or detergent acts as an emulsifier."}
{"id": 3795, "contents": "1285. Deepwater Horizon Oil Spill - \nThe blowout of the Deepwater Horizon oil drilling rig on April 20, 2010, in the Gulf of Mexico near Mississippi began the largest marine oil spill in the history of the petroleum industry. In the 87 days following the blowout, an estimated 4.9 million barrels ( 210 million gallons) of oil flowed from the ruptured well 5000 feet below the water's surface. The well was finally declared sealed on September 19, 2010.\n\nCrude oil is immiscible with and less dense than water, so the spilled oil rose to the surface of the water.\n\nFloating booms, skimmer ships, and controlled burns were used to remove oil from the water's surface in an attempt to protect beaches and wetlands along the Gulf coast. In addition to removal of the oil, attempts were also made to lessen its environmental impact by rendering it \"soluble\" (in the loose sense of the term) and thus allowing it to be diluted to hopefully less harmful levels by the vast volume of ocean water. This approach used 1.84 million gallons of the oil dispersant Corexit 9527, most of which was injected underwater at the site of the leak, with small amounts being sprayed on top of the spill. Corexit 9527 contains 2-butoxyethanol $\\left(\\mathrm{C}_{6} \\mathrm{H}_{14} \\mathrm{O}_{2}\\right)$, an amphiphilic molecule whose polar and nonpolar ends are useful for emulsifying oil into small droplets, increasing the surface area of the oil and making it more available to marine bacteria for digestion (Figure 11.34). While this approach avoids many of the immediate hazards that bulk oil poses to marine and coastal ecosystems, it introduces the possibility of long-term effects resulting from the introduction of the complex and potential toxic components of petroleum into the ocean's food chain. A number of organizations are involved in monitoring the extended impact of this oil spill, including the National Oceanic and Atmospheric Administration (visit this website (http://openstax.org/l/16gulfspill) for additional details)."}
{"id": 3796, "contents": "1285. Deepwater Horizon Oil Spill - \nFIGURE 11.34 (a) This NASA satellite image shows the oil slick from the Deepwater Horizon spill. (b) A US Air Force plane sprays Corexit, a dispersant. (c) The molecular structure of 2-butoxyethanol is shown. (credit a: modification of work by \"NASA, FT2, demis.nl\"/Wikimedia Commons; credit b: modification of work by \"NASA/ MODIS Rapid Response Team\"/Wikimedia Commons)"}
{"id": 3797, "contents": "1286. Electrical Properties of Colloidal Particles - \nDispersed colloidal particles are often electrically charged. A colloidal particle of iron(III) hydroxide, for example, does not contain enough hydroxide ions to compensate exactly for the positive charges on the iron(III) ions. Thus, each individual colloidal particle bears a positive charge, and the colloidal dispersion consists of charged colloidal particles and some free hydroxide ions, which keep the dispersion electrically neutral. Most metal hydroxide colloids have positive charges, whereas most metals and metal sulfides form negatively charged dispersions. All colloidal particles in any one system have charges of the same sign. This helps keep them dispersed because particles containing like charges repel each other.\n\nThe charged nature of some colloidal particles may be exploited to remove them from a variety of mixtures. For example, the particles comprising smoke are often colloidally dispersed and electrically charged.\nFrederick Cottrell, an American chemist, developed a process to remove these particles. The charged particles are attracted to highly charged electrodes, where they are neutralized and deposited as dust (Figure 11.36). This is one of the important methods used to clean up the smoke from a variety of industrial processes. The process is also important in the recovery of valuable products from the smoke and flue dust of smelters, furnaces, and kilns. There are also similar electrostatic air filters designed for home use to improve indoor air quality."}
{"id": 3798, "contents": "1288. Frederick Gardner Cottrell - \n(a)\n\n(b)\n\nFIGURE 11.35 (a) Frederick Cottrell developed (b) the electrostatic precipitator, a device designed to curb air pollution by removing colloidal particles from air. (credit b: modification of work by \"SpLot\"/Wikimedia Commons)\n\nBorn in Oakland, CA, in 1877, Frederick Cottrell devoured textbooks as if they were novels and graduated from high school at the age of 16. He then entered the University of California (UC), Berkeley, completing a Bachelor's degree in three years. He saved money from his \\$1200 annual salary as a chemistry teacher at Oakland High School to fund his studies in chemistry in Berlin with Nobel prize winner Jacobus Henricus van't Hoff, and in Leipzig with Wilhelm Ostwald, another Nobel awardee. After earning his PhD in physical chemistry, he returned to the United States to teach at UC Berkeley. He also consulted for the DuPont Company, where he developed the electrostatic precipitator, a device designed to curb air pollution by removing colloidal particles from air. Cottrell used the proceeds from his invention to fund a nonprofit research corporation to finance scientific research.\n\n\nFIGURE 11.36 In a Cottrell precipitator, positively and negatively charged particles are attracted to highly charged electrodes, where they are neutralized and deposited as dust."}
{"id": 3799, "contents": "1289. Gels - \nGelatin desserts, such as Jell-O, are a type of colloid (Figure 11.37). Gelatin sets on cooling because the hot aqueous mixture of gelatin coagulates as it cools, yielding an extremely viscous body known as a gel. A gel is a colloidal dispersion of a liquid phase throughout a solid phase. It appears that the fibers of the dispersing medium form a complex three-dimensional network, the interstices being filled with the liquid medium or a dilute solution of the dispersing medium.\n\n\nFIGURE 11.37 Gelatin desserts are colloids in which an aqueous solution of sweeteners and flavors is dispersed throughout a medium of solid proteins. (credit photo: modification of work by Steven Depolo)\n\nPectin, a carbohydrate from fruit juices, is a gel-forming substance important in jelly making. Silica gel, a colloidal dispersion of hydrated silicon dioxide, is formed when dilute hydrochloric acid is added to a dilute solution of sodium silicate. Canned Heat is a flammable gel made by mixing alcohol and a saturated aqueous solution of calcium acetate."}
{"id": 3800, "contents": "1290. Key Terms - \nalloy solid mixture of a metallic element and one or more additional elements\namphiphilic molecules possessing both hydrophobic (nonpolar) and a hydrophilic (polar) parts\nboiling point elevation elevation of the boiling point of a liquid by addition of a solute\nboiling point elevation constant the proportionality constant in the equation relating boiling point elevation to solute molality; also known as the ebullioscopic constant\ncolligative property property of a solution that depends only on the concentration of a solute species\ncolloid (also, colloidal dispersion) mixture in which relatively large solid or liquid particles are dispersed uniformly throughout a gas, liquid, or solid\ncrenation process whereby biological cells become shriveled due to loss of water by osmosis\ndispersed phase substance present as relatively large solid or liquid particles in a colloid\ndispersion medium solid, liquid, or gas in which colloidal particles are dispersed\ndissociation physical process accompanying the dissolution of an ionic compound in which the compound's constituent ions are solvated and dispersed throughout the solution\nelectrolyte substance that produces ions when dissolved in water\nemulsifying agent amphiphilic substance used to stabilize the particles of some emulsions\nemulsion colloid formed from immiscible liquids\nfreezing point depression lowering of the freezing point of a liquid by addition of a solute\nfreezing point depression constant (also, cryoscopic constant) proportionality constant in the equation relating freezing point depression to solute molality\ngel colloidal dispersion of a liquid in a solid\nhemolysis rupture of red blood cells due to the accumulation of excess water by osmosis\nHenry's law the proportional relationship between the concentration of dissolved gas in a solution and the partial pressure of the gas in contact with the solution\nhypertonic of greater osmotic pressure\nhypotonic of less osmotic pressure\nideal solution solution that forms with no accompanying energy change\nimmiscible of negligible mutual solubility; typically refers to liquid substances\nion pair solvated anion/cation pair held together by moderate electrostatic attraction\nion-dipole attraction electrostatic attraction between an ion and a polar molecule\nisotonic of equal osmotic pressure\nmiscible mutually soluble in all proportions; typically refers to liquid substances"}
{"id": 3801, "contents": "1290. Key Terms - \nion-dipole attraction electrostatic attraction between an ion and a polar molecule\nisotonic of equal osmotic pressure\nmiscible mutually soluble in all proportions; typically refers to liquid substances\nmolality ( $\\boldsymbol{m}$ ) a concentration unit defined as the ratio of the numbers of moles of solute to the mass of the solvent in kilograms\nnonelectrolyte substance that does not produce ions when dissolved in water\nosmosis diffusion of solvent molecules through a semipermeable membrane\nosmotic pressure (II) opposing pressure required to prevent bulk transfer of solvent molecules through a semipermeable membrane\npartially miscible of moderate mutual solubility; typically refers to liquid substances\nRaoult's law the relationship between a solution's vapor pressure and the vapor pressures and concentrations of its components\nsaturated of concentration equal to solubility; containing the maximum concentration of solute possible for a given temperature and pressure\nsemipermeable membrane a membrane that selectively permits passage of certain ions or molecules\nsolubility extent to which a solute may be dissolved in water, or any solvent\nsolvation exothermic process in which intermolecular attractive forces between the solute and solvent in a solution are established\nspontaneous process physical or chemical change that occurs without the addition of energy from an external source\nstrong electrolyte substance that dissociates or ionizes completely when dissolved in water\nsupersaturated of concentration that exceeds solubility; a nonequilibrium state\nsuspension heterogeneous mixture in which relatively large component particles are temporarily dispersed but settle out over time\nTyndall effect scattering of visible light by a colloidal dispersion\nunsaturated of concentration less than solubility\nvan't Hoff factor (i) the ratio of the number of moles of particles in a solution to the number of moles of formula units dissolved in the solution\nweak electrolyte substance that ionizes only partially when dissolved in water"}
{"id": 3802, "contents": "1290. Key Terms - \nKey Equations\n$C_{\\mathrm{g}}=k P_{\\mathrm{g}}$\n$\\left(P_{\\mathrm{A}}=X_{\\mathrm{A}} P_{\\mathrm{A}}^{*}\\right)$\n$P_{\\text {solution }}=\\sum_{i} P_{i}=\\sum_{i} X_{i} P_{i}^{*}$\n$P_{\\text {solution }}=X_{\\text {solvent }} P_{\\text {solvent }}^{*}$\n$\\Delta T_{\\mathrm{b}}=K_{\\mathrm{b}} m$\n$\\Delta T_{\\mathrm{f}}=K_{\\mathrm{f}} m$\n$\\Pi=M R T$"}
{"id": 3803, "contents": "1291. Summary - 1291.1. The Dissolution Process\nA solution forms when two or more substances combine physically to yield a mixture that is homogeneous at the molecular level. The solvent is the most concentrated component and determines the physical state of the solution. The solutes are the other components typically present at concentrations less than that of the solvent. Solutions may form endothermically or exothermically, depending upon the relative magnitudes of solute and solvent intermolecular attractive forces. Ideal solutions form with no appreciable change in energy."}
{"id": 3804, "contents": "1291. Summary - 1291.2. Electrolytes\nSubstances that dissolve in water to yield ions are called electrolytes. Electrolytes may be covalent compounds that chemically react with water to produce ions (for example, acids and bases), or they may be ionic compounds that dissociate to yield their constituent cations and anions, when dissolved. Dissolution of an ionic compound is facilitated by ion-dipole attractions between the ions of the compound and the polar water molecules. Soluble ionic substances and strong acids ionize completely and are strong electrolytes, while weak acids and bases ionize to only a small extent and are weak electrolytes. Nonelectrolytes are substances that do not produce ions when dissolved in water."}
{"id": 3805, "contents": "1291. Summary - 1291.3. Solubility\nThe extent to which one substance will dissolve in another is determined by several factors, including the types and relative strengths of intermolecular attractive forces that may exist between the substances' atoms, ions, or molecules. This tendency to dissolve is quantified as a substance's solubility, its maximum concentration in a solution at equilibrium under specified conditions. A\nsaturated solution contains solute at a concentration equal to its solubility. A supersaturated solution is one in which a solute's concentration exceeds its solubility-a nonequilibrium (unstable) condition that will result in solute precipitation when the solution is appropriately perturbed. Miscible liquids are soluble in all proportions, and immiscible liquids exhibit very low mutual solubility. Solubilities for gaseous solutes decrease with increasing temperature, while those for most, but not all, solid solutes increase with temperature. The concentration of a gaseous solute in a solution is proportional to the partial pressure of the gas to which the solution is exposed, a relation known as Henry's law."}
{"id": 3806, "contents": "1291. Summary - 1291.4. Colligative Properties\nProperties of a solution that depend only on the concentration of solute particles are called colligative properties. They include changes in the vapor pressure, boiling point, and freezing point of the solvent in the solution. The magnitudes of these properties depend only on the total concentration of solute particles in solution, not on the type of particles. The total concentration of solute particles in a solution also determines its osmotic pressure. This is the pressure that must be applied to the solution to prevent diffusion of molecules of pure solvent through a semipermeable membrane into the solution. Ionic compounds may not completely dissociate in solution due to activity effects, in which case observed colligative effects may be less than predicted."}
{"id": 3807, "contents": "1291. Summary - 1291.5. Colloids\nColloids are mixtures in which one or more substances are dispersed as relatively large solid particles or liquid droplets throughout a solid, liquid, or gaseous medium. The particles of a colloid remain dispersed and do not settle due to gravity,\nand they are often electrically charged. Colloids are widespread in nature and are involved in many\ntechnological applications."}
{"id": 3808, "contents": "1292. Exercises - 1292.1. The Dissolution Process\n1. How do solutions differ from compounds? From other mixtures?\n2. Which of the principal characteristics of solutions are evident in the solutions of $\\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$ shown in Figure 11.2?\n3. When $\\mathrm{KNO}_{3}$ is dissolved in water, the resulting solution is significantly colder than the water was originally.\n(a) Is the dissolution of $\\mathrm{KNO}_{3}$ an endothermic or an exothermic process?\n(b) What conclusions can you draw about the intermolecular attractions involved in the process?\n(c) Is the resulting solution an ideal solution?\n4. Give an example of each of the following types of solutions:\n(a) a gas in a liquid\n(b) a gas in a gas\n(c) a solid in a solid\n5. Indicate the most important types of intermolecular attractions in each of the following solutions:\n(a) The solution in Figure 11.2.\n(b) $\\mathrm{NO}(I)$ in $\\mathrm{CO}(I)$\n(c) $\\mathrm{Cl}_{2}(\\mathrm{~g})$ in $\\mathrm{Br}_{2}(\\mathrm{I})$\n(d) $\\mathrm{HCl}(g)$ in benzene $\\mathrm{C}_{6} \\mathrm{H}_{6}(\\mathrm{I})$\n(e) Methanol $\\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{I})$ in $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})$\n6. Predict whether each of the following substances would be more soluble in water (polar solvent) or in a hydrocarbon such as heptane ( $\\mathrm{C}_{7} \\mathrm{H}_{16}$, nonpolar solvent):\n(a) vegetable oil (nonpolar)\n(b) isopropyl alcohol (polar)\n(c) potassium bromide (ionic)\n7. Heat is released when some solutions form; heat is absorbed when other solutions form. Provide a molecular explanation for the difference between these two types of spontaneous processes."}
{"id": 3809, "contents": "1292. Exercises - 1292.1. The Dissolution Process\n(c) potassium bromide (ionic)\n7. Heat is released when some solutions form; heat is absorbed when other solutions form. Provide a molecular explanation for the difference between these two types of spontaneous processes.\n8. Solutions of hydrogen in palladium may be formed by exposing Pd metal to $\\mathrm{H}_{2}$ gas. The concentration of hydrogen in the palladium depends on the pressure of $\\mathrm{H}_{2}$ gas applied, but in a more complex fashion than can be described by Henry's law. Under certain conditions, 0.94 g of hydrogen gas is dissolved in 215 g of palladium metal (solution density $=10.8 \\mathrm{~g} \\mathrm{~cm}^{3}$ ).\n(a) Determine the molarity of this solution.\n(b) Determine the molality of this solution.\n(c) Determine the percent by mass of hydrogen atoms in this solution."}
{"id": 3810, "contents": "1292. Exercises - 1292.2. Electrolytes\n9. Explain why the ions $\\mathrm{Na}^{+}$and $\\mathrm{Cl}^{-}$are strongly solvated in water but not in hexane, a solvent composed of nonpolar molecules.\n10. Explain why solutions of HBr in benzene (a nonpolar solvent) are nonconductive, while solutions in water (a polar solvent) are conductive.\n11. Consider the solutions presented:\n(a) Which of the following sketches best represents the ions in a solution of $\\mathrm{Fe}\\left(\\mathrm{NO}_{3}\\right)_{3}(\\mathrm{aq})$ ?\n\n(x)\n\n(y)"}
{"id": 3811, "contents": "1292. Exercises - 1292.2. Electrolytes\n(x)\n\n(y)\n\n(z)\n(b) Write a balanced chemical equation showing the products of the dissolution of $\\mathrm{Fe}\\left(\\mathrm{NO}_{3}\\right)_{3}$.\n12. Compare the processes that occur when methanol $\\left(\\mathrm{CH}_{3} \\mathrm{OH}\\right)$, hydrogen chloride $(\\mathrm{HCl})$, and sodium hydroxide $(\\mathrm{NaOH})$ dissolve in water. Write equations and prepare sketches showing the form in which each of these compounds is present in its respective solution.\n13. What is the expected electrical conductivity of the following solutions?\n(a) $\\mathrm{NaOH}(a q)$\n(b) $\\mathrm{HCl}(a q)$\n(c) $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}(a q)$ (glucose)\n(d) $\\mathrm{NH}_{3}(\\mathrm{aq})$\n14. Why are most solid ionic compounds electrically nonconductive, whereas aqueous solutions of ionic compounds are good conductors? Would you expect a liquid (molten) ionic compound to be electrically conductive or nonconductive? Explain.\n15. Indicate the most important type of intermolecular attraction responsible for solvation in each of the following solutions:\n(a) the solutions in Figure 11.7\n(b) methanol, $\\mathrm{CH}_{3} \\mathrm{OH}$, dissolved in ethanol, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$\n(c) methane, $\\mathrm{CH}_{4}$, dissolved in benzene, $\\mathrm{C}_{6} \\mathrm{H}_{6}$\n(d) the polar halocarbon $\\mathrm{CF}_{2} \\mathrm{Cl}_{2}$ dissolved in the polar halocarbon $\\mathrm{CF}_{2} \\mathrm{ClCFCl}_{2}$\n(e) $\\mathrm{O}_{2}(I)$ in $\\mathrm{N}_{2}(I)$"}
{"id": 3812, "contents": "1292. Exercises - 1292.3. Solubility\n16. Suppose you are presented with a clear solution of sodium thiosulfate, $\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$. How could you determine whether the solution is unsaturated, saturated, or supersaturated?\n17. Supersaturated solutions of most solids in water are prepared by cooling saturated solutions. Supersaturated solutions of most gases in water are prepared by heating saturated solutions. Explain the reasons for the difference in the two procedures.\n18. Suggest an explanation for the observations that ethanol, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$, is completely miscible with water and that ethanethiol, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{SH}$, is soluble only to the extent of 1.5 g per 100 mL of water.\n19. Calculate the percent by mass of KBr in a saturated solution of KBr in water at $10^{\\circ} \\mathrm{C}$. See Figure 11.16 for useful data, and report the computed percentage to one significant digit.\n20. Which of the following gases is expected to be most soluble in water? Explain your reasoning.\n(a) $\\mathrm{CH}_{4}$\n(b) $\\mathrm{CCl}_{4}$\n(c) $\\mathrm{CHCl}_{3}$\n21. At $0^{\\circ} \\mathrm{C}$ and 1.00 atm , as much as 0.70 g of $\\mathrm{O}_{2}$ can dissolve in 1 L of water. At $0^{\\circ} \\mathrm{C}$ and 4.00 atm , how many grams of $\\mathrm{O}_{2}$ dissolve in 1 L of water?\n22. Refer to Figure 11.10 .\n(a) How did the concentration of dissolved $\\mathrm{CO}_{2}$ in the beverage change when the bottle was opened?\n(b) What caused this change?\n(c) Is the beverage unsaturated, saturated, or supersaturated with $\\mathrm{CO}_{2}$ ?"}
{"id": 3813, "contents": "1292. Exercises - 1292.3. Solubility\n(b) What caused this change?\n(c) Is the beverage unsaturated, saturated, or supersaturated with $\\mathrm{CO}_{2}$ ?\n23. The Henry's law constant for $\\mathrm{CO}_{2}$ is $3.4 \\times 10^{-2} \\mathrm{M} / \\mathrm{atm}$ at $25^{\\circ} \\mathrm{C}$. Assuming ideal solution behavior, what pressure of carbon dioxide is needed to maintain a $\\mathrm{CO}_{2}$ concentration of 0.10 M in a can of lemon-lime soda?\n24. The Henry's law constant for $\\mathrm{O}_{2}$ is $1.3 \\times 10^{-3} \\mathrm{M} / \\mathrm{atm}$ at $25^{\\circ} \\mathrm{C}$. Assuming ideal solution behavior, what mass of oxygen would be dissolved in a $40-\\mathrm{L}$ aquarium at $25^{\\circ} \\mathrm{C}$, assuming an atmospheric pressure of 1.00 atm , and that the partial pressure of $\\mathrm{O}_{2}$ is 0.21 atm ?\n25. Assuming ideal solution behavior, how many liters of HCl gas, measured at $30.0^{\\circ} \\mathrm{C}$ and 745 torr, are required to prepare 1.25 L of a $3.20-M$ solution of hydrochloric acid?"}
{"id": 3814, "contents": "1292. Exercises - 1292.4. Colligative Properties\n26. Which is/are part of the macroscopic domain of solutions and which is/are part of the microscopic domain: boiling point elevation, Henry's law, hydrogen bond, ion-dipole attraction, molarity, nonelectrolyte, nonstoichiometric compound, osmosis, solvated ion?\n27. What is the microscopic explanation for the macroscopic behavior illustrated in Figure 11.14?\n28. Sketch a qualitative graph of the pressure versus time for water vapor above a sample of pure water and a sugar solution, as the liquids evaporate to half their original volume.\n29. A solution of potassium nitrate, an electrolyte, and a solution of glycerin $\\left(\\mathrm{C}_{3} \\mathrm{H}_{5}(\\mathrm{OH})_{3}\\right)$, a nonelectrolyte, both boil at $100.3^{\\circ} \\mathrm{C}$. What other physical properties of the two solutions are identical?\n30. What are the mole fractions of $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ and water in a solution of 14.5 g of $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ in 125 g of water?\n(a) Outline the steps necessary to answer the question.\n(b) Answer the question.\n31. What are the mole fractions of $\\mathrm{HNO}_{3}$ and water in a concentrated solution of nitric acid $\\left(68.0 \\% \\mathrm{HNO}_{3}\\right.$ by mass)?\n(a) Outline the steps necessary to answer the question.\n(b) Answer the question.\n32. Calculate the mole fraction of each solute and solvent:\n(a) 583 g of $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ in 1.50 kg of water-the acid solution used in an automobile battery\n(b) 0.86 g of NaCl in $1.00 \\times 10^{2} \\mathrm{~g}$ of water-a solution of sodium chloride for intravenous injection"}
{"id": 3815, "contents": "1292. Exercises - 1292.4. Colligative Properties\n(b) 0.86 g of NaCl in $1.00 \\times 10^{2} \\mathrm{~g}$ of water-a solution of sodium chloride for intravenous injection\n(c) 46.85 g of codeine, $\\mathrm{C}_{18} \\mathrm{H}_{21} \\mathrm{NO}_{3}$, in 125.5 g of ethanol, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$\n(d) 25 g of $\\mathrm{I}_{2}$ in 125 g of ethanol, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$\n33. Calculate the mole fraction of each solute and solvent:\n(a) 0.710 kg of sodium carbonate (washing soda), $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$, in 10.0 kg of water-a saturated solution at $0{ }^{\\circ} \\mathrm{C}$\n(b) 125 g of $\\mathrm{NH}_{4} \\mathrm{NO}_{3}$ in 275 g of water-a mixture used to make an instant ice pack\n(c) 25 g of $\\mathrm{Cl}_{2}$ in 125 g of dichloromethane, $\\mathrm{CH}_{2} \\mathrm{Cl}_{2}$\n(d) 0.372 g of tetrahydropyridine, $\\mathrm{C}_{5} \\mathrm{H}_{9} \\mathrm{~N}$, in 125 g of chloroform, $\\mathrm{CHCl}_{3}$\n34. Calculate the mole fractions of methanol, $\\mathrm{CH}_{3} \\mathrm{OH}$; ethanol, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$; and water in a solution that is $40 \\%$ methanol, $40 \\%$ ethanol, and $20 \\%$ water by mass. (Assume the data are good to two significant figures.)\n35. What is the difference between a $1 M$ solution and a 1 m solution?"}
{"id": 3816, "contents": "1292. Exercises - 1292.4. Colligative Properties\n35. What is the difference between a $1 M$ solution and a 1 m solution?\n36. What is the molality of phosphoric acid, $\\mathrm{H}_{3} \\mathrm{PO}_{4}$, in a solution of 14.5 g of $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ in 125 g of water?\n(a) Outline the steps necessary to answer the question.\n(b) Answer the question.\n37. What is the molality of nitric acid in a concentrated solution of nitric acid ( $68.0 \\% \\mathrm{HNO}_{3}$ by mass)?\n(a) Outline the steps necessary to answer the question.\n(b) Answer the question.\n38. Calculate the molality of each of the following solutions:\n(a) 583 g of $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ in 1.50 kg of water-the acid solution used in an automobile battery\n(b) 0.86 g of NaCl in $1.00 \\times 10^{2} \\mathrm{~g}$ of water-a solution of sodium chloride for intravenous injection\n(c) 46.85 g of codeine, $\\mathrm{C}_{18} \\mathrm{H}_{21} \\mathrm{NO}_{3}$, in 125.5 g of ethanol, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$\n(d) 25 g of $\\mathrm{I}_{2}$ in 125 g of ethanol, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$\n39. Calculate the molality of each of the following solutions:\n(a) 0.710 kg of sodium carbonate (washing soda), $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$, in 10.0 kg of water-a saturated solution at $0^{\\circ} \\mathrm{C}$\n(b) 125 g of $\\mathrm{NH}_{4} \\mathrm{NO}_{3}$ in 275 g of water-a mixture used to make an instant ice pack\n(c) 25 g of $\\mathrm{Cl}_{2}$ in 125 g of dichloromethane, $\\mathrm{CH}_{2} \\mathrm{Cl}_{2}$"}
{"id": 3817, "contents": "1292. Exercises - 1292.4. Colligative Properties\n(c) 25 g of $\\mathrm{Cl}_{2}$ in 125 g of dichloromethane, $\\mathrm{CH}_{2} \\mathrm{Cl}_{2}$\n(d) 0.372 g of tetrahydropyridine, $\\mathrm{C}_{5} \\mathrm{H}_{9} \\mathrm{~N}$, in 125 g of chloroform, $\\mathrm{CHCl}_{3}$\n40. The concentration of glucose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$, in normal spinal fluid is $\\frac{75 \\mathrm{mg}}{100 \\mathrm{~g}}$. What is the molality of the solution?\n41. A $13.0 \\%$ solution of $\\mathrm{K}_{2} \\mathrm{CO}_{3}$ by mass has a density of $1.09 \\mathrm{~g} / \\mathrm{cm}^{3}$. Calculate the molality of the solution.\n42. Why does 1 mol of sodium chloride depress the freezing point of 1 kg of water almost twice as much as 1 mol of glycerin?\n43. Assuming ideal solution behavior, what is the boiling point of a solution of 115.0 g of nonvolatile sucrose, $\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}$, in 350.0 g of water?\n(a) Outline the steps necessary to answer the question\n(b) Answer the question\n44. Assuming ideal solution behavior, what is the boiling point of a solution of 9.04 g of $\\mathrm{I}_{2}$ in 75.5 g of benzene, assuming the $\\mathrm{I}_{2}$ is nonvolatile?\n(a) Outline the steps necessary to answer the question.\n(b) Answer the question.\n45. Assuming ideal solution behavior, what is the freezing temperature of a solution of 115.0 g of sucrose, $\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}$, in 350.0 g of water?\n(a) Outline the steps necessary to answer the question.\n(b) Answer the question."}
{"id": 3818, "contents": "1292. Exercises - 1292.4. Colligative Properties\n(a) Outline the steps necessary to answer the question.\n(b) Answer the question.\n46. Assuming ideal solution behavior, what is the freezing point of a solution of 9.04 g of $\\mathrm{I}_{2}$ in 75.5 g of benzene?\n(a) Outline the steps necessary to answer the following question.\n(b) Answer the question.\n47. Assuming ideal solution behavior, what is the osmotic pressure of an aqueous solution of 1.64 g of $\\mathrm{Ca}\\left(\\mathrm{NO}_{3}\\right)_{2}$ in water at $25^{\\circ} \\mathrm{C}$ ? The volume of the solution is 275 mL .\n(a) Outline the steps necessary to answer the question.\n(b) Answer the question.\n48. Assuming ideal solution behavior, what is osmotic pressure of a solution of bovine insulin (molar mass, $5700 \\mathrm{~g} \\mathrm{~mol}^{-1}$ ) at $18^{\\circ} \\mathrm{C}$ if 100.0 mL of the solution contains 0.103 g of the insulin?\n(a) Outline the steps necessary to answer the question.\n(b) Answer the question.\n49. Assuming ideal solution behavior, what is the molar mass of a solution of 5.00 g of a compound in 25.00 g of carbon tetrachloride (bp $76.8^{\\circ} \\mathrm{C} ; K_{\\mathrm{b}}=5.02^{\\circ} \\mathrm{C} / \\mathrm{m}$ ) that boils at $81.5^{\\circ} \\mathrm{C}$ at 1 atm ?\n(a) Outline the steps necessary to answer the question.\n(b) Solve the problem.\n50. A sample of an organic compound (a nonelectrolyte) weighing 1.35 g lowered the freezing point of 10.0 g of benzene by $3.66^{\\circ} \\mathrm{C}$. Assuming ideal solution behavior, calculate the molar mass of the compound.\n51. A 1.0 m solution of HCl in benzene has a freezing point of $0.4^{\\circ} \\mathrm{C}$. Is HCl an electrolyte in benzene? Explain."}
{"id": 3819, "contents": "1292. Exercises - 1292.4. Colligative Properties\n51. A 1.0 m solution of HCl in benzene has a freezing point of $0.4^{\\circ} \\mathrm{C}$. Is HCl an electrolyte in benzene? Explain.\n52. A solution contains 5.00 g of urea, $\\mathrm{CO}\\left(\\mathrm{NH}_{2}\\right)_{2}$, a nonvolatile compound, dissolved in 0.100 kg of water. If the vapor pressure of pure water at $25^{\\circ} \\mathrm{C}$ is 23.7 torr, what is the vapor pressure of the solution (assuming ideal solution behavior)?\n53. A $12.0-\\mathrm{g}$ sample of a nonelectrolyte is dissolved in 80.0 g of water. The solution freezes at $-1.94{ }^{\\circ} \\mathrm{C}$. Assuming ideal solution behavior, calculate the molar mass of the substance.\n54. Arrange the following solutions in order by their decreasing freezing points: $0.1 \\mathrm{~m} \\mathrm{Na}_{3} \\mathrm{PO}_{4}, 0.1 \\mathrm{~m}_{2} \\mathrm{H}_{5} \\mathrm{OH}$, $0.01 \\mathrm{~m} \\mathrm{CO}_{2}, 0.15 \\mathrm{~m} \\mathrm{NaCl}$, and $0.2 \\mathrm{~m} \\mathrm{CaCl}_{2}$.\n55. Calculate the boiling point elevation of 0.100 kg of water containing 0.010 mol of $\\mathrm{NaCl}, 0.020 \\mathrm{~mol}$ of $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$, and 0.030 mol of $\\mathrm{MgCl}_{2}$, assuming complete dissociation of these electrolytes and ideal solution behavior.\n56. How could you prepare a 3.08 m aqueous solution of glycerin, $\\mathrm{C}_{3} \\mathrm{H}_{8} \\mathrm{O}_{3}$ ? Assuming ideal solution behavior, what is the freezing point of this solution?"}
{"id": 3820, "contents": "1292. Exercises - 1292.4. Colligative Properties\n57. A sample of sulfur weighing 0.210 g was dissolved in 17.8 g of carbon disulfide, $\\mathrm{CS}_{2}\\left(K_{\\mathrm{b}}=2.34^{\\circ} \\mathrm{C} / \\mathrm{m}\\right)$. If the boiling point elevation was $0.107^{\\circ} \\mathrm{C}$, what is the formula of a sulfur molecule in carbon disulfide (assuming ideal solution behavior)?\n58. In a significant experiment performed many years ago, 5.6977 g of cadmium iodide in 44.69 g of water raised the boiling point $0.181{ }^{\\circ} \\mathrm{C}$. What does this suggest about the nature of a solution of $\\mathrm{CdI}_{2}$ ?\n59. Lysozyme is an enzyme that cleaves cell walls. A $0.100-\\mathrm{L}$ sample of a solution of lysozyme that contains 0.0750 g of the enzyme exhibits an osmotic pressure of $1.32 \\times 10^{-3}$ atm at $25^{\\circ} \\mathrm{C}$. Assuming ideal solution behavior, what is the molar mass of lysozyme?\n60. The osmotic pressure of a solution containing 7.0 g of insulin per liter is 23 torr at $25^{\\circ} \\mathrm{C}$. Assuming ideal solution behavior, what is the molar mass of insulin?\n61. The osmotic pressure of human blood is 7.6 atm at $37^{\\circ} \\mathrm{C}$. What mass of glucose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$, is required to make 1.00 L of aqueous solution for intravenous feeding if the solution must have the same osmotic pressure as blood at body temperature, $37^{\\circ} \\mathrm{C}$ (assuming ideal solution behavior)?"}
{"id": 3821, "contents": "1292. Exercises - 1292.4. Colligative Properties\n62. Assuming ideal solution behavior, what is the freezing point of a solution of dibromobenzene, $\\mathrm{C}_{6} \\mathrm{H}_{4} \\mathrm{Br}_{2}$, in 0.250 kg of benzene, if the solution boils at $83.5^{\\circ} \\mathrm{C}$ ?\n63. Assuming ideal solution behavior, what is the boiling point of a solution of NaCl in water if the solution freezes at $-0.93^{\\circ} \\mathrm{C}$ ?\n64. The sugar fructose contains $40.0 \\% \\mathrm{C}, 6.7 \\% \\mathrm{H}$, and $53.3 \\% \\mathrm{O}$ by mass. A solution of 11.7 g of fructose in 325 g of ethanol has a boiling point of $78.59^{\\circ} \\mathrm{C}$. The boiling point of ethanol is $78.35^{\\circ} \\mathrm{C}$, and $K_{\\mathrm{b}}$ for ethanol is $1.20^{\\circ} \\mathrm{C} / \\mathrm{m}$. Assuming ideal solution behavior, what is the molecular formula of fructose?\n65. The vapor pressure of methanol, $\\mathrm{CH}_{3} \\mathrm{OH}$, is 94 torr at $20^{\\circ} \\mathrm{C}$. The vapor pressure of ethanol, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$, is 44 torr at the same temperature.\n(a) Calculate the mole fraction of methanol and of ethanol in a solution of 50.0 g of methanol and 50.0 g of ethanol.\n(b) Ethanol and methanol form a solution that behaves like an ideal solution. Calculate the vapor pressure of methanol and of ethanol above the solution at $20^{\\circ} \\mathrm{C}$.\n(c) Calculate the mole fraction of methanol and of ethanol in the vapor above the solution.\n66. The triple point of air-free water is defined as 273.16 K . Why is it important that the water be free of air?"}
{"id": 3822, "contents": "1292. Exercises - 1292.4. Colligative Properties\n66. The triple point of air-free water is defined as 273.16 K . Why is it important that the water be free of air?\n67. Meat can be classified as fresh (not frozen) even though it is stored at $-1^{\\circ} \\mathrm{C}$. Why wouldn\u2019t meat freeze at this temperature?\n68. An organic compound has a composition of $93.46 \\% \\mathrm{C}$ and $6.54 \\% \\mathrm{H}$ by mass. A solution of 0.090 g of this compound in 1.10 g of camphor melts at $158.4^{\\circ} \\mathrm{C}$. The melting point of pure camphor is $178.4^{\\circ} \\mathrm{C} . K_{\\mathrm{f}}$ for camphor is $37.7^{\\circ} \\mathrm{C} / \\mathrm{m}$. Assuming ideal solution behavior, what is the molecular formula of the solute? Show your calculations.\n69. A sample of $\\mathrm{HgCl}_{2}$ weighing 9.41 g is dissolved in 32.75 g of ethanol, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}\\left(\\mathrm{K}_{\\mathrm{b}}=1.20^{\\circ} \\mathrm{C} / \\mathrm{m}\\right)$. The boiling point elevation of the solution is $1.27^{\\circ} \\mathrm{C}$. Is $\\mathrm{HgCl}_{2}$ an electrolyte in ethanol? Show your calculations.\n70. A salt is known to be an alkali metal fluoride. A quick approximate determination of freezing point indicates that 4 g of the salt dissolved in 100 g of water produces a solution that freezes at about $-1.4^{\\circ} \\mathrm{C}$. Assuming ideal solution behavior, what is the formula of the salt? Show your calculations."}
{"id": 3823, "contents": "1292. Exercises - 1292.5. Colloids\n71. Identify the dispersed phase and the dispersion medium in each of the following colloidal systems: starch dispersion, smoke, fog, pearl, whipped cream, floating soap, jelly, milk, and ruby.\n72. Distinguish between dispersion methods and condensation methods for preparing colloidal systems.\n73. How do colloids differ from solutions with regard to dispersed particle size and homogeneity?\n74. Explain the cleansing action of soap.\n75. How can it be demonstrated that colloidal particles are electrically charged?"}
{"id": 3824, "contents": "1293. CHAPTER 12 <br> Thermodynamics - \nFigure 12.1 Geysers are a dramatic display of thermodynamic principles in nature. Water deep within the underground channels of the geyser is under high pressure and heated to high temperature by magma. When a pocket of water near the surface reaches boiling point and is expelled, the resulting drop in pressure causes larger volumes of water to flash boil, forcefully ejecting steam and water in an impressive eruption. (credit: modification of work by Yellowstone National Park)"}
{"id": 3825, "contents": "1294. CHAPTER OUTLINE - 1294.1. Spontaneity\n12.2 Entropy\n12.3 The Second and Third Laws of Thermodynamics\n12.4 Free Energy\n\nINTRODUCTION Among the many capabilities of chemistry is its ability to predict if a process will occur under specified conditions. Thermodynamics, the study of relationships between the energy and work associated with chemical and physical processes, provides this predictive ability. Previous chapters in this text have described various applications of thermochemistry, an important aspect of thermodynamics concerned with the heat flow accompanying chemical reactions and phase transitions. This chapter will introduce additional thermodynamic concepts, including those that enable the prediction of any chemical or physical changes under a given set of conditions."}
{"id": 3826, "contents": "1295. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Distinguish between spontaneous and nonspontaneous processes\n- Describe the dispersal of matter and energy that accompanies certain spontaneous processes\n\nProcesses have a natural tendency to occur in one direction under a given set of conditions. Water will naturally flow downhill, but uphill flow requires outside intervention such as the use of a pump. Iron exposed to the earth's atmosphere will corrode, but rust is not converted to iron without intentional chemical treatment. A spontaneous process is one that occurs naturally under certain conditions. A nonspontaneous process, on the other hand, will not take place unless it is \"driven\" by the continual input of energy from an external source. A process that is spontaneous in one direction under a particular set of conditions is nonspontaneous in the reverse direction. At room temperature and typical atmospheric pressure, for example, ice will spontaneously melt, but water will not spontaneously freeze.\n\nThe spontaneity of a process is not correlated to the speed of the process. A spontaneous change may be so rapid that it is essentially instantaneous or so slow that it cannot be observed over any practical period of time. To illustrate this concept, consider the decay of radioactive isotopes, a topic more thoroughly treated in the chapter on nuclear chemistry. Radioactive decay is by definition a spontaneous process in which the nuclei of unstable isotopes emit radiation as they are converted to more stable nuclei. All the decay processes occur spontaneously, but the rates at which different isotopes decay vary widely. Technetium-99m is a popular radioisotope for medical imaging studies that undergoes relatively rapid decay and exhibits a half-life of about six hours. Uranium-238 is the most abundant isotope of uranium, and its decay occurs much more slowly, exhibiting a half-life of more than four billion years (Figure 12.2).\n\n\nFIGURE 12.2 Both U-238 and Tc-99m undergo spontaneous radioactive decay, but at drastically different rates. Over the course of one week, essentially all of a Tc-99m sample and none of a U-238 sample will have decayed.\n\nAs another example, consider the conversion of diamond into graphite (Figure 12.3).\n\n$$\n\\mathrm{C}(s, \\text { diamond }) \\longrightarrow \\mathrm{C}(s, \\text { graphite })\n$$"}
{"id": 3827, "contents": "1295. LEARNING OBJECTIVES - \nAs another example, consider the conversion of diamond into graphite (Figure 12.3).\n\n$$\n\\mathrm{C}(s, \\text { diamond }) \\longrightarrow \\mathrm{C}(s, \\text { graphite })\n$$\n\nThe phase diagram for carbon indicates that graphite is the stable form of this element under ambient atmospheric pressure, while diamond is the stable allotrope at very high pressures, such as those present during its geologic formation. Thermodynamic calculations of the sort described in the last section of this chapter indicate that the conversion of diamond to graphite at ambient pressure occurs spontaneously, yet diamonds are observed to exist, and persist, under these conditions. Though the process is spontaneous under typical ambient conditions, its rate is extremely slow; so, for all practical purposes diamonds are indeed \"forever.\" Situations such as these emphasize the important distinction between the thermodynamic and the kinetic aspects of a process. In this particular case, diamonds are said to be thermodynamically unstable but kinetically stable under ambient conditions.\n\n\nFIGURE 12.3 The conversion of carbon from the diamond allotrope to the graphite allotrope is spontaneous at ambient pressure, but its rate is immeasurably slow at low to moderate temperatures. This process is known as graphitization, and its rate can be increased to easily measurable values at temperatures in the 1000-2000 K range. (credit \"diamond\" photo: modification of work by \"Fancy Diamonds\"/Flickr; credit \"graphite\" photo: modification of work by images-of-elements.com/carbon.php)"}
{"id": 3828, "contents": "1296. Dispersal of Matter and Energy - \nExtending the discussion of thermodynamic concepts toward the objective of predicting spontaneity, consider now an isolated system consisting of two flasks connected with a closed valve. Initially there is an ideal gas in one flask and the other flask is empty $(P=0)$. (Figure 12.4). When the valve is opened, the gas spontaneously expands to fill both flasks equally. Recalling the definition of pressure-volume work from the chapter on thermochemistry, note that no work has been done because the pressure in a vacuum is zero.\n\n$$\nw=-P \\Delta V=0 \\quad(P=0 \\text { in a vacuum })\n$$\n\nNote as well that since the system is isolated, no heat has been exchanged with the surroundings ( $q=0$ ). The first law of thermodynamics confirms that there has been no change in the system's internal energy as a result of this process.\n\n$$\n\\Delta U=q+w=0+0=0\n$$\n\nThe spontaneity of this process is therefore not a consequence of any change in energy that accompanies the process. Instead, the driving force appears to be related to the greater, more uniform dispersal of matter that results when the gas is allowed to expand. Initially, the system was comprised of one flask containing matter and another flask containing nothing. After the spontaneous expansion took place, the matter was distributed both more widely (occupying twice its original volume) and more uniformly (present in equal amounts in each flask).\n\n\nFIGURE 12.4 An isolated system consists of an ideal gas in one flask that is connected by a closed valve to a\nsecond flask containing a vacuum. Once the valve is opened, the gas spontaneously becomes evenly distributed between the flasks.\n\nNow consider two objects at different temperatures: object X at temperature $T_{\\mathrm{X}}$ and object Y at temperature $T_{\\mathrm{Y}}$, with $T_{\\mathrm{X}}>T_{\\mathrm{Y}}$ (Figure 12.5). When these objects come into contact, heat spontaneously flows from the hotter object $(\\mathrm{X})$ to the colder one $(\\mathrm{Y})$. This corresponds to a loss of thermal energy by X and a gain of thermal energy by Y."}
{"id": 3829, "contents": "1296. Dispersal of Matter and Energy - \n$$\nq_{\\mathrm{X}}<0 \\quad \\text { and } \\quad q_{\\mathrm{Y}}=-q_{\\mathrm{X}}>0\n$$\n\nFrom the perspective of this two-object system, there was no net gain or loss of thermal energy, rather the available thermal energy was redistributed among the two objects. This spontaneous process resulted in a more uniform dispersal of energy.\n\n\nFIGURE 12.5 When two objects at different temperatures come in contact, heat spontaneously flows from the hotter to the colder object.\n\nAs illustrated by the two processes described, an important factor in determining the spontaneity of a process is the extent to which it changes the dispersal or distribution of matter and/or energy. In each case, a spontaneous process took place that resulted in a more uniform distribution of matter or energy."}
{"id": 3830, "contents": "1298. Redistribution of Matter during a Spontaneous Process - \nDescribe how matter is redistributed when the following spontaneous processes take place:\n(a) A solid sublimes.\n(b) A gas condenses.\n(c) A drop of food coloring added to a glass of water forms a solution with uniform color."}
{"id": 3831, "contents": "1299. Solution - \nFIGURE 12.6 (credit a: modification of work by Jenny Downing; credit b: modification of work by \"Fuzzy Gerdes\"/Flickr; credit c: modification of work by Paul A. Flowers)\n(a) Sublimation is the conversion of a solid (relatively high density) to a gas (much lesser density). This process yields a much greater dispersal of matter, since the molecules will occupy a much greater volume after the solid-to-gas transition.\n(b) Condensation is the conversion of a gas (relatively low density) to a liquid (much greater density). This process yields a much lesser dispersal of matter, since the molecules will occupy a much lesser volume after\nthe gas-to-liquid transition.\n(c) The process in question is diffusion. This process yields a more uniform dispersal of matter, since the initial state of the system involves two regions of different dye concentrations (high in the drop of dye, zero in the water), and the final state of the system contains a single dye concentration throughout."}
{"id": 3832, "contents": "1300. Check Your Learning - \nDescribe how energy is redistributed when a spoon at room temperature is placed in a cup of hot coffee."}
{"id": 3833, "contents": "1301. Answer: - \nHeat will spontaneously flow from the hotter object (coffee) to the colder object (spoon), resulting in a more uniform distribution of thermal energy as the spoon warms and the coffee cools."}
{"id": 3834, "contents": "1302. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Define entropy\n- Explain the relationship between entropy and the number of microstates\n- Predict the sign of the entropy change for chemical and physical processes\n\nIn 1824, at the age of 28, Nicolas L\u00e9onard Sadi Carnot (Figure 12.7) published the results of an extensive study regarding the efficiency of steam heat engines. A later review of Carnot's findings by Rudolf Clausius introduced a new thermodynamic property that relates the spontaneous heat flow accompanying a process to the temperature at which the process takes place. This new property was expressed as the ratio of the reversible heat ( $q_{\\mathrm{rev}}$ ) and the kelvin temperature ( $T$ ). In thermodynamics, a reversible process is one that takes place at such a slow rate that it is always at equilibrium and its direction can be changed (it can be \"reversed\") by an infinitesimally small change in some condition. Note that the idea of a reversible process is a formalism required to support the development of various thermodynamic concepts; no real processes are truly reversible, rather they are classified as irreversible.\n\n(a)\n\n(b)\n\nFIGURE 12.7 (a) Nicholas L\u00e9onard Sadi Carnot's research into steam-powered machinery and (b) Rudolf Clausius's later study of those findings led to groundbreaking discoveries about spontaneous heat flow processes.\n\nSimilar to other thermodynamic properties, this new quantity is a state function, so its change depends only upon the initial and final states of a system. In 1865, Clausius named this property entropy (S) and defined its change for any process as the following:\n\n$$\n\\Delta S=\\frac{q_{\\mathrm{rev}}}{T}\n$$\n\nThe entropy change for a real, irreversible process is then equal to that for the theoretical reversible process that involves the same initial and final states."}
{"id": 3835, "contents": "1303. Entropy and Microstates - \nFollowing the work of Carnot and Clausius, Ludwig Boltzmann developed a molecular-scale statistical model that related the entropy of a system to the number of microstates ( $W$ ) possible for the system. A microstate is a specific configuration of all the locations and energies of the atoms or molecules that make up a system. The relation between a system's entropy and the number of possible microstates is\n\n$$\nS=k \\ln W\n$$\n\nwhere $k$ is the Boltzmann constant, $1.38 \\times 10^{-23} \\mathrm{~J} / \\mathrm{K}$.\nAs for other state functions, the change in entropy for a process is the difference between its final ( $S_{\\mathrm{f}}$ ) and initial $\\left(S_{\\mathrm{i}}\\right)$ values:\n\n$$\n\\Delta S=S_{\\mathrm{f}}-S_{\\mathrm{i}}=k \\ln W_{\\mathrm{f}}-k \\ln W_{\\mathrm{i}}=k \\ln \\frac{W_{\\mathrm{f}}}{W_{\\mathrm{i}}}\n$$\n\nFor processes involving an increase in the number of microstates, $W_{\\mathrm{f}}>W_{\\mathrm{i}}$, the entropy of the system increases and $\\Delta S>0$. Conversely, processes that reduce the number of microstates, $W_{\\mathrm{f}}<W_{\\mathrm{i}}$, yield a decrease in system entropy, $\\Delta S<0$. This molecular-scale interpretation of entropy provides a link to the probability that a process will occur as illustrated in the next paragraphs.\n\nConsider the general case of a system comprised of $N$ particles distributed among $n$ boxes. The number of microstates possible for such a system is $n^{N}$. For example, distributing four particles among two boxes will result in $2^{4}=16$ different microstates as illustrated in Figure 12.8. Microstates with equivalent particle arrangements (not considering individual particle identities) are grouped together and are called distributions. The probability that a system will exist with its components in a given distribution is proportional to the number of microstates within the distribution. Since entropy increases logarithmically with the number of microstates, the most probable distribution is therefore the one of greatest entropy.\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)"}
{"id": 3836, "contents": "1303. Entropy and Microstates - \n(b)\n\n(c)\n\n(d)\n\n(e)\n\n\nFIGURE 12.8 The sixteen microstates associated with placing four particles in two boxes are shown. The microstates are collected into five distributions-(a), (b), (c), (d), and (e)-based on the numbers of particles in each box.\n\nFor this system, the most probable configuration is one of the six microstates associated with distribution (c) where the particles are evenly distributed between the boxes, that is, a configuration of two particles in each\nbox. The probability of finding the system in this configuration is $\\frac{6}{16}$ or $\\frac{3}{8}$. The least probable configuration of the system is one in which all four particles are in one box, corresponding to distributions (a) and (e), each with a probability of $\\frac{1}{16}$. The probability of finding all particles in only one box (either the left box or right box) is then $\\left(\\frac{1}{16}+\\frac{1}{16}\\right)=\\frac{2}{16}$ or $\\frac{1}{8}$.\nAs you add more particles to the system, the number of possible microstates increases exponentially $\\left(2^{N}\\right)$. A macroscopic (laboratory-sized) system would typically consist of moles of particles ( $N \\sim 10^{23}$ ), and the corresponding number of microstates would be staggeringly huge. Regardless of the number of particles in the system, however, the distributions in which roughly equal numbers of particles are found in each box are always the most probable configurations.\n\nThis matter dispersal model of entropy is often described qualitatively in terms of the disorder of the system. By this description, microstates in which all the particles are in a single box are the most ordered, thus possessing the least entropy. Microstates in which the particles are more evenly distributed among the boxes are more disordered, possessing greater entropy."}
{"id": 3837, "contents": "1303. Entropy and Microstates - \nThe previous description of an ideal gas expanding into a vacuum (Figure 12.4) is a macroscopic example of this particle-in-a-box model. For this system, the most probable distribution is confirmed to be the one in which the matter is most uniformly dispersed or distributed between the two flasks. Initially, the gas molecules are confined to just one of the two flasks. Opening the valve between the flasks increases the volume available to the gas molecules and, correspondingly, the number of microstates possible for the system. Since $W_{\\mathrm{f}}>W_{\\mathrm{i}}$, the expansion process involves an increase in entropy $(\\Delta S>0)$ and is spontaneous.\n\nA similar approach may be used to describe the spontaneous flow of heat. Consider a system consisting of two objects, each containing two particles, and two units of thermal energy (represented as \"*\") in Figure 12.9. The hot object is comprised of particles $\\mathbf{A}$ and $\\mathbf{B}$ and initially contains both energy units. The cold object is comprised of particles $\\mathbf{C}$ and $\\mathbf{D}$, which initially has no energy units. Distribution (a) shows the three microstates possible for the initial state of the system, with both units of energy contained within the hot object. If one of the two energy units is transferred, the result is distribution (b) consisting of four microstates. If both energy units are transferred, the result is distribution (c) consisting of three microstates. Thus, we may describe this system by a total of ten microstates. The probability that the heat does not flow when the two objects are brought into contact, that is, that the system remains in distribution (a), is $\\frac{3}{10}$. More likely is the flow of heat to yield one of the other two distribution, the combined probability being $\\frac{7}{10}$. The most likely result is the flow of heat to yield the uniform dispersal of energy represented by distribution (b), the probability of this configuration being $\\frac{4}{10}$. This supports the common observation that placing hot and cold objects in contact results in spontaneous heat flow that ultimately equalizes the objects' temperatures. And, again, this spontaneous process is also characterized by an increase in system entropy."}
{"id": 3838, "contents": "1303. Entropy and Microstates - \nFIGURE 12.9 This shows a microstate model describing the flow of heat from a hot object to a cold object. (a) Before the heat flow occurs, the object comprised of particles $\\mathbf{A}$ and $\\mathbf{B}$ contains both units of energy and as represented by a distribution of three microstates. (b) If the heat flow results in an even dispersal of energy (one energy unit transferred), a distribution of four microstates results. (c) If both energy units are transferred, the resulting distribution has three microstates."}
{"id": 3839, "contents": "1305. Determination of $\\Delta S$ - \nCalculate the change in entropy for the process depicted below."}
{"id": 3840, "contents": "1306. Solution - \nThe initial number of microstates is one, the final six:\n\n$$\n\\Delta S=k \\ln \\frac{W_{\\mathrm{c}}}{W_{\\mathrm{a}}}=1.38 \\times 10^{-23} \\mathrm{~J} / \\mathrm{K} \\times \\ln \\frac{6}{1}=2.47 \\times 10^{-23} \\mathrm{~J} / \\mathrm{K}\n$$\n\nThe sign of this result is consistent with expectation; since there are more microstates possible for the final state than for the initial state, the change in entropy should be positive."}
{"id": 3841, "contents": "1307. Check Your Learning - \nConsider the system shown in Figure 12.9. What is the change in entropy for the process where all the energy is transferred from the hot object ( $\\mathbf{A B}$ ) to the cold object ( $\\mathbf{C D}$ )?"}
{"id": 3842, "contents": "1308. Answer: - \n$0 \\mathrm{~J} / \\mathrm{K}$"}
{"id": 3843, "contents": "1309. Predicting the Sign of $\\Delta S$ - \nThe relationships between entropy, microstates, and matter/energy dispersal described previously allow us to make generalizations regarding the relative entropies of substances and to predict the sign of entropy changes for chemical and physical processes. Consider the phase changes illustrated in Figure 12.10. In the solid phase, the atoms or molecules are restricted to nearly fixed positions with respect to each other and are capable of only modest oscillations about these positions. With essentially fixed locations for the system's component particles, the number of microstates is relatively small. In the liquid phase, the atoms or molecules are free to move over and around each other, though they remain in relatively close proximity to one another. This increased freedom of motion results in a greater variation in possible particle locations, so the number of microstates is correspondingly greater than for the solid. As a result, $S_{\\text {liquid }}>S_{\\text {solid }}$ and the process of converting a substance from solid to liquid (melting) is characterized by an increase in entropy, $\\Delta S>0$. By the same logic, the reciprocal process (freezing) exhibits a decrease in entropy, $\\Delta S<0$.\n\n\nFIGURE 12.10 The entropy of a substance increases ( $\\Delta S>0$ ) as it transforms from a relatively ordered solid, to a less-ordered liquid, and then to a still less-ordered gas. The entropy decreases $(\\Delta S<0)$ as the substance transforms from a gas to a liquid and then to a solid.\n\nNow consider the gaseous phase, in which a given number of atoms or molecules occupy a much greater volume than in the liquid phase. Each atom or molecule can be found in many more locations, corresponding to a much greater number of microstates. Consequently, for any substance, $S_{\\text {gas }}>S_{\\text {liquid }}>S_{\\text {solid }}$, and the processes of vaporization and sublimation likewise involve increases in entropy, $\\Delta S>0$. Likewise, the reciprocal phase transitions, condensation and deposition, involve decreases in entropy, $\\Delta S<0$."}
{"id": 3844, "contents": "1309. Predicting the Sign of $\\Delta S$ - \nAccording to kinetic-molecular theory, the temperature of a substance is proportional to the average kinetic energy of its particles. Raising the temperature of a substance will result in more extensive vibrations of the particles in solids and more rapid translations of the particles in liquids and gases. At higher temperatures, the distribution of kinetic energies among the atoms or molecules of the substance is also broader (more dispersed) than at lower temperatures. Thus, the entropy for any substance increases with temperature (Figure 12.11).\n\n\nFIGURE 12.11 Entropy increases as the temperature of a substance is raised, which corresponds to the greater spread of kinetic energies. When a substance undergoes a phase transition, its entropy changes significantly."}
{"id": 3845, "contents": "1310. LINK TO LEARNING - \nTry this simulator (http://openstax.org/l/16freemotion) with interactive visualization of the dependence of\nparticle location and freedom of motion on physical state and temperature.\n\nThe entropy of a substance is influenced by the structure of the particles (atoms or molecules) that comprise the substance. With regard to atomic substances, heavier atoms possess greater entropy at a given temperature than lighter atoms, which is a consequence of the relation between a particle's mass and the spacing of quantized translational energy levels (a topic beyond the scope of this text). For molecules, greater numbers of atoms increase the number of ways in which the molecules can vibrate and thus the number of possible microstates and the entropy of the system.\n\nFinally, variations in the types of particles affects the entropy of a system. Compared to a pure substance, in which all particles are identical, the entropy of a mixture of two or more different particle types is greater. This is because of the additional orientations and interactions that are possible in a system comprised of nonidentical components. For example, when a solid dissolves in a liquid, the particles of the solid experience both a greater freedom of motion and additional interactions with the solvent particles. This corresponds to a more uniform dispersal of matter and energy and a greater number of microstates. The process of dissolution therefore involves an increase in entropy, $\\Delta S>0$.\n\nConsidering the various factors that affect entropy allows us to make informed predictions of the sign of $\\Delta S$ for various chemical and physical processes as illustrated in Example 12.3."}
{"id": 3846, "contents": "1312. Predicting the Sign of $\\Delta S$ - \nPredict the sign of the entropy change for the following processes. Indicate the reason for each of your predictions.\n(a) One mole liquid water at room temperature $\\longrightarrow$ one mole liquid water at $50^{\\circ} \\mathrm{C}$\n(b) $\\mathrm{Ag}^{+}(a q)+\\mathrm{Cl}^{-}(a q) \\longrightarrow \\mathrm{AgCl}(s)$\n(c) $\\mathrm{C}_{6} \\mathrm{H}_{6}(l)+\\frac{15}{2} \\mathrm{O}_{2}(g) \\longrightarrow 6 \\mathrm{CO}_{2}(g)+3 \\mathrm{H}_{2} \\mathrm{O}(l)$\n$(d) \\mathrm{NH}_{3}(s) \\longrightarrow \\mathrm{NH}_{3}(l)$"}
{"id": 3847, "contents": "1313. Solution - \n(a) positive, temperature increases\n(b) negative, reduction in the number of ions (particles) in solution, decreased dispersal of matter\n(c) negative, net decrease in the amount of gaseous species\n(d) positive, phase transition from solid to liquid, net increase in dispersal of matter"}
{"id": 3848, "contents": "1314. Check Your Learning - \nPredict the sign of the entropy change for the following processes. Give a reason for your prediction.\n(a) $\\mathrm{NaNO}_{3}(s) \\longrightarrow \\mathrm{Na}^{+}(a q)+\\mathrm{NO}_{3}{ }^{-}(a q)$\n(b) the freezing of liquid water\n(c) $\\mathrm{CO}_{2}(s) \\longrightarrow \\mathrm{CO}_{2}(g)$\n(d) $\\mathrm{CaCO}_{3}(s) \\longrightarrow \\mathrm{CaO}(s)+\\mathrm{CO}_{2}(g)$"}
{"id": 3849, "contents": "1315. Answer: - \n(a) Positive; The solid dissolves to give an increase of mobile ions in solution. (b) Negative; The liquid becomes a more ordered solid. (c) Positive; The relatively ordered solid becomes a gas. (d) Positive; There is a net increase in the amount of gaseous species."}
{"id": 3850, "contents": "1315. Answer: - 1315.1. The Second and Third Laws of Thermodynamics\nLEARNING OBJECTIVES\nBy the end of this section, you will be able to:\n\n- State and explain the second and third laws of thermodynamics\n- Calculate entropy changes for phase transitions and chemical reactions under standard conditions"}
{"id": 3851, "contents": "1316. The Second Law of Thermodynamics - \nIn the quest to identify a property that may reliably predict the spontaneity of a process, a promising candidate has been identified: entropy. Processes that involve an increase in entropy of the system ( $\\Delta S>0$ ) are very often spontaneous; however, examples to the contrary are plentiful. By expanding consideration of entropy changes to include the surroundings, we may reach a significant conclusion regarding the relation between this property and spontaneity. In thermodynamic models, the system and surroundings comprise everything, that is, the universe, and so the following is true:\n\n$$\n\\Delta S_{\\text {univ }}=\\Delta S_{\\text {sys }}+\\Delta S_{\\text {surr }}\n$$\n\nTo illustrate this relation, consider again the process of heat flow between two objects, one identified as the system and the other as the surroundings. There are three possibilities for such a process:\n\n1. The objects are at different temperatures, and heat flows from the hotter to the cooler object. This is always observed to occur spontaneously. Designating the hotter object as the system and invoking the definition of entropy yields the following:\n\n$$\n\\Delta S_{\\mathrm{sys}}=\\frac{-q_{\\mathrm{rev}}}{T_{\\mathrm{sys}}} \\quad \\text { and } \\quad \\Delta S_{\\mathrm{surr}}=\\frac{q_{\\mathrm{rev}}}{T_{\\mathrm{surr}}}\n$$\n\nThe magnitudes of $-q_{\\mathrm{rev}}$ and $q_{\\mathrm{rev}}$ are equal, their opposite arithmetic signs denoting loss of heat by the system and gain of heat by the surroundings. Since $T_{\\text {sys }}>T_{\\text {surr }}$ in this scenario, the entropy decrease of the system will be less than the entropy increase of the surroundings, and so the entropy of the universe will increase:\n\n$$\n\\begin{aligned}\n& \\left|\\Delta S_{\\text {sys }}\\right|<\\left|\\Delta S_{\\text {surr }}\\right| \\\\\n& \\Delta S_{\\text {univ }}=\\Delta S_{\\text {sys }}+\\Delta S_{\\text {surr }}>0\n\\end{aligned}\n$$"}
{"id": 3852, "contents": "1316. The Second Law of Thermodynamics - \n2. The objects are at different temperatures, and heat flows from the cooler to the hotter object. This is never observed to occur spontaneously. Again designating the hotter object as the system and invoking the definition of entropy yields the following:\n\n$$\n\\Delta S_{\\mathrm{sys}}=\\frac{q_{\\mathrm{rev}}}{T_{\\mathrm{sys}}} \\quad \\text { and } \\quad \\Delta S_{\\mathrm{surr}}=\\frac{-q_{\\mathrm{rev}}}{T_{\\mathrm{surr}}}\n$$\n\nThe arithmetic signs of $q_{\\mathrm{rev}}$ denote the gain of heat by the system and the loss of heat by the surroundings. The magnitude of the entropy change for the surroundings will again be greater than that for the system, but in this case, the signs of the heat changes (that is, the direction of the heat flow) will yield a negative value for $\\Delta S_{\\text {univ. }}$. This process involves a decrease in the entropy of the universe.\n3. The objects are at essentially the same temperature, $T_{\\text {sys }} \\approx T_{\\text {surr }}$, and so the magnitudes of the entropy changes are essentially the same for both the system and the surroundings. In this case, the entropy change of the universe is zero, and the system is at equilibrium.\n\n$$\n\\begin{aligned}\n& \\left|\\Delta S_{\\text {sys }}\\right| \\approx\\left|\\Delta S_{\\text {surr }}\\right| \\\\\n& \\Delta S_{\\text {univ }}=\\Delta S_{\\text {sys }}+\\Delta S_{\\text {surf }}=0\n\\end{aligned}\n$$\n\nThese results lead to a profound statement regarding the relation between entropy and spontaneity known as the second law of thermodynamics: all spontaneous changes cause an increase in the entropy of the universe. A summary of these three relations is provided in Table 12.1.\n\nThe Second Law of Thermodynamics\n\n| $\\Delta S_{\\text {univ }}>0$ | spontaneous |\n| :--- | :--- |\n| $\\Delta S_{\\text {univ }}<0$ | nonspontaneous (spontaneous in opposite direction) |\n| $\\Delta S_{\\text {univ }}=0$ | at equilibrium |"}
{"id": 3853, "contents": "1316. The Second Law of Thermodynamics - \nTABLE 12.1\n\nFor many realistic applications, the surroundings are vast in comparison to the system. In such cases, the heat gained or lost by the surroundings as a result of some process represents a very small, nearly infinitesimal, fraction of its total thermal energy. For example, combustion of a fuel in air involves transfer of heat from a system (the fuel and oxygen molecules undergoing reaction) to surroundings that are infinitely more massive (the earth's atmosphere). As a result, $q_{\\text {surr }}$ is a good approximation of $q_{\\text {rev }}$, and the second law may be stated as the following:\n\n$$\n\\Delta S_{\\mathrm{univ}}=\\Delta S_{\\mathrm{sys}}+\\Delta S_{\\mathrm{surr}}=\\Delta S_{\\mathrm{sys}}+\\frac{q_{\\mathrm{surr}}}{T}\n$$\n\nWe may use this equation to predict the spontaneity of a process as illustrated in Example 12.4."}
{"id": 3854, "contents": "1318. Will Ice Spontaneously Melt? - \nThe entropy change for the process\n\n$$\n\\mathrm{H}_{2} \\mathrm{O}(s) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nis $22.1 \\mathrm{~J} / \\mathrm{K}$ and requires that the surroundings transfer 6.00 kJ of heat to the system. Is the process spontaneous at $-10.00^{\\circ} \\mathrm{C}$ ? Is it spontaneous at $+10.00^{\\circ} \\mathrm{C}$ ?"}
{"id": 3855, "contents": "1319. Solution - \nWe can assess the spontaneity of the process by calculating the entropy change of the universe. If $\\Delta S_{\\text {univ }}$ is positive, then the process is spontaneous. At both temperatures, $\\Delta S_{\\text {sys }}=22.1 \\mathrm{~J} / \\mathrm{K}$ and $q_{\\text {surr }}=-6.00 \\mathrm{~kJ}$.\n\nAt $-10.00{ }^{\\circ} \\mathrm{C}(263.15 \\mathrm{~K})$, the following is true:\n\n$$\n\\begin{aligned}\n\\Delta S_{\\text {univ }} & =\\Delta S_{\\text {sys }}+\\Delta S_{\\text {surr }}=\\Delta S_{\\text {sys }}+\\frac{q_{\\text {surr }}}{T} \\\\\n& =22.1 \\mathrm{~J} / \\mathrm{K}+\\frac{-6.00 \\times 10^{3} \\mathrm{~J}}{263.15 \\mathrm{~K}}=-0.7 \\mathrm{~J} / \\mathrm{K}\n\\end{aligned}\n$$\n\n$S_{\\text {univ }}<0$, so melting is nonspontaneous (not spontaneous) at $-10.0^{\\circ} \\mathrm{C}$.\nAt $10.00^{\\circ} \\mathrm{C}(283.15 \\mathrm{~K})$, the following is true:\n\n$$\n\\begin{gathered}\n\\Delta S_{\\text {univ }}=\\Delta S_{\\text {sys }}+\\frac{q_{\\text {surr }}}{T} \\\\\n=22.1 \\mathrm{~J} / \\mathrm{K}+\\frac{-6.00 \\times 10^{3} \\mathrm{~J}}{283.15 \\mathrm{~K}}=+0.9 \\mathrm{~J} / \\mathrm{K}\n\\end{gathered}\n$$\n\n$S_{\\text {univ }}>0$, so melting is spontaneous at $10.00^{\\circ} \\mathrm{C}$."}
{"id": 3856, "contents": "1320. Check Your Learning - \nUsing this information, determine if liquid water will spontaneously freeze at the same temperatures. What can you say about the values of $S_{\\text {univ }}$ ?"}
{"id": 3857, "contents": "1321. Answer: - \nEntropy is a state function, so $\\Delta S_{\\text {freezing }}=-\\Delta S_{\\text {melting }}=-22.1 \\mathrm{~J} / \\mathrm{K}$ and $q_{\\text {surr }}=+6.00 \\mathrm{~kJ}$. At $-10.00^{\\circ} \\mathrm{C}$ spontaneous, $+0.7 \\mathrm{~J} / \\mathrm{K}$; at $+10.00^{\\circ} \\mathrm{C}$ nonspontaneous, $-0.9 \\mathrm{~J} / \\mathrm{K}$."}
{"id": 3858, "contents": "1322. The Third Law of Thermodynamics - \nThe previous section described the various contributions of matter and energy dispersal that contribute to the entropy of a system. With these contributions in mind, consider the entropy of a pure, perfectly crystalline solid possessing no kinetic energy (that is, at a temperature of absolute zero, 0 K ). This system may be described by a single microstate, as its purity, perfect crystallinity and complete lack of motion means there is but one possible location for each identical atom or molecule comprising the crystal ( $W=1$ ). According to the Boltzmann equation, the entropy of this system is zero.\n\n$$\nS=k \\ln W=k \\ln (1)=0\n$$\n\nThis limiting condition for a system's entropy represents the third law of thermodynamics: the entropy of a pure, perfect crystalline substance at 0 K is zero.\n\nCareful calorimetric measurements can be made to determine the temperature dependence of a substance's entropy and to derive absolute entropy values under specific conditions. Standard entropies ( $\\boldsymbol{S}^{\\circ}$ ) are for one mole of substance under standard conditions (a pressure of 1 bar and a temperature of 298.15 K ; see details regarding standard conditions in the thermochemistry chapter of this text). The standard entropy change ( $\\boldsymbol{\\Delta} \\boldsymbol{S}^{\\circ}$ ) for a reaction may be computed using standard entropies as shown below:\n\n$$\n\\Delta S^{\\circ}=\\sum v S^{\\circ}(\\text { products })-\\sum \\nu S^{\\circ}(\\text { reactants })\n$$\n\nwhere $\\nu$ represents stoichiometric coefficients in the balanced equation representing the process. For example, $\\Delta S^{\\circ}$ for the following reaction at room temperature\n\n$$\nm \\mathrm{~A}+n \\mathrm{~B} \\longrightarrow x \\mathrm{C}+y \\mathrm{D}\n$$\n\nis computed as:\n\n$$\n=\\left[x S^{\\circ}(\\mathrm{C})+y S^{\\circ}(\\mathrm{D})\\right]-\\left[m S^{\\circ}(\\mathrm{A})+n S^{\\circ}(\\mathrm{B})\\right]\n$$"}
{"id": 3859, "contents": "1322. The Third Law of Thermodynamics - \nA partial listing of standard entropies is provided in Table 12.2, and additional values are provided in Appendix G. The example exercises that follow demonstrate the use of $S^{\\circ}$ values in calculating standard entropy changes for physical and chemical processes.\n\n| Substance | $S^{\\circ}\\left(\\mathbf{J ~ m o l}^{\\mathbf{- 1}} \\mathbf{K}^{\\mathbf{- 1}}\\right)$ |  |\n| :--- | :--- | :---: |\n| carbon |  |  |\n| $\\mathrm{C}(s$, graphite $)$ | 5.740 |  |\n| $\\mathrm{C}(s$, diamond $)$ | 2.38 |  |\n| $\\mathrm{CO}(\\mathrm{g})$ | 197.7 |  |\n| $\\mathrm{CO}_{2}(g)$ | 213.8 |  |\n| $\\mathrm{CH}_{4}(\\mathrm{~g})$ | 186.3 |  |\n| $\\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{~g})$ | 219.5 |  |\n| $\\mathrm{C}_{2} \\mathrm{H}_{6}(\\mathrm{~g})$ | 229.5 |  |\n\n\n| $\\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{I})$ | 126.8 |\n| :--- | :---: |\n| $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(\\mathrm{l})$ | 160.7 |\n| hydrogen |  |\n| $\\mathrm{H}_{2}(g)$ | 130.57 |\n| $\\mathrm{H}(g)$ | 114.6 |\n| $\\mathrm{H}_{2} \\mathrm{O}(g)$ | 188.71 |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})$ | 69.91 |\n| $\\mathrm{HCI}(g)$ | 186.8 |\n| $\\mathrm{H}_{2} \\mathrm{~S}(g)$ | 205.7 |\n| oxygen | 205.03 |\n| $\\mathrm{O}_{2}(g)$ |  |"}
{"id": 3860, "contents": "1322. The Third Law of Thermodynamics - \nTABLE 12.2 Standard entropies for selected substances measured at 1 atm and 298.15 K. (Values are approximately equal to those measured at 1 bar, the currently accepted standard state pressure.)"}
{"id": 3861, "contents": "1324. Determination of $\\boldsymbol{\\Delta} \\boldsymbol{S}^{\\circ}$ - \nCalculate the standard entropy change for the following process:\n\n$$\n\\mathrm{H}_{2} \\mathrm{O}(g) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(l)\n$$"}
{"id": 3862, "contents": "1325. Solution - \nCalculate the entropy change using standard entropies as shown above:\n\n$$\n\\Delta S^{\\circ}=(1 \\mathrm{~mol})\\left(70.0 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}\\right)-(1 \\mathrm{~mol})\\left(188.8 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}\\right)=-118.8 \\mathrm{~J} / \\mathrm{K}\n$$\n\nThe value for $\\Delta S^{\\circ}$ is negative, as expected for this phase transition (condensation), which the previous section discussed."}
{"id": 3863, "contents": "1326. Check Your Learning - \nCalculate the standard entropy change for the following process:\n\n$$\n\\mathrm{H}_{2}(g)+\\mathrm{C}_{2} \\mathrm{H}_{4}(g) \\longrightarrow \\mathrm{C}_{2} \\mathrm{H}_{6}(g)\n$$"}
{"id": 3864, "contents": "1327. Answer: - \n$-120.6 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}$"}
{"id": 3865, "contents": "1329. Determination of $\\boldsymbol{\\Delta} \\boldsymbol{S}^{\\circ}$ - \nCalculate the standard entropy change for the combustion of methanol, $\\mathrm{CH}_{3} \\mathrm{OH}$ :\n\n$$\n2 \\mathrm{CH}_{3} \\mathrm{OH}(l)+3 \\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{CO}_{2}(g)+4 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$"}
{"id": 3866, "contents": "1330. Solution - \nCalculate the entropy change using standard entropies as shown above:\n\n$$\n\\begin{gathered}\n\\Delta S^{\\circ}=\\sum v S^{\\circ}(\\text { products })-\\sum \\nu S^{\\circ}(\\text { reactants }) \\\\\n{\\left[2 \\mathrm{~mol} \\times S^{\\circ}\\left(\\mathrm{CO}_{2}(g)\\right)+4 \\mathrm{~mol} \\times S^{\\circ}\\left(\\mathrm{H}_{2} \\mathrm{O}(l)\\right)\\right]-\\left[2 \\mathrm{~mol} \\times S^{\\circ}\\left(\\mathrm{CH}_{3} \\mathrm{OH}(l)\\right)+3 \\mathrm{~mol} \\times S^{\\circ}\\left(\\mathrm{O}_{2}(g)\\right)\\right]} \\\\\n=\\{[2(213.8)+4 \\times 70.0]-[2(126.8)+3(205.03)]\\}=-161.1 \\mathrm{~J} / \\mathrm{K}\n\\end{gathered}\n$$"}
{"id": 3867, "contents": "1331. Check Your Learning - \nCalculate the standard entropy change for the following reaction:\n\n$$\n\\mathrm{Ca}(\\mathrm{OH})_{2}(\\mathrm{~s}) \\longrightarrow \\mathrm{CaO}(s)+\\mathrm{H}_{2} \\mathrm{O}(l)\n$$"}
{"id": 3868, "contents": "1332. Answer: - \n$24.7 \\mathrm{~J} / \\mathrm{K}$"}
{"id": 3869, "contents": "1333. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Define Gibbs free energy, and describe its relation to spontaneity\n- Calculate free energy change for a process using free energies of formation for its reactants and products\n- Calculate free energy change for a process using enthalpies of formation and the entropies for its reactants and products\n\nOne of the challenges of using the second law of thermodynamics to determine if a process is spontaneous is that it requires measurements of the entropy change for the system and the entropy change for the surroundings. An alternative approach involving a new thermodynamic property defined in terms of system properties only was introduced in the late nineteenth century by American mathematician Josiah Willard Gibbs. This new property is called the Gibbs free energy ( $\\boldsymbol{G}$ ) (or simply the free energy), and it is defined in terms of a system's enthalpy and entropy as the following:\n\n$$\nG=H-T S\n$$\n\nFree energy is a state function, and at constant temperature and pressure, the free energy change ( $\\boldsymbol{\\Delta} \\boldsymbol{G}$ ) may be expressed as the following:\n\n$$\n\\Delta G=\\Delta H-T \\Delta S\n$$\n\n(For simplicity's sake, the subscript \"sys\" will be omitted henceforth.)\nThe relationship between this system property and the spontaneity of a process may be understood by recalling the previously derived second law expression:\n\n$$\n\\Delta S_{\\mathrm{univ}}=\\Delta S+\\frac{q_{\\mathrm{surr}}}{T}\n$$\n\nThe first law requires that $q_{\\text {surr }}=-q_{\\text {sys }}$, and at constant pressure $q_{\\text {sys }}=\\Delta H$, so this expression may be rewritten as:\n\n$$\n\\Delta S_{\\mathrm{univ}}=\\Delta S-\\frac{\\Delta H}{T}\n$$\n\nMultiplying both sides of this equation by $-T$, and rearranging yields the following:\n\n$$\n-T \\Delta S_{\\mathrm{univ}}=\\Delta H-T \\Delta S\n$$\n\nComparing this equation to the previous one for free energy change shows the following relation:\n\n$$\n\\Delta G=-T \\Delta S_{\\text {univ }}\n$$"}
{"id": 3870, "contents": "1333. LEARNING OBJECTIVES - \nComparing this equation to the previous one for free energy change shows the following relation:\n\n$$\n\\Delta G=-T \\Delta S_{\\text {univ }}\n$$\n\nThe free energy change is therefore a reliable indicator of the spontaneity of a process, being directly related to the previously identified spontaneity indicator, $\\Delta S_{\\text {univ }}$. Table 12.3 summarizes the relation between the spontaneity of a process and the arithmetic signs of these indicators.\n\nRelation between Process Spontaneity and Signs of Thermodynamic Properties\n\n| $\\Delta S_{\\text {univ }}>0$ | $\\Delta G<0$ | spontaneous |\n| :---: | :---: | :---: |\n| $\\Delta S_{\\text {univ }}<0$ | $\\Delta G>0$ | nonspontaneous |\n| $\\Delta S_{\\text {univ }}=0$ | $\\Delta G=0$ | at equilibrium |\n\nTABLE 12.3"}
{"id": 3871, "contents": "1334. What's \"Free\" about $\\Delta G$ ? - \nIn addition to indicating spontaneity, the free energy change also provides information regarding the amount of useful work ( $w$ ) that may be accomplished by a spontaneous process. Although a rigorous treatment of this subject is beyond the scope of an introductory chemistry text, a brief discussion is helpful for gaining a better perspective on this important thermodynamic property.\n\nFor this purpose, consider a spontaneous, exothermic process that involves a decrease in entropy. The free energy, as defined by\n\n$$\n\\Delta G=\\Delta H-T \\Delta S\n$$\n\nmay be interpreted as representing the difference between the energy produced by the process, $\\Delta H$, and the energy lost to the surroundings, $T \\Delta S$. The difference between the energy produced and the energy lost is the energy available (or \"free\") to do useful work by the process, $\\Delta G$. If the process somehow could be made to take place under conditions of thermodynamic reversibility, the amount of work that could be done would be maximal:\n\n$$\n\\Delta G=w_{\\max }\n$$\n\nHowever, as noted previously in this chapter, such conditions are not realistic. In addition, the technologies used to extract work from a spontaneous process (e.g., automobile engine, steam turbine) are never 100\\% efficient, and so the work done by these processes is always less than the theoretical maximum. Similar reasoning may be applied to a nonspontaneous process, for which the free energy change represents the minimum amount of work that must be done on the system to carry out the process."}
{"id": 3872, "contents": "1335. Calculating Free Energy Change - \nFree energy is a state function, so its value depends only on the conditions of the initial and final states of the system. A convenient and common approach to the calculation of free energy changes for physical and chemical reactions is by use of widely available compilations of standard state thermodynamic data. One method involves the use of standard enthalpies and entropies to compute standard free energy changes, $\\boldsymbol{\\Delta} \\boldsymbol{G}^{\\circ}$, according to the following relation:\n\n$$\n\\Delta G^{\\circ}=\\Delta H^{\\circ}-T \\Delta S^{\\circ}\n$$"}
{"id": 3873, "contents": "1337. Using Standard Enthalpy and Entropy Changes to Calculate $\\boldsymbol{\\Delta} \\boldsymbol{G}^{\\circ}$ - \nUse standard enthalpy and entropy data from Appendix G to calculate the standard free energy change for the vaporization of water at room temperature ( 298 K ). What does the computed value for $\\Delta G^{\\circ}$ say about the spontaneity of this process?"}
{"id": 3874, "contents": "1338. Solution - \nThe process of interest is the following:\n\n$$\n\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(g)\n$$\n\nThe standard change in free energy may be calculated using the following equation:\n\n$$\n\\Delta G^{\\circ}=\\Delta H^{\\circ}-T \\Delta S^{\\circ}\n$$\n\nFrom Appendix G:\n\n| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}(\\mathrm{kJ} / \\mathrm{mol})$ |  |\n| :---: | :---: | :---: |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{I})$ | -286.83 | 70.0 |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ | -241.82 | 188.8 |\n\nUsing the appendix data to calculate the standard enthalpy and entropy changes yields:"}
{"id": 3875, "contents": "1338. Solution - \nUsing the appendix data to calculate the standard enthalpy and entropy changes yields:\n\n$$\n\\begin{gathered}\n\\Delta H^{\\circ}=\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{H}_{2} \\mathrm{O}(g)\\right)-\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{H}_{2} \\mathrm{O}(l)\\right) \\\\\n=[-241.82 \\mathrm{~kJ} / \\mathrm{mol}-(-286.83)] \\mathrm{kJ} / \\mathrm{mol}=45.01 \\mathrm{~kJ} \\\\\n\\Delta S^{\\circ}=1 \\mathrm{~mol} \\times S^{\\circ}\\left(\\mathrm{H}_{2} \\mathrm{O}(g)\\right)-1 \\mathrm{~mol} \\times S^{\\circ}\\left(\\mathrm{H}_{2} \\mathrm{O}(l)\\right) \\\\\n=(1 \\mathrm{~mol}) 188.8 \\mathrm{~J} / \\mathrm{mol} \\cdot \\mathrm{~K}-(1 \\mathrm{~mol}) 70.0 \\mathrm{~J} / \\mathrm{mol} \\mathrm{~K}=118.8 \\mathrm{~J} / \\mathrm{mol} \\cdot \\mathrm{~K} \\\\\n\\Delta G^{\\circ}=\\Delta H^{\\circ}-T \\Delta S^{\\circ}\n\\end{gathered}\n$$\n\nSubstitution into the standard free energy equation yields:\n\n$$\n\\begin{gathered}\n\\Delta G^{\\circ}=\\Delta H^{\\circ}-T \\Delta S^{\\circ} \\\\\n=45.01 \\mathrm{~kJ}-(298 \\mathrm{~K} \\times 118.8 \\mathrm{~J} / \\mathrm{K}) \\times \\frac{1 \\mathrm{~kJ}}{1000 \\mathrm{~J}} \\\\\n45.01 \\mathrm{~kJ}-35.4 \\mathrm{~kJ}=9.6 \\mathrm{~kJ}\n\\end{gathered}\n$$"}
{"id": 3876, "contents": "1338. Solution - \nAt $298 \\mathrm{~K}\\left(25^{\\circ} \\mathrm{C}\\right) \\Delta G^{\\circ}>0$, so boiling is nonspontaneous (not spontaneous)."}
{"id": 3877, "contents": "1339. Check Your Learning - \nUse standard enthalpy and entropy data from Appendix G to calculate the standard free energy change for the reaction shown here ( 298 K ). What does the computed value for $\\Delta G^{\\circ}$ say about the spontaneity of this process?\n\n$$\n\\mathrm{C}_{2} \\mathrm{H}_{6}(g) \\longrightarrow \\mathrm{H}_{2}(g)+\\mathrm{C}_{2} \\mathrm{H}_{4}(g)\n$$"}
{"id": 3878, "contents": "1340. Answer: - \n$\\Delta G^{\\circ}=102.0 \\mathrm{~kJ} / \\mathrm{mol}$; the reaction is nonspontaneous (not spontaneous) at $25^{\\circ} \\mathrm{C}$.\n\nThe standard free energy change for a reaction may also be calculated from standard free energy of formation $\\Delta \\mathbf{G}^{\\mathbf{o}} \\mathbf{f}$ values of the reactants and products involved in the reaction. The standard free energy of formation is the free energy change that accompanies the formation of one mole of a substance from its elements in their standard states. Similar to the standard enthalpy of formation, $\\Delta G_{\\mathrm{f}}^{\\circ}$ is by definition zero for elemental substances in their standard states. The approach used to calculate $\\Delta G^{\\circ}$ for a reaction from $\\Delta G_{\\mathrm{f}}^{\\circ}$ values is the same as that demonstrated previously for enthalpy and entropy changes. For the reaction\n\n$$\nm \\mathrm{~A}+n \\mathrm{~B} \\longrightarrow x \\mathrm{C}+y \\mathrm{D}\n$$\n\nthe standard free energy change at room temperature may be calculated as\n\n$$\n\\begin{gathered}\n\\Delta G^{\\circ}=\\sum \\nu \\Delta G^{\\circ} \\text { (products) }-\\sum \\nu \\Delta G^{\\circ}(\\text { reactants }) \\\\\n=\\left[x \\Delta G_{\\mathrm{f}}^{\\circ}(\\mathrm{C})+y \\Delta G_{\\mathrm{f}}^{\\circ}(\\mathrm{D})\\right]-\\left[m \\Delta G_{\\mathrm{f}}^{\\circ}(\\mathrm{A})+n \\Delta G_{\\mathrm{f}}^{\\circ}(\\mathrm{B})\\right] .\n\\end{gathered}\n$$"}
{"id": 3879, "contents": "1342. Using Standard Free Energies of Formation to Calculate $\\Delta \\boldsymbol{G}^{\\circ}$ - \nConsider the decomposition of yellow mercury(II) oxide.\n\n$$\n\\mathrm{HgO}(s, \\text { yellow }) \\longrightarrow \\mathrm{Hg}(l)+\\frac{1}{2} \\mathrm{O}_{2}(g)\n$$\n\nCalculate the standard free energy change at room temperature, $\\Delta G^{\\circ}$, using (a) standard free energies of formation and (b) standard enthalpies of formation and standard entropies. Do the results indicate the reaction to be spontaneous or nonspontaneous under standard conditions?"}
{"id": 3880, "contents": "1343. Solution - \nThe required data are available in Appendix G and are shown here.\n\n| Compound | $\\Delta G_{\\mathrm{f}}^{\\circ}(\\mathrm{kJ} / \\mathrm{mol})$ | $\\Delta H_{\\mathrm{f}}^{\\circ}(\\mathrm{kJ} / \\mathrm{mol})$ | $S^{\\circ}(\\mathrm{J} / \\mathrm{K} \\cdot \\mathrm{mol})$ |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{HgO}(s$, yellow $)$ | -58.43 | -90.46 | 71.13 |\n| $\\mathrm{Hg}(\\mathrm{l})$ | 0 | 0 | 75.9 |\n| $\\mathrm{O}_{2}(g)$ | 0 | 0 | 205.2 |\n\n(a) Using free energies of formation:\n\n$$\n\\begin{gathered}\n\\Delta G^{\\circ}=\\sum \\nu G_{\\mathrm{f}}^{\\circ}(\\text { products })-\\sum \\nu \\Delta G_{\\mathrm{f}}^{\\circ}(\\text { reactants }) \\\\\n=\\left[1 \\Delta G_{\\mathrm{f}}^{\\circ} \\operatorname{Hg}(l)+\\frac{1}{2} \\Delta G_{\\mathrm{f}}^{\\circ} \\mathrm{O}_{2}(g)\\right]-1 \\Delta G_{\\mathrm{f}}^{\\circ} \\mathrm{HgO}(s, \\text { yellow }) \\\\\n=\\left[1 \\mathrm{~mol}(0 \\mathrm{~kJ} / \\mathrm{mol})+\\frac{1}{2} \\operatorname{mol}(0 \\mathrm{~kJ} / \\mathrm{mol})\\right]-1 \\mathrm{~mol}(-58.43 \\mathrm{~kJ} / \\mathrm{mol})=58.43 \\mathrm{~kJ} / \\mathrm{mol}\n\\end{gathered}\n$$\n\n(b) Using enthalpies and entropies of formation:"}
{"id": 3881, "contents": "1343. Solution - \n(b) Using enthalpies and entropies of formation:\n\n$$\n\\begin{gathered}\n\\Delta H^{\\circ}=\\sum \\nu \\Delta H_{\\mathrm{f}}^{\\circ}(\\text { products })-\\sum \\nu \\Delta H_{\\mathrm{f}}^{\\circ}(\\text { reactants }) \\\\\n=\\left[1 \\Delta H_{\\mathrm{f}}^{\\circ} \\mathrm{Hg}(l)+\\frac{1}{2} \\Delta H_{\\mathrm{f}}^{\\circ} \\mathrm{O}_{2}(g)\\right]-1 \\Delta H_{\\mathrm{f}}^{\\circ} \\mathrm{HgO}(s, \\text { yellow }) \\\\\n=\\left[1 \\mathrm{~mol}(0 \\mathrm{~kJ} / \\mathrm{mol})+\\frac{1}{2} \\operatorname{mol}(0 \\mathrm{~kJ} / \\mathrm{mol})\\right]-1 \\mathrm{~mol}(-90.46 \\mathrm{~kJ} / \\mathrm{mol})=90.46 \\mathrm{~kJ} / \\mathrm{mol} \\\\\n\\Delta S^{\\circ}=\\sum \\nu \\Delta S^{\\circ}(\\text { products })-\\sum \\nu \\Delta S^{\\circ}(\\text { reactants }) \\\\\n=\\left[1 \\Delta S^{\\circ} \\mathrm{Hg}(l)+\\frac{1}{2} \\Delta S^{\\circ} \\mathrm{O}_{2}(g)\\right]-1 \\Delta S^{\\circ} \\mathrm{HgO}(s, \\text { yellow })\n\\end{gathered}\n$$"}
{"id": 3882, "contents": "1343. Solution - \n$$\n\\begin{gathered}\n=\\left[1 \\mathrm{~mol}(75.9 \\mathrm{~J} / \\mathrm{mol} \\mathrm{~K})+\\frac{1}{2} \\operatorname{mol}(205.2 \\mathrm{~J} / \\mathrm{mol} \\mathrm{~K})\\right]-1 \\mathrm{~mol}(71.13 \\mathrm{~J} / \\mathrm{mol} \\mathrm{~K})=107.4 \\mathrm{~J} / \\mathrm{mol} \\mathrm{~K} \\\\\n\\Delta G^{\\circ}=\\Delta H^{\\circ}-T \\Delta S^{\\circ}=90.46 \\mathrm{~kJ}-298.15 \\mathrm{~K} \\times 107.4 \\mathrm{~J} / \\mathrm{K} \\cdot \\mathrm{~mol} \\times \\frac{1 \\mathrm{~kJ}}{1000 \\mathrm{~J}} \\\\\n\\Delta G^{\\circ}=(90.46-32.01) \\mathrm{kJ} / \\mathrm{mol}=58.45 \\mathrm{~kJ} / \\mathrm{mol}\n\\end{gathered}\n$$\n\nBoth ways to calculate the standard free energy change at $25^{\\circ} \\mathrm{C}$ give the same numerical value (to three significant figures), and both predict that the process is nonspontaneous (not spontaneous) at room temperature."}
{"id": 3883, "contents": "1344. Check Your Learning - \nCalculate $\\Delta G^{\\circ}$ using (a) free energies of formation and (b) enthalpies of formation and entropies (Appendix G). Do the results indicate the reaction to be spontaneous or nonspontaneous at $25^{\\circ} \\mathrm{C}$ ?\n\n$$\n\\mathrm{C}_{2} \\mathrm{H}_{4}(g) \\longrightarrow \\mathrm{H}_{2}(g)+\\mathrm{C}_{2} \\mathrm{H}_{2}(g)\n$$"}
{"id": 3884, "contents": "1345. Answer: - \n(a) $140.8 \\mathrm{~kJ} / \\mathrm{mol}$, nonspontaneous\n(b) $141.5 \\mathrm{~kJ} / \\mathrm{mol}$, nonspontaneous"}
{"id": 3885, "contents": "1346. Free Energy Changes for Coupled Reactions - \nThe use of free energies of formation to compute free energy changes for reactions as described above is possible because $\\Delta \\mathrm{G}$ is a state function, and the approach is analogous to the use of Hess' Law in computing enthalpy changes (see the chapter on thermochemistry). Consider the vaporization of water as an example:\n\n$$\n\\mathrm{H}_{2} \\mathrm{O}(l) \\rightarrow \\mathrm{H}_{2} \\mathrm{O}(g)\n$$\n\nAn equation representing this process may be derived by adding the formation reactions for the two phases of water (necessarily reversing the reaction for the liquid phase). The free energy change for the sum reaction is the sum of free energy changes for the two added reactions:\n\n$$\n\\begin{array}{ll}\n\\mathrm{H}_{2}(g)+1 / 2 \\mathrm{O}_{2}(g) \\rightarrow \\mathrm{H}_{2} \\mathrm{O}(g) & \\Delta G_{\\mathrm{f}, \\mathrm{gas}}^{\\circ} \\\\\n\\mathrm{H}_{2} \\mathrm{O}(l) \\rightarrow \\mathrm{H}_{2}(g)+1 / 2 \\mathrm{O}_{2}(g) & -\\Delta G_{\\mathrm{f}, \\text { liquid }}^{\\circ} \\\\\n\\hline \\mathrm{H}_{2} \\mathrm{O}(l) \\rightarrow \\mathrm{H}_{2} \\mathrm{O}(g) & \\Delta G=\\Delta G_{\\mathrm{f}, \\mathrm{gas}}^{\\circ}-\\Delta G_{\\mathrm{f}, \\text { liquid }}^{\\circ}\n\\end{array}\n$$\n\nThis approach may also be used in cases where a nonspontaneous reaction is enabled by coupling it to a spontaneous reaction. For example, the production of elemental zinc from zinc sulfide is thermodynamically unfavorable, as indicated by a positive value for $\\Delta G^{\\circ}$ :\n\n$$\n\\mathrm{ZnS}(s) \\rightarrow \\mathrm{Zn}(s)+\\mathrm{S}(s) \\quad \\Delta G_{1}^{\\circ}=201.3 \\mathrm{~kJ}\n$$\n\nThe industrial process for production of zinc from sulfidic ores involves coupling this decomposition reaction to the thermodynamically favorable oxidation of sulfur:"}
{"id": 3886, "contents": "1346. Free Energy Changes for Coupled Reactions - \nThe industrial process for production of zinc from sulfidic ores involves coupling this decomposition reaction to the thermodynamically favorable oxidation of sulfur:\n\n$$\n\\mathrm{S}(s)+\\mathrm{O}_{2}(g) \\rightarrow \\mathrm{SO}_{2}(g) \\quad \\Delta G_{2}^{\\circ}=-300.1 \\mathrm{~kJ}\n$$\n\nThe coupled reaction exhibits a negative free energy change and is spontaneous:\n\n$$\n\\mathrm{ZnS}(s)+\\mathrm{O}_{2}(g) \\rightarrow \\mathrm{Zn}(s)+\\mathrm{SO}(g) \\quad \\Delta G^{\\circ}=201.3 \\mathrm{~kJ}+-300.1 \\mathrm{~kJ}=-98.8 \\mathrm{~kJ}\n$$\n\nThis process is typically carried out at elevated temperatures, so this result obtained using standard free energy values is just an estimate. The gist of the calculation, however, holds true."}
{"id": 3887, "contents": "1348. Calculating Free Energy Change for a Coupled Reaction - \nIs a reaction coupling the decomposition of ZnS to the formation of $\\mathrm{H}_{2} \\mathrm{~S}$ expected to be spontaneous under standard conditions?"}
{"id": 3888, "contents": "1349. Solution - \nFollowing the approach outlined above and using free energy values from Appendix G:\n\n$$\n\\begin{array}{lll}\n\\text { Decomposition of zinc sulfide: } & \\mathrm{Zn}(s) \\rightarrow \\mathrm{Zn}(s)+\\mathrm{S}(s) & \\Delta G_{1}^{\\circ}=201.3 \\mathrm{~kJ} \\\\\n\\text { Formation of hydrogen sulfide: } & \\mathrm{S}(s)+\\mathrm{H}_{2}(g) \\rightarrow \\mathrm{H}_{2} \\mathrm{~S}(g) & \\Delta G_{2}^{\\circ}=-33.4 \\mathrm{~kJ} \\\\\n\\text { Coupled reaction: } & \\mathrm{ZnS}(s)+\\mathrm{H}_{2}(g) \\rightarrow \\mathrm{Zn}(s)+\\mathrm{H}_{2} \\mathrm{~S}(g) & \\Delta G^{\\circ}=201.3 \\mathrm{~kJ}+-33.4 \\mathrm{~kJ}=167.9 \\mathrm{~kJ}\n\\end{array}\n$$\n\nThe coupled reaction exhibits a positive free energy change and is thus nonspontaneous."}
{"id": 3889, "contents": "1350. Check Your Learning - \nWhat is the standard free energy change for the reaction below? Is the reaction expected to be spontaneous under standard conditions?\n\n$$\n\\mathrm{FeS}(s)+\\mathrm{O}_{2}(g) \\rightarrow \\mathrm{Fe}(s)+\\mathrm{SO}_{2}(g)\n$$"}
{"id": 3890, "contents": "1351. Answer: - \n-199.7 kJ; spontaneous"}
{"id": 3891, "contents": "1352. Key Terms - \nentropy ( $\\boldsymbol{S}$ ) state function that is a measure of the matter and/or energy dispersal within a system, determined by the number of system microstates; often described as a measure of the disorder of the system\nGibbs free energy change ( $\\boldsymbol{G}$ ) thermodynamic property defined in terms of system enthalpy and entropy; all spontaneous processes involve a decrease in $G$\nmicrostate possible configuration or arrangement of matter and energy within a system\nnonspontaneous process process that requires continual input of energy from an external source\nreversible process process that takes place so slowly as to be capable of reversing direction in response to an infinitesimally small change in conditions; hypothetical construct that can only be approximated by real processes\nsecond law of thermodynamics all spontaneous processes involve an increase in the entropy of the universe"}
{"id": 3892, "contents": "1353. Key Equations - \n$\\Delta S=\\frac{q_{\\mathrm{rev}}}{T}$\n$S=k \\ln W$\n$\\Delta S=k \\ln \\frac{W_{\\mathrm{f}}}{W_{\\mathrm{i}}}$\n$\\Delta S^{\\circ}=\\sum \\nu S^{\\circ}($ products $)-\\sum \\nu S^{\\circ}($ reactants $)$\n$\\Delta S=\\frac{q_{\\mathrm{rev}}}{T}$\n$\\Delta S_{\\text {univ }}=\\Delta S_{\\text {sys }}+\\Delta S_{\\text {surr }}$\n$\\Delta S_{\\text {univ }}=\\Delta S_{\\text {sys }}+\\Delta S_{\\text {surr }}=\\Delta S_{\\text {sys }}+\\frac{q_{\\text {surr }}}{T}$\n$\\Delta G=\\Delta H-T \\Delta S$"}
{"id": 3893, "contents": "1354. Summary - 1354.1. Spontaneity\nChemical and physical processes have a natural tendency to occur in one direction under certain conditions. A spontaneous process occurs without the need for a continual input of energy from some external source, while a nonspontaneous process requires such. Systems undergoing a spontaneous process may or may not experience a gain or loss of energy, but they will experience a change in the way matter and/or energy is distributed within the system."}
{"id": 3894, "contents": "1354. Summary - 1354.2. Entropy\nEntropy ( $S$ ) is a state function that can be related to\nspontaneous change process that takes place without a continuous input of energy from an external source\nstandard entropy ( $\\boldsymbol{S}^{\\circ}$ ) entropy for one mole of a substance at 1 bar pressure; tabulated values are usually determined at 298.15 K\nstandard entropy change ( $\\boldsymbol{\\Delta} \\boldsymbol{S}^{\\circ}$ ) change in entropy for a reaction calculated using the standard entropies\nstandard free energy change ( $\\boldsymbol{\\Delta} \\boldsymbol{G}^{\\mathbf{0}}$ ) change in free energy for a process occurring under standard conditions (1 bar pressure for gases, 1 M concentration for solutions)\nstandard free energy of formation ( $\\Delta \\boldsymbol{G}_{\\mathrm{f}}^{\\circ}$ ) change in free energy accompanying the formation of one mole of substance from its elements in their standard states\nthird law of thermodynamics entropy of a perfect crystal at absolute zero ( 0 K ) is zero\nthe number of microstates for a system (the number of ways the system can be arranged) and to the ratio of reversible heat to kelvin temperature. It may be interpreted as a measure of the dispersal or distribution of matter and/or energy in a system, and it is often described as representing the \"disorder\" of the system.\n\nFor a given substance, entropy depends on phase with $S_{\\text {solid }}<S_{\\text {liquid }}<S_{\\text {gas }}$. For different substances in the same physical state at a given temperature, entropy is typically greater for heavier atoms or more complex molecules. Entropy increases when a system is heated and when solutions form. Using these guidelines, the sign of entropy changes for\nsome chemical reactions and physical changes may be reliably predicted."}
{"id": 3895, "contents": "1354. Summary - 1354.3. The Second and Third Laws of Thermodynamics\nThe second law of thermodynamics states that a spontaneous process increases the entropy of the universe, $S_{\\text {univ }}>0$. If $\\Delta S_{\\text {univ }}<0$, the process is nonspontaneous, and if $\\Delta S_{\\text {univ }}=0$, the system is at equilibrium. The third law of thermodynamics establishes the zero for entropy as that of a perfect, pure crystalline solid at 0 K . With only one possible\nmicrostate, the entropy is zero. We may compute the standard entropy change for a process by using standard entropy values for the reactants and products involved in the process."}
{"id": 3896, "contents": "1354. Summary - 1354.4. Free Energy\nGibbs free energy $(G)$ is a state function defined with regard to system quantities only and may be used to predict the spontaneity of a process. A number of approaches to the computation of free energy changes are possible."}
{"id": 3897, "contents": "1355. Exercises - 1355.1. Spontaneity\n1. What is a spontaneous reaction?\n2. What is a nonspontaneous reaction?\n3. Indicate whether the following processes are spontaneous or nonspontaneous.\n(a) Liquid water freezing at a temperature below its freezing point\n(b) Liquid water freezing at a temperature above its freezing point\n(c) The combustion of gasoline\n(d) A ball thrown into the air\n(e) A raindrop falling to the ground\n(f) Iron rusting in a moist atmosphere\n4. A helium-filled balloon spontaneously deflates overnight as He atoms diffuse through the wall of the balloon. Describe the redistribution of matter and/or energy that accompanies this process.\n5. Many plastic materials are organic polymers that contain carbon and hydrogen. The oxidation of these plastics in air to form carbon dioxide and water is a spontaneous process; however, plastic materials tend to persist in the environment. Explain."}
{"id": 3898, "contents": "1355. Exercises - 1355.2. Entropy\n6. In Figure 12.8 all possible distributions and microstates are shown for four different particles shared between two boxes. Determine the entropy change, $\\Delta S$, if the particles are initially evenly distributed between the two boxes, but upon redistribution all end up in Box (b).\n7. In Figure 12.8 all of the possible distributions and microstates are shown for four different particles shared between two boxes. Determine the entropy change, $\\Delta S$, for the system when it is converted from distribution (b) to distribution (d).\n8. How does the process described in the previous item relate to the system shown in Figure 12.4?\n9. Consider a system similar to the one in Figure 12.8, except that it contains six particles instead of four. What is the probability of having all the particles in only one of the two boxes in the case? Compare this with the similar probability for the system of four particles that we have derived to be equal to $\\frac{1}{8}$. What does this comparison tell us about even larger systems?\n10. Consider the system shown in Figure 12.9. What is the change in entropy for the process where the energy is initially associated only with particle A, but in the final state the energy is distributed between two different particles?\n11. Consider the system shown in Figure 12.9. What is the change in entropy for the process where the energy is initially associated with particles A and B, and the energy is distributed between two particles in different boxes (one in A-B, the other in C-D)?\n12. Arrange the following sets of systems in order of increasing entropy. Assume one mole of each substance and the same temperature for each member of a set.\n(a) $\\mathrm{H}_{2}(g), \\mathrm{HBrO}_{4}(g), \\mathrm{HBr}(g)$\n(b) $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{I}), \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}), \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$\n(c) $\\mathrm{He}(g), \\mathrm{Cl}_{2}(g), \\mathrm{P}_{4}(g)$"}
{"id": 3899, "contents": "1355. Exercises - 1355.2. Entropy\n(c) $\\mathrm{He}(g), \\mathrm{Cl}_{2}(g), \\mathrm{P}_{4}(g)$\n13. At room temperature, the entropy of the halogens increases from $I_{2}$ to $B r_{2}$ to $\\mathrm{Cl}_{2}$. Explain.\n14. Consider two processes: sublimation of $I_{2}(s)$ and melting of $I_{2}(S)$ (Note: the latter process can occur at the same temperature but somewhat higher pressure).\n$\\mathrm{I}_{2}(s) \\longrightarrow \\mathrm{I}_{2}(g)$\n$\\mathrm{I}_{2}(s) \\longrightarrow \\mathrm{I}_{2}(l)$\nIs $\\Delta S$ positive or negative in these processes? In which of the processes will the magnitude of the entropy change be greater?\n15. Indicate which substance in the given pairs has the higher entropy value. Explain your choices.\n(a) $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(I)$ or $\\mathrm{C}_{3} \\mathrm{H}_{7} \\mathrm{OH}(I)$\n(b) $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(\\mathrm{I})$ or $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(g)$\n(c) $2 \\mathrm{H}(\\mathrm{g})$ or $\\mathrm{H}(\\mathrm{g})$\n16. Predict the sign of the entropy change for the following processes.\n(a) An ice cube is warmed to near its melting point.\n(b) Exhaled breath forms fog on a cold morning.\n(c) Snow melts.\n17. Predict the sign of the entropy change for the following processes. Give a reason for your prediction.\n(a) $\\mathrm{Na}^{+}(a q)+\\mathrm{Cl}^{-}(a q) \\longrightarrow \\mathrm{NaCl}(s)$\n(b) $2 \\mathrm{Fe}(s)+\\frac{3}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{Fe}_{2} \\mathrm{O}_{2}(s)$"}
{"id": 3900, "contents": "1355. Exercises - 1355.2. Entropy\n(b) $2 \\mathrm{Fe}(s)+\\frac{3}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{Fe}_{2} \\mathrm{O}_{2}(s)$\n(c) $2 \\mathrm{C}_{6} \\mathrm{H}_{14}(l)+19 \\mathrm{O}_{2}(g) \\longrightarrow 14 \\mathrm{H}_{2} \\mathrm{O}(g)+12 \\mathrm{CO}_{2}(g)$\n18. Write the balanced chemical equation for the combustion of methane, $\\mathrm{CH}_{4}(g)$, to give carbon dioxide and water vapor. Explain why it is difficult to predict whether $\\Delta S$ is positive or negative for this chemical reaction.\n19. Write the balanced chemical equation for the combustion of benzene, $\\mathrm{C}_{6} \\mathrm{H}_{6}(\\mathrm{l})$, to give carbon dioxide and water vapor. Would you expect $\\Delta S$ to be positive or negative in this process?"}
{"id": 3901, "contents": "1355. Exercises - 1355.3. The Second and Third Laws of Thermodynamics\n20. What is the difference between $\\Delta S$ and $\\Delta S^{\\circ}$ for a chemical change?\n21. Calculate $\\Delta S^{\\circ}$ for the following changes.\n(a) $\\mathrm{SnCl}_{4}(l) \\longrightarrow \\mathrm{SnCl}_{4}(g)$\n(b) $\\mathrm{CS}_{2}(g) \\longrightarrow \\mathrm{CS}_{2}(l)$\n(c) $\\mathrm{Cu}(s) \\longrightarrow \\mathrm{Cu}(g)$\n(d) $\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(g)$\n(e) $2 \\mathrm{H}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n(f) $2 \\mathrm{HCl}(g)+\\mathrm{Pb}(s) \\longrightarrow \\mathrm{PbCl}_{2}(s)+\\mathrm{H}_{2}(g)$\n$(\\mathrm{g}) \\mathrm{Zn}(s)+\\mathrm{CuSO}_{4}(s) \\longrightarrow \\mathrm{Cu}(s)+\\mathrm{ZnSO}_{4}(s)$\n22. Determine the entropy change for the combustion of liquid ethanol, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$, under the standard conditions to give gaseous carbon dioxide and liquid water.\n23. Determine the entropy change for the combustion of gaseous propane, $\\mathrm{C}_{3} \\mathrm{H}_{8}$, under the standard conditions to give gaseous carbon dioxide and water.\n24. \"Thermite\" reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite reaction is $\\mathrm{Fe}_{2} \\mathrm{O}_{3}(s)+2 \\mathrm{Al}(s) \\longrightarrow \\mathrm{Al}_{2} \\mathrm{O}_{3}(s)+2 \\mathrm{Fe}(s)$. Is the reaction spontaneous at room temperature under standard conditions? During the reaction, the surroundings absorb $851.8 \\mathrm{~kJ} / \\mathrm{mol}$ of heat."}
{"id": 3902, "contents": "1355. Exercises - 1355.3. The Second and Third Laws of Thermodynamics\n25. Using the relevant $S^{\\circ}$ values listed in Appendix G, calculate $\\Delta S^{\\circ}{ }_{298}$ for the following changes:\n(a) $\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\longrightarrow 2 \\mathrm{NH}_{3}(g)$\n(b) $\\mathrm{N}_{2}(g)+\\frac{5}{2} \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{N}_{2} \\mathrm{O}_{5}(g)$\n26. From the following information, determine $\\Delta S^{\\circ}$ for the following:\n$\\mathrm{N}(g)+\\mathrm{O}(g) \\longrightarrow \\mathrm{NO}(g) \\quad \\Delta S^{\\circ}=$ ?\n$\\mathrm{N}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{NO}(g) \\quad \\Delta S^{\\circ}=24.8 \\mathrm{~J} / \\mathrm{K}$\n$\\mathrm{N}_{2}(g) \\longrightarrow 2 \\mathrm{~N}(g) \\quad \\Delta S^{\\circ}=115.0 \\mathrm{~J} / \\mathrm{K}$\n$\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{O}(g) \\quad \\Delta S^{\\circ}=117.0 \\mathrm{~J} / \\mathrm{K}$\n27. By calculating $\\Delta S_{\\text {univ }}$ at each temperature, determine if the melting of 1 mole of $\\mathrm{NaCl}(s)$ is spontaneous at $500^{\\circ} \\mathrm{C}$ and at $700^{\\circ} \\mathrm{C}$.\n$S_{\\mathrm{NaCl}(s)}^{\\circ}=72.11 \\frac{\\mathrm{~J}}{\\mathrm{~mol} \\cdot \\mathrm{~K}} \\quad S_{\\mathrm{NaCl}(l)}^{\\circ}=95.06 \\frac{\\mathrm{~J}}{\\mathrm{~mol} \\cdot \\mathrm{~K}} \\quad \\Delta H_{\\text {fusion }}^{\\circ}=27.95 \\mathrm{~kJ} / \\mathrm{mol}$"}
{"id": 3903, "contents": "1355. Exercises - 1355.3. The Second and Third Laws of Thermodynamics\nWhat assumptions are made about the thermodynamic information (entropy and enthalpy values) used to solve this problem?\n28. Use the standard entropy data in Appendix $G$ to determine the change in entropy for each of the following reactions. All the processes occur at the standard conditions and $25^{\\circ} \\mathrm{C}$.\n(a) $\\mathrm{MnO}_{2}(s) \\longrightarrow \\mathrm{Mn}(s)+\\mathrm{O}_{2}(g)$\n(b) $\\mathrm{H}_{2}(g)+\\mathrm{Br}_{2}(l) \\longrightarrow 2 \\mathrm{HBr}(g)$\n(c) $\\mathrm{Cu}(s)+\\mathrm{S}(g) \\longrightarrow \\mathrm{CuS}(s)$\n(d) $2 \\mathrm{LiOH}(s)+\\mathrm{CO}_{2}(g) \\longrightarrow \\mathrm{Li}_{2} \\mathrm{CO}_{3}(s)+\\mathrm{H}_{2} \\mathrm{O}(g)$\n(e) $\\mathrm{CH}_{4}(g)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{C}(s$, graphite $)+2 \\mathrm{H}_{2} \\mathrm{O}(g)$\n(f) $\\mathrm{CS}_{2}(g)+3 \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{CCl}_{4}(g)+\\mathrm{S}_{2} \\mathrm{Cl}_{2}(g)$\n29. Use the standard entropy data in Appendix $G$ to determine the change in entropy for each of the following reactions. All the processes occur at the standard conditions and $25^{\\circ} \\mathrm{C}$.\n(a) $\\mathrm{C}(s$, graphite $)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)$\n(b) $\\mathrm{O}_{2}(g)+\\mathrm{N}_{2}(g) \\longrightarrow 2 \\mathrm{NO}(g)$\n(c) $2 \\mathrm{Cu}(s)+\\mathrm{S}(g) \\longrightarrow \\mathrm{Cu}_{2} \\mathrm{~S}(s)$"}
{"id": 3904, "contents": "1355. Exercises - 1355.3. The Second and Third Laws of Thermodynamics\n(c) $2 \\mathrm{Cu}(s)+\\mathrm{S}(g) \\longrightarrow \\mathrm{Cu}_{2} \\mathrm{~S}(s)$\n(d) $\\mathrm{CaO}(s)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Ca}(\\mathrm{OH})_{2}(s)$\n(e) $\\mathrm{Fe}_{2} \\mathrm{O}_{3}(s)+3 \\mathrm{CO}(g) \\longrightarrow 2 \\mathrm{Fe}(s)+3 \\mathrm{CO}_{2}(g)$\n(f) $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(s) \\longrightarrow \\mathrm{CaSO}_{4}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(g)$"}
{"id": 3905, "contents": "1355. Exercises - 1355.4. Free Energy\n30. What is the difference between $\\Delta G$ and $\\Delta G^{\\circ}$ for a chemical change?\n31. A reaction has $\\Delta H^{\\circ}=100 \\mathrm{~kJ} / \\mathrm{mol}$ and $\\Delta S^{\\circ}=250 \\mathrm{~J} / \\mathrm{mol} \\cdot \\mathrm{K}$. Is the reaction spontaneous at room temperature? If not, under what temperature conditions will it become spontaneous?\n32. Explain what happens as a reaction starts with $\\Delta G<0$ (negative) and reaches the point where $\\Delta G=0$.\n33. Use the standard free energy of formation data in Appendix $G$ to determine the free energy change for each of the following reactions, which are run under standard state conditions and $25^{\\circ} \\mathrm{C}$. Identify each as either spontaneous or nonspontaneous at these conditions.\n(a) $\\mathrm{MnO}_{2}(s) \\longrightarrow \\mathrm{Mn}(s)+\\mathrm{O}_{2}(g)$\n(b) $\\mathrm{H}_{2}(g)+\\mathrm{Br}_{2}(l) \\longrightarrow 2 \\mathrm{HBr}(g)$\n(c) $\\mathrm{Cu}(s)+\\mathrm{S}(g) \\longrightarrow \\mathrm{CuS}(s)$\n(d) $2 \\mathrm{LiOH}(s)+\\mathrm{CO}_{2}(g) \\longrightarrow \\mathrm{Li}_{2} \\mathrm{CO}_{3}(s)+\\mathrm{H}_{2} \\mathrm{O}(g)$\n(e) $\\mathrm{CH}_{4}(g)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{C}(s$, graphite $)+2 \\mathrm{H}_{2} \\mathrm{O}(g)$\n(f) $\\mathrm{CS}_{2}(g)+3 \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{CCl}_{4}(g)+\\mathrm{S}_{2} \\mathrm{Cl}_{2}(g)$\n34. Use the standard free energy data in Appendix $G$ to determine the free energy change for each of the following reactions, which are run under standard state conditions and $25^{\\circ} \\mathrm{C}$. Identify each as either spontaneous or nonspontaneous at these conditions."}
{"id": 3906, "contents": "1355. Exercises - 1355.4. Free Energy\n(a) $\\mathrm{C}(s$, graphite $)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)$\n(b) $\\mathrm{O}_{2}(g)+\\mathrm{N}_{2}(g) \\longrightarrow 2 \\mathrm{NO}(g)$\n(c) $2 \\mathrm{Cu}(s)+\\mathrm{S}(g) \\longrightarrow \\mathrm{Cu}_{2} \\mathrm{~S}(s)$\n(d) $\\mathrm{CaO}(s)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Ca}(\\mathrm{OH})_{2}(s)$\n(e) $\\mathrm{Fe}_{2} \\mathrm{O}_{3}(s)+3 \\mathrm{CO}(g) \\longrightarrow 2 \\mathrm{Fe}(s)+3 \\mathrm{CO}_{2}(g)$\n(f) $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(s) \\longrightarrow \\mathrm{CaSO}_{4}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(g)$\n35. Given:\n$\\mathrm{P}_{4}(s)+5 \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{P}_{4} \\mathrm{O}_{10}(s)$\n$\\Delta G^{\\circ}=-2697.0 \\mathrm{~kJ} / \\mathrm{mol}$\n$2 \\mathrm{H}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(g)$\n$\\Delta G^{\\circ}=-457.18 \\mathrm{~kJ} / \\mathrm{mol}$\n$6 \\mathrm{H}_{2} \\mathrm{O}(g)+\\mathrm{P}_{4} \\mathrm{O}_{10}(s) \\longrightarrow 4 \\mathrm{H}_{3} \\mathrm{PO}_{4}(l)$\n$\\Delta G^{\\circ}=-428.66 \\mathrm{~kJ} / \\mathrm{mol}$\n(a) Determine the standard free energy of formation, $\\Delta G_{\\mathrm{f}}^{\\circ}$, for phosphoric acid."}
{"id": 3907, "contents": "1355. Exercises - 1355.4. Free Energy\n(a) Determine the standard free energy of formation, $\\Delta G_{\\mathrm{f}}^{\\circ}$, for phosphoric acid.\n(b) How does your calculated result compare to the value in Appendix G? Explain.\n36. Is the formation of ozone $\\left(\\mathrm{O}_{3}(\\mathrm{~g})\\right)$ from oxygen $\\left(\\mathrm{O}_{2}(\\mathrm{~g})\\right)$ spontaneous at room temperature under standard state conditions?\n37. Consider the decomposition of red mercury(II) oxide under standard state conditions.\n$2 \\mathrm{HgO}(s$, red $) \\longrightarrow 2 \\mathrm{Hg}(l)+\\mathrm{O}_{2}(g)$\n(a) Is the decomposition spontaneous under standard state conditions?\n(b) Above what temperature does the reaction become spontaneous?\n38. Among other things, an ideal fuel for the control thrusters of a space vehicle should decompose in a spontaneous exothermic reaction when exposed to the appropriate catalyst. Evaluate the following substances under standard state conditions as suitable candidates for fuels.\n(a) Ammonia: $2 \\mathrm{NH}_{3}(g) \\longrightarrow \\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g)$\n(b) Diborane: $\\mathrm{B}_{2} \\mathrm{H}_{6}(g) \\longrightarrow 2 \\mathrm{~B}(g)+3 \\mathrm{H}_{2}(g)$\n(c) Hydrazine: $\\mathrm{N}_{2} \\mathrm{H}_{4}(g) \\longrightarrow \\mathrm{N}_{2}(g)+2 \\mathrm{H}_{2}(g)$\n(d) Hydrogen peroxide: $\\mathrm{H}_{2} \\mathrm{O}_{2}(l) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(g)+\\frac{1}{2} \\mathrm{O}_{2}(g)$\n39. Calculate $\\Delta G^{\\circ}$ for each of the following reactions from the equilibrium constant at the temperature given.\n(a) $\\mathrm{N}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{NO}(g)$"}
{"id": 3908, "contents": "1355. Exercises - 1355.4. Free Energy\n(a) $\\mathrm{N}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{NO}(g)$\n$\\mathrm{T}=2000^{\\circ} \\mathrm{C} \\quad K_{p}=4.1 \\times 10^{-4}$\n(b) $\\mathrm{H}_{2}(g)+\\mathrm{I}_{2}(g) \\longrightarrow 2 \\mathrm{HI}(g)$\n$\\mathrm{T}=400^{\\circ} \\mathrm{C} \\quad K_{p}=50.0$\n(c) $\\mathrm{CO}_{2}(g)+\\mathrm{H}_{2}(g) \\longrightarrow \\mathrm{CO}(g)+\\mathrm{H}_{2} \\mathrm{O}(g)$ $\\mathrm{T}=980^{\\circ} \\mathrm{C} \\quad K_{p}=1.67$\n$(\\mathrm{d}) \\mathrm{CaCO}_{3}(s) \\longrightarrow \\mathrm{CaO}(s)+\\mathrm{CO}_{2}(g)$\n$\\mathrm{T}=900^{\\circ} \\mathrm{C} \\quad K_{p}=1.04$\n(e) $\\mathrm{HF}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{F}^{-}(a q)$\n$\\mathrm{T}=25^{\\circ} \\mathrm{C}$\n$K_{p}=7.2 \\times 10^{-4}$\n(f) $\\mathrm{AgBr}(s) \\longrightarrow \\mathrm{Ag}^{+}(a q)+\\mathrm{Br}^{-}(a q)$ $\\mathrm{T}=25^{\\circ} \\mathrm{C} \\quad K_{p}=3.3 \\times 10^{-13}$\n40. Calculate $\\Delta G^{\\circ}$ for each of the following reactions from the equilibrium constant at the temperature given.\n(a) $\\mathrm{Cl}_{2}(g)+\\mathrm{Br}_{2}(g) \\longrightarrow 2 \\mathrm{BrCl}(g)$\n$\\mathrm{T}=25^{\\circ} \\mathrm{C} \\quad K_{p}=4.7 \\times 10^{-2}$"}
{"id": 3909, "contents": "1355. Exercises - 1355.4. Free Energy\n$\\mathrm{T}=25^{\\circ} \\mathrm{C} \\quad K_{p}=4.7 \\times 10^{-2}$\n(b) $2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{SO}_{3}(g)$\n$\\mathrm{T}=500^{\\circ} \\mathrm{C}$\n$K_{p}=48.2$\n(c) $\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{O}(g)$\n$\\mathrm{T}=60^{\\circ} \\mathrm{C}$\n$K_{p}=0.196$\n(d) $\\mathrm{CoO}(s)+\\mathrm{CO}(g) \\rightleftharpoons \\mathrm{Co}(\\mathrm{s})+\\mathrm{CO}_{2}(g)$\n$\\mathrm{T}=550^{\\circ} \\mathrm{C} \\quad K_{p}=4.90 \\times 10^{2}$\n(e) $\\mathrm{CH}_{3} \\mathrm{NH}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{CH}_{3} \\mathrm{NH}_{3}{ }^{+}(a q)+\\mathrm{OH}^{-}(a q)$\n$\\mathrm{T}=25^{\\circ} \\mathrm{C} \\quad K_{p}=4.4 \\times 10^{-4}$\n(f) $\\mathrm{PbI}_{2}(s) \\longrightarrow \\mathrm{Pb}^{2+}(a q)+2 \\mathrm{I}^{-}(a q)$\n$\\mathrm{T}=25^{\\circ} \\mathrm{C}$\n$K_{p}=8.7 \\times 10^{-9}$\n41. Calculate the equilibrium constant at $25^{\\circ} \\mathrm{C}$ for each of the following reactions from the value of $\\Delta G^{\\circ}$ given.\n(a) $\\mathrm{O}_{2}(g)+2 \\mathrm{~F}_{2}(g) \\longrightarrow 2 \\mathrm{OF}_{2}(g)$\n$\\Delta G^{\\circ}=-9.2 \\mathrm{~kJ}$"}
{"id": 3910, "contents": "1355. Exercises - 1355.4. Free Energy\n$\\Delta G^{\\circ}=-9.2 \\mathrm{~kJ}$\n(b) $\\mathrm{I}_{2}(s)+\\mathrm{Br}_{2}(l) \\longrightarrow 2 \\operatorname{IBr}(g)$\n$\\Delta G^{\\circ}=7.3 \\mathrm{~kJ}$\n(c) $2 \\mathrm{LiOH}(s)+\\mathrm{CO}_{2}(g) \\longrightarrow \\mathrm{Li}_{2} \\mathrm{CO}_{3}(s)+\\mathrm{H}_{2} \\mathrm{O}(g)$\n$\\Delta G^{\\circ}=-79 \\mathrm{~kJ}$\n$(d) \\mathrm{N}_{2} \\mathrm{O}_{3}(g) \\longrightarrow \\mathrm{NO}(g)+\\mathrm{NO}_{2}(g)$\n$\\Delta G^{\\circ}=-1.6 \\mathrm{~kJ}$\n(e) $\\mathrm{SnCl}_{4}(l) \\longrightarrow \\mathrm{SnCl}_{4}(l)$\n$\\Delta G^{\\circ}=8.0 \\mathrm{~kJ}$\n42. Determine $\\Delta G^{\\circ}$ for the following reactions.\n(a) Antimony pentachloride decomposes at $448^{\\circ} \\mathrm{C}$. The reaction is:\n$\\mathrm{SbCl}_{5}(\\mathrm{~g}) \\longrightarrow \\mathrm{SbCl}_{3}(\\mathrm{~g})+\\mathrm{Cl}_{2}(\\mathrm{~g})$\nAn equilibrium mixture in a 5.00 L flask at $448{ }^{\\circ} \\mathrm{C}$ contains 3.85 g of $\\mathrm{SbCl}_{5}, 9.14 \\mathrm{~g}$ of $\\mathrm{SbCl}_{3}$, and 2.84 g of $\\mathrm{Cl}_{2}$.\n(b) Chlorine molecules dissociate according to this reaction:\n$\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{Cl}(g)$\n$1.00 \\%$ of $\\mathrm{Cl}_{2}$ molecules dissociate at 975 K and a pressure of 1.00 atm ."}
{"id": 3911, "contents": "1355. Exercises - 1355.4. Free Energy\n$1.00 \\%$ of $\\mathrm{Cl}_{2}$ molecules dissociate at 975 K and a pressure of 1.00 atm .\n43. Given that the $\\Delta G_{\\mathrm{f}}^{\\circ}$ for $\\mathrm{Pb}^{2+}(\\mathrm{aq})$ and $\\mathrm{Cl}^{-}(\\mathrm{aq})$ is $-24.3 \\mathrm{~kJ} / \\mathrm{mole}$ and $-131.2 \\mathrm{~kJ} /$ mole respectively, determine the solubility product, $K_{\\mathrm{sp}}$, for $\\mathrm{PbCl}_{2}(s)$.\n44. Determine the standard free energy change, $\\Delta \\boldsymbol{G}_{\\mathrm{f}}^{\\circ}$, for the formation of $S^{2-}(a q)$ given that the $\\Delta \\boldsymbol{G}_{\\mathrm{f}}^{\\circ}$ for $\\mathrm{Ag}^{+}(\\mathrm{aq})$ and $\\mathrm{Ag}_{2} \\mathrm{~S}(s)$ are $77.1 \\mathrm{~kJ} /$ mole and $-39.5 \\mathrm{~kJ} /$ mole respectively, and the solubility product for $\\mathrm{Ag}_{2} \\mathrm{~S}(s)$ is $8 \\times 10^{-51}$.\n45. Determine the standard enthalpy change, entropy change, and free energy change for the conversion of diamond to graphite. Discuss the spontaneity of the conversion with respect to the enthalpy and entropy changes. Explain why diamond spontaneously changing into graphite is not observed.\n46. The evaporation of one mole of water at 298 K has a standard free energy change of 8.58 kJ .\n$\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{O}(g) \\quad \\Delta G^{\\circ}=8.58 \\mathrm{~kJ}$\n(a) Is the evaporation of water under standard thermodynamic conditions spontaneous?\n(b) Determine the equilibrium constant, $K_{P}$, for this physical process."}
{"id": 3912, "contents": "1355. Exercises - 1355.4. Free Energy\n(a) Is the evaporation of water under standard thermodynamic conditions spontaneous?\n(b) Determine the equilibrium constant, $K_{P}$, for this physical process.\n(c) By calculating $\\Delta G$, determine if the evaporation of water at 298 K is spontaneous when the partial pressure of water, $P_{\\mathrm{H}_{2} \\mathrm{O}}$, is 0.011 atm .\n(d) If the evaporation of water were always nonspontaneous at room temperature, wet laundry would never dry when placed outside. In order for laundry to dry, what must be the value of $P_{\\mathrm{H}_{2} \\mathrm{O}}$ in the air?\n47. In glycolysis, the reaction of glucose (Glu) to form glucose-6-phosphate (G6P) requires ATP to be present as described by the following equation:\n$\\mathrm{Glu}+\\mathrm{ATP} \\longrightarrow \\mathrm{G} 6 \\mathrm{P}+\\mathrm{ADP} \\quad \\Delta G^{\\circ}=-17 \\mathrm{~kJ}$\nIn this process, ATP becomes ADP summarized by the following equation:\nATP $\\longrightarrow$ ADP $\\Delta G^{\\circ}=-30 \\mathrm{~kJ}$\nDetermine the standard free energy change for the following reaction, and explain why ATP is necessary to drive this process:\nGlu $\\longrightarrow$ G6P $\\quad \\Delta G^{\\circ}=$ ?\n48. One of the important reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to form fructose-6-phosphate (F6P):\n$\\mathrm{G6P} \\rightleftharpoons \\mathrm{~F} 6 \\mathrm{P} \\quad \\Delta G^{\\circ}=1.7 \\mathrm{~kJ}$\n(a) Is the reaction spontaneous or nonspontaneous under standard thermodynamic conditions?"}
{"id": 3913, "contents": "1355. Exercises - 1355.4. Free Energy\n(a) Is the reaction spontaneous or nonspontaneous under standard thermodynamic conditions?\n(b) Standard thermodynamic conditions imply the concentrations of G6P and F6P to be $1 M$, however, in a typical cell, they are not even close to these values. Calculate $\\Delta G$ when the concentrations of G6P and F6P are $120 \\mu M$ and $28 \\mu M$ respectively, and discuss the spontaneity of the forward reaction under these conditions. Assume the temperature is $37^{\\circ} \\mathrm{C}$.\n49. Without doing a numerical calculation, determine which of the following will reduce the free energy change for the reaction, that is, make it less positive or more negative, when the temperature is increased. Explain.\n(a) $\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\longrightarrow 2 \\mathrm{NH}_{3}(g)$\n(b) $\\mathrm{HCl}(g)+\\mathrm{NH}_{3}(g) \\longrightarrow \\mathrm{NH}_{4} \\mathrm{Cl}(s)$\n(c) $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}(s) \\longrightarrow \\mathrm{Cr}_{2} \\mathrm{O}_{3}(s)+4 \\mathrm{H}_{2} \\mathrm{O}(g)+\\mathrm{N}_{2}(g)$\n(d) $2 \\mathrm{Fe}(s)+3 \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{Fe}_{2} \\mathrm{O}_{3}(s)$\n50. When ammonium chloride is added to water and stirred, it dissolves spontaneously and the resulting solution feels cold. Without doing any calculations, deduce the signs of $\\Delta G, \\Delta H$, and $\\Delta S$ for this process, and justify your choices.\n51. An important source of copper is from the copper ore, chalcocite, a form of copper(I) sulfide. When heated, the $\\mathrm{Cu}_{2} \\mathrm{~S}$ decomposes to form copper and sulfur described by the following equation:\n$\\mathrm{Cu}_{2} \\mathrm{~S}(s) \\longrightarrow \\mathrm{Cu}(s)+\\mathrm{S}(s)$"}
{"id": 3914, "contents": "1355. Exercises - 1355.4. Free Energy\n$\\mathrm{Cu}_{2} \\mathrm{~S}(s) \\longrightarrow \\mathrm{Cu}(s)+\\mathrm{S}(s)$\n(a) Determine $\\Delta \\boldsymbol{G}^{\\circ}$ for the decomposition of $\\mathrm{Cu}_{2} \\mathrm{~S}(S)$.\n(b) The reaction of sulfur with oxygen yields sulfur dioxide as the only product. Write an equation that describes this reaction, and determine $\\Delta G^{\\circ}$ for the process.\n(c) The production of copper from chalcocite is performed by roasting the $\\mathrm{Cu}_{2} \\mathrm{~S}$ in air to produce the Cu . By combining the equations from Parts (a) and (b), write the equation that describes the roasting of the chalcocite, and explain why coupling these reactions together makes for a more efficient process for the production of the copper.\n52. What happens to $\\Delta G^{\\circ}$ (becomes more negative or more positive) for the following chemical reactions when the partial pressure of oxygen is increased?\n(a) $\\mathrm{S}(\\mathrm{s})+\\mathrm{O}_{2}(\\mathrm{~g}) \\longrightarrow \\mathrm{SO}_{2}(\\mathrm{~g})$\n(b) $2 \\mathrm{SO}_{2}(\\mathrm{~g})+\\mathrm{O}_{2}(\\mathrm{~g}) \\longrightarrow \\mathrm{SO}_{3}(\\mathrm{~g})$\n(c) $\\mathrm{HgO}(s) \\longrightarrow \\mathrm{Hg}(l)+\\mathrm{O}_{2}(g)$"}
{"id": 3915, "contents": "1355. Exercises - 1355.4. Free Energy\n626 12\u2022Exercises"}
{"id": 3916, "contents": "1356. CHAPTER 13 <br> Fundamental Equilibrium Concepts - \nFigure 13.1 Transport of carbon dioxide in the body involves several reversible chemical reactions, including hydrolysis and acid ionization (among others)."}
{"id": 3917, "contents": "1357. CHAPTER OUTLINE - 1357.1. Chemical Equilibria\n13.2 Equilibrium Constants\n13.3 Shifting Equilibria: Le Ch\u00e2telier's Principle\n13.4 Equilibrium Calculations\n\nINTRODUCTION Imagine a beach populated with sunbathers and swimmers. As those basking in the sun get too hot, they enter the surf to swim and cool off. As the swimmers tire, they return to the beach to rest. If the rate at which sunbathers enter the surf were to equal the rate at which swimmers return to the sand, then the numbers (though not the identities) of sunbathers and swimmers would remain constant. This scenario illustrates a dynamic phenomenon known as equilibrium, in which opposing processes occur at equal rates. Chemical and physical processes are subject to this phenomenon; these processes are at equilibrium when the forward and reverse reaction rates are equal. Equilibrium systems are pervasive in nature; the various reactions involving carbon dioxide dissolved in blood are examples (see Figure 13.1). This chapter provides a thorough introduction to the essential aspects of chemical equilibria.\n\nWe now have a good understanding of chemical and physical change that allow us to determine, for any given process:\n\n1. Whether the process is endothermic or exothermic\n2. Whether the process is accompanied by an increase of decrease in entropy\n3. Whether a process will be spontaneous, non-spontaneous, or what we have called an equilibrium process\n\nRecall that when the value $\\Delta G$ for a reaction is zero, we consider there to be no free energy change-that is, no\nfree energy available to do useful work. Does this mean a reaction where $\\Delta G=0$ comes to a complete halt? No, it does not. Just as a liquid exists in equilibrium with its vapor in a closed container, where the rates of evaporation and condensation are equal, there is a connection to the state of equilibrium for a phase change or a chemical reaction. That is, at equilibrium, the forward and reverse rates of reaction are equal. We will develop that concept and extend it to a relationship between equilibrium and free energy later in this chapter.\n\nIn the explanation that follows, we will use the term $Q$ to refer to any reactant or product concentration or pressure. When the concentrations or pressure of reactants and products are at equilibrium, the term $K$ will be used. This will be more clearly explained as we go along in this chapter."}
{"id": 3918, "contents": "1357. CHAPTER OUTLINE - 1357.1. Chemical Equilibria\nNow we will consider the connection between the free energy change and the equilibrium constant. The fundamental relationship is:\n$\\Delta G^{\\circ}=-\\mathrm{R} T \\ln K$-this can be for $K_{c}$ or $K_{p}$ (and we will see later, any equilibrium constant we encounter).\nWe also know that the form of $K$ can be used in non-equilibrium conditions as the reaction quotient, $Q$. The defining relationship here is\n\n$$\n\\Delta G=\\Delta G^{\\circ}+R T \\ln Q\n$$\n\nWithout the superscript, the value of $\\Delta G$ can be calculated for any set of concentrations.\nNote that since $Q$ is a mass-action reaction of productions/reactants, as a reaction proceeds from left to right, product concentrations increase as reactant concentrations decrease, until $Q=K$, and at which time $\\Delta G$ becomes zero:\n$0=\\Delta G^{\\circ}+R T \\ln K$, a relationship that reduces to our defining connection between $Q$ and $K$.\nThus, we can see clearly that as a reaction moves toward equilibrium, the value of $\\Delta G$ goes to zero.\nNow, think back to the connection between the signs of $\\Delta G^{\\circ}$ and $\\Delta H^{\\circ}$\n$\\Delta H^{\\circ} \\quad \\Delta S^{\\circ} \\quad$ Result\nNegative Positive Always spontaneous\nPositive Negative Never spontaneous\nPositive Positive Spontaneous at high temperatures\nNegative Negative Spontaneous at low temperatures\nOnly in the last two cases is there a point at which the process swings from spontaneous to non-spontaneous (or the reverse); in these cases, the process must pass through equilibrium when the change occurs. The concept of the connection between the free energy change and the equilibrium constant is an important one that we will expand upon in future sections. The fact that the change in free energy for an equilibrium process is zero, and that displacement of a process from that zero point results in a drive to re-establish equilibrium is fundamental to understanding the behavior of chemical reactions and phase changes."}
{"id": 3919, "contents": "1358. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the nature of equilibrium systems\n- Explain the dynamic nature of a chemical equilibrium\n\nThe convention for writing chemical equations involves placing reactant formulas on the left side of a reaction arrow and product formulas on the right side. By this convention, and the definitions of \"reactant\" and \"product,\" a chemical equation represents the reaction in question as proceeding from left to right. Reversible reactions, however, may proceed in both forward (left to right) and reverse (right to left) directions. When the rates of the forward and reverse reactions are equal, the concentrations of the reactant and product species remain constant over time and the system is at equilibrium. The relative concentrations of reactants and products in equilibrium systems vary greatly; some systems contain mostly products at equilibrium, some contain mostly reactants, and some contain appreciable amounts of both.\n\nFigure 13.2 illustrates fundamental equilibrium concepts using the reversible decomposition of colorless dinitrogen tetroxide to yield brown nitrogen dioxide, an elementary reaction described by the equation:\n\n$$\n\\mathrm{N}_{2} \\mathrm{O}_{4}(g) \\rightleftharpoons 2 \\mathrm{NO}_{2}(g)\n$$\n\nNote that a special double arrow is used to emphasize the reversible nature of the reaction.\n\n\nFIGURE 13.2 (a) A sealed tube containing colorless $\\mathrm{N}_{2} \\mathrm{O}_{4}$ darkens as it decomposes to yield brown $\\mathrm{NO}_{2}$. (b) Changes in concentration over time as the decomposition reaction achieves equilibrium. (c) At equilibrium, the forward and reverse reaction rates are equal.\n\nFor this elementary process, rate laws for the forward and reverse reactions may be derived directly from the reaction stoichiometry:\n\n$$\n\\begin{aligned}\n\\operatorname{rate}_{f} & =k_{f}\\left[\\mathrm{~N}_{-2} \\mathrm{O}_{4}\\right] \\\\\n\\operatorname{rate}_{r} & =k_{r}\\left[\\mathrm{NO}_{2}\\right]^{2}\n\\end{aligned}\n$$"}
{"id": 3920, "contents": "1358. LEARNING OBJECTIVES - \nAs the reaction begins ( $t=0$ ), the concentration of the $\\mathrm{N}_{2} \\mathrm{O}_{4}$ reactant is finite and that of the $\\mathrm{NO}_{2}$ product is zero, so the forward reaction proceeds at a finite rate while the reverse reaction rate is zero. As time passes, N.\n${ }_{2} \\mathrm{O}_{4}$ is consumed and its concentration falls, while $\\mathrm{NO}_{2}$ is produced and its concentration increases (Figure $13.2 \\mathbf{b})$. The decreasing concentration of the reactant slows the forward reaction rate, and the increasing product concentration speeds the reverse reaction rate (Figure 13.2c). This process continues until the forward and reverse reaction rates become equal, at which time the reaction has reached equilibrium, as characterized by constant concentrations of its reactants and products (shaded areas of Figure $13.2 \\mathbf{b}$ and Figure 13.2c). It's important to emphasize that chemical equilibria are dynamic; a reaction at equilibrium has not \"stopped,\" but is proceeding in the forward and reverse directions at the same rate. This dynamic nature is essential to understanding equilibrium behavior as discussed in this and subsequent chapters of the text.\n\n\nFIGURE 13.3 A two-person juggling act illustrates the dynamic aspect of chemical equilibria. Each person is throwing and catching clubs at the same rate, and each holds a (approximately) constant number of clubs.\n\nPhysical changes, such as phase transitions, are also reversible and may establish equilibria. This concept was introduced in another chapter of this text through discussion of the vapor pressure of a condensed phase (liquid or solid). As one example, consider the vaporization of bromine:\n\n$$\n\\operatorname{Br}_{2}(l) \\rightleftharpoons \\mathrm{Br}_{2}(g)\n$$\n\nWhen liquid bromine is added to an otherwise empty container and the container is sealed, the forward process depicted above (vaporization) will commence and continue at a roughly constant rate as long as the exposed surface area of the liquid and its temperature remain constant. As increasing amounts of gaseous bromine are produced, the rate of the reverse process (condensation) will increase until it equals the rate of vaporization and equilibrium is established. A photograph showing this phase transition equilibrium is provided in Figure 13.4."}
{"id": 3921, "contents": "1358. LEARNING OBJECTIVES - \nFIGURE 13.4 A sealed tube containing an equilibrium mixture of liquid and gaseous bromine. (credit: http://images-of-elements.com/bromine.php)"}
{"id": 3922, "contents": "1359. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Derive reaction quotients from chemical equations representing homogeneous and heterogeneous reactions\n- Calculate values of reaction quotients and equilibrium constants, using concentrations and pressures\n- Relate the magnitude of an equilibrium constant to properties of the chemical system\n\nThe status of a reversible reaction is conveniently assessed by evaluating its reaction quotient ( $\\boldsymbol{Q}$ ). For a reversible reaction described by\n\n$$\nm \\mathrm{~A}+n \\mathrm{~B} \\rightleftharpoons x \\mathrm{C}+y \\mathrm{D}\n$$\n\nthe reaction quotient is derived directly from the stoichiometry of the balanced equation as\n\n$$\nQ_{c}=\\frac{[\\mathrm{C}]^{x}[\\mathrm{D}]^{y}}{[\\mathrm{~A}]^{m}[\\mathrm{~B}]^{n}}\n$$\n\nwhere the subscript $c$ denotes the use of molar concentrations in the expression. If the reactants and products are gaseous, a reaction quotient may be similarly derived using partial pressures:\n\n$$\nQ_{p}=\\frac{P_{\\mathrm{C}}^{x} P_{\\mathrm{D}}^{y}}{P_{\\mathrm{A}}^{m} P_{\\mathrm{B}}^{n}}\n$$\n\nNote that the reaction quotient equations above are a simplification of more rigorous expressions that use relative values for concentrations and pressures rather than absolute values. These relative concentration and pressure values are dimensionless (they have no units); consequently, so are the reaction quotients. For purposes of this introductory text, it will suffice to use the simplified equations and to disregard units when computing $Q$. In most cases, this will introduce only modest errors in calculations involving reaction quotients."}
{"id": 3923, "contents": "1361. Writing Reaction Quotient Expressions - \nWrite the concentration-based reaction quotient expression for each of the following reactions:\n(a) $3 \\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{O}_{3}(g)$\n(b) $\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g)$\n(c) $4 \\mathrm{NH}_{3}(g)+7 \\mathrm{O}_{2}(g) \\rightleftharpoons 4 \\mathrm{NO}_{2}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(g)$"}
{"id": 3924, "contents": "1362. Solution - \n(a) $Q_{c}=\\frac{\\left[\\mathrm{O}_{3}\\right]^{2}}{\\left[\\mathrm{O}_{2}\\right]^{3}}$\n(b) $Q_{c}=\\frac{\\left[\\mathrm{NH}_{3}\\right]^{2}}{\\left[\\mathrm{~N}_{2}\\right]\\left[\\mathrm{H}_{2}\\right]^{3}}$\n(c) $Q_{c}=\\frac{\\left[\\mathrm{NO}_{2}\\right]^{4}\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]^{6}}{\\left[\\mathrm{NH}_{3}\\right]^{4}\\left[\\mathrm{O}_{2}\\right]^{7}}$"}
{"id": 3925, "contents": "1363. Check Your Learning - \nWrite the concentration-based reaction quotient expression for each of the following reactions:\n(a) $2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{SO}_{3}(g)$\n(b) $\\mathrm{C}_{4} \\mathrm{H}_{8}(g) \\rightleftharpoons 2 \\mathrm{C}_{2} \\mathrm{H}_{4}(g)$\n(c) $2 \\mathrm{C}_{4} \\mathrm{H}_{10}(g)+13 \\mathrm{O}_{2}(g) \\rightleftharpoons 8 \\mathrm{CO}_{2}(g)+10 \\mathrm{H}_{2} \\mathrm{O}(g)$"}
{"id": 3926, "contents": "1364. Answer: - \n(a) $Q_{c}=\\frac{\\left[\\mathrm{SO}_{3}\\right]^{2}}{\\left[\\mathrm{SO}_{2}\\right]^{2}\\left[\\mathrm{O}_{2}\\right]}$; (b) $Q_{c}=\\frac{\\left[\\mathrm{C}_{2} \\mathrm{H}_{4}\\right]^{2}}{\\left[\\mathrm{C}_{4} \\mathrm{H}_{8}\\right]} ;$ (c) $Q_{c}=\\frac{\\left[\\mathrm{CO}_{2}\\right]^{8}\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]{ }^{10}}{\\left[\\mathrm{C}_{4} \\mathrm{H}_{10}\\right]^{2}\\left[\\mathrm{O}_{2}\\right]^{13}}$\n\n\nFIGURE 13.5 Changes in concentrations and $Q_{C}$ for a chemical equilibrium achieved beginning with (a) a mixture of reactants only and (b) products only.\n\nThe numerical value of $Q$ varies as a reaction proceeds towards equilibrium; therefore, it can serve as a useful indicator of the reaction's status. To illustrate this point, consider the oxidation of sulfur dioxide:\n\n$$\n2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{SO}_{3}(g)\n$$\n\nTwo different experimental scenarios are depicted in Figure 13.5, one in which this reaction is initiated with a mixture of reactants only, $\\mathrm{SO}_{2}$ and $\\mathrm{O}_{2}$, and another that begins with only product, $\\mathrm{SO}_{3}$. For the reaction that begins with a mixture of reactants only, $Q$ is initially equal to zero:\n\n$$\nQ_{c}=\\frac{\\left[\\mathrm{SO}_{3}\\right]^{2}}{\\left[\\mathrm{SO}_{2}\\right]^{2}\\left[\\mathrm{O}_{2}\\right]}=\\frac{0^{2}}{\\left[\\mathrm{SO}_{2}\\right]^{2}\\left[\\mathrm{O}_{2}\\right]}=0\n$$"}
{"id": 3927, "contents": "1364. Answer: - \nAs the reaction proceeds toward equilibrium in the forward direction, reactant concentrations decrease (as does the denominator of $Q_{c}$ ), product concentration increases (as does the numerator of $Q_{c}$ ), and the reaction quotient consequently increases. When equilibrium is achieved, the concentrations of reactants and product remain constant, as does the value of $Q_{c}$.\n\nIf the reaction begins with only product present, the value of $Q_{c}$ is initially undefined (immeasurably large, or infinite):\n\n$$\nQ_{c}=\\frac{\\left[\\mathrm{SO}_{3}\\right]^{2}}{\\left[\\mathrm{SO}_{2}\\right]^{2}\\left[\\mathrm{O}_{2}\\right]}=\\frac{\\left[\\mathrm{SO}_{3}\\right]^{2}}{0} \\rightarrow \\infty\n$$\n\nIn this case, the reaction proceeds toward equilibrium in the reverse direction. The product concentration and the numerator of $Q_{c}$ decrease with time, the reactant concentrations and the denominator of $Q_{c}$ increase, and the reaction quotient consequently decreases until it becomes constant at equilibrium.\n\nThe constant value of $Q$ exhibited by a system at equilibrium is called the equilibrium constant, $\\boldsymbol{K}$ :\n\n$$\nK \\equiv Q \\text { at equilibrium }\n$$\n\nComparison of the data plots in Figure 13.5 shows that both experimental scenarios resulted in the same value for the equilibrium constant. This is a general observation for all equilibrium systems, known as the law of mass action: At a given temperature, the reaction quotient for a system at equilibrium is constant."}
{"id": 3928, "contents": "1366. Evaluating a Reaction Quotient - \nGaseous nitrogen dioxide forms dinitrogen tetroxide according to this equation:\n\n$$\n2 \\mathrm{NO}_{2}(g) \\rightleftharpoons \\mathrm{N}_{2} \\mathrm{O}_{4}(g)\n$$\n\nWhen $0.10 \\mathrm{~mol} \\mathrm{NO}_{2}$ is added to a 1.0-L flask at $25^{\\circ} \\mathrm{C}$, the concentration changes so that at equilibrium, $\\left[\\mathrm{NO}_{2}\\right]=$ 0.016 M and $\\left[\\mathrm{N}_{2} \\mathrm{O}_{4}\\right]=0.042 \\mathrm{M}$.\n(a) What is the value of the reaction quotient before any reaction occurs?\n(b) What is the value of the equilibrium constant for the reaction?"}
{"id": 3929, "contents": "1367. Solution - \nAs for all equilibrium calculations in this text, use the simplified equations for $Q$ and $K$ and disregard any concentration or pressure units, as noted previously in this section.\n(a) Before any product is formed, $\\left[\\mathrm{NO}_{2}\\right]=\\frac{0.10 \\mathrm{~mol}}{1.0 \\mathrm{~L}}=0.10 M$, and $\\left[\\mathrm{N}_{2} \\mathrm{O}_{4}\\right]=0 \\mathrm{M}$. Thus,\n\n$$\nQ_{c}=\\frac{\\left[\\mathrm{N}_{2} \\mathrm{O}_{4}\\right]}{\\left[\\mathrm{NO}_{2}\\right]^{2}}=\\frac{0}{0.10^{2}}=0\n$$\n\n(b) At equilibrium, $K_{c}=Q_{c}=\\frac{\\left[\\mathrm{N}_{2} \\mathrm{O}_{4}\\right]}{\\left[\\mathrm{NO}_{2}\\right]^{2}}=\\frac{0.042}{0.016^{2}}=1.6 \\times 10^{2}$. The equilibrium constant is $1.6 \\times 10^{2}$."}
{"id": 3930, "contents": "1368. Check Your Learning - \nFor the reaction $2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{SO}_{3}(g)$, the concentrations at equilibrium are $\\left[\\mathrm{SO}_{2}\\right]=0.90 \\mathrm{M},\\left[\\mathrm{O}_{2}\\right]=0.35$ $M$, and $\\left[\\mathrm{SO}_{3}\\right]=1.1 \\mathrm{M}$. What is the value of the equilibrium constant, $K_{c}$ ?"}
{"id": 3931, "contents": "1369. Answer: - \n$K_{c}=4.3$\n\nBy its definition, the magnitude of an equilibrium constant explicitly reflects the composition of a reaction mixture at equilibrium, and it may be interpreted with regard to the extent of the forward reaction. A reaction exhibiting a large $K$ will reach equilibrium when most of the reactant has been converted to product, whereas a small $K$ indicates the reaction achieves equilibrium after very little reactant has been converted. It's important to keep in mind that the magnitude of $K$ does not indicate how rapidly or slowly equilibrium will be reached. Some equilibria are established so quickly as to be nearly instantaneous, and others so slowly that no perceptible change is observed over the course of days, years, or longer.\n\nThe equilibrium constant for a reaction can be used to predict the behavior of mixtures containing its reactants and/or products. As demonstrated by the sulfur dioxide oxidation process described above, a chemical reaction will proceed in whatever direction is necessary to achieve equilibrium. Comparing $Q$ to $K$ for\nan equilibrium system of interest allows prediction of what reaction (forward or reverse), if any, will occur. To further illustrate this important point, consider the reversible reaction shown below:\n\n$$\n\\mathrm{CO}(g)+\\mathrm{H}_{2} \\mathrm{O}(g) \\rightleftharpoons \\mathrm{CO}_{2}(g)+\\mathrm{H}_{2}(g) \\quad K_{c}=0.640 \\quad \\mathrm{~T}=800^{\\circ} \\mathrm{C}\n$$\n\nThe bar charts in Figure 13.6 represent changes in reactant and product concentrations for three different reaction mixtures. The reaction quotients for mixtures 1 and 3 are initially lesser than the reaction's equilibrium constant, so each of these mixtures will experience a net forward reaction to achieve equilibrium. The reaction quotient for mixture 2 is initially greater than the equilibrium constant, so this mixture will proceed in the reverse direction until equilibrium is established."}
{"id": 3932, "contents": "1369. Answer: - \nFIGURE 13.6 Compositions of three mixtures before $\\left(Q_{c} \\neq K_{c}\\right)$ and after $\\left(Q_{c}=K_{c}\\right)$ equilibrium is established for the reaction $\\mathrm{CO}(g)+\\mathrm{H}_{2} \\mathrm{O}(g) \\rightleftharpoons \\mathrm{CO}_{2}(g)+\\mathrm{H}_{2}(g)$."}
{"id": 3933, "contents": "1371. Predicting the Direction of Reaction - \nGiven here are the starting concentrations of reactants and products for three experiments involving this reaction:\n\n$$\n\\begin{gathered}\n\\mathrm{CO}(g)+\\mathrm{H}_{2} \\mathrm{O}(g) \\rightleftharpoons \\mathrm{CO}_{2}(g)+\\mathrm{H}_{2}(g) \\\\\nK_{c}=0.64\n\\end{gathered}\n$$\n\nDetermine in which direction the reaction proceeds as it goes to equilibrium in each of the three experiments shown.\n\n| Reactants/Products |  | Experiment 1 | Experiment 2 |\n| :--- | :--- | :--- | :--- |\n| $[\\mathrm{CO}]_{\\mathrm{i}}$ | 0.020 M | 0.011 M | 0.0094 M |\n| $\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]_{\\mathrm{i}}$ | 0.020 M | 0.0011 M | 0.0025 M |\n| $\\left[\\mathrm{CO}_{2}\\right]_{\\mathrm{i}}$ | 0.0040 M | 0.037 M | 0.0015 M |\n| $\\left[\\mathrm{H}_{2}\\right]_{\\mathrm{i}}$ | 0.0040 M | 0.046 M | 0.0076 M |"}
{"id": 3934, "contents": "1372. Solution - \nExperiment 1:\n\n$$\nQ_{c}=\\frac{\\left[\\mathrm{CO}_{2}\\right]\\left[\\mathrm{H}_{2}\\right]}{[\\mathrm{CO}]\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]}=\\frac{(0.0040)(0.0040)}{(0.020)(0.020)}=0.040\n$$\n\n$Q_{c}<K_{C}(0.040<0.64)$\nThe reaction will proceed in the forward direction.\nExperiment 2:\n\n$$\nQ_{c}=\\frac{\\left[\\mathrm{CO}_{2}\\right]\\left[\\mathrm{H}_{2}\\right]}{[\\mathrm{CO}]\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]}=\\frac{(0.037)(0.046)}{(0.011)(0.0011)}=1.4 \\times 10^{2}\n$$\n\n$Q_{c}>K_{c}(140>0.64)$\nThe reaction will proceed in the reverse direction.\nExperiment 3:\n\n$$\nQ_{c}=\\frac{\\left[\\mathrm{CO}_{2}\\right]\\left[\\mathrm{H}_{2}\\right]}{[\\mathrm{CO}]\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]}=\\frac{(0.0015)(0.0076)}{(0.0094)(0.0025)}=0.48\n$$\n\n$Q_{c}<K_{c}(0.48<0.64)$\nThe reaction will proceed in the forward direction."}
{"id": 3935, "contents": "1373. Check Your Learning - \nCalculate the reaction quotient and determine the direction in which each of the following reactions will proceed to reach equilibrium.\n(a) A 1.00-L flask containing 0.0500 mol of $\\mathrm{NO}(\\mathrm{g}), 0.0155 \\mathrm{~mol}^{2} \\mathrm{Cl}_{2}(\\mathrm{~g})$, and 0.500 mol of NOCl :\n\n$$\n2 \\mathrm{NO}(g)+\\mathrm{Cl}_{2}(g) \\rightleftharpoons 2 \\mathrm{NOCl}(g) \\quad K_{c}=4.6 \\times 10^{4}\n$$\n\n(b) A 5.0-L flask containing 17 g of $\\mathrm{NH}_{3}, 14 \\mathrm{~g}$ of $\\mathrm{N}_{2}$, and 12 g of $\\mathrm{H}_{2}$ :\n\n$$\n\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g) \\quad K_{c}=0.060\n$$\n\n(c) A 2.00-L flask containing 230 g of $\\mathrm{SO}_{3}(\\mathrm{~g})$ :\n\n$$\n2 \\mathrm{SO}_{3}(g) \\rightleftharpoons 2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\quad K_{c}=0.230\n$$"}
{"id": 3936, "contents": "1374. Answer: - \n(a) $Q_{c}=6.45 \\times 10^{3}$, forward. (b) $Q_{c}=0.23$, reverse. (c) $Q_{c}=0$, forward."}
{"id": 3937, "contents": "1375. Homogeneous Equilibria - \nA homogeneous equilibrium is one in which all reactants and products (and any catalysts, if applicable) are present in the same phase. By this definition, homogeneous equilibria take place in solutions. These solutions are most commonly either liquid or gaseous phases, as shown by the examples below:\n\n$$\n\\begin{aligned}\n\\mathrm{C}_{2} \\mathrm{H}_{2}(a q)+2 \\mathrm{Br}_{2}(a q) & \\rightleftharpoons \\mathrm{C}_{2} \\mathrm{H}_{2} \\mathrm{Br}_{4}(a q) & K_{c} & =\\frac{\\left[\\mathrm{C}_{2} \\mathrm{H}_{2} \\mathrm{Br}_{4}\\right]}{\\left[\\mathrm{C}_{2} \\mathrm{H}_{2}\\right]\\left[\\mathrm{Br}_{2}\\right]^{2}} \\\\\n\\mathrm{I}_{2}(a q)+\\mathrm{I}^{-}(a q) & \\rightleftharpoons \\mathrm{I}_{3}^{-}(a q) & K_{c} & =\\frac{\\left[\\mathrm{I}_{3}^{-}\\right]}{\\left[\\mathrm{I}_{2}\\right]\\left[\\mathrm{I}^{-}\\right]} \\\\\n\\mathrm{HF}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) & \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{F}^{-}(a q) & K_{c} & =\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{F}^{-}\\right]}{[\\mathrm{HF}]} \\\\\n\\mathrm{NH}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) & \\rightleftharpoons \\mathrm{NH}_{4}^{+}(a q)+\\mathrm{OH}^{-}(a q) & K_{c} & =\\frac{\\left[\\mathrm{NH}_{4}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]}{\\left[\\mathrm{NH}_{3}\\right]}\n\\end{aligned}\n$$"}
{"id": 3938, "contents": "1375. Homogeneous Equilibria - \nThese examples all involve aqueous solutions, those in which water functions as the solvent. In the last two\nexamples, water also functions as a reactant, but its concentration is not included in the reaction quotient. The reason for this omission is related to the more rigorous form of the $Q$ (or $K$ ) expression mentioned previously in this chapter, in which relative concentrations for liquids and solids are equal to 1 and needn't be included. Consequently, reaction quotients include concentration or pressure terms only for gaseous and solute species.\n\nThe equilibria below all involve gas-phase solutions:"}
{"id": 3939, "contents": "1375. Homogeneous Equilibria - \nThe equilibria below all involve gas-phase solutions:\n\n$$\n\\begin{aligned}\n\\mathrm{C}_{2} \\mathrm{H}_{6}(g) & \\rightleftharpoons \\mathrm{C}_{2} \\mathrm{H}_{4}(g)+\\mathrm{H}_{2}(g) & K_{c} & =\\frac{\\left[\\mathrm{C}_{2} \\mathrm{H}_{4}\\right]\\left[\\mathrm{H}_{2}\\right]}{\\left[\\mathrm{C}_{2} \\mathrm{H}_{6}\\right]} \\\\\n3 \\mathrm{O}_{2}(g) & \\rightleftharpoons 2 \\mathrm{O}_{3}(g) & K_{c} & =\\frac{\\left[\\mathrm{O}_{3}\\right]^{2}}{\\left[\\mathrm{O}_{2}\\right]^{3}} \\\\\n\\mathrm{~N}_{2}(g)+3 \\mathrm{H}_{2}(g) & \\rightleftharpoons 2 \\mathrm{NH}_{3}(g) & K_{c} & =\\frac{\\left[\\mathrm{NH}_{3}\\right]^{2}}{\\left[\\mathrm{~N}_{2}\\right]\\left[\\mathrm{H}_{2}\\right]^{3}} \\\\\n\\mathrm{C}_{3} \\mathrm{H}_{8}(g)+5 \\mathrm{O}_{2}(g) & \\rightleftharpoons 3 \\mathrm{CO}_{2}(g)+4 \\mathrm{H}_{2} \\mathrm{O}(g) & K_{c} & =\\frac{\\left[\\mathrm{CO}_{2}\\right]^{3}\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]^{4}}{\\left[\\mathrm{C}_{3} \\mathrm{H}_{8}\\right]\\left[\\mathrm{O}_{2}\\right]^{5}}\n\\end{aligned}\n$$\n\nFor gas-phase solutions, the equilibrium constant may be expressed in terms of either the molar concentrations ( $K_{c}$ ) or partial pressures $\\left(K_{p}\\right)$ of the reactants and products. A relation between these two $K$ values may be simply derived from the ideal gas equation and the definition of molarity:"}
{"id": 3940, "contents": "1375. Homogeneous Equilibria - \n$$\n\\begin{gathered}\nP V=n R T \\\\\n\\begin{array}{c}\nP=\\left(\\frac{n}{V}\\right) R T \\\\\n=M R T\n\\end{array}\n\\end{gathered}\n$$\n\nwhere $P$ is partial pressure, $V$ is volume, $n$ is molar amount, $R$ is the gas constant, $T$ is temperature, and $M$ is molar concentration.\n\nFor the gas-phase reaction $m \\mathrm{~A}+n \\mathrm{~B} \\rightleftharpoons x \\mathrm{C}+y \\mathrm{D}$ :\n\n$$\n\\begin{gathered}\nK_{P}=\\frac{\\left(P_{C}\\right)^{x}\\left(P_{D}\\right)^{y}}{\\left(P_{A}\\right)^{m}\\left(P_{B}\\right)^{n}} \\\\\n=\\frac{([\\mathrm{C}] \\times R T)^{x}([\\mathrm{D}] \\times R T)^{y}}{([\\mathrm{~A}] \\times R T)^{m}([\\mathrm{~B}] \\times R T)^{n}} \\\\\n=\\frac{[\\mathrm{C}]^{x}[\\mathrm{D}]^{y}}{[\\mathrm{~A}]^{m}[\\mathrm{~B}]^{n}} \\times \\frac{(R T)^{x+y}}{(R T)^{m+n}} \\\\\n=K_{c}(R T)^{(x+y)-(m+n)} \\\\\n=K_{c}(R T)^{\\Delta n}\n\\end{gathered}\n$$\n\nAnd so, the relationship between $K_{c}$ and $K_{P}$ is\n\n$$\nK_{P}=K_{c}(R T)^{\\Delta n}\n$$\n\nwhere $\\Delta n$ is the difference in the molar amounts of product and reactant gases, in this case:\n\n$$\n\\Delta n=(x+y)-(m+n)\n$$"}
{"id": 3941, "contents": "1377. Calculation of $\\boldsymbol{K}_{\\boldsymbol{P}}$ - \nWrite the equations relating $K_{c}$ to $K_{P}$ for each of the following reactions:\n(a) $\\mathrm{C}_{2} \\mathrm{H}_{6}(g) \\rightleftharpoons \\mathrm{C}_{2} \\mathrm{H}_{4}(g)+\\mathrm{H}_{2}(g)$\n(b) $\\mathrm{CO}(g)+\\mathrm{H}_{2} \\mathrm{O}(g) \\rightleftharpoons \\mathrm{CO}_{2}(g)+\\mathrm{H}_{2}(g)$\n(c) $\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g)$\n(d) $K_{c}$ is equal to 0.28 for the following reaction at $900^{\\circ} \\mathrm{C}$ :\n\n$$\n\\mathrm{CS}_{2}(g)+4 \\mathrm{H}_{2}(g) \\rightleftharpoons \\mathrm{CH}_{4}(g)+2 \\mathrm{H}_{2} \\mathrm{~S}(g)\n$$\n\nWhat is $K_{P}$ at this temperature?"}
{"id": 3942, "contents": "1378. Solution - \n(a) $\\Delta n=(2)-(1)=1$\n$K_{P}=K_{c}(R T)^{\\Delta n}=K_{c}(R T)^{1}=K_{c}(R T)$\n(b) $\\Delta n=(2)-(2)=0$\n$K_{P}=K_{c}(R T)^{\\Delta n}=K_{c}(R T)^{0}=K_{c}$\n(c) $\\Delta n=(2)-(1+3)=-2$\n$K_{P}=K_{c}(R T)^{\\Delta n}=K_{c}(R T)^{-2}=\\frac{K_{c}}{(R T)^{2}}$\n(d) $K_{P}=K_{c}(\\mathrm{RT})^{\\Delta n}=(0.28)[(0.0821)(1173)]^{-2}=3.0 \\times 10^{-5}$"}
{"id": 3943, "contents": "1379. Check Your Learning - \nWrite the equations relating $K_{c}$ to $K_{P}$ for each of the following reactions:\n(a) $2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{SO}_{3}(g)$\n(b) $\\mathrm{N}_{2} \\mathrm{O}_{4}(g) \\rightleftharpoons 2 \\mathrm{NO}_{2}(g)$\n(c) $\\mathrm{C}_{3} \\mathrm{H}_{8}(g)+5 \\mathrm{O}_{2}(g) \\rightleftharpoons 3 \\mathrm{CO}_{2}(g)+4 \\mathrm{H}_{2} \\mathrm{O}(g)$\n(d) At $227^{\\circ} \\mathrm{C}$, the following reaction has $K_{c}=0.0952$ :\n\n$$\n\\mathrm{CH}_{3} \\mathrm{OH}(g) \\rightleftharpoons \\mathrm{CO}(g)+2 \\mathrm{H}_{2}(g)\n$$\n\nWhat would be the value of $K_{P}$ at this temperature?\nAnswer:\n(a) $K_{P}=K_{c}(R T)^{-1}$; (b) $K_{P}=K_{c}(R T)$; (c) $K_{P}=K_{c}(R T)$; (d) 160 or $1.6 \\times 10^{2}$"}
{"id": 3944, "contents": "1380. Heterogeneous Equilibria - \nA heterogeneous equilibrium involves reactants and products in two or more different phases, as illustrated by the following examples:\n\n$$\n\\begin{aligned}\n\\mathrm{PbCl}_{2}(s) & \\rightleftharpoons \\mathrm{Pb}^{2+}(a q)+2 \\mathrm{Cl}^{-}(a q) & K_{c} & =\\left[\\mathrm{Pb}^{2+}\\right]\\left[\\mathrm{Cl}^{-}\\right]^{2} \\\\\n\\mathrm{CaO}(s)+\\mathrm{CO}_{2}(g) & \\rightleftharpoons \\mathrm{CaCO}_{3}(s) & K_{c} & =\\frac{1}{\\left[\\mathrm{CO}_{2}\\right]} \\\\\n\\mathrm{C}(s)+2 \\mathrm{~S}(g) & \\rightleftharpoons \\mathrm{CS}_{2}(g) & K_{c} & =\\frac{\\left[\\mathrm{CS}_{2}\\right]}{[\\mathrm{S}]^{2}} \\\\\n\\mathrm{Br}_{2}(l) & \\rightleftharpoons \\mathrm{Br}_{2}(g) & K_{c} & =\\left[\\mathrm{Br}_{2}(g)\\right]\n\\end{aligned}\n$$\n\nAgain, note that concentration terms are only included for gaseous and solute species, as discussed previously.\nTwo of the above examples include terms for gaseous species only in their equilibrium constants, and so $K_{p}$ expressions may also be written:\n\n$$\n\\begin{aligned}\n\\mathrm{CaO}(s)+\\mathrm{CO}_{2}(g) & \\rightleftharpoons \\mathrm{CaCO}_{3}(s) & K_{P} & =\\frac{1}{P_{\\mathrm{CO}_{2}}} \\\\\n\\mathrm{C}(s)+2 \\mathrm{~S}(g) & \\rightleftharpoons \\mathrm{CS}_{2}(g) & K_{P} & =\\frac{P_{\\mathrm{CS}_{2}}}{\\left(P_{\\mathrm{S}}\\right)^{2}}\n\\end{aligned}\n$$"}
{"id": 3945, "contents": "1381. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the ways in which an equilibrium system can be stressed\n- Predict the response of a stressed equilibrium using Le Ch\u00e2telier's principle\n\nA system at equilibrium is in a state of dynamic balance, with forward and reverse reactions taking place at equal rates. If an equilibrium system is subjected to a change in conditions that affects these reaction rates differently (a stress), then the rates are no longer equal and the system is not at equilibrium. The system will subsequently experience a net reaction in the direction of greater rate (a shift) that will re-establish the equilibrium. This phenomenon is summarized by Le Ch\u00e2telier's principle: if an equilibrium system is stressed, the system will experience a shift in response to the stress that re-establishes equilibrium.\n\nReaction rates are affected primarily by concentrations, as described by the reaction's rate law, and temperature, as described by the Arrhenius equation. Consequently, changes in concentration and temperature are the two stresses that can shift an equilibrium."}
{"id": 3946, "contents": "1382. Effect of a Change in Concentration - \nIf an equilibrium system is subjected to a change in the concentration of a reactant or product species, the rate of either the forward or the reverse reaction will change. As an example, consider the equilibrium reaction\n\n$$\n\\mathrm{H}_{2}(g)+\\mathrm{I}_{2}(g) \\rightleftharpoons 2 \\mathrm{HI}(g) \\quad K_{c}=50.0 \\text { at } 400^{\\circ} \\mathrm{C}\n$$\n\nThe rate laws for the forward and reverse reactions are\n\n$$\n\\begin{array}{lll}\n\\text { forward } & \\mathrm{H}_{2}(g)+\\mathrm{I}_{2}(g) \\rightarrow 2 \\mathrm{HI}(g) & \\operatorname{rate}_{f}=k_{f}\\left[\\mathrm{H}_{2}\\right]^{m}\\left[\\mathrm{I}_{2}\\right]^{n} \\\\\n\\text { reverse } & 2 \\mathrm{HI}(g) \\rightarrow \\mathrm{H}_{2}(g)+\\mathrm{I}_{2}(g) & \\operatorname{rate}_{r}=k_{r}[\\mathrm{HI}]^{x}\n\\end{array}\n$$\n\nWhen this system is at equilibrium, the forward and reverse reaction rates are equal.\n\n$$\n\\operatorname{rate}_{f}=\\operatorname{rate}_{r}\n$$\n\nIf the system is stressed by adding reactant, either $\\mathrm{H}_{2}$ or $\\mathrm{I}_{2}$, the resulting increase in concentration causes the rate of the forward reaction to increase, exceeding that of the reverse reaction:\n\n$$\n\\operatorname{rate}_{f}>\\operatorname{rate}_{r}\n$$\n\nThe system will experience a temporary net reaction in the forward direction to re-establish equilibrium (the equilibrium will shift right). This same shift will result if some product HI is removed from the system, which decreases the rate of the reverse reaction, again resulting in the same imbalance in rates.\n\nThe same logic can be used to explain the left shift that results from either removing reactant or adding product to an equilibrium system. These stresses both result in an increased rate for the reverse reaction\n\n$$\n\\operatorname{rate}_{f}<\\operatorname{rate}_{r}\n$$"}
{"id": 3947, "contents": "1382. Effect of a Change in Concentration - \n$$\n\\operatorname{rate}_{f}<\\operatorname{rate}_{r}\n$$\n\nand a temporary net reaction in the reverse direction to re-establish equilibrium.\nAs an alternative to this kinetic interpretation, the effect of changes in concentration on equilibria can be rationalized in terms of reaction quotients. When the system is at equilibrium,\n\n$$\nQ_{c}=\\frac{[\\mathrm{HI}]^{2}}{\\left[\\mathrm{H}_{2}\\right]\\left[\\mathrm{I}_{2}\\right]}=K_{c}\n$$\n\nIf reactant is added (increasing the denominator of the reaction quotient) or product is removed (decreasing the numerator), then $Q_{c}<K_{c}$ and the equilibrium will shift right. Note that the three different ways of inducing this stress result in three different changes in the composition of the equilibrium mixture. If $\\mathrm{H}_{2}$ is added, the right shift will consume $\\mathrm{I}_{2}$ and produce HI as equilibrium is re-established, yielding a mixture with a greater concentrations of $\\mathrm{H}_{2}$ and HI and a lesser concentration of $\\mathrm{I}_{2}$ than was present before. If $\\mathrm{I}_{2}$ is added, the new equilibrium mixture will have greater concentrations of $\\mathrm{I}_{2}$ and HI and a lesser concentration of $\\mathrm{H}_{2}$. Finally, if HI\nis removed, the new equilibrium mixture will have greater concentrations of $\\mathrm{H}_{2}$ and $\\mathrm{I}_{2}$ and a lesser concentration of HI. Despite these differences in composition, the value of the equilibrium constant will be the same after the stress as it was before (per the law of mass action). The same logic may be applied for stresses involving removing reactants or adding product, in which case $Q_{c}>K_{c}$ and the equilibrium will shift left.\n\nFor gas-phase equilibria such as this one, some additional perspectives on changing the concentrations of reactants and products are worthy of mention. The partial pressure $P$ of an ideal gas is proportional to its molar concentration $M$,\n\n$$\nM=\\frac{n}{V}=\\frac{P}{R T}\n$$"}
{"id": 3948, "contents": "1382. Effect of a Change in Concentration - \n$$\nM=\\frac{n}{V}=\\frac{P}{R T}\n$$\n\nand so changes in the partial pressures of any reactant or product are essentially changes in concentrations and thus yield the same effects on equilibria. Aside from adding or removing reactant or product, the pressures (concentrations) of species in a gas-phase equilibrium can also be changed by changing the volume occupied by the system. Since all species of a gas-phase equilibrium occupy the same volume, a given change in volume will cause the same change in concentration for both reactants and products. In order to discern what shift, if any, this type of stress will induce the stoichiometry of the reaction must be considered.\n\nAt equilibrium, the reaction $\\mathrm{H}_{2}(g)+\\mathrm{I}_{2}(g) \\rightleftharpoons 2 \\mathrm{HI}(g)$ is described by the reaction quotient\n\n$$\nQ_{p}=\\frac{P_{\\mathrm{HI}^{2}}}{P_{\\mathrm{H}^{2}} P_{\\mathrm{I}^{2}}}=K_{p}\n$$\n\nIf the volume occupied by an equilibrium mixture of these species is decreased by a factor of 3 , the partial pressures of all three species will be increased by a factor of 3:\n\n$$\n\\begin{aligned}\n& Q_{p}^{\\prime}=\\frac{\\left(3 P_{\\mathrm{HI}^{2}}\\right)^{2}}{3 P_{\\mathrm{H}^{2}}{ }^{3 \\mathrm{I}^{2}}}=\\frac{9 P_{\\mathrm{HI}^{2}}}{9 P_{\\mathrm{H}^{2}} P_{\\mathrm{I}^{2}}}=\\frac{P_{\\mathrm{HI}^{2}}}{P_{\\mathrm{H}^{2}} P_{\\mathrm{I}^{2}}}=Q_{p}=K_{p} \\\\\n& Q_{p}{ }^{\\prime}=Q_{p}=K_{p}\n\\end{aligned}\n$$\n\nAnd so, changing the volume of this gas-phase equilibrium mixture does not result in a shift of the equilibrium."}
{"id": 3949, "contents": "1382. Effect of a Change in Concentration - \nAnd so, changing the volume of this gas-phase equilibrium mixture does not result in a shift of the equilibrium.\n\nA similar treatment of a different system, $2 \\mathrm{NO}_{2}(g) \\rightleftharpoons 2 \\mathrm{NO}(g)+\\mathrm{O}_{2}(g)$, however, yields a different result:\n\n$$\n\\begin{aligned}\n& Q_{p}=\\frac{P_{\\mathrm{NO}^{2}} P_{\\mathrm{O}^{2}}}{P}=K_{p} \\\\\n& \\left(\\mathrm{NO}^{2}\\right)^{2} \\\\\n& Q_{p}{ }^{\\prime}=\\frac{\\left(3 P_{\\mathrm{NO}^{2}}\\right)^{2} 3 P_{\\mathrm{O}^{2}}}{\\left(3 P_{\\mathrm{NO}^{2}}\\right)^{2}}=\\frac{9 P_{\\mathrm{NO}^{2}}{ }^{3 P} \\mathrm{O}^{2}}{9 P}\\left(\\mathrm{NO}^{2}\\right)^{2}\n\\end{aligned} \\frac{27 P_{\\mathrm{NO}^{2} P_{\\mathrm{O}^{2}}}^{P}\\left(\\mathrm{NO}^{2}\\right)^{2}}{}=3 Q_{p}>K_{p} .\n$$\n\nIn this case, the change in volume results in a reaction quotient greater than the equilibrium constant, and so the equilibrium will shift left."}
{"id": 3950, "contents": "1382. Effect of a Change in Concentration - \nIn this case, the change in volume results in a reaction quotient greater than the equilibrium constant, and so the equilibrium will shift left.\n\nThese results illustrate the relationship between the stoichiometry of a gas-phase equilibrium and the effect of a volume-induced pressure (concentration) change. If the total molar amounts of reactants and products are equal, as in the first example, a change in volume does not shift the equilibrium. If the molar amounts of reactants and products are different, a change in volume will shift the equilibrium in a direction that better \"accommodates\" the volume change. In the second example, two moles of reactant $\\left(\\mathrm{NO}_{2}\\right)$ yield three moles of product ( $2 \\mathrm{NO}+\\mathrm{O}_{2}$ ), and so decreasing the system volume causes the equilibrium to shift left since the reverse reaction produces less gas ( 2 mol ) than the forward reaction ( 3 mol ). Conversely, increasing the volume of this equilibrium system would result in a shift towards products."}
{"id": 3951, "contents": "1383. LINK TO LEARNING - \nCheck out this link (http://openstax.org/l/16equichange) to see a dramatic visual demonstration of how\nequilibrium changes with pressure changes."}
{"id": 3952, "contents": "1385. Equilibrium and Soft Drinks - \nThe connection between chemistry and carbonated soft drinks goes back to 1767, when Joseph Priestley (1733-1804) developed a method of infusing water with carbon dioxide to make carbonated water. Priestly's approach involved production of carbon dioxidey reacting oil of vitriol (sulfuric acid) with chalk (calcium carbonate).\n\nThe carbon dioxide was then dissolved in water, reacting to produce hydrogen carbonate, a weak acid that subsequently ionized to yield bicarbonate and hydrogen ions:\n\n$$\n\\begin{array}{ll}\n\\text { dissolution } & \\mathrm{CO}_{2}(g)=\\mathrm{CO}_{2}(a q) \\\\\n\\text { hydrolysis } & \\mathrm{CO}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)=\\mathrm{H}_{2} \\mathrm{CO}_{3}(a q) \\\\\n\\text { ionization } & \\mathrm{H}_{2} \\mathrm{CO}_{3}(a q)=\\mathrm{HCO}_{3}-(a q)+\\mathrm{H}^{+}(a q)\n\\end{array}\n$$\n\nThese same equilibrium reactions are the basis of today's soft-drink carbonation process. Beverages are exposed to a high pressure of gaseous carbon dioxide during the process to shift the first equilibrium above to the right, resulting in desirably high concentrations of dissolved carbon dioxide and, per similar shifts in the other two equilibria, its hydrolysis and ionization products. A bottle or can is then nearly filled with the carbonated beverage, leaving a relatively small volume of air in the container above the beverage surface (the headspace) before it is sealed. The pressure of carbon dioxide in the container headspace is very low immediately after sealing, but it rises as the dissolution equilibrium is re-established by shifting to the left. Since the volume of the beverage is significantly greater than the volume of the headspace, only a relatively small amount of dissolved carbon dioxide is lost to the headspace."}
{"id": 3953, "contents": "1385. Equilibrium and Soft Drinks - \nWhen a carbonated beverage container is opened, a hissing sound is heard as pressurized $\\mathrm{CO}_{2}$ escapes from the headspace. This causes the dissolution equilibrium to shift left, resulting in a decrease in the concentration of dissolved $\\mathrm{CO}_{2}$ and subsequent left-shifts of the hydrolysis and ionization equilibria. Fortunately for the consumer, the dissolution equilibrium is usually re-established slowly, and so the beverage may be enjoyed while its dissolved carbon dioxide concentration remains palatably high. Once the equilibria are re-established, the $\\mathrm{CO}_{2}(\\mathrm{aq})$ concentration will be significantly lowered, and the beverage acquires a characteristic taste referred to as \"flat.\"\n\n\nFIGURE 13.7 Opening a soft-drink bottle lowers the $\\mathrm{CO}_{2}$ pressure above the beverage, shifting the dissolution equilibrium and releasing dissolved $\\mathrm{CO}_{2}$ from the beverage. (credit: modification of work by \"D Coetzee\"/Flickr)"}
{"id": 3954, "contents": "1386. Effect of a Change in Temperature - \nConsistent with the law of mass action, an equilibrium stressed by a change in concentration will shift to reestablish equilibrium without any change in the value of the equilibrium constant, $K$. When an equilibrium shifts in response to a temperature change, however, it is re-established with a different relative composition that exhibits a different value for the equilibrium constant.\n\nTo understand this phenomenon, consider the elementary reaction\n\n$$\n\\mathrm{A} \\rightleftharpoons \\mathrm{~B}\n$$\n\nSince this is an elementary reaction, the rates laws for the forward and reverse may be derived directly from the balanced equation's stoichiometry:\n\n$$\n\\begin{aligned}\n& \\operatorname{rate}_{f}=k_{f}[\\mathrm{~A}] \\\\\n& \\operatorname{rate}_{r}=k_{r}[\\mathrm{~B}]\n\\end{aligned}\n$$\n\nWhen the system is at equilibrium,\n\n$$\n\\operatorname{rate}_{r}=\\operatorname{rate}_{f}\n$$\n\nSubstituting the rate laws into this equality and rearranging gives\n\n$$\n\\begin{aligned}\nk_{f}[\\mathrm{~A}] & =k_{r}[\\mathrm{~B}] \\\\\n\\frac{[\\mathrm{B}]}{[\\mathrm{A}]} & =\\frac{k_{f}}{k_{r}}=K_{c}\n\\end{aligned}\n$$\n\nThe equilibrium constant is seen to be a mathematical function of the rate constants for the forward and reverse reactions. Since the rate constants vary with temperature as described by the Arrhenius equation, is stands to reason that the equilibrium constant will likewise vary with temperature (assuming the rate constants are affected to different extents by the temperature change). For more complex reactions involving multistep reaction mechanisms, a similar but more complex mathematical relation exists between the equilibrium constant and the rate constants of the steps in the mechanism. Regardless of how complex the reaction may be, the temperature-dependence of its equilibrium constant persists.\n\nPredicting the shift an equilibrium will experience in response to a change in temperature is most conveniently accomplished by considering the enthalpy change of the reaction. For example, the decomposition of dinitrogen tetroxide is an endothermic (heat-consuming) process:"}
{"id": 3955, "contents": "1386. Effect of a Change in Temperature - \n$$\n\\mathrm{N}_{2} \\mathrm{O}_{4}(g) \\rightleftharpoons 2 \\mathrm{NO}_{2}(g) \\quad \\Delta H=57.20 \\mathrm{~kJ}\n$$\n\nFor purposes of applying Le Chatelier's principle, heat $(q)$ may be viewed as a reactant:\n\n$$\n\\text { heat }+\\mathrm{N}_{2} \\mathrm{O}_{4}(g) \\rightleftharpoons 2 \\mathrm{NO}_{2}(g)\n$$\n\nRaising the temperature of the system is akin to increasing the amount of a reactant, and so the equilibrium will shift to the right. Lowering the system temperature will likewise cause the equilibrium to shift left. For exothermic processes, heat is viewed as a product of the reaction and so the opposite temperature dependence is observed."}
{"id": 3956, "contents": "1387. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Identify the changes in concentration or pressure that occur for chemical species in equilibrium systems\n- Calculate equilibrium concentrations or pressures and equilibrium constants, using various algebraic approaches\n- Explain how temperature affects the spontaneity of some proceses\n- Relate standard free energy changes to equilibrium constants\n\nHaving covered the essential concepts of chemical equilibria in the preceding sections of this chapter, this final section will demonstrate the more practical aspect of using these concepts and appropriate mathematical strategies to perform various equilibrium calculations. These types of computations are essential to many areas of science and technology-for example, in the formulation and dosing of pharmaceutical products. After a drug is ingested or injected, it is typically involved in several chemical equilibria that affect its ultimate concentration in the body system of interest. Knowledge of the quantitative aspects of these equilibria is required to compute a dosage amount that will solicit the desired therapeutic effect.\n\nMany of the useful equilibrium calculations that will be demonstrated here require terms representing changes in reactant and product concentrations. These terms are derived from the stoichiometry of the reaction, as illustrated by decomposition of ammonia:\n\n$$\n2 \\mathrm{NH}_{3}(g) \\rightleftharpoons \\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g)\n$$\n\nAs shown earlier in this chapter, this equilibrium may be established within a sealed container that initially contains either $\\mathrm{NH}_{3}$ only, or a mixture of any two of the three chemical species involved in the equilibrium. Regardless of its initial composition, a reaction mixture will show the same relationships between changes in the concentrations of the three species involved, as dictated by the reaction stoichiometry (see also the related content on expressing reaction rates in the chapter on kinetics). For example, if the nitrogen concentration increases by an amount $x$ :\n\n$$\n\\Delta\\left[\\mathrm{N}_{2}\\right]=+x\n$$\n\nthe corresponding changes in the other species concentrations are"}
{"id": 3957, "contents": "1387. LEARNING OBJECTIVES - \n$$\n\\Delta\\left[\\mathrm{N}_{2}\\right]=+x\n$$\n\nthe corresponding changes in the other species concentrations are\n\n$$\n\\begin{aligned}\n\\Delta\\left[\\mathrm{H}_{2}\\right] & =\\Delta\\left[\\mathrm{N}_{2}\\right]\\left(\\frac{3 \\mathrm{~mol} \\mathrm{H}_{2}}{1 \\mathrm{~mol} \\mathrm{~N}_{2}}\\right)=+3 x \\\\\n\\Delta\\left[\\mathrm{NH}_{3}\\right] & =-\\Delta\\left[\\mathrm{N}_{2}\\right]\\left(\\frac{2 \\mathrm{~mol} \\mathrm{NH}_{3}}{1 \\mathrm{~mol} \\mathrm{~N}_{2}}\\right)=-2 x\n\\end{aligned}\n$$\n\nwhere the negative sign indicates a decrease in concentration."}
{"id": 3958, "contents": "1389. Determining Relative Changes in Concentration - \nDerive the missing terms representing concentration changes for each of the following reactions.\n(a)\n\n(b)\n$\\mathrm{I}_{2}(a q)+\\mathrm{I}^{-}(a q) \\rightleftharpoons \\mathrm{I}_{3}^{-}(a q)$\n$-\\quad x$\n(c)\n$\\begin{array}{ll}\\mathrm{C}_{3} \\mathrm{H}_{8}(g)+5 \\mathrm{O}_{2}(g) \\rightleftharpoons 3 \\mathrm{CO}_{2}(g)+4 \\mathrm{H}_{2} \\mathrm{O}(g) \\\\ x & -\\end{array}$"}
{"id": 3959, "contents": "1390. Solution - \n(a) $\\begin{array}{rl}\\mathrm{C}_{2} \\mathrm{H}_{2}(g)+\\underset{2}{2} & 2 \\mathrm{Br}_{2}(g)\\end{array} \\underset{2 x}{ } \\mathrm{C}_{2} \\mathrm{H}_{2} \\mathrm{Br}_{4}(g)$\n(b) $\\begin{array}{lll}\\mathrm{I}_{2}(a q)+ & \\mathrm{I}^{-}(a q) & \\rightleftharpoons \\\\ -x & -x & \\mathrm{I}_{3}^{-}(a q) \\\\ x\\end{array}$\n(c)"}
{"id": 3960, "contents": "1391. Check Your Learning - \nComplete the changes in concentrations for each of the following reactions:\n(a)\n\n(b)\n\n$4 \\mathrm{NH}_{3}(g)+7 \\mathrm{O}_{2}(g) \\rightleftharpoons 4 \\mathrm{NO}_{2}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(g)$\n(c)"}
{"id": 3961, "contents": "1392. Answer: - \n(a) $2 x, x,-2 x$; (b) $x,-2 x$; (c) $4 x, 7 x,-4 x,-6 x$ or $-4 x,-7 x, 4 x, 6 x$"}
{"id": 3962, "contents": "1393. Calculation of an Equilibrium Constant - \nThe equilibrium constant for a reaction is calculated from the equilibrium concentrations (or pressures) of its reactants and products. If these concentrations are known, the calculation simply involves their substitution into the $K$ expression, as was illustrated by Example 13.2. A slightly more challenging example is provided next, in which the reaction stoichiometry is used to derive equilibrium concentrations from the information provided. The basic strategy of this computation is helpful for many types of equilibrium computations and relies on the use of terms for the reactant and product concentrations initially present, for how they change as the reaction proceeds, and for what they are when the system reaches equilibrium. The acronym ICE is commonly used to refer to this mathematical approach, and the concentrations terms are usually gathered in a tabular format called an ICE table."}
{"id": 3963, "contents": "1395. Calculation of an Equilibrium Constant - \nIodine molecules react reversibly with iodide ions to produce triiodide ions.\n\n$$\n\\mathrm{I}_{2}(a q)+\\mathrm{I}^{-}(a q) \\rightleftharpoons \\mathrm{I}_{3}^{-}(a q)\n$$\n\nIf a solution with the concentrations of $\\mathrm{I}_{2}$ and $\\mathrm{I}^{-}$both equal to $1.000 \\times 10^{-3} \\mathrm{M}$ before reaction gives an equilibrium concentration of $\\mathrm{I}_{2}$ of $6.61 \\times 10^{-4} \\mathrm{M}$, what is the equilibrium constant for the reaction?"}
{"id": 3964, "contents": "1396. Solution - \nTo calculate the equilibrium constants, equilibrium concentrations are needed for all the reactants and products:\n\n$$\n\\begin{aligned}\nK_{C}= & {\\left[\\mathrm{I}_{3}{ }^{-}\\right] } \\\\\n& {\\left[\\mathrm{I}_{2}\\right]\\left[\\mathrm{I}^{-}\\right] }\n\\end{aligned}\n$$\n\nProvided are the initial concentrations of the reactants and the equilibrium concentration of the product. Use this information to derive terms for the equilibrium concentrations of the reactants, presenting all the information in an ICE table.\n\n|  |  |  | $\\mathrm{I}_{3}{ }^{-}$ |\n| :---: | :---: | :---: | :---: |\n| Initial concentration (M) | $1.000 \\times 10^{-3}$ | $1.000 \\times 10^{-3}$ | 0 |\n| Change ( $M$ ) | -x | -x | $+x$ |\n| Equilibrium concentration ( $M$ ) | $1.000 \\times 10^{-3}-x$ | $1.000 \\times 10^{-3}-x$ | $x$ |\n\nAt equilibrium the concentration of $\\mathrm{I}_{2}$ is $6.61 \\times 10^{-4} \\mathrm{M}$ so that\n\n$$\n\\begin{gathered}\n1.000 \\times 10^{-3}-x=6.61 \\times 10^{-4} \\\\\nx=1.000 \\times 10^{-3}-6.61 \\times 10^{-4} \\\\\n=3.39 \\times 10^{-4} M\n\\end{gathered}\n$$\n\nThe ICE table may now be updated with numerical values for all its concentrations:"}
{"id": 3965, "contents": "1396. Solution - \nThe ICE table may now be updated with numerical values for all its concentrations:\n\n|  |  | r | $\\mathrm{I}_{3}{ }^{-}$ |\n| :---: | :---: | :---: | :---: |\n| Initial concentration (M) | $1.000 \\times 10^{-3}$ | $1.000 \\times 10^{-3}$ | 0 |\n| Change ( $M$ ) | $-3.39 \\times 10^{-4}$ | $-3.39 \\times 10^{-4}$ | $+3.39 \\times 10^{-4}$ |\n| Equilibrium concentration ( $M$ ) | $6.61 \\times 10^{-4}$ | $6.61 \\times 10^{-4}$ | $3.39 \\times 10^{-4}$ |\n\nFinally, substitute the equilibrium concentrations into the $K$ expression and solve:\n\n$$\n\\begin{gathered}\nK_{c}=\\frac{\\left[\\mathrm{I}_{3}-\\right]}{\\left[\\mathrm{I}_{2}\\right]\\left[\\mathrm{I}^{-}\\right]} \\\\\n=\\frac{3.39 \\times 10^{-4} M}{\\left(6.61 \\times 10^{-4} M\\right)\\left(6.61 \\times 10^{-4} M\\right)}=776\n\\end{gathered}\n$$"}
{"id": 3966, "contents": "1397. Check Your Learning - \nEthanol and acetic acid react and form water and ethyl acetate, the solvent responsible for the odor of some\nnail polish removers.\n\n$$\n\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}+\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H} \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{C}_{2} \\mathrm{H}_{5}+\\mathrm{H}_{2} \\mathrm{O}\n$$\n\nWhen 1 mol each of $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$ and $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$ are allowed to react in 1 L of the solvent dioxane, equilibrium is established when $\\frac{1}{3} \\mathrm{~mol}$ of each of the reactants remains. Calculate the equilibrium constant for the reaction. (Note: Water is a solute in this reaction.)"}
{"id": 3967, "contents": "1398. Answer: - \n$K_{C}=4$"}
{"id": 3968, "contents": "1399. Calculation of a Missing Equilibrium Concentration - \nWhen the equilibrium constant and all but one equilibrium concentration are provided, the other equilibrium concentration(s) may be calculated. A computation of this sort is illustrated in the next example exercise."}
{"id": 3969, "contents": "1401. Calculation of a Missing Equilibrium Concentration - \nNitrogen oxides are air pollutants produced by the reaction of nitrogen and oxygen at high temperatures. At $2000^{\\circ} \\mathrm{C}$, the value of the $K_{c}$ for the reaction, $\\mathrm{N}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{NO}(g)$, is $4.1 \\times 10^{-4}$. Calculate the equilibrium concentration of $\\mathrm{NO}(\\mathrm{g})$ in air at 1 atm pressure and $2000^{\\circ} \\mathrm{C}$. The equilibrium concentrations of $\\mathrm{N}_{2}$ and $\\mathrm{O}_{2}$ at this pressure and temperature are 0.036 M and 0.0089 M , respectively."}
{"id": 3970, "contents": "1402. Solution - \nSubstitute the provided quantities into the equilibrium constant expression and solve for [NO]:\n\n$$\n\\begin{gathered}\nK_{c}=\\frac{[\\mathrm{NO}]^{2}}{\\left[\\mathrm{~N}_{2}\\right]\\left[\\mathrm{O}_{2}\\right]} \\\\\n{[\\mathrm{NO}]^{2}=K_{c}\\left[\\mathrm{~N}_{2}\\right]\\left[\\mathrm{O}_{2}\\right]} \\\\\n{[\\mathrm{NO}]=\\sqrt{K_{c}\\left[\\mathrm{~N}_{2}\\right]\\left[\\mathrm{O}_{2}\\right]}} \\\\\n=\\sqrt{\\left(4.1 \\times 10^{-4}\\right)(0.036)(0.0089)} \\\\\n=\\sqrt{1.31 \\times 10^{-7}} \\\\\n=3.6 \\times 10^{-4}\n\\end{gathered}\n$$\n\nThus [NO] is $3.6 \\times 10^{-4} \\mathrm{~mol} / \\mathrm{L}$ at equilibrium under these conditions.\nTo confirm this result, it may be used along with the provided equilibrium concentrations to calculate a value for $K$ :\n\n$$\n\\begin{aligned}\nK_{c} & =\\frac{[\\mathrm{NO}]^{2}}{\\left[\\mathrm{~N}_{2}\\right]\\left[\\mathrm{O}_{2}\\right]} \\\\\n& =\\frac{\\left(3.6 \\times 10^{-4}\\right)^{2}}{(0.036)(0.0089)} \\\\\n& =4.0 \\times 10^{-4}\n\\end{aligned}\n$$\n\nThis result is consistent with the provided value for $K$ within nominal uncertainty, differing by just 1 in the least significant digit's place."}
{"id": 3971, "contents": "1403. Check Your Learning - \nThe equilibrium constant $K_{c}$ for the reaction of nitrogen and hydrogen to produce ammonia at a certain temperature is $6.00 \\times 10^{-2}$. Calculate the equilibrium concentration of ammonia if the equilibrium\nconcentrations of nitrogen and hydrogen are $4.26 M$ and $2.09 M$, respectively."}
{"id": 3972, "contents": "1404. Answer: - \n$1.53 \\mathrm{~mol} / \\mathrm{L}$"}
{"id": 3973, "contents": "1405. Calculation of Equilibrium Concentrations from Initial Concentrations - \nPerhaps the most challenging type of equilibrium calculation can be one in which equilibrium concentrations are derived from initial concentrations and an equilibrium constant. For these calculations, a four-step approach is typically useful:\n\n1. Identify the direction in which the reaction will proceed to reach equilibrium.\n2. Develop an ICE table.\n3. Calculate the concentration changes and, subsequently, the equilibrium concentrations.\n4. Confirm the calculated equilibrium concentrations.\n\nThe last two example exercises of this chapter demonstrate the application of this strategy."}
{"id": 3974, "contents": "1407. Calculation of Equilibrium Concentrations - \nUnder certain conditions, the equilibrium constant $K_{c}$ for the decomposition of $\\mathrm{PCl}_{5}(g)$ into $\\mathrm{PCl}_{3}(g)$ and $\\mathrm{Cl}_{2}(g)$ is 0.0211. What are the equilibrium concentrations of $\\mathrm{PCl}_{5}, \\mathrm{PCl}_{3}$, and $\\mathrm{Cl}_{2}$ in a mixture that initially contained only $\\mathrm{PCl}_{5}$ at a concentration of 1.00 M ?"}
{"id": 3975, "contents": "1408. Solution - \nUse the stepwise process described earlier.\nStep 1.\nDetermine the direction the reaction proceeds.\nThe balanced equation for the decomposition of $\\mathrm{PCl}_{5}$ is\n\n$$\n\\mathrm{PCl}_{5}(g) \\rightleftharpoons \\mathrm{PCl}_{3}(g)+\\mathrm{Cl}_{2}(g)\n$$\n\nBecause only the reactant is present initially $Q_{c}=0$ and the reaction will proceed to the right.\nStep 2.\nDevelop an ICE table.\n\n|  | $\\mathbf{P C l}_{\\mathbf{5}}$ |  |  |\n| :---: | :---: | :---: | :---: |\n|  | $\\mathrm{PCl}_{3}$ | + | $\\mathbf{C l}_{\\mathbf{2}}$ |\n| Initial concentration $(M)$ | 1.00 | 0 | 0 |\n| Change $(M)$ | $-x$ | $+x$ | $+x$ |\n| Equilibrium concentration $(M)$ | $1.00-x$ | $x$ | $x$ |"}
{"id": 3976, "contents": "1409. Step 3. - \nSolve for the change and the equilibrium concentrations.\nSubstituting the equilibrium concentrations into the equilibrium constant equation gives\n\n$$\n\\begin{gathered}\nK_{c}=\\frac{\\left[\\mathrm{PCl}_{3}\\right]\\left[\\mathrm{Cl}_{2}\\right]}{\\left[\\mathrm{PCl}_{5}\\right]}=0.0211 \\\\\n=\\frac{(x)(x)}{(1.00-x)}\n\\end{gathered}\n$$\n\n$$\n\\begin{gathered}\n0.0211=\\frac{(x)(x)}{(1.00-x)} \\\\\n0.0211(1.00-x)=x^{2} \\\\\nx^{2}+0.0211 x-0.0211=0\n\\end{gathered}\n$$\n\nAppendix B shows an equation of the form $a x^{2}+b x+c=0$ can be rearranged to solve for $x$ :\n\n$$\nx=\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}\n$$\n\nIn this case, $a=1, b=0.0211$, and $c=-0.0211$. Substituting the appropriate values for $a, b$, and $c$ yields:\n\n$$\n\\begin{gathered}\nx=\\frac{-0.0211 \\pm \\sqrt{(0.0211)^{2}-4(1)(-0.0211)}}{2(1)} \\\\\n=\\frac{-0.0211 \\pm \\sqrt{\\left(4.45 \\times 10^{-4}\\right)+\\left(8.44 \\times 10^{-2}\\right)}}{2} \\\\\n=\\frac{-0.0211 \\pm 0.291}{2}\n\\end{gathered}\n$$\n\nThe two roots of the quadratic are, therefore,\n\n$$\nx=\\frac{-0.0211+0.291}{2}=0.135\n$$\n\nand\n\n$$\nx=\\frac{-0.0211-0.291}{2}=-0.156\n$$\n\nFor this scenario, only the positive root is physically meaningful (concentrations are either zero or positive), and so $x=0.135 \\mathrm{M}$.\n\nThe equilibrium concentrations are"}
{"id": 3977, "contents": "1409. Step 3. - \nFor this scenario, only the positive root is physically meaningful (concentrations are either zero or positive), and so $x=0.135 \\mathrm{M}$.\n\nThe equilibrium concentrations are\n\n$$\n\\begin{gathered}\n{\\left[\\mathrm{PCl}_{5}\\right]=1.00-0.135=0.87 \\mathrm{M}} \\\\\n{\\left[\\mathrm{PCl}_{3}\\right]=x=0.135 \\mathrm{M}} \\\\\n{\\left[\\mathrm{Cl}_{2}\\right]=x=0.135 \\mathrm{M}}\n\\end{gathered}\n$$\n\nStep 4.\nConfirm the calculated equilibrium concentrations.\nSubstitution into the expression for $K_{c}$ (to check the calculation) gives\n\n$$\nK_{c}=\\frac{\\left[\\mathrm{PCl}_{3}\\right]\\left[\\mathrm{Cl}_{2}\\right]}{\\left[\\mathrm{PCl}_{5}\\right]}=\\frac{(0.135)(0.135)}{0.87}=0.021\n$$\n\nThe equilibrium constant calculated from the equilibrium concentrations is equal to the value of $K_{c}$ given in the problem (when rounded to the proper number of significant figures)."}
{"id": 3978, "contents": "1410. Check Your Learning - \nAcetic acid, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$, reacts with ethanol, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$, to form water and ethyl acetate, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{C}_{2} \\mathrm{H}_{5}$.\n\n$$\n\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}+\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH} \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{C}_{2} \\mathrm{H}_{5}+\\mathrm{H}_{2} \\mathrm{O}\n$$\n\nThe equilibrium constant for this reaction with dioxane as a solvent is 4.0. What are the equilibrium concentrations for a mixture that is initially 0.15 M in $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}, 0.15 \\mathrm{M}$ in $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}, 0.40 \\mathrm{M}$ in $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{C}_{2} \\mathrm{H}_{5}$, and $0.40 \\mathrm{Min}_{2} \\mathrm{O}$ ?"}
{"id": 3979, "contents": "1411. Answer: - \n$\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right]=0.18 \\mathrm{M},\\left[\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}\\right]=0.18 \\mathrm{M},\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{C}_{2} \\mathrm{H}_{5}\\right]=0.37 \\mathrm{M},\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]=0.37 \\mathrm{M}$\nCheck Your Learning\nA 1.00-L flask is filled with 1.00 moles of $\\mathrm{H}_{2}$ and 2.00 moles of $\\mathrm{I}_{2}$. The value of the equilibrium constant for the reaction of hydrogen and iodine reacting to form hydrogen iodide is 50.5 under the given conditions. What are the equilibrium concentrations of $\\mathrm{H}_{2}, \\mathrm{I}_{2}$, and HI in moles/L?\n\n$$\n\\mathrm{H}_{2}(g)+\\mathrm{I}_{2}(g) \\rightleftharpoons 2 \\mathrm{HI}(g)\n$$"}
{"id": 3980, "contents": "1412. Answer: - \n$\\left[\\mathrm{H}_{2}\\right]=0.06 \\mathrm{M},\\left[\\mathrm{I}_{2}\\right]=1.06 \\mathrm{M},[\\mathrm{HI}]=1.88 \\mathrm{M}$"}
{"id": 3981, "contents": "1414. Calculation of Equilibrium Concentrations Using an Algebra-Simplifying Assumption - \nWhat are the concentrations at equilibrium of a 0.15 M solution of HCN?\n\n$$\n\\mathrm{HCN}(a q) \\rightleftharpoons \\mathrm{H}^{+}(a q)+\\mathrm{CN}^{-}(a q) \\quad K_{c}=4.9 \\times 10^{-10}\n$$"}
{"id": 3982, "contents": "1415. Solution - \nUsing \" $x$ \" to represent the concentration of each product at equilibrium gives this ICE table.\n\n|  | $\\mathrm{HCN}(\\mathrm{aq}) \\rightleftharpoons$ |  | $\\mathrm{H}^{+}(\\mathrm{aq})$ |\n| :---: | :---: | :---: | :---: |\n|  | + | $\\mathrm{CN}^{-}(\\mathrm{aq})$ |  |\n| Initial concentration $(M)$ | 0.15 | 0 | 0 |\n| Change $(M)$ | $-x$ | $+x$ | $+x$ |\n| Equilibrium concentration $(M)$ | $0.15-x$ | $x$ | $x$ |\n\nSubstitute the equilibrium concentration terms into the $K_{c}$ expression\n\n$$\nK_{c}=\\frac{(x)(x)}{0.15-x}\n$$\n\nRearrange to the quadratic form and solve for $x$\n\n$$\n\\begin{gathered}\nx^{2}+4.9 \\times 10^{-10}-7.35 \\times 10^{-11}=0 \\\\\nx=8.56 \\times 10^{-6} M(3 \\text { sig. figs. })=8.6 \\times 10^{-6} M(2 \\text { sig. figs. })\n\\end{gathered}\n$$\n\nThus $\\left[\\mathrm{H}^{+}\\right]=\\left[\\mathrm{CN}^{-}\\right]=x=8.6 \\times 10^{-6} \\mathrm{M}$ and $[\\mathrm{HCN}]=0.15-x=0.15 \\mathrm{M}$.\nNote in this case that the change in concentration is significantly less than the initial concentration (a consequence of the small $K$ ), and so the initial concentration experiences a negligible change:\n\n$$\n\\text { if } x \\ll 0.15 \\mathrm{M} \\text {, then }(0.15-x) \\approx 0.15\n$$\n\nThis approximation allows for a more expedient mathematical approach to the calculation that avoids the need to solve for the roots of a quadratic equation:"}
{"id": 3983, "contents": "1415. Solution - \nThis approximation allows for a more expedient mathematical approach to the calculation that avoids the need to solve for the roots of a quadratic equation:\n\n$$\n\\begin{gathered}\nK_{c}=\\frac{(x)(x)}{0.15-x} \\approx \\frac{x^{2}}{0.15} \\\\\n4.9 \\times 10^{-10}=\\frac{x^{2}}{0.15} \\\\\nx^{2}=(0.15)\\left(4.9 \\times 10^{-10}\\right)=7.4 \\times 10^{-11}\n\\end{gathered}\n$$\n\n$$\nx=\\sqrt{7.4 \\times 10^{-11}}=8.6 \\times 10^{-6} M\n$$\n\nThe value of $x$ calculated is, indeed, much less than the initial concentration\n\n$$\n8.6 \\times 10^{-6} \\ll 0.15\n$$\n\nand so the approximation was justified. If this simplified approach were to yield a value for $x$ that did not justify the approximation, the calculation would need to be repeated without making the approximation."}
{"id": 3984, "contents": "1416. Check Your Learning - \nWhat are the equilibrium concentrations in a $0.25 \\mathrm{MNH}_{3}$ solution?\n\n$$\n\\mathrm{NH}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{NH}_{4}^{+}(a q)+\\mathrm{OH}^{-}(a q) \\quad K_{\\mathrm{c}}=1.8 \\times 10^{-5}\n$$"}
{"id": 3985, "contents": "1417. Answer: - \n$\\left[\\mathrm{OH}^{-}\\right]=\\left[\\mathrm{NH}_{4}{ }^{+}\\right]=0.0021 \\mathrm{M} ;\\left[\\mathrm{NH}_{3}\\right]=0.25 \\mathrm{M}$"}
{"id": 3986, "contents": "1418. Temperature Dependence of Spontaneity - \nAs was previously demonstrated in the section on entropy in an earlier chapter, the spontaneity of a process may depend upon the temperature of the system. Phase transitions, for example, will proceed spontaneously in one direction or the other depending upon the temperature of the substance in question. Likewise, some chemical reactions can also exhibit temperature dependent spontaneities. To illustrate this concept, the equation relating free energy change to the enthalpy and entropy changes for the process is considered:\n\n$$\n\\Delta G=\\Delta H-T \\Delta S\n$$\n\nThe spontaneity of a process, as reflected in the arithmetic sign of its free energy change, is then determined by the signs of the enthalpy and entropy changes and, in some cases, the absolute temperature. Since $T$ is the absolute (kelvin) temperature, it can only have positive values. Four possibilities therefore exist with regard to the signs of the enthalpy and entropy changes:"}
{"id": 3987, "contents": "1418. Temperature Dependence of Spontaneity - \n1. Both $\\boldsymbol{\\Delta H}$ and $\\boldsymbol{\\Delta S}$ are positive. This condition describes an endothermic process that involves an increase in system entropy. In this case, $\\Delta G$ will be negative if the magnitude of the $T \\Delta S$ term is greater than $\\Delta H$. If the $T \\Delta S$ term is less than $\\Delta H$, the free energy change will be positive. Such a process is spontaneous at high temperatures and nonspontaneous at low temperatures.\n2. Both $\\boldsymbol{\\Delta H}$ and $\\boldsymbol{\\Delta S}$ are negative. This condition describes an exothermic process that involves a decrease in system entropy. In this case, $\\Delta G$ will be negative if the magnitude of the $T \\Delta S$ term is less than $\\Delta H$. If the $T \\Delta S$ term's magnitude is greater than $\\Delta H$, the free energy change will be positive. Such a process is spontaneous at low temperatures and nonspontaneous at high temperatures.\n3. $\\boldsymbol{\\Delta H}$ is positive and $\\boldsymbol{\\Delta} \\boldsymbol{S}$ is negative. This condition describes an endothermic process that involves a decrease in system entropy. In this case, $\\Delta G$ will be positive regardless of the temperature. Such a process is nonspontaneous at all temperatures.\n4. $\\boldsymbol{\\Delta} \\boldsymbol{H}$ is negative and $\\boldsymbol{\\Delta} \\boldsymbol{S}$ is positive. This condition describes an exothermic process that involves an increase in system entropy. In this case, $\\Delta G$ will be negative regardless of the temperature. Such a process is spontaneous at all temperatures.\n\nThese four scenarios are summarized in Figure 13.8."}
{"id": 3988, "contents": "1418. Temperature Dependence of Spontaneity - \nThese four scenarios are summarized in Figure 13.8.\n\n|  | $\\begin{gathered} \\Delta \\mathrm{H}>0 \\\\ \\text { (endothermic) } \\end{gathered}$ | $\\begin{gathered} \\Delta \\mathrm{H}<0 \\\\ \\text { (exothermic) } \\end{gathered}$ |\n| :---: | :---: | :---: |\n| $\\Delta S>0$ <br> (increase in entropy) | $\\Delta \\mathrm{G}<0$ at high temperature $\\Delta G>0$ at low temperature Process is spontaneous at high temperature | $\\Delta G<0$ at any temperature <br> Process is spontaneous at any temperature |\n| $\\Delta \\mathrm{S}<0$ (decrease in entropy) | $\\Delta \\mathrm{G}>0$ at any temperature <br> Process is nonspontaneous at any temperature | $\\Delta \\mathrm{G}<0$ at low temperature $\\Delta G>0$ at high temperature Process is spontaneous at low temperature |\n\nFIGURE 13.8 There are four possibilities regarding the signs of enthalpy and entropy changes."}
{"id": 3989, "contents": "1420. Predicting the Temperature Dependence of Spontaneity - \nThe incomplete combustion of carbon is described by the following equation:\n\n$$\n2 \\mathrm{C}(s)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{CO}(g)\n$$\n\nHow does the spontaneity of this process depend upon temperature?"}
{"id": 3990, "contents": "1421. Solution - \nCombustion processes are exothermic ( $\\Delta H<0$ ). This particular reaction involves an increase in entropy due to the accompanying increase in the amount of gaseous species (net gain of one mole of gas, $\\Delta S>0$ ). The reaction is therefore spontaneous $(\\Delta G<0)$ at all temperatures."}
{"id": 3991, "contents": "1422. Check Your Learning - \nPopular chemical hand warmers generate heat by the air-oxidation of iron:\n\n$$\n4 \\mathrm{Fe}(s)+3 \\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{Fe}_{2} \\mathrm{O}_{3}(s)\n$$\n\nHow does the spontaneity of this process depend upon temperature?"}
{"id": 3992, "contents": "1423. Answer: - \n$\\Delta H$ and $\\Delta S$ are negative; the reaction is spontaneous at low temperatures.\n\nWhen considering the conclusions drawn regarding the temperature dependence of spontaneity, it is important to keep in mind what the terms \"high\" and \"low\" mean. Since these terms are adjectives, the temperatures in question are deemed high or low relative to some reference temperature. A process that is nonspontaneous at one temperature but spontaneous at another will necessarily undergo a change in \"spontaneity\" (as reflected by its $\\Delta G$ ) as temperature varies. This is clearly illustrated by a graphical presentation of the free energy change equation, in which $\\Delta G$ is plotted on the $y$ axis versus $T$ on the $x$ axis:\n\n$$\n\\begin{gathered}\n\\Delta G=\\Delta H-T \\Delta S \\\\\ny=b+m x\n\\end{gathered}\n$$\n\nSuch a plot is shown in Figure 13.9. A process whose enthalpy and entropy changes are of the same arithmetic sign will exhibit a temperature-dependent spontaneity as depicted by the two yellow lines in the plot. Each line crosses from one spontaneity domain (positive or negative $\\Delta G$ ) to the other at a temperature that is characteristic of the process in question. This temperature is represented by the $x$-intercept of the line, that is, the value of $T$ for which $\\Delta G$ is zero:\n\n$$\n\\Delta G=0=\\Delta H-T \\Delta S\n$$\n\n$$\nT=\\frac{\\Delta H}{\\Delta S}\n$$\n\nSo, saying a process is spontaneous at \"high\" or \"low\" temperatures means the temperature is above or below, respectively, that temperature at which $\\Delta G$ for the process is zero. As noted earlier, the condition of $\\Delta \\mathrm{G}=0$ describes a system at equilibrium.\n\n\nFIGURE 13.9 These plots show the variation in $\\Delta G$ with temperature for the four possible combinations of arithmetic sign for $\\Delta H$ and $\\Delta S$."}
{"id": 3993, "contents": "1425. Equilibrium Temperature for a Phase Transition - \nAs defined in the chapter on liquids and solids, the boiling point of a liquid is the temperature at which its liquid and gaseous phases are in equilibrium (that is, when vaporization and condensation occur at equal rates). Use the information in Appendix G to estimate the boiling point of water."}
{"id": 3994, "contents": "1426. Solution - \nThe process of interest is the following phase change:\n\n$$\n\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(g)\n$$\n\nWhen this process is at equilibrium, $\\Delta G=0$, so the following is true:\n\n$$\n0=\\Delta H^{\\circ}-T \\Delta S^{\\circ} \\quad \\text { or } \\quad T=\\frac{\\Delta H^{\\circ}}{\\Delta S^{\\circ}}\n$$\n\nUsing the standard thermodynamic data from Appendix G,"}
{"id": 3995, "contents": "1426. Solution - \nUsing the standard thermodynamic data from Appendix G,\n\n$$\n\\begin{aligned}\n\\Delta H^{\\circ} & =1 \\mathrm{~mol} \\times \\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{H}_{2} \\mathrm{O}(g)\\right)-1 \\mathrm{~mol} \\times \\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{H}_{2} \\mathrm{O}(l)\\right) \\\\\n& =(1 \\mathrm{~mol})-241.82 \\mathrm{~kJ} / \\mathrm{mol}-(1 \\mathrm{~mol})(-241.82 \\mathrm{~kJ} / \\mathrm{mol})=44.01 \\mathrm{~kJ} \\\\\n\\Delta S^{\\circ} & =1 \\mathrm{~mol} \\times \\Delta S^{\\circ}\\left(\\mathrm{H}_{2} \\mathrm{O}(g)\\right)-1 \\mathrm{~mol} \\times \\Delta S^{\\circ}\\left(\\mathrm{H}_{2} \\mathrm{O}(l)\\right) \\\\\n& =(1 \\mathrm{~mol}) 188.8 \\mathrm{~J} / \\mathrm{K} \\cdot \\mathrm{~mol}-(1 \\mathrm{~mol}) 70.0 \\mathrm{~J} / \\mathrm{K} \\cdot \\mathrm{~mol}=118.8 \\mathrm{~J} / \\mathrm{K} \\\\\nT & =\\frac{\\Delta H^{\\circ}}{\\Delta S^{\\circ}}=\\frac{44.01 \\times 10^{3} \\mathrm{~J}}{118.8 \\mathrm{~J} / \\mathrm{K}}=370.5 \\mathrm{~K}=97.3^{\\circ} \\mathrm{C}\n\\end{aligned}\n$$"}
{"id": 3996, "contents": "1426. Solution - \nThe accepted value for water's normal boiling point is $373.2 \\mathrm{~K}\\left(100.0^{\\circ} \\mathrm{C}\\right)$, and so this calculation is in reasonable agreement. Note that the values for enthalpy and entropy changes data used were derived from standard data at 298 K (Appendix G). If desired, you could obtain more accurate results by using enthalpy and entropy changes determined at (or at least closer to) the actual boiling point."}
{"id": 3997, "contents": "1427. Check Your Learning - \nUse the information in Appendix G to estimate the boiling point of $\\mathrm{CS}_{2}$."}
{"id": 3998, "contents": "1428. Answer: - \n313 K (accepted value 319 K )"}
{"id": 3999, "contents": "1429. Free Energy and Equilibrium - \nThe free energy change for a process may be viewed as a measure of its driving force. A negative value for $\\Delta G$ represents a driving force for the process in the forward direction, while a positive value represents a driving force for the process in the reverse direction. When $\\Delta G$ is zero, the forward and reverse driving forces are equal, and the process occurs in both directions at the same rate (the system is at equilibrium).\n\nIn the section on equilibrium, the reaction quotient, $Q$, was introduced as a convenient measure of the status of an equilibrium system. Recall that $Q$ is the numerical value of the mass action expression for the system, and that you may use its value to identify the direction in which a reaction will proceed in order to achieve equilibrium. When $Q$ is lesser than the equilibrium constant, $K$, the reaction will proceed in the forward direction until equilibrium is reached and $Q=K$. Conversely, if $Q>K$, the process will proceed in the reverse direction until equilibrium is achieved.\n\nThe free energy change for a process taking place with reactants and products present under nonstandard conditions (pressures other than 1 bar; concentrations other than 1 M ) is related to the standard free energy change, according to this equation:\n\n$$\n\\Delta G=\\Delta G^{\\circ}+R T \\ln Q\n$$\n\n$R$ is the gas constant ( $8.314 \\mathrm{~J} / \\mathrm{K} \\mathrm{mol}$ ), $T$ is the kelvin or absolute temperature, and $Q$ is the reaction quotient. This equation may be used to predict the spontaneity for a process under any given set of conditions as illustrated in Example 13.12."}
{"id": 4000, "contents": "1431. Calculating $\\boldsymbol{\\Delta} \\boldsymbol{G}$ under Nonstandard Conditions - \nWhat is the free energy change for the process shown here under the specified conditions?\n\n$$\n\\begin{aligned}\nT=25^{\\circ} \\mathrm{C}, P_{\\mathrm{N}_{2}}=0.870 \\mathrm{~atm}, P_{\\mathrm{H}_{2}} & =0.250 \\mathrm{~atm}, \\text { and } P_{\\mathrm{NH}_{3}}=12.9 \\mathrm{~atm} \\\\\n2 \\mathrm{NH}_{3}(g) & \\longrightarrow 3 \\mathrm{H}_{2}(g)+\\mathrm{N}_{2}(g) \\quad \\Delta G^{\\circ}=33.0 \\mathrm{~kJ} / \\mathrm{mol}\n\\end{aligned}\n$$"}
{"id": 4001, "contents": "1432. Solution - \nThe equation relating free energy change to standard free energy change and reaction quotient may be used directly:\n\n$$\n\\Delta G=\\Delta G^{\\circ}+R T \\ln Q=33.0 \\frac{\\mathrm{~kJ}}{\\mathrm{~mol}}+\\left(8.314 \\frac{\\mathrm{~J}}{\\mathrm{~mol} \\mathrm{~K}} \\times 298 \\mathrm{~K} \\times \\ln \\frac{\\left(0.250^{3}\\right) \\times 0.870}{12.9^{2}}\\right)=9680 \\frac{\\mathrm{~J}}{\\mathrm{~mol}} \\text { or } 9.68 \\mathrm{~kJ} / \\mathrm{mol}\n$$\n\nSince the computed value for $\\Delta G$ is positive, the reaction is nonspontaneous under these conditions."}
{"id": 4002, "contents": "1433. Check Your Learning - \nCalculate the free energy change for this same reaction at $875^{\\circ} \\mathrm{C}$ in a 5.00 L mixture containing 0.100 mol of each gas. Is the reaction spontaneous under these conditions?"}
{"id": 4003, "contents": "1434. Answer: - \n$\\Delta G=-47 \\mathrm{~kJ} / \\mathrm{mol}$; yes\n\nFor a system at equilibrium, $Q=K$ and $\\Delta G=0$, and the previous equation may be written as\n\n$$\n\\begin{array}{rlrr}\n0 & =\\Delta G^{\\circ}+R T \\ln K & & \\text { (at equilibrium) } \\\\\n\\Delta G^{\\circ} & =-R T \\ln K & \\text { or } & K=e^{-\\frac{\\Delta G^{\\circ}}{R T}}\n\\end{array}\n$$\n\nThis form of the equation provides a useful link between these two essential thermodynamic properties, and it can be used to derive equilibrium constants from standard free energy changes and vice versa. The relations between standard free energy changes and equilibrium constants are summarized in Table 13.1.\n\nRelations between Standard Free Energy Changes and Equilibrium Constants\n\n| $\\boldsymbol{K}$ | $\\Delta \\boldsymbol{G}^{\\circ}$ | Composition of an Equilibrium Mixture |\n| :--- | :--- | :--- |\n| $>1$ | $<0$ | Products are more abundant |\n| $<1$ | $>0$ | Reactants are more abundant |\n| $=1$ | $=0$ | Reactants and products are comparably abundant |\n\nTABLE 13.1"}
{"id": 4004, "contents": "1436. Calculating an Equilibrium Constant using Standard Free Energy Change - \nGiven that the standard free energies of formation of $\\mathrm{Ag}^{+}(\\mathrm{aq}), \\mathrm{Cl}^{-}(\\mathrm{aq})$, and $\\mathrm{AgCl}(s)$ are $77.1 \\mathrm{~kJ} / \\mathrm{mol},-131.2 \\mathrm{~kJ} /$ mol , and $-109.8 \\mathrm{~kJ} / \\mathrm{mol}$, respectively, calculate the solubility product, $K_{\\mathrm{sp}}$, for AgCl ."}
{"id": 4005, "contents": "1437. Solution - \nThe reaction of interest is the following:\n\n$$\n\\mathrm{AgCl}(s) \\rightleftharpoons \\mathrm{Ag}^{+}(a q)+\\mathrm{Cl}^{-}(a q) \\quad K_{\\mathrm{sp}}=\\left[\\mathrm{Ag}^{+}\\right]\\left[\\mathrm{Cl}^{-}\\right]\n$$\n\nThe standard free energy change for this reaction is first computed using standard free energies of formation for its reactants and products:\n\n$$\n\\begin{aligned}\n& \\Delta G^{\\circ}=\\left[\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{Ag}^{+}(a q)\\right)+\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{Cl}^{-}(a q)\\right)\\right]-\\left[\\Delta G_{\\mathrm{f}}^{\\circ}(\\mathrm{AgCl}(s))\\right] \\\\\n& =[77.1 \\mathrm{~kJ} / \\mathrm{mol}-131.2 \\mathrm{~kJ} / \\mathrm{mol}]-[-109.8 \\mathrm{~kJ} / \\mathrm{mol}]=55.7 \\mathrm{~kJ} / \\mathrm{mol}\n\\end{aligned}\n$$\n\nThe equilibrium constant for the reaction may then be derived from its standard free energy change:\n\n$$\nK_{\\mathrm{sp}}=e^{-\\frac{\\Delta G^{\\circ}}{R T}}=\\exp \\left(-\\frac{\\Delta G^{\\circ}}{R T}\\right)=\\exp \\left(-\\frac{55.7 \\times 10^{3} \\mathrm{~J} / \\mathrm{mol}}{8.314 \\mathrm{~J} / \\mathrm{mol} \\cdot \\mathrm{~K} \\times 298.15 \\mathrm{~K}}\\right)=\\exp (-22.470)=e^{-22.470}=1.74 \\times 10^{-10}\n$$\n\nThis result is in reasonable agreement with the value provided in Appendix J."}
{"id": 4006, "contents": "1438. Check Your Learning - \nUse the thermodynamic data provided in Appendix G to calculate the equilibrium constant for the dissociation of dinitrogen tetroxide at $25^{\\circ} \\mathrm{C}$.\n\n$$\n\\mathrm{N}_{2} \\mathrm{O}_{4}(g) \\rightleftharpoons 2 \\mathrm{NO}_{2}(g)\n$$"}
{"id": 4007, "contents": "1439. Answer: - \n$K=0.32$\n\nTo further illustrate the relation between these two essential thermodynamic concepts, consider the observation that reactions spontaneously proceed in a direction that ultimately establishes equilibrium. As may be shown by plotting the free energy change versus the extent of the reaction (for example, as reflected in the value of $Q$ ), equilibrium is established when the system's free energy is minimized (Figure 13.10). If a system consists of reactants and products in nonequilibrium amounts $(Q \\neq K)$, the reaction will proceed spontaneously in the direction necessary to establish equilibrium.\n\n\nFIGURE 13.10 These plots show the free energy versus reaction progress for systems whose standard free changes are (a) negative, (b) positive, and (c) zero. Nonequilibrium systems will proceed spontaneously in whatever direction is necessary to minimize free energy and establish equilibrium."}
{"id": 4008, "contents": "1440. Key Terms - \nequilibrium state of a reversible reaction in which the forward and reverse processes occur at equal rates\nequilibrium constant ( $\\boldsymbol{K}$ ) value of the reaction quotient for a system at equilibrium; may be expressed using concentrations ( $K_{c}$ ) or partial pressures ( $K_{p}$ )\nheterogeneous equilibria equilibria in which reactants and products occupy two or more different phases\nhomogeneous equilibria equilibria in which all reactants and products occupy the same phase\nlaw of mass action when a reversible reaction has\nattained equilibrium at a given temperature, the reaction quotient remains constant\nLe Ch\u00e2telier's principle an equilibrium subjected to stress will shift in a way to counter the stress and re-establish equilibrium\nreaction quotient $(Q)$ mathematical function describing the relative amounts of reactants and products in a reaction mixture; may be expressed in terms of concentrations $\\left(Q_{c}\\right)$ or pressures $\\left(Q_{p}\\right)$\nreversible reaction chemical reaction that can proceed in both the forward and reverse directions under given conditions"}
{"id": 4009, "contents": "1441. Key Equations - \n$$\n\\begin{array}{lr}\nQ_{c}=\\frac{[\\mathrm{C}]^{x}[\\mathrm{D}]^{y}}{[\\mathrm{~A}]^{m}[\\mathrm{~B}]^{n}} & \\text { for the reaction } m \\mathrm{~A}+n \\mathrm{~B} \\rightleftharpoons x \\mathrm{C}+y \\mathrm{D} \\\\\nQ_{P}=\\frac{\\left(P_{C}\\right)^{x}\\left(P_{D}\\right)^{y}}{\\left(P_{A}\\right)^{m}\\left(P_{B}\\right)^{n}} & \\text { for the reaction } m \\mathrm{~A}+n \\mathrm{~B} \\rightleftharpoons x \\mathrm{C}+y \\mathrm{D} \\\\\nP=M R T & \\\\\nK_{c}=Q_{c} \\text { at equilibrium } & \\\\\nK_{p}=Q_{p} \\text { at equilibrium } & \\\\\nK_{P}=K_{c}(R T)^{\\Delta n} &\n\\end{array}\n$$"}
{"id": 4010, "contents": "1442. Summary - 1442.1. Chemical Equilibria\nA reversible reaction is at equilibrium when the forward and reverse processes occur at equal rates. Chemical equilibria are dynamic processes characterized by constant amounts of reactant and product species."}
{"id": 4011, "contents": "1442. Summary - 1442.2. Equilibrium Constants\nThe composition of a reaction mixture may be represented by a mathematical function known as the reaction quotient, $Q$. For a reaction at equilibrium, the composition is constant, and $Q$ is called the equilibrium constant, $K$.\n\nA homogeneous equilibrium is an equilibrium in which all components are in the same phase. A heterogeneous equilibrium is an equilibrium in which components are in two or more phases."}
{"id": 4012, "contents": "1442. Summary - 1442.3. Shifting Equilibria: Le Ch\u00e2telier's Principle\nSystems at equilibrium can be disturbed by changes to temperature, concentration, and, in some cases, volume and pressure. The system's response to\nthese disturbances is described by Le Ch\u00e2telier's principle: An equilibrium system subjected to a disturbance will shift in a way that counters the disturbance and re-establishes equilibrium. A catalyst will increase the rate of both the forward and reverse reactions of a reversible process, increasing the rate at which equilibrium is reached but not altering the equilibrium mixture's composition ( $K$ does not change)."}
{"id": 4013, "contents": "1442. Summary - 1442.4. Equilibrium Calculations\nCalculating values for equilibrium constants and/or equilibrium concentrations is of practical benefit to many applications. A mathematical strategy that uses initial concentrations, changes in concentrations, and equilibrium concentrations (and goes by the acronym ICE) is useful for several types of equilibrium calculations. We also learned that a negative value for $\\Delta G$ indicates a spontaneous process; a positive $\\Delta G$ indicates a nonspontaneous process; and a $\\Delta G$ of zero indicates that the system is at equilibrium. We also saw how free energy, spontaneity, and equilibrium relate."}
{"id": 4014, "contents": "1443. Exercises - 1443.1. Chemical Equilibria\n1. What does it mean to describe a reaction as \"reversible\"?\n2. When writing an equation, how is a reversible reaction distinguished from a nonreversible reaction?\n3. If a reaction is reversible, when can it be said to have reached equilibrium?\n4. Is a system at equilibrium if the rate constants of the forward and reverse reactions are equal?\n5. If the concentrations of products and reactants are equal, is the system at equilibrium?"}
{"id": 4015, "contents": "1443. Exercises - 1443.2. Equilibrium Constants\n6. Explain why there may be an infinite number of values for the reaction quotient of a reaction at a given temperature but there can be only one value for the equilibrium constant at that temperature.\n7. Explain why an equilibrium between $\\mathrm{Br}_{2}(I)$ and $\\mathrm{Br}_{2}(g)$ would not be established if the container were not a closed vessel shown in Figure 13.4.\n8. If you observe the following reaction at equilibrium, is it possible to tell whether the reaction started with pure $\\mathrm{NO}_{2}$ or with pure $\\mathrm{N}_{2} \\mathrm{O}_{4}$ ? $2 \\mathrm{NO}_{2}(g) \\rightleftharpoons \\mathrm{N}_{2} \\mathrm{O}_{4}(g)$\n9. Among the solubility rules previously discussed is the statement: All chlorides are soluble except $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$, $\\mathrm{AgCl}, \\mathrm{PbCl}_{2}$, and CuCl .\n(a) Write the expression for the equilibrium constant for the reaction represented by the equation\n$\\mathrm{AgCl}(s) \\rightleftharpoons \\mathrm{Ag}^{+}(a q)+\\mathrm{Cl}^{-}(a q)$. Is $K_{c}>1,<1$, or $\\approx 1$ ? Explain your answer.\n(b) Write the expression for the equilibrium constant for the reaction represented by the equation $\\mathrm{Pb}^{2+}(a q)+2 \\mathrm{Cl}^{-}(a q) \\rightleftharpoons \\mathrm{PbCl}_{2}(s)$. Is $K_{c}>1$, <1, or $\\approx 1$ ? Explain your answer.\n10. Among the solubility rules previously discussed is the statement: Carbonates, phosphates, borates, and arsenates-except those of the ammonium ion and the alkali metals-are insoluble.\n(a) Write the expression for the equilibrium constant for the reaction represented by the equation"}
{"id": 4016, "contents": "1443. Exercises - 1443.2. Equilibrium Constants\n(a) Write the expression for the equilibrium constant for the reaction represented by the equation\n$\\mathrm{CaCO}_{3}(s) \\rightleftharpoons \\mathrm{Ca}^{2+}(a q)+\\mathrm{CO}_{3}^{2-}(a q)$. Is $K_{c}>1,<1$, or $\\approx 1$ ? Explain your answer.\n(b) Write the expression for the equilibrium constant for the reaction represented by the equation $3 \\mathrm{Ba}^{2+}(a q)+2 \\mathrm{PO}_{4}{ }^{3-}(a q) \\rightleftharpoons \\mathrm{Ba}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}(s)$. Is $K_{c}>1,<1$, or $\\approx 1$ ? Explain your answer.\n11. Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene: $3 \\mathrm{C}_{2} \\mathrm{H}_{2}(g) \\rightleftharpoons \\mathrm{C}_{6} \\mathrm{H}_{6}(g)$. Which value of $K_{c}$ would make this reaction most useful commercially? $K_{c} \\approx 0.01, K_{c} \\approx 1$, or $K_{c} \\approx 10$. Explain your answer.\n12. Show that the complete chemical equation, the total ionic equation, and the net ionic equation for the reaction represented by the equation $\\mathrm{KI}(a q)+\\mathrm{I}_{2}(a q) \\rightleftharpoons \\mathrm{KI}_{3}(a q)$ give the same expression for the reaction quotient. $\\mathrm{KI}_{3}$ is composed of the ions $\\mathrm{K}^{+}$and $\\mathrm{I}_{3}{ }^{-}$.\n13. For a titration to be effective, the reaction must be rapid and the yield of the reaction must essentially be $100 \\%$. Is $K_{c}>1,<1$, or $\\approx 1$ for a titration reaction?"}
{"id": 4017, "contents": "1443. Exercises - 1443.2. Equilibrium Constants\n14. For a precipitation reaction to be useful in a gravimetric analysis, the product of the reaction must be insoluble. Is $K_{c}>1,<1$, or $\\approx 1$ for a useful precipitation reaction?\n15. Write the mathematical expression for the reaction quotient, $Q_{c}$, for each of the following reactions:\n(a) $\\mathrm{CH}_{4}(g)+\\mathrm{Cl}_{2}(g) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{Cl}(g)+\\mathrm{HCl}(g)$\n(b) $\\mathrm{N}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{NO}(g)$\n(c) $2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{SO}_{3}(g)$\n(d) $\\mathrm{BaSO}_{3}(s) \\rightleftharpoons \\mathrm{BaO}(s)+\\mathrm{SO}_{2}(g)$\n(e) $\\mathrm{P}_{4}(g)+5 \\mathrm{O}_{2}(g) \\rightleftharpoons \\mathrm{P}_{4} \\mathrm{O}_{10}(s)$\n(f) $\\mathrm{Br}_{2}(g) \\rightleftharpoons 2 \\mathrm{Br}(g)$\n(g) $\\mathrm{CH}_{4}(g)+2 \\mathrm{O}_{2}(g) \\rightleftharpoons \\mathrm{CO}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n(h) $\\mathrm{CuSO}_{4} \\cdot 5 \\mathrm{H}_{2} \\mathrm{O}(s) \\rightleftharpoons \\mathrm{CuSO}_{4}(s)+5 \\mathrm{H}_{2} \\mathrm{O}(g)$\n16. Write the mathematical expression for the reaction quotient, $Q_{c}$, for each of the following reactions:\n(a) $\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g)$"}
{"id": 4018, "contents": "1443. Exercises - 1443.2. Equilibrium Constants\n(a) $\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g)$\n(b) $4 \\mathrm{NH}_{3}(g)+5 \\mathrm{O}_{2}(g) \\rightleftharpoons 4 \\mathrm{NO}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(g)$\n(c) $\\mathrm{N}_{2} \\mathrm{O}_{4}(g) \\rightleftharpoons 2 \\mathrm{NO}_{2}(g)$\n(d) $\\mathrm{CO}_{2}(g)+\\mathrm{H}_{2}(g) \\rightleftharpoons \\mathrm{CO}(g)+\\mathrm{H}_{2} \\mathrm{O}(g)$\n(e) $\\mathrm{NH}_{4} \\mathrm{Cl}(s) \\rightleftharpoons \\mathrm{NH}_{3}(g)+\\mathrm{HCl}(g)$\n(f) $2 \\mathrm{~Pb}\\left(\\mathrm{NO}_{3}\\right)_{2}(s) \\rightleftharpoons 2 \\mathrm{PbO}(s)+4 \\mathrm{NO}_{2}(g)+\\mathrm{O}_{2}(g)$\n(g) $2 \\mathrm{H}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n(h) $\\mathrm{S}_{8}(g) \\rightleftharpoons 8 \\mathrm{~S}(g)$\n17. The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.\n(a) $2 \\mathrm{NH}_{3}(g) \\rightleftharpoons \\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g)$\n$K_{c}=17 ;\\left[\\mathrm{NH}_{3}\\right]=0.20 M,\\left[\\mathrm{~N}_{2}\\right]=1.00 M,\\left[\\mathrm{H}_{2}\\right]=1.00 M$"}
{"id": 4019, "contents": "1443. Exercises - 1443.2. Equilibrium Constants\n(b) $2 \\mathrm{NH}_{3}(g) \\rightleftharpoons \\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g)$\n$K_{P}=6.8 \\times 10^{4} ; \\mathrm{NH}_{3}=3.0 \\mathrm{~atm}, \\mathrm{~N}_{2}=2.0 \\mathrm{~atm}, \\mathrm{H}_{2}=1.0 \\mathrm{~atm}$\n(c) $2 \\mathrm{SO}_{3}(g) \\rightleftharpoons 2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g)$\n$K_{c}=0.230 ;\\left[\\mathrm{SO}_{3}\\right]=0.00 \\mathrm{M},\\left[\\mathrm{SO}_{2}\\right]=1.00 \\mathrm{M},\\left[\\mathrm{O}_{2}\\right]=1.00 \\mathrm{M}$\n(d) $2 \\mathrm{SO}_{3}(g) \\rightleftharpoons 2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g)$ $K_{P}=16.5 ; \\mathrm{SO}_{3}=1.00 \\mathrm{~atm}, \\mathrm{SO}_{2}=1.00 \\mathrm{~atm}, \\mathrm{O}_{2}=1.00 \\mathrm{~atm}$\n(e) $2 \\mathrm{NO}(g)+\\mathrm{Cl}_{2}(g) \\rightleftharpoons 2 \\mathrm{NOCl}(g)$ $K_{c}=4.6 \\times 10^{4} ;[\\mathrm{NO}]=1.00 M,\\left[\\mathrm{Cl}_{2}\\right]=1.00 M,[\\mathrm{NOCl}]=0 M$\n(f) $\\mathrm{N}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{NO}(g)$\n$K_{P}=0.050 ; \\mathrm{NO}=10.0 \\mathrm{~atm}, \\mathrm{~N}_{2}=\\mathrm{O}_{2}=5 \\mathrm{~atm}$"}
{"id": 4020, "contents": "1443. Exercises - 1443.2. Equilibrium Constants\n$K_{P}=0.050 ; \\mathrm{NO}=10.0 \\mathrm{~atm}, \\mathrm{~N}_{2}=\\mathrm{O}_{2}=5 \\mathrm{~atm}$\n18. The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.\n(a) $2 \\mathrm{NH}_{3}(g) \\rightleftharpoons \\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\quad K_{c}=17 ;\\left[\\mathrm{NH}_{3}\\right]=0.50 \\mathrm{M},\\left[\\mathrm{N}_{2}\\right]=0.15 \\mathrm{M},\\left[\\mathrm{H}_{2}\\right]=0.12 \\mathrm{M}$\n(b) $2 \\mathrm{NH}_{3}(g) \\rightleftharpoons \\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g)$\n$K_{P}=6.8 \\times 10^{4} ; \\mathrm{NH}_{3}=2.00 \\mathrm{~atm}, \\mathrm{~N}_{2}=10.00 \\mathrm{~atm}, \\mathrm{H}_{2}=10.00 \\mathrm{~atm}$\n(c) $2 \\mathrm{SO}_{3}(g) \\rightleftharpoons 2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g)$\n$K_{c}=0.230 ;\\left[\\mathrm{SO}_{3}\\right]=2.00 \\mathrm{M},\\left[\\mathrm{SO}_{2}\\right]=2.00 \\mathrm{M},\\left[\\mathrm{O}_{2}\\right]=2.00 \\mathrm{M}$\n(d) $2 \\mathrm{SO}_{3}(g) \\rightleftharpoons 2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g)$"}
{"id": 4021, "contents": "1443. Exercises - 1443.2. Equilibrium Constants\n(d) $2 \\mathrm{SO}_{3}(g) \\rightleftharpoons 2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g)$\n$K_{P}=6.5 \\mathrm{~atm} ; \\mathrm{SO}_{2}=1.00 \\mathrm{~atm}, \\mathrm{O}_{2}=1.130 \\mathrm{~atm}, \\mathrm{SO}_{3}=0 \\mathrm{~atm}$\n(e) $2 \\mathrm{NO}(g)+\\mathrm{Cl}_{2}(g) \\rightleftharpoons 2 \\mathrm{NOCl}(g)$\n$K_{P}=2.5 \\times 10^{3} ; \\mathrm{NO}=1.00 \\mathrm{~atm}, \\mathrm{Cl}_{2}=1.00 \\mathrm{~atm}, \\mathrm{NOCl}=0 \\mathrm{~atm}$\n(f) $\\mathrm{N}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{NO}(g)$\n$K_{c}=0.050 ;\\left[\\mathrm{N}_{2}\\right]=0.100 \\mathrm{M},\\left[\\mathrm{O}_{2}\\right]=0.200 \\mathrm{M},[\\mathrm{NO}]=1.00 \\mathrm{M}$\n19. The following reaction has $K_{P}=4.50 \\times 10^{-5}$ at 720 K .\n$\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g)$\nIf a reaction vessel is filled with each gas to the partial pressures listed, in which direction will it shift to reach equilibrium? $P\\left(\\mathrm{NH}_{3}\\right)=93 \\mathrm{~atm}, P\\left(\\mathrm{~N}_{2}\\right)=48 \\mathrm{~atm}$, and $P\\left(\\mathrm{H}_{2}\\right)=52$ atm\n20. Determine if the following system is at equilibrium. If not, in which direction will the system need to shift to reach equilibrium?"}
{"id": 4022, "contents": "1443. Exercises - 1443.2. Equilibrium Constants\n20. Determine if the following system is at equilibrium. If not, in which direction will the system need to shift to reach equilibrium?\n$\\mathrm{SO}_{2} \\mathrm{Cl}_{2}(g) \\rightleftharpoons \\mathrm{SO}_{2}(g)+\\mathrm{Cl}_{2}(g)$\n$\\left[\\mathrm{SO}_{2} \\mathrm{Cl}_{2}\\right]=0.12 \\mathrm{M},\\left[\\mathrm{Cl}_{2}\\right]=0.16 \\mathrm{M}$ and $\\left[\\mathrm{SO}_{2}\\right]=0.050 \\mathrm{M} . K_{c}$ for the reaction is 0.078 .\n21. Which of the systems described in Exercise 13.15 are homogeneous equilibria? Which are heterogeneous equilibria?\n22. Which of the systems described in Exercise 13.16 are homogeneous equilibria? Which are heterogeneous equilibria?\n23. For which of the reactions in Exercise 13.15 does $K_{c}$ (calculated using concentrations) equal $K_{P}$ (calculated using pressures)?\n24. For which of the reactions in Exercise 13.16 does $K_{c}$ (calculated using concentrations) equal $K_{P}$ (calculated using pressures)?\n25. Convert the values of $K_{c}$ to values of $K_{P}$ or the values of $K_{P}$ to values of $K_{c}$.\n(a) $\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g)$\n$K_{c}=0.50$ at $400^{\\circ} \\mathrm{C}$\n(b) $\\mathrm{H}_{2}(g)+\\mathrm{I}_{2}(g) \\rightleftharpoons 2 \\mathrm{HI}(g)$\n$K_{c}=50.2$ at $448^{\\circ} \\mathrm{C}$"}
{"id": 4023, "contents": "1443. Exercises - 1443.2. Equilibrium Constants\n$K_{c}=50.2$ at $448^{\\circ} \\mathrm{C}$\n(c) $\\mathrm{Na}_{2} \\mathrm{SO}_{4} \\cdot 10 \\mathrm{H}_{2} \\mathrm{O}(s) \\rightleftharpoons \\mathrm{Na}_{2} \\mathrm{SO}_{4}(s)+10 \\mathrm{H}_{2} \\mathrm{O}(g)$\n$K_{P}=4.08 \\times 10^{-25}$ at $25^{\\circ} \\mathrm{C}$\n(d) $\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{O}(g)$\n$K_{P}=0.122$ at $50^{\\circ} \\mathrm{C}$\n26. Convert the values of $K_{c}$ to values of $K_{P}$ or the values of $K_{P}$ to values of $K_{c}$.\n(a) $\\mathrm{Cl}_{2}(g)+\\mathrm{Br}_{2}(g) \\rightleftharpoons 2 \\mathrm{BrCl}(g)$\n$K_{c}=4.7 \\times 10^{-2}$ at $25^{\\circ} \\mathrm{C}$\n(b) $2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{SO}_{3}(g)$\n$K_{P}=48.2$ at $500^{\\circ} \\mathrm{C}$\n(c) $\\mathrm{CaCl}_{2} \\cdot 6 \\mathrm{H}_{2} \\mathrm{O}(s) \\rightleftharpoons \\mathrm{CaCl}_{2}(s)+6 \\mathrm{H}_{2} \\mathrm{O}(g)$\n$K_{P}=5.09 \\times 10^{-44}$ at $25^{\\circ} \\mathrm{C}$\n(d) $\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{O}(g)$\n$K_{P}=0.196$ at $60^{\\circ} \\mathrm{C}$"}
{"id": 4024, "contents": "1443. Exercises - 1443.2. Equilibrium Constants\n$K_{P}=0.196$ at $60^{\\circ} \\mathrm{C}$\n27. What is the value of the equilibrium constant expression for the change $\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{O}(g)$ at $30^{\\circ} \\mathrm{C}$ ? (See Appendix E.)\n28. Write the expression of the reaction quotient for the ionization of HOCN in water.\n29. Write the reaction quotient expression for the ionization of $\\mathrm{NH}_{3}$ in water.\n30. What is the approximate value of the equilibrium constant $K_{P}$ for the change $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OC}_{2} \\mathrm{H}_{5}(l) \\rightleftharpoons \\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OC}_{2} \\mathrm{H}_{5}(g)$ at $25^{\\circ} \\mathrm{C}$. (The equilibrium vapor pressure for this substance is 570 torr at $25^{\\circ} \\mathrm{C}$.)"}
{"id": 4025, "contents": "1443. Exercises - 1443.3. Shifting Equilibria: Le Ch\u00e2telier's Principle\n31. The following equation represents a reversible decomposition:\n$\\mathrm{CaCO}_{3}(s) \\rightleftharpoons \\mathrm{CaO}(s)+\\mathrm{CO}_{2}(g)$\nUnder what conditions will decomposition in a closed container proceed to completion so that no $\\mathrm{CaCO}_{3}$ remains?\n32. Explain how to recognize the conditions under which changes in volume will affect gas-phase systems at equilibrium.\n33. What property of a reaction can we use to predict the effect of a change in temperature on the value of an equilibrium constant?\n34. The following reaction occurs when a burner on a gas stove is lit:\n$\\mathrm{CH}_{4}(g)+2 \\mathrm{O}_{2}(g) \\rightleftharpoons \\mathrm{CO}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(g)$\nIs an equilibrium among $\\mathrm{CH}_{4}, \\mathrm{O}_{2}, \\mathrm{CO}_{2}$, and $\\mathrm{H}_{2} \\mathrm{O}$ established under these conditions? Explain your answer.\n35. A necessary step in the manufacture of sulfuric acid is the formation of sulfur trioxide, $\\mathrm{SO}_{3}$, from sulfur dioxide, $\\mathrm{SO}_{2}$, and oxygen, $\\mathrm{O}_{2}$, shown here. At high temperatures, the rate of formation of $\\mathrm{SO}_{3}$ is higher, but the equilibrium amount (concentration or partial pressure) of $\\mathrm{SO}_{3}$ is lower than it would be at lower temperatures.\n$2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{SO}_{3}(g)$\n(a) Does the equilibrium constant for the reaction increase, decrease, or remain about the same as the temperature increases?\n(b) Is the reaction endothermic or exothermic?\n36. Suggest four ways in which the concentration of hydrazine, $\\mathrm{N}_{2} \\mathrm{H}_{4}$, could be increased in an equilibrium described by the following equation:"}
{"id": 4026, "contents": "1443. Exercises - 1443.3. Shifting Equilibria: Le Ch\u00e2telier's Principle\n36. Suggest four ways in which the concentration of hydrazine, $\\mathrm{N}_{2} \\mathrm{H}_{4}$, could be increased in an equilibrium described by the following equation:\n$\\mathrm{N}_{2}(g)+2 \\mathrm{H}_{2}(g) \\rightleftharpoons \\mathrm{N}_{2} \\mathrm{H}_{4}(g) \\quad \\Delta H=95 \\mathrm{~kJ}$\n37. Suggest four ways in which the concentration of $\\mathrm{PH}_{3}$ could be increased in an equilibrium described by the following equation:"}
{"id": 4027, "contents": "1443. Exercises - 1443.3. Shifting Equilibria: Le Ch\u00e2telier's Principle\n$$\n\\mathrm{P}_{4}(g)+6 \\mathrm{H}_{2}(g) \\rightleftharpoons 4 \\mathrm{PH}_{3}(g) \\quad \\Delta H=110.5 \\mathrm{~kJ}\n$$\n\n38. How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?\n(a) $2 \\mathrm{NH}_{3}(g) \\rightleftharpoons \\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g)$\n$\\Delta H=92 \\mathrm{~kJ}$\n(b) $\\mathrm{N}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{NO}(g)$\n$\\Delta H=181 \\mathrm{~kJ}$\n(c) $2 \\mathrm{O}_{3}(g) \\rightleftharpoons 3 \\mathrm{O}_{2}(\\mathrm{~g}) \\quad \\Delta H=-285 \\mathrm{~kJ}$\n(d) $\\mathrm{CaO}(s)+\\mathrm{CO}_{2}(g) \\rightleftharpoons \\mathrm{CaCO}_{3}(s)$\n$\\Delta H=-176 \\mathrm{~kJ}$\n39. How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?\n(a) $2 \\mathrm{H}_{2} \\mathrm{O}(g) \\rightleftharpoons 2 \\mathrm{H}_{2}(g)+\\mathrm{O}_{2}(g)$\n$\\Delta H=484 \\mathrm{~kJ}$\n(b) $\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g)$\n$\\Delta H=-92.2 \\mathrm{~kJ}$\n(c) $2 \\mathrm{Br}(g) \\rightleftharpoons \\mathrm{Br}_{2}(g)$\n(d) $\\mathrm{H}_{2}(g)+\\mathrm{I}_{2}(s) \\rightleftharpoons 2 \\mathrm{HI}(g)$"}
{"id": 4028, "contents": "1443. Exercises - 1443.3. Shifting Equilibria: Le Ch\u00e2telier's Principle\n$$\n\\begin{aligned}\n\\Delta H=-224 \\mathrm{~kJ} \\\\\n\\Delta H=53 \\mathrm{~kJ}\n\\end{aligned}\n$$"}
{"id": 4029, "contents": "1443. Exercises - 1443.3. Shifting Equilibria: Le Ch\u00e2telier's Principle\n40. Methanol can be prepared from carbon monoxide and hydrogen at high temperature and pressure in the presence of a suitable catalyst.\n(a) Write the expression for the equilibrium constant $\\left(K_{c}\\right)$ for the reversible reaction\n$2 \\mathrm{H}_{2}(g)+\\mathrm{CO}(g) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{OH}(g) \\quad \\Delta H=-90.2 \\mathrm{~kJ}$\n(b) What will happen to the concentrations of $\\mathrm{H}_{2}, \\mathrm{CO}$, and $\\mathrm{CH}_{3} \\mathrm{OH}$ at equilibrium if more $\\mathrm{H}_{2}$ is added?\n(c) What will happen to the concentrations of $\\mathrm{H}_{2}, \\mathrm{CO}$, and $\\mathrm{CH}_{3} \\mathrm{OH}$ at equilibrium if CO is removed?\n(d) What will happen to the concentrations of $\\mathrm{H}_{2}, \\mathrm{CO}$, and $\\mathrm{CH}_{3} \\mathrm{OH}$ at equilibrium if $\\mathrm{CH}_{3} \\mathrm{OH}$ is added?\n(e) What will happen to the concentrations of $\\mathrm{H}_{2}, \\mathrm{CO}$, and $\\mathrm{CH}_{3} \\mathrm{OH}$ at equilibrium if the temperature of the system is increased?\n41. Nitrogen and oxygen react at high temperatures.\n(a) Write the expression for the equilibrium constant $\\left(K_{c}\\right)$ for the reversible reaction\n$\\mathrm{N}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{NO}(g) \\quad \\Delta H=181 \\mathrm{~kJ}$\n(b) What will happen to the concentrations of $\\mathrm{N}_{2}, \\mathrm{O}_{2}$, and NO at equilibrium if more $\\mathrm{O}_{2}$ is added?\n(c) What will happen to the concentrations of $\\mathrm{N}_{2}, \\mathrm{O}_{2}$, and NO at equilibrium if $\\mathrm{N}_{2}$ is removed?\n(d) What will happen to the concentrations of $\\mathrm{N}_{2}, \\mathrm{O}_{2}$, and NO at equilibrium if NO is added?"}
{"id": 4030, "contents": "1443. Exercises - 1443.3. Shifting Equilibria: Le Ch\u00e2telier's Principle\n(d) What will happen to the concentrations of $\\mathrm{N}_{2}, \\mathrm{O}_{2}$, and NO at equilibrium if NO is added?\n(e) What will happen to the concentrations of $\\mathrm{N}_{2}, \\mathrm{O}_{2}$, and NO at equilibrium if the volume of the reaction vessel is decreased?\n(f) What will happen to the concentrations of $\\mathrm{N}_{2}, \\mathrm{O}_{2}$, and NO at equilibrium if the temperature of the system is increased?\n42. Water gas, a mixture of $\\mathrm{H}_{2}$ and CO , is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon.\n(a) Write the expression for the equilibrium constant for the reversible reaction\n$\\mathrm{C}(s)+\\mathrm{H}_{2} \\mathrm{O}(g) \\rightleftharpoons \\mathrm{CO}(g)+\\mathrm{H}_{2}(g) \\quad \\Delta H=131.30 \\mathrm{~kJ}$\n(b) What will happen to the concentration of each reactant and product at equilibrium if more C is added?\n(c) What will happen to the concentration of each reactant and product at equilibrium if $\\mathrm{H}_{2} \\mathrm{O}$ is removed?\n(d) What will happen to the concentration of each reactant and product at equilibrium if CO is added?\n(e) What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?\n43. Pure iron metal can be produced by the reduction of iron(III) oxide with hydrogen gas.\n(a) Write the expression for the equilibrium constant $\\left(K_{c}\\right)$ for the reversible reaction\n$\\mathrm{Fe}_{2} \\mathrm{O}_{3}(s)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{Fe}(s)+3 \\mathrm{H}_{2} \\mathrm{O}(g) \\quad \\Delta H=98.7 \\mathrm{~kJ}$\n(b) What will happen to the concentration of each reactant and product at equilibrium if more Fe is added?\n(c) What will happen to the concentration of each reactant and product at equilibrium if $\\mathrm{H}_{2} \\mathrm{O}$ is removed?"}
{"id": 4031, "contents": "1443. Exercises - 1443.3. Shifting Equilibria: Le Ch\u00e2telier's Principle\n(c) What will happen to the concentration of each reactant and product at equilibrium if $\\mathrm{H}_{2} \\mathrm{O}$ is removed?\n(d) What will happen to the concentration of each reactant and product at equilibrium if $\\mathrm{H}_{2}$ is added?\n(e) What will happen to the concentration of each reactant and product at equilibrium if the volume of the reaction vessel is decreased?\n(f) What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?\n44. Ammonia is a weak base that reacts with water according to this equation:\n$\\mathrm{NH}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{NH}_{4}{ }^{+}(a q)+\\mathrm{OH}^{-}(a q)$\nWill any of the following increase the percent of ammonia that is converted to the ammonium ion in water?\n(a) Addition of NaOH\n(b) Addition of HCl\n(c) Addition of $\\mathrm{NH}_{4} \\mathrm{Cl}$\n45. Acetic acid is a weak acid that reacts with water according to this equation:\n$\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(a q)+\\mathrm{H}_{2} \\mathrm{O}(a q) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}(a q)$\nWill any of the following increase the percent of acetic acid that reacts and produces $\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}$ion?\n(a) Addition of HCl\n(b) Addition of NaOH\n(c) Addition of $\\mathrm{NaCH}_{3} \\mathrm{CO}_{2}$\n46. Suggest two ways in which the equilibrium concentration of $\\mathrm{Ag}^{+}$can be reduced in a solution of $\\mathrm{Na}^{+}, \\mathrm{Cl}^{-}$, $\\mathrm{Ag}^{+}$, and $\\mathrm{NO}_{3}{ }^{-}$, in contact with solid AgCl ."}
{"id": 4032, "contents": "1443. Exercises - 1443.3. Shifting Equilibria: Le Ch\u00e2telier's Principle\n$\\mathrm{Na}^{+}(a q)+\\mathrm{Cl}^{-}(a q)+\\mathrm{Ag}^{+}(a q)+\\mathrm{NO}_{3}{ }^{-}(a q) \\rightleftharpoons \\mathrm{AgCl}(s)+\\mathrm{Na}^{+}(a q)+\\mathrm{NO}_{3}{ }^{-}(a q)$\n$\\Delta H=-65.9 \\mathrm{~kJ}$\n47. How can the pressure of water vapor be increased in the following equilibrium?\n$\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{O}(g) \\quad \\Delta H=41 \\mathrm{~kJ}$\n48. A solution is saturated with silver sulfate and contains excess solid silver sulfate:\n$\\mathrm{Ag}_{2} \\mathrm{SO}_{4}(s) \\rightleftharpoons 2 \\mathrm{Ag}^{+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q)$\nA small amount of solid silver sulfate containing a radioactive isotope of silver is added to this solution. Within a few minutes, a portion of the solution phase is sampled and tests positive for radioactive $\\mathrm{Ag}^{+}$ ions. Explain this observation.\n49. When equal molar amounts of HCl and HOCl are dissolved separately in equal amounts of water, the solution of HCl freezes at a lower temperature. Which compound has the larger equilibrium constant for acid ionization?\n(a) HCl\n(b) $\\mathrm{H}^{+}+\\mathrm{Cl}^{-}$\n(c) HOCl\n(d) $\\mathrm{H}^{+}+\\mathrm{OCl}^{-}$"}
{"id": 4033, "contents": "1443. Exercises - 1443.4. Equilibrium Calculations\n50. A reaction is represented by this equation: $\\mathrm{A}(a q)+2 \\mathrm{~B}(a q) \\rightleftharpoons 2 \\mathrm{C}(a q) \\quad K_{c}=1 \\times 10^{3}$\n(a) Write the mathematical expression for the equilibrium constant.\n(b) Using concentrations $\\leq 1 M$, identify two sets of concentrations that describe a mixture of A, B, and C at equilibrium.\n51. A reaction is represented by this equation: $2 \\mathrm{~W}(a q) \\rightleftharpoons \\mathrm{X}(a q)+2 \\mathrm{Y}(a q)$\n\n$$\nK_{c}=5 \\times 10^{-4}\n$$"}
{"id": 4034, "contents": "1443. Exercises - 1443.4. Equilibrium Calculations\n(a) Write the mathematical expression for the equilibrium constant.\n(b) Using concentrations of $\\leq 1 M$, identify two sets of concentrations that describe a mixture of $\\mathrm{W}, \\mathrm{X}$, and Y at equilibrium.\n52. What is the value of the equilibrium constant at $500^{\\circ} \\mathrm{C}$ for the formation of $\\mathrm{NH}_{3}$ according to the following equation?\n$\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g)$\nAn equilibrium mixture of $\\mathrm{NH}_{3}(g), \\mathrm{H}_{2}(g)$, and $\\mathrm{N}_{2}(g)$ at $500^{\\circ} \\mathrm{C}$ was found to contain $1.35 \\mathrm{MH}_{2}, 1.15 \\mathrm{M}_{2}$, and $4.12 \\times 10^{-1} \\mathrm{MNH}_{3}$.\n53. Hydrogen is prepared commercially by the reaction of methane and water vapor at elevated temperatures. $\\mathrm{CH}_{4}(g)+\\mathrm{H}_{2} \\mathrm{O}(g) \\rightleftharpoons 3 \\mathrm{H}_{2}(g)+\\mathrm{CO}(g)$\nWhat is the equilibrium constant for the reaction if a mixture at equilibrium contains gases with the following concentrations: $\\mathrm{CH}_{4}, 0.126 \\mathrm{M} ; \\mathrm{H}_{2} \\mathrm{O}, 0.242 \\mathrm{M} ; \\mathrm{CO}, 0.126 \\mathrm{M} ; \\mathrm{H}_{2} 1.15 \\mathrm{M}$, at a temperature of $760^{\\circ} \\mathrm{C}$ ?"}
{"id": 4035, "contents": "1443. Exercises - 1443.4. Equilibrium Calculations\n54. A $0.72-\\mathrm{mol} \\mathrm{sample}$ of $\\mathrm{PCl}_{5}$ is put into a $1.00-\\mathrm{L}$ vessel and heated. At equilibrium, the vessel contains 0.40 mol of $\\mathrm{PCl}_{3}(g)$ and 0.40 mol of $\\mathrm{Cl}_{2}(g)$. Calculate the value of the equilibrium constant for the decomposition of $\\mathrm{PCl}_{5}$ to $\\mathrm{PCl}_{3}$ and $\\mathrm{Cl}_{2}$ at this temperature.\n55. At 1 atm and $25^{\\circ} \\mathrm{C}, \\mathrm{NO}_{2}$ with an initial concentration of 1.00 M is $0.0033 \\%$ decomposed into NO and $\\mathrm{O}_{2}$. Calculate the value of the equilibrium constant for the reaction.\n$2 \\mathrm{NO}_{2}(g) \\rightleftharpoons 2 \\mathrm{NO}(g)+\\mathrm{O}_{2}(g)$\n56. Calculate the value of the equilibrium constant $K_{P}$ for the reaction $2 \\mathrm{NO}(g)+\\mathrm{Cl}_{2}(g) \\rightleftharpoons 2 \\mathrm{NOCl}(g)$ from these equilibrium pressures: NO, $0.050 \\mathrm{~atm} ; \\mathrm{Cl}_{2}, 0.30 \\mathrm{~atm} ; \\mathrm{NOCl}, 1.2 \\mathrm{~atm}$.\n57. When heated, iodine vapor dissociates according to this equation:\n$\\mathrm{I}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{I}(\\mathrm{g})$\nAt 1274 K , a sample exhibits a partial pressure of $\\mathrm{I}_{2}$ of 0.1122 atm and a partial pressure due to I atoms of 0.1378 atm . Determine the value of the equilibrium constant, $K_{P}$, for the decomposition at 1274 K .\n58. A sample of ammonium chloride was heated in a closed container."}
{"id": 4036, "contents": "1443. Exercises - 1443.4. Equilibrium Calculations\n58. A sample of ammonium chloride was heated in a closed container.\n$\\mathrm{NH}_{4} \\mathrm{Cl}(s) \\rightleftharpoons \\mathrm{NH}_{3}(g)+\\mathrm{HCl}(g)$\nAt equilibrium, the pressure of $\\mathrm{NH}_{3}(g)$ was found to be 1.75 atm . What is the value of the equilibrium constant $K_{P}$ for the decomposition at this temperature?\n59. At a temperature of $60^{\\circ} \\mathrm{C}$, the vapor pressure of water is 0.196 atm . What is the value of the equilibrium constant $K_{P}$ for the vaporization equilibrium at $60^{\\circ} \\mathrm{C}$ ?\n$\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{O}(g)$\n60. Complete the following partial ICE tables.\n(a)"}
{"id": 4037, "contents": "1443. Exercises - 1443.4. Equilibrium Calculations\n$$\n2 \\mathrm{SO}_{3}(g) \\rightleftharpoons 2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g)\n$$\n\nchange $-\\quad+x$\n(b)\n\n$$\n4 \\mathrm{NH}_{3}(g)+3 \\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{~N}_{2}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(g)\n$$\n\nchange $\\quad+x$\n(c)\n\n$$\n2 \\mathrm{CH}_{4}(g) \\rightleftharpoons \\mathrm{C}_{2} \\mathrm{H}_{2}(g)+3 \\mathrm{H}_{2}(g)\n$$\n\nchange\n(d)\n\n$$\n\\begin{array}{ll}\n\\mathrm{CH}_{4}(g)+\\mathrm{H}_{2} \\mathrm{O}(g) & \\rightleftharpoons \\mathrm{CO}(g)+3 \\mathrm{H}_{2}(g) \\\\\n\\text { change_ } & \\rightleftharpoons-\n\\end{array}\n$$\n\n(e)\n\n$$\n\\mathrm{NH}_{4} \\mathrm{Cl}(s) \\rightleftharpoons \\mathrm{NH}_{3}(g)+\\mathrm{HCl}(g)\n$$\n\nchange $+x$\n(f)\n\n$$\n\\mathrm{Ni}(s)+4 \\mathrm{CO}(g) \\rightleftharpoons \\mathrm{Ni}(\\mathrm{CO})_{4}(g)\n$$\n\nchange $+x$\n61. Complete the following partial ICE tables.\n(a)\n\n$$\n2 \\mathrm{H}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{H}_{2} \\mathrm{O}(g)\n$$\n\nchange\n\n$$\n\\begin{array}{llll}\n& - & +x\n\\end{array}\n$$\n\n(b)\n\n$$\n\\mathrm{CS}_{2}(g)+4 \\mathrm{H}_{2}(g) \\rightleftharpoons \\mathrm{CH}_{4}(g)+2 \\mathrm{H}_{2} \\mathrm{~S}(g)\n$$"}
{"id": 4038, "contents": "1443. Exercises - 1443.4. Equilibrium Calculations\nchange $+x$\n(c)\n$\\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g) \\rightleftharpoons 2 \\mathrm{HCl}(g)$\nchange $+x$\n(d)\n$2 \\mathrm{NH}_{3}(g)+2 \\mathrm{O}_{2}(g) \\rightleftharpoons \\mathrm{N}_{2} \\mathrm{O}(g)+3 \\mathrm{H}_{2} \\mathrm{O}(g)$\nchange $\\qquad$\n$\\qquad$ $+x$\n(e)\n\n$$\n\\mathrm{NH}_{4} \\mathrm{HS}(s) \\rightleftharpoons \\mathrm{NH}_{3}(g)+\\mathrm{H}_{2} \\mathrm{~S}(g)\n$$\n\nchange\n\n$$\n+x\n$$\n\n(f)\n\n$$\n\\mathrm{Fe}(s)+5 \\mathrm{CO}(g) \\rightleftharpoons \\mathrm{Fe}(\\mathrm{CO})_{5}(g)\n$$"}
{"id": 4039, "contents": "1443. Exercises - 1443.4. Equilibrium Calculations\nchange $+x$\n62. Why are there no changes specified for Ni in Exercise 13.60, part (f)? What property of Ni does change?\n63. Why are there no changes specified for $\\mathrm{NH}_{4} \\mathrm{HS}$ in Exercise 13.61, part (e)? What property of $\\mathrm{NH}_{4} \\mathrm{HS}$ does change?\n64. Analysis of the gases in a sealed reaction vessel containing $\\mathrm{NH}_{3}, \\mathrm{~N}_{2}$, and $\\mathrm{H}_{2}$ at equilibrium at $400^{\\circ} \\mathrm{C}$ established the concentration of $\\mathrm{N}_{2}$ to be 1.2 M and the concentration of $\\mathrm{H}_{2}$ to be 0.24 M .\n$\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g) \\quad K_{c}=0.50$ at $400^{\\circ} \\mathrm{C}$\nCalculate the equilibrium molar concentration of $\\mathrm{NH}_{3}$.\n65. Calculate the number of moles of HI that are at equilibrium with 1.25 mol of $\\mathrm{H}_{2}$ and $1.25 \\mathrm{~mol}^{\\text {of }} \\mathrm{I}_{2}$ in a 5.00-L flask at $448^{\\circ} \\mathrm{C}$.\n$\\mathrm{H}_{2}+\\mathrm{I}_{2} \\rightleftharpoons 2 \\mathrm{HI} \\quad K_{c}=50.2$ at $448^{\\circ} \\mathrm{C}$\n66. What is the pressure of BrCl in an equilibrium mixture of $\\mathrm{Cl}_{2}, \\mathrm{Br}_{2}$, and BrCl if the pressure of $\\mathrm{Cl}_{2}$ in the mixture is 0.115 atm and the pressure of $\\mathrm{Br}_{2}$ in the mixture is 0.450 atm ?\n$\\mathrm{Cl}_{2}(g)+\\mathrm{Br}_{2}(g) \\rightleftharpoons 2 \\mathrm{BrCl}(g) \\quad K_{P}=4.7 \\times 10^{-2}$"}
{"id": 4040, "contents": "1443. Exercises - 1443.4. Equilibrium Calculations\n67. What is the pressure of $\\mathrm{CO}_{2}$ in a mixture at equilibrium that contains 0.50 atm $\\mathrm{H}_{2}, 2.0$ atm of $\\mathrm{H}_{2} \\mathrm{O}$, and 1.0 atm of CO at $990^{\\circ} \\mathrm{C}$ ?\n$\\mathrm{H}_{2}(g)+\\mathrm{CO}_{2}(g) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{O}(g)+\\mathrm{CO}(g) \\quad K_{P}=1.6$ at $990^{\\circ} \\mathrm{C}$\n68. Cobalt metal can be prepared by reducing cobalt(II) oxide with carbon monoxide.\n$\\mathrm{CoO}(s)+\\mathrm{CO}(g) \\rightleftharpoons \\mathrm{Co}(s)+\\mathrm{CO}_{2}(g) \\quad K_{c}=4.90 \\times 10^{2}$ at $550^{\\circ} \\mathrm{C}$\nWhat concentration of CO remains in an equilibrium mixture with $\\left[\\mathrm{CO}_{2}\\right]=0.100 \\mathrm{M}$ ?\n69. Carbon reacts with water vapor at elevated temperatures.\n$\\mathrm{C}(s)+\\mathrm{H}_{2} \\mathrm{O}(g) \\rightleftharpoons \\mathrm{CO}(g)+\\mathrm{H}_{2}(g) \\quad K_{c}=0.2$ at $1000^{\\circ} \\mathrm{C}$\nAssuming a reaction mixture initially contains only reactants, what is the concentration of CO in an equilibrium mixture with $\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]=0.500 \\mathrm{M}$ at $1000^{\\circ} \\mathrm{C}$ ?\n70. Sodium sulfate 10 -hydrate, $\\mathrm{Na}_{2} \\mathrm{SO}_{4} \\cdot 10 \\mathrm{H}_{2} \\mathrm{O}$, dehydrates according to the equation"}
{"id": 4041, "contents": "1443. Exercises - 1443.4. Equilibrium Calculations\n70. Sodium sulfate 10 -hydrate, $\\mathrm{Na}_{2} \\mathrm{SO}_{4} \\cdot 10 \\mathrm{H}_{2} \\mathrm{O}$, dehydrates according to the equation\n$\\mathrm{Na}_{2} \\mathrm{SO}_{4} \\cdot 10 \\mathrm{H}_{2} \\mathrm{O}(s) \\rightleftharpoons \\mathrm{Na}_{2} \\mathrm{SO}_{4}(s)+10 \\mathrm{H}_{2} \\mathrm{O}(g) \\quad K_{P}=4.08 \\times 10^{-25}$ at $25^{\\circ} \\mathrm{C}$\nWhat is the pressure of water vapor at equilibrium with a mixture of $\\mathrm{Na}_{2} \\mathrm{SO}_{4} \\cdot 10 \\mathrm{H}_{2} \\mathrm{O}$ and $\\mathrm{NaSO}_{4}$ ?\n71. Calcium chloride 6-hydrate, $\\mathrm{CaCl}_{2} \\cdot 6 \\mathrm{H}_{2} \\mathrm{O}$, dehydrates according to the equation\n$\\mathrm{CaCl}_{2} \\cdot 6 \\mathrm{H}_{2} \\mathrm{O}(s) \\rightleftharpoons \\mathrm{CaCl}_{2}(s)+6 \\mathrm{H}_{2} \\mathrm{O}(g) \\quad K_{P}=5.09 \\times 10^{-44}$ at $25^{\\circ} \\mathrm{C}$\nWhat is the pressure of water vapor at equilibrium with a mixture of $\\mathrm{CaCl}_{2} \\cdot 6 \\mathrm{H}_{2} \\mathrm{O}$ and $\\mathrm{CaCl}_{2}$ at $25{ }^{\\circ} \\mathrm{C}$ ?"}
{"id": 4042, "contents": "1443. Exercises - 1443.4. Equilibrium Calculations\n72. A student solved the following problem and found the equilibrium concentrations to be $\\left[\\mathrm{SO}_{2}\\right]=0.590 \\mathrm{M}$, $\\left[\\mathrm{O}_{2}\\right]=0.0450 \\mathrm{M}$, and $\\left[\\mathrm{SO}_{3}\\right]=0.260 \\mathrm{M}$. How could this student check the work without reworking the problem? The problem was: For the following reaction at $600^{\\circ} \\mathrm{C}$ :\n$2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{SO}_{3}(g) \\quad K_{c}=4.32$\nWhat are the equilibrium concentrations of all species in a mixture that was prepared with $\\left[\\mathrm{SO}_{3}\\right]=0.500$ $\\mathrm{M},\\left[\\mathrm{SO}_{2}\\right]=0 \\mathrm{M}$, and $\\left[\\mathrm{O}_{2}\\right]=0.350 \\mathrm{M}$ ?\n73. A student solved the following problem and found $\\left[\\mathrm{N}_{2} \\mathrm{O}_{4}\\right]=0.16 \\mathrm{M}$ at equilibrium. How could this student recognize that the answer was wrong without reworking the problem? The problem was: What is the equilibrium concentration of $\\mathrm{N}_{2} \\mathrm{O}_{4}$ in a mixture formed from a sample of $\\mathrm{NO}_{2}$ with a concentration of 0.10\n$M$ ?\n$2 \\mathrm{NO}_{2}(g) \\rightleftharpoons \\mathrm{N}_{2} \\mathrm{O}_{4}(g) \\quad K_{c}=160$\n74. Assume that the change in concentration of $\\mathrm{N}_{2} \\mathrm{O}_{4}$ is small enough to be neglected in the following problem.\n(a) Calculate the equilibrium concentration of both species in 1.00 L of a solution prepared from 0.129 mol of $\\mathrm{N}_{2} \\mathrm{O}_{4}$ with chloroform as the solvent."}
{"id": 4043, "contents": "1443. Exercises - 1443.4. Equilibrium Calculations\n(a) Calculate the equilibrium concentration of both species in 1.00 L of a solution prepared from 0.129 mol of $\\mathrm{N}_{2} \\mathrm{O}_{4}$ with chloroform as the solvent.\n$\\mathrm{N}_{2} \\mathrm{O}_{4}(g) \\rightleftharpoons 2 \\mathrm{NO}_{2}(g) \\quad K_{c}=1.07 \\times 10^{-5}$ in chloroform\n(b) Confirm that the change is small enough to be neglected.\n75. Assume that the change in concentration of $\\mathrm{COCl}_{2}$ is small enough to be neglected in the following problem.\n(a) Calculate the equilibrium concentration of all species in an equilibrium mixture that results from the decomposition of $\\mathrm{COCl}_{2}$ with an initial concentration of 0.3166 M .\n$\\mathrm{COCl}_{2}(g) \\rightleftharpoons \\mathrm{CO}(g)+\\mathrm{Cl}_{2}(g) \\quad K_{c}=2.2 \\times 10^{-10}$\n(b) Confirm that the change is small enough to be neglected.\n76. Assume that the change in pressure of $\\mathrm{H}_{2} \\mathrm{~S}$ is small enough to be neglected in the following problem. (a) Calculate the equilibrium pressures of all species in an equilibrium mixture that results from the decomposition of $\\mathrm{H}_{2} \\mathrm{~S}$ with an initial pressure of 0.824 atm ."}
{"id": 4044, "contents": "1443. Exercises - 1443.4. Equilibrium Calculations\n$$\n2 \\mathrm{H}_{2} \\mathrm{~S}(g) \\rightleftharpoons 2 \\mathrm{H}_{2}(g)+\\mathrm{S}_{2}(g) \\quad K_{P}=2.2 \\times 10^{-6}\n$$\n\n(b) Confirm that the change is small enough to be neglected.\n77. What are all concentrations after a mixture that contains $\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]=1.00 \\mathrm{M}$ and $\\left[\\mathrm{Cl}_{2} \\mathrm{O}\\right]=1.00 \\mathrm{M}$ comes to equilibrium at $25^{\\circ} \\mathrm{C}$ ?\n$\\mathrm{H}_{2} \\mathrm{O}(g)+\\mathrm{Cl}_{2} \\mathrm{O}(g) \\rightleftharpoons 2 \\mathrm{HOCl}(g) \\quad K_{c}=0.0900$\n78. Calculate the number of grams of HI that are at equilibrium with 1.25 mol of $\\mathrm{H}_{2}$ and 63.5 g of iodine at 448 ${ }^{\\circ} \\mathrm{C}$.\n$\\mathrm{H}_{2}+\\mathrm{I}_{2} \\rightleftharpoons 2 \\mathrm{HI} \\quad K_{c}=50.2$ at $448^{\\circ} \\mathrm{C}$\n79. Butane exists as two isomers, $n$-butane and isobutane."}
{"id": 4045, "contents": "1443. Exercises - 1443.4. Equilibrium Calculations\n$K_{P}=2.5$ at $25^{\\circ} \\mathrm{C}$\nWhat is the pressure of isobutane in a container of the two isomers at equilibrium with a total pressure of 1.22 atm?\n80. What is the minimum mass of $\\mathrm{CaCO}_{3}$ required to establish equilibrium at a certain temperature in a 6.50-L container if the equilibrium constant $\\left(K_{c}\\right)$ is 0.50 for the decomposition reaction of $\\mathrm{CaCO}_{3}$ at that temperature?\n$\\mathrm{CaCO}_{3}(s) \\rightleftharpoons \\mathrm{CaO}(s)+\\mathrm{CO}_{2}(g)$\n81. The equilibrium constant $\\left(K_{c}\\right)$ for this reaction is 1.60 at $990^{\\circ} \\mathrm{C}$ :\n$\\mathrm{H}_{2}(g)+\\mathrm{CO}_{2}(g) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{O}(g)+\\mathrm{CO}(g)$\nCalculate the number of moles of each component in the final equilibrium mixture obtained from adding 1.00 mol of $\\mathrm{H}_{2}, 2.00 \\mathrm{~mol}$ of $\\mathrm{CO}_{2}, 0.750 \\mathrm{~mol}$ of $\\mathrm{H}_{2} \\mathrm{O}$, and 1.00 mol of CO to a $5.00-\\mathrm{L}$ container at $990^{\\circ} \\mathrm{C}$.\n82. In a 3.0-L vessel, the following equilibrium partial pressures are measured: $\\mathrm{N}_{2}, 190$ torr; $\\mathrm{H}_{2}, 317$ torr; $\\mathrm{NH}_{3}, 1.00 \\times 10^{3}$ torr.\n$\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g)$"}
{"id": 4046, "contents": "1443. Exercises - 1443.4. Equilibrium Calculations\n$\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g)$\n(a) How will the partial pressures of $\\mathrm{H}_{2}, \\mathrm{~N}_{2}$, and $\\mathrm{NH}_{3}$ change if $\\mathrm{H}_{2}$ is removed from the system? Will they increase, decrease, or remain the same?\n(b) Hydrogen is removed from the vessel until the partial pressure of nitrogen, at equilibrium, is 250 torr. Calculate the partial pressures of the other substances under the new conditions.\n83. The equilibrium constant $\\left(K_{c}\\right)$ for this reaction is 5.0 at a given temperature.\n$\\mathrm{CO}(g)+\\mathrm{H}_{2} \\mathrm{O}(g) \\rightleftharpoons \\mathrm{CO}_{2}(g)+\\mathrm{H}_{2}(g)$\n(a) On analysis, an equilibrium mixture of the substances present at the given temperature was found to contain 0.20 mol of $\\mathrm{CO}, 0.30 \\mathrm{~mol}$ of water vapor, and 0.90 mol of $\\mathrm{H}_{2}$ in a liter. How many moles of $\\mathrm{CO}_{2}$ were there in the equilibrium mixture?\n(b) Maintaining the same temperature, additional $\\mathrm{H}_{2}$ was added to the system, and some water vapor was removed by drying. A new equilibrium mixture was thereby established containing 0.40 mol of $\\mathrm{CO}, 0.30$ mol of water vapor, and 1.2 mol of $\\mathrm{H}_{2}$ in a liter. How many moles of $\\mathrm{CO}_{2}$ were in the new equilibrium mixture? Compare this with the quantity in part (a), and discuss whether the second value is reasonable. Explain how it is possible for the water vapor concentration to be the same in the two equilibrium solutions even though some vapor was removed before the second equilibrium was established.\n84. Antimony pentachloride decomposes according to this equation:\n$\\mathrm{SbCl}_{5}(g) \\rightleftharpoons \\mathrm{SbCl}_{3}(g)+\\mathrm{Cl}_{2}(g)$"}
{"id": 4047, "contents": "1443. Exercises - 1443.4. Equilibrium Calculations\n$\\mathrm{SbCl}_{5}(g) \\rightleftharpoons \\mathrm{SbCl}_{3}(g)+\\mathrm{Cl}_{2}(g)$\nAn equilibrium mixture in a 5.00 -L flask at $448{ }^{\\circ} \\mathrm{C}$ contains 3.85 g of $\\mathrm{SbCl}_{5}, 9.14 \\mathrm{~g}$ of $\\mathrm{SbCl}_{3}$, and 2.84 g of $\\mathrm{Cl}_{2}$. How many grams of each will be found if the mixture is transferred into a $2.00-\\mathrm{L}$ flask at the same temperature?\n85. Consider the equilibrium\n$4 \\mathrm{NO}_{2}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(g) \\rightleftharpoons 4 \\mathrm{NH}_{3}(g)+7 \\mathrm{O}_{2}(g)$\n(a) What is the expression for the equilibrium constant $\\left(K_{c}\\right)$ of the reaction?\n(b) How must the concentration of $\\mathrm{NH}_{3}$ change to reach equilibrium if the reaction quotient is less than the equilibrium constant?\n(c) If the reaction were at equilibrium, how would an increase in the volume of the reaction vessel affect the pressure of $\\mathrm{NO}_{2}$ ?\n(d) If the change in the pressure of $\\mathrm{NO}_{2}$ is 28 torr as a mixture of the four gases reaches equilibrium, how much will the pressure of $\\mathrm{O}_{2}$ change?\n86. The binding of oxygen by hemoglobin $(\\mathrm{Hb})$, giving oxyhemoglobin $\\left(\\mathrm{HbO}_{2}\\right)$, is partially regulated by the concentration of $\\mathrm{H}_{3} \\mathrm{O}^{+}$and dissolved $\\mathrm{CO}_{2}$ in the blood. Although the equilibrium is complicated, it can be summarized as"}
{"id": 4048, "contents": "1443. Exercises - 1443.4. Equilibrium Calculations\n$\\mathrm{HbO}_{2}(a q)+\\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{CO}_{2}(g) \\rightleftharpoons \\mathrm{CO}_{2}-\\mathrm{Hb}-\\mathrm{H}^{+}+\\mathrm{O}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l)$\n(a) Write the equilibrium constant expression for this reaction.\n(b) Explain why the production of lactic acid and $\\mathrm{CO}_{2}$ in a muscle during exertion stimulates release of $\\mathrm{O}_{2}$ from the oxyhemoglobin in the blood passing through the muscle.\n87. Liquid $\\mathrm{N}_{2} \\mathrm{O}_{3}$ is dark blue at low temperatures, but the color fades and becomes greenish at higher temperatures as the compound decomposes to NO and $\\mathrm{NO}_{2}$. At $25^{\\circ} \\mathrm{C}$, a value of $K_{P}=1.91$ has been established for this decomposition. If 0.236 moles of $\\mathrm{N}_{2} \\mathrm{O}_{3}$ are placed in a 1.52 - L vessel at $25^{\\circ} \\mathrm{C}$, calculate the equilibrium partial pressures of $\\mathrm{N}_{2} \\mathrm{O}_{3}(\\mathrm{~g}), \\mathrm{NO}_{2}(\\mathrm{~g})$, and $\\mathrm{NO}(\\mathrm{g})$.\n88. A 1.00-L vessel at $400^{\\circ} \\mathrm{C}$ contains the following equilibrium concentrations: $\\mathrm{N}_{2}, 1.00 \\mathrm{M} ; \\mathrm{H}_{2}, 0.50 \\mathrm{M}$; and $\\mathrm{NH}_{3}, 0.25 \\mathrm{M}$. How many moles of hydrogen must be removed from the vessel to increase the concentration of nitrogen to 1.1 $M$ ? The equilibrium reaction is $\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g)$"}
{"id": 4049, "contents": "1443. Exercises - 1443.4. Equilibrium Calculations\n89. Calculate the equilibrium constant at $25^{\\circ} \\mathrm{C}$ for each of the following reactions from the value of $\\Delta G^{\\circ}$ given.\n(a) $\\mathrm{I}_{2}(s)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{ICl}(g)$\n$\\Delta G^{\\circ}=-10.88 \\mathrm{~kJ}$\n(b) $\\mathrm{H}_{2}(g)+\\mathrm{I}_{2}(s) \\longrightarrow 2 \\mathrm{HI}(g)$\n$\\Delta G^{\\circ}=3.4 \\mathrm{~kJ}$\n(c) $\\mathrm{CS}_{2}(g)+3 \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{CCl}_{4}(g)+\\mathrm{S}_{2} \\mathrm{Cl}_{2}(g) \\quad \\Delta G^{\\circ}=-39 \\mathrm{~kJ}$\n(d) $2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{SO}_{3}(g) \\quad \\Delta G^{\\circ}=-141.82 \\mathrm{~kJ}$\n(e) $\\mathrm{CS}_{2}(g) \\longrightarrow \\mathrm{CS}_{2}(l) \\quad \\Delta G^{\\circ}=-1.88 \\mathrm{~kJ}$\n90. Calculate the equilibrium constant at the temperature given.\n(a) $\\mathrm{O}_{2}(g)+2 \\mathrm{~F}_{2}(g) \\longrightarrow 2 \\mathrm{~F}_{2} \\mathrm{O}(g) \\quad\\left(\\mathrm{T}=100^{\\circ} \\mathrm{C}\\right)$\n(b) $\\mathrm{I}_{2}(s)+\\mathrm{Br}_{2}(l) \\longrightarrow 2 \\mathrm{IBr}(g) \\quad\\left(\\mathrm{T}=0.0^{\\circ} \\mathrm{C}\\right)$"}
{"id": 4050, "contents": "1443. Exercises - 1443.4. Equilibrium Calculations\n(c) $2 \\mathrm{LiOH}(s)+\\mathrm{CO}_{2}(g) \\longrightarrow \\mathrm{Li}_{2} \\mathrm{CO}_{3}(s)+\\mathrm{H}_{2} \\mathrm{O}(g) \\quad\\left(\\mathrm{T}=575^{\\circ} \\mathrm{C}\\right)$\n$(d) \\mathrm{N}_{2} \\mathrm{O}_{3}(g) \\longrightarrow \\mathrm{NO}(g)+\\mathrm{NO}_{2}(g) \\quad\\left(\\mathrm{T}=-10.0^{\\circ} \\mathrm{C}\\right)$\n(e) $\\mathrm{SnCl}_{4}(l) \\longrightarrow \\mathrm{SnCl}_{4}(g) \\quad\\left(\\mathrm{T}=200^{\\circ} \\mathrm{C}\\right)$\n91. Calculate the equilibrium constant at the temperature given.\n(a) $\\mathrm{I}_{2}(s)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{ICl}(g) \\quad\\left(\\mathrm{T}=100^{\\circ} \\mathrm{C}\\right)$\n(b) $\\mathrm{H}_{2}(g)+\\mathrm{I}_{2}(s) \\longrightarrow 2 \\mathrm{HI}(g) \\quad\\left(\\mathrm{T}=0.0^{\\circ} \\mathrm{C}\\right)$\n(c) $\\mathrm{CS}_{2}(g)+3 \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{CCl}_{4}(g)+\\mathrm{S}_{2} \\mathrm{Cl}_{2}(g) \\quad\\left(\\mathrm{T}=125^{\\circ} \\mathrm{C}\\right)$\n(d) $2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{SO}_{3}(g) \\quad\\left(\\mathrm{T}=675^{\\circ} \\mathrm{C}\\right)$\n(e) $\\mathrm{CS}_{2}(g) \\longrightarrow \\mathrm{CS}_{2}(l) \\quad\\left(\\mathrm{T}=90^{\\circ} \\mathrm{C}\\right)$\n92. Consider the following reaction at 298 K :"}
{"id": 4051, "contents": "1443. Exercises - 1443.4. Equilibrium Calculations\n92. Consider the following reaction at 298 K :\n$\\mathrm{N}_{2} \\mathrm{O}_{4}(g) \\rightleftharpoons 2 \\mathrm{NO}_{2}(g) \\quad K_{P}=0.142$\nWhat is the standard free energy change at this temperature? Describe what happens to the initial system, where the reactants and products are in standard states, as it approaches equilibrium.\n93. Determine the normal boiling point (in kelvin) of dichloroethane, $\\mathrm{CH}_{2} \\mathrm{Cl}_{2}$. Find the actual boiling point using the Internet or some other source, and calculate the percent error in the temperature. Explain the differences, if any, between the two values.\n94. Under what conditions is $\\mathrm{N}_{2} \\mathrm{O}_{3}(g) \\longrightarrow \\mathrm{NO}(g)+\\mathrm{NO}_{2}(g)$ spontaneous?\n95. At room temperature, the equilibrium constant $\\left(K_{W}\\right)$ for the self-ionization of water is $1.00 \\times 10^{-14}$. Using this information, calculate the standard free energy change for the aqueous reaction of hydrogen ion with hydroxide ion to produce water. (Hint: The reaction is the reverse of the self-ionization reaction.)\n96. Hydrogen sulfide is a pollutant found in natural gas. Following its removal, it is converted to sulfur by the reaction $2 \\mathrm{H}_{2} \\mathrm{~S}(g)+\\mathrm{SO}_{2}(g) \\rightleftharpoons \\frac{3}{8} \\mathrm{~S}_{8}(s$, rhombic $)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$. What is the equilibrium constant for this reaction? Is the reaction endothermic or exothermic?\n97. Consider the decomposition of $\\mathrm{CaCO}_{3}(s)$ into $\\mathrm{CaO}(s)$ and $\\mathrm{CO}_{2}(g)$. What is the equilibrium partial pressure of $\\mathrm{CO}_{2}$ at room temperature?"}
{"id": 4052, "contents": "1443. Exercises - 1443.4. Equilibrium Calculations\n98. In the laboratory, hydrogen chloride $(\\mathrm{HCl}(g))$ and ammonia $\\left(\\mathrm{NH}_{3}(g)\\right)$ often escape from bottles of their solutions and react to form the ammonium chloride $\\left(\\mathrm{NH}_{4} \\mathrm{Cl}(s)\\right)$, the white glaze often seen on glassware. Assuming that the number of moles of each gas that escapes into the room is the same, what is the maximum partial pressure of HCl and $\\mathrm{NH}_{3}$ in the laboratory at room temperature? (Hint: The partial pressures will be equal and are at their maximum value when at equilibrium.)\n99. Benzene can be prepared from acetylene. $3 \\mathrm{C}_{2} \\mathrm{H}_{2}(g) \\rightleftharpoons \\mathrm{C}_{6} \\mathrm{H}_{6}(g)$. Determine the equilibrium constant at $25^{\\circ} \\mathrm{C}$ and at $850^{\\circ} \\mathrm{C}$. Is the reaction spontaneous at either of these temperatures? Why is all acetylene not found as benzene?\n100. Carbon dioxide decomposes into CO and $\\mathrm{O}_{2}$ at elevated temperatures. What is the equilibrium partial pressure of oxygen in a sample at $1000{ }^{\\circ} \\mathrm{C}$ for which the initial pressure of $\\mathrm{CO}_{2}$ was 1.15 atm?\n101. Carbon tetrachloride, an important industrial solvent, is prepared by the chlorination of methane at 850 K.\n$\\mathrm{CH}_{4}(g)+4 \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{CCl}_{4}(g)+4 \\mathrm{HCl}(g)$\nWhat is the equilibrium constant for the reaction at 850 K ? Would the reaction vessel need to be heated or cooled to keep the temperature of the reaction constant?"}
{"id": 4053, "contents": "1443. Exercises - 1443.4. Equilibrium Calculations\nWhat is the equilibrium constant for the reaction at 850 K ? Would the reaction vessel need to be heated or cooled to keep the temperature of the reaction constant?\n102. Acetic acid, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$, can form a dimer, $\\left(\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right)_{2}$, in the gas phase. $2 \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(\\mathrm{g}) \\longrightarrow\\left(\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right)_{2}(\\mathrm{~g})$\nThe dimer is held together by two hydrogen bonds with a total strength of 66.5 kJ per mole of dimer."}
{"id": 4054, "contents": "1443. Exercises - 1443.4. Equilibrium Calculations\nAt $25^{\\circ} \\mathrm{C}$, the equilibrium constant for the dimerization is $1.3 \\times 10^{3}$ (pressure in atm). What is $\\Delta S^{\\circ}$ for the reaction?"}
{"id": 4055, "contents": "1444. CHAPTER 14 <br> Acid-Base Equilibria - \nFigure 14.1 Sinkholes such as this are the result of reactions between acidic groundwaters and basic rock formations, like limestone. (credit: modification of work by Emil Kehnel)"}
{"id": 4056, "contents": "1446. 2 pH and pOH - \n14.3 Relative Strengths of Acids and Bases\n14.4 Hydrolysis of Salts\n14.5 Polyprotic Acids\n14.6 Buffers\n14.7 Acid-Base Titrations\n\nINTRODUCTION Liquid water is essential to life on our planet, and chemistry involving the characteristic ions of water, $\\mathrm{H}^{+}$and $\\mathrm{OH}^{-}$, is widely encountered in nature and society. As introduced in another chapter of this text, acid-base chemistry involves the transfer of hydrogen ions from donors (acids) to acceptors (bases). These H+ transfer reactions are reversible, and the equilibria established by acid-base systems are essential aspects of phenomena ranging from sinkhole formation (Figure 14.1) to oxygen transport in the human body. This chapter will further explore acid-base chemistry with an emphasis on the equilibrium aspects of this important reaction class."}
{"id": 4057, "contents": "1447. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Identify acids, bases, and conjugate acid-base pairs according to the Br\u00f8nsted-Lowry definition\n- Write equations for acid and base ionization reactions\n- Use the ion-product constant for water to calculate hydronium and hydroxide ion concentrations\n- Describe the acid-base behavior of amphiprotic substances\n\nThe acid-base reaction class has been studied for quite some time. In 1680, Robert Boyle reported traits of acid solutions that included their ability to dissolve many substances, to change the colors of certain natural dyes, and to lose these traits after coming in contact with alkali (base) solutions. In the eighteenth century, it was recognized that acids have a sour taste, react with limestone to liberate a gaseous substance (now known to be $\\mathrm{CO}_{2}$ ), and interact with alkalis to form neutral substances. In 1815, Humphry Davy contributed greatly to the development of the modern acid-base concept by demonstrating that hydrogen is the essential constituent of acids. Around that same time, Joseph Louis Gay-Lussac concluded that acids are substances that can neutralize bases and that these two classes of substances can be defined only in terms of each other. The significance of hydrogen was reemphasized in 1884 when Svante Arrhenius defined an acid as a compound that dissolves in water to yield hydrogen cations (now recognized to be hydronium ions) and a base as a compound that dissolves in water to yield hydroxide anions.\n\nJohannes Br\u00f8nsted and Thomas Lowry proposed a more general description in 1923 in which acids and bases were defined in terms of the transfer of hydrogen ions, $\\mathrm{H}^{+}$. (Note that these hydrogen ions are often referred to simply as protons, since that subatomic particle is the only component of cations derived from the most abundant hydrogen isotope, ${ }^{1} \\mathrm{H}$.) A compound that donates a proton to another compound is called a Br\u00f8nsted-Lowry acid, and a compound that accepts a proton is called a Br\u00f8nsted-Lowry base. An acid-base reaction is, thus, the transfer of a proton from a donor (acid) to an acceptor (base)."}
{"id": 4058, "contents": "1447. LEARNING OBJECTIVES - \nThe concept of conjugate pairs is useful in describing Br\u00f8nsted-Lowry acid-base reactions (and other reversible reactions, as well). When an acid donates $\\mathrm{H}^{+}$, the species that remains is called the conjugate base of the acid because it reacts as a proton acceptor in the reverse reaction. Likewise, when a base accepts $\\mathrm{H}^{+}$, it is converted to its conjugate acid. The reaction between water and ammonia illustrates this idea. In the forward direction, water acts as an acid by donating a proton to ammonia and subsequently becoming a hydroxide ion, $\\mathrm{OH}^{-}$, the conjugate base of water. The ammonia acts as a base in accepting this proton, becoming an ammonium ion, $\\mathrm{NH}_{4}{ }^{+}$, the conjugate acid of ammonia. In the reverse direction, a hydroxide ion acts as a base in accepting a proton from ammonium ion, which acts as an acid.\n\n\nThe reaction between a Br\u00f8nsted-Lowry acid and water is called acid ionization. For example, when hydrogen fluoride dissolves in water and ionizes, protons are transferred from hydrogen fluoride molecules to water molecules, yielding hydronium ions and fluoride ions:\n\n\nBase ionization of a species occurs when it accepts protons from water molecules. In the example below, pyridine molecules, $\\mathrm{C}_{5} \\mathrm{NH}_{5}$, undergo base ionization when dissolved in water, yielding hydroxide and pyridinium ions:\n\n\nThe preceding ionization reactions suggest that water may function as both a base (as in its reaction with hydrogen fluoride) and an acid (as in its reaction with ammonia). Species capable of either donating or accepting protons are called amphiprotic, or more generally, amphoteric, a term that may be used for acids and bases per definitions other than the Br\u00f8nsted-Lowry one. The equations below show the two possible acidbase reactions for two amphiprotic species, bicarbonate ion and water:"}
{"id": 4059, "contents": "1447. LEARNING OBJECTIVES - \n$$\n\\begin{aligned}\n\\mathrm{HCO}_{3}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) & \\mathrm{CO}_{3}^{2-}(a q)+\\mathrm{H}_{3} \\mathrm{O}^{+}(a q) \\\\\n\\mathrm{HCO}_{3}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) & \\mathrm{H}_{2} \\mathrm{CO}_{3}(a q)+\\mathrm{OH}^{-}(a q)\n\\end{aligned}\n$$\n\nThe first equation represents the reaction of bicarbonate as an acid with water as a base, whereas the second represents reaction of bicarbonate as a base with water as an acid. When bicarbonate is added to water, both these equilibria are established simultaneously and the composition of the resulting solution may be determined through appropriate equilibrium calculations, as described later in this chapter.\n\nIn the liquid state, molecules of an amphiprotic substance can react with one another as illustrated for water in the equations below:\n\n\nThe process in which like molecules react to yield ions is called autoionization. Liquid water undergoes autoionization to a very slight extent; at $25^{\\circ} \\mathrm{C}$, approximately two out of every billion water molecules are ionized. The extent of the water autoionization process is reflected in the value of its equilibrium constant, the ion-product constant for water, $\\boldsymbol{K}_{\\mathbf{w}}$ :\n\n$$\n\\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{OH}^{-}(a q) \\quad K_{\\mathrm{w}}=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]\n$$"}
{"id": 4060, "contents": "1447. LEARNING OBJECTIVES - \nThe slight ionization of pure water is reflected in the small value of the equilibrium constant; at $25^{\\circ} \\mathrm{C}, K_{\\mathrm{w}}$ has a value of $1.0 \\times 10^{-14}$. The process is endothermic, and so the extent of ionization and the resulting concentrations of hydronium ion and hydroxide ion increase with temperature. For example, at $100^{\\circ} \\mathrm{C}$, the value for $K_{\\mathrm{w}}$ is about $5.6 \\times 10^{-13}$, roughly 50 times larger than the value at $25^{\\circ} \\mathrm{C}$."}
{"id": 4061, "contents": "1449. Ion Concentrations in Pure Water - \nWhat are the hydronium ion concentration and the hydroxide ion concentration in pure water at $25^{\\circ} \\mathrm{C}$ ?"}
{"id": 4062, "contents": "1450. Solution - \nThe autoionization of water yields the same number of hydronium and hydroxide ions. Therefore, in pure water, $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=\\left[\\mathrm{OH}^{-}\\right]=x$. At $25^{\\circ} \\mathrm{C}$ :\n\n$$\nK_{\\mathrm{w}}=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]=(x)(x)=x^{2}=1.0 \\times 10^{-14}\n$$\n\nSo:\n\n$$\nx=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=\\left[\\mathrm{OH}^{-}\\right]=\\sqrt{1.0 \\times 10^{-14}}=1.0 \\times 10^{-7} \\mathrm{M}\n$$\n\nThe hydronium ion concentration and the hydroxide ion concentration are the same, $1.0 \\times 10^{-7} \\mathrm{M}$."}
{"id": 4063, "contents": "1451. Check Your Learning - \nThe ion product of water at $80^{\\circ} \\mathrm{C}$ is $2.4 \\times 10^{-13}$. What are the concentrations of hydronium and hydroxide ions in pure water at $80^{\\circ} \\mathrm{C}$ ?"}
{"id": 4064, "contents": "1452. Answer: - \n$\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=\\left[\\mathrm{OH}^{-}\\right]=4.9 \\times 10^{-7} \\mathrm{M}$"}
{"id": 4065, "contents": "1454. The Inverse Relation between $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right.$] and $\\left[\\mathrm{OH}^{-}\\right.$] - \nA solution of an acid in water has a hydronium ion concentration of $2.0 \\times 10^{-6} \\mathrm{M}$. What is the concentration of hydroxide ion at $25^{\\circ} \\mathrm{C}$ ?"}
{"id": 4066, "contents": "1455. Solution - \nUse the value of the ion-product constant for water at $25^{\\circ} \\mathrm{C}$\n\n$$\n2 \\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{OH}^{-}(a q) \\quad K_{\\mathrm{w}}=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]=1.0 \\times 10^{-14}\n$$\n\nto calculate the missing equilibrium concentration.\nRearrangement of the $K_{\\mathrm{w}}$ expression shows that $\\left[\\mathrm{OH}^{-}\\right]$is inversely proportional to $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]$:\n\n$$\n\\left[\\mathrm{OH}^{-}\\right]=\\frac{K_{\\mathrm{w}}}{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]}=\\frac{1.0 \\times 10^{-14}}{2.0 \\times 10^{-6}}=5.0 \\times 10^{-9}\n$$\n\nCompared with pure water, a solution of acid exhibits a higher concentration of hydronium ions (due to ionization of the acid) and a proportionally lower concentration of hydroxide ions. This may be explained via Le Ch\u00e2telier's principle as a left shift in the water autoionization equilibrium resulting from the stress of increased hydronium ion concentration.\n\nSubstituting the ion concentrations into the $K_{\\mathrm{w}}$ expression confirms this calculation, resulting in the expected value:\n\n$$\nK_{\\mathrm{w}}=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]=\\left(2.0 \\times 10^{-6}\\right)\\left(5.0 \\times 10^{-9}\\right)=1.0 \\times 10^{-14}\n$$"}
{"id": 4067, "contents": "1456. Check Your Learning - \nWhat is the hydronium ion concentration in an aqueous solution with a hydroxide ion concentration of 0.001 $M$ at $25^{\\circ} \\mathrm{C}$ ?\n\nAnswer:\n$\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=1 \\times 10^{-11} \\mathrm{M}$"}
{"id": 4068, "contents": "1458. Representing the Acid-Base Behavior of an Amphoteric Substance - \nWrite separate equations representing the reaction of $\\mathrm{HSO}_{3}{ }^{-}$\n(a) as an acid with $\\mathrm{OH}^{-}$\n(b) as a base with HI"}
{"id": 4069, "contents": "1459. Solution - \n(a) $\\mathrm{HSO}_{3}{ }^{-}(a q)+\\mathrm{OH}^{-}(a q) \\rightleftharpoons \\mathrm{SO}_{3}{ }^{2-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)$\n(b) $\\mathrm{HSO}_{3}{ }^{-}(a q)+\\mathrm{HI}(a q) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{SO}_{3}(a q)+\\mathrm{I}^{-}(a q)$"}
{"id": 4070, "contents": "1460. Check Your Learning - \nWrite separate equations representing the reaction of $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}$\n(a) as a base with HBr\n(b) as an acid with $\\mathrm{OH}^{-}$"}
{"id": 4071, "contents": "1461. Answer: - \n(a) $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}(a q)+\\mathrm{HBr}(a q) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{PO}_{4}(a q)+\\mathrm{Br}^{-}(a q)$; (b)\n$\\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{-}(a q)+\\mathrm{OH}^{-}(a q) \\rightleftharpoons \\mathrm{HPO}_{4}{ }^{2-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)$"}
{"id": 4072, "contents": "1463. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Explain the characterization of aqueous solutions as acidic, basic, or neutral\n- Express hydronium and hydroxide ion concentrations on the pH and pOH scales\n- Perform calculations relating pH and pOH\n\nAs discussed earlier, hydronium and hydroxide ions are present both in pure water and in all aqueous solutions, and their concentrations are inversely proportional as determined by the ion product of water ( $K_{\\mathrm{w}}$ ). The concentrations of these ions in a solution are often critical determinants of the solution's properties and the chemical behaviors of its other solutes, and specific vocabulary has been developed to describe these concentrations in relative terms. A solution is neutral if it contains equal concentrations of hydronium and hydroxide ions; acidic if it contains a greater concentration of hydronium ions than hydroxide ions; and basic if it contains a lesser concentration of hydronium ions than hydroxide ions.\n\nA common means of expressing quantities that may span many orders of magnitude is to use a logarithmic scale. One such scale that is very popular for chemical concentrations and equilibrium constants is based on the p-function, defined as shown where \" X \" is the quantity of interest and \"log\" is the base-10 logarithm:\n\n$$\n\\mathrm{pX}=-\\log \\mathrm{X}\n$$\n\nThe $\\mathbf{p H}$ of a solution is therefore defined as shown here, where $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]$is the molar concentration of hydronium ion in the solution:\n\n$$\n\\mathrm{pH}=-\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\n$$\n\nRearranging this equation to isolate the hydronium ion molarity yields the equivalent expression:\n\n$$\n\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=10^{-\\mathrm{pH}}\n$$\n\nLikewise, the hydroxide ion molarity may be expressed as a p-function, or $\\mathbf{p O H}$ :\n\n$$\n\\mathrm{pOH}=-\\log \\left[\\mathrm{OH}^{-}\\right]\n$$\n\nor"}
{"id": 4073, "contents": "1463. LEARNING OBJECTIVES - \n$$\n\\mathrm{pOH}=-\\log \\left[\\mathrm{OH}^{-}\\right]\n$$\n\nor\n\n$$\n\\left[\\mathrm{OH}^{-}\\right]=10^{-\\mathrm{pOH}}\n$$\n\nFinally, the relation between these two ion concentration expressed as p-functions is easily derived from the $K_{\\mathrm{w}}$ expression:\n\n$$\nK_{\\mathrm{w}}=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]\n$$\n\n$$\n\\begin{gathered}\n-\\log K_{\\mathrm{w}}=-\\log \\left(\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]\\right)=-\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]+-\\log \\left[\\mathrm{OH}^{-}\\right] \\\\\n\\mathrm{p} K_{\\mathrm{w}}=\\mathrm{pH}+\\mathrm{pOH}\n\\end{gathered}\n$$\n\nAt $25^{\\circ} \\mathrm{C}$, the value of $K_{\\mathrm{W}}$ is $1.0 \\times 10^{-14}$, and so:\n\n$$\n14.00=\\mathrm{pH}+\\mathrm{pOH}\n$$\n\nAs was shown in Example 14.1, the hydronium ion molarity in pure water (or any neutral solution) is $1.0 \\times$ $10^{-7} \\mathrm{M}$ at $25^{\\circ} \\mathrm{C}$. The pH and pOH of a neutral solution at this temperature are therefore:\n\n$$\n\\begin{aligned}\n& \\mathrm{pH}=-\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=-\\log \\left(1.0 \\times 10^{-7}\\right)=7.00 \\\\\n& \\mathrm{pOH}=-\\log \\left[\\mathrm{OH}^{-}\\right]=-\\log \\left(1.0 \\times 10^{-7}\\right)=7.00\n\\end{aligned}\n$$"}
{"id": 4074, "contents": "1463. LEARNING OBJECTIVES - \nAnd so, at this temperature, acidic solutions are those with hydronium ion molarities greater than $1.0 \\times 10^{-7} \\mathrm{M}$ and hydroxide ion molarities less than $1.0 \\times 10^{-7} \\mathrm{M}$ (corresponding to pH values less than 7.00 and pOH values greater than 7.00). Basic solutions are those with hydronium ion molarities less than $1.0 \\times 10^{-7} \\mathrm{M}$ and hydroxide ion molarities greater than $1.0 \\times 10^{-7} M$ (corresponding to pH values greater than 7.00 and pOH values less than 7.00).\n\nSince the autoionization constant $K_{\\mathrm{w}}$ is temperature dependent, these correlations between pH values and the acidic/neutral/basic adjectives will be different at temperatures other than $25^{\\circ} \\mathrm{C}$. For example, the \"Check Your Learning\" exercise accompanying Example 14.1 showed the hydronium molarity of pure water at $80^{\\circ} \\mathrm{C}$ is $4.9 \\times 10^{-7} \\mathrm{M}$, which corresponds to pH and pOH values of:\n\n$$\n\\begin{aligned}\n& \\mathrm{pH}=-\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=-\\log \\left(4.9 \\times 10^{-7}\\right)=6.31 \\\\\n& \\mathrm{pOH}=-\\log \\left[\\mathrm{OH}^{-}\\right]=-\\log \\left(4.9 \\times 10^{-7}\\right)=6.31\n\\end{aligned}\n$$"}
{"id": 4075, "contents": "1463. LEARNING OBJECTIVES - \nAt this temperature, then, neutral solutions exhibit $\\mathrm{pH}=\\mathrm{pOH}=6.31$, acidic solutions exhibit pH less than 6.31 and pOH greater than 6.31, whereas basic solutions exhibit pH greater than 6.31 and pOH less than 6.31 . This distinction can be important when studying certain processes that occur at other temperatures, such as enzyme reactions in warm-blooded organisms at a temperature around $36-40^{\\circ} \\mathrm{C}$. Unless otherwise noted, references to pH values are presumed to be those at $25^{\\circ} \\mathrm{C}$ (Table 14.1).\n\n| Summary of Relations for Acidic, Basic and Neutral Solutions |  |  |\n| :---: | :---: | :---: |\n| Classification | Relative lon Concentrations | pH at $\\mathbf{2 5}^{\\circ} \\mathrm{C}$ |\n| acidic | $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]>\\left[\\mathrm{OH}^{-}\\right]$ | $\\mathrm{pH}<7$ |\n| neutral | $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=\\left[\\mathrm{OH}^{-}\\right]$ | $\\mathrm{pH}=7$ |\n| basic | $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]<\\left[\\mathrm{OH}^{-}\\right]$ | $\\mathrm{pH}>7$ |\n\nTABLE 14.1\n\nFigure 14.2 shows the relationships between $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right],\\left[\\mathrm{OH}^{-}\\right], \\mathrm{pH}$, and pOH for solutions classified as acidic, basic, and neutral.\n\n\nFIGURE 14.2 The pH and pOH scales represent concentrations of $\\mathrm{H}_{3} \\mathrm{O}^{+}$and $\\mathrm{OH}^{-}$, respectively. The pH and pOH values of some common substances at $25^{\\circ} \\mathrm{C}$ are shown in this chart."}
{"id": 4076, "contents": "1465. Calculation of pH from $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]$ - \nWhat is the pH of stomach acid, a solution of HCl with a hydronium ion concentration of $1.2 \\times 10^{-3} \\mathrm{M}$ ?"}
{"id": 4077, "contents": "1466. Solution - \n$$\n\\begin{gathered}\n\\mathrm{pH}=-\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right] \\\\\n=-\\log \\left(1.2 \\times 10^{-3}\\right) \\\\\n=-(-2.92)=2.92\n\\end{gathered}\n$$\n\n(The use of logarithms is explained in Appendix B. When taking the log of a value, keep as many decimal places in the result as there are significant figures in the value.)"}
{"id": 4078, "contents": "1467. Check Your Learning - \nWater exposed to air contains carbonic acid, $\\mathrm{H}_{2} \\mathrm{CO}_{3}$, due to the reaction between carbon dioxide and water:\n\n$$\n\\mathrm{CO}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{CO}_{3}(a q)\n$$\n\nAir-saturated water has a hydronium ion concentration caused by the dissolved $\\mathrm{CO}_{2}$ of $2.0 \\times 10^{-6} \\mathrm{M}$, about 20-times larger than that of pure water. Calculate the pH of the solution at $25^{\\circ} \\mathrm{C}$."}
{"id": 4079, "contents": "1470. Calculation of Hydronium Ion Concentration from pH - \nCalculate the hydronium ion concentration of blood, the pH of which is 7.3."}
{"id": 4080, "contents": "1471. Solution - \n$$\n\\begin{gathered}\n\\mathrm{pH}=-\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=7.3 \\\\\n\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=-7.3 \\\\\n{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=10^{-7.3} \\text { or }\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=\\text {antilog of }-7.3} \\\\\n{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=5 \\times 10^{-8} \\mathrm{M}}\n\\end{gathered}\n$$\n\n(On a calculator take the antilog, or the \"inverse\" log, of -7.3 , or calculate $10^{-7.3}$.)"}
{"id": 4081, "contents": "1472. Check Your Learning - \nCalculate the hydronium ion concentration of a solution with a pH of -1.07 ."}
{"id": 4082, "contents": "1473. Answer: - \n12 M"}
{"id": 4083, "contents": "1475. Environmental Science - \nNormal rainwater has a pH between 5 and 6 due to the presence of dissolved $\\mathrm{CO}_{2}$ which forms carbonic acid:\n\n$$\n\\begin{gathered}\n\\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{CO}_{2}(g) \\longrightarrow \\mathrm{H}_{2} \\mathrm{CO}_{3}(a q) \\\\\n\\mathrm{H}_{2} \\mathrm{CO}_{3}(a q) \\rightleftharpoons \\mathrm{H}^{+}(a q)+\\mathrm{HCO}_{3}^{-}(a q)\n\\end{gathered}\n$$\n\nAcid rain is rainwater that has a pH of less than 5 , due to a variety of nonmetal oxides, including $\\mathrm{CO}_{2}, \\mathrm{SO}_{2}, \\mathrm{SO}_{3}$, NO , and $\\mathrm{NO}_{2}$ being dissolved in the water and reacting with it to form not only carbonic acid, but sulfuric acid and nitric acid. The formation and subsequent ionization of sulfuric acid are shown here:\n\n$$\n\\begin{gathered}\n\\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{SO}_{3}(g) \\longrightarrow \\mathrm{H}_{2} \\mathrm{SO}_{4}(a q) \\\\\n\\mathrm{H}_{2} \\mathrm{SO}_{4}(a q) \\longrightarrow \\mathrm{H}^{+}(a q)+\\mathrm{HSO}_{4}^{-}(a q)\n\\end{gathered}\n$$\n\nCarbon dioxide is naturally present in the atmosphere because most organisms produce it as a waste product of metabolism. Carbon dioxide is also formed when fires release carbon stored in vegetation or fossil fuels. Sulfur trioxide in the atmosphere is naturally produced by volcanic activity, but it also originates from burning fossil fuels, which have traces of sulfur, and from the process of \"roasting\" ores of metal sulfides in metalrefining processes. Oxides of nitrogen are formed in internal combustion engines where the high temperatures make it possible for the nitrogen and oxygen in air to chemically combine."}
{"id": 4084, "contents": "1475. Environmental Science - \nAcid rain is a particular problem in industrial areas where the products of combustion and smelting are released into the air without being stripped of sulfur and nitrogen oxides. In North America and Europe until\nthe 1980s, it was responsible for the destruction of forests and freshwater lakes, when the acidity of the rain actually killed trees, damaged soil, and made lakes uninhabitable for all but the most acid-tolerant species. Acid rain also corrodes statuary and building facades that are made of marble and limestone (Figure 14.3). Regulations limiting the amount of sulfur and nitrogen oxides that can be released into the atmosphere by industry and automobiles have reduced the severity of acid damage to both natural and manmade environments in North America and Europe. It is now a growing problem in industrial areas of China and India.\n\nFor further information on acid rain, visit this website (http://openstax.org/l/16EPA) hosted by the US Environmental Protection Agency.\n\n\nFIGURE 14.3 (a) Acid rain makes trees more susceptible to drought and insect infestation, and depletes nutrients in the soil. (b) It also is corrodes statues that are carved from marble or limestone. (credit a: modification of work by Chris M Morris; credit b: modification of work by \"Eden, Janine and Jim\"/Flickr)"}
{"id": 4085, "contents": "1477. Calculation of pOH - \nWhat are the pOH and the pH of a $0.0125-M$ solution of potassium hydroxide, KOH ?"}
{"id": 4086, "contents": "1478. Solution - \nPotassium hydroxide is a highly soluble ionic compound and completely dissociates when dissolved in dilute solution, yielding $\\left[\\mathrm{OH}^{-}\\right]=0.0125 \\mathrm{M}$ :\n\n$$\n\\begin{aligned}\n\\mathrm{pOH}= & =-\\log \\left[\\mathrm{OH}^{-}\\right]=-\\log 0.0125 \\\\\n& =-(-1.903)=1.903\n\\end{aligned}\n$$\n\nThe pH can be found from the pOH :\n\n$$\n\\begin{gathered}\n\\mathrm{pH}+\\mathrm{pOH}=14.00 \\\\\n\\mathrm{pH}=14.00-\\mathrm{pOH}=14.00-1.903=12.10\n\\end{gathered}\n$$"}
{"id": 4087, "contents": "1479. Check Your Learning - \nThe hydronium ion concentration of vinegar is approximately $4 \\times 10^{-3} \\mathrm{M}$. What are the corresponding values of pOH and pH ?"}
{"id": 4088, "contents": "1480. Answer: - \n$\\mathrm{pOH}=11.6, \\mathrm{pH}=2.4$\n\nThe acidity of a solution is typically assessed experimentally by measurement of its pH . The pOH of a solution is not usually measured, as it is easily calculated from an experimentally determined pH value. The pH of a solution can be directly measured using a pH meter (Figure 14.4).\n\n\nFIGURE 14.4 (a) A research-grade pH meter used in a laboratory can have a resolution of 0.001 pH units, an accuracy of $\\pm 0.002 \\mathrm{pH}$ units, and may cost in excess of $\\$ 1000$. (b) A portable pH meter has lower resolution ( 0.01 pH units), lower accuracy ( $\\pm 0.2 \\mathrm{pH}$ units), and a far lower price tag. (credit b: modification of work by Jacopo Werther)\n\nThe pH of a solution may also be visually estimated using colored indicators (Figure 14.5). The acid-base equilibria that enable use of these indicator dyes for pH measurements are described in a later section of this chapter.\n\n\nFIGURE 14.5 (a) A solution containing a dye mixture, called universal indicator, takes on different colors depending upon its pH . (b) Convenient test strips, called pH paper, contain embedded indicator dyes that yield pH dependent color changes on contact with aqueous solutions.(credit: modification of work by Sahar Atwa)"}
{"id": 4089, "contents": "1481. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Assess the relative strengths of acids and bases according to their ionization constants\n- Rationalize trends in acid-base strength in relation to molecular structure\n- Carry out equilibrium calculations for weak acid-base systems"}
{"id": 4090, "contents": "1482. Acid and Base Ionization Constants - \nThe relative strength of an acid or base is the extent to which it ionizes when dissolved in water. If the ionization reaction is essentially complete, the acid or base is termed strong; if relatively little ionization occurs, the acid or base is weak. As will be evident throughout the remainder of this chapter, there are many more weak acids and bases than strong ones. The most common strong acids and bases are listed in Figure 14.6.\n\n| 6 Strong Acids |  | 6 Strong Bases |  |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{HClO}_{4}$ | perchloric acid | LiOH | lithium hydroxide |\n| HCl | hydrochloric acid | NaOH | sodium hydroxide |\n| HBr | hydrobromic acid | KOH | potassium hydroxide |\n| HI | hydroiodic acid | $\\mathrm{Ca}(\\mathrm{OH})_{2}$ | calcium hydroxide |\n| $\\mathrm{HNO}_{3}$ | nitric acid | $\\mathrm{Sr}(\\mathrm{OH})_{2}$ | strontium hydroxide |\n| $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ | sulfuric acid | $\\mathrm{Ba}(\\mathrm{OH})_{2}$ | barium hydroxide |\n\nFIGURE 14.6 Some of the common strong acids and bases are listed here.\nThe relative strengths of acids may be quantified by measuring their equilibrium constants in aqueous solutions. In solutions of the same concentration, stronger acids ionize to a greater extent, and so yield higher concentrations of hydronium ions than do weaker acids. The equilibrium constant for an acid is called the acid-ionization constant, $\\boldsymbol{K}_{\\mathbf{a}}$. For the reaction of an acid HA:\n\n$$\n\\mathrm{HA}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{A}^{-}(a q)\n$$\n\nthe acid ionization constant is written"}
{"id": 4091, "contents": "1482. Acid and Base Ionization Constants - \nthe acid ionization constant is written\n\n$$\nK_{\\mathrm{a}}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]}\n$$\n\nwhere the concentrations are those at equilibrium. Although water is a reactant in the reaction, it is the solvent as well, so we do not include $\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]$ in the equation. The larger the $K_{\\mathrm{a}}$ of an acid, the larger the concentration of $\\mathrm{H}_{3} \\mathrm{O}^{+}$and $\\mathrm{A}^{-}$relative to the concentration of the nonionized acid, HA , in an equilibrium mixture, and the stronger the acid. An acid is classified as \"strong\" when it undergoes complete ionization, in which case the concentration of HA is zero and the acid ionization constant is immeasurably large ( $K_{\\mathrm{a}} \\approx \\infty$ ). Acids that are partially ionized are called \"weak,\" and their acid ionization constants may be experimentally measured. A table of ionization constants for weak acids is provided in Appendix H.\n\nTo illustrate this idea, three acid ionization equations and $K_{\\mathrm{a}}$ values are shown below. The ionization constants increase from first to last of the listed equations, indicating the relative acid strength increases in the order $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}<\\mathrm{HNO}_{2}<\\mathrm{HSO}_{4}{ }^{-}$:"}
{"id": 4092, "contents": "1482. Acid and Base Ionization Constants - \n$$\n\\begin{array}{ccc}\n\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) & \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}(a q) & K_{\\mathrm{a}}=1.8 \\times 10^{-5} \\\\\n\\mathrm{HNO}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{NO}_{2}{ }^{-}(a q) & K_{\\mathrm{a}}=4.6 \\times 10^{-4} \\\\\n\\mathrm{HSO}_{4}{ }^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(a q) & \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q) & K_{\\mathrm{a}}=1.2 \\times 10^{-2}\n\\end{array}\n$$\n\nAnother measure of the strength of an acid is its percent ionization. The percent ionization of a weak acid is defined in terms of the composition of an equilibrium mixture:\n\n$$\n\\% \\text { ionization }=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]_{\\mathrm{eq}}}{[\\mathrm{HA}]_{0}} \\times 100\n$$\n\nwhere the numerator is equivalent to the concentration of the acid's conjugate base (per stoichiometry, $\\left[\\mathrm{A}^{-}\\right]=$\n$\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]$). Unlike the $K_{\\mathrm{a}}$ value, the percent ionization of a weak acid varies with the initial concentration of acid, typically decreasing as concentration increases. Equilibrium calculations of the sort described later in this chapter can be used to confirm this behavior."}
{"id": 4093, "contents": "1484. Calculation of Percent Ionization from pH - \nCalculate the percent ionization of a $0.125-M$ solution of nitrous acid (a weak acid), with a pH of 2.09 ."}
{"id": 4094, "contents": "1485. Solution - \nThe percent ionization for an acid is:\n\n$$\n\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]_{\\mathrm{eq}}}{\\left[\\mathrm{HNO}_{2}\\right]_{0}} \\times 100\n$$\n\nConverting the provided pH to hydronium ion molarity yields\n\n$$\n\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=10^{-2.09}=0.0081 \\mathrm{M}\n$$\n\nSubstituting this value and the provided initial acid concentration into the percent ionization equation gives\n\n$$\n\\frac{8.1 \\times 10^{-3}}{0.125} \\times 100=6.5 \\%\n$$\n\n(Recall the provided pH value of 2.09 is logarithmic, and so it contains just two significant digits, limiting the certainty of the computed percent ionization.)"}
{"id": 4095, "contents": "1486. Check Your Learning - \nCalculate the percent ionization of a $0.10-M$ solution of acetic acid with a pH of 2.89 ."}
{"id": 4096, "contents": "1487. Answer: - \n1.3\\% ionized"}
{"id": 4097, "contents": "1488. LINK TO LEARNING - \nView the simulation (http://openstax.org/l/16AcidBase) of strong and weak acids and bases at the molecular level.\n\nJust as for acids, the relative strength of a base is reflected in the magnitude of its base-ionization constant $\\left(K_{\\mathbf{b}}\\right)$ in aqueous solutions. In solutions of the same concentration, stronger bases ionize to a greater extent, and so yield higher hydroxide ion concentrations than do weaker bases. A stronger base has a larger ionization constant than does a weaker base. For the reaction of a base, B:\n\n$$\n\\mathrm{B}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{HB}^{+}(a q)+\\mathrm{OH}^{-}(a q)\n$$\n\nthe ionization constant is written as\n\n$$\nK_{\\mathrm{b}}=\\frac{\\left[\\mathrm{HB}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]}{[\\mathrm{B}]}\n$$\n\nInspection of the data for three weak bases presented below shows the base strength increases in the order $\\mathrm{NO}_{2}{ }^{-}<\\mathrm{CH}_{2} \\mathrm{CO}_{2}^{-}<\\mathrm{NH}_{3}$."}
{"id": 4098, "contents": "1488. LINK TO LEARNING - \n$$\n\\begin{array}{ll}\n\\mathrm{NO}_{2}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{HNO}_{2}(a q)+\\mathrm{OH}^{-}(a q) & K_{\\mathrm{b}}=2.17 \\times 10^{-11} \\\\\n\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(a q)+\\mathrm{OH}^{-}(a q) & K_{\\mathrm{b}}=5.6 \\times 10^{-10} \\\\\n\\mathrm{NH}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{NH}_{4}^{+}(a q)+\\mathrm{OH}^{-}(a q) & K_{\\mathrm{b}}=1.8 \\times 10^{-5}\n\\end{array}\n$$\n\nA table of ionization constants for weak bases appears in Appendix I. As for acids, the relative strength of a\nbase is also reflected in its percent ionization, computed as\n\n$$\n\\% \\text { ionization }=\\left[\\mathrm{OH}^{-}\\right]_{e q} /[\\mathrm{B}]_{0} \\times 100 \\%\n$$\n\nbut will vary depending on the base ionization constant and the initial concentration of the solution."}
{"id": 4099, "contents": "1489. Relative Strengths of Conjugate Acid-Base Pairs - \nBr\u00f8nsted-Lowry acid-base chemistry is the transfer of protons; thus, logic suggests a relation between the relative strengths of conjugate acid-base pairs. The strength of an acid or base is quantified in its ionization constant, $K_{\\mathrm{a}}$ or $K_{\\mathrm{b}}$, which represents the extent of the acid or base ionization reaction. For the conjugate acidbase pair HA / $\\mathrm{A}^{-}$, ionization equilibrium equations and ionization constant expressions are\n\n$$\n\\begin{aligned}\n\\mathrm{HA}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{A}^{-}(a q) & K_{\\mathrm{a}}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]} \\\\\n\\mathrm{A}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{OH}^{-}(a q)+\\mathrm{HA}(a q) & K_{\\mathrm{b}}=\\frac{[\\mathrm{HA}]\\left[\\mathrm{OH}^{-}\\right]}{\\left[\\mathrm{A}^{-}\\right]}\n\\end{aligned}\n$$\n\nAdding these two chemical equations yields the equation for the autoionization for water:\n\n$$\n\\begin{gathered}\n\\mathrm{HA}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{A}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{A}^{-}(a q)+\\mathrm{OH}^{-}(a q)+\\mathrm{HA}(a q) \\\\\n2 \\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{OH}^{-}(a q)\n\\end{gathered}\n$$"}
{"id": 4100, "contents": "1489. Relative Strengths of Conjugate Acid-Base Pairs - \nAs discussed in another chapter on equilibrium, the equilibrium constant for a summed reaction is equal to the mathematical product of the equilibrium constants for the added reactions, and so\n\n$$\nK_{\\mathrm{a}} \\times K_{\\mathrm{b}}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]} \\times \\frac{[\\mathrm{HA}]\\left[\\mathrm{OH}^{-}\\right]}{\\left[\\mathrm{A}^{-}\\right]}=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]=K_{\\mathrm{w}}\n$$\n\nThis equation states the relation between ionization constants for any conjugate acid-base pair, namely, their mathematical product is equal to the ion product of water, $K_{\\mathrm{w}}$. By rearranging this equation, a reciprocal relation between the strengths of a conjugate acid-base pair becomes evident:\n\n$$\nK_{\\mathrm{a}}=K_{\\mathrm{w}} / K_{\\mathrm{b}} \\text { or } K_{\\mathrm{b}}=K_{\\mathrm{w}} / K_{\\mathrm{a}}\n$$\n\nThe inverse proportional relation between $K_{\\mathrm{a}}$ and $K_{\\mathrm{b}}$ means the stronger the acid or base, the weaker its conjugate partner. Figure 14.7 illustrates this relation for several conjugate acid-base pairs.\n\n\nRelative conjugate base strength\n\n\nFIGURE 14.7 Relative strengths of several conjugate acid-base pairs are shown.\n\n\nFIGURE 14.8 This figure shows strengths of conjugate acid-base pairs relative to the strength of water as the reference substance."}
{"id": 4101, "contents": "1489. Relative Strengths of Conjugate Acid-Base Pairs - \nFIGURE 14.7 Relative strengths of several conjugate acid-base pairs are shown.\n\n\nFIGURE 14.8 This figure shows strengths of conjugate acid-base pairs relative to the strength of water as the reference substance.\n\nThe listing of conjugate acid-base pairs shown in Figure 14.8 is arranged to show the relative strength of each species as compared with water, whose entries are highlighted in each of the table's columns. In the acid column, those species listed below water are weaker acids than water. These species do not undergo acid ionization in water; they are not Bronsted-Lowry acids. All the species listed above water are stronger acids, transferring protons to water to some extent when dissolved in an aqueous solution to generate hydronium ions. Species above water but below hydronium ion are weak acids, undergoing partial acid ionization, wheres those above hydronium ion are strong acids that are completely ionized in aqueous solution.\n\nIf all these strong acids are completely ionized in water, why does the column indicate they vary in strength, with nitric acid being the weakest and perchloric acid the strongest? Notice that the sole acid species present in an aqueous solution of any strong acid is $\\mathrm{H}_{3} \\mathrm{O}^{+}(a q)$, meaning that hydronium ion is the strongest acid that may exist in water; any stronger acid will react completely with water to generate hydronium ions. This limit on the acid strength of solutes in a solution is called a leveling effect. To measure the differences in acid strength for \"strong\" acids, the acids must be dissolved in a solvent that is less basic than water. In such solvents, the acids will be \"weak,\" and so any differences in the extent of their ionization can be determined. For example, the binary hydrogen halides $\\mathrm{HCl}, \\mathrm{HBr}$, and HI are strong acids in water but weak acids in ethanol (strength increasing $\\mathrm{HCl}<\\mathrm{HBr}<\\mathrm{HI}$ )."}
{"id": 4102, "contents": "1489. Relative Strengths of Conjugate Acid-Base Pairs - \nThe right column of Figure 14.8 lists a number of substances in order of increasing base strength from top to bottom. Following the same logic as for the left column, species listed above water are weaker bases and so they don't undergo base ionization when dissolved in water. Species listed between water and its conjugate base, hydroxide ion, are weak bases that partially ionize. Species listed below hydroxide ion are strong bases that completely ionize in water to yield hydroxide ions (i.e., they are leveled to hydroxide). A comparison of the acid and base columns in this table supports the reciprocal relation between the strengths of conjugate acidbase pairs. For example, the conjugate bases of the strong acids (top of table) are all of negligible strength. A strong acid exhibits an immeasurably large $K_{\\mathrm{a}}$, and so its conjugate base will exhibit a $K_{\\mathrm{b}}$ that is essentially zero:\nstrong acid: $\\quad K_{\\mathrm{a}} \\approx \\infty$\nconjugate base : $K_{\\mathrm{b}}=K_{\\mathrm{W}} / K_{\\mathrm{a}}=K_{\\mathrm{W}} / \\infty \\approx 0$\nA similar approach can be used to support the observation that conjugate acids of strong bases ( $K_{\\mathrm{b}} \\approx \\infty$ ) are of negligible strength ( $K_{\\mathrm{a}} \\approx 0$ )."}
{"id": 4103, "contents": "1491. Calculating Ionization Constants for Conjugate Acid-Base Pairs - \nUse the $K_{\\mathrm{b}}$ for the nitrite ion, $\\mathrm{NO}_{2}{ }^{-}$, to calculate the $K_{\\mathrm{a}}$ for its conjugate acid."}
{"id": 4104, "contents": "1492. Solution - \n$K_{\\mathrm{b}}$ for $\\mathrm{NO}_{2}{ }^{-}$is given in this section as $2.17 \\times 10^{-11}$. The conjugate acid of $\\mathrm{NO}_{2}{ }^{-}$is $\\mathrm{HNO}_{2} ; K_{\\mathrm{a}}$ for $\\mathrm{HNO}_{2}$ can be calculated using the relationship:\n\n$$\nK_{\\mathrm{a}} \\times K_{\\mathrm{b}}=1.0 \\times 10^{-14}=K_{\\mathrm{w}}\n$$\n\nSolving for $K_{\\mathrm{a}}$ yields\n\n$$\nK_{\\mathrm{a}}=\\frac{K_{\\mathrm{w}}}{K_{\\mathrm{b}}}=\\frac{1.0 \\times 10^{-14}}{2.17 \\times 10^{-11}}=4.6 \\times 10^{-4}\n$$\n\nThis answer can be verified by finding the $K_{\\mathrm{a}}$ for $\\mathrm{HNO}_{2}$ in Appendix H."}
{"id": 4105, "contents": "1493. Check Your Learning - \nDetermine the relative acid strengths of $\\mathrm{NH}_{4}{ }^{+}$and HCN by comparing their ionization constants. The ionization constant of HCN is given in Appendix H as $4.9 \\times 10^{-10}$. The ionization constant of $\\mathrm{NH}_{4}{ }^{+}$is not listed, but the ionization constant of its conjugate base, $\\mathrm{NH}_{3}$, is listed as $1.8 \\times 10^{-5}$."}
{"id": 4106, "contents": "1494. Answer: - \n$\\mathrm{NH}_{4}{ }^{+}$is the slightly stronger acid ( $K_{\\mathrm{a}}$ for $\\mathrm{NH}_{4}{ }^{+}=5.6 \\times 10^{-10}$ )."}
{"id": 4107, "contents": "1495. Acid-Base Equilibrium Calculations - \nThe chapter on chemical equilibria introduced several types of equilibrium calculations and the various mathematical strategies that are helpful in performing them. These strategies are generally useful for equilibrium systems regardless of chemical reaction class, and so they may be effectively applied to acid-base equilibrium problems. This section presents several example exercises involving equilibrium calculations for acid-base systems.\n\n(8) EXAMPLE 14.9\n\nDetermination of $\\boldsymbol{K}_{\\mathrm{a}}$ from Equilibrium Concentrations\nAcetic acid is the principal ingredient in vinegar (Figure 14.9) that provides its sour taste. At equilibrium, a solution contains $\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right]=0.0787 \\mathrm{M}$ and $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}\\right]=0.00118 \\mathrm{M}$. What is the value of $K_{\\mathrm{a}}$ for acetic acid?\n\n\nFIGURE 14.9 Vinegar contains acetic acid, a weak acid. (credit: modification of work by \"HomeSpot HQ\"/Flickr)"}
{"id": 4108, "contents": "1496. Solution - \nThe relevant equilibrium equation and its equilibrium constant expression are shown below. Substitution of the provided equilibrium concentrations permits a straightforward calculation of the $K_{\\mathrm{a}}$ for acetic acid.\n\n$$\n\\begin{gathered}\n\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}(a q) \\\\\nK_{\\mathrm{a}}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}\\right]}{\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right]}=\\frac{(0.00118)(0.00118)}{0.0787}=1.77 \\times 10^{-5}\n\\end{gathered}\n$$"}
{"id": 4109, "contents": "1497. Check Your Learning - \nThe $\\mathrm{HSO}_{4}{ }^{-}$ion, weak acid used in some household cleansers:\n\n$$\n\\mathrm{HSO}_{4}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q)\n$$\n\nWhat is the acid ionization constant for this weak acid if an equilibrium mixture has the following composition: $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=0.027 \\mathrm{M} ;\\left[\\mathrm{HSO}_{4}{ }^{-}\\right]=0.29 \\mathrm{M}$; and $\\left[\\mathrm{SO}_{4}{ }^{2-}\\right]=0.13 \\mathrm{M}$ ?"}
{"id": 4110, "contents": "1498. Answer: - \n$K_{\\mathrm{a}}$ for $\\mathrm{HSO}_{4}^{-}=1.2 \\times 10^{-2}$"}
{"id": 4111, "contents": "1500. Determination of $\\boldsymbol{K}_{\\mathbf{b}}$ from Equilibrium Concentrations - \nCaffeine, $\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2}$ is a weak base. What is the value of $K_{\\mathrm{b}}$ for caffeine if a solution at equilibrium has $\\left[\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2}\\right]=0.050 \\mathrm{M},\\left[\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2} \\mathrm{H}^{+}\\right]=5.0 \\times 10^{-3} \\mathrm{M}$, and $\\left[\\mathrm{OH}^{-}\\right]=2.5 \\times 10^{-3} \\mathrm{M}$ ?"}
{"id": 4112, "contents": "1501. Solution - \nThe relevant equilibrium equation and its equilibrium constant expression are shown below. Substitution of the provided equilibrium concentrations permits a straightforward calculation of the $K_{\\mathrm{b}}$ for caffeine.\n\n$$\n\\begin{gathered}\n\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2} \\mathrm{H}^{+}(a q)+\\mathrm{OH}^{-}(a q) \\\\\nK_{\\mathrm{b}}=\\frac{\\left[\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2} \\mathrm{H}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]}{\\left[\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2}\\right]}=\\frac{\\left(5.0 \\times 10^{-3}\\right)\\left(2.5 \\times 10^{-3}\\right)}{0.050}=2.5 \\times 10^{-4}\n\\end{gathered}\n$$"}
{"id": 4113, "contents": "1502. Check Your Learning - \nWhat is the equilibrium constant for the ionization of the $\\mathrm{HPO}_{4}{ }^{2-}$ ion, a weak base\n\n$$\n\\mathrm{HPO}_{4}{ }^{2-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}(a q)+\\mathrm{OH}^{-}(a q)\n$$\n\nif the composition of an equilibrium mixture is as follows: $\\left[\\mathrm{OH}^{-}\\right]=1.3 \\times 10^{-6} \\mathrm{M}$; $\\left[\\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{-}\\right]=0.042 \\mathrm{M}$; and $\\left[\\mathrm{HPO}_{4}{ }^{2-}\\right]=0.341 \\mathrm{M}$ ?\n\nAnswer:\n$K_{\\mathrm{b}}$ for $\\mathrm{HPO}_{4}{ }^{2-}=1.6 \\times 10^{-7}$"}
{"id": 4114, "contents": "1504. Determination of $\\boldsymbol{K}_{\\mathbf{a}}$ or $\\boldsymbol{K}_{\\mathbf{b}}$ from $\\mathbf{p H}$ - \nThe pH of a $0.0516-M$ solution of nitrous acid, $\\mathrm{HNO}_{2}$, is 2.34 . What is its $K_{\\mathrm{a}}$ ?\n\n$$\n\\mathrm{HNO}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{NO}_{2}^{-}(a q)\n$$"}
{"id": 4115, "contents": "1505. Solution - \nThe nitrous acid concentration provided is a formal concentration, one that does not account for any chemical equilibria that may be established in solution. Such concentrations are treated as \"initial\" values for equilibrium calculations using the ICE table approach. Notice the initial value of hydronium ion is listed as approximately zero because a small concentration of $\\mathrm{H}_{3} \\mathrm{O}^{+}$is present $\\left(1 \\times 10^{-7} \\mathrm{M}\\right)$ due to the autoprotolysis of water. In many cases, such as all the ones presented in this chapter, this concentration is much less than that generated by ionization of the acid (or base) in question and may be neglected.\n\nThe pH provided is a logarithmic measure of the hydronium ion concentration resulting from the acid ionization of the nitrous acid, and so it represents an \"equilibrium\" value for the ICE table:\n\n$$\n\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=10^{-2.34}=0.0046 \\mathrm{M}\n$$\n\nThe ICE table for this system is then\n\n|  | $\\mathrm{HNO}_{2}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathbf{H}_{3} \\mathrm{O}^{+}+\\mathrm{NO}_{2}^{-}$ |  |  |  |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial concentration $(M)$ | 0.0516 |  | $\\sim 0$ | 0 |\n| Change $(M)$ | -0.0046 |  | +0.0046 | +0.0046 |\n| Equilibrium concentration $(M)$ | 0.0470 |  | 0.0046 | 0.0046 |\n\nFinally, calculate the value of the equilibrium constant using the data in the table:"}
{"id": 4116, "contents": "1505. Solution - \nFinally, calculate the value of the equilibrium constant using the data in the table:\n\n$$\nK_{\\mathrm{a}}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{NO}_{2}^{-}\\right]}{\\left[\\mathrm{HNO}_{2}\\right]}=\\frac{(0.0046)(0.0046)}{(0.0470)}=4.6 \\times 10^{-4}\n$$"}
{"id": 4117, "contents": "1506. Check Your Learning. - \nThe pH of a solution of household ammonia, a $0.950-M$ solution of $\\mathrm{NH}_{3}$, is 11.612 . What is $K_{\\mathrm{b}}$ for $\\mathrm{NH}_{3}$."}
{"id": 4118, "contents": "1507. Answer: - \n$K_{\\mathrm{b}}=1.8 \\times 10^{-5}$"}
{"id": 4119, "contents": "1508. Calculating Equilibrium Concentrations in a Weak Acid Solution - \nFormic acid, $\\mathrm{HCO}_{2} \\mathrm{H}$, is one irritant that causes the body's reaction to some ant bites and stings (Figure 14.10).\n\n\nFIGURE 14.10 The pain of some ant bites and stings is caused by formic acid. (credit: John Tann)\nWhat is the concentration of hydronium ion and the pH of a $0.534-M$ solution of formic acid?\n\n$$\n\\mathrm{HCO}_{2} \\mathrm{H}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{HCO}_{2}^{-}(a q) \\quad K_{\\mathrm{a}}=1.8 \\times 10^{-4}\n$$"}
{"id": 4120, "contents": "1509. Solution - \nThe ICE table for this system is\n\n\nSubstituting the equilibrium concentration terms into the $K_{\\mathrm{a}}$ expression gives\n\n$$\n\\begin{aligned}\nK_{\\mathrm{a}}= & 1.8 \\times 10^{-4}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{HCO}_{2}^{-}\\right]}{\\left[\\mathrm{HCO}_{2} \\mathrm{H}\\right]} \\\\\n& =\\frac{(x)(x)}{0.534-x}=1.8 \\times 10^{-4}\n\\end{aligned}\n$$\n\nThe relatively large initial concentration and small equilibrium constant permits the simplifying assumption that $x$ will be much lesser than 0.534 , and so the equation becomes\n\n$$\nK_{\\mathrm{a}}=1.8 \\times 10^{-4}=\\frac{x^{2}}{0.534}\n$$\n\nSolving the equation for $x$ yields\n\n$$\n\\begin{gathered}\nx^{2}=0.534 \\times\\left(1.8 \\times 10^{-4}\\right)=9.6 \\times 10^{-5} \\\\\nx=\\sqrt{9.6 \\times 10^{-5}}\n\\end{gathered}\n$$\n\n$$\n=9.8 \\times 10^{-3} M\n$$\n\nTo check the assumption that $x$ is small compared to 0.534 , its relative magnitude can be estimated:\n\n$$\n\\frac{x}{0.534}=\\frac{9.8 \\times 10^{-3}}{0.534}=1.8 \\times 10^{-2}(1.8 \\% \\text { of } 0.534)\n$$\n\nBecause $x$ is less than $5 \\%$ of the initial concentration, the assumption is valid.\nAs defined in the ICE table, $x$ is equal to the equilibrium concentration of hydronium ion:\n\n$$\nx=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=0.0098 \\mathrm{M}\n$$\n\nFinally, the pH is calculated to be"}
{"id": 4121, "contents": "1509. Solution - \n$$\nx=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=0.0098 \\mathrm{M}\n$$\n\nFinally, the pH is calculated to be\n\n$$\n\\mathrm{pH}=-\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=-\\log (0.0098)=2.01\n$$"}
{"id": 4122, "contents": "1510. Check Your Learning - \nOnly a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of a $0.100-M$ solution of acetic acid, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$ ?\n\n$$\n\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}(a q) \\quad K_{\\mathrm{a}}=1.8 \\times 10^{-5}\n$$"}
{"id": 4123, "contents": "1511. Answer: - \npercent ionization $=1.3 \\%$"}
{"id": 4124, "contents": "1513. Calculating Equilibrium Concentrations in a Weak Base Solution - \nFind the concentration of hydroxide ion, the pOH , and the pH of a $0.25-M$ solution of trimethylamine, a weak base:\n\n$$\n\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{~N}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{NH}^{+}(a q)+\\mathrm{OH}^{-}(a q) \\quad K_{\\mathrm{b}}=6.3 \\times 10^{-5}\n$$"}
{"id": 4125, "contents": "1514. Solution - \nThe ICE table for this system is\n\n|  | $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathbf{N}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{NH}^{+}+\\mathrm{OH}^{-}$ |  |  |  |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial concentration $(M)$ | 0.25 |  | 0 | $\\sim 0$ |\n| Change $(M)$ | $-x$ |  | $x$ | $x$ |\n| Equilibrium concentration $(M)$ | $0.25+(-x)$ |  | $0+x$ | $\\sim 0+x$ |\n\nSubstituting the equilibrium concentration terms into the $K_{\\mathrm{b}}$ expression gives\n\n$$\nK_{\\mathrm{b}}=\\frac{\\left[\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{NH}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]}{\\left[\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{~N}\\right]}=\\frac{(x)(x)}{0.25-x}=6.3 \\times 10^{-5}\n$$\n\nAssuming $x \\ll 0.25$ and solving for $x$ yields\n\n$$\nx=4.0 \\times 10^{-3} M\n$$\n\nThis value is less than $5 \\%$ of the initial concentration ( 0.25 ), so the assumption is justified.\n\nAs defined in the ICE table, $x$ is equal to the equilibrium concentration of hydroxide ion:\n\n$$\n\\begin{aligned}\n{\\left[\\mathrm{OH}^{-}\\right]=} & \\sim 0+x=x=4.0 \\times 10^{-3} \\mathrm{M} \\\\\n& =4.0 \\times 10^{-3} \\mathrm{M}\n\\end{aligned}\n$$\n\nThe pOH is calculated to be\n\n$$\n\\mathrm{pOH}=-\\log \\left(4.0 \\times 10^{-3}\\right)=2.40\n$$"}
{"id": 4126, "contents": "1514. Solution - \nThe pOH is calculated to be\n\n$$\n\\mathrm{pOH}=-\\log \\left(4.0 \\times 10^{-3}\\right)=2.40\n$$\n\nUsing the relation introduced in the previous section of this chapter:\n\n$$\n\\mathrm{pH}+\\mathrm{pOH}=\\mathrm{p} K_{\\mathrm{w}}=14.00\n$$\n\npermits the computation of pH :\n\n$$\n\\mathrm{pH}=14.00-\\mathrm{pOH}=14.00-2.40=11.60\n$$"}
{"id": 4127, "contents": "1515. Check Your Learning - \nCalculate the hydroxide ion concentration and the percent ionization of a $0.0325-M$ solution of ammonia, a weak base with a $K_{\\mathrm{b}}$ of $1.76 \\times 10^{-5}$."}
{"id": 4128, "contents": "1516. Answer: - \n$7.56 \\times 10^{-4} \\mathrm{M}, 2.33 \\%$\n\nIn some cases, the strength of the weak acid or base and its formal (initial) concentration result in an appreciable ionization. Though the ICE strategy remains effective for these systems, the algebra is a bit more involved because the simplifying assumption that $x$ is negligible cannot be made. Calculations of this sort are demonstrated in Example 14.14 below."}
{"id": 4129, "contents": "1518. Calculating Equilibrium Concentrations without Simplifying Assumptions - \nSodium bisulfate, $\\mathrm{NaHSO}_{4}$, is used in some household cleansers as a source of the $\\mathrm{HSO}_{4}{ }^{-}$ion, a weak acid. What is the pH of a $0.50-M$ solution of $\\mathrm{HSO}_{4}{ }^{-}$?\n\n$$\n\\mathrm{HSO}_{4}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q) \\quad K_{\\mathrm{a}}=1.2 \\times 10^{-2}\n$$"}
{"id": 4130, "contents": "1519. Solution - \nThe ICE table for this system is\n\n|  | $\\mathrm{HSO}_{4}{ }^{-}+\\mathbf{H}_{\\mathbf{2}} \\mathrm{O} \\rightleftharpoons \\mathbf{H}_{3} \\mathbf{O}^{+}+\\mathbf{S O}_{4}{ }^{\\mathbf{2 -}}$ |  |  |  |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial concentration $(M)$ | 0.50 |  | $\\sim 0$ | 0 |\n| Change $(M)$ | $-x$ |  | $+x$ | $+x$ |\n| Equilibrium <br> concentration $(M)$ | $0.50-x$ |  | $x$ | $x$ |\n\nSubstituting the equilibrium concentration terms into the $K_{\\mathrm{a}}$ expression gives\n\n$$\nK_{\\mathrm{a}}=1.2 \\times 10^{-2}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{SO}_{4}{ }^{2-}\\right]}{\\left[\\mathrm{HSO}_{4}^{-}\\right]}=\\frac{(x)(x)}{0.50-x}\n$$\n\nIf the assumption that $x \\ll 0.5$ is made, simplifying and solving the above equation yields\n\n$$\nx=0.077 M\n$$\n\nThis value of $x$ is clearly not significantly less than $0.50 M$; rather, it is approximately $15 \\%$ of the initial concentration:\nWhen we check the assumption, we calculate:\n\n$$\n\\begin{gathered}\n\\frac{x}{\\left[\\mathrm{HSO}_{4}^{-}\\right]_{\\mathrm{i}}} \\\\\n\\frac{x}{0.50}=\\frac{7.7 \\times 10^{-2}}{0.50}=0.15(15 \\%)\n\\end{gathered}\n$$\n\nBecause the simplifying assumption is not valid for this system, the equilibrium constant expression is solved as follows:\n\n$$\nK_{\\mathrm{a}}=1.2 \\times 10^{-2}=\\frac{(x)(x)}{0.50-x}\n$$"}
{"id": 4131, "contents": "1519. Solution - \n$$\nK_{\\mathrm{a}}=1.2 \\times 10^{-2}=\\frac{(x)(x)}{0.50-x}\n$$\n\nRearranging this equation yields\n\n$$\n6.0 \\times 10^{-3}-1.2 \\times 10^{-2} x=x^{2}\n$$\n\nWriting the equation in quadratic form gives\n\n$$\nx^{2}+1.2 \\times 10^{-2} x-6.0 \\times 10^{-3}=0\n$$\n\nSolving for the two roots of this quadratic equation results in a negative value that may be discarded as physically irrelevant and a positive value equal to $x$. As defined in the ICE table, $x$ is equal to the hydronium concentration.\n\n$$\n\\begin{aligned}\n& x=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=0.072 M \\\\\n& \\mathrm{pH}=-\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=-\\log (0.072)=1.14\n\\end{aligned}\n$$"}
{"id": 4132, "contents": "1520. Check Your Learning - \nCalculate the pH in a $0.010-M$ solution of caffeine, a weak base:\n\n$$\n\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2} \\mathrm{H}^{+}(a q)+\\mathrm{OH}^{-}(a q) \\quad K_{\\mathrm{b}}=2.5 \\times 10^{-4}\n$$"}
{"id": 4133, "contents": "1521. Answer: - \npH 11.16"}
{"id": 4134, "contents": "1523. Binary Acids and Bases - \nIn the absence of any leveling effect, the acid strength of binary compounds of hydrogen with nonmetals (A) increases as the $\\mathrm{H}-\\mathrm{A}$ bond strength decreases down a group in the periodic table. For group 17, the order of increasing acidity is $\\mathrm{HF}<\\mathrm{HCl}<\\mathrm{HBr}<\\mathrm{HI}$. Likewise, for group 16, the order of increasing acid strength is $\\mathrm{H}_{2} \\mathrm{O}<$ $\\mathrm{H}_{2} \\mathrm{~S}<\\mathrm{H}_{2} \\mathrm{Se}<\\mathrm{H}_{2} \\mathrm{Te}$.\n\nAcross a row in the periodic table, the acid strength of binary hydrogen compounds increases with increasing electronegativity of the nonmetal atom because the polarity of the $\\mathrm{H}-\\mathrm{A}$ bond increases. Thus, the order of increasing acidity (for removal of one proton) across the second row is $\\mathrm{CH}_{4}<\\mathrm{NH}_{3}<\\mathrm{H}_{2} \\mathrm{O}<\\mathrm{HF}$; across the third row, it is $\\mathrm{SiH}_{4}<\\mathrm{PH}_{3}<\\mathrm{H}_{2} \\mathrm{~S}<\\mathrm{HCl}$ (see Figure 14.11).\n\n\nFIGURE 14.11 The figure shows trends in the strengths of binary acids and bases."}
{"id": 4135, "contents": "1524. Ternary Acids and Bases - \nTernary compounds composed of hydrogen, oxygen, and some third element (\"E\") may be structured as depicted in the image below. In these compounds, the central E atom is bonded to one or more O atoms, and at least one of the O atoms is also bonded to an H atom, corresponding to the general molecular formula $\\mathrm{O}_{\\mathrm{m}} \\mathrm{E}(\\mathrm{OH})_{\\mathrm{n}}$. These compounds may be acidic, basic, or amphoteric depending on the properties of the central E atom. Examples of such compounds include sulfuric acid, $\\mathrm{O}_{2} \\mathrm{~S}(\\mathrm{OH})_{2}$, sulfurous acid, $\\mathrm{OS}(\\mathrm{OH})_{2}$, nitric acid, $\\mathrm{O}_{2} \\mathrm{NOH}$, perchloric acid, $\\mathrm{O}_{3} \\mathrm{ClOH}$, aluminum hydroxide, $\\mathrm{Al}(\\mathrm{OH})_{3}$, calcium hydroxide, $\\mathrm{Ca}(\\mathrm{OH})_{2}$, and potassium hydroxide, KOH :\n\n\nIf the central atom, E , has a low electronegativity, its attraction for electrons is low. Little tendency exists for the central atom to form a strong covalent bond with the oxygen atom, and bond a between the element and oxygen is more readily broken than bond $b$ between oxygen and hydrogen. Hence bond $a$ is ionic, hydroxide ions are released to the solution, and the material behaves as a base-this is the case with $\\mathrm{Ca}(\\mathrm{OH})_{2}$ and KOH . Lower electronegativity is characteristic of the more metallic elements; hence, the metallic elements form ionic hydroxides that are by definition basic compounds."}
{"id": 4136, "contents": "1524. Ternary Acids and Bases - \nIf, on the other hand, the atom E has a relatively high electronegativity, it strongly attracts the electrons it shares with the oxygen atom, making bond a relatively strongly covalent. The oxygen-hydrogen bond, bond $b$, is thereby weakened because electrons are displaced toward E. Bond $b$ is polar and readily releases hydrogen ions to the solution, so the material behaves as an acid. High electronegativities are characteristic of the more nonmetallic elements. Thus, nonmetallic elements form covalent compounds containing acidic -OH groups that are called oxyacids.\n\nIncreasing the oxidation number of the central atom $E$ also increases the acidity of an oxyacid because this increases the attraction of E for the electrons it shares with oxygen and thereby weakens the $\\mathrm{O}-\\mathrm{H}$ bond. Sulfuric acid, $\\mathrm{H}_{2} \\mathrm{SO}_{4}$, or $\\mathrm{O}_{2} \\mathrm{~S}(\\mathrm{OH})_{2}$ (with a sulfur oxidation number of +6 ), is more acidic than sulfurous acid,\n$\\mathrm{H}_{2} \\mathrm{SO}_{3}$, or $\\mathrm{OS}(\\mathrm{OH})_{2}$ (with a sulfur oxidation number of +4 ). Likewise nitric acid, $\\mathrm{HNO}_{3}$, or $\\mathrm{O}_{2} \\mathrm{NOH}$ ( N oxidation number $=+5$ ), is more acidic than nitrous acid, $\\mathrm{HNO}_{2}$, or ONOH ( N oxidation number $=+3$ ). In each of these pairs, the oxidation number of the central atom is larger for the stronger acid (Figure 14.12)."}
{"id": 4137, "contents": "1524. Ternary Acids and Bases - \nFIGURE 14.12 As the oxidation number of the central atom $E$ increases, the acidity also increases.\nHydroxy compounds of elements with intermediate electronegativities and relatively high oxidation numbers (for example, elements near the diagonal line separating the metals from the nonmetals in the periodic table) are usually amphoteric. This means that the hydroxy compounds act as acids when they react with strong bases and as bases when they react with strong acids. The amphoterism of aluminum hydroxide, which commonly exists as the hydrate $\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{3}(\\mathrm{OH})_{3}$, is reflected in its solubility in both strong acids and strong bases. In strong bases, the relatively insoluble hydrated aluminum hydroxide, $\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{3}(\\mathrm{OH})_{3}$, is converted into the soluble ion, $\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{2}(\\mathrm{OH})_{4}\\right]^{-}$, by reaction with hydroxide ion:\n\n$$\n\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{3}(\\mathrm{OH})_{3}(a q)+\\mathrm{OH}^{-}(a q) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{O}(l)+\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{2}(\\mathrm{OH})_{4}\\right]^{-}(a q)\n$$"}
{"id": 4138, "contents": "1524. Ternary Acids and Bases - \nIn this reaction, a proton is transferred from one of the aluminum-bound $\\mathrm{H}_{2} \\mathrm{O}$ molecules to a hydroxide ion in solution. The $\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{3}(\\mathrm{OH})_{3}$ compound thus acts as an acid under these conditions. On the other hand, when dissolved in strong acids, it is converted to the soluble ion $\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{3+}$ by reaction with hydronium ion:\n\n$$\n3 \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{3}(\\mathrm{OH})_{3}(a q) \\rightleftharpoons \\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{3+}(a q)+3 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nIn this case, protons are transferred from hydronium ions in solution to $\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{3}(\\mathrm{OH})_{3}$, and the compound functions as a base."}
{"id": 4139, "contents": "1525. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Predict whether a salt solution will be acidic, basic, or neutral\n- Calculate the concentrations of the various species in a salt solution\n- Describe the acid ionization of hydrated metal ions"}
{"id": 4140, "contents": "1526. Salts with Acidic Ions - \nSalts are ionic compounds composed of cations and anions, either of which may be capable of undergoing an acid or base ionization reaction with water. Aqueous salt solutions, therefore, may be acidic, basic, or neutral, depending on the relative acid-base strengths of the salt's constituent ions. For example, dissolving ammonium chloride in water results in its dissociation, as described by the equation\n\n$$\n\\mathrm{NH}_{4} \\mathrm{Cl}(s) \\rightleftharpoons \\mathrm{NH}_{4}^{+}(a q)+\\mathrm{Cl}^{-}(a q)\n$$\n\nThe ammonium ion is the conjugate acid of the base ammonia, $\\mathrm{NH}_{3}$; its acid ionization (or acid hydrolysis)\nreaction is represented by\n\n$$\n\\mathrm{NH}_{4}^{+}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{NH}_{3}(a q) \\quad K_{\\mathrm{a}}=K_{\\mathrm{w}} / K_{\\mathrm{b}}\n$$\n\nSince ammonia is a weak base, $K_{\\mathrm{b}}$ is measurable and $K_{\\mathrm{a}}>0$ (ammonium ion is a weak acid).\nThe chloride ion is the conjugate base of hydrochloric acid, and so its base ionization (or base hydrolysis) reaction is represented by\n\n$$\n\\mathrm{Cl}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{HCl}(a q)+\\mathrm{OH}^{-}(a q) \\quad K_{\\mathrm{b}}=K_{\\mathrm{w}} / K_{\\mathrm{a}}\n$$\n\nSince HCl is a strong acid, $K_{\\mathrm{a}}$ is immeasurably large and $K_{\\mathrm{b}} \\approx 0$ (chloride ions don't undergo appreciable hydrolysis)."}
{"id": 4141, "contents": "1526. Salts with Acidic Ions - \nThus, dissolving ammonium chloride in water yields a solution of weak acid cations $\\left(\\mathrm{NH}_{4}{ }^{+}\\right)$and inert anions $\\left(\\mathrm{Cl}^{-}\\right)$, resulting in an acidic solution."}
{"id": 4142, "contents": "1528. Calculating the pH of an Acidic Salt Solution - \nAniline is an amine that is used to manufacture dyes. It is isolated as anilinium chloride, $\\left[\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{3}\\right] \\mathrm{Cl}$, a salt prepared by the reaction of the weak base aniline and hydrochloric acid. What is the pH of a 0.233 M solution of anilinium chloride\n\n$$\n\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{3}+(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{2}(a q)\n$$"}
{"id": 4143, "contents": "1529. Solution - \nThe $K_{\\mathrm{a}}$ for anilinium ion is derived from the $K_{\\mathrm{b}}$ for its conjugate base, aniline (see Appendix H):\n\n$$\nK_{\\mathrm{a}}=\\frac{K_{\\mathrm{w}}}{K_{\\mathrm{b}}}=\\frac{1.0 \\times 10^{-14}}{4.3 \\times 10^{-10}}=2.3 \\times 10^{-5}\n$$\n\nUsing the provided information, an ICE table for this system is prepared:\n\n|  | $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{3}{ }^{+}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{2}+\\mathrm{H}_{3} \\mathrm{O}^{+}$ |  |  |  |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial concentration $(M)$ | 0.233 |  | 0 | $\\sim 0$ |\n| Change $(M)$ | $-x$ |  | $+x$ | $+x$ |\n| Equilibrium concentration $(M)$ | $0.233-x$ |  | $x$ | $x$ |\n\nSubstituting these equilibrium concentration terms into the $K_{\\mathrm{a}}$ expression gives\n\n$$\n\\begin{aligned}\n& \\mathrm{K}_{a}=\\left[\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{2}\\right]\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right] /\\left[\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{3}{ }^{+}\\right] \\\\\n& \\left.2.3 \\times 10^{-5}=(x)(x) / 0.233-x\\right)\n\\end{aligned}\n$$\n\nAssuming $x \\ll 0.233$, the equation is simplified and solved for $x$ :"}
{"id": 4144, "contents": "1529. Solution - \nAssuming $x \\ll 0.233$, the equation is simplified and solved for $x$ :\n\n$$\n\\begin{aligned}\n& 2.3 \\times 10^{-5}=x^{2} / 0.233 \\\\\n& x=0.0023 M\n\\end{aligned}\n$$\n\nThe ICE table defines x as the hydronium ion molarity, and so the pH is computed as\n\n$$\n\\mathrm{pH}=-\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=-\\log (0.0023)=2.64\n$$"}
{"id": 4145, "contents": "1530. Check Your Learning - \nWhat is the hydronium ion concentration in a $0.100-M$ solution of ammonium nitrate, $\\mathrm{NH}_{4} \\mathrm{NO}_{3}$, a salt\ncomposed of the ions $\\mathrm{NH}_{4}{ }^{+}$and $\\mathrm{NO}_{3}{ }^{-}$. Which is the stronger acid $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{3}{ }^{+}$or $\\mathrm{NH}_{4}{ }^{+}$?"}
{"id": 4146, "contents": "1531. Answer: - \n$\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=7.5 \\times 10^{-6} \\mathrm{M} ; \\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{3}{ }^{+}$is the stronger acid."}
{"id": 4147, "contents": "1532. Salts with Basic lons - \nAs another example, consider dissolving sodium acetate in water:\n\n$$\n\\mathrm{NaCH}_{3} \\mathrm{CO}_{2}(s) \\leftrightharpoons \\mathrm{Na}^{+}(a q)+\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}(a q)\n$$\n\nThe sodium ion does not undergo appreciable acid or base ionization and has no effect on the solution pH . This may seem obvious from the ion's formula, which indicates no hydrogen or oxygen atoms, but some dissolved metal ions function as weak acids, as addressed later in this section.\n\nThe acetate ion, $\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}$, is the conjugate base of acetic acid, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$, and so its base ionization (or base hydrolysis) reaction is represented by\n\n$$\n\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(a q)+\\mathrm{OH}-(a q) \\quad K_{\\mathrm{b}}=K_{\\mathrm{w}} / K_{\\mathrm{a}}\n$$\n\nBecause acetic acid is a weak acid, its $K_{\\mathrm{a}}$ is measurable and $K_{\\mathrm{b}}>0$ (acetate ion is a weak base).\nDissolving sodium acetate in water yields a solution of inert cations $\\left(\\mathrm{Na}^{+}\\right)$and weak base anions $\\left(\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}\\right)$, resulting in a basic solution."}
{"id": 4148, "contents": "1534. Equilibrium in a Solution of a Salt of a Weak Acid and a Strong Base - \nDetermine the acetic acid concentration in a solution with $\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}\\right]=0.050 \\mathrm{M}$ and $\\left[\\mathrm{OH}^{-}\\right]=2.5 \\times 10^{-6} \\mathrm{M}$ at equilibrium. The reaction is:\n\n$$\n\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(a q)+\\mathrm{OH}^{-}(a q)\n$$"}
{"id": 4149, "contents": "1535. Solution - \nThe provided equilibrium concentrations and a value for the equilibrium constant will permit calculation of the missing equilibrium concentration. The process in question is the base ionization of acetate ion, for which\n\n$$\nK_{\\mathrm{b}}\\left(\\text { for } \\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}\\right)=\\frac{K_{\\mathrm{w}}}{K_{\\mathrm{a}}\\left(\\text { for } \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right)}=\\frac{1.0 \\times 10^{-14}}{1.8 \\times 10^{-5}}=5.6 \\times 10^{-10}\n$$\n\nSubstituting the available values into the $K_{\\mathrm{b}}$ expression gives\n\n$$\n\\begin{aligned}\n& K_{\\mathrm{b}}=\\frac{\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right]\\left[\\mathrm{OH}^{-}\\right]}{\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2}-\\right]}=5.6 \\times 10^{-10} \\\\\n= & \\frac{\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right]\\left(2.5 \\times 10^{-6}\\right)}{(0.050)}=5.6 \\times 10^{-10}\n\\end{aligned}\n$$\n\nSolving the above equation for the acetic acid molarity yields $\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right]=1.1 \\times 10^{-5} \\mathrm{M}$."}
{"id": 4150, "contents": "1536. Check Your Learning - \nWhat is the pH of a $0.083-M$ solution of NaCN ?"}
{"id": 4151, "contents": "1537. Answer: - \n11.11"}
{"id": 4152, "contents": "1538. Salts with Acidic and Basic Ions - \nSome salts are composed of both acidic and basic ions, and so the pH of their solutions will depend on the\nrelative strengths of these two species. Likewise, some salts contain a single ion that is amphiprotic, and so the relative strengths of this ion's acid and base character will determine its effect on solution pH . For both types of salts, a comparison of the $K_{\\mathrm{a}}$ and $K_{\\mathrm{b}}$ values allows prediction of the solution's acid-base status, as illustrated in the following example exercise."}
{"id": 4153, "contents": "1540. Determining the Acidic or Basic Nature of Salts - \nDetermine whether aqueous solutions of the following salts are acidic, basic, or neutral:\n(a) KBr\n(b) $\\mathrm{NaHCO}_{3}$\n(c) $\\mathrm{Na}_{2} \\mathrm{HPO}_{4}$\n(d) $\\mathrm{NH}_{4} \\mathrm{~F}$"}
{"id": 4154, "contents": "1541. Solution - \nConsider each of the ions separately in terms of its effect on the pH of the solution, as shown here:\n(a) The $\\mathrm{K}^{+}$cation is inert and will not affect pH . The bromide ion is the conjugate base of a strong acid, and so it is of negligible base strength (no appreciable base ionization). The solution is neutral.\n(b) The $\\mathrm{Na}^{+}$cation is inert and will not affect the pH of the solution; while the $\\mathrm{HCO}_{3}{ }^{-}$anion is amphiprotic. The $K_{\\mathrm{a}}$ of $\\mathrm{HCO}_{3}{ }^{-}$is $4.7 \\times 10^{-11}$, and its $K_{\\mathrm{b}}$ is $\\frac{1.0 \\times 10^{-14}}{4.3 \\times 10^{-7}}=2.3 \\times 10^{-8}$."}
{"id": 4155, "contents": "1541. Solution - \nSince $K_{\\mathrm{b}} \\gg K_{\\mathrm{a}}$, the solution is basic.\n(c) The $\\mathrm{Na}^{+}$cation is inert and will not affect the pH of the solution, while the $\\mathrm{HPO}_{4}{ }^{2-}$ anion is amphiprotic. The $K_{\\mathrm{a}}$ of $\\mathrm{HPO}_{4}{ }^{2-}$ is $4.2 \\times 10^{-13}$,\nand its $K_{\\mathrm{b}}$ is $\\frac{1.0 \\times 10^{-14}}{6.2 \\times 10^{-8}}=1.6 \\times 10^{-7}$. Because $K_{\\mathrm{b}} \\gg K_{\\mathrm{a}}$, the solution is basic.\n(d) The $\\mathrm{NH}_{4}{ }^{+}$ion is acidic (see above discussion) and the $\\mathrm{F}^{-}$ion is basic (conjugate base of the weak acid HF). Comparing the two ionization constants: $K_{\\mathrm{a}}$ of $\\mathrm{NH}_{4}{ }^{+}$is $5.6 \\times 10^{-10}$ and the $K_{\\mathrm{b}}$ of $\\mathrm{F}^{-}$is $1.6 \\times 10^{-11}$, so the solution is acidic, since $K_{\\mathrm{a}}>K_{\\mathrm{b}}$."}
{"id": 4156, "contents": "1542. Check Your Learning - \nDetermine whether aqueous solutions of the following salts are acidic, basic, or neutral:\n(a) $\\mathrm{K}_{2} \\mathrm{CO}_{3}$\n(b) $\\mathrm{CaCl}_{2}$\n(c) $\\mathrm{KH}_{2} \\mathrm{PO}_{4}$\n(d) $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{CO}_{3}$"}
{"id": 4157, "contents": "1543. Answer: - \n(a) basic; (b) neutral; (c) acidic; (d) basic"}
{"id": 4158, "contents": "1544. The Ionization of Hydrated Metal Ions - \nUnlike the group 1 and 2 metal ions of the preceding examples ( $\\mathrm{Na}^{+}, \\mathrm{Ca}^{2+}$, etc.), some metal ions function as acids in aqueous solutions. These ions are not just loosely solvated by water molecules when dissolved, instead they are covalently bonded to a fixed number of water molecules to yield a complex ion (see chapter on coordination chemistry). As an example, the dissolution of aluminum nitrate in water is typically represented as\n\n$$\n\\mathrm{Al}\\left(\\mathrm{NO}_{3}\\right)(s) \\rightleftharpoons \\mathrm{Al}^{3}+(a q)+3 \\mathrm{NO}_{3}^{-}(a q)\n$$\n\nHowever, the aluminum(III) ion actually reacts with six water molecules to form a stable complex ion, and so the more explicit representation of the dissolution process is\n\n$$\n\\mathrm{Al}\\left(\\mathrm{NO}_{3}\\right)_{3}(s)+6 \\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{3+}(a q)+3 \\mathrm{NO}_{3}{ }^{-}(a q)\n$$\n\nAs shown in Figure 14.13, the $\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{3+}$ ions involve bonds between a central Al atom and the O atoms of the six water molecules. Consequently, the bonded water molecules' $\\mathrm{O}-\\mathrm{H}$ bonds are more polar than in nonbonded water molecules, making the bonded molecules more prone to donation of a hydrogen ion:"}
{"id": 4159, "contents": "1544. The Ionization of Hydrated Metal Ions - \n$$\n\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{3+}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{5}(\\mathrm{OH})^{2+}(a q) \\quad K_{\\mathrm{a}}=1.4 \\times 10^{-5}\n$$\n\nThe conjugate base produced by this process contains five other bonded water molecules capable of acting as acids, and so the sequential or step-wise transfer of protons is possible as depicted in few equations below:"}
{"id": 4160, "contents": "1544. The Ionization of Hydrated Metal Ions - \nThe conjugate base produced by this process contains five other bonded water molecules capable of acting as acids, and so the sequential or step-wise transfer of protons is possible as depicted in few equations below:\n\n$$\n\\begin{gathered}\n\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{3+}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{5}(\\mathrm{OH})^{2+}(a q) \\\\\n\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{5}(\\mathrm{OH})^{2+}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{4}(\\mathrm{OH})_{2}^{+}(a q) \\\\\n\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{4}(\\mathrm{OH})_{2}^{+}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{3}(\\mathrm{OH})_{3}(a q)\n\\end{gathered}\n$$\n\nThis is an example of a polyprotic acid, the topic of discussion in a later section of this chapter."}
{"id": 4161, "contents": "1544. The Ionization of Hydrated Metal Ions - \nThis is an example of a polyprotic acid, the topic of discussion in a later section of this chapter.\n\n\nFIGURE 14.13 When an aluminum ion reacts with water, the hydrated aluminum ion becomes a weak acid.\nAside from the alkali metals (group 1) and some alkaline earth metals (group 2), most other metal ions will undergo acid ionization to some extent when dissolved in water. The acid strength of these complex ions typically increases with increasing charge and decreasing size of the metal ions. The first-step acid ionization equations for a few other acidic metal ions are shown below:\n\n$$\n\\begin{array}{lc}\n\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{3+}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{5}(\\mathrm{OH})^{2+}(a q) & p K_{\\mathrm{a}}=2.74 \\\\\n\\mathrm{Cu}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{2+}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{Cu}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{5}(\\mathrm{OH})^{+}(a q) & p K_{\\mathrm{a}}=\\sim 6.3 \\\\\n\\mathrm{Zn}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{4}{ }^{2+}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{Zn}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{3}(\\mathrm{OH})^{+}(a q) & p K_{\\mathrm{a}}=9.6\n\\end{array}\n$$"}
{"id": 4162, "contents": "1546. Hydrolysis of $\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{3+}$ - \nCalculate the pH of a $0.10-M$ solution of aluminum chloride, which dissolves completely to give the hydrated aluminum ion $\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{3+}$ in solution."}
{"id": 4163, "contents": "1547. Solution - \nThe equation for the reaction and $K_{\\mathrm{a}}$ are:\n\n$$\n\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{3+}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{5}(\\mathrm{OH})^{2+}(a q) \\quad K_{\\mathrm{a}}=1.4 \\times 10^{-5}\n$$\n\nAn ICE table with the provided information is\n\n|  | $\\mathrm{Al}_{1}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{3+}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}+\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{5}(\\mathrm{OH})^{2+}$ |  |  |\n| :---: | :---: | :---: | :---: |\n| Initial concentration $(M)$ | 0.10 | $\\sim 0$ | 0 |\n| Change $(M)$ | $-x$ | $+x$ | $+x$ |\n| Equilibrium concentration $(M)$ | $0.10-x$ | $x$ | $x$ |\n\nSubstituting the expressions for the equilibrium concentrations into the equation for the ionization constant yields:\n\n$$\n\\begin{aligned}\nK_{\\mathrm{a}} & =\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{5}(\\mathrm{OH})^{2+}\\right]}{\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{3+}\\right]} \\\\\n& =\\frac{(x)(x)}{0.10-x}=1.4 \\times 10^{-5}\n\\end{aligned}\n$$"}
{"id": 4164, "contents": "1547. Solution - \nAssuming $x \\ll 0.10$ and solving the simplified equation gives:\n\n$$\nx=1.2 \\times 10^{-3} M\n$$\n\nThe ICE table defined $x$ as equal to the hydronium ion concentration, and so the pH is calculated to be\n\n$$\n\\begin{gathered}\n{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=0+x=1.2 \\times 10^{-3} M} \\\\\n\\mathrm{pH}=-\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=2.92(\\text { an acidic solution })\n\\end{gathered}\n$$"}
{"id": 4165, "contents": "1548. Check Your Learning - \nWhat is $\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{5}(\\mathrm{OH})^{2+}\\right]$ in a $0.15-M$ solution of $\\mathrm{Al}\\left(\\mathrm{NO}_{3}\\right)_{3}$ that contains enough of the strong acid $\\mathrm{HNO}_{3}$ to bring $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]$to 0.10 M ?"}
{"id": 4166, "contents": "1549. Answer: - \n$2.1 \\times 10^{-5} \\mathrm{M}$"}
{"id": 4167, "contents": "1550. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Extend previously introduced equilibrium concepts to acids and bases that may donate or accept more than one proton\n\nAcids are classified by the number of protons per molecule that they can give up in a reaction. Acids such as $\\mathrm{HCl}, \\mathrm{HNO}_{3}$, and HCN that contain one ionizable hydrogen atom in each molecule are called monoprotic acids. Their reactions with water are:\n\n$$\n\\begin{gathered}\n\\mathrm{HCl}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{Cl}^{-}(a q) \\\\\n\\mathrm{HNO}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{NO}_{3}^{-}(a q) \\\\\n\\mathrm{HCN}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{CN}^{-}(a q)\n\\end{gathered}\n$$\n\nEven though it contains four hydrogen atoms, acetic acid, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$, is also monoprotic because only the hydrogen atom from the carboxyl group $(\\mathrm{COOH})$ reacts with bases:\n\n\nSimilarly, monoprotic bases are bases that will accept a single proton.\nDiprotic acids contain two ionizable hydrogen atoms per molecule; ionization of such acids occurs in two steps. The first ionization always takes place to a greater extent than the second ionization. For example, sulfuric acid, a strong acid, ionizes as follows:"}
{"id": 4168, "contents": "1550. LEARNING OBJECTIVES - \nFirst ionization: $\\mathrm{H}_{2} \\mathrm{SO}_{4}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{HSO}_{4}{ }^{-}(a q)$\n$K_{\\mathrm{a} 1}=$ more than $10^{2} ;$ complete dissociation\n$\\quad K_{\\mathrm{a} 2}=1.2 \\times 10^{-2}$\n\nSecond ionization: $\\mathrm{HSO}_{4}{ }^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q)$\nThis stepwise ionization process occurs for all polyprotic acids. Carbonic acid, $\\mathrm{H}_{2} \\mathrm{CO}_{3}$, is an example of a weak diprotic acid. The first ionization of carbonic acid yields hydronium ions and bicarbonate ions in small amounts.\n\nFirst ionization:\n\n$$\n\\mathrm{H}_{2} \\mathrm{CO}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{HCO}_{3}^{-}(a q) \\quad K_{\\mathrm{H}_{2} \\mathrm{CO}_{3}}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{HCO}_{3}^{-}\\right]}{\\left[\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right]}=4.3 \\times 10^{-7}\n$$\n\nThe bicarbonate ion can also act as an acid. It ionizes and forms hydronium ions and carbonate ions in even smaller quantities."}
{"id": 4169, "contents": "1551. Second ionization: - \n$$\n\\mathrm{HCO}_{3}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{CO}_{3}^{2-}(a q) \\quad K_{\\mathrm{HCO}_{3}-}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{CO}_{3}^{2-}\\right]}{\\left[\\mathrm{HCO}_{3}^{-}\\right]}=4.7 \\times 10^{-11}\n$$\n\n$K_{\\mathrm{H}_{2} \\mathrm{CO}_{3}}$ is larger than $K_{\\mathrm{HCO}_{3}}$ - by a factor of $10^{4}$, so $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ is the dominant producer of hydronium ion in the solution. This means that little of the $\\mathrm{HCO}_{3}{ }^{-}$formed by the ionization of $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ ionizes to give hydronium ions (and carbonate ions), and the concentrations of $\\mathrm{H}_{3} \\mathrm{O}^{+}$and $\\mathrm{HCO}_{3}{ }^{-}$are practically equal in a pure aqueous solution of $\\mathrm{H}_{2} \\mathrm{CO}_{3}$.\n\nIf the first ionization constant of a weak diprotic acid is larger than the second by a factor of at least 20, it is appropriate to treat the first ionization separately and calculate concentrations resulting from it before calculating concentrations of species resulting from subsequent ionization. This approach is demonstrated in the following example exercise."}
{"id": 4170, "contents": "1553. Ionization of a Diprotic Acid - \n\"Carbonated water\" contains a palatable amount of dissolved carbon dioxide. The solution is acidic because $\\mathrm{CO}_{2}$ reacts with water to form carbonic acid, $\\mathrm{H}_{2} \\mathrm{CO}_{3}$. What are $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right],\\left[\\mathrm{HCO}_{3}{ }^{-}\\right]$, and $\\left[\\mathrm{CO}_{3}{ }^{2-}\\right]$ in a saturated solution of $\\mathrm{CO}_{2}$ with an initial $\\left[\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right]=0.033 \\mathrm{M}$ ?\n\n$$\n\\begin{array}{ll}\n\\mathrm{H}_{2} \\mathrm{CO}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{HCO}_{3}^{-}(a q) & K_{\\mathrm{a} 1}=4.3 \\times 10^{-7} \\\\\n\\mathrm{HCO}_{3}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{CO}_{3}^{2-}(a q) & K_{\\mathrm{a} 2}=4.7 \\times 10^{-11}\n\\end{array}\n$$"}
{"id": 4171, "contents": "1554. Solution - \nAs indicated by the ionization constants, $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ is a much stronger acid than $\\mathrm{HCO}_{3}{ }^{-}$, so the stepwise\nionization reactions may be treated separately.\nThe first ionization reaction is\n\n$$\n\\mathrm{H}_{2} \\mathrm{CO}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{HCO}_{3}^{-}(a q) \\quad K_{\\mathrm{a} 1}=4.3 \\times 10^{-7}\n$$\n\nUsing provided information, an ICE table for this first step is prepared:\n\n|  | $\\mathrm{H}_{2} \\mathrm{CO}_{3}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathrm{H}_{3} \\mathbf{O}^{+}+\\mathbf{H C O}{ }^{-}$ |  |  |  |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial concentration (M) | 0.033 |  | $\\sim 0$ | 0 |\n| Change $(M)$ | $-x$ |  | $+x$ | $+x$ |\n| Equilibrium concentration $(M)$ | $0.033-x$ |  | $x$ | $x$ |\n\nSubstituting the equilibrium concentrations into the equilibrium equation gives\n\n$$\nK_{\\mathrm{H}_{2} \\mathrm{CO}_{3}}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{HCO}_{3}^{-}\\right]}{\\left[\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right]}=\\frac{(x)(x)}{0.033-x}=4.3 \\times 10^{-7}\n$$\n\nAssuming $x \\ll 0.033$ and solving the simplified equation yields\n\n$$\nx=1.2 \\times 10^{-4}\n$$\n\nThe ICE table defined $x$ as equal to the bicarbonate ion molarity and the hydronium ion molarity:"}
{"id": 4172, "contents": "1554. Solution - \n$$\nx=1.2 \\times 10^{-4}\n$$\n\nThe ICE table defined $x$ as equal to the bicarbonate ion molarity and the hydronium ion molarity:\n\n$$\n\\begin{gathered}\n{\\left[\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right]=0.033 \\mathrm{M}} \\\\\n{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=\\left[\\mathrm{HCO}_{3}^{-}\\right]=1.2 \\times 10^{-4} \\mathrm{M}}\n\\end{gathered}\n$$\n\nUsing the bicarbonate ion concentration computed above, the second ionization is subjected to a similar equilibrium calculation:\n\n$$\n\\begin{gathered}\n\\mathrm{HCO}_{3}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{CO}_{3}^{2-}(a q) \\\\\nK_{\\mathrm{HCO}_{3}-}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{CO}_{3}^{2-}\\right]}{\\left[\\mathrm{HCO}_{3}^{-}\\right]}=\\frac{\\left(1.2 \\times 10^{-4}\\right)\\left[\\mathrm{CO}_{3}^{2-}\\right]}{1.2 \\times 10^{-4}} \\\\\n{\\left[\\mathrm{CO}_{3}^{2-}\\right]=\\frac{\\left(4.7 \\times 10^{-11}\\right)\\left(1.2 \\times 10^{-4}\\right)}{1.2 \\times 10^{-4}}=4.7 \\times 10^{-11} \\mathrm{M}}\n\\end{gathered}\n$$"}
{"id": 4173, "contents": "1554. Solution - \nTo summarize: at equilibrium $\\left[\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right]=0.033 \\mathrm{M} ;\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=1.2 \\times 10^{-4} ;\\left[\\mathrm{HCO}_{3}{ }^{-}\\right]=1.2 \\times 10^{-4} \\mathrm{M}$; $\\left[\\mathrm{CO}_{3}{ }^{2-}\\right]=4.7 \\times 10^{-11} \\mathrm{M}$."}
{"id": 4174, "contents": "1555. Check Your Learning - \nThe concentration of $\\mathrm{H}_{2} \\mathrm{~S}$ in a saturated aqueous solution at room temperature is approximately 0.1 M . Calculate $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]$, $\\left[\\mathrm{HS}^{-}\\right]$, and $\\left[\\mathrm{S}^{2-}\\right]$ in the solution:\n\n$$\n\\begin{aligned}\n\\mathrm{H}_{2} \\mathrm{~S}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) & \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{HS}^{-}(a q) & K_{\\mathrm{a} 1}=8.9 \\times 10^{-8} \\\\\n\\mathrm{HS}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) & \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{S}^{2-}(a q) & K_{\\mathrm{a} 2}=1.0 \\times 10^{-19}\n\\end{aligned}\n$$"}
{"id": 4175, "contents": "1556. Answer: - \n$\\left[\\mathrm{H}_{2} \\mathrm{~S}\\right]=0.1 \\mathrm{M} ;\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=\\left[\\mathrm{HS}^{-}\\right]=0.000094 \\mathrm{M} ;\\left[\\mathrm{S}^{2-}\\right]=1 \\times 10^{-19} \\mathrm{M}$\n\nA triprotic acid is an acid that has three ionizable H atoms. Phosphoric acid is one example:\n\n$$\n\\begin{array}{lc}\n\\text { First ionization: } \\mathrm{H}_{3} \\mathrm{PO}_{4}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{-}(a q) & K_{\\mathrm{a} 1}=7.5 \\times 10^{-3} \\\\\n\\text { Second ionization: } \\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{HPO}_{4}{ }^{2-}(a q) & K_{\\mathrm{a} 2}=6.2 \\times 10^{-8} \\\\\n\\text { Third ionization: } \\mathrm{HPO}_{4}{ }^{2-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{PO}_{4}{ }^{3-}(a q) & K_{\\mathrm{a} 3}=4.2 \\times 10^{-13}\n\\end{array}\n$$"}
{"id": 4176, "contents": "1556. Answer: - \nAs for the diprotic acid examples, each successive ionization reaction is less extensive than the former, reflected in decreasing values for the stepwise acid ionization constants. This is a general characteristic of polyprotic acids and successive ionization constants often differ by a factor of about $10^{5}$ to $10^{6}$.\n\nThis set of three dissociation reactions may appear to make calculations of equilibrium concentrations in a solution of $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ complicated. However, because the successive ionization constants differ by a factor of $10^{5}$ to $10^{6}$, large differences exist in the small changes in concentration accompanying the ionization reactions. This allows the use of math-simplifying assumptions and processes, as demonstrated in the examples above.\n\nPolyprotic bases are capable of accepting more than one hydrogen ion. The carbonate ion is an example of a diprotic base, because it can accept two protons, as shown below. Similar to the case for polyprotic acids, note the ionization constants decrease with ionization step. Likewise, equilibrium calculations involving polyprotic bases follow the same approaches as those for polyprotic acids.\n\n$$\n\\begin{array}{lc}\n\\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{CO}_{3}^{2-}(a q) \\rightleftharpoons \\mathrm{HCO}_{3}^{-}(a q)+\\mathrm{OH}^{-}(a q) & K_{\\mathrm{b} 1}=2.1 \\times 10^{-4} \\\\\n\\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{HCO}_{3}^{-}(a q) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{CO}_{3}(a q)+\\mathrm{OH}^{-}(a q) & K_{\\mathrm{b} 2}=2.3 \\times 10^{-8}\n\\end{array}\n$$"}
{"id": 4177, "contents": "1557. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the composition and function of acid-base buffers\n- Calculate the pH of a buffer before and after the addition of added acid or base\n\nA solution containing appreciable amounts of a weak conjugate acid-base pair is called a buffer solution, or a buffer. Buffer solutions resist a change in pH when small amounts of a strong acid or a strong base are added (Figure 14.14). A solution of acetic acid and sodium acetate ( $\\mathrm{CH}_{3} \\mathrm{COOH}+\\mathrm{CH}_{3} \\mathrm{COONa}$ ) is an example of a buffer that consists of a weak acid and its salt. An example of a buffer that consists of a weak base and its salt is a solution of ammonia and ammonium chloride $\\left(\\mathrm{NH}_{3}(a q)+\\mathrm{NH}_{4} \\mathrm{Cl}(a q)\\right)$.\n\n(a)\n\n(b)\n\nFIGURE 14.14 (a) The unbuffered solution on the left and the buffered solution on the right have the same pH ( pH 8); they are basic, showing the yellow color of the indicator methyl orange at this pH . (b) After the addition of 1 mL of a $0.01-\\mathrm{M} \\mathrm{HCl}$ solution, the buffered solution has not detectably changed its pH but the unbuffered solution has become acidic, as indicated by the change in color of the methyl orange, which turns red at a pH of about 4. (credit: modification of work by Mark Ott)"}
{"id": 4178, "contents": "1558. How Buffers Work - \nTo illustrate the function of a buffer solution, consider a mixture of roughly equal amounts of acetic acid and sodium acetate. The presence of a weak conjugate acid-base pair in the solution imparts the ability to neutralize modest amounts of added strong acid or base. For example, strong base added to this solution will neutralize hydronium ion, causing the acetic acid ionization equilibrium to shift to the right and generate additional amounts of the weak conjugate base (acetate ion):\n\n$$\n\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}(a q)\n$$\n\nLikewise, strong acid added to this buffer solution will shift the above ionization equilibrium left, producing additional amounts of the weak conjugate acid (acetic acid). Figure 14.15 provides a graphical illustration of the changes in conjugate-partner concentration that occur in this buffer solution when strong acid and base are added. The buffering action of the solution is essentially a result of the added strong acid and base being converted to the weak acid and base that make up the buffer's conjugate pair. The weaker acid and base undergo only slight ionization, as compared with the complete ionization of the strong acid and base, and the solution pH , therefore, changes much less drastically than it would in an unbuffered solution.\n\n$$\n\\mathrm{CH}_{3} \\mathrm{COOH}(a q)+\\mathrm{H}_{2} \\mathrm{O}(I) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{CH}_{3} \\mathrm{COO}^{-}(a q)\n$$\n\n\n\nFIGURE 14.15 Buffering action in a mixture of acetic acid and acetate salt."}
{"id": 4179, "contents": "1560. pH Changes in Buffered and Unbuffered Solutions - \nAcetate buffers are used in biochemical studies of enzymes and other chemical components of cells to prevent pH changes that might affect the biochemical activity of these compounds.\n(a) Calculate the pH of an acetate buffer that is a mixture with 0.10 M acetic acid and 0.10 M sodium acetate.\n(b) Calculate the pH after 1.0 mL of 0.10 NaOH is added to 100 mL of this buffer.\n(c) For comparison, calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of a solution of an unbuffered solution with a pH of 4.74."}
{"id": 4180, "contents": "1561. Solution - \n(a) Following the ICE approach to this equilibrium calculation yields the following:\n\n\nSubstituting the equilibrium concentration terms into the $K_{\\mathrm{a}}$ expression, assuming $x \\ll 0.10$, and solving the simplified equation for $x$ yields\n\n$$\n\\begin{gathered}\nx=1.8 \\times 10^{-5} M \\\\\n{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=0+x=1.8 \\times 10^{-5} M} \\\\\n\\mathrm{pH}=-\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=-\\log \\left(1.8 \\times 10^{-5}\\right) \\\\\n=4.74\n\\end{gathered}\n$$\n\n(b) Calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of this buffer.\n\nAdding strong base will neutralize some of the acetic acid, yielding the conjugate base acetate ion. Compute the new concentrations of these two buffer components, then repeat the equilibrium calculation of part (a) using these new concentrations.\n\n$$\n0.0010 \\mathrm{~L} \\times\\left(\\frac{0.10 \\mathrm{~mol} \\mathrm{NaOH}}{1 \\mathrm{~L}}\\right)=1.0 \\times 10^{-4} \\mathrm{~mol} \\mathrm{NaOH}\n$$\n\nThe initial molar amount of acetic acid is\n\n$$\n0.100 \\mathrm{~L} \\times\\left(\\frac{0.100 \\mathrm{~mol} \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}}{1 \\mathrm{\\leftarrow}}\\right)=1.00 \\times 10^{-2} \\mathrm{~mol} \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\n$$\n\nThe amount of acetic acid remaining after some is neutralized by the added base is"}
{"id": 4181, "contents": "1561. Solution - \nThe amount of acetic acid remaining after some is neutralized by the added base is\n\n$$\n\\left(1.0 \\times 10^{-2}\\right)-\\left(0.01 \\times 10^{-2}\\right)=0.99 \\times 10^{-2} \\mathrm{~mol} \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\n$$\n\nThe newly formed acetate ion, along with the initially present acetate, gives a final acetate concentration of\n\n$$\n\\left(1.0 \\times 10^{-2}\\right)+\\left(0.01 \\times 10^{-2}\\right)=1.01 \\times 10^{-2} \\mathrm{~mol} \\mathrm{NaCH}_{3} \\mathrm{CO}_{2}\n$$\n\nCompute molar concentrations for the two buffer components:\n\n$$\n\\begin{aligned}\n{\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right] } & =\\frac{9.9 \\times 10^{-3} \\mathrm{~mol}}{0.101 \\mathrm{~L}}=0.098 \\mathrm{M} \\\\\n{\\left[\\mathrm{NaCH}_{3} \\mathrm{CO}_{2}\\right] } & =\\frac{1.01 \\times 10^{-2} \\mathrm{~mol}}{0.101 \\mathrm{~L}}=0.100 \\mathrm{M}\n\\end{aligned}\n$$\n\nUsing these concentrations, the pH of the solution may be computed as in part (a) above, yielding $\\mathrm{pH}=4.75$ (only slightly different from that prior to adding the strong base).\n(c) For comparison, calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of a solution of an unbuffered solution with a pH of 4.74 .\n\nThe amount of hydronium ion initially present in the solution is"}
{"id": 4182, "contents": "1561. Solution - \nThe amount of hydronium ion initially present in the solution is\n\n$$\n\\begin{gathered}\n{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=10^{-4.74}=1.8 \\times 10^{-5} M} \\\\\n\\mathrm{~mol} \\mathrm{H}_{3} \\mathrm{O}^{+}=(0.100 L)\\left(1.8 \\times 10^{-5} M\\right)=1.8 \\times 10^{-6} \\mathrm{~mol} \\mathrm{H}_{3} \\mathrm{O}^{+}\n\\end{gathered}\n$$\n\nThe amount of hydroxide ion added to the solution is\n\n$$\n\\mathrm{mol} \\mathrm{OH}^{-}=(0.0010 L)(0.10 M)=1.0 \\times 10^{-4} \\mathrm{~mol} \\mathrm{OH}^{-}\n$$\n\nThe added hydroxide will neutralize hydronium ion via the reaction\n\n$$\n\\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{OH}^{-}(a q) \\leftrightharpoons 2 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nThe 1:1 stoichiometry of this reaction shows that an excess of hydroxide has been added (greater molar amount than the initially present hydronium ion).\n\nThe amount of hydroxide ion remaining is\n\n$$\n1.0 \\times 10^{-4} \\mathrm{~mol}-1.8 \\times 10^{-6} \\mathrm{~mol}=9.8 \\times 10^{-5} \\mathrm{~mol} \\mathrm{OH}^{-}\n$$\n\ncorresponding to a hydroxide molarity of\n\n$$\n9.8 \\times 10^{-5} \\mathrm{~mol} \\mathrm{OH}^{-} / 0.101 L=9.7 \\times 10^{-4} M\n$$\n\nThe pH of the solution is then calculated to be\n\n$$\n\\mathrm{pH}=14.00-\\mathrm{pOH}=14.00--\\log \\left(9.7 \\times 10^{-4}\\right)=10.99\n$$"}
{"id": 4183, "contents": "1561. Solution - \n$$\n\\mathrm{pH}=14.00-\\mathrm{pOH}=14.00--\\log \\left(9.7 \\times 10^{-4}\\right)=10.99\n$$\n\nIn this unbuffered solution, addition of the base results in a significant rise in pH (from 4.74 to 10.99) compared with the very slight increase observed for the buffer solution in part (b) (from 4.74 to 4.75)."}
{"id": 4184, "contents": "1562. Check Your Learning - \nShow that adding 1.0 mL of 0.10 M HCl changes the pH of 100 mL of a $1.8 \\times 10^{-5} \\mathrm{M} \\mathrm{HCl}$ solution from 4.74 to 3.00."}
{"id": 4185, "contents": "1563. Answer: - \nInitial pH of $1.8 \\times 10^{-5} \\mathrm{M} \\mathrm{HCl} ; \\mathrm{pH}=-\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=-\\log \\left[1.8 \\times 10^{-5}\\right]=4.74$\nMoles of $\\mathrm{H}_{3} \\mathrm{O}^{+}$in $100 \\mathrm{~mL} 1.8 \\times 10^{-5} \\mathrm{M} \\mathrm{HCl} ; 1.8 \\times 10^{-5} \\mathrm{moles} / \\mathrm{L} \\times 0.100 \\mathrm{~L}=1.8 \\times 10^{-6}$\nMoles of $\\mathrm{H}_{3} \\mathrm{O}^{+}$added by addition of 1.0 mL of 0.10 M HCl : $0.10 \\mathrm{moles} / \\mathrm{L} \\times 0.0010 \\mathrm{~L}=1.0 \\times 10^{-4} \\mathrm{moles}$; final pH after addition of 1.0 mL of 0.10 M HCl :\n\n$$\n\\mathrm{pH}=-\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=-\\log \\left(\\frac{\\text { total moles } \\mathrm{H}_{3} \\mathrm{O}^{+}}{\\text {total volume }}\\right)=-\\log \\left(\\frac{1.0 \\times 10^{-4} \\mathrm{~mol}+1.8 \\times 10^{-6} \\mathrm{~mol}}{101 \\mathrm{~mL}\\left(\\frac{1 \\mathrm{~L}}{1000 \\mathrm{~mL}}\\right)}\\right)=3.00\n$$"}
{"id": 4186, "contents": "1564. Buffer Capacity - \nBuffer solutions do not have an unlimited capacity to keep the pH relatively constant (Figure 14.16). Instead, the ability of a buffer solution to resist changes in pH relies on the presence of appreciable amounts of its conjugate weak acid-base pair. When enough strong acid or base is added to substantially lower the concentration of either member of the buffer pair, the buffering action within the solution is compromised.\n\n\nFIGURE 14.16 The indicator color (methyl orange) shows that a small amount of acid added to a buffered solution of pH 8 (beaker on the left) has little affect on the buffered system (middle beaker). However, a large amount of acid exhausts the buffering capacity of the solution and the pH changes dramatically (beaker on the right). (credit: modification of work by Mark Ott)\n\nThe buffer capacity is the amount of acid or base that can be added to a given volume of a buffer solution before the pH changes significantly, usually by one unit. Buffer capacity depends on the amounts of the weak\nacid and its conjugate base that are in a buffer mixture. For example, 1 L of a solution that is 1.0 M in acetic acid and 1.0 M in sodium acetate has a greater buffer capacity than 1 L of a solution that is 0.10 M in acetic acid and 0.10 M in sodium acetate even though both solutions have the same pH . The first solution has more buffer capacity because it contains more acetic acid and acetate ion."}
{"id": 4187, "contents": "1565. Selection of Suitable Buffer Mixtures - \nThere are two useful rules of thumb for selecting buffer mixtures:\n\n1. A good buffer mixture should have about equal concentrations of both of its components. A buffer solution has generally lost its usefulness when one component of the buffer pair is less than about $10 \\%$ of the other. Figure 14.17 shows how pH changes for an acetic acid-acetate ion buffer as base is added. The initial pH is 4.74. A change of 1 pH unit occurs when the acetic acid concentration is reduced to $11 \\%$ of the acetate ion concentration.\n\n\nFIGURE 14.17 Change in pH as an increasing amount of a $0.10-\\mathrm{M} \\mathrm{NaOH}$ solution is added to 100 mL of a buffer solution in which, initially, $\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right]=0.10 \\mathrm{M}$ and $\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}\\right]=0.10 \\mathrm{M}$. Note the greatly diminished buffering action occurring after the buffer capacity has been reached, resulting in drastic rises in pH on adding more strong base.\n2. Weak acids and their salts are better as buffers for pHs less than 7; weak bases and their salts are better as buffers for pHs greater than 7.\n\nBlood is an important example of a buffered solution, with the principal acid and ion responsible for the buffering action being carbonic acid, $\\mathrm{H}_{2} \\mathrm{CO}_{3}$, and the bicarbonate ion, $\\mathrm{HCO}_{3}{ }^{-}$. When a hydronium ion is introduced to the blood stream, it is removed primarily by the reaction:\n\n$$\n\\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{HCO}_{3}^{-}(a q) \\longrightarrow \\mathrm{H}_{2} \\mathrm{CO}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nAn added hydroxide ion is removed by the reaction:"}
{"id": 4188, "contents": "1565. Selection of Suitable Buffer Mixtures - \nAn added hydroxide ion is removed by the reaction:\n\n$$\n\\mathrm{OH}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{CO}_{3}(a q) \\longrightarrow \\mathrm{HCO}_{3}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nThe added strong acid or base is thus effectively converted to the much weaker acid or base of the buffer pair $\\left(\\mathrm{H}_{3} \\mathrm{O}^{+}\\right.$is converted to $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ and $\\mathrm{OH}^{-}$is converted to $\\mathrm{HCO}_{3}{ }^{-}$). The pH of human blood thus remains very near the value determined by the buffer pairs pKa , in this case, 7.35. Normal variations in blood pH are usually less than 0.1 , and pH changes of 0.4 or greater are likely to be fatal."}
{"id": 4189, "contents": "1566. The Henderson-Hasselbalch Equation - \nThe ionization-constant expression for a solution of a weak acid can be written as:\n\n$$\nK_{\\mathrm{a}}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]}\n$$\n\nRearranging to solve for $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]$yields:\n\n$$\n\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=K_{\\mathrm{a}} \\times \\frac{[\\mathrm{HA}]}{\\left[\\mathrm{A}^{-}\\right]}\n$$\n\nTaking the negative logarithm of both sides of this equation gives\n\n$$\n-\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=-\\log K_{\\mathrm{a}}-\\log \\frac{[\\mathrm{HA}]}{\\left[\\mathrm{A}^{-}\\right]}\n$$\n\nwhich can be written as\n\n$$\n\\mathrm{pH}=\\mathrm{p} K_{\\mathrm{a}}+\\log \\frac{\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]}\n$$\n\nwhere $\\mathrm{p} K_{\\mathrm{a}}$ is the negative of the logarithm of the ionization constant of the weak acid ( $\\mathrm{p} K_{\\mathrm{a}}=-\\log K_{\\mathrm{a}}$ ). This equation relates the pH , the ionization constant of a weak acid, and the concentrations of the weak conjugate acid-base pair in a buffered solution. Scientists often use this expression, called the Henderson-Hasselbalch equation, to calculate the pH of buffer solutions. It is important to note that the \" $x$ is small\" assumption must be valid to use this equation."}
{"id": 4190, "contents": "1568. Lawrence Joseph Henderson and Karl Albert Hasselbalch - \nLawrence Joseph Henderson (1878-1942) was an American physician, biochemist and physiologist, to name only a few of his many pursuits. He obtained a medical degree from Harvard and then spent 2 years studying in Strasbourg, then a part of Germany, before returning to take a lecturer position at Harvard. He eventually became a professor at Harvard and worked there his entire life. He discovered that the acid-base balance in human blood is regulated by a buffer system formed by the dissolved carbon dioxide in blood. He wrote an equation in 1908 to describe the carbonic acid-carbonate buffer system in blood. Henderson was broadly knowledgeable; in addition to his important research on the physiology of blood, he also wrote on the adaptations of organisms and their fit with their environments, on sociology and on university education. He also founded the Fatigue Laboratory, at the Harvard Business School, which examined human physiology with specific focus on work in industry, exercise, and nutrition.\n\nIn 1916, Karl Albert Hasselbalch (1874-1962), a Danish physician and chemist, shared authorship in a paper with Christian Bohr in 1904 that described the Bohr effect, which showed that the ability of hemoglobin in the blood to bind with oxygen was inversely related to the acidity of the blood and the concentration of carbon dioxide. The pH scale was introduced in 1909 by another Dane, S\u00f8rensen, and in 1912, Hasselbalch published measurements of the pH of blood. In 1916, Hasselbalch expressed Henderson's equation in logarithmic terms, consistent with the logarithmic scale of pH , and thus the Henderson-Hasselbalch equation was born."}
{"id": 4191, "contents": "1570. Medicine: The Buffer System in Blood - \nThe normal pH of human blood is about 7.4. The carbonate buffer system in the blood uses the following equilibrium reaction:\n\n$$\n\\mathrm{CO}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{CO}_{3}(a q) \\rightleftharpoons \\mathrm{HCO}_{3}^{-}(a q)+\\mathrm{H}_{3} \\mathrm{O}^{+}(a q)\n$$\n\nThe concentration of carbonic acid, $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ is approximately 0.0012 M , and the concentration of the hydrogen carbonate ion, $\\mathrm{HCO}_{3}{ }^{-}$, is around 0.024 M . Using the Henderson-Hasselbalch equation and the $\\mathrm{p} K_{\\mathrm{a}}$ of carbonic acid at body temperature, we can calculate the pH of blood:\n\n$$\n\\mathrm{pH}=\\mathrm{p} K_{\\mathrm{a}}+\\log \\frac{[\\text { base }]}{[\\text { acid }]}=6.4+\\log \\frac{0.024}{0.0012}=7.7\n$$\n\nThe fact that the $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ concentration is significantly lower than that of the $\\mathrm{HCO}_{3}{ }^{-}$ion may seem unusual, but this imbalance is due to the fact that most of the by-products of our metabolism that enter our bloodstream are acidic. Therefore, there must be a larger proportion of base than acid, so that the capacity of the buffer will not be exceeded."}
{"id": 4192, "contents": "1570. Medicine: The Buffer System in Blood - \nLactic acid is produced in our muscles when we exercise. As the lactic acid enters the bloodstream, it is neutralized by the $\\mathrm{HCO}_{3}{ }^{-}$ion, producing $\\mathrm{H}_{2} \\mathrm{CO}_{3}$. An enzyme then accelerates the breakdown of the excess carbonic acid to carbon dioxide and water, which can be eliminated by breathing. In fact, in addition to the regulating effects of the carbonate buffering system on the pH of blood, the body uses breathing to regulate blood pH . If the pH of the blood decreases too far, an increase in breathing removes $\\mathrm{CO}_{2}$ from the blood through the lungs driving the equilibrium reaction such that $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]$is lowered. If the blood is too alkaline, a lower breath rate increases $\\mathrm{CO}_{2}$ concentration in the blood, driving the equilibrium reaction the other way, increasing $\\left[\\mathrm{H}^{+}\\right]$and restoring an appropriate pH ."}
{"id": 4193, "contents": "1571. LINK TO LEARNING - \nView information (http://openstax.org/1/16BufferSystem) on the buffer system encountered in natural waters."}
{"id": 4194, "contents": "1572. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Interpret titration curves for strong and weak acid-base systems\n- Compute sample pH at important stages of a titration\n- Explain the function of acid-base indicators\n\nAs seen in the chapter on the stoichiometry of chemical reactions, titrations can be used to quantitatively analyze solutions for their acid or base concentrations. In this section, we will explore the underlying chemical equilibria that make acid-base titrimetry a useful analytical technique."}
{"id": 4195, "contents": "1573. Titration Curves - \nA titration curve is a plot of some solution property versus the amount of added titrant. For acid-base titrations, solution pH is a useful property to monitor because it varies predictably with the solution composition and, therefore, may be used to monitor the titration's progress and detect its end point. The following example exercise demonstrates the computation of pH for a titration solution after additions of several specified titrant volumes. The first example involves a strong acid titration that requires only stoichiometric calculations to derive the solution pH . The second example addresses a weak acid titration requiring equilibrium calculations."}
{"id": 4196, "contents": "1575. Calculating pH for Titration Solutions: Strong Acid/Strong Base - \nA titration is carried out for 25.00 mL of 0.100 M HCl (strong acid) with 0.100 M of a strong base NaOH (the titration curve is shown in Figure 14.18). Calculate the pH at these volumes of added base solution:\n(a) 0.00 mL\n(b) 12.50 mL\n(c) 25.00 mL\n(d) 37.50 mL"}
{"id": 4197, "contents": "1576. Solution - \n(a) Titrant volume $=0 \\mathrm{~mL}$. The solution pH is due to the acid ionization of HCl . Because this is a strong acid, the ionization is complete and the hydronium ion molarity is 0.100 M . The pH of the solution is then\n\n$$\n\\mathrm{pH}=-\\log (0.100)=1.000\n$$\n\n(b) Titrant volume $=12.50 \\mathrm{~mL}$. Since the acid sample and the base titrant are both monoprotic and equally concentrated, this titrant addition involves less than a stoichiometric amount of base, and so it is completely consumed by reaction with the excess acid in the sample. The concentration of acid remaining is computed by subtracting the consumed amount from the intial amount and then dividing by the solution volume:\n\n$$\n\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=\\frac{\\mathrm{n}\\left(\\mathrm{H}^{+}\\right)}{V}=\\frac{0.002500 \\mathrm{~mol} \\times\\left(\\frac{1000 \\mathrm{~mL}}{1 \\mathrm{~L}}\\right)-0.100 M \\times 12.50 \\mathrm{~mL}}{25.00 \\mathrm{~mL}+12.50 \\mathrm{~mL}}=0.0333 \\mathrm{M}\n$$\n\n(c) Titrant volume $=25.00 \\mathrm{~mL}$. This titrant addition involves a stoichiometric amount of base (the equivalence point), and so only products of the neutralization reaction are in solution (water and NaCl ). Neither the cation nor the anion of this salt undergo acid-base ionization; the only process generating hydronium ions is the autoprotolysis of water. The solution is neutral, having a $\\mathrm{pH}=7.00$.\n(d) Titrant volume $=37.50 \\mathrm{~mL}$. This involves the addition of titrant in excess of the equivalence point. The solution pH is then calculated using the concentration of hydroxide ion:"}
{"id": 4198, "contents": "1576. Solution - \n$$\n\\begin{gathered}\n\\mathrm{n}\\left(\\mathrm{OH}^{-}\\right)_{0}>\\mathrm{n}\\left(\\mathrm{H}^{+}\\right)_{0} \\\\\n{\\left[\\mathrm{OH}^{-}\\right]=\\frac{\\mathrm{n}\\left(\\mathrm{OH}^{-}\\right)}{V}=\\frac{0.100 M \\times 37.50 \\mathrm{~mL}-0.002500 \\mathrm{~mol} \\times\\left(\\frac{1000 \\mathrm{~mL}}{1 \\mathrm{~L}}\\right)}{25.00 \\mathrm{~mL}+37.50 \\mathrm{~mL}}=0.0200 \\mathrm{M}} \\\\\n\\mathrm{pH}=14-\\mathrm{pOH}=14+\\log \\left(\\left[\\mathrm{OH}^{-}\\right]\\right)=14+\\log (0.0200)=12.30\n\\end{gathered}\n$$"}
{"id": 4199, "contents": "1577. Check Your Learning - \nCalculate the pH for the strong acid/strong base titration between 50.0 mL of $0.100 \\mathrm{M} \\mathrm{HNO}_{3}(\\mathrm{aq})$ and 0.200 M NaOH (titrant) at the listed volumes of added base: $0.00 \\mathrm{~mL}, 15.0 \\mathrm{~mL}, 25.0 \\mathrm{~mL}$, and 40.0 mL ."}
{"id": 4200, "contents": "1578. Answer: - \n0.00: 1.000; 15.0: 1.5111; 25.0: 7; 40.0: 12.523"}
{"id": 4201, "contents": "1580. Titration of a Weak Acid with a Strong Base - \nConsider the titration of 25.00 mL of $0.100 \\mathrm{MCH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$ with 0.100 M NaOH . The reaction can be represented as:\n\n$$\n\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}+\\mathrm{OH}^{-} \\longrightarrow \\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}+\\mathrm{H}_{2} \\mathrm{O}\n$$\n\nCalculate the pH of the titration solution after the addition of the following volumes of NaOH titrant:\n(a) 0.00 mL\n(b) 25.00 mL\n(c) 12.50 mL\n(d) 37.50 mL"}
{"id": 4202, "contents": "1581. Solution - \n(a) The initial pH is computed for the acetic acid solution in the usual ICE approach:\n\n$$\n\\begin{aligned}\n& K_{\\mathrm{a}}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2}-\\right]}{\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right]} \\approx \\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]^{2}}{\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right]_{0}}, \\text { and } \\\\\n& {\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=\\sqrt{K_{a} \\times\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right]}=\\sqrt{1.8 \\times 10^{-5} \\times 0.100}=1.3 \\times 10^{-3}} \\\\\n& \\quad \\mathrm{pH}=-\\log \\left(1.3 \\times 10^{-3}\\right)=2.87\n\\end{aligned}\n$$\n\n(b) The acid and titrant are both monoprotic and the sample and titrant solutions are equally concentrated; thus, this volume of titrant represents the equivalence point. Unlike the strong-acid example above, however, the reaction mixture in this case contains a weak conjugate base (acetate ion). The solution pH is computed considering the base ionization of acetate, which is present at a concentration of\n\n$$\n\\frac{0.00250 \\mathrm{~mol}}{0.0500 \\mathrm{~L}}=0.0500 \\mathrm{MCH}_{3} \\mathrm{CO}_{2}^{-}\n$$\n\nBase ionization of acetate is represented by the equation"}
{"id": 4203, "contents": "1581. Solution - \nBase ionization of acetate is represented by the equation\n\n$$\n\\begin{gathered}\n\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(a q)+\\mathrm{OH}^{-}(a q) \\\\\nK_{\\mathrm{b}}=\\frac{\\left[\\mathrm{H}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]}{K_{\\mathrm{a}}}=\\frac{K_{\\mathrm{w}}}{K_{\\mathrm{a}}}=\\frac{1.0 \\times 10^{-14}}{1.8 \\times 10^{-5}}=5.6 \\times 10^{-10}\n\\end{gathered}\n$$\n\nAssuming $x \\ll 0.0500$, the pH may be calculated via the usual ICE approach: $K_{\\mathrm{b}}=\\frac{x^{2}}{0.0500 \\mathrm{M}}$\n\n$$\n\\begin{aligned}\nx & =\\left[\\mathrm{OH}^{-}\\right]=5.3 \\times 10^{-6} \\\\\n\\mathrm{pOH} & =-\\log \\left(5.3 \\times 10^{-6}\\right)=5.28 \\\\\n\\mathrm{pH} & =14.00-5.28=8.72\n\\end{aligned}\n$$\n\nNote that the pH at the equivalence point of this titration is significantly greater than 7, as expected when titrating a weak acid with a strong base.\n(c) Titrant volume $=12.50 \\mathrm{~mL}$. This volume represents one-half of the stoichiometric amount of titrant, and so one-half of the acetic acid has been neutralized to yield an equivalent amount of acetate ion. The concentrations of these conjugate acid-base partners, therefore, are equal. A convenient approach to computing the pH is use of the Henderson-Hasselbalch equation:"}
{"id": 4204, "contents": "1581. Solution - \n$$\n\\begin{gathered}\n\\mathrm{pH}=p K_{\\mathrm{a}}+\\log \\frac{[\\text { Base }]}{[\\text { Acid }]}=-\\log \\left(K_{\\mathrm{a}}\\right)+\\log \\frac{\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}\\right]}{\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right]}=-\\log \\left(1.8 \\times 10^{-5}\\right)+\\log (1) \\\\\n\\mathrm{pH}=-\\log \\left(1.8 \\times 10^{-5}\\right)=4.74\n\\end{gathered}\n$$\n\n( $\\mathrm{pH}=\\mathrm{p} K_{\\mathrm{a}}$ at the half-equivalence point in a titration of a weak acid)\n(d) Titrant volume $=37.50 \\mathrm{~mL}$. This volume represents a stoichiometric excess of titrant, and a reaction solution containing both the titration product, acetate ion, and the excess strong titrant. In such solutions, the solution pH is determined primarily by the amount of excess strong base:\n\n$$\n\\begin{gathered}\n{\\left[\\mathrm{OH}^{-}\\right]=\\frac{(0.003750 \\mathrm{~mol}-0.00250 \\mathrm{~mol})}{0.06250 \\mathrm{~L}}=2.00 \\times 10^{-2} M} \\\\\n\\mathrm{pOH}=-\\log \\left(2.00 \\times 10^{-2}\\right)=1.70, \\text { and } \\mathrm{pH}=14.00-1.70=12.30\n\\end{gathered}\n$$"}
{"id": 4205, "contents": "1582. Check Your Learning - \nCalculate the pH for the weak acid/strong base titration between 50.0 mL of $0.100 \\mathrm{MHCOOH}(\\mathrm{aq})$ (formic acid) and 0.200 M NaOH (titrant) at the listed volumes of added base: $0.00 \\mathrm{~mL}, 15.0 \\mathrm{~mL}, 25.0 \\mathrm{~mL}$, and 30.0 mL ."}
{"id": 4206, "contents": "1583. Answer: - \n$0.00 \\mathrm{~mL}: 2.37 ; 15.0 \\mathrm{~mL}: 3.92 ; 25.00 \\mathrm{~mL}: 8.29 ; 30.0 \\mathrm{~mL}: 12.097$\n\nPerforming additional calculations similar to those in the preceding example permits a more full assessment of titration curves. A summary of $\\mathrm{pH} /$ volume data pairs for the strong and weak acid titrations is provided in Table 14.2 and plotted as titration curves in Figure 14.18. A comparison of these two curves illustrates several important concepts that are best addressed by identifying the four stages of a titration:\ninitial state (added titrant volume $=0 \\mathrm{~mL}$ ): pH is determined by the acid being titrated; because the two acid samples are equally concentrated, the weak acid will exhibit a greater initial pH\npre-equivalence point ( $0 \\mathrm{~mL}<V<25 \\mathrm{~mL}$ ): solution pH increases gradually and the acid is consumed by reaction with added titrant; composition includes unreacted acid and the reaction product, its conjugate base equivalence point ( $V=25 \\mathrm{~mL}$ ): a drastic rise in pH is observed as the solution composition transitions from acidic to either neutral (for the strong acid sample) or basic (for the weak acid sample), with pH determined by ionization of the conjugate base of the acid\npostequivalence point ( $V>25 \\mathrm{~mL}$ ): pH is determined by the amount of excess strong base titrant added; since both samples are titrated with the same titrant, both titration curves appear similar at this stage.\npH Values in the Titrations of a Strong Acid and of a Weak Acid"}
{"id": 4207, "contents": "1583. Answer: - \n| Volume of 0.100 M NaOH Added (mL) | Moles of NaOH Added | $\\begin{aligned} & \\mathrm{pH} \\text { Values } 0.100 \\mathrm{M} \\\\ & \\mathrm{HCl}^{1} \\end{aligned}$ | pH Values 0.100 M $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}^{2}$ |\n| :---: | :---: | :---: | :---: |\n| 0.0 | 0.0 | 1.00 | 2.87 |\n| 5.0 | 0.00050 | 1.18 | 4.14 |\n| 10.0 | 0.00100 | 1.37 | 4.57 |\n| 15.0 | 0.00150 | 1.60 | 4.92 |\n| 20.0 | 0.00200 | 1.95 | 5.35 |\n| 22.0 | 0.00220 | 2.20 | 5.61 |\n| 24.0 | 0.00240 | 2.69 | 6.13 |\n| 24.5 | 0.00245 | 3.00 | 6.44 |\n| 24.9 | 0.00249 | 3.70 | 7.14 |\n| 25.0 | 0.00250 | 7.00 | 8.72 |\n| 25.1 | 0.00251 | 10.30 | 10.30 |\n| 25.5 | 0.00255 | 11.00 | 11.00 |\n| 26.0 | 0.00260 | 11.29 | 11.29 |"}
{"id": 4208, "contents": "1583. Answer: - \n[^9]| Volume of 0.100 M NaOH <br> Added (mL) | Moles of NaOH <br> Added | pH Values 0.100 M <br> $\\mathrm{HCl}^{1}$ | pH Values 0.100 M <br> $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}^{2}$ |\n| :--- | :--- | :--- | :--- |\n| 28.0 | 0.00280 | 11.75 | 11.75 |\n| 30.0 | 0.00300 | 11.96 | 11.96 |\n| 35.0 | 0.00350 | 12.22 | 12.22 |\n| 40.0 | 0.00400 | 12.36 | 12.36 |\n| 45.0 | 0.00450 | 12.46 | 12.46 |\n| 50.0 | 0.00500 | 12.52 | 12.52 |\n\nTABLE 14.2\n\n\nFIGURE 14.18 (a) The titration curve for the titration of 25.00 mL of 0.100 M HCl (strong acid) with 0.100 M NaOH (strong base) has an equivalence point of 7.00 pH . (b) The titration curve for the titration of 25.00 mL of 0.100 M acetic acid (weak acid) with 0.100 M NaOH (strong base) has an equivalence point of 8.72 pH ."}
{"id": 4209, "contents": "1584. Acid-Base Indicators - \nCertain organic substances change color in dilute solution when the hydronium ion concentration reaches a particular value. For example, phenolphthalein is a colorless substance in any aqueous solution with a hydronium ion concentration greater than $5.0 \\times 10^{-9} M(\\mathrm{pH}<8.3)$. In more basic solutions where the hydronium ion concentration is less than $5.0 \\times 10^{-9} M(\\mathrm{pH}>8.3)$, it is red or pink. Substances such as phenolphthalein, which can be used to determine the pH of a solution, are called acid-base indicators. Acidbase indicators are either weak organic acids or weak organic bases.\n\nThe equilibrium in a solution of the acid-base indicator methyl orange, a weak acid, can be represented by an equation in which we use HIn as a simple representation for the complex methyl orange molecule:\n\n$$\n\\begin{aligned}\n& \\underset{\\text { red }}{\\mathrm{HIn}(a q)}+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{In}^{-}(a q) \\\\\n& \\text { yellow } \\\\\n& \\quad K_{a}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{In}^{-}\\right]}{[\\mathrm{HIn}]}=4.0 \\times 10^{-4}\n\\end{aligned}\n$$\n\nThe anion of methyl orange, $\\mathrm{In}^{-}$, is yellow, and the nonionized form, HIn, is red. When we add acid to a solution of methyl orange, the increased hydronium ion concentration shifts the equilibrium toward the nonionized red form, in accordance with Le Ch\u00e2telier's principle. If we add base, we shift the equilibrium towards the yellow form. This behavior is completely analogous to the action of buffers."}
{"id": 4210, "contents": "1584. Acid-Base Indicators - \nThe perceived color of an indicator solution is determined by the ratio of the concentrations of the two species $\\mathrm{In}^{-}$and HIn. If most of the indicator (typically about $60-90 \\%$ or more) is present as $\\mathrm{In}^{-}$, the perceived color of the solution is yellow. If most is present as HIn, then the solution color appears red. The HendersonHasselbalch equation is useful for understanding the relationship between the pH of an indicator solution and its composition (thus, perceived color):\n\n$$\n\\mathrm{pH}=\\mathrm{p} K \\mathrm{a}+\\log \\left(\\frac{\\left[\\mathrm{In}^{-}\\right]}{[\\mathrm{HIn}]}\\right)\n$$\n\nIn solutions where $\\mathrm{pH}>\\mathrm{p} K_{\\mathrm{a}}$, the logarithmic term must be positive, indicating an excess of the conjugate base form of the indicator (yellow solution). When $\\mathrm{pH}<\\mathrm{p} K_{\\mathrm{a}}$, the log term must be negative, indicating an excess of the conjugate acid (red solution). When the solution pH is close to the indicator pKa , appreciable amounts of both conjugate partners are present, and the solution color is that of an additive combination of each (yellow and red, yielding orange). The color change interval (or pH interval) for an acid-base indicator is defined as the range of pH values over which a change in color is observed, and for most indicators this range is approximately $\\mathrm{p} K_{\\mathrm{a}} \\pm 1$.\n\nThere are many different acid-base indicators that cover a wide range of pH values and can be used to determine the approximate pH of an unknown solution by a process of elimination. Universal indicators and pH paper contain a mixture of indicators and exhibit different colors at different pHs. Figure 14.19 presents several indicators, their colors, and their color-change intervals.\n\n\nFIGURE 14.19 This chart illustrates the color change intervals for several acid-base indicators."}
{"id": 4211, "contents": "1584. Acid-Base Indicators - \nFIGURE 14.19 This chart illustrates the color change intervals for several acid-base indicators.\n\n\nFIGURE 14.20 Titration curves for strong and weak acids illustrating the proper choice of acid-base indicator. Any of the three indicators will exhibit a reasonably sharp color change at the equivalence point of the strong acid titration, but only phenolphthalein is suitable for use in the weak acid titration.\n\nThe titration curves shown in Figure 14.20 illustrate the choice of a suitable indicator for specific titrations. In the strong acid titration, use of any of the three indicators should yield reasonably sharp color changes and accurate end point determinations. For this titration, the solution pH reaches the lower limit of the methyl orange color change interval after addition of $\\sim 24 \\mathrm{~mL}$ of titrant, at which point the initially red solution would begin to appear orange. When 25 mL of titrant has been added (the equivalence point), the pH is well above the upper limit and the solution will appear yellow. The titration's end point may then be estimated as the volume of titrant that yields a distinct orange-to-yellow color change. This color change would be challenging for most human eyes to precisely discern. More-accurate estimates of the titration end point are possible using either litmus or phenolphthalein, both of which exhibit color change intervals that are encompassed by the steep rise in pH that occurs around the 25.00 mL equivalence point.\n\nThe weak acid titration curve in Figure 14.20 shows that only one of the three indicators is suitable for end point detection. If methyl orange is used in this titration, the solution will undergo a gradual red-to-orange-toyellow color change over a relatively large volume interval ( $0-6 \\mathrm{~mL}$ ), completing the color change well before the equivalence point ( 25 mL ) has been reached. Use of litmus would show a color change that begins after adding $7-8 \\mathrm{~mL}$ of titrant and ends just before the equivalence point. Phenolphthalein, on the other hand, exhibits a color change interval that nicely brackets the abrupt change in pH occurring at the titration's equivalence point. A sharp color change from colorless to pink will be observed within a very small volume interval around the equivalence point."}
{"id": 4212, "contents": "1585. Key Terms - \nacid ionization reaction involving the transfer of a proton from an acid to water, yielding hydronium ions and the conjugate base of the acid\nacid ionization constant ( $\\boldsymbol{K}_{\\mathrm{a}}$ ) equilibrium constant for an acid ionization reaction\nacid-base indicator weak acid or base whose conjugate partner imparts a different solution color; used in visual assessments of solution pH\nacidic a solution in which $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]>\\left[\\mathrm{OH}^{-}\\right]$\namphiprotic species that may either donate or accept a proton in a Bronsted-Lowry acid-base reaction\namphoteric species that can act as either an acid or a base\nautoionization reaction between identical species yielding ionic products; for water, this reaction involves transfer of protons to yield hydronium and hydroxide ions\nbase ionization reaction involving the transfer of a proton from water to a base, yielding hydroxide ions and the conjugate acid of the base\nbase ionization constant ( $\\boldsymbol{K}_{\\mathbf{b}}$ ) equilibrium constant for a base ionization reaction\nbasic a solution in which $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]<\\left[\\mathrm{OH}^{-}\\right]$\nBr\u00f8nsted-Lowry acid proton donor\nBr\u00f8nsted-Lowry base proton acceptor\nbuffer mixture of appreciable amounts of a weak acid-base pair the pH of a buffer resists change when small amounts of acid or base are added\nbuffer capacity amount of an acid or base that can be added to a volume of a buffer solution before its pH changes significantly (usually by one pH unit)\ncolor-change interval range in pH over which the color change of an indicator is observed\nconjugate acid substance formed when a base gains a proton\nconjugate base substance formed when an acid"}
{"id": 4213, "contents": "1586. Key Equations - \n$K_{\\mathrm{w}}=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]=1.0 \\times 10^{-14}\\left(\\right.$ at $\\left.25^{\\circ} \\mathrm{C}\\right)$\n$\\mathrm{pH}=-\\log \\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]$\n$\\mathrm{pOH}=-\\log \\left[\\mathrm{OH}^{-}\\right]$\n$\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=10^{-\\mathrm{pH}}$\n$\\left[\\mathrm{OH}^{-}\\right]=10^{-\\mathrm{pOH}}$\n$\\mathrm{pH}+\\mathrm{pOH}=\\mathrm{p} K_{\\mathrm{w}}=14.00$ at $25^{\\circ} \\mathrm{C}$\n$K_{\\mathrm{a}}=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]}$\n$K_{\\mathrm{b}}=\\frac{\\left[\\mathrm{HB}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]}{[\\mathrm{B}]}$\nloses a proton\ndiprotic acid acid containing two ionizable hydrogen atoms per molecule\ndiprotic base base capable of accepting two protons\nHenderson-Hasselbalch equation logarithmic version of the acid ionization constant expression, conveniently formatted for calculating the pH of buffer solutions\nion-product constant for water ( $\\boldsymbol{K}_{\\mathrm{w}}$ ) equilibrium constant for the autoionization of water\nleveling effect observation that acid-base strength of solutes in a given solvent is limited to that of the solvent's characteristic acid and base species (in water, hydronium and hydroxide ions, respectively)\nmonoprotic acid acid containing one ionizable hydrogen atom per molecule\nneutral describes a solution in which $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=$ [ $\\mathrm{OH}^{-}$]"}
{"id": 4214, "contents": "1586. Key Equations - \nmonoprotic acid acid containing one ionizable hydrogen atom per molecule\nneutral describes a solution in which $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=$ [ $\\mathrm{OH}^{-}$]\noxyacid ternary compound with acidic properties, molecules of which contain a central nonmetallic atom bonded to one or more O atoms, at least one of which is bonded to an ionizable H atom\npercent ionization ratio of the concentration of ionized acid to initial acid concentration expressed as a percentage\npH logarithmic measure of the concentration of hydronium ions in a solution\n$\\mathbf{p O H}$ logarithmic measure of the concentration of hydroxide ions in a solution\nstepwise ionization process in which a polyprotic acid is ionized by losing protons sequentially\ntitration curve plot of some sample property (such as pH ) versus volume of added titrant\ntriprotic acid acid that contains three ionizable hydrogen atoms per molecule\n$K_{\\mathrm{a}} \\times K_{\\mathrm{b}}=1.0 \\times 10^{-14}=K_{\\mathrm{w}}$\nPercent ionization $=\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]_{\\mathrm{eq}}}{[\\mathrm{HA}]_{0}} \\times 100$\n$\\mathrm{p} K_{\\mathrm{a}}=-\\log K_{\\mathrm{a}}$\n$\\mathrm{p} K_{\\mathrm{b}}=-\\log K_{\\mathrm{b}}$\n$\\mathrm{pH}=\\mathrm{p} K_{\\mathrm{a}}+\\log \\frac{\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]}$"}
{"id": 4215, "contents": "1587. Summary - 1587.1. Br\u00f8nsted-Lowry Acids and Bases\nA compound that can donate a proton (a hydrogen ion) to another compound is called a Br\u00f8nstedLowry acid. The compound that accepts the proton is called a Br\u00f8nsted-Lowry base. The species remaining after a Br\u00f8nsted-Lowry acid has lost a proton is the conjugate base of the acid. The species formed when a Br\u00f8nsted-Lowry base gains a proton is the conjugate acid of the base. Thus, an acid-base reaction occurs when a proton is transferred from an acid to a base, with formation of the conjugate base of the reactant acid and formation of the conjugate acid of the reactant base. Amphiprotic species can act as both proton donors and proton acceptors. Water is the most important amphiprotic species. It can form both the hydronium ion, $\\mathrm{H}_{3} \\mathrm{O}^{+}$, and the hydroxide ion, $\\mathrm{OH}^{-}$when it undergoes autoionization:\n\n$$\n2 \\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{OH}^{-}(a q)\n$$\n\nThe ion product of water, $K_{\\mathrm{w}}$ is the equilibrium constant for the autoionization reaction:\n\n$$\nK_{\\mathrm{w}}=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]=1.0 \\times 10^{-14} \\text { at } 25^{\\circ} \\mathrm{C}\n$$"}
{"id": 4216, "contents": "1588. 2 pH and pOH - \nConcentrations of hydronium and hydroxide ions in aqueous media are often represented as logarithmic pH and pOH values, respectively. At $25^{\\circ} \\mathrm{C}$, the autoprotolysis equilibrium for water requires the sum of pH and pOH to equal 14 for any aqueous solution. The relative concentrations of hydronium and hydroxide ion in a solution define its status as acidic $\\left(\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]>\\left[\\mathrm{OH}^{-}\\right]\\right)$, basic $\\left(\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]<\\left[\\mathrm{OH}^{-}\\right]\\right)$, or neutral $\\left(\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=\\left[\\mathrm{OH}^{-}\\right]\\right)$. At $25^{\\circ} \\mathrm{C}$, a pH $<7$ indicates an acidic solution, a $\\mathrm{pH}>7$ a basic solution, and a $\\mathrm{pH}=7$ a neutral solution."}
{"id": 4217, "contents": "1588. 2 pH and pOH - 1588.1. Relative Strengths of Acids and Bases\nThe relative strengths of acids and bases are reflected in the magnitudes of their ionization constants; the stronger the acid or base, the larger its ionization constant. A reciprocal relation exists\nbetween the strengths of a conjugate acid-base pair: the stronger the acid, the weaker its conjugate base. Water exerts a leveling effect on dissolved acids or bases, reacting completely to generate its characteristic hydronium and hydroxide ions (the strongest acid and base that may exist in water). The strengths of the binary acids increase from left to right across a period of the periodic table $\\left(\\mathrm{CH}_{4}<\\mathrm{NH}_{3}\\right.$ $<\\mathrm{H}_{2} \\mathrm{O}<\\mathrm{HF}$ ), and they increase down a group ( $\\mathrm{HF}<$ $\\mathrm{HCl}<\\mathrm{HBr}<\\mathrm{HI})$. The strengths of oxyacids that contain the same central element increase as the oxidation number of the element increases $\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}<\\right.$ $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ ). The strengths of oxyacids also increase as the electronegativity of the central element increases $\\left[\\mathrm{H}_{2} \\mathrm{SeO}_{4}<\\mathrm{H}_{2} \\mathrm{SO}_{4}\\right]$."}
{"id": 4218, "contents": "1588. 2 pH and pOH - 1588.2. Hydrolysis of Salts\nThe ions composing salts may possess acidic or basic character, ionizing when dissolved in water to yield acidic or basic solutions. Acidic cations are typically the conjugate partners of weak bases, and basic anions are the conjugate partners of weak acids. Many metal ions bond to water molecules when dissolved to yield complex ions that may function as acids."}
{"id": 4219, "contents": "1588. 2 pH and pOH - 1588.3. Polyprotic Acids\nAn acid that contains more than one ionizable proton is a polyprotic acid. These acids undergo stepwise ionization reactions involving the transfer of single protons. The ionization constants for polyprotic acids decrease with each subsequent step; these decreases typically are large enough to permit simple equilibrium calculations that treat each step separately."}
{"id": 4220, "contents": "1588. 2 pH and pOH - 1588.4. Buffers\nSolutions that contain appreciable amounts of a weak conjugate acid-base pair are called buffers. A buffered solution will experience only slight changes in pH when small amounts of acid or base are added. Addition of large amounts of acid or base can exceed the buffer capacity, consuming most of one\nconjugate partner and preventing further buffering action."}
{"id": 4221, "contents": "1588. 2 pH and pOH - 1588.5. Acid-Base Titrations\nThe titration curve for an acid-base titration is\ntypically a plot of pH versus volume of added titrant. These curves are useful in selecting appropriate acid-base indicators that will permit accurate determinations of titration end points."}
{"id": 4222, "contents": "1589. Exercises - 1589.1. Br\u00f8nsted-Lowry Acids and Bases\n1. Write equations that show $\\mathrm{NH}_{3}$ as both a conjugate acid and a conjugate base.\n2. Write equations that show $\\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{-}$acting both as an acid and as a base.\n3. Show by suitable net ionic equations that each of the following species can act as a Br\u00f8nsted-Lowry acid:\n(a) $\\mathrm{H}_{3} \\mathrm{O}^{+}$\n(b) HCl\n(c) $\\mathrm{NH}_{3}$\n(d) $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$\n(e) $\\mathrm{NH}_{4}{ }^{+}$\n(f) $\\mathrm{HSO}_{4}{ }^{-}$\n4. Show by suitable net ionic equations that each of the following species can act as a Br\u00f8nsted-Lowry acid:\n(a) $\\mathrm{HNO}_{3}$\n(b) $\\mathrm{PH}_{4}^{+}$\n(c) $\\mathrm{H}_{2} \\mathrm{~S}$\n(d) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COOH}$\n(e) $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}$\n(f) $\\mathrm{HS}^{-}$\n5. Show by suitable net ionic equations that each of the following species can act as a Br\u00f8nsted-Lowry base:\n(a) $\\mathrm{H}_{2} \\mathrm{O}$\n(b) $\\mathrm{OH}^{-}$\n(c) $\\mathrm{NH}_{3}$\n(d) $\\mathrm{CN}^{-}$\n(e) $\\mathrm{S}^{2-}$\n(f) $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}$\n6. Show by suitable net ionic equations that each of the following species can act as a Br\u00f8nsted-Lowry base:\n(a) $\\mathrm{HS}^{-}$\n(b) $\\mathrm{PO}_{4}{ }^{3-}$\n(c) $\\mathrm{NH}_{2}{ }^{-}$\n(d) $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$\n(e) $\\mathrm{O}^{2-}$"}
{"id": 4223, "contents": "1589. Exercises - 1589.1. Br\u00f8nsted-Lowry Acids and Bases\n(c) $\\mathrm{NH}_{2}{ }^{-}$\n(d) $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$\n(e) $\\mathrm{O}^{2-}$\n(f) $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}$\n7. What is the conjugate acid of each of the following? What is the conjugate base of each?\n(a) $\\mathrm{OH}^{-}$\n(b) $\\mathrm{H}_{2} \\mathrm{O}$\n(c) $\\mathrm{HCO}_{3}{ }^{-}$\n(d) $\\mathrm{NH}_{3}$\n(e) $\\mathrm{HSO}_{4}^{-}$\n(f) $\\mathrm{H}_{2} \\mathrm{O}_{2}$\n(g) $\\mathrm{HS}^{-}$\n(h) $\\mathrm{H}_{5} \\mathrm{~N}_{2}{ }^{+}$\n8. What is the conjugate acid of each of the following? What is the conjugate base of each?\n(a) $\\mathrm{H}_{2} \\mathrm{~S}$\n(b) $\\mathrm{H}_{2} \\mathrm{PO}_{4}-$\n(c) $\\mathrm{PH}_{3}$\n(d) $\\mathrm{HS}^{-}$\n(e) $\\mathrm{HSO}_{3}{ }^{-}$\n(f) $\\mathrm{H}_{3} \\mathrm{O}_{2}{ }^{+}$\n(g) $\\mathrm{H}_{4} \\mathrm{~N}_{2}$\n(h) $\\mathrm{CH}_{3} \\mathrm{OH}$\n9. Identify and label the Br\u00f8nsted-Lowry acid, its conjugate base, the Br\u00f8nsted-Lowry base, and its conjugate acid in each of the following equations:\n(a) $\\mathrm{HNO}_{3}+\\mathrm{H}_{2} \\mathrm{O} \\longrightarrow \\mathrm{H}_{3} \\mathrm{O}^{+}+\\mathrm{NO}_{3}{ }^{-}$\n(b) $\\mathrm{CN}^{-}+\\mathrm{H}_{2} \\mathrm{O} \\longrightarrow \\mathrm{HCN}+\\mathrm{OH}^{-}$"}
{"id": 4224, "contents": "1589. Exercises - 1589.1. Br\u00f8nsted-Lowry Acids and Bases\n(b) $\\mathrm{CN}^{-}+\\mathrm{H}_{2} \\mathrm{O} \\longrightarrow \\mathrm{HCN}+\\mathrm{OH}^{-}$\n(c) $\\mathrm{H}_{2} \\mathrm{SO}_{4}+\\mathrm{Cl}^{-} \\longrightarrow \\mathrm{HCl}+\\mathrm{HSO}_{4}^{-}$\n(d) $\\mathrm{HSO}_{4}{ }^{-}+\\mathrm{OH}^{-} \\longrightarrow \\mathrm{SO}_{4}{ }^{2-}+\\mathrm{H}_{2} \\mathrm{O}$\n(e) $\\mathrm{O}^{2-}+\\mathrm{H}_{2} \\mathrm{O} \\longrightarrow 2 \\mathrm{OH}^{-}$\n(f) $\\left[\\mathrm{Cu}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{3}(\\mathrm{OH})\\right]^{+}+\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{3+} \\longrightarrow\\left[\\mathrm{Cu}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{4}\\right]^{2+}+\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{5}(\\mathrm{OH})\\right]^{2+}$\n(g) $\\mathrm{H}_{2} \\mathrm{~S}+\\mathrm{NH}_{2}{ }^{-} \\longrightarrow \\mathrm{HS}^{-}+\\mathrm{NH}_{3}$\n10. Identify and label the Br\u00f8nsted-Lowry acid, its conjugate base, the Br\u00f8nsted-Lowry base, and its conjugate acid in each of the following equations:\n(a) $\\mathrm{NO}_{2}^{-}+\\mathrm{H}_{2} \\mathrm{O} \\longrightarrow \\mathrm{HNO}_{2}+\\mathrm{OH}^{-}$\n(b) $\\mathrm{HBr}+\\mathrm{H}_{2} \\mathrm{O} \\longrightarrow \\mathrm{H}_{3} \\mathrm{O}^{+}+\\mathrm{Br}^{-}$"}
{"id": 4225, "contents": "1589. Exercises - 1589.1. Br\u00f8nsted-Lowry Acids and Bases\n(b) $\\mathrm{HBr}+\\mathrm{H}_{2} \\mathrm{O} \\longrightarrow \\mathrm{H}_{3} \\mathrm{O}^{+}+\\mathrm{Br}^{-}$\n(c) $\\mathrm{HS}^{-}+\\mathrm{H}_{2} \\mathrm{O} \\longrightarrow \\mathrm{H}_{2} \\mathrm{~S}+\\mathrm{OH}^{-}$\n(d) $\\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{-}+\\mathrm{OH}^{-} \\longrightarrow \\mathrm{HPO}_{4}{ }^{2-}+\\mathrm{H}_{2} \\mathrm{O}$\n(e) $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}+\\mathrm{HCl} \\longrightarrow \\mathrm{H}_{3} \\mathrm{PO}_{4}+\\mathrm{Cl}^{-}$\n(f) $\\left[\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{5}(\\mathrm{OH})\\right]^{2+}+\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{3+} \\longrightarrow\\left[\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{3+}+\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{5}(\\mathrm{OH})\\right]^{2+}$\n(g) $\\mathrm{CH}_{3} \\mathrm{OH}+\\mathrm{H}^{-} \\longrightarrow \\mathrm{CH}_{3} \\mathrm{O}^{-}+\\mathrm{H}_{2}$\n11. What are amphiprotic species? Illustrate with suitable equations.\n12. State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species:\n(a) $\\mathrm{H}_{2} \\mathrm{O}$\n(b) $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}$\n(c) $\\mathrm{S}^{2-}$\n(d) $\\mathrm{CO}_{3}{ }^{2-}$\n(e) $\\mathrm{HSO}_{4}{ }^{-}$"}
{"id": 4226, "contents": "1589. Exercises - 1589.1. Br\u00f8nsted-Lowry Acids and Bases\n(c) $\\mathrm{S}^{2-}$\n(d) $\\mathrm{CO}_{3}{ }^{2-}$\n(e) $\\mathrm{HSO}_{4}{ }^{-}$\n13. State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species.\n(a) $\\mathrm{NH}_{3}$\n(b) $\\mathrm{HPO}_{4}^{-}$\n(c) $\\mathrm{Br}^{-}$\n(d) $\\mathrm{NH}_{4}{ }^{+}$\n(e) $\\mathrm{ASO}_{4}{ }^{3-}$\n14. Is the self-ionization of water endothermic or exothermic? The ionization constant for water $\\left(K_{\\mathrm{w}}\\right)$ is $2.9 \\times$ $10^{-14}$ at $40^{\\circ} \\mathrm{C}$ and $9.3 \\times 10^{-14}$ at $60^{\\circ} \\mathrm{C}$."}
{"id": 4227, "contents": "1590. 2 pH and pOH - \n15. Explain why a sample of pure water at $40^{\\circ} \\mathrm{C}$ is neutral even though $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=1.7 \\times 10^{-7} \\mathrm{M}$. $K_{\\mathrm{w}}$ is $2.9 \\times 10^{-14}$ at $40^{\\circ} \\mathrm{C}$.\n16. The ionization constant for water $\\left(K_{\\mathrm{w}}\\right)$ is $2.9 \\times 10^{-14}$ at $40^{\\circ} \\mathrm{C}$. Calculate $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right],\\left[\\mathrm{OH}^{-}\\right], \\mathrm{pH}$, and pOH for pure water at $40^{\\circ} \\mathrm{C}$.\n17. The ionization constant for water $\\left(K_{\\mathrm{w}}\\right)$ is $9.311 \\times 10^{-14}$ at $60^{\\circ} \\mathrm{C}$. Calculate $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right],\\left[\\mathrm{OH}^{-}\\right], \\mathrm{pH}$, and pOH for pure water at $60^{\\circ} \\mathrm{C}$.\n18. Calculate the pH and the pOH of each of the following solutions at $25^{\\circ} \\mathrm{C}$ for which the substances ionize completely:\n(a) 0.200 M HCl\n(b) 0.0143 M NaOH\n(c) $3.0 \\mathrm{M} \\mathrm{HNO}_{3}$\n(d) $0.0031 \\mathrm{M} \\mathrm{Ca}(\\mathrm{OH})_{2}$\n19. Calculate the pH and the pOH of each of the following solutions at $25^{\\circ} \\mathrm{C}$ for which the substances ionize completely:\n(a) $0.000259 \\mathrm{M} \\mathrm{HClO}_{4}$\n(b) 0.21 M NaOH\n(c) $0.000071 \\mathrm{MBa}(\\mathrm{OH})_{2}$"}
{"id": 4228, "contents": "1590. 2 pH and pOH - \n(b) 0.21 M NaOH\n(c) $0.000071 \\mathrm{MBa}(\\mathrm{OH})_{2}$\n(d) 2.5 M KOH\n20. What are the pH and pOH of a solution of 2.0 M HCl , which ionizes completely?\n21. What are the hydronium and hydroxide ion concentrations in a solution whose pH is 6.52 ?\n22. Calculate the hydrogen ion concentration and the hydroxide ion concentration in wine from its pH . See Figure 14.2 for useful information.\n23. Calculate the hydronium ion concentration and the hydroxide ion concentration in lime juice from its pH . See Figure 14.2 for useful information.\n24. The hydronium ion concentration in a sample of rainwater is found to be $1.7 \\times 10^{-6} \\mathrm{M}$ at $25^{\\circ} \\mathrm{C}$. What is the concentration of hydroxide ions in the rainwater?\n25. The hydroxide ion concentration in household ammonia is $3.2 \\times 10^{-3} \\mathrm{M}$ at $25^{\\circ} \\mathrm{C}$. What is the concentration of hydronium ions in the solution?"}
{"id": 4229, "contents": "1590. 2 pH and pOH - 1590.1. Relative Strengths of Acids and Bases\n26. Explain why the neutralization reaction of a strong acid and a weak base gives a weakly acidic solution.\n27. Explain why the neutralization reaction of a weak acid and a strong base gives a weakly basic solution.\n28. Use this list of important industrial compounds (and Figure 14.8) to answer the following questions regarding: $\\mathrm{Ca}(\\mathrm{OH})_{2}, \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}, \\mathrm{HCl}, \\mathrm{H}_{2} \\mathrm{CO}_{3}$, $\\mathrm{HF}, \\mathrm{HNO}_{2}, \\mathrm{HNO}_{3}, \\mathrm{H}_{3} \\mathrm{PO}_{4}, \\mathrm{H}_{2} \\mathrm{SO}_{4}, \\mathrm{NH}_{3}, \\mathrm{NaOH}, \\mathrm{Na}_{2} \\mathrm{CO}_{3}$. (a) Identify the strong Br\u00f8nsted-Lowry acids and strong Br\u00f8nsted-Lowry bases.\n(b) Identify the compounds that can behave as Br\u00f8nsted-Lowry acids with strengths lying between those of $\\mathrm{H}_{3} \\mathrm{O}^{+}$and $\\mathrm{H}_{2} \\mathrm{O}$.\n(c) Identify the compounds that can behave as Br\u00f8nsted-Lowry bases with strengths lying between those of $\\mathrm{H}_{2} \\mathrm{O}$ and $\\mathrm{OH}^{-}$.\n29. The odor of vinegar is due to the presence of acetic acid, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$, a weak acid. List, in order of descending concentration, all of the ionic and molecular species present in a 1-M aqueous solution of this acid.\n30. Household ammonia is a solution of the weak base $\\mathrm{NH}_{3}$ in water. List, in order of descending concentration, all of the ionic and molecular species present in a 1-M aqueous solution of this base."}
{"id": 4230, "contents": "1590. 2 pH and pOH - 1590.1. Relative Strengths of Acids and Bases\n31. Explain why the ionization constant, $K_{\\mathrm{a}}$, for $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ is larger than the ionization constant for $\\mathrm{H}_{2} \\mathrm{SO}_{3}$.\n32. Explain why the ionization constant, $K_{\\mathrm{a}}$, for HI is larger than the ionization constant for HF.\n33. Gastric juice, the digestive fluid produced in the stomach, contains hydrochloric acid, HCl. Milk of Magnesia, a suspension of solid $\\mathrm{Mg}(\\mathrm{OH})_{2}$ in an aqueous medium, is sometimes used to neutralize excess stomach acid. Write a complete balanced equation for the neutralization reaction, and identify the conjugate acid-base pairs.\n34. Nitric acid reacts with insoluble copper(II) oxide to form soluble copper(II) nitrate, $\\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}$, a compound that has been used to prevent the growth of algae in swimming pools. Write the balanced chemical equation for the reaction of an aqueous solution of $\\mathrm{HNO}_{3}$ with CuO .\n35. What is the ionization constant at $25^{\\circ} \\mathrm{C}$ for the weak acid $\\mathrm{CH}_{3} \\mathrm{NH}_{3}{ }^{+}$, the conjugate acid of the weak base $\\mathrm{CH}_{3} \\mathrm{NH}_{2}, K_{\\mathrm{b}}=4.4 \\times 10^{-4}$.\n36. What is the ionization constant at $25^{\\circ} \\mathrm{C}$ for the weak acid $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NH}_{2}{ }^{+}$, the conjugate acid of the weak base $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NH}, K_{\\mathrm{b}}=5.9 \\times 10^{-4}$ ?"}
{"id": 4231, "contents": "1590. 2 pH and pOH - 1590.1. Relative Strengths of Acids and Bases\n37. Which base, $\\mathrm{CH}_{3} \\mathrm{NH}_{2}$ or $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NH}$, is the stronger base? Which conjugate acid, $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NH}_{2}{ }^{+}$or $\\mathrm{CH}_{3} \\mathrm{NH}_{3}{ }^{+}$, is the stronger acid?\n38. Which is the stronger acid, $\\mathrm{NH}_{4}{ }^{+}$or HBrO ?\n39. Which is the stronger base, $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{~N}^{2}$ or $\\mathrm{H}_{2} \\mathrm{BO}_{3}-$ ?\n40. Predict which acid in each of the following pairs is the stronger and explain your reasoning for each.\n(a) $\\mathrm{H}_{2} \\mathrm{O}$ or HF\n(b) $\\mathrm{B}(\\mathrm{OH})_{3}$ or $\\mathrm{Al}(\\mathrm{OH})_{3}$\n(c) $\\mathrm{HSO}_{3}{ }^{-}$or $\\mathrm{HSO}_{4}^{-}$\n(d) $\\mathrm{NH}_{3}$ or $\\mathrm{H}_{2} \\mathrm{~S}$\n(e) $\\mathrm{H}_{2} \\mathrm{O}$ or $\\mathrm{H}_{2} \\mathrm{Te}$\n41. Predict which compound in each of the following pairs of compounds is more acidic and explain your reasoning for each.\n(a) $\\mathrm{HSO}_{4}^{-}$or $\\mathrm{HSeO}_{4}^{-}$\n(b) $\\mathrm{NH}_{3}$ or $\\mathrm{H}_{2} \\mathrm{O}$\n(c) $\\mathrm{PH}_{3}$ or HI\n(d) $\\mathrm{NH}_{3}$ or $\\mathrm{PH}_{3}$\n(e) $\\mathrm{H}_{2} \\mathrm{~S}$ or HBr\n42. Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign.\n(a) acidity: $\\mathrm{HCl}, \\mathrm{HBr}, \\mathrm{HI}$"}
{"id": 4232, "contents": "1590. 2 pH and pOH - 1590.1. Relative Strengths of Acids and Bases\n(a) acidity: $\\mathrm{HCl}, \\mathrm{HBr}, \\mathrm{HI}$\n(b) basicity: $\\mathrm{H}_{2} \\mathrm{O}, \\mathrm{OH}^{-}, \\mathrm{H}^{-}, \\mathrm{Cl}^{-}$\n(c) basicity: $\\mathrm{Mg}(\\mathrm{OH})_{2}, \\mathrm{Si}(\\mathrm{OH})_{4}, \\mathrm{ClO}_{3}(\\mathrm{OH})$ (Hint: Formula could also be written as $\\mathrm{HClO}_{4}$.)\n(d) acidity: $\\mathrm{HF}, \\mathrm{H}_{2} \\mathrm{O}, \\mathrm{NH}_{3}, \\mathrm{CH}_{4}$\n43. Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign.\n(a) acidity: $\\mathrm{NaHSO}_{3}, \\mathrm{NaHSeO}_{3}, \\mathrm{NaHSO}_{4}$\n(b) basicity: $\\mathrm{BrO}_{2}{ }^{-}, \\mathrm{ClO}_{2}{ }^{-}, \\mathrm{IO}_{2}{ }^{-}$\n(c) acidity: $\\mathrm{HOCl}, \\mathrm{HOBr}, \\mathrm{HOI}$\n(d) acidity: $\\mathrm{HOCl}, \\mathrm{HOClO}, \\mathrm{HOClO}_{2}, \\mathrm{HOClO}_{3}$\n(e) basicity: $\\mathrm{NH}_{2}{ }^{-}, \\mathrm{HS}^{-}, \\mathrm{HTe}^{-}, \\mathrm{PH}_{2}{ }^{-}$\n(f) basicity: $\\mathrm{BrO}^{-}, \\mathrm{BrO}_{2}^{-}, \\mathrm{BrO}_{3}^{-}, \\mathrm{BrO}_{4}^{-}$\n44. Both HF and HCN ionize in water to a limited extent. Which of the conjugate bases, $\\mathrm{F}^{-}$or $\\mathrm{CN}^{-}$, is the stronger base?"}
{"id": 4233, "contents": "1590. 2 pH and pOH - 1590.1. Relative Strengths of Acids and Bases\n44. Both HF and HCN ionize in water to a limited extent. Which of the conjugate bases, $\\mathrm{F}^{-}$or $\\mathrm{CN}^{-}$, is the stronger base?\n45. The active ingredient formed by aspirin in the body is salicylic acid, $\\mathrm{C}_{6} \\mathrm{H}_{4} \\mathrm{OH}\\left(\\mathrm{CO}_{2} \\mathrm{H}\\right)$. The carboxyl group $\\left(-\\mathrm{CO}_{2} \\mathrm{H}\\right)$ acts as a weak acid. The phenol group (an OH group bonded to an aromatic ring) also acts as an acid but a much weaker acid. List, in order of descending concentration, all of the ionic and molecular species present in a $0.001-M$ aqueous solution of $\\mathrm{C}_{6} \\mathrm{H}_{4} \\mathrm{OH}\\left(\\mathrm{CO}_{2} \\mathrm{H}\\right)$.\n46. Are the concentrations of hydronium ion and hydroxide ion in a solution of an acid or a base in water directly proportional or inversely proportional? Explain your answer.\n47. What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak acid or base?\n48. Which of the following will increase the percent of $\\mathrm{NH}_{3}$ that is converted to the ammonium ion in water?\n(a) addition of NaOH\n(b) addition of HCl\n(c) addition of $\\mathrm{NH}_{4} \\mathrm{Cl}$\n49. Which of the following will increase the percentage of HF that is converted to the fluoride ion in water?\n(a) addition of NaOH\n(b) addition of HCl\n(c) addition of NaF\n50. What is the effect on the concentrations of $\\mathrm{NO}_{2}{ }^{-}, \\mathrm{HNO}_{2}$, and $\\mathrm{OH}^{-}$when the following are added to a solution of $\\mathrm{KNO}_{2}$ in water:\n(a) HCl\n(b) $\\mathrm{HNO}_{2}$\n(c) NaOH\n(d) NaCl\n(e) KNO"}
{"id": 4234, "contents": "1590. 2 pH and pOH - 1590.1. Relative Strengths of Acids and Bases\n(a) HCl\n(b) $\\mathrm{HNO}_{2}$\n(c) NaOH\n(d) NaCl\n(e) KNO\n51. What is the effect on the concentration of hydrofluoric acid, hydronium ion, and fluoride ion when the following are added to separate solutions of hydrofluoric acid?\n(a) HCl\n(b) KF\n(c) NaCl\n(d) KOH\n(e) HF\n52. Why is the hydronium ion concentration in a solution that is 0.10 Min HCl and 0.10 Min HCOOH determined by the concentration of HCl ?\n53. From the equilibrium concentrations given, calculate $K_{\\mathrm{a}}$ for each of the weak acids and $K_{\\mathrm{b}}$ for each of the weak bases.\n(a) $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}:\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=1.34 \\times 10^{-3} \\mathrm{M}$;\n$\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}\\right]=1.34 \\times 10^{-3} \\mathrm{M}$;\n$\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right]=9.866 \\times 10^{-2} \\mathrm{M}$;\n(b) $\\mathrm{ClO}^{-}:\\left[\\mathrm{OH}^{-}\\right]=4.0 \\times 10^{-4} \\mathrm{M}$;\n[ HClO ] $=2.38 \\times 10^{-4} \\mathrm{M}$;\n$\\left[\\mathrm{ClO}^{-}\\right]=0.273 \\mathrm{M}$;\n(c) $\\mathrm{HCO}_{2} \\mathrm{H}:\\left[\\mathrm{HCO}_{2} \\mathrm{H}\\right]=0.524 \\mathrm{M}$;\n$\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=9.8 \\times 10^{-3} \\mathrm{M}$;"}
{"id": 4235, "contents": "1590. 2 pH and pOH - 1590.1. Relative Strengths of Acids and Bases\n$\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=9.8 \\times 10^{-3} \\mathrm{M}$;\n$\\left[\\mathrm{HCO}_{2}{ }^{-}\\right]=9.8 \\times 10^{-3} \\mathrm{M}$;\n(d) $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{3}{ }^{+}:\\left[\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{3}{ }^{+}\\right]=0.233 \\mathrm{M}$;\n$\\left[\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{2}\\right]=2.3 \\times 10^{-3} \\mathrm{M}$;\n$\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=2.3 \\times 10^{-3} \\mathrm{M}$\n54. From the equilibrium concentrations given, calculate $K_{\\mathrm{a}}$ for each of the weak acids and $K_{\\mathrm{b}}$ for each of the weak bases.\n(a) $\\mathrm{NH}_{3}:\\left[\\mathrm{OH}^{-}\\right]=3.1 \\times 10^{-3} \\mathrm{M}$;\n$\\left[\\mathrm{NH}_{4}{ }^{+}\\right]=3.1 \\times 10^{-3} \\mathrm{M}$;\n$\\left[\\mathrm{NH}_{3}\\right]=0.533 \\mathrm{M}$;\n(b) $\\mathrm{HNO}_{2}:\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=0.011 \\mathrm{M}$;\n$\\left[\\mathrm{NO}_{2}{ }^{-}\\right]=0.0438 \\mathrm{M}$;\n$\\left[\\mathrm{HNO}_{2}\\right]=1.07 \\mathrm{M}$;\n(c) $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{~N}:\\left[\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{~N}\\right]=0.25 \\mathrm{M}$;"}
{"id": 4236, "contents": "1590. 2 pH and pOH - 1590.1. Relative Strengths of Acids and Bases\n$\\left[\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{NH}^{+}\\right]=4.3 \\times 10^{-3} \\mathrm{M}$;\n$\\left[\\mathrm{OH}^{-}\\right]=3.7 \\times 10^{-3} \\mathrm{M}$;\n(d) $\\mathrm{NH}_{4}{ }^{+}:\\left[\\mathrm{NH}_{4}{ }^{+}\\right]=0.100 \\mathrm{M}$;\n$\\left[\\mathrm{NH}_{3}\\right]=7.5 \\times 10^{-6} \\mathrm{M}$;\n$\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=7.5 \\times 10^{-6} \\mathrm{M}$\n55. Determine $K_{\\mathrm{b}}$ for the nitrite ion, $\\mathrm{NO}_{2}{ }^{-}$. In a $0.10-M$ solution this base is $0.0015 \\%$ ionized.\n56. Determine $K_{\\mathrm{a}}$ for hydrogen sulfate ion, $\\mathrm{HSO}_{4}{ }^{-}$. In a $0.10-M$ solution the acid is $29 \\%$ ionized.\n57. Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:\n(a) $\\mathrm{F}^{-}$\n(b) $\\mathrm{NH}_{4}{ }^{+}$\n(c) $\\mathrm{AsO}_{4}{ }^{3-}$\n(d) $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NH}_{2}{ }^{+}$\n(e) $\\mathrm{NO}_{2}{ }^{-}$\n(f) $\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}$(as a base)\n58. Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:\n(a) $\\mathrm{HTe}^{-}$(as a base)\n(b) $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{NH}^{+}$"}
{"id": 4237, "contents": "1590. 2 pH and pOH - 1590.1. Relative Strengths of Acids and Bases\n(a) $\\mathrm{HTe}^{-}$(as a base)\n(b) $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{NH}^{+}$\n(c) $\\mathrm{HAsO}_{4}{ }^{2-}$ (as a base)\n(d) $\\mathrm{HO}_{2}{ }^{-}$(as a base)\n(e) $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{3}+$\n(f) $\\mathrm{HSO}_{3}{ }^{-}$(as a base)\n59. Using the $K_{\\mathrm{a}}$ value of $1.4 \\times 10^{-5}$, place $\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{3+}$ in the correct location in Figure 14.7.\n60. Calculate the concentration of all solute species in each of the following solutions of acids or bases. Assume that the ionization of water can be neglected, and show that the change in the initial concentrations can be neglected.\n(a) 0.0092 M HClO , a weak acid\n(b) $0.0784 \\mathrm{MC}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{2}$, a weak base\n(c) 0.0810 M HCN , a weak acid\n(d) $0.11 \\mathrm{M}\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{~N}$, a weak base\n(e) $0.120 \\mathrm{MFe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{2+}$ a weak acid, $K_{\\mathrm{a}}=1.6 \\times 10^{-7}$"}
{"id": 4238, "contents": "1590. 2 pH and pOH - 1590.1. Relative Strengths of Acids and Bases\n61. Propionic acid, $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{CO}_{2} \\mathrm{H}\\left(K_{\\mathrm{a}}=1.34 \\times 10^{-5}\\right)$, is used in the manufacture of calcium propionate, a food preservative. What is the pH of a $0.698-M$ solution of $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{CO}_{2} \\mathrm{H}$ ?\n62. White vinegar is a $5.0 \\%$ by mass solution of acetic acid in water. If the density of white vinegar is $1.007 \\mathrm{~g} /$ $\\mathrm{cm}^{3}$, what is the pH ?\n63. The ionization constant of lactic acid, $\\mathrm{CH}_{3} \\mathrm{CH}(\\mathrm{OH}) \\mathrm{CO}_{2} \\mathrm{H}$, an acid found in the blood after strenuous exercise, is $1.36 \\times 10^{-4}$. If 20.0 g of lactic acid is used to make a solution with a volume of 1.00 L , what is the concentration of hydronium ion in the solution?\n64. Nicotine, $\\mathrm{C}_{10} \\mathrm{H}_{14} \\mathrm{~N}_{2}$, is a base that will accept two protons ( $K_{\\mathrm{b} 1}=7 \\times 10^{-7}, K_{\\mathrm{b} 2}=1.4 \\times 10^{-11}$ ). What is the concentration of each species present in a $0.050-M$ solution of nicotine?\n65. The pH of a $0.23-M$ solution of HF is 1.92 . Determine $K_{\\mathrm{a}}$ for HF from these data.\n66. The pH of a $0.15-M$ solution of $\\mathrm{HSO}_{4}{ }^{-}$is 1.43. Determine $K_{\\mathrm{a}}$ for $\\mathrm{HSO}_{4}{ }^{-}$from these data."}
{"id": 4239, "contents": "1590. 2 pH and pOH - 1590.1. Relative Strengths of Acids and Bases\n67. The pH of a $0.10-M$ solution of caffeine is 11.70 . Determine $K_{\\mathrm{b}}$ for caffeine from these data: $\\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{C}_{8} \\mathrm{H}_{10} \\mathrm{~N}_{4} \\mathrm{O}_{2} \\mathrm{H}^{+}(a q)+\\mathrm{OH}^{-}(a q)$\n68. The pH of a solution of household ammonia, a 0.950 M solution of $\\mathrm{NH}_{3}$, is 11.612 . Determine $K_{\\mathrm{b}}$ for $\\mathrm{NH}_{3}$ from these data."}
{"id": 4240, "contents": "1590. 2 pH and pOH - 1590.2. Hydrolysis of Salts\n69. Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:\n(a) $\\mathrm{Al}\\left(\\mathrm{NO}_{3}\\right)_{3}$\n(b) RbI\n(c) $\\mathrm{KHCO}_{2}$\n(d) $\\mathrm{CH}_{3} \\mathrm{NH}_{3} \\mathrm{Br}$\n70. Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:\n(a) $\\mathrm{FeCl}_{3}$\n(b) $\\mathrm{K}_{2} \\mathrm{CO}_{3}$\n(c) $\\mathrm{NH}_{4} \\mathrm{Br}$\n(d) $\\mathrm{KClO}_{4}$\n71. Novocaine, $\\mathrm{C}_{13} \\mathrm{H}_{21} \\mathrm{O}_{2} \\mathrm{~N}_{2} \\mathrm{Cl}$, is the salt of the base procaine and hydrochloric acid. The ionization constant for procaine is $7 \\times 10^{-6}$. Is a solution of novocaine acidic or basic? What are $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right],\\left[\\mathrm{OH}^{-}\\right]$, and pH of a $2.0 \\%$ solution by mass of novocaine, assuming that the density of the solution is $1.0 \\mathrm{~g} / \\mathrm{mL}$."}
{"id": 4241, "contents": "1590. 2 pH and pOH - 1590.3. Polyprotic Acids\n72. Which of the following concentrations would be practically equal in a calculation of the equilibrium concentrations in a $0.134-M$ solution of $\\mathrm{H}_{2} \\mathrm{CO}_{3}$, a diprotic acid: $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right],\\left[\\mathrm{OH}^{-}\\right],\\left[\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right],\\left[\\mathrm{HCO}_{3}{ }^{-}\\right]$, $\\left[\\mathrm{CO}_{3}{ }^{2-}\\right]$ ? No calculations are needed to answer this question.\n73. Calculate the concentration of each species present in a $0.050-M$ solution of $\\mathrm{H}_{2} \\mathrm{~S}$.\n74. Calculate the concentration of each species present in a $0.010-M$ solution of phthalic acid, $\\mathrm{C}_{6} \\mathrm{H}_{4}\\left(\\mathrm{CO}_{2} \\mathrm{H}\\right)_{2}$."}
{"id": 4242, "contents": "1590. 2 pH and pOH - 1590.3. Polyprotic Acids\n$$\n\\begin{array}{lr}\n\\mathrm{C}_{6} \\mathrm{H}_{4}\\left(\\mathrm{CO}_{2} \\mathrm{H}\\right)_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{C}_{6} \\mathrm{H}_{4}\\left(\\mathrm{CO}_{2} \\mathrm{H}\\right)\\left(\\mathrm{CO}_{2}\\right)^{-}(a q) & K_{\\mathrm{a}}=1.1 \\times 10^{-3} \\\\\n\\mathrm{C}_{6} \\mathrm{H}_{4}\\left(\\mathrm{CO}_{2} \\mathrm{H}\\right)\\left(\\mathrm{CO}_{2}\\right)(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{C}_{6} \\mathrm{H}_{4}\\left(\\mathrm{CO}_{2}\\right)_{2}^{2-}(a q) & K_{\\mathrm{a}}=3.9 \\times 10^{-6}\n\\end{array}\n$$"}
{"id": 4243, "contents": "1590. 2 pH and pOH - 1590.3. Polyprotic Acids\n75. Salicylic acid, $\\mathrm{HOC}_{6} \\mathrm{H}_{4} \\mathrm{CO}_{2} \\mathrm{H}$, and its derivatives have been used as pain relievers for a long time. Salicylic acid occurs in small amounts in the leaves, bark, and roots of some vegetation (most notably historically in the bark of the willow tree). Extracts of these plants have been used as medications for centuries. The acid was first isolated in the laboratory in 1838.\n(a) Both functional groups of salicylic acid ionize in water, with $K_{\\mathrm{a}}=1.0 \\times 10^{-3}$ for the $-\\mathrm{CO}_{2} \\mathrm{H}$ group and $4.2 \\times 10^{-13}$ for the -OH group. What is the pH of a saturated solution of the acid (solubility $=1.8 \\mathrm{~g} / \\mathrm{L}$ ).\n(b) Aspirin was discovered as a result of efforts to produce a derivative of salicylic acid that would not be irritating to the stomach lining. Aspirin is acetylsalicylic acid, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{C}_{6} \\mathrm{H}_{4} \\mathrm{CO}_{2} \\mathrm{H}$. The $-\\mathrm{CO}_{2} \\mathrm{H}$ functional group is still present, but its acidity is reduced, $K_{\\mathrm{a}}=3.0 \\times 10^{-4}$. What is the pH of a solution of aspirin with the same concentration as a saturated solution of salicylic acid (See Part a).\n76. The ion $\\mathrm{HTe}^{-}$is an amphiprotic species; it can act as either an acid or a base.\n(a) What is $K_{\\mathrm{a}}$ for the acid reaction of $\\mathrm{HTe}^{-}$with $\\mathrm{H}_{2} \\mathrm{O}$ ?\n(b) What is $K_{\\mathrm{b}}$ for the reaction in which $\\mathrm{HTe}^{-}$functions as a base in water?"}
{"id": 4244, "contents": "1590. 2 pH and pOH - 1590.3. Polyprotic Acids\n(b) What is $K_{\\mathrm{b}}$ for the reaction in which $\\mathrm{HTe}^{-}$functions as a base in water?\n(c) Demonstrate whether or not the second ionization of $\\mathrm{H}_{2} \\mathrm{Te}$ can be neglected in the calculation of [ $\\mathrm{HTe}^{-}$] in a 0.10 M solution of $\\mathrm{H}_{2} \\mathrm{Te}$."}
{"id": 4245, "contents": "1590. 2 pH and pOH - 1590.4. Buffers\n77. Explain why a buffer can be prepared from a mixture of $\\mathrm{NH}_{4} \\mathrm{Cl}$ and NaOH but not from $\\mathrm{NH}_{3}$ and NaOH .\n78. Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the acid $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ and a salt of its conjugate base $\\mathrm{NaH}_{2} \\mathrm{PO}_{4}$.\n79. Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the base $\\mathrm{NH}_{3}$ and a salt of its conjugate acid $\\mathrm{NH}_{4} \\mathrm{Cl}$.\n80. What is $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]$in a solution of $0.25 \\mathrm{MCH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$ and $0.030 \\mathrm{M} \\mathrm{NaCH} \\mathrm{NO}_{2}$ ? $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}(a q) \\quad K_{\\mathrm{a}}=1.8 \\times 10^{-5}$\n81. What is $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]$in a solution of $0.075 \\mathrm{M} \\mathrm{HNO}_{2}$ and $0.030 M \\mathrm{NaNO}_{2}$ ?\n$\\mathrm{HNO}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{NO}_{2}^{-}(a q) \\quad K_{\\mathrm{a}}=4.5 \\times 10^{-5}$"}
{"id": 4246, "contents": "1590. 2 pH and pOH - 1590.4. Buffers\n82. What is $\\left[\\mathrm{OH}^{-}\\right]$in a solution of $0.125 \\mathrm{MCH}_{3} \\mathrm{NH}_{2}$ and $0.130 \\mathrm{MCH}_{3} \\mathrm{NH}_{3} \\mathrm{Cl}$ ?\n$\\mathrm{CH}_{3} \\mathrm{NH}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{NH}_{3}+(a q)+\\mathrm{OH}^{-}(a q) \\quad K_{\\mathrm{b}}=4.4 \\times 10^{-4}$\n83. What is $\\left[\\mathrm{OH}^{-}\\right]$in a solution of $1.25 \\mathrm{MNH}_{3}$ and $0.78 \\mathrm{MNH}_{4} \\mathrm{NO}_{3}$ ?\n$\\mathrm{NH}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{NH}_{4}{ }^{( }(a q)+\\mathrm{OH}^{-}(a q) \\quad K_{\\mathrm{b}}=1.8 \\times 10^{-5}$\n84. What is the effect on the concentration of acetic acid, hydronium ion, and acetate ion when the following are added to an acidic buffer solution of equal concentrations of acetic acid and sodium acetate:\n(a) HCl\n(b) $\\mathrm{KCH}_{3} \\mathrm{CO}_{2}$\n(c) NaCl\n(d) KOH\n(e) $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$\n85. What is the effect on the concentration of ammonia, hydroxide ion, and ammonium ion when the following are added to a basic buffer solution of equal concentrations of ammonia and ammonium nitrate:\n(a) KI\n(b) $\\mathrm{NH}_{3}$\n(c) HI\n(d) NaOH\n(e) $\\mathrm{NH}_{4} \\mathrm{Cl}$"}
{"id": 4247, "contents": "1590. 2 pH and pOH - 1590.4. Buffers\n(a) KI\n(b) $\\mathrm{NH}_{3}$\n(c) HI\n(d) NaOH\n(e) $\\mathrm{NH}_{4} \\mathrm{Cl}$\n86. What will be the pH of a buffer solution prepared from $0.20 \\mathrm{~mol} \\mathrm{NH}_{3}, 0.40 \\mathrm{~mol} \\mathrm{NH}_{4} \\mathrm{NO}_{3}$, and just enough water to give 1.00 L of solution?\n87. Calculate the pH of a buffer solution prepared from 0.155 mol of phosphoric acid, 0.250 mole of $\\mathrm{KH}_{2} \\mathrm{PO}_{4}$, and enough water to make 0.500 L of solution.\n88. How much solid $\\mathrm{NaCH}_{3} \\mathrm{CO}_{2} \\cdot 3 \\mathrm{H}_{2} \\mathrm{O}$ must be added to 0.300 L of a $0.50-M$ acetic acid solution to give a buffer with a pH of 5.00 ? (Hint: Assume a negligible change in volume as the solid is added.)\n89. What mass of $\\mathrm{NH}_{4} \\mathrm{Cl}$ must be added to 0.750 L of a $0.100-\\mathrm{M}$ solution of $\\mathrm{NH}_{3}$ to give a buffer solution with a pH of 9.26 ? (Hint: Assume a negligible change in volume as the solid is added.)\n90. A buffer solution is prepared from equal volumes of 0.200 M acetic acid and 0.600 M sodium acetate. Use $1.80 \\times 10^{-5}$ as $K_{\\mathrm{a}}$ for acetic acid.\n(a) What is the pH of the solution?\n(b) Is the solution acidic or basic?\n(c) What is the pH of a solution that results when 3.00 mL of 0.034 MHCl is added to 0.200 L of the original buffer?\n91. A $5.36-\\mathrm{g}$ sample of $\\mathrm{NH}_{4} \\mathrm{Cl}$ was added to 25.0 mL of 1.00 M NaOH and the resulting solution diluted to 0.100 L ."}
{"id": 4248, "contents": "1590. 2 pH and pOH - 1590.4. Buffers\n(a) What is the pH of this buffer solution?\n(b) Is the solution acidic or basic?\n(c) What is the pH of a solution that results when 3.00 mL of 0.034 MHCl is added to the solution?"}
{"id": 4249, "contents": "1590. 2 pH and pOH - 1590.5. Acid-Base Titrations\n92. Explain how to choose the appropriate acid-base indicator for the titration of a weak base with a strong acid.\n93. Explain why an acid-base indicator changes color over a range of pH values rather than at a specific pH .\n94. Calculate the pH at the following points in a titration of $40 \\mathrm{~mL}(0.040 \\mathrm{~L})$ of 0.100 M barbituric acid $\\left(K_{\\mathrm{a}}=9.8\\right.$ $\\times 10^{-5}$ ) with 0.100 M KOH .\n(a) no KOH added\n(b) 20 mL of KOH solution added\n(c) 39 mL of KOH solution added\n(d) 40 mL of KOH solution added\n(e) 41 mL of KOH solution added\n95. The indicator dinitrophenol is an acid with a $K_{\\mathrm{a}}$ of $1.1 \\times 10^{-4}$. In a $1.0 \\times 10^{-4}-M$ solution, it is colorless in acid and yellow in base. Calculate the pH range over which it goes from $10 \\%$ ionized (colorless) to $90 \\%$ ionized (yellow).\n\n724 14\u2022Exercises"}
{"id": 4250, "contents": "1591. CHAPTER 15 Equilibria of Other Reaction Classes - \nFigure 15.1 The mineral fluorite $\\left(\\mathrm{CaF}_{2}\\right)$ is formed when dissolved calcium and fluoride ions precipitate from groundwater within the Earth's crust. Note that pure fluorite is colorless, and that the color in this sample is due to the presence of other metal ions in the crystal."}
{"id": 4251, "contents": "1592. CHAPTER OUTLINE - 1592.1. Precipitation and Dissolution\n15.2 Lewis Acids and Bases\n15.3 Coupled Equilibria\n\nINTRODUCTION The mineral fluorite, $\\mathrm{CaF}_{2}$ Figure 15.1, is commonly used as a semiprecious stone in many types of jewelry because of its striking appearance. Deposits of fluorite are formed through a process called hydrothermal precipitation in which calcium and fluoride ions dissolved in groundwater combine to produce insoluble $\\mathrm{CaF}_{2}$ in response to some change in solution conditions. For example, a decrease in temperature may trigger fluorite precipitation if its solubility is exceeded at the lower temperature. Because fluoride ion is a weak base, its solubility is also affected by solution pH , and so geologic or other processes that change groundwater pH will also affect the precipitation of fluorite. This chapter extends the equilibrium discussion of other chapters by addressing some additional reaction classes (including precipitation) and systems involving coupled equilibrium reactions."}
{"id": 4252, "contents": "1593. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Write chemical equations and equilibrium expressions representing solubility equilibria\n- Carry out equilibrium computations involving solubility, equilibrium expressions, and solute concentrations\n\nSolubility equilibria are established when the dissolution and precipitation of a solute species occur at equal\nrates. These equilibria underlie many natural and technological processes, ranging from tooth decay to water purification. An understanding of the factors affecting compound solubility is, therefore, essential to the effective management of these processes. This section applies previously introduced equilibrium concepts and tools to systems involving dissolution and precipitation."}
{"id": 4253, "contents": "1594. The Solubility Product - \nRecall from the chapter on solutions that the solubility of a substance can vary from essentially zero (insoluble or sparingly soluble) to infinity (miscible). A solute with finite solubility can yield a saturated solution when it is added to a solvent in an amount exceeding its solubility, resulting in a heterogeneous mixture of the saturated solution and the excess, undissolved solute. For example, a saturated solution of silver chloride is one in which the equilibrium shown below has been established.\n\nIn this solution, an excess of solid AgCl dissolves and dissociates to produce aqueous $\\mathrm{Ag}^{+}$and $\\mathrm{Cl}^{-}$ions at the same rate that these aqueous ions combine and precipitate to form solid AgCl (Figure 15.2). Because silver chloride is a sparingly soluble salt, the equilibrium concentration of its dissolved ions in the solution is relatively low.\n\n\nFIGURE 15.2 Silver chloride is a sparingly soluble ionic solid. When it is added to water, it dissolves slightly and produces a mixture consisting of a very dilute solution of $\\mathrm{Ag}^{+}$and $\\mathrm{Cl}^{-}$ions in equilibrium with undissolved silver chloride.\n\nThe equilibrium constant for solubility equilibria such as this one is called the solubility product constant, $\\boldsymbol{K}_{\\mathbf{s p}}$, in this case\n\n$$\n\\mathrm{AgCl}(s) \\rightleftharpoons \\mathrm{Ag}^{+}(a q)+\\mathrm{Cl}^{-}(a q) \\quad K_{\\mathrm{sp}}=\\left[\\mathrm{Ag}^{+}(a q)\\right]\\left[\\mathrm{Cl}^{-}(a q)\\right]\n$$\n\nRecall that only gases and solutes are represented in equilibrium constant expressions, so the $K_{\\text {sp }}$ does not include a term for the undissolved AgCl . A listing of solubility product constants for several sparingly soluble compounds is provided in Appendix J."}
{"id": 4254, "contents": "1596. Writing Equations and Solubility Products - \nWrite the dissolution equation and the solubility product expression for each of the following slightly soluble ionic compounds:\n(a) AgI, silver iodide, a solid with antiseptic properties\n(b) $\\mathrm{CaCO}_{3}$, calcium carbonate, the active ingredient in many over-the-counter chewable antacids\n(c) $\\mathrm{Mg}(\\mathrm{OH})_{2}$, magnesium hydroxide, the active ingredient in Milk of Magnesia\n(d) $\\mathrm{Mg}\\left(\\mathrm{NH}_{4}\\right) \\mathrm{PO}_{4}$, magnesium ammonium phosphate, an essentially insoluble substance used in tests for\nmagnesium\n(e) $\\mathrm{Ca}_{5}\\left(\\mathrm{PO}_{4}\\right)_{3} \\mathrm{OH}$, the mineral apatite, a source of phosphate for fertilizers"}
{"id": 4255, "contents": "1596. Writing Equations and Solubility Products - \nSolution\n(a) $\\mathrm{AgI}(s) \\rightleftharpoons \\mathrm{Ag}^{+}(a q)+\\mathrm{I}^{-}(a q) \\quad K_{\\text {sp }}=\\left[\\mathrm{Ag}^{+}\\right]\\left[\\mathrm{I}^{-}\\right]$\n(b) $\\mathrm{CaCO}_{3}(s) \\rightleftharpoons \\mathrm{Ca}^{2+}(a q)+\\mathrm{CO}_{3}{ }^{2-}(a q)$\n$K_{\\mathrm{sp}}=\\left[\\mathrm{Ca}^{2+}\\right]\\left[\\mathrm{CO}_{3}{ }^{2-}\\right]$\n(c) $\\mathrm{Mg}(\\mathrm{OH})_{2}(s) \\rightleftharpoons \\mathrm{Mg}^{2+}(a q)+2 \\mathrm{OH}^{-}(a q)$\n$K_{\\mathrm{sp}}=\\left[\\mathrm{Mg}^{2+}\\right]\\left[\\mathrm{OH}^{-}\\right]^{2}$\n(d) $\\mathrm{Mg}\\left(\\mathrm{NH}_{4}\\right) \\mathrm{PO}_{4}(s) \\rightleftharpoons \\mathrm{Mg}^{2+}(a q)+\\mathrm{NH}_{4}{ }^{+}(a q)+\\mathrm{PO}_{4}{ }^{3-}(a q)$\n$K_{\\text {sp }}=\\left[\\mathrm{Mg}^{2+}\\right]\\left[\\mathrm{NH}_{4}^{+}\\right]\\left[\\mathrm{PO}_{4}{ }^{3-}\\right]$\n(e) $\\mathrm{Ca}_{5}\\left(\\mathrm{PO}_{4}\\right) 3 \\mathrm{OH}(s) \\rightleftharpoons 5 \\mathrm{Ca}^{2+}(a q)+3 \\mathrm{PO}_{4}{ }^{3-}(a q)+\\mathrm{OH}^{-}(a q) \\quad K_{\\mathrm{sp}}=\\left[\\mathrm{Ca}^{2+}\\right]^{5}\\left[\\mathrm{PO}_{4}^{3-}\\right]^{3}\\left[\\mathrm{OH}^{-}\\right]$"}
{"id": 4256, "contents": "1596. Writing Equations and Solubility Products - \nCheck Your Learning\nWrite the dissolution equation and the solubility product for each of the following slightly soluble compounds:\n(a) $\\mathrm{BaSO}_{4}$\n(b) $\\mathrm{Ag}_{2} \\mathrm{SO}_{4}$\n(c) $\\mathrm{Al}(\\mathrm{OH})_{3}$\n(d) $\\mathrm{Pb}(\\mathrm{OH}) \\mathrm{Cl}$"}
{"id": 4257, "contents": "1597. Answer: - \n(a) $\\mathrm{BaSO}_{4}(s) \\rightleftharpoons \\mathrm{Ba}^{2+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q) \\quad K_{\\mathrm{sp}}=\\left[\\mathrm{Ba}^{2+}\\right]\\left[\\mathrm{SO}_{4}{ }^{2-}\\right]$;\n(b) $\\mathrm{Ag}_{2} \\mathrm{SO}_{4}(s) \\rightleftharpoons 2 \\mathrm{Ag}^{+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q)$\n$K_{\\mathrm{sp}}=\\left[\\mathrm{Ag}^{+}\\right]^{2}\\left[\\mathrm{SO}_{4}{ }^{2-}\\right] ;$\n(c) $\\mathrm{Al}(\\mathrm{OH})_{3}(s) \\rightleftharpoons \\mathrm{Al}^{3+}(a q)+3 \\mathrm{OH}^{-}(a q)$\n$K_{\\mathrm{sp}}=\\left[\\mathrm{Al}^{3+}\\right]\\left[\\mathrm{OH}^{-}\\right]^{3}$;\n(d) $\\mathrm{Pb}(\\mathrm{OH}) \\mathrm{Cl}(s) \\rightleftharpoons \\mathrm{Pb}^{2+}(a q)+\\mathrm{OH}^{-}(a q)+\\mathrm{Cl}^{-}(a q)$\n$K_{\\mathrm{sp}}=\\left[\\mathrm{Pb}^{2+}\\right]\\left[\\mathrm{OH}^{-}\\right]\\left[\\mathrm{Cl}^{-}\\right]$"}
{"id": 4258, "contents": "1598. $K_{\\text {sp }}$ and Solubility - \nThe $K_{\\text {sp }}$ of a slightly soluble ionic compound may be simply related to its measured solubility provided the dissolution process involves only dissociation and solvation, for example:\n\n$$\n\\mathrm{M}_{p} \\mathrm{X}_{q}(s) \\rightleftharpoons p \\mathrm{M}^{\\mathrm{m}+}(a q)+q \\mathrm{X}^{\\mathrm{n}-}(a q)\n$$\n\nFor cases such as these, one may derive $K_{\\text {sp }}$ values from provided solubilities, or vice-versa. Calculations of this sort are most conveniently performed using a compound's molar solubility, measured as moles of dissolved solute per liter of saturated solution."}
{"id": 4259, "contents": "1600. Calculation of $\\boldsymbol{K}_{\\text {sp }}$ from Equilibrium Concentrations - \nFluorite, $\\mathrm{CaF}_{2}$, is a slightly soluble solid that dissolves according to the equation:\n\n$$\n\\mathrm{CaF}_{2}(s) \\rightleftharpoons \\mathrm{Ca}^{2+}(a q)+2 \\mathrm{~F}^{-}(a q)\n$$\n\nThe concentration of $\\mathrm{Ca}^{2+}$ in a saturated solution of $\\mathrm{CaF}_{2}$ is $2.15 \\times 10^{-4} \\mathrm{M}$. What is the solubility product of fluorite?"}
{"id": 4260, "contents": "1601. Solution - \nAccording to the stoichiometry of the dissolution equation, the fluoride ion molarity of a $\\mathrm{CaF}_{2}$ solution is equal to twice its calcium ion molarity:\n\n$$\n\\left[\\mathrm{F}^{-}\\right]=\\left(2 \\mathrm{~mol} \\mathrm{~F}^{-} / 1 \\mathrm{~mol} \\mathrm{Ca}^{2+}\\right)=(2)\\left(2.15 \\times 10^{-4} M\\right)=4.30 \\times 10^{-4} M\n$$\n\nSubstituting the ion concentrations into the $K_{\\mathrm{sp}}$ expression gives\n\n$$\nK_{\\mathrm{sp}}=\\left[\\mathrm{Ca}^{2+}\\right]\\left[\\mathrm{F}^{-}\\right]^{2}=\\left(2.15 \\times 10^{-4}\\right)\\left(4.30 \\times 10^{-4}\\right)^{2}=3.98 \\times 10^{-11}\n$$"}
{"id": 4261, "contents": "1602. Check Your Learning - \nIn a saturated solution of $\\mathrm{Mg}(\\mathrm{OH})_{2}$, the concentration of $\\mathrm{Mg}^{2+}$ is $1.31 \\times 10^{-4} \\mathrm{M}$. What is the solubility product for $\\mathrm{Mg}(\\mathrm{OH})_{2}$ ?\n\n$$\n\\mathrm{Mg}(\\mathrm{OH})_{2}(s) \\rightleftharpoons \\mathrm{Mg}^{2+}(a q)+2 \\mathrm{OH}^{-}(a q)\n$$"}
{"id": 4262, "contents": "1603. Answer: - \n$8.99 \\times 10^{-12}$"}
{"id": 4263, "contents": "1605. Determination of Molar Solubility from $\\boldsymbol{K}_{\\mathbf{s p}}$ - \nThe $K_{\\mathrm{sp}}$ of copper(I) bromide, CuBr , is $6.3 \\times 10^{-9}$. Calculate the molar solubility of copper bromide."}
{"id": 4264, "contents": "1606. Solution - \nThe dissolution equation and solubility product expression are\n\n$$\n\\begin{gathered}\n\\mathrm{CuBr}(s) \\rightleftharpoons \\mathrm{Cu}^{+}(a q)+\\mathrm{Br}^{-}(a q) \\\\\nK_{\\mathrm{sp}}=\\left[\\mathrm{Cu}^{+}\\right]\\left[\\mathrm{Br}^{-}\\right]\n\\end{gathered}\n$$\n\nFollowing the ICE approach to this calculation yields the table\n\n|  | $\\mathrm{CuBr}(s) \\rightleftharpoons \\mathrm{Cu}^{+}(a q) \\quad+\\mathrm{Br}^{-}(\\mathrm{aq})$ |  |  |\n| :---: | :---: | :---: | :---: |\n| Initial concentration $(M)$ |  | 0 | 0 |\n| Change $(M)$ | $+x$ | $+x$ |  |\n| Equilibrium concentration $(M)$ |  | $x$ | $x$ |\n\nSubstituting the equilibrium concentration terms into the solubility product expression and solving for $x$ yields\n\n$$\n\\begin{gathered}\nK_{\\mathrm{sp}}=\\left[\\mathrm{Cu}^{+}\\right]\\left[\\mathrm{Br}^{-}\\right] \\\\\n6.3 \\times 10^{-9}=(x)(x)=x^{2} \\\\\nx=\\sqrt{\\left(6.3 \\times 10^{-9}\\right)}=7.9 \\times 10^{-5} M\n\\end{gathered}\n$$\n\nSince the dissolution stoichiometry shows one mole of copper(I) ion and one mole of bromide ion are produced for each moles of Br dissolved, the molar solubility of CuBr is $7.9 \\times 10^{-5} \\mathrm{M}$."}
{"id": 4265, "contents": "1607. Check Your Learning - \nThe $K_{\\mathrm{sp}}$ of AgI is $1.5 \\times 10^{-16}$. Calculate the molar solubility of silver iodide."}
{"id": 4266, "contents": "1608. Answer: - \n$1.2 \\times 10^{-8} \\mathrm{M}$"}
{"id": 4267, "contents": "1610. Determination of Molar Solubility from $\\boldsymbol{K}_{\\text {sp }}$ - \nThe $K_{\\mathrm{sp}}$ of calcium hydroxide, $\\mathrm{Ca}(\\mathrm{OH})_{2}$, is $1.3 \\times 10^{-6}$. Calculate the molar solubility of calcium hydroxide."}
{"id": 4268, "contents": "1611. Solution - \nThe dissolution equation and solubility product expression are\n\n$$\n\\begin{aligned}\n\\mathrm{Ca}(\\mathrm{OH})_{2}(s) & \\rightleftharpoons \\mathrm{Ca}^{2+}(a q)+2 \\mathrm{OH}^{-}(a q) \\\\\nK_{\\text {sp }} & =\\left[\\mathrm{Ca}^{2+}\\right]\\left[\\mathrm{OH}^{-}\\right]^{2}\n\\end{aligned}\n$$\n\nThe ICE table for this system is\n\n|  | $\\mathbf{C a}(\\mathbf{O H})_{2}(s) \\rightleftharpoons \\mathrm{Ca}^{2+}(\\mathrm{aq}) \\mathbf{+} \\mathbf{2 O H}^{-}(\\mathrm{aq})$ |  |  |\n| :---: | :---: | :---: | :---: |\n| Initial concentration $(M)$ |  | 0 | 0 |\n| Change $(M)$ |  | $+x$ | $+2 x$ |\n| Equilibrium concentration $(M)$ |  | $x$ | $2 x$ |\n\nSubstituting terms for the equilibrium concentrations into the solubility product expression and solving for $x$ gives\n\n$$\n\\begin{gathered}\nK_{\\mathrm{sp}}=\\left[\\mathrm{Ca}^{2+}\\right]\\left[\\mathrm{OH}^{-}\\right]^{2} \\\\\n1.3 \\times 10^{-6}=(x)(2 x)^{2}=(x)\\left(4 x^{2}\\right)=4 x^{3} \\\\\nx=\\sqrt[3]{\\frac{1.3 \\times 10^{-6}}{4}}=6.9 \\times 10^{-3} M\n\\end{gathered}\n$$\n\nAs defined in the ICE table, $x$ is the molarity of calcium ion in the saturated solution. The dissolution stoichiometry shows a 1:1 relation between moles of calcium ion in solution and moles of compound dissolved, and so, the molar solubility of $\\mathrm{Ca}(\\mathrm{OH})_{2}$ is $6.9 \\times 10^{-3} \\mathrm{M}$."}
{"id": 4269, "contents": "1612. Check Your Learning - \nThe $K_{\\mathrm{sp}}$ of $\\mathrm{PbI}_{2}$ is $1.4 \\times 10^{-8}$. Calculate the molar solubility of lead(II) iodide."}
{"id": 4270, "contents": "1613. Answer: - \n$1.5 \\times 10^{-3} \\mathrm{M}$"}
{"id": 4271, "contents": "1615. Determination of $\\boldsymbol{K}_{\\text {sp }}$ from Gram Solubility - \nMany of the pigments used by artists in oil-based paints (Figure 15.3) are sparingly soluble in water. For example, the solubility of the artist's pigment chrome yellow, $\\mathrm{PbCrO}_{4}$, is $4.6 \\times 10^{-6} \\mathrm{~g} / \\mathrm{L}$. Determine the solubility product for $\\mathrm{PbCrO}_{4}$.\n\n\nFIGURE 15.3 Oil paints contain pigments that are very slightly soluble in water. In addition to chrome yellow $\\left(\\mathrm{PbCrO}_{4}\\right)$, examples include Prussian blue $\\left(\\mathrm{Fe}_{7}(\\mathrm{CN})_{18}\\right)$, the reddish-orange color vermilion $(\\mathrm{HgS})$, and green color veridian $\\left(\\mathrm{Cr}_{2} \\mathrm{O}_{3}\\right)$. (credit: Sonny Abesamis)"}
{"id": 4272, "contents": "1616. Solution - \nBefore calculating the solubility product, the provided solubility must be converted to molarity:\n\n$$\n\\begin{gathered}\n{\\left[\\mathrm{PbCrO}_{4}\\right]=\\frac{4.6 \\times 10^{-6} \\mathrm{~g} \\mathrm{PbCrO}_{4}}{1 \\mathrm{~L}} \\times \\frac{1 \\mathrm{~mol} \\mathrm{PbCrO}_{4}}{323.2 \\mathrm{~g} \\mathrm{PbCrO}_{4}}} \\\\\n=\\frac{1.4 \\times 10^{-8} \\mathrm{~mol} \\mathrm{PbCrO}_{4}}{1 \\mathrm{~L}} \\\\\n=1.4 \\times 10^{-8} \\mathrm{M}\n\\end{gathered}\n$$\n\nThe dissolution equation for this compound is\n\n$$\n\\mathrm{PbCrO}_{4}(s) \\rightleftharpoons \\mathrm{Pb}^{2+}(a q)+\\mathrm{CrO}_{4}^{2-}(a q)\n$$\n\nThe dissolution stoichiometry shows a 1:1 relation between the molar amounts of compound and its two ions, and so both $\\left[\\mathrm{Pb}^{2+}\\right]$ and $\\left[\\mathrm{CrO}_{4}{ }^{2-}\\right]$ are equal to the molar solubility of $\\mathrm{PbCrO}_{4}$ :\n\n$$\n\\left[\\mathrm{Pb}^{2+}\\right]=\\left[\\mathrm{CrO}_{4}{ }^{2-}\\right]=1.4 \\times 10^{-8} \\mathrm{M}\n$$\n\n$K_{\\mathrm{sp}}=\\left[\\mathrm{Pb}^{2+}\\right]\\left[\\mathrm{CrO}_{4}{ }^{2-}\\right]=\\left(1.4 \\times 10^{-8}\\right)\\left(1.4 \\times 10^{-8}\\right)=2.0 \\times 10^{-16}$"}
{"id": 4273, "contents": "1617. Check Your Learning - \nThe solubility of TlCl [thallium(I) chloride], an intermediate formed when thallium is being isolated from ores, is 3.12 grams per liter at $20^{\\circ} \\mathrm{C}$. What is its solubility product?"}
{"id": 4274, "contents": "1618. Answer: - \n$1.69 \\times 10^{-4}$"}
{"id": 4275, "contents": "1620. Calculating the Solubility of $\\mathrm{Hg}_{2} \\mathbf{C l}_{\\mathbf{2}}$ - \nCalomel, $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$, is a compound composed of the diatomic ion of mercury(I), $\\mathrm{Hg}_{2}{ }^{2+}$, and chloride ions, $\\mathrm{Cl}^{-}$. Although most mercury compounds are now known to be poisonous, eighteenth-century physicians used calomel as a medication. Their patients rarely suffered any mercury poisoning from the treatments because calomel has a very low solubility, as suggested by its very small $K_{\\mathrm{sp}}$ :\n\n$$\n\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}(s) \\rightleftharpoons \\mathrm{Hg}_{2}^{2+}(a q)+2 \\mathrm{Cl}^{-}(a q) \\quad K_{\\mathrm{sp}}=1.1 \\times 10^{-18}\n$$\n\nCalculate the molar solubility of $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$."}
{"id": 4276, "contents": "1621. Solution - \nThe dissolution stoichiometry shows a 1:1 relation between the amount of compound dissolved and the amount of mercury(I) ions, and so the molar solubility of $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$ is equal to the concentration of $\\mathrm{Hg}_{2}{ }^{2+}$ ions\n\nFollowing the ICE approach results in\n\n|  | $\\mathbf{H g}_{2} \\mathrm{Cl}_{2}(\\mathbf{s}) \\rightleftharpoons \\mathbf{H g}_{2}{ }^{2+}(a q) \\mathbf{+} \\mathbf{2 C l}(a q)$ |  |  |\n| :---: | :---: | :---: | :---: |\n| Initial concentration $(M)$ |  | 0 | 0 |\n| Change $(M)$ |  | $+x$ | $+2 x$ |\n| Equilibrium concentration $(M)$ | $x$ | $2 x$ |  |\n\nSubstituting the equilibrium concentration terms into the solubility product expression and solving for $x$ gives\n\n$$\n\\begin{gathered}\nK_{\\mathrm{sp}}=\\left[\\mathrm{Hg}_{2}^{2+}\\right]\\left[\\mathrm{Cl}^{-}\\right]^{2} \\\\\n1.1 \\times 10^{-18}=(x)(2 x)^{2} \\\\\n4 x^{3}=1.1 \\times 10^{-18} \\\\\nx=\\sqrt[3]{\\left(\\frac{1.1 \\times 10^{-18}}{4}\\right)}=6.5 \\times 10^{-7} M \\\\\n{\\left[\\mathrm{Hg}_{2}^{2+}\\right]=6.5 \\times 10^{-7} M=6.5 \\times 10^{-7} M} \\\\\n{\\left[\\mathrm{Cl}^{-}\\right]=2 x=2\\left(6.5 \\times 10^{-7}\\right)=1.3 \\times 10^{-6} M}\n\\end{gathered}\n$$"}
{"id": 4277, "contents": "1621. Solution - \nThe dissolution stoichiometry shows the molar solubility of $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$ is equal to $\\left[\\mathrm{Hg}_{2}{ }^{2+}\\right]$, or $6.5 \\times 10^{-7} \\mathrm{M}$."}
{"id": 4278, "contents": "1622. Check Your Learning - \nDetermine the molar solubility of $\\mathrm{MgF}_{2}$ from its solubility product: $K_{\\mathrm{sp}}=6.4 \\times 10^{-9}$."}
{"id": 4279, "contents": "1623. Answer: - \n$1.2 \\times 10^{-3} \\mathrm{M}$"}
{"id": 4280, "contents": "1625. Using Barium Sulfate for Medical Imaging - \nVarious types of medical imaging techniques are used to aid diagnoses of illnesses in a noninvasive manner. One such technique utilizes the ingestion of a barium compound before taking an X-ray image. A suspension of barium sulfate, a chalky powder, is ingested by the patient. Since the $K_{\\text {sp }}$ of barium sulfate is $2.3 \\times 10^{-8}$, very little of it dissolves as it coats the lining of the patient's intestinal tract. Barium-coated areas of the digestive tract then appear on an X-ray as white, allowing for greater visual detail than a traditional X-ray (Figure 15.4).\n\n\nFIGURE 15.4 A suspension of barium sulfate coats the intestinal tract, permitting greater visual detail than a traditional X-ray. (credit modification of work by \"glitzy queen00\"/Wikimedia Commons)\n\nMedical imaging using barium sulfate can be used to diagnose acid reflux disease, Crohn's disease, and ulcers in addition to other conditions.\n\nVisit this website (http://openstax.org/l/16barium) for more information on how barium is used in medical diagnoses and which conditions it is used to diagnose."}
{"id": 4281, "contents": "1626. Predicting Precipitation - \nThe equation that describes the equilibrium between solid calcium carbonate and its solvated ions is:\n\n$$\n\\mathrm{CaCO}_{3}(s) \\rightleftharpoons \\mathrm{Ca}^{2+}(a q)+\\mathrm{CO}_{3}^{2-}(a q) \\quad K_{s p}=\\left[\\mathrm{Ca}^{2+}\\right]\\left[\\mathrm{CO}_{3}^{2-}\\right]=8.7 \\times 10^{-9}\n$$\n\nIt is important to realize that this equilibrium is established in any aqueous solution containing $\\mathrm{Ca}^{2+}$ and $\\mathrm{CO}_{3}{ }^{2-}$ ions, not just in a solution formed by saturating water with calcium carbonate. Consider, for example, mixing aqueous solutions of the soluble compounds sodium carbonate and calcium nitrate. If the concentrations of calcium and carbonate ions in the mixture do not yield a reaction quotient, $Q_{s p}$, that exceeds the solubility product, $K_{\\text {sp }}$, then no precipitation will occur. If the ion concentrations yield a reaction quotient greater than the solubility product, then precipitation will occur, lowering those concentrations until equilibrium is established ( $Q_{s p}=K_{\\mathrm{sp}}$ ). The comparison of $Q_{s p}$ to $K_{\\mathrm{sp}}$ to predict precipitation is an example of the general approach to predicting the direction of a reaction first introduced in the chapter on equilibrium. For the specific case of solubility equilibria:\n$Q_{s p}<K_{\\mathrm{sp}}$ : the reaction proceeds in the forward direction (solution is not saturated; no precipitation observed)\n$Q_{s p}>K_{\\mathrm{sp}}$ : the reaction proceeds in the reverse direction (solution is supersaturated; precipitation will occur)\nThis predictive strategy and related calculations are demonstrated in the next few example exercises."}
{"id": 4282, "contents": "1628. Precipitation of $\\mathrm{Mg}(\\mathrm{OH})_{\\mathbf{2}}$ - \nThe first step in the preparation of magnesium metal is the precipitation of $\\mathrm{Mg}(\\mathrm{OH})_{2}$ from sea water by the addition of lime, $\\mathrm{Ca}(\\mathrm{OH})_{2}$, a readily available inexpensive source of $\\mathrm{OH}^{-}$ion:\n\n$$\n\\mathrm{Mg}(\\mathrm{OH})_{2}(s) \\rightleftharpoons \\mathrm{Mg}^{2+}(a q)+2 \\mathrm{OH}^{-}(a q) \\quad K_{\\mathrm{sp}}=8.9 \\times 10^{-12}\n$$\n\nThe concentration of $\\mathrm{Mg}^{2+}(\\mathrm{aq})$ in sea water is 0.0537 M . Will $\\mathrm{Mg}(\\mathrm{OH})_{2}$ precipitate when enough $\\mathrm{Ca}(\\mathrm{OH})_{2}$ is added to give a $\\left[\\mathrm{OH}^{-}\\right]$of 0.0010 M ?"}
{"id": 4283, "contents": "1629. Solution - \nCalculation of the reaction quotient under these conditions is shown here:\n\n$$\nQ=\\left[\\mathrm{Mg}^{2+}\\right]\\left[\\mathrm{OH}^{-}\\right]^{2}=(0.0537)(0.0010)^{2}=5.4 \\times 10^{-8}\n$$\n\nBecause $Q$ is greater than $K_{\\mathrm{sp}}\\left(Q=5.4 \\times 10^{-8}\\right.$ is larger than $K_{\\mathrm{sp}}=8.9 \\times 10^{-12}$ ), the reverse reaction will proceed, precipitating magnesium hydroxide until the dissolved ion concentrations have been sufficiently lowered, so that $Q_{s p}=K_{\\mathrm{sp}}$."}
{"id": 4284, "contents": "1630. Check Your Learning - \nPredict whether $\\mathrm{CaHPO}_{4}$ will precipitate from a solution with $\\left[\\mathrm{Ca}^{2+}\\right]=0.0001 \\mathrm{M}$ and $\\left[\\mathrm{HPO}_{4}{ }^{2-}\\right]=0.001 \\mathrm{M}$."}
{"id": 4285, "contents": "1631. Answer: - \nNo precipitation of $\\mathrm{CaHPO}_{4} ; Q=1 \\times 10^{-7}$, which is less than $K_{\\text {sp }}\\left(7 \\times 10^{-7}\\right)$"}
{"id": 4286, "contents": "1633. Precipitation of AgCl - \nDoes silver chloride precipitate when equal volumes of a $2.0 \\times 10^{-4}-M$ solution of $\\mathrm{AgNO}_{3}$ and a $2.0 \\times 10^{-4}-M$ solution of NaCl are mixed?"}
{"id": 4287, "contents": "1634. Solution - \nThe equation for the equilibrium between solid silver chloride, silver ion, and chloride ion is:\n\n$$\n\\operatorname{AgCl}(s) \\rightleftharpoons \\mathrm{Ag}^{+}(a q)+\\mathrm{Cl}^{-}(a q)\n$$\n\nThe solubility product is $1.6 \\times 10^{-10}$ (see Appendix J).\nAgCl will precipitate if the reaction quotient calculated from the concentrations in the mixture of $\\mathrm{AgNO}_{3}$ and NaCl is greater than $K_{\\text {sp }}$. Because the volume doubles when equal volumes of $\\mathrm{AgNO}_{3}$ and NaCl solutions are mixed, each concentration is reduced to half its initial value\n\n$$\n\\frac{1}{2}\\left(2.0 \\times 10^{-4}\\right) M=1.0 \\times 10^{-4} M\n$$\n\nThe reaction quotient, $Q$, is greater than $K_{\\mathrm{sp}}$ for AgCl , so a supersaturated solution is formed:\n\n$$\nQ=\\left[\\mathrm{Ag}^{+}\\right]\\left[\\mathrm{Cl}^{-}\\right]=\\left(1.0 \\times 10^{-4}\\right)\\left(1.0 \\times 10^{-4}\\right)=1.0 \\times 10^{-8}>K_{\\mathrm{sp}}\n$$\n\nAgCl will precipitate from the mixture until the dissolution equilibrium is established, with $Q$ equal to $K_{\\mathrm{sp}}$."}
{"id": 4288, "contents": "1635. Check Your Learning - \nWill $\\mathrm{KClO}_{4}$ precipitate when 20 mL of a $0.050-M$ solution of $\\mathrm{K}^{+}$is added to 80 mL of a $0.50-M$ solution of $\\mathrm{ClO}_{4}{ }^{-}$? (Hint: Use the dilution equation to calculate the concentrations of potassium and perchlorate ions in the mixture.)"}
{"id": 4289, "contents": "1636. Answer: - \nNo, $Q=4.0 \\times 10^{-3}$, which is less than $K_{\\text {sp }}=1.05 \\times 10^{-2}$"}
{"id": 4290, "contents": "1638. Precipitation of Calcium Oxalate - \nBlood will not clot if calcium ions are removed from its plasma. Some blood collection tubes contain salts of the oxalate ion, $\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}$, for this purpose (Figure 15.5). At sufficiently high concentrations, the calcium and oxalate ions form solid, $\\mathrm{CaC}_{2} \\mathrm{O}_{4} \\cdot \\mathrm{H}_{2} \\mathrm{O}$ (calcium oxalate monohydrate). The concentration of $\\mathrm{Ca}^{2+}$ in a sample of blood serum is $2.2 \\times 10^{-3} \\mathrm{M}$. What concentration of $\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}$ ion must be established before $\\mathrm{CaC}_{2} \\mathrm{O}_{4} \\cdot \\mathrm{H}_{2} \\mathrm{O}$ begins to precipitate?\n\n\nFIGURE 15.5 Anticoagulants can be added to blood that will combine with the $\\mathrm{Ca}^{2+}$ ions in blood serum and prevent the blood from clotting. (credit: modification of work by Neeta Lind)"}
{"id": 4291, "contents": "1639. Solution - \nThe equilibrium expression is:\n\n$$\n\\mathrm{CaC}_{2} \\mathrm{O}_{4}(s) \\rightleftharpoons \\mathrm{Ca}^{2+}(a q)+\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}(a q)\n$$\n\nFor this reaction:\n\n$$\nK_{\\mathrm{sp}}=\\left[\\mathrm{Ca}^{2+}\\right]\\left[\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right]=1.96 \\times 10^{-8}\n$$\n\n(see Appendix J)\nSubstitute the provided calcium ion concentration into the solubility product expression and solve for oxalate concentration:\n\n$$\n\\begin{gathered}\nQ=K_{\\text {sp }}=\\left[\\mathrm{Ca}^{2+}\\right]\\left[\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right]=1.96 \\times 10^{-8} \\\\\n\\left(2.2 \\times 10^{-3}\\right)\\left[\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right]=1.96 \\times 10^{-8} \\\\\n{\\left[\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right]=\\frac{1.96 \\times 10^{-8}}{2.2 \\times 10^{-3}}=8.9 \\times 10^{-6} \\mathrm{M}}\n\\end{gathered}\n$$\n\nA concentration of $\\left[\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right]=8.9 \\times 10^{-6} M$ is necessary to initiate the precipitation of $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$ under these\nconditions."}
{"id": 4292, "contents": "1640. Check Your Learning - \nIf a solution contains 0.0020 mol of $\\mathrm{CrO}_{4}{ }^{2-}$ per liter, what concentration of $\\mathrm{Ag}^{+}$ion must be reached by adding solid $\\mathrm{AgNO}_{3}$ before $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}$ begins to precipitate? Neglect any increase in volume upon adding the solid silver nitrate."}
{"id": 4293, "contents": "1641. Answer: - \n$6.7 \\times 10^{-5} \\mathrm{M}$"}
{"id": 4294, "contents": "1643. Concentrations Following Precipitation - \nClothing washed in water that has a manganese $\\left[\\mathrm{Mn}^{2+}(\\mathrm{aq})\\right]$ concentration exceeding $0.1 \\mathrm{mg} / \\mathrm{L}\\left(1.8 \\times 10^{-6} \\mathrm{M}\\right)$ may be stained by the manganese upon oxidation, but the amount of $\\mathrm{Mn}^{2+}$ in the water can be decreased by adding a base to precipitate $\\mathrm{Mn}(\\mathrm{OH})_{2}$. What pH is required to keep $\\left[\\mathrm{Mn}^{2+}\\right.$ ] equal to $1.8 \\times 10^{-6} \\mathrm{M}$ ?"}
{"id": 4295, "contents": "1644. Solution - \nThe dissolution of $\\mathrm{Mn}(\\mathrm{OH})_{2}$ is described by the equation:\n\n$$\n\\mathrm{Mn}(\\mathrm{OH})_{2}(s) \\rightleftharpoons \\mathrm{Mn}^{2+}(a q)+2 \\mathrm{OH}^{-}(a q) \\quad K_{\\mathrm{sp}}=2 \\times 10^{-13}\n$$\n\nAt equilibrium:\n\n$$\nK_{\\mathrm{sp}}=\\left[\\mathrm{Mn}^{2+}\\right]\\left[\\mathrm{OH}^{-}\\right]^{2}\n$$\n\nor\n\n$$\n\\left(1.8 \\times 10^{-6}\\right)\\left[\\mathrm{OH}^{-}\\right]^{2}=2 \\times 10^{-13}\n$$\n\nso\n\n$$\n\\left[\\mathrm{OH}^{-}\\right]=3.3 \\times 10^{-4} \\mathrm{M}\n$$\n\nCalculate the pH from the pOH :\n\n$$\n\\begin{gathered}\n\\mathrm{pOH}=-\\log \\left[\\mathrm{OH}^{-}\\right]=-\\log (3.3 \\times 10-4)=3.48 \\\\\n\\mathrm{pH}=14.00-\\mathrm{pOH}=14.00-3.48=10.52\n\\end{gathered}\n$$\n\n(final result rounded to one significant digit, limited by the certainty of the $K_{\\mathrm{sp}}$ )"}
{"id": 4296, "contents": "1645. Check Your Learning - \nThe first step in the preparation of magnesium metal is the precipitation of $\\mathrm{Mg}(\\mathrm{OH})_{2}$ from sea water by the addition of $\\mathrm{Ca}(\\mathrm{OH})_{2}$. The concentration of $\\mathrm{Mg}^{2+}(\\mathrm{aq})$ in sea water is $5.37 \\times 10^{-2} \\mathrm{M}$. Calculate the pH at which $\\left[\\mathrm{Mg}^{2+}\\right]$ is decreased to $1.0 \\times 10^{-5} \\mathrm{M}$"}
{"id": 4297, "contents": "1646. Answer: - \n10.97\n\nIn solutions containing two or more ions that may form insoluble compounds with the same counter ion, an experimental strategy called selective precipitation may be used to remove individual ions from solution. By increasing the counter ion concentration in a controlled manner, ions in solution may be precipitated individually, assuming their compound solubilities are adequately different. In solutions with equal concentrations of target ions, the ion forming the least soluble compound will precipitate first (at the lowest concentration of counter ion), with the other ions subsequently precipitating as their compound's solubilities are reached. As an illustration of this technique, the next example exercise describes separation of a two halide ions via precipitation of one as a silver salt."}
{"id": 4298, "contents": "1648. The Role of Precipitation in Wastewater Treatment - \nSolubility equilibria are useful tools in the treatment of wastewater carried out in facilities that may treat the municipal water in your city or town (Figure 15.6). Specifically, selective precipitation is used to remove contaminants from wastewater before it is released back into natural bodies of water. For example, phosphate ions $\\left(\\mathrm{PO}_{4}{ }^{3-}\\right)$ are often present in the water discharged from manufacturing facilities. An abundance of phosphate causes excess algae to grow, which impacts the amount of oxygen available for marine life as well as making water unsuitable for human consumption.\n\n\nFIGURE 15.6 Wastewater treatment facilities, such as this one, remove contaminants from wastewater before the water is released back into the natural environment. (credit: \"eutrophication\\&hypoxia\"/Wikimedia Commons)\n\nOne common way to remove phosphates from water is by the addition of calcium hydroxide, or lime, $\\mathrm{Ca}(\\mathrm{OH})_{2}$. As the water is made more basic, the calcium ions react with phosphate ions to produce hydroxylapatite, $\\mathrm{Ca}_{5}(\\mathrm{PO} 4)_{3} \\mathrm{OH}$, which then precipitates out of the solution:\n\n$$\n5 \\mathrm{Ca}^{2+}+3 \\mathrm{PO}_{4}{ }^{3-}+\\mathrm{OH}^{-} \\rightleftharpoons \\mathrm{Ca}_{5}\\left(\\mathrm{PO}_{4}\\right)_{3} \\cdot \\mathrm{OH}(s)\n$$\n\nBecause the amount of calcium ion added does not result in exceeding the solubility products for other calcium salts, the anions of those salts remain behind in the wastewater. The precipitate is then removed by filtration and the water is brought back to a neutral pH by the addition of $\\mathrm{CO}_{2}$ in a recarbonation process. Other chemicals can also be used for the removal of phosphates by precipitation, including iron(III) chloride and aluminum sulfate.\n\nView this site (http://openstax.org/l/16Wastewater) for more information on how phosphorus is removed from wastewater."}
{"id": 4299, "contents": "1650. Precipitation of Silver Halides - \nA solution contains 0.00010 mol of KBr and 0.10 mol of KCl per liter. $\\mathrm{AgNO}_{3}$ is gradually added to this solution. Which forms first, solid AgBr or solid AgCl ?"}
{"id": 4300, "contents": "1651. Solution - \nThe two equilibria involved are:\n\n$$\n\\begin{array}{lll}\n\\mathrm{AgCl}(s) & \\mathrm{Ag}^{+}(a q)+\\mathrm{Cl}^{-}(a q) & K_{\\mathrm{sp}}=1.6 \\times 10^{-10} \\\\\n\\mathrm{AgBr}(s) \\rightleftharpoons \\mathrm{Ag}^{+}(a q)+\\mathrm{Br}^{-}(a q) & K_{\\mathrm{sp}}=5.0 \\times 10^{-13}\n\\end{array}\n$$\n\nIf the solution contained about equal concentrations of $\\mathrm{Cl}^{-}$and $\\mathrm{Br}^{-}$, then the silver salt with the smaller $K_{\\text {sp }}$ ( AgBr ) would precipitate first. The concentrations are not equal, however, so the $\\left[\\mathrm{Ag}^{+}\\right]$at which AgCl begins to precipitate and the $\\left[\\mathrm{Ag}^{+}\\right]$at which AgBr begins to precipitate must be calculated. The salt that forms at the lower $\\left[\\mathrm{Ag}^{+}\\right]$precipitates first.\n\nAgBr precipitates when $Q$ equals $K_{\\mathrm{sp}}$ for AgBr\n\n$$\n\\begin{gathered}\nQ_{\\mathrm{sp}}=K_{\\mathrm{sp}}=\\left[\\mathrm{Ag}^{+}\\right]\\left[\\mathrm{Br}^{-}\\right]=\\left\\mathrm{Ag}^{+}\\right=5.0 \\times 10^{-13} \\\\\n{\\left[\\mathrm{Ag}^{+}\\right]=\\frac{5.0 \\times 10^{-13}}{0.00010}=5.0 \\times 10^{-9} M}\n\\end{gathered}\n$$"}
{"id": 4301, "contents": "1651. Solution - \nAgBr begins to precipitate when $\\left[\\mathrm{Ag}^{+}\\right]$is $5.0 \\times 10^{-9} \\mathrm{M}$.\nFor AgCl : AgCl precipitates when $Q$ equals $K_{\\text {sp }}$ for $\\mathrm{AgCl}\\left(1.6 \\times 10^{-10}\\right)$. When $\\left[\\mathrm{Cl}^{-}\\right]=0.10 \\mathrm{M}$ :\n\n$$\n\\begin{gathered}\nQ_{\\mathrm{sp}}=K_{\\mathrm{sp}}=\\left[\\mathrm{Ag}^{+}\\right]\\left[\\mathrm{Cl}^{-}\\right]=\\left\\mathrm{Ag}^{+}\\right=1.6 \\times 10^{-10} \\\\\n{\\left[\\mathrm{Ag}^{+}\\right]=\\frac{1.6 \\times 10^{-10}}{0.10}=1.6 \\times 10^{-9} M}\n\\end{gathered}\n$$\n\nAgCl begins to precipitate when $\\left[\\mathrm{Ag}^{+}\\right]$is $1.6 \\times 10^{-9} \\mathrm{M}$.\nAgCl begins to precipitate at a lower $\\left[\\mathrm{Ag}^{+}\\right]$than AgBr , so AgCl begins to precipitate first. Note the chloride ion concentration of the initial mixture was significantly greater than the bromide ion concentration, and so silver chloride precipitated first despite having a $K_{\\mathrm{sp}}$ greater than that of silver bromide."}
{"id": 4302, "contents": "1652. Check Your Learning - \nIf silver nitrate solution is added to a solution which is 0.050 M in both $\\mathrm{Cl}^{-}$and $\\mathrm{Br}^{-}$ions, at what $\\left[\\mathrm{Ag}^{+}\\right]$would precipitation begin, and what would be the formula of the precipitate?"}
{"id": 4303, "contents": "1653. Answer: - \n$\\left[\\mathrm{Ag}^{+}\\right]=1.0 \\times 10^{-11} \\mathrm{M} ; \\mathrm{AgBr}$ precipitates first"}
{"id": 4304, "contents": "1654. Common Ion Effect - \nCompared with pure water, the solubility of an ionic compound is less in aqueous solutions containing a common ion (one also produced by dissolution of the ionic compound). This is an example of a phenomenon known as the common ion effect, which is a consequence of the law of mass action that may be explained using Le Ch\u00c2telier's principle. Consider the dissolution of silver iodide:\n\n$$\n\\operatorname{AgI}(s) \\rightleftharpoons \\mathrm{Ag}^{+}(a q)+\\mathrm{I}^{-}(a q)\n$$\n\nThis solubility equilibrium may be shifted left by the addition of either silver(I) or iodide ions, resulting in the precipitation of AgI and lowered concentrations of dissolved $\\mathrm{Ag}^{+}$and $\\mathrm{I}^{-}$. In solutions that already contain either of these ions, less AgI may be dissolved than in solutions without these ions.\n\nThis effect may also be explained in terms of mass action as represented in the solubility product expression:\n\n$$\nK_{\\mathrm{sp}}=\\left[\\mathrm{Ag}^{+}\\right]\\left[\\mathrm{I}^{-}\\right]\n$$\n\nThe mathematical product of silver(I) and iodide ion molarities is constant in an equilibrium mixture regardless of the source of the ions, and so an increase in one ion's concentration must be balanced by a proportional decrease in the other."}
{"id": 4305, "contents": "1655. LINK TO LEARNING - \nView this simulation (http://openstax.org/l/16solublesalts) to explore various aspects of the common ion effect."}
{"id": 4306, "contents": "1657. Common Ion Effect on Solubility - \nWhat is the effect on the amount of solid $\\mathrm{Mg}(\\mathrm{OH})_{2}$ and the concentrations of $\\mathrm{Mg}^{2+}$ and $\\mathrm{OH}^{-}$when each of the following are added to a saturated solution of $\\mathrm{Mg}(\\mathrm{OH})_{2}$ ?\n(a) $\\mathrm{MgCl}_{2}$\n(b) KOH\n(c) $\\mathrm{NaNO}_{3}$\n(d) $\\mathrm{Mg}(\\mathrm{OH})_{2}$"}
{"id": 4307, "contents": "1658. Solution - \nThe solubility equilibrium is\n\n$$\n\\mathrm{Mg}(\\mathrm{OH})_{2}(s) \\rightleftharpoons \\mathrm{Mg}^{2+}(a q)+2 \\mathrm{OH}^{-}(a q)\n$$\n\n(a) Adding a common ion, $\\mathrm{Mg}^{2+}$, will increase the concentration of this ion and shift the solubility equilibrium to the left, decreasing the concentration of hydroxide ion and increasing the amount of undissolved magnesium hydroxide.\n(b) Adding a common ion, $\\mathrm{OH}^{-}$, will increase the concentration of this ion and shift the solubility equilibrium to the left, decreasing the concentration of magnesium ion and increasing the amount of undissolved magnesium hydroxide.\n(c) The added compound does not contain a common ion, and no effect on the magnesium hydroxide solubility equilibrium is expected.\n(d) Adding more solid magnesium hydroxide will increase the amount of undissolved compound in the mixture. The solution is already saturated, though, so the concentrations of dissolved magnesium and hydroxide ions will remain the same.\n\n$$\nQ=\\left[\\mathrm{Mg}^{2+}\\right]\\left[\\mathrm{OH}^{-}\\right]^{2}\n$$\n\nThus, changing the amount of solid magnesium hydroxide in the mixture has no effect on the value of $Q$, and no shift is required to restore $Q$ to the value of the equilibrium constant."}
{"id": 4308, "contents": "1659. Check Your Learning - \nWhat is the effect on the amount of solid $\\mathrm{NiCO}_{3}$ and the concentrations of $\\mathrm{Ni}^{2+}$ and $\\mathrm{CO}_{3}{ }^{2-}$ when each of the following are added to a saturated solution of $\\mathrm{NiCO}_{3}$\n(a) $\\mathrm{Ni}\\left(\\mathrm{NO}_{3}\\right)_{2}$\n(b) $\\mathrm{KClO}_{4}$\n(c) $\\mathrm{NiCO}_{3}$\n(d) $\\mathrm{K}_{2} \\mathrm{CO}_{3}$"}
{"id": 4309, "contents": "1660. Answer: - \n(a) mass of $\\mathrm{NiCO}_{3}(s)$ increases, $\\left[\\mathrm{Ni}^{2+}\\right]$ increases, $\\left[\\mathrm{CO}_{3}{ }^{2-}\\right]$ decreases; (b) no appreciable effect; (c) no effect except to increase the amount of solid $\\mathrm{NiCO}_{3}$; (d) mass of $\\mathrm{NiCO}_{3}(s)$ increases, $\\left[\\mathrm{Ni}^{2+}\\right]$ decreases, $\\left[\\mathrm{CO}_{3}{ }^{2-}\\right]$ increases;"}
{"id": 4310, "contents": "1662. Common Ion Effect - \nCalculate the molar solubility of cadmium sulfide (CdS) in a $0.010-M$ solution of cadmium bromide $\\left(\\mathrm{CdBr}_{2}\\right)$. The $K_{\\mathrm{sp}}$ of CdS is $1.0 \\times 10^{-28}$."}
{"id": 4311, "contents": "1663. Solution - \nThis calculation can be performed using the ICE approach:\n\n| $\\mathrm{CdS}(s) \\rightleftharpoons \\mathrm{Cd}^{2+}(a q)+\\mathrm{S}^{2-}(a q)$ |  |  |  |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{CdS}(\\mathrm{s}) \\rightleftharpoons \\mathrm{Cd}^{2+}(\\mathrm{aq})+\\mathrm{s}^{2-}(\\mathrm{aq})$ |  |  |  |\n| Initial concentration $(M)$ |  | 0.010 | 0 |\n| Change (M) |  | $+x$ | $+x$ |\n| Equilibrium concentration $(M)$ |  | $0.010+x$ | $x$ |\n\nBecause $K_{\\text {sp }}$ is very small, assume $x \\ll 0.010$ and solve the simplified equation for $x$ :\n\n$$\n\\begin{gathered}\n(0.010)(x)=1.0 \\times 10^{-28} \\\\\nx=1.0 \\times 10^{-26} M\n\\end{gathered}\n$$\n\nThe molar solubility of CdS in this solution is $1.0 \\times 10^{-26} \\mathrm{M}$."}
{"id": 4312, "contents": "1664. Check Your Learning - \nCalculate the molar solubility of aluminum hydroxide, $\\mathrm{Al}(\\mathrm{OH})_{3}$, in a $0.015-M$ solution of aluminum nitrate, $\\mathrm{Al}\\left(\\mathrm{NO}_{3}\\right)_{3}$. The $K_{\\text {sp }}$ of $\\mathrm{Al}(\\mathrm{OH})_{3}$ is $2 \\times 10^{-32}$.\n\nAnswer:\n$4 \\times 10^{-11} \\mathrm{M}$"}
{"id": 4313, "contents": "1665. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Explain the Lewis model of acid-base chemistry\n- Write equations for the formation of adducts and complex ions\n- Perform equilibrium calculations involving formation constants\n\nIn 1923, G. N. Lewis proposed a generalized definition of acid-base behavior in which acids and bases are identified by their ability to accept or to donate a pair of electrons and form a coordinate covalent bond.\n\nA coordinate covalent bond (or dative bond) occurs when one of the atoms in the bond provides both bonding electrons. For example, a coordinate covalent bond occurs when a water molecule combines with a hydrogen ion to form a hydronium ion. A coordinate covalent bond also results when an ammonia molecule combines with a hydrogen ion to form an ammonium ion. Both of these equations are shown here.\n\n\n\nReactions involving the formation of coordinate covalent bonds are classified as Lewis acid-base chemistry. The species donating the electron pair that compose the bond is a Lewis base, the species accepting the electron pair is a Lewis acid, and the product of the reaction is a Lewis acid-base adduct. As the two examples above illustrate, Br\u00f8nsted-Lowry acid-base reactions represent a subcategory of Lewis acid reactions, specifically, those in which the acid species is $\\mathrm{H}^{+}$. A few examples involving other Lewis acids and bases are described below.\n\nThe boron atom in boron trifluoride, $\\mathrm{BF}_{3}$, has only six electrons in its valence shell. Being short of the preferred octet, $\\mathrm{BF}_{3}$ is a very good Lewis acid and reacts with many Lewis bases; a fluoride ion is the Lewis base in this reaction, donating one of its lone pairs:\n\n\nIn the following reaction, each of two ammonia molecules, Lewis bases, donates a pair of electrons to a silver ion, the Lewis acid:\n\n\nNonmetal oxides act as Lewis acids and react with oxide ions, Lewis bases, to form oxyanions:\n\n\nMany Lewis acid-base reactions are displacement reactions in which one Lewis base displaces another Lewis base from an acid-base adduct, or in which one Lewis acid displaces another Lewis acid:"}
{"id": 4314, "contents": "1665. LEARNING OBJECTIVES - \nMany Lewis acid-base reactions are displacement reactions in which one Lewis base displaces another Lewis base from an acid-base adduct, or in which one Lewis acid displaces another Lewis acid:\n\n\n\nAnother type of Lewis acid-base chemistry involves the formation of a complex ion (or a coordination complex) comprising a central atom, typically a transition metal cation, surrounded by ions or molecules called ligands. These ligands can be neutral molecules like $\\mathrm{H}_{2} \\mathrm{O}$ or $\\mathrm{NH}_{3}$, or ions such as $\\mathrm{CN}^{-}$or $\\mathrm{OH}^{-}$. Often, the ligands act as Lewis bases, donating a pair of electrons to the central atom. These types of Lewis acid-base reactions are examples of a broad subdiscipline called coordination chemistry-the topic of another chapter in this text.\n\nThe equilibrium constant for the reaction of a metal ion with one or more ligands to form a coordination complex is called a formation constant ( $\\boldsymbol{K}_{\\mathbf{f}}$ ) (sometimes called a stability constant). For example, the complex ion $\\mathrm{Cu}(\\mathrm{CN})_{2}{ }^{-}$\n\n```\n[:N\u4e09C-Cu-C\\equivN:]-\n```\n\nis produced by the reaction\n\n$$\n\\mathrm{Cu}^{+}(a q)+2 \\mathrm{CN}^{-}(a q) \\rightleftharpoons \\mathrm{Cu}(\\mathrm{CN})_{2}^{-}(a q)\n$$\n\nThe formation constant for this reaction is\n\n$$\nK_{\\mathrm{f}}=\\frac{\\left[\\mathrm{Cu}(\\mathrm{CN})_{2}^{-}\\right]}{\\left[\\mathrm{Cu}^{+}\\right]\\left[\\mathrm{CN}^{-}\\right]^{2}}\n$$\n\nAlternatively, the reverse reaction (decomposition of the complex ion) can be considered, in which case the equilibrium constant is a dissociation constant ( $\\boldsymbol{K}_{\\mathbf{d}}$ ). Per the relation between equilibrium constants for reciprocal reactions described, the dissociation constant is the mathematical inverse of the formation constant, $K_{d}=\\mathrm{K}_{\\mathrm{f}}{ }^{-1}$. A tabulation of formation constants is provided in Appendix K."}
{"id": 4315, "contents": "1665. LEARNING OBJECTIVES - \nAs an example of dissolution by complex ion formation, let us consider what happens when we add aqueous ammonia to a mixture of silver chloride and water. Silver chloride dissolves slightly in water, giving a small concentration of $\\mathrm{Ag}^{+}\\left(\\left[\\mathrm{Ag}^{+}\\right]=1.3 \\times 10^{-5} \\mathrm{M}\\right)$ :\n\n$$\n\\operatorname{AgCl}(s) \\rightleftharpoons \\mathrm{Ag}^{+}(a q)+\\mathrm{Cl}^{-}(a q)\n$$\n\nHowever, if $\\mathrm{NH}_{3}$ is present in the water, the complex ion, $\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}{ }^{+}$, can form according to the equation:\n\n$$\n\\mathrm{Ag}^{+}(a q)+2 \\mathrm{NH}_{3}(a q) \\rightleftharpoons \\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}^{+}(a q)\n$$\n\nwith\n\n$$\nK_{\\mathrm{f}}=\\frac{\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}^{+}\\right]}{\\left[\\mathrm{Ag}^{+}\\right]\\left[\\mathrm{NH}_{3}\\right]^{2}}=1.7 \\times 10^{7}\n$$\n\nThe large size of this formation constant indicates that most of the free silver ions produced by the dissolution of AgCl combine with $\\mathrm{NH}_{3}$ to form $\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}{ }^{+}$. As a consequence, the concentration of silver ions, $\\left[\\mathrm{Ag}^{+}\\right.$], is reduced, and the reaction quotient for the dissolution of silver chloride, $\\left[\\mathrm{Ag}^{+}\\right]\\left[\\mathrm{Cl}^{-}\\right]$, falls below the solubility\nproduct of AgCl :\n\n$$\nQ=\\left[\\mathrm{Ag}^{+}\\right]\\left[\\mathrm{Cl}^{-}\\right]<K_{\\mathrm{sp}}\n$$\n\nMore silver chloride then dissolves. If the concentration of ammonia is great enough, all of the silver chloride dissolves."}
{"id": 4316, "contents": "1667. Dissociation of a Complex Ion - \nCalculate the concentration of the silver ion in a solution that initially is $0.10 M$ with respect to $\\operatorname{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}{ }^{+}$."}
{"id": 4317, "contents": "1668. Solution - \nApplying the standard ICE approach to this reaction yields the following:\n\n|  | $\\mathbf{A g}^{+}+\\mathbf{2 N H} \\mathbf{3}^{c \\mid} \\rightleftharpoons \\mathbf{A g}\\left(\\mathbf{N H}_{3}\\right)_{2}{ }^{+}$ |  |  |\n| :---: | :---: | :---: | :---: |\n| Initial concentration $(M)$ | 0 | 0 | 0.10 |\n| Change $(M)$ | $+x$ | $+2 x$ | $-x$ |\n| Equilibrium concentration $(M)$ | $x$ | $2 x$ | $0.10-x$ |\n\nSubstituting these equilibrium concentration terms into the $K_{\\mathrm{f}}$ expression gives\n\n$$\n\\begin{aligned}\nK_{\\mathrm{f}} & =\\frac{\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}^{+}\\right]}{\\left[\\mathrm{Ag}^{+}\\right]\\left[\\mathrm{NH}_{3}\\right]^{2}} \\\\\n1.7 & \\times 10^{7}=\\frac{0.10-x}{(x)(2 x)^{2}}\n\\end{aligned}\n$$\n\nThe very large equilibrium constant means the amount of the complex ion that will dissociate, $x$, will be very small. Assuming $x \\ll 0.1$ permits simplifying the above equation:\n\n$$\n\\begin{gathered}\n1.7 \\times 10^{7}=\\frac{0.10}{(x)(2 x)^{2}} \\\\\nx^{3}=\\frac{0.10}{4\\left(1.7 \\times 10^{7}\\right)}=1.5 \\times 10^{-9} \\\\\nx=\\sqrt[3]{1.5 \\times 10^{-9}}=1.1 \\times 10^{-3}\n\\end{gathered}\n$$"}
{"id": 4318, "contents": "1668. Solution - \nBecause only $1.1 \\%$ of the $\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}{ }^{+}$dissociates into $\\mathrm{Ag}^{+}$and $\\mathrm{NH}_{3}$, the assumption that $x$ is small is justified. Using this value of $x$ and the relations in the above ICE table allows calculation of all species' equilibrium concentrations:\n\n$$\n\\begin{gathered}\n{\\left[\\mathrm{Ag}^{+}\\right]=0+x=1.1 \\times 10^{-3} M} \\\\\n{\\left[\\mathrm{NH}_{3}\\right]=0+2 x=2.2 \\times 10^{-3} M} \\\\\n{\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}^{+}\\right]=0.10-x=0.10-0.0011=0.099}\n\\end{gathered}\n$$\n\nThe concentration of free silver ion in the solution is 0.0011 M ."}
{"id": 4319, "contents": "1669. Check Your Learning - \nCalculate the silver ion concentration, $\\left[\\mathrm{Ag}^{+}\\right.$], of a solution prepared by dissolving 1.00 g of $\\mathrm{AgNO}_{3}$ and 10.0 g of KCN in sufficient water to make 1.00 L of solution. (Hint: Because $K_{\\mathrm{f}}$ is very large, assume the reaction goes to completion then calculate the $\\left[\\mathrm{Ag}^{+}\\right]$produced by dissociation of the complex.)"}
{"id": 4320, "contents": "1670. Answer: - \n$2.9 \\times 10^{-22} \\mathrm{M}$"}
{"id": 4321, "contents": "1671. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe examples of systems involving two (or more) coupled chemical equilibria\n- Calculate reactant and product concentrations for coupled equilibrium systems\n\nAs discussed in preceding chapters on equilibrium, coupled equilibria involve two or more separate chemical reactions that share one or more reactants or products. This section of this chapter will address solubility equilibria coupled with acid-base and complex-formation reactions.\n\nAn environmentally relevant example illustrating the coupling of solubility and acid-base equilibria is the impact of ocean acidification on the health of the ocean's coral reefs. These reefs are built upon skeletons of sparingly soluble calcium carbonate excreted by colonies of corals (small marine invertebrates). The relevant dissolution equilibrium is\n\n$$\n\\mathrm{CaCO}_{3}(s) \\rightleftharpoons \\mathrm{Ca}^{2+}(a q)+\\mathrm{CO}_{3}^{-2}(a q) \\quad K_{\\mathrm{sp}}=8.7 \\times 10^{-9}\n$$\n\nRising concentrations of atmospheric carbon dioxide contribute to an increased acidity of ocean waters due to the dissolution, hydrolysis, and acid ionization of carbon dioxide:"}
{"id": 4322, "contents": "1671. LEARNING OBJECTIVES - \nRising concentrations of atmospheric carbon dioxide contribute to an increased acidity of ocean waters due to the dissolution, hydrolysis, and acid ionization of carbon dioxide:\n\n$$\n\\begin{gathered}\n\\mathrm{CO}_{2}(g) \\rightleftharpoons \\mathrm{CO}_{2}(a q) \\\\\n\\mathrm{CO}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{CO}_{3}(a q) \\\\\n\\mathrm{H}_{2} \\mathrm{CO}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{HCO}_{3}^{-}(a q)+\\mathrm{H}_{3} \\mathrm{O}^{+}(a q) \\quad K_{\\mathrm{a} 1}=4.3 \\times 10^{-7} \\\\\n\\mathrm{HCO}_{3}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{CO}_{3}^{2-}(a q)+\\mathrm{H}_{3} \\mathrm{O}^{+}(a q) \\quad K_{\\mathrm{a} 2}=4.7 \\times 10^{-11}\n\\end{gathered}\n$$\n\nInspection of these equilibria shows the carbonate ion is involved in the calcium carbonate dissolution and the acid hydrolysis of bicarbonate ion. Combining the dissolution equation with the reverse of the acid hydrolysis equation yields\n\n$$\n\\mathrm{CaCO}_{3}(s)+\\mathrm{H}_{3} \\mathrm{O}^{+}(a q) \\rightleftharpoons \\mathrm{Ca}^{2+}(a q)+\\mathrm{HCO}_{3}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\quad K=K_{\\mathrm{sp}} / K_{\\mathrm{a} 2}=180\n$$"}
{"id": 4323, "contents": "1671. LEARNING OBJECTIVES - \nThe equilibrium constant for this net reaction is much greater than the $K_{s p}$ for calcium carbonate, indicating its solubility is markedly increased in acidic solutions. As rising carbon dioxide levels in the atmosphere increase the acidity of ocean waters, the calcium carbonate skeletons of coral reefs become more prone to dissolution and subsequently less healthy (Figure 15.7).\n\n\nFIGURE 15.7 Healthy coral reefs (a) support a dense and diverse array of sea life across the ocean food chain. But when coral are unable to adequately build and maintain their calcium carbonate skeletons because of excess ocean acidification, the unhealthy reef (b) is only capable of hosting a small fraction of the species as before, and the local food chain starts to collapse. (credit a: modification of work by NOAA Photo Library; credit b: modification of work by\n\"prilfish\"/Flickr)"}
{"id": 4324, "contents": "1672. LINK TO LEARNING - \nLearn more about ocean acidification (http://openstax.org/l/16acidicocean) and how it affects other marine creatures.\n\nThis site (http://openstax.org/l/16coralreef) has detailed information about how ocean acidification specifically affects coral reefs.\n\nThe dramatic increase in solubility with increasing acidity described above for calcium carbonate is typical of salts containing basic anions (e.g., carbonate, fluoride, hydroxide, sulfide). Another familiar example is the formation of dental cavities in tooth enamel. The major mineral component of enamel is calcium hydroxyapatite (Figure 15.8), a sparingly soluble ionic compound whose dissolution equilibrium is\n\n\nFIGURE 15.8 Crystal of the mineral hydroxyapatite, $\\mathrm{Ca}_{5}\\left(\\mathrm{PO}_{4}\\right)_{3} \\mathrm{OH}$, is shown here. The pure compound is white, but like many other minerals, this sample is colored because of the presence of impurities.\n\nThis compound dissolved to yield two different basic ions: triprotic phosphate ions\n\n$$\n\\begin{gathered}\n\\mathrm{PO}_{4}{ }^{3-}(a q)+\\mathrm{H}_{3} \\mathrm{O}^{+}(a q) \\longrightarrow \\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{2-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\\\\n\\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{2-}(a q)+\\mathrm{H}_{3} \\mathrm{O}^{+}(a q) \\longrightarrow \\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\\\\n\\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{-}(a q)+\\mathrm{H}_{3} \\mathrm{O}^{+}(a q) \\longrightarrow \\mathrm{H}_{3} \\mathrm{PO}_{4}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)\n\\end{gathered}\n$$\n\nand monoprotic hydroxide ions:"}
{"id": 4325, "contents": "1672. LINK TO LEARNING - \nand monoprotic hydroxide ions:\n\n$$\n\\mathrm{OH}^{-}(a q)+\\mathrm{H}_{3} \\mathrm{O}^{+} \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}\n$$\n\nOf the two basic productions, the hydroxide is, of course, by far the stronger base (it's the strongest base that can exist in aqueous solution), and so it is the dominant factor providing the compound an acid-dependent solubility. Dental cavities form when the acid waste of bacteria growing on the surface of teeth hastens the dissolution of tooth enamel by reacting completely with the strong base hydroxide, shifting the hydroxyapatite solubility equilibrium to the right. Some toothpastes and mouth rinses contain added NaF or $\\mathrm{SnF}_{2}$ that make enamel more acid resistant by replacing the strong base hydroxide with the weak base fluoride:\n\n$$\n\\mathrm{NaF}+\\mathrm{Ca}_{5}\\left(\\mathrm{PO}_{4}\\right)_{3} \\mathrm{OH} \\rightleftharpoons \\mathrm{Ca}_{5}\\left(\\mathrm{PO}_{4}\\right)_{3} \\mathrm{~F}+\\mathrm{Na}^{+}+\\mathrm{OH}^{-}\n$$\n\nThe weak base fluoride ion reacts only partially with the bacterial acid waste, resulting in a less extensive shift in the solubility equilibrium and an increased resistance to acid dissolution. See the Chemistry in Everyday Life feature on the role of fluoride in preventing tooth decay for more information."}
{"id": 4326, "contents": "1674. Role of Fluoride in Preventing Tooth Decay - \nAs we saw previously, fluoride ions help protect our teeth by reacting with hydroxylapatite to form fluorapatite, $\\mathrm{Ca}_{5}\\left(\\mathrm{PO}_{4}\\right)_{3} \\mathrm{~F}$. Since it lacks a hydroxide ion, fluorapatite is more resistant to attacks by acids in our mouths and is thus less soluble, protecting our teeth. Scientists discovered that naturally fluorinated water could be beneficial to your teeth, and so it became common practice to add fluoride to drinking water. Toothpastes and mouthwashes also contain amounts of fluoride (Figure 15.9).\n\n\nFIGURE 15.9 Fluoride, found in many toothpastes, helps prevent tooth decay (credit: Kerry Ceszyk).\nUnfortunately, excess fluoride can negate its advantages. Natural sources of drinking water in various parts of the world have varying concentrations of fluoride, and places where that concentration is high are prone to certain health risks when there is no other source of drinking water. The most serious side effect of excess fluoride is the bone disease, skeletal fluorosis. When excess fluoride is in the body, it can cause the joints to stiffen and the bones to thicken. It can severely impact mobility and can negatively affect the thyroid gland. Skeletal fluorosis is a condition that over 2.7 million people suffer from across the world. So while fluoride can protect our teeth from decay, the US Environmental Protection Agency sets a maximum level of $4 \\mathrm{ppm}(4 \\mathrm{mg} / \\mathrm{L})$ of fluoride in drinking water in the US. Fluoride levels in water are not regulated in all countries, so fluorosis is a problem in areas with high levels of fluoride in the groundwater.\n\nThe solubility of ionic compounds may also be increased when dissolution is coupled to the formation of a complex ion. For example, aluminum hydroxide dissolves in a solution of sodium hydroxide or another strong base because of the formation of the complex ion $\\mathrm{Al}(\\mathrm{OH})_{4}{ }^{-}$."}
{"id": 4327, "contents": "1674. Role of Fluoride in Preventing Tooth Decay - \nThe equations for the dissolution of aluminum hydroxide, the formation of the complex ion, and the combined (net) equation are shown below. As indicated by the relatively large value of K for the net reaction, coupling complex formation with dissolution drastically increases the solubility of $\\mathrm{Al}(\\mathrm{OH})_{3}$.\n\n$$\n\\begin{array}{cl}\n\\mathrm{Al}(\\mathrm{OH})_{3}(s) \\rightleftharpoons \\mathrm{Al}^{3+}(a q)+3 \\mathrm{OH}^{-}(a q) & K_{\\mathrm{sp}}=2 \\times 10^{-32} \\\\\n\\mathrm{Al}^{3+}(a q)+4 \\mathrm{OH}^{-}(a q) \\rightleftharpoons \\mathrm{Al}(\\mathrm{OH})_{4}^{-}(a q) & K_{\\mathrm{f}}=1.1 \\times 10^{33} \\\\\n\\mathrm{Net}: \\mathrm{Al}(\\mathrm{OH})_{3}(s)+\\mathrm{OH}^{-}(a q) \\rightleftharpoons \\mathrm{Al}(\\mathrm{OH})_{4}^{-}(a q) & K=K_{\\mathrm{sp}} K_{\\mathrm{f}}=22\n\\end{array}\n$$"}
{"id": 4328, "contents": "1676. Increased Solubility in Acidic Solutions - \nCompute and compare the molar solublities for aluminum hydroxide, $\\mathrm{Al}(\\mathrm{OH})_{3}$, dissolved in (a) pure water and (b) a buffer containing $0.100 M$ acetic acid and $0.100 M$ sodium acetate."}
{"id": 4329, "contents": "1677. Solution - \n(a) The molar solubility of aluminum hydroxide in water is computed considering the dissolution equilibrium only as demonstrated in several previous examples:\n\n$$\n\\begin{aligned}\n& \\mathrm{Al}(\\mathrm{OH})_{3}(s) \\rightleftharpoons \\mathrm{Al}^{3+}(a q)+3 \\mathrm{OH}^{-}(a q) \\quad K_{\\mathrm{sp}}=2 \\times 10^{-32} \\\\\n& \\text { molar solubility in water }=\\left[\\mathrm{Al}^{3+}\\right]=\\left(2 \\times 10^{-32} / 27\\right)^{1 / 4}=5 \\times 10^{-9} M\n\\end{aligned}\n$$\n\n(b) The concentration of hydroxide ion of the buffered solution is conveniently calculated by the HendersonHasselbalch equation:\n\n$$\n\\begin{aligned}\n& \\mathrm{pH}=\\mathrm{pK}_{\\mathrm{a}}+\\log \\left[\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right] /\\left[\\mathrm{CH}_{3} \\mathrm{COOH}\\right] \\\\\n& \\mathrm{pH}=4.74+\\log (0.100 / 0.100)=4.74\n\\end{aligned}\n$$\n\nAt this pH , the concentration of hydroxide ion is\n\n$$\n\\begin{aligned}\n& \\mathrm{pOH}=14.00-4.74=9.26 \\\\\n& {\\left[\\mathrm{OH}^{-}\\right]=10^{-9.26}=5.5 \\times 10^{-10}}\n\\end{aligned}\n$$\n\nThe solubility of $\\mathrm{Al}(\\mathrm{OH})_{3}$ in this buffer is then calculated from its solubility product expressions:\n\n$$\nK_{\\mathrm{sp}}=\\left[\\mathrm{Al}^{3+}\\right]\\left[\\mathrm{OH}^{-}\\right]^{3}\n$$"}
{"id": 4330, "contents": "1677. Solution - \n$$\nK_{\\mathrm{sp}}=\\left[\\mathrm{Al}^{3+}\\right]\\left[\\mathrm{OH}^{-}\\right]^{3}\n$$\n\nmolar solubility in buffer $=\\left[\\mathrm{Al}^{3+}\\right]=K_{\\text {sp }} /\\left[\\mathrm{OH}^{-}\\right]^{3}=\\left(2 \\times 10^{-32}\\right) /\\left(5.5 \\times 10^{-10}\\right)^{3}=1.2 \\times 10^{-4} \\mathrm{M}$\nCompared to pure water, the solubility of aluminum hydroxide in this mildly acidic buffer is approximately ten million times greater (though still relatively low)."}
{"id": 4331, "contents": "1678. Check Your Learning - \nWhat is the solubility of aluminum hydroxide in a buffer comprised of 0.100 M formic acid and 0.100 M sodium formate?"}
{"id": 4332, "contents": "1679. Answer: - \n0.1 M"}
{"id": 4333, "contents": "1681. Multiple Equilibria - \nUnexposed silver halides are removed from photographic film when they react with sodium thiosulfate $\\left(\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}\\right.$, called hypo) to form the complex ion $\\mathrm{Ag}\\left(\\mathrm{S}_{2} \\mathrm{O}_{3}\\right)_{2}{ }^{3-}\\left(K_{\\mathrm{f}}=4.7 \\times 10^{13}\\right)$.\n\n\nWhat mass of $\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$ is required to prepare 1.00 L of a solution that will dissolve 1.00 g of AgBr by the formation of $\\mathrm{Ag}\\left(\\mathrm{S}_{2} \\mathrm{O}_{3}\\right)_{2}{ }^{3-}$ ?"}
{"id": 4334, "contents": "1682. Solution - \nTwo equilibria are involved when silver bromide dissolves in an aqueous thiosulfate solution containing the $\\mathrm{S}_{2} \\mathrm{O}_{3}{ }^{2-}$ ion:\n\n$$\n\\begin{aligned}\n& \\text { dissolution: } \\mathrm{AgBr}(s) \\rightleftharpoons \\mathrm{Ag}^{+}(a q)+\\mathrm{Br}^{-}(a q) \\quad K_{\\mathrm{sp}}=5.0 \\times 10^{-13} \\\\\n& \\text { complexation: } \\mathrm{Ag}^{+}(a q)+2 \\mathrm{~S}_{2} \\mathrm{O}_{3}^{2-}(a q) \\rightleftharpoons \\mathrm{Ag}\\left(\\mathrm{~S}_{2} \\mathrm{O}_{3}\\right)_{2}^{3-}(a q) \\quad K_{\\mathrm{f}}=4.7 \\times 10^{13}\n\\end{aligned}\n$$\n\nCombining these two equilibrium equations yields\n\n$$\n\\begin{array}{cc}\n\\mathrm{AgBr}(s)+2 \\mathrm{~S}_{2} \\mathrm{O}_{3}{ }^{2-}(a q) & \\mathrm{Ag}\\left(\\mathrm{~S}_{2} \\mathrm{O}_{3}\\right)_{2}{ }^{3-}(a q) \\\\\n\\frac{K=\\left[\\mathrm{Ag}\\left(\\mathrm{~S}_{2} \\mathrm{O}_{3}\\right)_{2}{ }^{3-}\\right]\\left[\\mathrm{Br}^{-}\\right]}{\\left[\\mathrm{S}_{2} \\mathrm{O}_{3}{ }^{2-}\\right]^{2}}=K_{\\mathrm{sp}} K_{\\mathrm{f}}=24\n\\end{array}\n$$\n\nThe concentration of bromide resulting from dissolution of 1.00 g of AgBr in 1.00 L of solution is"}
{"id": 4335, "contents": "1682. Solution - \nThe concentration of bromide resulting from dissolution of 1.00 g of AgBr in 1.00 L of solution is\n\n$$\n\\left[\\mathrm{Br}^{-}\\right]=\\frac{1.00 \\mathrm{~g} \\mathrm{AgBr} \\times \\frac{1 \\mathrm{~mol} \\mathrm{AgBr}}{187.77 \\mathrm{~g} / \\mathrm{mol}} \\times \\frac{1 \\mathrm{~mol} \\mathrm{Br}^{-}}{1 \\mathrm{~mol} \\mathrm{AgBr}}}{1.00 \\mathrm{~L}}=0.00532 \\mathrm{M}\n$$\n\nThe stoichiometry of the dissolution equilibrium indicates the same concentration of aqueous silver ion will result, 0.00532 M , and the very large value of $K_{\\mathrm{f}}$ ensures that essentially all the dissolved silver ion will be complexed by thiosulfate ion:\n\n$$\n\\left[\\mathrm{Ag}\\left(\\mathrm{~S}_{2} \\mathrm{O}_{3}\\right)_{2}{ }^{3-}\\right]=0.00532 \\mathrm{M}\n$$\n\nRearranging the K expression for the combined equilibrium equations and solving for the concentration of thiosulfate ion yields\n\n$$\n\\left[\\mathrm{S}_{2} \\mathrm{O}_{3}{ }^{2-}\\right]=\\frac{\\left[\\mathrm{Ag}\\left(\\mathrm{~S}_{2} \\mathrm{O}_{3}\\right)_{2}{ }^{3-}\\right]\\left[\\mathrm{Br}^{-}\\right]}{K}=\\frac{(0.00532 \\mathrm{M})(0.00532 \\mathrm{M})}{24}=0.0011 \\mathrm{M}\n$$\n\nFinally, the total mass of $\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$ required to provide enough thiosulfate to yield the concentrations cited above can be calculated."}
{"id": 4336, "contents": "1682. Solution - \nFinally, the total mass of $\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$ required to provide enough thiosulfate to yield the concentrations cited above can be calculated.\n\nMass of $\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$ required to yield $0.00532 \\mathrm{M} \\mathrm{Ag}\\left(\\mathrm{S}_{2} \\mathrm{O}_{2}\\right)_{2}{ }^{3-}$\n\n$$\n0.00532 \\frac{\\mathrm{~mol} \\mathrm{Ag}\\left(\\mathrm{~S}_{2} \\mathrm{O}_{3}\\right)_{2}{ }^{3-}}{1.00 \\mathrm{~L}} \\times \\frac{2 \\mathrm{~mol} \\mathrm{~S}_{2} \\mathrm{O}_{3}{ }^{2-}}{1 \\mathrm{~mol} \\mathrm{Ag}\\left(\\mathrm{~S}_{2} \\mathrm{O}_{3}\\right)_{2}{ }^{3-}} \\times \\frac{1 \\mathrm{~mol} \\mathrm{NaS}_{2} \\mathrm{O}_{3}}{1 \\mathrm{~mol} \\mathrm{~S}_{2} \\mathrm{O}_{3}{ }^{2-}} \\times \\frac{158.1 \\mathrm{~g} \\mathrm{NaS}_{2} \\mathrm{O}_{3}}{1 \\mathrm{~mol} \\mathrm{NaS}_{2} \\mathrm{O}_{3}}=1.68 \\mathrm{~g}\n$$"}
{"id": 4337, "contents": "1682. Solution - \nMass of $\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$ required to yield $0.00110 \\mathrm{M} \\mathrm{S}_{2} \\mathrm{O}_{3}{ }^{2-}$\n$0.0011 \\frac{\\mathrm{~mol} \\mathrm{~S}_{2} \\mathrm{O}_{3}{ }^{2-}}{1.00 \\mathrm{~L}} \\times \\frac{1 \\mathrm{~mol} \\mathrm{~S}_{2} \\mathrm{O}_{3}{ }^{2-}}{1 \\mathrm{~mol} \\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}} \\times \\frac{158.1 \\mathrm{~g} \\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}}{1 \\mathrm{~mol} \\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}}=0.17 \\mathrm{~g}$\nThe mass of $\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$ required to dissolve 1.00 g of AgBr in 1.00 L of water is thus $1.68 \\mathrm{~g}+0.17 \\mathrm{~g}=1.85 \\mathrm{~g}$"}
{"id": 4338, "contents": "1683. Check Your Learning - \n$\\mathrm{AgCl}(s)$, silver chloride, has a very low solubility: $\\mathrm{AgCl}(s) \\rightleftharpoons \\mathrm{Ag}^{+}(a q)+\\mathrm{Cl}^{-}(a q), K_{\\mathrm{sp}}=1.6 \\times 10^{-10}$. Adding ammonia significantly increases the solubility of AgCl because a complex ion is formed:\n$\\mathrm{Ag}^{+}(a q)+2 \\mathrm{NH}_{3}(a q) \\rightleftharpoons \\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}{ }^{+}(a q), K_{\\mathrm{f}}=1.7 \\times 10^{7}$. What mass of $\\mathrm{NH}_{3}$ is required to prepare 1.00 L of solution that will dissolve 2.00 g of AgCl by formation of $\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}{ }^{+}$?"}
{"id": 4339, "contents": "1684. Answer: - \n1.00 L of a solution prepared with $4.81 \\mathrm{~g} \\mathrm{NH}_{3}$ dissolves 2.0 g of AgCl ."}
{"id": 4340, "contents": "1685. Key Terms - \ncommon ion effect effect on equilibrium when a substance with an ion in common with the dissolved species is added to the solution; causes a decrease in the solubility of an ionic species, or a decrease in the ionization of a weak acid or base\ncomplex ion ion consisting of a central atom surrounding molecules or ions called ligands via coordinate covalent bonds\ncoordinate covalent bond (also, dative bond) covalent bond in which both electrons originated from the same atom\ncoupled equilibria system characterized the simultaneous establishment of two or more equilibrium reactions sharing one or more reactant or product\ndissociation constant ( $\\boldsymbol{K}_{\\mathbf{d}}$ ) equilibrium constant for the decomposition of a complex ion into its components\nformation constant ( $\\boldsymbol{K}_{\\mathbf{f}}$ ) (also, stability constant) equilibrium constant for the formation of a\ncomplex ion from its components\nLewis acid any species that can accept a pair of electrons and form a coordinate covalent bond\nLewis acid-base adduct compound or ion that contains a coordinate covalent bond between a Lewis acid and a Lewis base\nLewis acid-base chemistry reactions involving the formation of coordinate covalent bonds\nLewis base any species that can donate a pair of electrons and form a coordinate covalent bond\nligand molecule or ion acting as a Lewis base in complex ion formation; bonds to the central atom of the complex\nmolar solubility solubility of a compound expressed in units of moles per liter ( $\\mathrm{mol} / \\mathrm{L}$ )\nselective precipitation process in which ions are separated using differences in their solubility with a given precipitating reagent\nsolubility product constant ( $\\boldsymbol{K}_{\\text {sp }}$ ) equilibrium constant for the dissolution of an ionic compound"}
{"id": 4341, "contents": "1686. Key Equations - \n$\\mathrm{M}_{p} \\mathrm{X}_{q}(s) \\rightleftharpoons p \\mathrm{M}^{\\mathrm{m}+}(a q)+q \\mathrm{X}^{\\mathrm{n}-}(a q) \\quad K_{\\mathrm{sp}}=\\left[\\mathrm{M}^{\\mathrm{m}+}\\right]^{p}\\left[\\mathrm{X}^{\\mathrm{n}-}\\right]^{q}$"}
{"id": 4342, "contents": "1687. Summary - 1687.1. Precipitation and Dissolution\nThe equilibrium constant for an equilibrium involving the precipitation or dissolution of a slightly soluble ionic solid is called the solubility product, $K_{\\mathrm{sp}}$, of the solid. For a heterogeneous equilibrium involving the slightly soluble solid $\\mathrm{M}_{p} \\mathrm{X}_{q}$ and its ions $\\mathrm{M}^{\\mathrm{m}+}$ and $\\mathrm{X}^{\\mathrm{n}-}$ :\n\n$$\n\\mathrm{M}_{p} \\mathrm{X}_{q}(s) \\rightleftharpoons p \\mathrm{M}^{\\mathrm{m}+}(a q)+q \\mathrm{X}^{\\mathrm{n}-}(a q)\n$$\n\nthe solubility product expression is:\n\n$$\nK_{\\mathrm{sp}}=\\left[\\mathrm{M}^{\\mathrm{m}+}\\right]^{p}\\left[\\mathrm{X}^{\\mathrm{n}-}\\right]^{q}\n$$\n\nThe solubility product of a slightly soluble electrolyte can be calculated from its solubility; conversely, its solubility can be calculated from its $K_{\\mathrm{sp}}$, provided the only significant reaction that occurs when the solid dissolves is the formation of its ions.\n\nA slightly soluble electrolyte begins to precipitate when the magnitude of the reaction quotient for the dissolution reaction exceeds the magnitude of the solubility product. Precipitation continues until the reaction quotient equals the solubility product."}
{"id": 4343, "contents": "1687. Summary - 1687.2. Lewis Acids and Bases\nA Lewis acid is a species that can accept an electron pair, whereas a Lewis base has an electron pair available for donation to a Lewis acid. Complex ions are examples of Lewis acid-base adducts and comprise central metal atoms or ions acting as Lewis acids bonded to molecules or ions called ligands that act as Lewis bases. The equilibrium constant for the reaction between a metal ion and ligands produces a complex ion called a formation constant; for the reverse reaction, it is called a dissociation constant."}
{"id": 4344, "contents": "1687. Summary - 1687.3. Coupled Equilibria\nSystems involving two or more chemical equilibria that share one or more reactant or product are called coupled equilibria. Common examples of coupled equilibria include the increased solubility of some compounds in acidic solutions (coupled dissolution and neutralization equilibria) and in solutions containing ligands (coupled dissolution and complex formation). The equilibrium tools from other chapters may be applied to describe and perform calculations on these systems."}
{"id": 4345, "contents": "1688. Exercises - 1688.1. Precipitation and Dissolution\n1. Complete the changes in concentrations for each of the following reactions:\n(a)\n$\\mathrm{AgI}(s) \\longrightarrow \\mathrm{Ag}^{+}(a q)+\\mathrm{I}^{-}(a q)$\n$x$\n(b)\n$\\mathrm{CaCO}_{3}(s) \\longrightarrow \\mathrm{Ca}^{2+}(a q)+\\mathrm{CO}_{3}{ }^{2-}(a q)$\n$\\qquad$\n(c)\n$\\operatorname{Mg}(\\mathrm{OH})_{2}(s) \\longrightarrow \\mathrm{Mg}^{2+}(a q)+2 \\mathrm{OH}^{-}(a q)$\n$x$\n(d)\n$\\mathrm{Mg}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}(s) \\longrightarrow 3 \\mathrm{Mg}^{2+}(a q)+2 \\mathrm{PO}_{4}{ }^{3-}(a q)$\n\n$x$\n2. Complete the changes in concentrations for each of the following reactions:\n(a)\n$\\mathrm{BaSO}_{4}(s) \\longrightarrow \\mathrm{Ba}^{2+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q)$\n$x$\n(b\n$\\mathrm{Ag}_{2} \\mathrm{SO}_{4}(s) \\longrightarrow 2 \\mathrm{Ag}^{+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q)$\n(c)\n$\\mathrm{Al}(\\mathrm{OH})_{3}(s) \\longrightarrow \\mathrm{Al}^{3+}(a q)+3 \\mathrm{OH}^{-}(a q)$\n$x$\n(d)\n$\\mathrm{Pb}(\\mathrm{OH}) \\mathrm{Cl}(s) \\longrightarrow \\mathrm{Pb}^{2+}(a q)+\\mathrm{OH}^{-}(a q)+\\mathrm{Cl}^{-}(a q)$\n(e)\n$\\mathrm{Ca}_{3}\\left(\\mathrm{AsO}_{4}\\right)_{2}(s) \\longrightarrow 3 \\mathrm{Ca}^{2+}(a q)+2 \\mathrm{AsO}_{4}{ }^{3-}(a q)$"}
{"id": 4346, "contents": "1689. $3 x$ - \n3. How do the concentrations of $\\mathrm{Ag}^{+}$and $\\mathrm{CrO}_{4}{ }^{2-}$ in a saturated solution above 1.0 g of solid $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}$ change when 100 g of solid $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}$ is added to the system? Explain.\n4. How do the concentrations of $\\mathrm{Pb}^{2+}$ and $\\mathrm{S}^{2-}$ change when $\\mathrm{K}_{2} \\mathrm{~S}$ is added to a saturated solution of PbS ?\n5. What additional information do we need to answer the following question: How is the equilibrium of solid silver bromide with a saturated solution of its ions affected when the temperature is raised?\n6. Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: $\\mathrm{CoSO}_{3}, \\mathrm{CuI}, \\mathrm{PbCO}_{3}, \\mathrm{PbCl}_{2}, \\mathrm{Tl}_{2} \\mathrm{~S}, \\mathrm{KClO}_{4}$ ?\n7. Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: $\\mathrm{AgCl}, \\mathrm{BaSO}_{4}, \\mathrm{CaF}_{2}, \\mathrm{Hg}_{2} \\mathrm{I}_{2}, \\mathrm{MnCO}_{3}$, and ZnS ?\n8. Write the ionic equation for dissolution and the solubility product ( $K_{\\mathrm{sp}}$ ) expression for each of the following slightly soluble ionic compounds:\n(a) $\\mathrm{PbCl}_{2}$\n(b) $\\mathrm{Ag}_{2} \\mathrm{~S}$\n(c) $\\mathrm{Sr}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$\n(d) $\\mathrm{SrSO}_{4}$\n9. Write the ionic equation for the dissolution and the $K_{\\mathrm{sp}}$ expression for each of the following slightly soluble ionic compounds:"}
{"id": 4347, "contents": "1689. $3 x$ - \n(d) $\\mathrm{SrSO}_{4}$\n9. Write the ionic equation for the dissolution and the $K_{\\mathrm{sp}}$ expression for each of the following slightly soluble ionic compounds:\n(a) $\\mathrm{LaF}_{3}$\n(b) $\\mathrm{CaCO}_{3}$\n(c) $\\mathrm{Ag}_{2} \\mathrm{SO}_{4}$\n(d) $\\mathrm{Pb}(\\mathrm{OH})_{2}$\n10. The Handbook of Chemistry and Physics (http://openstax.org/l/16Handbook) gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.\n(a) $\\mathrm{BaSiF}_{6}, 0.026 \\mathrm{~g} / 100 \\mathrm{~mL}$ (contains $\\mathrm{SiF}_{6}{ }^{2-}$ ions)\n(b) $\\mathrm{Ce}\\left(\\mathrm{IO}_{3}\\right)_{4}, 1.5 \\times 10^{-2} \\mathrm{~g} / 100 \\mathrm{~mL}$\n(c) $\\mathrm{Gd}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}, 3.98 \\mathrm{~g} / 100 \\mathrm{~mL}$\n(d) $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{PtBr}_{6}, 0.59 \\mathrm{~g} / 100 \\mathrm{~mL}$ (contains $\\mathrm{PtBr}_{6}{ }^{2-}$ ions)\n11. The Handbook of Chemistry and Physics (http://openstax.org/l/16Handbook) gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.\n(a) $\\mathrm{BaSeO}_{4}, 0.0118 \\mathrm{~g} / 100 \\mathrm{~mL}$"}
{"id": 4348, "contents": "1689. $3 x$ - \n(a) $\\mathrm{BaSeO}_{4}, 0.0118 \\mathrm{~g} / 100 \\mathrm{~mL}$\n(b) $\\mathrm{Ba}\\left(\\mathrm{BrO}_{3}\\right)_{2} \\cdot \\mathrm{H}_{2} \\mathrm{O}, 0.30 \\mathrm{~g} / 100 \\mathrm{~mL}$\n(c) $\\mathrm{NH}_{4} \\mathrm{MgAsO}_{4} \\cdot 6 \\mathrm{H}_{2} \\mathrm{O}, 0.038 \\mathrm{~g} / 100 \\mathrm{~mL}$\n(d) $\\mathrm{La}_{2}\\left(\\mathrm{MoO}_{4}\\right)_{3}, 0.00179 \\mathrm{~g} / 100 \\mathrm{~mL}$\n12. Use solubility products and predict which of the following salts is the most soluble, in terms of moles per liter, in pure water: $\\mathrm{CaF}_{2}, \\mathrm{Hg}_{2} \\mathrm{Cl}_{2}, \\mathrm{PbI}_{2}$, or $\\mathrm{Sn}(\\mathrm{OH})_{2}$.\n13. Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product:\n(a) $\\mathrm{KHC}_{4} \\mathrm{H}_{4} \\mathrm{O}_{6}$\n(b) $\\mathrm{PbI}_{2}$\n(c) $\\mathrm{Ag}_{4}\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]$, a salt containing the $\\mathrm{Fe}(\\mathrm{CN})_{6}{ }^{4-}$ ion\n(d) $\\mathrm{Hg}_{2} \\mathrm{I}_{2}$\n14. Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product:\n(a) $\\mathrm{Ag}_{2} \\mathrm{SO}_{4}$\n(b) $\\mathrm{PbBr}_{2}$\n(c) AgI"}
{"id": 4349, "contents": "1689. $3 x$ - \n(a) $\\mathrm{Ag}_{2} \\mathrm{SO}_{4}$\n(b) $\\mathrm{PbBr}_{2}$\n(c) AgI\n(d) $\\mathrm{CaC}_{2} \\mathrm{O}_{4} \\cdot \\mathrm{H}_{2} \\mathrm{O}$\n15. Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected.\n(a) $\\mathrm{AgCl}(s)$ in 0.025 M NaCl\n(b) $\\mathrm{CaF}_{2}(s)$ in 0.00133 MKF\n(c) $\\mathrm{Ag}_{2} \\mathrm{SO}_{4}(s)$ in 0.500 L of a solution containing 19.50 g of $\\mathrm{K}_{2} \\mathrm{SO}_{4}$\n(d) $\\mathrm{Zn}(\\mathrm{OH})_{2}(s)$ in a solution buffered at a pH of 11.45\n16. Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected.\n(a) $\\mathrm{TlCl}(s)$ in 1.250 M HCl\n(b) $\\mathrm{PbI}_{2}(s)$ in $0.0355 \\mathrm{MCaI}_{2}$\n(c) $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}(s)$ in 0.225 L of a solution containing 0.856 g of $\\mathrm{K}_{2} \\mathrm{CrO}_{4}$\n(d) $\\mathrm{Cd}(\\mathrm{OH})_{2}(s)$ in a solution buffered at a pH of 10.995\n17. Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that it is not appropriate to neglect the changes in the initial concentrations of the common ions."}
{"id": 4350, "contents": "1689. $3 x$ - \n(a) $\\mathrm{TlCl}(s)$ in $0.025 \\mathrm{M} \\mathrm{TlNO}_{3}$\n(b) $\\mathrm{BaF}_{2}(s)$ in 0.0313 M KF\n(c) $\\mathrm{MgC}_{2} \\mathrm{O}_{4}$ in 2.250 L of a solution containing 8.156 g of $\\mathrm{Mg}\\left(\\mathrm{NO}_{3}\\right)_{2}$\n(d) $\\mathrm{Ca}(\\mathrm{OH})_{2}(s)$ in an unbuffered solution initially with a pH of 12.700\n18. Explain why the changes in concentrations of the common ions in Exercise 15.17 can be neglected.\n19. Explain why the changes in concentrations of the common ions in Exercise 15.18 cannot be neglected.\n20. Calculate the solubility of aluminum hydroxide, $\\mathrm{Al}(\\mathrm{OH})_{3}$, in a solution buffered at pH 11.00 .\n21. Refer to Appendix J for solubility products for calcium salts. Determine which of the calcium salts listed is most soluble in moles per liter and which is most soluble in grams per liter.\n22. Most barium compounds are very poisonous; however, barium sulfate is often administered internally as an aid in the X-ray examination of the lower intestinal tract (Figure 15.4). This use of $\\mathrm{BaSO}_{4}$ is possible because of its low solubility. Calculate the molar solubility of $\\mathrm{BaSO}_{4}$ and the mass of barium present in 1.00 L of water saturated with $\\mathrm{BaSO}_{4}$."}
{"id": 4351, "contents": "1689. $3 x$ - \n23. Public Health Service standards for drinking water set a maximum of $250 \\mathrm{mg} / \\mathrm{L}\\left(2.60 \\times 10^{-3} \\mathrm{M}\\right)$ of $\\mathrm{SO}_{4}{ }^{2-}$ because of its cathartic action (it is a laxative). Does natural water that is saturated with $\\mathrm{CaSO}_{4}$ (\"gyp\" water) as a result or passing through soil containing gypsum, $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$, meet these standards? What is the concentration of $\\mathrm{SO}_{4}{ }^{2-}$ in such water?\n24. Perform the following calculations:\n(a) Calculate $\\left[\\mathrm{Ag}^{+}\\right]$in a saturated aqueous solution of AgBr .\n(b) What will $\\left[\\mathrm{Ag}^{+}\\right]$be when enough KBr has been added to make $\\left[\\mathrm{Br}^{-}\\right]=0.050 \\mathrm{M}$ ?\n(c) What will $\\left[\\mathrm{Br}^{-}\\right]$be when enough $\\mathrm{AgNO}_{3}$ has been added to make $\\left[\\mathrm{Ag}^{+}\\right]=0.020 \\mathrm{M}$ ?\n25. The solubility product of $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$ is $2.4 \\times 10^{-5}$. What mass of this salt will dissolve in 1.0 L of 0.010 M $\\mathrm{SO} 4^{2-}$ ?\n26. Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Appendix J for solubility products).\n(a) TlCl\n(b) $\\mathrm{BaF}_{2}$\n(c) $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}$\n(d) $\\mathrm{CaC}_{2} \\mathrm{O}_{4} \\cdot \\mathrm{H}_{2} \\mathrm{O}$\n(e) the mineral anglesite, $\\mathrm{PbSO}_{4}$"}
{"id": 4352, "contents": "1689. $3 x$ - \n(d) $\\mathrm{CaC}_{2} \\mathrm{O}_{4} \\cdot \\mathrm{H}_{2} \\mathrm{O}$\n(e) the mineral anglesite, $\\mathrm{PbSO}_{4}$\n27. Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Appendix J for solubility products):\n(a) AgI\n(b) $\\mathrm{Ag}_{2} \\mathrm{SO}_{4}$\n(c) $\\mathrm{Mn}(\\mathrm{OH})_{2}$\n(d) $\\mathrm{Sr}(\\mathrm{OH})_{2} \\cdot 8 \\mathrm{H}_{2} \\mathrm{O}$\n(e) the mineral brucite, $\\mathrm{Mg}(\\mathrm{OH})_{2}$\n28. The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate $K_{\\text {sp }}$ for each of the slightly soluble solids indicated:\n(a) $\\mathrm{AgBr}:\\left[\\mathrm{Ag}^{+}\\right]=5.7 \\times 10^{-7} \\mathrm{M},\\left[\\mathrm{Br}^{-}\\right]=5.7 \\times 10^{-7} \\mathrm{M}$\n(b) $\\mathrm{CaCO}_{3}:\\left[\\mathrm{Ca}^{2+}\\right]=5.3 \\times 10^{-3} \\mathrm{M},\\left[\\mathrm{CO}_{3}{ }^{2-}\\right]=9.0 \\times 10^{-7} \\mathrm{M}$\n(c) $\\mathrm{PbF}_{2}:\\left[\\mathrm{Pb}^{2+}\\right]=2.1 \\times 10^{-3} \\mathrm{M},\\left[\\mathrm{F}^{-}\\right]=4.2 \\times 10^{-3} \\mathrm{M}$\n(d) $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}:\\left[\\mathrm{Ag}^{+}\\right]=5.3 \\times 10^{-5} \\mathrm{M}, 3.2 \\times 10^{-3} \\mathrm{M}$"}
{"id": 4353, "contents": "1689. $3 x$ - \n(e) $\\mathrm{InF}_{3}:\\left[\\mathrm{In}^{3+}\\right]=2.3 \\times 10^{-3} \\mathrm{M},\\left[\\mathrm{F}^{-}\\right]=7.0 \\times 10^{-3} \\mathrm{M}$\n29. The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate $K_{\\text {sp }}$ for each of the slightly soluble solids indicated:\n(a) $\\mathrm{TlCl}:\\left[\\mathrm{Tl}^{+}\\right]=1.21 \\times 10^{-2} \\mathrm{M},\\left[\\mathrm{Cl}^{-}\\right]=1.2 \\times 10^{-2} \\mathrm{M}$\n(b) $\\mathrm{Ce}\\left(\\mathrm{IO}_{3}\\right)_{4}:\\left[\\mathrm{Ce}^{4+}\\right]=1.8 \\times 10^{-4} \\mathrm{M},\\left[\\mathrm{IO}_{3}{ }^{-}\\right]=2.6 \\times 10^{-13} \\mathrm{M}$\n(c) $\\mathrm{Gd}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}:\\left[\\mathrm{Gd}^{3+}\\right]=0.132 \\mathrm{M},\\left[\\mathrm{SO}_{4}{ }^{2-}\\right]=0.198 \\mathrm{M}$\n(d) $\\mathrm{Ag}_{2} \\mathrm{SO}_{4}:\\left[\\mathrm{Ag}^{+}\\right]=2.40 \\times 10^{-2} \\mathrm{M},\\left[\\mathrm{SO}_{4}{ }^{2-}\\right]=2.05 \\times 10^{-2} \\mathrm{M}$\n(e) $\\mathrm{BaSO}_{4}:\\left[\\mathrm{Ba}^{2+}\\right]=0.500 \\mathrm{M},\\left[\\mathrm{SO}_{4}{ }^{2-}\\right]=4.6 \\times 10^{-8} \\mathrm{M}$\n30. Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Appendix J for $K_{\\text {sp }}$ values.)"}
{"id": 4354, "contents": "1689. $3 x$ - \n30. Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Appendix J for $K_{\\text {sp }}$ values.)\n(a) $\\mathrm{KClO}_{4}:\\left[\\mathrm{K}^{+}\\right]=0.01 \\mathrm{M},\\left[\\mathrm{ClO}_{4}{ }^{-}\\right]=0.01 \\mathrm{M}$\n(b) $\\mathrm{K}_{2} \\mathrm{PtCl}_{6}:\\left[\\mathrm{K}^{+}\\right]=0.01 \\mathrm{M},\\left[\\mathrm{PtCl}_{6}{ }^{2-}\\right]=0.01 \\mathrm{M}$\n(c) $\\mathrm{PbI}_{2}:\\left[\\mathrm{Pb}^{2+}\\right]=0.003 \\mathrm{M},\\left[\\mathrm{I}^{-}\\right]=1.3 \\times 10^{-3} \\mathrm{M}$\n(d) $\\mathrm{Ag}_{2} \\mathrm{~S}:\\left[\\mathrm{Ag}^{+}\\right]=1 \\times 10^{-10} \\mathrm{M},\\left[\\mathrm{S}^{2-}\\right]=1 \\times 10^{-13} \\mathrm{M}$\n31. Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Appendix J for $K_{\\mathrm{sp}}$ values.)\n(a) $\\mathrm{CaCO}_{3}:\\left[\\mathrm{Ca}^{2+}\\right]=0.003 \\mathrm{M},\\left[\\mathrm{CO}_{3}{ }^{2-}\\right]=0.003 \\mathrm{M}$\n(b) $\\mathrm{Co}(\\mathrm{OH})_{2}:\\left[\\mathrm{Co}^{2+}\\right]=0.01 \\mathrm{M},\\left[\\mathrm{OH}^{-}\\right]=1 \\times 10^{-7} \\mathrm{M}$\n(c) $\\mathrm{CaHPO}_{4}:\\left[\\mathrm{Ca}^{2+}\\right]=0.01 \\mathrm{M},\\left[\\mathrm{HPO}_{4}{ }^{2-}\\right]=2 \\times 10^{-6} \\mathrm{M}$"}
{"id": 4355, "contents": "1689. $3 x$ - \n(d) $\\mathrm{Pb}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}:\\left[\\mathrm{Pb}^{2+}\\right]=0.01 \\mathrm{M},\\left[\\mathrm{PO}_{4}{ }^{3-}\\right]=1 \\times 10^{-13} \\mathrm{M}$\n32. Calculate the concentration of $\\mathrm{Tl}^{+}$when TlCl just begins to precipitate from a solution that is 0.0250 M in $\\mathrm{Cl}^{-}$.\n33. Calculate the concentration of sulfate ion when $\\mathrm{BaSO}_{4}$ just begins to precipitate from a solution that is 0.0758 M in $\\mathrm{Ba}^{2+}$.\n34. Calculate the concentration of $\\mathrm{Sr}^{2+}$ when $\\mathrm{SrCrO}_{4}$ starts to precipitate from a solution that is 0.0025 M in $\\mathrm{CrO}_{4}{ }^{2-}$.\n35. Calculate the concentration of $\\mathrm{PO}_{4}{ }^{3-}$ when $\\mathrm{Ag}_{3} \\mathrm{PO}_{4}$ starts to precipitate from a solution that is 0.0125 M in $\\mathrm{Ag}^{+}$.\n36. Calculate the concentration of $\\mathrm{F}^{-}$required to begin precipitation of $\\mathrm{CaF}_{2}$ in a solution that is 0.010 M in $\\mathrm{Ca}^{2+}$.\n37. Calculate the concentration of $\\mathrm{Ag}^{+}$required to begin precipitation of $\\mathrm{Ag}_{2} \\mathrm{CO}_{3}$ in a solution that is $2.50 \\times$ $10^{-6} \\mathrm{M}$ in $\\mathrm{CO}_{3}{ }^{2-}$.\n38. What $\\left[\\mathrm{Ag}^{+}\\right]$is required to reduce $\\left[\\mathrm{CO}_{3}^{2-}\\right]$ to $8.2 \\times 10^{-4} M$ by precipitation of $\\mathrm{Ag}_{2} \\mathrm{CO}_{3}$ ?"}
{"id": 4356, "contents": "1689. $3 x$ - \n39. What $\\left[\\mathrm{F}^{-}\\right]$is required to reduce $\\left[\\mathrm{Ca}^{2+}\\right]$ to $1.0 \\times 10^{-4} \\mathrm{M}$ by precipitation of $\\mathrm{CaF}_{2}$ ?\n40. A volume of 0.800 L of a $2 \\times 10^{-4}-\\mathrm{MBa}\\left(\\mathrm{NO}_{3}\\right)_{2}$ solution is added to 0.200 L of $5 \\times 10^{-4} \\mathrm{MLi}_{2} \\mathrm{SO}_{4}$. Does $\\mathrm{BaSO}_{4}$ precipitate? Explain your answer.\n41. Perform these calculations for nickel(II) carbonate. (a) With what volume of water must a precipitate containing $\\mathrm{NiCO}_{3}$ be washed to dissolve 0.100 g of this compound? Assume that the wash water becomes saturated with $\\mathrm{NiCO}_{3}\\left(K_{\\mathrm{sp}}=1.36 \\times 10^{-7}\\right)$.\n(b) If the $\\mathrm{NiCO}_{3}$ were a contaminant in a sample of $\\mathrm{CoCO}_{3}\\left(K_{\\text {sp }}=1.0 \\times 10^{-12}\\right)$, what mass of $\\mathrm{CoCO}_{3}$ would have been lost? Keep in mind that both $\\mathrm{NiCO}_{3}$ and $\\mathrm{CoCO}_{3}$ dissolve in the same solution.\n42. Iron concentrations greater than $5.4 \\times 10^{-6} M$ in water used for laundry purposes can cause staining. What $\\left[\\mathrm{OH}^{-}\\right]$is required to reduce $\\left[\\mathrm{Fe}^{2+}\\right]$ to this level by precipitation of $\\mathrm{Fe}(\\mathrm{OH})_{2}$ ?\n43. A solution is 0.010 M in both $\\mathrm{Cu}^{2+}$ and $\\mathrm{Cd}^{2+}$. What percentage of $\\mathrm{Cd}^{2+}$ remains in the solution when $99.9 \\%$ of the $\\mathrm{Cu}^{2+}$ has been precipitated as CuS by adding sulfide?"}
{"id": 4357, "contents": "1689. $3 x$ - \n44. A solution is 0.15 M in both $\\mathrm{Pb}^{2+}$ and $\\mathrm{Ag}^{+}$. If $\\mathrm{Cl}^{-}$is added to this solution, what is $\\left[\\mathrm{Ag}^{+}\\right]$when $\\mathrm{PbCl}_{2}$ begins to precipitate?\n45. What reagent might be used to separate the ions in each of the following mixtures, which are 0.1 M with respect to each ion? In some cases it may be necessary to control the pH . (Hint: Consider the $K_{\\mathrm{sp}}$ values given in Appendix J.)\n(a) $\\mathrm{Hg}_{2}{ }^{2+}$ and $\\mathrm{Cu}^{2+}$\n(b) $\\mathrm{SO}_{4}{ }^{2-}$ and $\\mathrm{Cl}^{-}$\n(c) $\\mathrm{Hg}^{2+}$ and $\\mathrm{Co}^{2+}$\n(d) $\\mathrm{Zn}^{2+}$ and $\\mathrm{Sr}^{2+}$\n(e) $\\mathrm{Ba}^{2+}$ and $\\mathrm{Mg}^{2+}$\n(f) $\\mathrm{CO}_{3}{ }^{2-}$ and $\\mathrm{OH}^{-}$\n46. A solution contains $1.0 \\times 10^{-5} \\mathrm{~mol}$ of KBr and 0.10 mol of KCl per liter. $\\mathrm{AgNO}_{3}$ is gradually added to this solution. Which forms first, solid AgBr or solid AgCl ?\n47. A solution contains $1.0 \\times 10^{-2} \\mathrm{~mol}$ of KI and 0.10 mol of KCl per liter. $\\mathrm{AgNO}_{3}$ is gradually added to this solution. Which forms first, solid AgI or solid AgCl ?"}
{"id": 4358, "contents": "1689. $3 x$ - \n48. The calcium ions in human blood serum are necessary for coagulation (Figure 15.5). Potassium oxalate, $\\mathrm{K}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$, is used as an anticoagulant when a blood sample is drawn for laboratory tests because it removes the calcium as a precipitate of $\\mathrm{CaC}_{2} \\mathrm{O}_{4} \\cdot \\mathrm{H}_{2} \\mathrm{O}$. It is necessary to remove all but $1.0 \\%$ of the $\\mathrm{Ca}^{2+}$ in serum in order to prevent coagulation. If normal blood serum with a buffered pH of 7.40 contains 9.5 mg of $\\mathrm{Ca}^{2+}$ per 100 mL of serum, what mass of $\\mathrm{K}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ is required to prevent the coagulation of a 10 mL blood sample that is $55 \\%$ serum by volume? (All volumes are accurate to two significant figures. Note that the volume of serum in a $10-\\mathrm{mL}$ blood sample is 5.5 mL . Assume that the $\\mathrm{K}_{\\mathrm{sp}}$ value for $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$ in serum is the same as in water.)\n49. About $50 \\%$ of urinary calculi (kidney stones) consist of calcium phosphate, $\\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$. The normal mid range calcium content excreted in the urine is 0.10 g of $\\mathrm{Ca}^{2+}$ per day. The normal mid range amount of urine passed may be taken as 1.4 L per day. What is the maximum concentration of phosphate ion that urine can contain before a calculus begins to form?"}
{"id": 4359, "contents": "1689. $3 x$ - \n50. The pH of normal urine is 6.30 , and the total phosphate concentration $\\left(\\left[\\mathrm{PO}_{4}{ }^{3-}\\right]+\\left[\\mathrm{HPO}_{4}{ }^{2-}\\right]+\\left[\\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{-}\\right]\\right.$ $\\left.+\\left[\\mathrm{H}_{3} \\mathrm{PO}_{4}\\right]\\right)$ is 0.020 M . What is the minimum concentration of $\\mathrm{Ca}^{2+}$ necessary to induce kidney stone formation? (See Exercise 15.49 for additional information.)\n51. Magnesium metal (a component of alloys used in aircraft and a reducing agent used in the production of uranium, titanium, and other active metals) is isolated from sea water by the following sequence of reactions:\n$\\mathrm{Mg}^{2+}(a q)+\\mathrm{Ca}(\\mathrm{OH})_{2}(a q) \\longrightarrow \\mathrm{Mg}(\\mathrm{OH})_{2}(s)+\\mathrm{Ca}^{2+}(a q)$\n$\\mathrm{Mg}(\\mathrm{OH})_{2}(s)+2 \\mathrm{HCl}(a q) \\longrightarrow \\mathrm{MgCl}_{2}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n$\\mathrm{MgCl}_{2}(l) \\xrightarrow{\\text { electrolysis }} \\mathrm{Mg}(s)+\\mathrm{Cl}_{2}(g)$\nSea water has a density of $1.026 \\mathrm{~g} / \\mathrm{cm}^{3}$ and contains 1272 parts per million of magnesium as $\\mathrm{Mg}^{2+}(\\mathrm{aq})$ by mass. What mass, in kilograms, of $\\mathrm{Ca}(\\mathrm{OH})_{2}$ is required to precipitate $99.9 \\%$ of the magnesium in $1.00 \\times$ $10^{3} \\mathrm{~L}$ of sea water?"}
{"id": 4360, "contents": "1689. $3 x$ - \n52. Hydrogen sulfide is bubbled into a solution that is $0.10 M$ in both $\\mathrm{Pb}^{2+}$ and $\\mathrm{Fe}^{2+}$ and 0.30 M in HCl . After the solution has come to equilibrium it is saturated with $\\mathrm{H}_{2} \\mathrm{~S}\\left(\\left[\\mathrm{H}_{2} \\mathrm{~S}\\right]=0.10 \\mathrm{M}\\right)$. What concentrations of $\\mathrm{Pb}^{2+}$ and $\\mathrm{Fe}^{2+}$ remain in the solution? For a saturated solution of $\\mathrm{H}_{2} \\mathrm{~S}$ we can use the equilibrium:"}
{"id": 4361, "contents": "1689. $3 x$ - \n$$\n\\mathrm{H}_{2} \\mathrm{~S}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons 2 \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{S}^{2-}(a q) \\quad K=1.0 \\times 10^{-26}\n$$"}
{"id": 4362, "contents": "1689. $3 x$ - \n(Hint: The $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]$changes as metal sulfides precipitate.)\n53. Perform the following calculations involving concentrations of iodate ions:\n(a) The iodate ion concentration of a saturated solution of $\\mathrm{La}\\left(\\mathrm{IO}_{3}\\right)_{3}$ was found to be $3.1 \\times 10^{-3} \\mathrm{~mol} / \\mathrm{L}$. Find the $K_{\\mathrm{sp}}$.\n(b) Find the concentration of iodate ions in a saturated solution of $\\mathrm{Cu}\\left(\\mathrm{IO}_{3}\\right)_{2}\\left(K_{\\text {sp }}=7.4 \\times 10^{-8}\\right)$.\n54. Calculate the molar solubility of AgBr in $0.035 \\mathrm{M} \\mathrm{NaBr}\\left(K_{\\mathrm{sp}}=5 \\times 10^{-13}\\right)$.\n55. How many grams of $\\mathrm{Pb}(\\mathrm{OH})_{2}$ will dissolve in 500 mL of a $0.050-M \\mathrm{PbCl}_{2}$ solution $\\left(K_{\\text {sp }}=1.2 \\times 10^{-15}\\right)$ ?\n56. Use the simulation (http://openstax.org/l/16solublesalts) from the earlier Link to Learning to complete the following exercise. Using $0.01 \\mathrm{~g} \\mathrm{CaF}_{2}$, give the $\\mathrm{K}_{\\mathrm{sp}}$ values found in a $0.2-M$ solution of each of the salts. Discuss why the values change as you change soluble salts."}
{"id": 4363, "contents": "1689. $3 x$ - \n57. How many grams of Milk of Magnesia, $\\mathrm{Mg}(\\mathrm{OH})_{2}(S)(58.3 \\mathrm{~g} / \\mathrm{mol})$, would be soluble in 200 mL of water. $K_{\\mathrm{sp}}=$ $7.1 \\times 10^{-12}$. Include the ionic reaction and the expression for $K_{\\text {sp }}$ in your answer. ( $K_{\\mathrm{w}}=1 \\times 10^{-14}=$ $\\left.\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]\\right)$\n58. Two hypothetical salts, $\\mathrm{LM}_{2}$ and LQ , have the same molar solubility in $\\mathrm{H}_{2} \\mathrm{O}$. If $K_{\\text {sp }}$ for $\\mathrm{LM}_{2}$ is $3.20 \\times 10^{-5}$, what is the $K_{\\mathrm{sp}}$ value for LQ ?\n59. The carbonate ion concentration is gradually increased in a solution containing equal concentrations of the divalent cations of magnesium, calcium, strontium, barium, and manganese. Which of the following carbonates will precipitate first? Which will precipitate last? Explain.\n(a) $\\mathrm{MgCO}_{3} \\cdot 3 \\mathrm{H}_{2} \\mathrm{O} \\quad K_{\\text {sp }}=1 \\times 10^{-5}$\n(b) $\\mathrm{CaCO}_{3}$\n(c) $\\mathrm{SrCO}_{3}$\n$K_{\\mathrm{sp}}=8.7 \\times 10^{-9}$\n(d) $\\mathrm{BaCO}_{3}$\n$K_{\\mathrm{sp}}=7 \\times 10^{-10}$\n(e) $\\mathrm{MnCO}_{3}$\n$K_{\\text {sp }}=1.6 \\times 10^{-9}$\n$K_{\\text {sp }}=8.8 \\times 10^{-11}$"}
{"id": 4364, "contents": "1689. $3 x$ - \n$K_{\\text {sp }}=1.6 \\times 10^{-9}$\n$K_{\\text {sp }}=8.8 \\times 10^{-11}$\n60. How many grams of $\\mathrm{Zn}(\\mathrm{CN})_{2}(S)(117.44 \\mathrm{~g} / \\mathrm{mol})$ would be soluble in 100 mL of $\\mathrm{H}_{2} \\mathrm{O}$ ? Include the balanced reaction and the expression for $K_{\\text {sp }}$ in your answer. The $K_{\\text {sp }}$ value for $\\mathrm{Zn}(\\mathrm{CN})_{2}(s)$ is $3.0 \\times 10^{-16}$."}
{"id": 4365, "contents": "1689. $3 x$ - 1689.1. Lewis Acids and Bases\n61. Even though $\\mathrm{Ca}(\\mathrm{OH})_{2}$ is an inexpensive base, its limited solubility restricts its use. What is the pH of a saturated solution of $\\mathrm{Ca}(\\mathrm{OH})_{2}$ ?\n62. Under what circumstances, if any, does a sample of solid AgCl completely dissolve in pure water?\n63. Explain why the addition of $\\mathrm{NH}_{3}$ or $\\mathrm{HNO}_{3}$ to a saturated solution of $\\mathrm{Ag}_{2} \\mathrm{CO}_{3}$ in contact with solid $\\mathrm{Ag}_{2} \\mathrm{CO}_{3}$ increases the solubility of the solid.\n64. Calculate the cadmium ion concentration, $\\left[\\mathrm{Cd}^{2+}\\right]$, in a solution prepared by mixing 0.100 L of 0.0100 M $\\mathrm{Cd}\\left(\\mathrm{NO}_{3}\\right)_{2}$ with 0.150 L of $0.100 \\mathrm{NH}_{3}(\\mathrm{aq})$.\n65. Explain why addition of $\\mathrm{NH}_{3}$ or $\\mathrm{HNO}_{3}$ to a saturated solution of $\\mathrm{Cu}(\\mathrm{OH})_{2}$ in contact with solid $\\mathrm{Cu}(\\mathrm{OH})_{2}$ increases the solubility of the solid.\n66. Sometimes equilibria for complex ions are described in terms of dissociation constants, $K_{d}$. For the complex ion $\\mathrm{AlF}_{6}{ }^{3-}$ the dissociation reaction is:\n$\\mathrm{AlF}_{6}{ }^{3-} \\rightleftharpoons \\mathrm{Al}^{3+}+6 \\mathrm{~F}^{-}$and $K_{\\mathrm{d}}=\\frac{\\left[\\mathrm{Al}^{3+}\\right]\\left[\\mathrm{F}^{-}\\right]^{6}}{\\left[\\mathrm{AlF}_{6}{ }^{3-}\\right]}=2 \\times 10^{-24}$\nCalculate the value of the formation constant, $K_{\\mathrm{f}}$, for $\\mathrm{AlF}_{6}{ }^{3-}$."}
{"id": 4366, "contents": "1689. $3 x$ - 1689.1. Lewis Acids and Bases\nCalculate the value of the formation constant, $K_{\\mathrm{f}}$, for $\\mathrm{AlF}_{6}{ }^{3-}$.\n67. Using the value of the formation constant for the complex ion $\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}{ }^{2+}$, calculate the dissociation constant.\n68. Using the dissociation constant, $K_{\\mathrm{d}}=7.8 \\times 10^{-18}$, calculate the equilibrium concentrations of $\\mathrm{Cd}^{2+}$ and $\\mathrm{CN}^{-}$in a $0.250-M$ solution of $\\mathrm{Cd}(\\mathrm{CN})_{4}{ }^{2-}$.\n69. Using the dissociation constant, $K_{\\mathrm{d}}=3.4 \\times 10^{-15}$, calculate the equilibrium concentrations of $\\mathrm{Zn}^{2+}$ and $\\mathrm{OH}^{-}$in a $0.0465-M$ solution of $\\mathrm{Zn}(\\mathrm{OH})_{4}{ }^{2-}$.\n70. Using the dissociation constant, $K_{\\mathrm{d}}=2.2 \\times 10^{-34}$, calculate the equilibrium concentrations of $\\mathrm{Co}^{3+}$ and $\\mathrm{NH}_{3}$ in a $0.500-M$ solution of $\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}{ }^{3+}$.\n71. Using the dissociation constant, $K_{\\mathrm{d}}=1 \\times 10^{-44}$, calculate the equilibrium concentrations of $\\mathrm{Fe}^{3+}$ and $\\mathrm{CN}^{-}$ in a 0.333 M solution of $\\mathrm{Fe}(\\mathrm{CN})_{6}{ }^{3-}$.\n72. Calculate the mass of potassium cyanide ion that must be added to 100 mL of solution to dissolve $2.0 \\times$ $10^{-2} \\mathrm{~mol}$ of silver cyanide, AgCN .\n73. Calculate the minimum concentration of ammonia needed in 1.0 L of solution to dissolve $3.0 \\times 10^{-3} \\mathrm{~mol}$ of silver bromide."}
{"id": 4367, "contents": "1689. $3 x$ - 1689.1. Lewis Acids and Bases\n73. Calculate the minimum concentration of ammonia needed in 1.0 L of solution to dissolve $3.0 \\times 10^{-3} \\mathrm{~mol}$ of silver bromide.\n74. A roll of $35-\\mathrm{mm}$ black and white photographic film contains about 0.27 g of unexposed AgBr before developing. What mass of $\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3} \\cdot 5 \\mathrm{H}_{2} \\mathrm{O}$ (sodium thiosulfate pentahydrate or hypo) in 1.0 L of developer is required to dissolve the AgBr as $\\mathrm{Ag}\\left(\\mathrm{S}_{2} \\mathrm{O}_{3}\\right)_{2}{ }^{3-}\\left(K_{\\mathrm{f}}=4.7 \\times 10^{13}\\right)$ ?\n75. We have seen an introductory definition of an acid: An acid is a compound that reacts with water and increases the amount of hydronium ion present. In the chapter on acids and bases, we saw two more definitions of acids: a compound that donates a proton (a hydrogen ion, $\\mathrm{H}^{+}$) to another compound is called a Br\u00f8nsted-Lowry acid, and a Lewis acid is any species that can accept a pair of electrons. Explain why the introductory definition is a macroscopic definition, while the Br\u00f8nsted-Lowry definition and the Lewis definition are microscopic definitions.\n76. Write the Lewis structures of the reactants and product of each of the following equations, and identify the Lewis acid and the Lewis base in each:\n(a) $\\mathrm{CO}_{2}+\\mathrm{OH}^{-} \\longrightarrow \\mathrm{HCO}_{3}{ }^{-}$\n(b) $\\mathrm{B}(\\mathrm{OH})_{3}+\\mathrm{OH}^{-} \\longrightarrow \\mathrm{B}(\\mathrm{OH})_{4}{ }^{-}$\n(c) $\\mathrm{I}^{-}+\\mathrm{I}_{2} \\longrightarrow \\mathrm{I}_{3}{ }^{-}$"}
{"id": 4368, "contents": "1689. $3 x$ - 1689.1. Lewis Acids and Bases\n(c) $\\mathrm{I}^{-}+\\mathrm{I}_{2} \\longrightarrow \\mathrm{I}_{3}{ }^{-}$\n(d) $\\mathrm{AlCl}_{3}+\\mathrm{Cl}^{-} \\longrightarrow \\mathrm{AlCl}_{4}{ }^{-}$(use Al-Cl single bonds)\n(e) $\\mathrm{O}^{2-}+\\mathrm{SO}_{3} \\longrightarrow \\mathrm{SO}_{4}{ }^{2-}$\n77. Write the Lewis structures of the reactants and product of each of the following equations, and identify the Lewis acid and the Lewis base in each:\n(a) $\\mathrm{CS}_{2}+\\mathrm{SH}^{-} \\longrightarrow \\mathrm{HCS}_{3}{ }^{-}$\n(b) $\\mathrm{BF}_{3}+\\mathrm{F}^{-} \\longrightarrow \\mathrm{BF}_{4}^{-}$\n(c) $\\mathrm{I}^{-}+\\mathrm{SnI}_{2} \\longrightarrow \\mathrm{SnI}_{3}{ }^{-}$\n(d) $\\mathrm{Al}(\\mathrm{OH})_{3}+\\mathrm{OH}^{-} \\longrightarrow \\mathrm{Al}(\\mathrm{OH})_{4}{ }^{-}$\n(e) $\\mathrm{F}^{-}+\\mathrm{SO}_{3} \\longrightarrow \\mathrm{SFO}_{3}{ }^{-}$\n78. Using Lewis structures, write balanced equations for the following reactions:\n(a) $\\mathrm{HCl}(g)+\\mathrm{PH}_{3}(g) \\longrightarrow$\n(b) $\\mathrm{H}_{3} \\mathrm{O}^{+}+\\mathrm{CH}_{3}^{-} \\longrightarrow$\n(c) $\\mathrm{CaO}+\\mathrm{SO}_{3} \\longrightarrow$\n(d) $\\mathrm{NH}_{4}^{+}+\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{O}^{-} \\longrightarrow$\n79. Calculate $\\left[\\mathrm{HgCl}_{4}{ }^{2-}\\right.$ ] in a solution prepared by adding 0.0200 mol of NaCl to 0.250 L of a $0.100-\\mathrm{M} \\mathrm{HgCl} 2$ solution."}
{"id": 4369, "contents": "1689. $3 x$ - 1689.1. Lewis Acids and Bases\n80. In a titration of cyanide ion, 28.72 mL of $0.0100 \\mathrm{MANO}_{3}$ is added before precipitation begins. [The reaction of $\\mathrm{Ag}^{+}$with $\\mathrm{CN}^{-}$goes to completion, producing the $\\mathrm{Ag}(\\mathrm{CN})_{2}{ }^{-}$complex.] Precipitation of solid AgCN takes place when excess $\\mathrm{Ag}^{+}$is added to the solution, above the amount needed to complete the formation of $\\mathrm{Ag}(\\mathrm{CN})_{2}{ }^{-}$. How many grams of NaCN were in the original sample?\n81. What are the concentrations of $\\mathrm{Ag}^{+}, \\mathrm{CN}^{-}$, and $\\mathrm{Ag}(\\mathrm{CN})_{2}{ }^{-}$in a saturated solution of AgCN ?\n82. In dilute aqueous solution HF acts as a weak acid. However, pure liquid HF (boiling point $=19.5^{\\circ} \\mathrm{C}$ ) is a strong acid. In liquid $\\mathrm{HF}, \\mathrm{HNO}_{3}$ acts like a base and accepts protons. The acidity of liquid HF can be increased by adding one of several inorganic fluorides that are Lewis acids and accept $\\mathrm{F}^{-}$ion (for example, $\\mathrm{BF}_{3}$ or $\\mathrm{SbF}_{5}$ ). Write balanced chemical equations for the reaction of pure $\\mathrm{HNO}_{3}$ with pure HF and of pure HF with $\\mathrm{BF}_{3}$."}
{"id": 4370, "contents": "1689. $3 x$ - 1689.1. Lewis Acids and Bases\n83. The simplest amino acid is glycine, $\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$. The common feature of amino acids is that they contain the functional groups: an amine group, $-\\mathrm{NH}_{2}$, and a carboxylic acid group, $-\\mathrm{CO}_{2} \\mathrm{H}$. An amino acid can function as either an acid or a base. For glycine, the acid strength of the carboxyl group is about the same as that of acetic acid, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$, and the base strength of the amino group is slightly greater than that of ammonia, $\\mathrm{NH}_{3}$.\n(a) Write the Lewis structures of the ions that form when glycine is dissolved in 1 MHCl and in 1 MKOH .\n(b) Write the Lewis structure of glycine when this amino acid is dissolved in water. (Hint: Consider the relative base strengths of the $-\\mathrm{NH}_{2}$ and $-\\mathrm{CO}_{2}{ }^{-}$groups.)\n84. Boric acid, $\\mathrm{H}_{3} \\mathrm{BO}_{3}$, is not a Br\u00f8nsted-Lowry acid but a Lewis acid.\n(a) Write an equation for its reaction with water.\n(b) Predict the shape of the anion thus formed.\n(c) What is the hybridization on the boron consistent with the shape you have predicted?"}
{"id": 4371, "contents": "1689. $3 x$ - 1689.2. Coupled Equilibria\n85. A saturated solution of a slightly soluble electrolyte in contact with some of the solid electrolyte is said to be a system in equilibrium. Explain. Why is such a system called a heterogeneous equilibrium?\n86. Calculate the equilibrium concentration of $\\mathrm{Ni}^{2+}$ in a $1.0-M$ solution $\\left[\\mathrm{Ni}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]\\left(\\mathrm{NO}_{3}\\right)_{2}$.\n87. Calculate the equilibrium concentration of $\\mathrm{Zn}^{2+}$ in a $0.30-M$ solution of $\\mathrm{Zn}(\\mathrm{CN})_{4}{ }^{2-}$.\n88. Calculate the equilibrium concentration of $\\mathrm{Cu}^{2+}$ in a solution initially with $0.050 \\mathrm{MCu}^{2+}$ and $1.00 \\mathrm{MNH}_{3}$.\n89. Calculate the equilibrium concentration of $\\mathrm{Zn}^{2+}$ in a solution initially with $0.150 \\mathrm{M} \\mathrm{Zn}^{2+}$ and $2.50 \\mathrm{MCN}^{-}$.\n90. Calculate the $\\mathrm{Fe}^{3+}$ equilibrium concentration when 0.0888 mole of $\\mathrm{K}_{3}\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]$ is added to a solution with 0.0.00010 $\\mathrm{MCN}^{-}$.\n91. Calculate the $\\mathrm{Co}^{2+}$ equilibrium concentration when 0.010 mole of $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]\\left(\\mathrm{NO}_{3}\\right)_{2}$ is added to a solution with $0.25 \\mathrm{M} \\mathrm{NH}_{3}$. Assume the volume is 1.00 L .\n92. Calculate the molar solubility of $\\mathrm{Sn}(\\mathrm{OH})_{2}$ in a buffer solution containing equal concentrations of $\\mathrm{NH}_{3}$ and $\\mathrm{NH}_{4}{ }^{+}$."}
{"id": 4372, "contents": "1689. $3 x$ - 1689.2. Coupled Equilibria\n93. Calculate the molar solubility of $\\mathrm{Al}(\\mathrm{OH})_{3}$ in a buffer solution with $0.100 \\mathrm{MNH}_{3}$ and $0.400 \\mathrm{MNH}_{4}{ }^{+}$.\n94. What is the molar solubility of $\\mathrm{CaF}_{2}$ in a $0.100-M$ solution of HF ? $K_{\\mathrm{a}}$ for $\\mathrm{HF}=6.4 \\times 10^{-4}$.\n95. What is the molar solubility of $\\mathrm{BaSO}_{4}$ in a $0.250-M$ solution of $\\mathrm{NaHSO}_{4}$ ? $K_{\\mathrm{a}}$ for $\\mathrm{HSO}_{4}^{-}=1.2 \\times 10^{-2}$.\n96. What is the molar solubility of $\\mathrm{Tl}(\\mathrm{OH})_{3}$ in a $0.10-M$ solution of $\\mathrm{NH}_{3}$ ?\n97. What is the molar solubility of $\\mathrm{Pb}(\\mathrm{OH})_{2}$ in a $0.138-M$ solution of $\\mathrm{CH}_{3} \\mathrm{NH}_{2}$ ?\n98. A solution of $0.075 \\mathrm{MCoBr}_{2}$ is saturated with $\\mathrm{H}_{2} \\mathrm{~S}\\left(\\left[\\mathrm{H}_{2} \\mathrm{~S}\\right]=0.10 \\mathrm{M}\\right)$. What is the minimum pH at which CoS begins to precipitate?\n$\\mathrm{CoS}(s) \\rightleftharpoons \\mathrm{Co}^{2+}(a q)+\\mathrm{S}^{2-}(a q) \\quad K_{\\mathrm{sp}}=2.3 \\times 10^{27}$"}
{"id": 4373, "contents": "1689. $3 x$ - 1689.2. Coupled Equilibria\n$\\mathrm{H}_{2} \\mathrm{~S}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons 2 \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{S}^{2-}(a q) \\quad K=8.9 \\times 10^{-27}$\n99. A $0.125-M$ solution of $\\mathrm{Mn}\\left(\\mathrm{NO}_{3}\\right)_{2}$ is saturated with $\\mathrm{H}_{2} \\mathrm{~S}\\left(\\left[\\mathrm{H}_{2} \\mathrm{~S}\\right]=0.10 \\mathrm{M}\\right)$. At what pH does MnS begin to precipitate?\n$\\operatorname{MnS}(s) \\rightleftharpoons \\mathrm{Mn}^{2+}(a q)+\\mathrm{S}^{2-}(a q) \\quad K_{\\mathrm{sp}}=2.3 \\times 10^{-13}$\n$\\mathrm{H}_{2} \\mathrm{~S}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons 2 \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{S}^{2-}(a q) \\quad K=1.0 \\times 10^{-26}$\n100. Both AgCl and AgI dissolve in $\\mathrm{NH}_{3}$.\n(a) What mass of AgI dissolves in 1.0 L of $1.0 \\mathrm{MNH}_{3}$ ?\n(b) What mass of AgCl dissolves in 1.0 L of $1.0 \\mathrm{M} \\mathrm{NH}_{3}$ ?\n101. The following question is taken from a Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service.\nSolve the following problem:\n$\\operatorname{MgF}_{2}(s) \\rightleftharpoons \\mathrm{Mg}^{2+}(a q)+2 \\mathrm{~F}^{-}(a q)$"}
{"id": 4374, "contents": "1689. $3 x$ - 1689.2. Coupled Equilibria\n$\\operatorname{MgF}_{2}(s) \\rightleftharpoons \\mathrm{Mg}^{2+}(a q)+2 \\mathrm{~F}^{-}(a q)$\nIn a saturated solution of $\\mathrm{MgF}_{2}$ at $18{ }^{\\circ} \\mathrm{C}$, the concentration of $\\mathrm{Mg}^{2+}$ is $1.21 \\times 10^{-3} \\mathrm{M}$. The equilibrium is represented by the preceding equation.\n(a) Write the expression for the solubility-product constant, $K_{\\text {sp }}$, and calculate its value at $18{ }^{\\circ} \\mathrm{C}$.\n(b) Calculate the equilibrium concentration of $\\mathrm{Mg}^{2+}$ in 1.000 L of saturated $\\mathrm{MgF}_{2}$ solution at $18{ }^{\\circ} \\mathrm{C}$ to which 0.100 mol of solid KF has been added. The KF dissolves completely. Assume the volume change is negligible.\n(c) Predict whether a precipitate of $\\mathrm{MgF}_{2}$ will form when 100.0 mL of a $3.00 \\times 10^{-3}-M$ solution of $\\mathrm{Mg}\\left(\\mathrm{NO}_{3}\\right)_{2}$ is mixed with 200.0 mL of a $2.00 \\times 10^{-3}-\\mathrm{M}$ solution of NaF at $18^{\\circ} \\mathrm{C}$. Show the calculations to support your prediction.\n(d) At $27{ }^{\\circ} \\mathrm{C}$ the concentration of $\\mathrm{Mg}^{2+}$ in a saturated solution of $\\mathrm{MgF}_{2}$ is $1.17 \\times 10^{-3} \\mathrm{M}$. Is the dissolving of $\\mathrm{MgF}_{2}$ in water an endothermic or an exothermic process? Give an explanation to support your conclusion."}
{"id": 4375, "contents": "1689. $3 x$ - 1689.2. Coupled Equilibria\n102. Which of the following compounds, when dissolved in a $0.01-M$ solution of $\\mathrm{HClO}_{4}$, has a solubility greater than in pure water: $\\mathrm{CuCl}, \\mathrm{CaCO}_{3}, \\mathrm{MnS}, \\mathrm{PbBr}_{2}, \\mathrm{CaF}_{2}$ ? Explain your answer.\n103. Which of the following compounds, when dissolved in a $0.01-M$ solution of $\\mathrm{HClO}_{4}$, has a solubility greater than in pure water: $\\mathrm{AgBr}, \\mathrm{BaF}_{2}, \\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}, \\mathrm{ZnS}, \\mathrm{PbI}_{2}$ ? Explain your answer.\n104. What is the effect on the amount of solid $\\mathrm{Mg}(\\mathrm{OH})_{2}$ that dissolves and the concentrations of $\\mathrm{Mg}^{2+}$ and $\\mathrm{OH}^{-}$ when each of the following are added to a mixture of solid $\\mathrm{Mg}(\\mathrm{OH})_{2}$ and water at equilibrium?\n(a) $\\mathrm{MgCl}_{2}$\n(b) KOH\n(c) $\\mathrm{HClO}_{4}$\n(d) $\\mathrm{NaNO}_{3}$\n(e) $\\mathrm{Mg}(\\mathrm{OH})_{2}$\n105. What is the effect on the amount of $\\mathrm{CaHPO}_{4}$ that dissolves and the concentrations of $\\mathrm{Ca}^{2+}$ and $\\mathrm{HPO}_{4}{ }^{2-}$ when each of the following are added to a mixture of solid $\\mathrm{CaHPO}_{4}$ and water at equilibrium?\n(a) $\\mathrm{CaCl}_{2}$\n(b) HCl\n(c) $\\mathrm{KClO}_{4}$\n(d) NaOH\n(e) $\\mathrm{CaHPO}_{4}$"}
{"id": 4376, "contents": "1689. $3 x$ - 1689.2. Coupled Equilibria\n(a) $\\mathrm{CaCl}_{2}$\n(b) HCl\n(c) $\\mathrm{KClO}_{4}$\n(d) NaOH\n(e) $\\mathrm{CaHPO}_{4}$\n106. Identify all chemical species present in an aqueous solution of $\\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$ and list these species in decreasing order of their concentrations. (Hint: Remember that the $\\mathrm{PO}_{4}{ }^{3-}$ ion is a weak base.)"}
{"id": 4377, "contents": "1689. $3 x$ - 1689.2. Coupled Equilibria\n758 15 \u2022 Exercises"}
{"id": 4378, "contents": "1690. CHAPTER 16 <br> Electrochemistry - \nFigure 16.1 Electric vehicles are powered by batteries, devices that harness the energy of spontaneous redox reactions. (credit: modification of work by Robert Couse-Baker)"}
{"id": 4379, "contents": "1691. CHAPTER OUTLINE - 1691.1. Review of Redox Chemistry\n16.2 Galvanic Cells\n16.3 Electrode and Cell Potentials\n16.4 Potential, Free Energy, and Equilibrium\n16.5 Batteries and Fuel Cells\n16.6 Corrosion\n16.7 Electrolysis\n\nINTRODUCTION Another chapter in this text introduced the chemistry of reduction-oxidation (redox) reactions. This important reaction class is defined by changes in oxidation states for one or more reactant elements, and it includes a subset of reactions involving the transfer of electrons between reactant species. Around the turn of the nineteenth century, chemists began exploring ways these electrons could be transferred indirectly via an external circuit rather than directly via intimate contact of redox reactants. In the two centuries since, the field of electrochemistry has evolved to yield significant insights on the fundamental aspects of redox chemistry as well as a wealth of technologies ranging from industrial-scale metallurgical processes to robust, rechargeable batteries for electric vehicles (Figure 16.1). In this chapter, the essential concepts of electrochemistry will be addressed."}
{"id": 4380, "contents": "1692. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe defining traits of redox chemistry\n- Identify the oxidant and reductant of a redox reaction\n- Balance chemical equations for redox reactions using the half-reaction method\n\nSince reactions involving electron transfer are essential to the topic of electrochemistry, a brief review of redox chemistry is provided here that summarizes and extends the content of an earlier text chapter (see chapter on reaction stoichiometry). Readers wishing additional review are referred to the text chapter on reaction stoichiometry."}
{"id": 4381, "contents": "1693. Oxidation Numbers - \nBy definition, a redox reaction is one that entails changes in oxidation number (or oxidation state) for one or more of the elements involved. The oxidation number of an element in a compound is essentially an assessment of how the electronic environment of its atoms is different in comparison to atoms of the pure element. By this description, the oxidation number of an atom in an element is equal to zero. For an atom in a compound, the oxidation number is equal to the charge the atom would have in the compound if the compound were ionic. Consequential to these rules, the sum of oxidation numbers for all atoms in a molecule is equal to the charge on the molecule. To illustrate this formalism, examples from the two compound classes, ionic and covalent, will be considered.\n\nSimple ionic compounds present the simplest examples to illustrate this formalism, since by definition the elements' oxidation numbers are numerically equivalent to ionic charges. Sodium chloride, NaCl , is comprised of $\\mathrm{Na}^{+}$cations and $\\mathrm{Cl}^{-}$anions, and so oxidation numbers for sodium and chlorine are, +1 and -1 , respectively. Calcium fluoride, $\\mathrm{CaF}_{2}$, is comprised of $\\mathrm{Ca}^{2+}$ cations and $\\mathrm{F}^{-}$anions, and so oxidation numbers for calcium and fluorine are, +2 and -1 , respectively.\n\nCovalent compounds require a more challenging use of the formalism. Water is a covalent compound whose molecules consist of two H atoms bonded separately to a central O atom via polar covalent $\\mathrm{O}-\\mathrm{H}$ bonds. The shared electrons comprising an $\\mathrm{O}-\\mathrm{H}$ bond are more strongly attracted to the more electronegative O atom, and so it acquires a partial negative charge in the water molecule (relative to an O atom in elemental oxygen). Consequently, H atoms in a water molecule exhibit partial positive charges compared to H atoms in elemental hydrogen. The sum of the partial negative and partial positive charges for each water molecule is zero, and the water molecule is neutral."}
{"id": 4382, "contents": "1693. Oxidation Numbers - \nImagine that the polarization of shared electrons within the $\\mathrm{O}-\\mathrm{H}$ bonds of water were $100 \\%$ complete-the result would be transfer of electrons from H to O , and water would be an ionic compound comprised of $\\mathrm{O}^{2-}$ anions and $\\mathrm{H}^{+}$cations. And so, the oxidations numbers for oxygen and hydrogen in water are -2 and +1 , respectively. Applying this same logic to carbon tetrachloride, $\\mathrm{CCl}_{4}$, yields oxidation numbers of +4 for carbon and -1 for chlorine. In the nitrate ion, $\\mathrm{NO}_{3}^{-}$, the oxidation number for nitrogen is +5 and that for oxygen is -2 , summing to equal the 1 - charge on the molecule:\n\n$$\n(1 \\mathrm{~N} \\text { atom })\\left(\\frac{+5}{\\mathrm{~N} \\text { atom }}\\right)+(3 \\mathrm{O} \\text { atoms })\\left(\\frac{-2}{\\mathrm{O} \\text { atom }}\\right)=+5+-6=-1\n$$"}
{"id": 4383, "contents": "1694. Balancing Redox Equations - \nThe unbalanced equation below describes the decomposition of molten sodium chloride:\n\n$$\n\\mathrm{NaCl}(l) \\longrightarrow \\mathrm{Na}(l)+\\mathrm{Cl}_{2}(g) \\quad \\text { unbalanced }\n$$\n\nThis reaction satisfies the criterion for redox classification, since the oxidation number for Na is decreased from +1 to 0 (it undergoes reduction) and that for Cl is increased from -1 to 0 (it undergoes oxidation). The equation in this case is easily balanced by inspection, requiring stoichiometric coefficients of 2 for the NaCl and Na :\n\n$$\n2 \\mathrm{NaCl}(l) \\longrightarrow 2 \\mathrm{Na}(l)+\\mathrm{Cl}_{2}(g) \\quad \\text { balanced }\n$$\n\nRedox reactions that take place in aqueous solutions are commonly encountered in electrochemistry, and many involve water or its characteristic ions, $\\mathrm{H}^{+}(a q)$ and $\\mathrm{OH}^{-}(a q)$, as reactants or products. In these cases, equations representing the redox reaction can be very challenging to balance by inspection, and the use of a systematic approach called the half-reaction method is helpful. This approach involves the following steps:\n\n1. Write skeletal equations for the oxidation and reduction half-reactions.\n2. Balance each half-reaction for all elements except H and O .\n3. Balance each half-reaction for O by adding $\\mathrm{H}_{2} \\mathrm{O}$.\n4. Balance each half-reaction for H by adding $\\mathrm{H}^{+}$.\n5. Balance each half-reaction for charge by adding electrons.\n6. If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other.\n7. Add the two half-reactions and simplify.\n8. If the reaction takes place in a basic medium, add $\\mathrm{OH}^{-}$ions the equation obtained in step 7 to neutralize the $\\mathrm{H}^{+}$ions (add in equal numbers to both sides of the equation) and simplify.\n\nThe examples below demonstrate the application of this method to balancing equations for aqueous redox reactions."}
{"id": 4384, "contents": "1696. Balancing Equations for Redox Reactions in Acidic Solutions - \nWrite the balanced equation representing reaction between solid copper and nitric acid to yield aqueous copper(II) ions and nitrogen monoxide gas."}
{"id": 4385, "contents": "1697. Solution - \nFollowing the steps of the half-reaction method:\n\n1. Write skeletal equations for the oxidation and reduction half-reactions.\n\n$$\n\\begin{array}{ll}\n\\text { oxidation: } & \\mathrm{Cu}(s) \\longrightarrow \\mathrm{Cu}^{2+}(a q) \\\\\n\\text { reduction: } & \\mathrm{HNO}_{3}(a q) \\longrightarrow \\mathrm{NO}(g)\n\\end{array}\n$$\n\n2. Balance each half-reaction for all elements except $H$ and $O$.\noxidation: $\\quad \\mathrm{Cu}(s) \\longrightarrow \\mathrm{Cu}^{2+}(a q)$\nreduction: $\\quad \\mathrm{HNO}_{3}(a q) \\longrightarrow \\mathrm{NO}(g)$\n3. Balance each half-reaction for O by adding $\\mathrm{H}_{2} \\mathrm{O}$.\noxidation: $\\quad \\mathrm{Cu}(s) \\longrightarrow \\mathrm{Cu}^{2+}(a q)$\nreduction: $\\quad \\mathrm{HNO}_{3}(a q) \\longrightarrow \\mathrm{NO}(g)+\\mathbf{2 H}_{\\mathbf{2}} \\mathbf{O}(l)$\n4. Balance each half-reaction for $H$ by adding $H^{+}$.\n\n$$\n\\begin{array}{ll}\n\\text { oxidation: } & \\mathrm{Cu}(s) \\longrightarrow \\mathrm{Cu}^{2+}(a q) \\\\\n\\text { reduction: } & \\mathbf{3 H}^{+}(a q)+\\mathrm{HNO}_{3}(a q) \\longrightarrow \\mathrm{NO}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)\n\\end{array}\n$$"}
{"id": 4386, "contents": "1697. Solution - \n5. Balance each half-reaction for charge by adding electrons.\noxidation: $\\quad \\mathrm{Cu}(s) \\longrightarrow \\mathrm{Cu}^{2+}(a q)+2 \\mathbf{e}^{-}$\nreduction: $\\quad 3 \\mathrm{e}^{-}+3 \\mathrm{H}^{+}(a q)+\\mathrm{HNO}_{3}(a q) \\longrightarrow \\mathrm{NO}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n6. If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other.\noxidation $(\\times 3): \\quad 3 \\mathrm{Cu}(s) \\longrightarrow 3 \\mathrm{Cu}^{2+}(a q)+62 \\mathrm{e}^{-}$\nreduction $(\\times 2): \\quad \\quad 63-\\mathrm{e}^{-}+63-\\mathrm{H}^{+}(a q)+2 \\mathrm{HNO}_{3}(a q) \\longrightarrow 2 \\mathrm{NO}(g)+42 \\mathrm{H}_{2} \\mathrm{O}(l)$\n7. Add the two half-reactions and simplify.\n$3 \\mathrm{Cu}(s)+\\mathbf{6} \\mathrm{e}^{-}+6 \\mathrm{H}^{+}(a q)+2 \\mathrm{HNO}_{3}(a q) \\longrightarrow 3 \\mathrm{Cu}^{2+}(a q)+\\mathbf{6} \\mathrm{e}^{-}+2 \\mathrm{NO}(g)+4 \\mathrm{H}_{2} \\mathrm{O}(l)$\n$3 \\mathrm{Cu}(s)+6 \\mathrm{H}^{+}(a q)+2 \\mathrm{HNO}_{3}(a q) \\longrightarrow 3 \\mathrm{Cu}^{2+}(a q)+2 \\mathrm{NO}(g)+4 \\mathrm{H}_{2} \\mathrm{O}(l)$"}
{"id": 4387, "contents": "1697. Solution - \n8. If the reaction takes place in a basic medium, add $\\mathrm{OH}^{-}$ions the equation obtained in step 7 to neutralize the $\\mathrm{H}^{+}$ions (add in equal numbers to both sides of the equation) and simplify."}
{"id": 4388, "contents": "1697. Solution - \nThis step not necessary since the solution is stipulated to be acidic.\nThe balanced equation for the reaction in an acidic solution is then\n\n$$\n3 \\mathrm{Cu}(s)+6 \\mathrm{H}^{+}(a q)+2 \\mathrm{HNO}_{3}(a q) \\longrightarrow 3 \\mathrm{Cu}^{2+}(a q)+2 \\mathrm{NO}(g)+4 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$"}
{"id": 4389, "contents": "1698. Check Your Learning - \nThe reaction above results when using relatively diluted nitric acid. If concentrated nitric acid is used, nitrogen dioxide is produced instead of nitrogen monoxide. Write a balanced equation for this reaction.\n\nAnswer:\n$\\mathrm{Cu}(s)+2 \\mathrm{H}^{+}(a q)+2 \\mathrm{HNO}_{3}(a q) \\longrightarrow \\mathrm{Cu}^{2+}(a q)+2 \\mathrm{NO}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$"}
{"id": 4390, "contents": "1700. Balancing Equations for Redox Reactions in Basic Solutions - \nWrite the balanced equation representing reaction between aqueous permanganate ion, $\\mathrm{MnO}_{4}{ }^{-}$, and solid chromium(III) hydroxide, $\\mathrm{Cr}(\\mathrm{OH})_{3}$, to yield solid manganese(IV) oxide, $\\mathrm{MnO}_{2}$, and aqueous chromate ion, $\\mathrm{CrO}_{4}{ }^{2-}$ The reaction takes place in a basic solution."}
{"id": 4391, "contents": "1701. Solution - \nFollowing the steps of the half-reaction method:"}
{"id": 4392, "contents": "1701. Solution - \n1. Write skeletal equations for the oxidation and reduction half-reactions.\noxidation: $\\quad \\mathrm{Cr}(\\mathrm{OH})_{3}(s) \\longrightarrow \\mathrm{CrO}_{4}{ }^{2-}(a q)$\nreduction: $\\quad \\mathrm{MnO}_{4}{ }^{-}(a q) \\longrightarrow \\mathrm{MnO}_{2}(s)$\n2. Balance each half-reaction for all elements except $H$ and $O$.\noxidation: $\\quad \\mathrm{Cr}(\\mathrm{OH})_{3}(s) \\longrightarrow \\mathrm{CrO}_{4}{ }^{2-}(a q)$\nreduction: $\\quad \\mathrm{MnO}_{4}{ }^{-}(a q) \\longrightarrow \\mathrm{MnO}_{2}(s)$\n3. Balance each half-reaction for O by adding $\\mathrm{H}_{2} \\mathrm{O}$.\noxidation: $\\quad \\mathbf{H}_{2} \\mathbf{O}(l)+\\mathrm{Cr}(\\mathrm{OH})_{3}(s) \\longrightarrow \\mathrm{CrO}_{4}{ }^{2-}(a q)$\nreduction: $\\quad \\mathrm{MnO}_{4}{ }^{-}(a q) \\longrightarrow \\mathrm{MnO}_{2}(s)+2 \\mathbf{H}_{\\mathbf{2}} \\mathbf{O}(l)$\n4. Balance each half-reaction for $H$ by adding $H^{+}$.\noxidation: $\\quad \\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{Cr}(\\mathrm{OH})_{3}(s) \\longrightarrow \\mathrm{CrO}_{4}{ }^{2-}(a q)+\\mathbf{5 H}^{+}(a q)$\nreduction: $\\quad 4 \\mathbf{H}^{+}(a q)+\\mathrm{MnO}_{4}^{-}(a q) \\longrightarrow \\mathrm{MnO}_{2}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n5. Balance each half-reaction for charge by adding electrons."}
{"id": 4393, "contents": "1701. Solution - \n5. Balance each half-reaction for charge by adding electrons.\noxidation: $\\quad \\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{Cr}(\\mathrm{OH})_{3}(s) \\longrightarrow \\mathrm{CrO}_{4}{ }^{2-}(a q)+5 \\mathrm{H}^{+}(a q)+\\mathbf{3 e}^{-}$\nreduction: $\\quad 3 \\mathbf{e}^{-}+4 \\mathrm{H}^{+}(a q)+\\mathrm{MnO}_{4}{ }^{-}(a q) \\longrightarrow \\mathrm{MnO}_{2}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n6. If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other.\nThis step is not necessary since the number of electrons is already in balance.\n7. Add the two half-reactions and simplify.\n$\\mathbf{H}_{\\mathbf{2}} \\mathbf{O}(\\boldsymbol{(})+\\mathrm{Cr}(\\mathrm{OH})_{3}(s)+\\mathbf{3} \\mathbf{e}^{-}+\\mathbf{- 4 \\mathbf { H } ^ { + }}(\\mathrm{aq})+\\mathrm{MnO}_{4}{ }^{-}(a q) \\longrightarrow \\mathrm{CrO}_{4}{ }^{2-}(a q)+\\mathbf{5}^{-} \\mathrm{H}^{+}(a q)$ $+3 \\mathrm{e}^{-}+\\mathrm{MnO}_{2}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n$\\mathrm{Cr}(\\mathrm{OH})_{3}(s)+\\mathrm{MnO}_{4}{ }^{-}(a q) \\longrightarrow \\mathrm{CrO}_{4}{ }^{2-}(a q)+\\mathrm{H}^{+}(a q)+\\mathrm{MnO}_{2}(s)+\\mathrm{H}_{2} \\mathrm{O}(l)$"}
{"id": 4394, "contents": "1701. Solution - \n8. If the reaction takes place in a basic medium, add $\\mathrm{OH}^{-}$ions the equation obtained in step 7 to neutralize the $\\mathrm{H}^{+}$ions (add in equal numbers to both sides of the equation) and simplify.\n$\\mathbf{O H}^{-}(a q)+\\mathrm{Cr}(\\mathrm{OH})_{3}(s)+\\mathrm{MnO}_{4}{ }^{-}(a q) \\longrightarrow \\mathrm{CrO}_{4}{ }^{2-}(a q)+\\mathrm{H}^{+}(a q)+\\mathbf{O H}^{-}(a q)+\\mathrm{MnO}_{2}(s)+\\mathrm{H}_{2} \\mathrm{O}(l)$\n$\\mathrm{OH}^{-}(a q)+\\mathrm{Cr}(\\mathrm{OH})_{3}(s)+\\mathrm{MnO}_{4}{ }^{-}(a q) \\longrightarrow \\mathrm{CrO}_{4}{ }^{2-}(a q)+\\mathrm{MnO}_{2}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$"}
{"id": 4395, "contents": "1702. Check Your Learning - \nAqueous permanganate ion may also be reduced using aqueous bromide ion, $\\mathrm{Br}^{-}$, the products of this reaction being solid manganese(IV) oxide and aqueous bromate ion, $\\mathrm{BrO}_{3}{ }^{-}$. Write the balanced equation for this\nreaction occurring in a basic medium.\nAnswer:\n$\\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{MnO}_{4}^{-}(a q)+\\mathrm{Br}^{-}(a q) \\longrightarrow 2 \\mathrm{MnO}_{2}(s)+\\mathrm{BrO}_{3}^{-}(a q)+2 \\mathrm{OH}^{-}(a q)$"}
{"id": 4396, "contents": "1703. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the function of a galvanic cell and its components\n- Use cell notation to symbolize the composition and construction of galvanic cells\n\nAs demonstration of spontaneous chemical change, Figure 16.2 shows the result of immersing a coiled wire of copper into an aqueous solution of silver nitrate. A gradual but visually impressive change spontaneously occurs as the initially colorless solution becomes increasingly blue, and the initially smooth copper wire becomes covered with a porous gray solid.\n\n\nFIGURE 16.2 A copper wire and an aqueous solution of silver nitrate (left) are brought into contact (center) and a spontaneous transfer of electrons occurs, creating blue $\\mathrm{Cu}^{2+}(a q)$ and gray $\\operatorname{Ag}(s)$ (right).\n\nThese observations are consistent with (i) the oxidation of elemental copper to yield copper(II) ions, $\\mathrm{Cu}^{2+}(a q)$, which impart a blue color to the solution, and (ii) the reduction of silver(I) ions to yield elemental silver, which deposits as a fluffy solid on the copper wire surface. And so, the direct transfer of electrons from the copper wire to the aqueous silver ions is spontaneous under the employed conditions. A summary of this redox system is provided by these equations:\n\n$$\n\\begin{array}{ll}\n\\text { overall reaction: } & \\mathrm{Cu}(s)+2 \\mathrm{Ag}^{+}(a q) \\longrightarrow \\mathrm{Cu}^{2+}(a q)+2 \\mathrm{Ag}(s) \\\\\n\\text { oxidation half-reaction: } & \\mathrm{Cu}(s) \\longrightarrow \\mathrm{Cu}^{2+}(a q)+2 \\mathrm{e}^{-} \\\\\n\\text {reduction half-reaction: } & 2 \\mathrm{Ag}^{+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Ag}(s)\n\\end{array}\n$$"}
{"id": 4397, "contents": "1703. LEARNING OBJECTIVES - \nConsider the construction of a device that contains all the reactants and products of a redox system like the one here, but prevents physical contact between the reactants. Direct transfer of electrons is, therefore, prevented; transfer, instead, takes place indirectly through an external circuit that contacts the separated reactants. Devices of this sort are generally referred to as electrochemical cells, and those in which a spontaneous redox reaction takes place are called galvanic cells (or voltaic cells).\n\nA galvanic cell based on the spontaneous reaction between copper and silver(I) is depicted in Figure 16.3. The cell is comprised of two half-cells, each containing the redox conjugate pair (\"couple\") of a single reactant. The half-cell shown at the left contains the $\\mathrm{Cu}(0) / \\mathrm{Cu}(\\mathrm{II})$ couple in the form of a solid copper foil and an aqueous solution of copper nitrate. The right half-cell contains the $\\mathrm{Ag}(\\mathrm{I}) / \\mathrm{Ag}(0)$ couple as solid silver foil and an aqueous silver nitrate solution. An external circuit is connected to each half-cell at its solid foil, meaning the Cu and Ag foil each function as an electrode. By definition, the anode of an electrochemical cell is the electrode at which oxidation occurs (in this case, the Cu foil) and the cathode is the electrode where reduction occurs (the Ag foil). The redox reactions in a galvanic cell occur only at the interface between each half-cell's reaction mixture and its electrode. To keep the reactants separate while maintaining charge-balance, the two half-cell solutions are connected by a tube filled with inert electrolyte solution called a salt bridge. The spontaneous reaction in this cell produces $\\mathrm{Cu}^{2+}$ cations in the anode half-cell and consumes $\\mathrm{Ag}^{+}$ions in the cathode half-cell, resulting in a compensatory flow of inert ions from the salt bridge that maintains charge balance. Increasing concentrations of $\\mathrm{Cu}^{2+}$ in the anode half-cell are balanced by an influx of $\\mathrm{NO}_{3}{ }^{-}$from the salt bridge, while a flow of $\\mathrm{Na}^{+}$into the cathode half-cell compensates for the decreasing $\\mathrm{Ag}^{+}$concentration."}
{"id": 4398, "contents": "1703. LEARNING OBJECTIVES - \nFIGURE 16.3 A galvanic cell based on the spontaneous reaction between copper and silver(I) ions."}
{"id": 4399, "contents": "1704. Cell Notation - \nAbbreviated symbolism is commonly used to represent a galvanic cell by providing essential information on its composition and structure. These symbolic representations are called cell notations or cell schematics, and they are written following a few guidelines:\n\n- The relevant components of each half-cell are represented by their chemical formulas or element symbols\n- All interfaces between component phases are represented by vertical parallel lines; if two or more components are present in the same phase, their formulas are separated by commas\n- By convention, the schematic begins with the anode and proceeds left-to-right identifying phases and interfaces encountered within the cell, ending with the cathode\n\nA verbal description of the cell as viewed from anode-to-cathode is often a useful first-step in writing its schematic. For example, the galvanic cell shown in Figure 16.3 consists of a solid copper anode immersed in an aqueous solution of copper(II) nitrate that is connected via a salt bridge to an aqueous silver(I) nitrate solution, immersed in which is a solid silver cathode. Converting this statement to symbolism following the above guidelines results in the cell schematic:\n\n$$\n\\mathrm{Cu}(s)\\left|1 M \\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q) \\| 1 M \\mathrm{AgNO}_{3}(a q)\\right| \\mathrm{Ag}(s)\n$$\n\nConsider a different galvanic cell (see Figure 16.4) based on the spontaneous reaction between solid magnesium and aqueous iron(III) ions:\nnet cell reaction:\noxidation half-reaction:\nreduction half-reaction:\n\n$$\n\\begin{aligned}\n& \\mathrm{Mg}(s)+2 \\mathrm{Fe}^{3+}(a q) \\longrightarrow \\mathrm{Mg}^{2+}(a q)+2 \\mathrm{Fe}^{2+}(a q) \\\\\n& \\operatorname{Mg}(s) \\longrightarrow \\mathrm{Mg}^{2+}(a q)+2 \\mathrm{e}^{-} \\\\\n& 2 \\mathrm{Fe}^{3+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Fe}^{2+}(a q)\n\\end{aligned}\n$$"}
{"id": 4400, "contents": "1704. Cell Notation - \nIn this cell, a solid magnesium anode is immersed in an aqueous solution of magnesium chloride that is connected via a salt bridge to an aqueous solution containing a mixture of iron(III) chloride and iron(II)\nchloride, immersed in which is a platinum cathode. The cell schematic is then written as\n\n$$\n\\operatorname{Mg}(s)\\left|0.1 M \\mathrm{MgCl}_{2}(a q) \\| 0.2 M \\mathrm{FeCl}_{3}(a q), 0.3 M \\mathrm{FeCl}_{2}(a q)\\right| \\operatorname{Pt}(s)\n$$\n\nNotice the cathode half-cell is different from the others considered thus far in that its electrode is comprised of a substance ( Pt ) that is neither a reactant nor a product of the cell reaction. This is required when neither member of the half-cell's redox couple can reasonably function as an electrode, which must be electrically conductive and in a phase separate from the half-cell solution. In this case, both members of the redox couple are solute species, and so Pt is used as an inert electrode that can simply provide or accept electrons to redox species in solution. Electrodes constructed from a member of the redox couple, such as the Mg anode in this cell, are called active electrodes.\n\n\nFIGURE 16.4 A galvanic cell based on the spontaneous reaction between magnesium and iron(III) ions."}
{"id": 4401, "contents": "1706. Writing Galvanic Cell Schematics - \nA galvanic cell is fabricated by connecting two half-cells with a salt bridge, one in which a chromium wire is immersed in a $1 \\mathrm{M} \\mathrm{CrCl}_{3}$ solution and another in which a copper wire is immersed in $1 \\mathrm{M} \\mathrm{CuCl}_{2}$. Assuming the chromium wire functions as an anode, write the schematic for this cell along with equations for the anode halfreaction, the cathode half-reaction, and the overall cell reaction."}
{"id": 4402, "contents": "1707. Solution - \nSince the chromium wire is stipulated to be the anode, the schematic begins with it and proceeds left-to-right, symbolizing the other cell components until ending with the copper wire cathode:\n\n$$\n\\mathrm{Cr}(s)\\left|1 M \\mathrm{CrCl}_{3}(a q) \\| 1 M \\mathrm{CuCl}_{2}(a q)\\right| \\mathrm{Cu}(s)\n$$\n\nThe half-reactions for this cell are\n\n$$\n\\begin{array}{ll}\n\\text { anode (oxidation): } & \\mathrm{Cr}(s) \\longrightarrow \\mathrm{Cr}^{3+}(a q)+3 \\mathrm{e}^{-} \\\\\n\\text {cathode (reduction): } & \\mathrm{Cu}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Cu}(s)\n\\end{array}\n$$\n\nMultiplying to make the number of electrons lost by Cr and gained by $\\mathrm{Cu}^{2+}$ equal yields\n\n$$\n\\begin{array}{ll}\n\\text { anode (oxidation): } & 2 \\mathrm{Cr}(s) \\longrightarrow 2 \\mathrm{Cr}^{3+}(a q)+6 \\mathrm{e}^{-} \\\\\n\\text {cathode (reduction): } & 3 \\mathrm{Cu}^{2+}(a q)+6 \\mathrm{e}^{-} \\longrightarrow 3 \\mathrm{Cu}(s)\n\\end{array}\n$$\n\nAdding the half-reaction equations and simplifying yields an equation for the cell reaction:\n\n$$\n2 \\mathrm{Cr}(s)+3 \\mathrm{Cu}^{2+}(a q) \\longrightarrow 2 \\mathrm{Cr}^{3+}(a q)+3 \\mathrm{Cu}(s)\n$$"}
{"id": 4403, "contents": "1708. Check Your Learning - \nOmitting solute concentrations and spectator ion identities, write the schematic for a galvanic cell whose net cell reaction is shown below.\n\n$$\n\\mathrm{Sn}^{4+}(a q)+\\mathrm{Zn}(s) \\longrightarrow \\mathrm{Sn}^{2+}(a q)+\\mathrm{Zn}^{2+}(a q)\n$$"}
{"id": 4404, "contents": "1709. Answer: - \n$\\mathrm{Zn}(s)\\left|\\mathrm{Zn}^{2+}(a q) \\| \\mathrm{Sn}^{4+}(a q), \\mathrm{Sn}^{2+}(a q)\\right| \\operatorname{Pt}(s)$"}
{"id": 4405, "contents": "1710. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe and relate the definitions of electrode and cell potentials\n- Interpret electrode potentials in terms of relative oxidant and reductant strengths\n- Calculate cell potentials and predict redox spontaneity using standard electrode potentials\n\nUnlike the spontaneous oxidation of copper by aqueous silver(I) ions described in section 17.2, immersing a copper wire in an aqueous solution of lead(II) ions yields no reaction. The two species, $\\mathrm{Ag}^{+}(\\mathrm{aq})$ and $\\mathrm{Pb}^{2+}(\\mathrm{aq})$, thus show a distinct difference in their redox activity towards copper: the silver ion spontaneously oxidized copper, but the lead ion did not. Electrochemical cells permit this relative redox activity to be quantified by an easily measured property, potential. This property is more commonly called voltage when referenced in regard to electrical applications, and it is a measure of energy accompanying the transfer of charge. Potentials are measured in the volt unit, defined as one joule of energy per one coulomb of charge, $\\mathrm{V}=\\mathrm{J} / \\mathrm{C}$.\n\nWhen measured for purposes of electrochemistry, a potential reflects the driving force for a specific type of charge transfer process, namely, the transfer of electrons between redox reactants. Considering the nature of potential in this context, it is clear that the potential of a single half-cell or a single electrode can't be measured; \"transfer\" of electrons requires both a donor and recipient, in this case a reductant and an oxidant, respectively. Instead, a half-cell potential may only be assessed relative to that of another half-cell. It is only the difference in potential between two half-cells that may be measured, and these measured potentials are called\ncell potentials, E $_{\\text {cell }}$, defined as\n\n$$\n\\mathrm{E}_{\\text {cell }}=\\mathrm{E}_{\\text {cathode }}-\\mathrm{E}_{\\text {anode }}\n$$"}
{"id": 4406, "contents": "1710. LEARNING OBJECTIVES - \n$$\n\\mathrm{E}_{\\text {cell }}=\\mathrm{E}_{\\text {cathode }}-\\mathrm{E}_{\\text {anode }}\n$$\n\nwhere $\\mathrm{E}_{\\text {cathode }}$ and $\\mathrm{E}_{\\text {anode }}$ are the potentials of two different half-cells functioning as specified in the subscripts. As for other thermodynamic quantities, the standard cell potential, $\\mathbf{E}^{\\circ}{ }_{\\text {cell }}$, is a cell potential measured when both half-cells are under standard-state conditions (1 M concentrations, 1 bar pressures, 298 K ):\n\n$$\n\\mathrm{E}_{\\text {cell }}^{\\circ}=\\mathrm{E}_{\\text {cathode }}^{\\circ}-\\mathrm{E}_{\\text {anode }}^{\\circ}\n$$\n\nTo simplify the collection and sharing of potential data for half-reactions, the scientific community has designated one particular half-cell to serve as a universal reference for cell potential measurements, assigning it a potential of exactly 0 V . This half-cell is the standard hydrogen electrode (SHE) and it is based on halfreaction below:\n\n$$\n2 \\mathrm{H}^{+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{H}_{2}(g)\n$$\n\nA typical SHE contains an inert platinum electrode immersed in precisely $1 M$ aqueous $\\mathrm{H}^{+}$and a stream of bubbling $\\mathrm{H}_{2}$ gas at 1 bar pressure, all maintained at a temperature of 298 K (see Figure 16.5).\n\n\nFIGURE 16.5 A standard hydrogen electrode (SHE).\nThe assigned potential of the SHE permits the definition of a conveniently measured potential for a single halfcell. The electrode potential ( $\\mathbf{E}_{\\mathbf{X}}$ ) for a half-cell $X$ is defined as the potential measured for a cell comprised of $X$ acting as cathode and the SHE acting as anode:"}
{"id": 4407, "contents": "1710. LEARNING OBJECTIVES - \n$$\n\\begin{aligned}\n& E_{\\text {cell }}=E_{X}-E_{\\text {SHE }} \\\\\n& E_{S H E}=0 V(\\text { defined }) \\\\\n& E_{\\text {cell }}=E_{X}\n\\end{aligned}\n$$\n\nWhen the half-cell $X$ is under standard-state conditions, its potential is the standard electrode potential, $\\mathbf{E}^{\\circ} \\mathbf{x}$. Since the definition of cell potential requires the half-cells function as cathodes, these potentials are sometimes called standard reduction potentials.\n\nThis approach to measuring electrode potentials is illustrated in Figure 16.6, which depicts a cell comprised of an SHE connected to a copper(II)/copper(0) half-cell under standard-state conditions. A voltmeter in the external circuit allows measurement of the potential difference between the two half-cells. Since the Cu halfcell is designated as the cathode in the definition of cell potential, it is connected to the red (positive) input of the voltmeter, while the designated SHE anode is connected to the black (negative) input. These connections insure that the sign of the measured potential will be consistent with the sign conventions of electrochemistry per the various definitions discussed above. A cell potential of +0.337 V is measured, and so\n\n$$\n\\mathrm{E}_{\\text {cell }}^{\\circ}=\\mathrm{E}_{\\mathrm{Cu}}^{\\circ}=+0.337 \\mathrm{~V}\n$$\n\nTabulations of $\\mathrm{E}^{\\circ}$ values for other half-cells measured in a similar fashion are available as reference literature to permit calculations of cell potentials and the prediction of the spontaneity of redox processes.\n\n\nFIGURE 16.6 A cell permitting experimental measurement of the standard electrode potential for the half-reaction $\\mathrm{Cu}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Cu}(s)$\n\nTable 16.1 provides a listing of standard electrode potentials for a selection of half-reactions in numerical order, and a more extensive alphabetical listing is given in Appendix L.\n\nSelected Standard Reduction Potentials at $25^{\\circ} \\mathrm{C}$"}
{"id": 4408, "contents": "1710. LEARNING OBJECTIVES - \nSelected Standard Reduction Potentials at $25^{\\circ} \\mathrm{C}$\n\n| Half-Reaction | $E^{\\circ}(\\mathrm{V})$ |\n| :--- | :--- |\n| $\\mathrm{F}_{2}(g)+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{~F}^{-}(a q)$ | +2.866 |\n| $\\mathrm{PbO}_{2}(s)+\\mathrm{SO}_{4}{ }^{2-}(a q)+4 \\mathrm{H}^{+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{PbSO}_{4}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$ | +1.69 |\n| $\\mathrm{MnO}_{4}^{-}(a q)+8 \\mathrm{H}^{+}(a q)+5 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Mn}^{2+}(a q)+4 \\mathrm{H}_{2} \\mathrm{O}(l)$ | +1.507 |\n| $\\mathrm{Au}^{3+}(a q)+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Au}(s)$ | +1.498 |\n| $\\mathrm{Cl}_{2}(g)+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Cl}^{-}(a q)$ | +1.35827 |\n| $\\mathrm{O}_{2}(g)+4 \\mathrm{H}^{+}(a q)+4 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(l)$ | +1.229 |\n| $\\mathrm{Pt}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Pt}(s)$ | +1.20 |\n| $\\mathrm{Br}_{2}(a q)+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Br}(a q)$ | +1.0873 |\n| $\\mathrm{Ag}^{+}(a q)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Ag}(s)$ | +0.7996 |\n\nTABLE 16.1"}
{"id": 4409, "contents": "1710. LEARNING OBJECTIVES - \n| Half-Reaction | $E^{\\circ}(\\mathrm{V})$ |\n| :---: | :---: |\n| $\\mathrm{Hg}_{2}{ }^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Hg}(l)$ | +0.7973 |\n| $\\mathrm{Fe}^{3+}(a q)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Fe}^{2+}(a q)$ | +0.771 |\n| $\\mathrm{MnO}_{4}{ }^{-}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l)+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{MnO}_{2}(s)+4 \\mathrm{OH}^{-}(a q)$ | +0.558 |\n| $\\mathrm{I}_{2}(s)+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{I}^{-}(a q)$ | +0.5355 |\n| $\\mathrm{NiO}_{2}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Ni}(\\mathrm{OH})_{2}(s)+2 \\mathrm{OH}^{-}(a q)$ | +0.49 |\n| $\\mathrm{Cu}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Cu}(s)$ | +0.34 |\n| $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}(s)+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Hg}(l)+2 \\mathrm{Cl}^{-}(a q)$ | +0.26808 |\n| $\\mathrm{AgCl}(s)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Ag}(s)+\\mathrm{Cl}^{-}(a q)$ | +0.22233 |\n| $\\mathrm{Sn}^{4+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Sn}^{2+}(a q)$ | +0.151 |"}
{"id": 4410, "contents": "1710. LEARNING OBJECTIVES - \n| $\\mathrm{Sn}^{4+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Sn}^{2+}(a q)$ | +0.151 |\n| $2 \\mathrm{H}^{+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{H}_{2}(g)$ | 0.00 |\n| $\\mathrm{Pb}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Pb}(s)$ | -0.1262 |\n| $\\mathrm{Sn}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Sn}(s)$ | -0.1375 |\n| $\\mathrm{Ni}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Ni}(s)$ | -0.257 |\n| $\\mathrm{Co}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Co}(s)$ | -0.28 |\n| $\\mathrm{PbSO}_{4}(s)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Pb}(s)+\\mathrm{SO}_{4}{ }^{2-}(a q)$ | -0.3505 |\n| $\\mathrm{Cd}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Cd}(s)$ | -0.4030 |\n| $\\mathrm{Fe}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Fe}(s)$ | -0.447 |\n| $\\mathrm{Cr}^{3+}(a q)+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Cr}(s)$ | -0.744 |\n| $\\mathrm{Mn}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Mn}(s)$ | -1.185 |\n| $\\mathrm{Zn}(\\mathrm{OH})_{2}(s)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Zn}(s)+2 \\mathrm{OH}^{-}(a q)$ | -1.245 |"}
{"id": 4411, "contents": "1710. LEARNING OBJECTIVES - \n| $\\mathrm{Zn}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Zn}(s)$ | -0.7618 |\n| $\\mathrm{Al}^{3+}(a q)+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Al}(s)$ | -1.662 |\n| $\\mathrm{Mg}^{2}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Mg}(s)$ | -2.372 |\n| $\\mathrm{Na}^{+}(a q)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Na}(s)$ | -2.71 |"}
{"id": 4412, "contents": "1710. LEARNING OBJECTIVES - \nTABLE 16.1\n\n| Half-Reaction | $E^{0}(\\mathrm{~V})$ |\n| :--- | :--- |\n| $\\mathrm{Ca}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Ca}(s)$ | -2.868 |\n| $\\mathrm{Ba}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Ba}(s)$ | -2.912 |\n| $\\mathrm{~K}^{+}(a q)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{K}(s)$ | -2.931 |\n| $\\mathrm{Li}^{+}(a q)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Li}(s)$ | -3.04 |\n\nTABLE 16.1"}
{"id": 4413, "contents": "1712. Calculating Standard Cell Potentials - \nWhat is the standard potential of the galvanic cell shown in Figure 16.3?"}
{"id": 4414, "contents": "1713. Solution - \nThe cell in Figure 16.3 is galvanic, the spontaneous cell reaction involving oxidation of its copper anode and reduction of silver(I) ions at its silver cathode:\ncell reaction:\nanode half-reaction:\ncathode half-reaction:\n\n$$\n\\begin{aligned}\n& \\mathrm{Cu}(s)+2 \\mathrm{Ag}^{+}(a q) \\longrightarrow \\mathrm{Cu}^{2+}(a q)+2 \\mathrm{Ag}(s) \\\\\n& \\mathrm{Cu}(s) \\longrightarrow \\mathrm{Cu}^{2+}(a q)+2 \\mathrm{e}^{-} \\\\\n& 2 \\mathrm{Ag}^{+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Ag}(s)\n\\end{aligned}\n$$\n\nThe standard cell potential computed as\n\n$$\n\\begin{aligned}\n\\mathrm{E}_{\\text {cell }}^{\\circ} & =\\mathrm{E}_{\\text {cathode }}^{\\circ}-\\mathrm{E}_{\\text {anode }}^{\\circ} \\\\\n& =\\mathrm{E}_{\\mathrm{Ag}}^{\\circ}-\\mathrm{E}_{\\mathrm{Cu}}^{\\circ} \\\\\n& =0.7996 \\mathrm{~V}-0.34 \\mathrm{~V} \\\\\n& =+0.46 \\mathrm{~V}\n\\end{aligned}\n$$"}
{"id": 4415, "contents": "1714. Check Your Learning - \nWhat is the standard cell potential expected if the silver cathode half-cell in Figure 16.3 is replaced with a lead half-cell: $\\mathrm{Pb}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Pb}(s)$ ?"}
{"id": 4416, "contents": "1715. Answer: - \n-0. 47 V"}
{"id": 4417, "contents": "1716. Intrepreting Electrode and Cell Potentials - \nThinking carefully about the definitions of cell and electrode potentials and the observations of spontaneous redox change presented thus far, a significant relation is noted. The previous section described the spontaneous oxidation of copper by aqueous silver(I) ions, but no observed reaction with aqueous lead(II) ions. Results of the calculations in Example 16.4 have just shown the spontaneous process is described by a positive cell potential while the nonspontaneous process exhibits a negative cell potential. And so, with regard to the relative effectiveness (\"strength\") with which aqueous $\\mathrm{Ag}^{+}$and $\\mathrm{Pb}^{2+}$ ions oxidize Cu under standard conditions, the stronger oxidant is the one exhibiting the greater standard electrode potential, $E^{\\circ}$. Since by convention electrode potentials are for reduction processes, an increased value of $E^{\\circ}$ corresponds to an increased driving force behind the reduction of the species (hence increased effectiveness of its action as an oxidizing agent on some other species). Negative values for electrode potentials are simply a consequence of assigning a value of 0 V to the SHE, indicating the reactant of the half-reaction is a weaker oxidant than aqueous hydrogen ions.\n\nApplying this logic to the numerically ordered listing of standard electrode potentials in Table 16.1 shows this listing to be likewise in order of the oxidizing strength of the half-reaction's reactant species, decreasing from strongest oxidant (most positive $E^{\\circ}$ ) to weakest oxidant (most negative $E^{\\circ}$ ). Predictions regarding the spontaneity of redox reactions under standard state conditions can then be easily made by simply comparing the relative positions of their table entries. By definition, $E^{\\circ}{ }_{\\text {cell }}$ is positive when $E^{\\circ}{ }_{\\text {cathode }}>E^{\\circ}{ }_{\\text {anode }}$, and so any redox reaction in which the oxidant's entry is above the reductant's entry is predicted to be spontaneous."}
{"id": 4418, "contents": "1716. Intrepreting Electrode and Cell Potentials - \nReconsideration of the two redox reactions in Example 16.4 provides support for this fact. The entry for the silver(I)/silver(0) half-reaction is above that for the copper(II)/copper(0) half-reaction, and so the oxidation of Cu by $\\mathrm{Ag}^{+}$is predicted to be spontaneous ( $E^{\\circ}{ }_{\\text {cathode }}>E^{\\circ}$ anode and so $E^{\\circ}{ }_{\\text {cell }}>0$ ). Conversely, the entry for the lead(II)/lead(0) half-cell is beneath that for copper(II)/copper(0), and the oxidation of Cu by $\\mathrm{Pb}^{2+}$ is nonspontaneous ( $E^{\\circ}{ }_{\\text {cathode }}<E^{\\circ}$ anode and so $E^{\\circ}{ }_{\\text {cell }}<0$ ).\n\nRecalling the chapter on thermodynamics, the spontaneities of the forward and reverse reactions of a reversible process show a reciprocal relationship: if a process is spontaneous in one direction, it is nonspontaneous in the opposite direction. As an indicator of spontaneity for redox reactions, the potential of a cell reaction shows a consequential relationship in its arithmetic sign. The spontaneous oxidation of copper by lead(II) ions is not observed,\n\n$$\n\\mathrm{Cu}(s)+\\mathrm{Pb}^{2+}(a q) \\longrightarrow \\mathrm{Cu}^{2+}(a q)+\\mathrm{Pb}(s) E_{\\text {forward }}^{\\circ}=-0.47 \\mathrm{~V} \\text { (negative, non-spontaneous) }\n$$\n\nand so the reverse reaction, the oxidation of lead by copper(II) ions, is predicted to occur spontaneously:\n\n$$\n\\mathrm{Pb}(s)+\\mathrm{Cu}^{2+}(a q) \\longrightarrow \\mathrm{Pb}^{2+}(a q)+\\mathrm{Cu}(s) E_{\\text {forward }}^{\\circ}=+0.47 \\mathrm{~V} \\text { (positive, spontaneous) }\n$$"}
{"id": 4419, "contents": "1716. Intrepreting Electrode and Cell Potentials - \nNote that reversing the direction of a redox reaction effectively interchanges the identities of the cathode and anode half-reactions, and so the cell potential is calculated from electrode potentials in the reverse subtraction order than that for the forward reaction. In practice, a voltmeter would report a potential of -0.47 V with its red and black inputs connected to the Pb and Cu electrodes, respectively. If the inputs were swapped, the reported voltage would be +0.47 V ."}
{"id": 4420, "contents": "1718. Predicting Redox Spontaneity - \nAre aqueous iron(II) ions predicted to spontaneously oxidize elemental chromium under standard state conditions? Assume the half-reactions to be those available in Table 16.1."}
{"id": 4421, "contents": "1719. Solution - \nReferring to the tabulated half-reactions, the redox reaction in question can be represented by the equations below:\n\n$$\n\\mathrm{Cr}(s)+\\mathrm{Fe}^{2+}(a q) \\longrightarrow \\mathrm{Cr}^{3+}(a q)+\\mathrm{Fe}(s)\n$$\n\nThe entry for the putative oxidant, $\\mathrm{Fe}^{2+}$, appears above the entry for the reductant, Cr , and so a spontaneous reaction is predicted per the quick approach described above. Supporting this predication by calculating the standard cell potential for this reaction gives\n\n$$\n\\begin{aligned}\n\\mathrm{E}_{\\text {cell }}^{\\circ} & =\\mathrm{E}_{\\text {cathode }}^{\\circ}-\\mathrm{E}_{\\text {anode }}^{\\circ} \\\\\n& =\\mathrm{E}_{\\mathrm{Fe}(\\mathrm{II})}^{\\circ}-\\mathrm{E}^{\\circ} \\mathrm{Cr} \\\\\n& =-0.447 \\mathrm{~V}--0.744 \\mathrm{~V}=+0.297 \\mathrm{~V}\n\\end{aligned}\n$$\n\nThe positive value for the standard cell potential indicates the process is spontaneous under standard state conditions."}
{"id": 4422, "contents": "1720. Check Your Learning - \nUse the data in Table 16.1 to predict the spontaneity of the oxidation of bromide ion by molecular iodine under standard state conditions, supporting the prediction by calculating the standard cell potential for the reaction.\n\nRepeat for the oxidation of iodide ion by molecular bromine."}
{"id": 4423, "contents": "1721. Answer: - \n$$\n\\begin{array}{ll}\n\\mathrm{I}_{2}(s)+2 \\mathrm{Br}^{-}(a q) \\longrightarrow 2 \\mathrm{I}^{-}(a q)+\\mathrm{Br}_{2}(l) & E_{\\text {cell }}^{\\circ}=-0.5518 \\mathrm{~V} \\text { (nonspontaneous) } \\\\\n\\mathrm{Br}_{2}(s)+2 \\mathrm{I}^{-}(a q) \\longrightarrow 2 \\mathrm{Br}^{-}(a q)+\\mathrm{I}_{2}(l) & E_{\\text {cell }}^{\\circ}=+0.5518 \\mathrm{~V} \\text { (spontaneous) }\n\\end{array}\n$$"}
{"id": 4424, "contents": "1722. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Explain the relations between potential, free energy change, and equilibrium constants\n- Perform calculations involving the relations between cell potentials, free energy changes, and equilibrium\n- Use the Nernst equation to determine cell potentials under nonstandard conditions\n\nSo far in this chapter, the relationship between the cell potential and reaction spontaneity has been described, suggesting a link to the free energy change for the reaction (see chapter on thermodynamics). The interpretation of potentials as measures of oxidant strength was presented, bringing to mind similar measures of acid-base strength as reflected in equilibrium constants (see the chapter on acid-base equilibria). This section provides a summary of the relationships between potential and the related thermodynamic properties $\\Delta \\mathrm{G}$ and K ."}
{"id": 4425, "contents": "1723. $\\mathrm{E}^{\\circ}$ and $\\Delta \\mathrm{G}^{\\circ}$ - \nThe standard free energy change of a process, $\\Delta G^{\\circ}$, was defined in a previous chapter as the maximum work that could be performed by a system, $w_{\\text {max }}$. In the case of a redox reaction taking place within a galvanic cell under standard state conditions, essentially all the work is associated with transferring the electrons from reductant-to-oxidant, $w_{\\text {elec }}$ :\n\n$$\n\\Delta G^{\\circ}=w_{\\max }=w_{\\text {elec }}\n$$\n\nThe work associated with transferring electrons is determined by the total amount of charge (coulombs) transferred and the cell potential:\n\n$$\n\\begin{aligned}\n\\Delta G^{\\circ}=w_{\\mathrm{elec}} & =-n F E_{\\mathrm{cell}}^{\\circ} \\\\\n\\Delta G^{\\circ} & =-n F E_{\\mathrm{cell}}^{\\circ}\n\\end{aligned}\n$$\n\nwhere $n$ is the number of moles of electrons transferred, $F$ is Faraday's constant, and $E^{\\circ}$ cell is the standard cell potential. The relation between free energy change and standard cell potential confirms the sign conventions and spontaneity criteria previously discussed for both of these properties: spontaneous redox reactions exhibit positive potentials and negative free energy changes."}
{"id": 4426, "contents": "1724. $E^{\\circ}$ and $K$ - \nCombining a previously derived relation between $\\Delta \\mathrm{G}^{\\circ}$ and K (see the chapter on thermodynamics) and the equation above relating $\\Delta \\mathrm{G}^{\\circ}$ and $E^{\\circ}$ cell yields the following:\n\n$$\n\\begin{aligned}\n\\Delta G^{\\circ} & =-R T \\ln K=-n F E_{\\text {cell }}^{\\circ} \\\\\nE_{\\text {cell }}^{\\circ} & =\\left(\\frac{R T}{n F}\\right) \\ln K\n\\end{aligned}\n$$\n\nThis equation indicates redox reactions with large (positive) standard cell potentials will proceed far towards completion, reaching equilibrium when the majority of reactant has been converted to product. A summary of the relations between $E^{\\circ}, \\Delta G^{\\circ}$ and $K$ is depicted in Figure 16.7, and a table correlating reaction spontaneity to values of these properties is provided in Table 16.2.\n\n\nFIGURE 16.7 Graphic depicting the relation between three important thermodynamic properties.\n\n| $K$ | $\\Delta G^{\\circ}$ | $E^{\\circ}$ cell |  |\n| :--- | :--- | :--- | :--- |\n| $>1$ | $<0$ | $>0$ | Reaction is spontaneous under standard conditions <br> Products more abundant at equilibrium |\n| $<1$ | $>0$ | $<0$ | Reaction is non-spontaneous under standard conditions <br> Reactants more abundant at equilibrium |\n| $=1$ | $=0$ | $=0$ | Reaction is at equilibrium under standard conditions <br> Reactants and products equally abundant |\n\nTABLE 16.2"}
{"id": 4427, "contents": "1726. Equilibrium Constants, Standard Cell Potentials, and Standard Free Energy Changes - \nUse data from Appendix L to calculate the standard cell potential, standard free energy change, and equilibrium constant for the following reaction at $25^{\\circ} \\mathrm{C}$. Comment on the spontaneity of the forward reaction and the composition of an equilibrium mixture of reactants and products.\n\n$$\n2 \\mathrm{Ag}^{+}(a q)+\\mathrm{Fe}(s) \\rightleftharpoons 2 \\mathrm{Ag}(s)+\\mathrm{Fe}^{2+}(a q)\n$$"}
{"id": 4428, "contents": "1727. Solution - \nThe reaction involves an oxidation-reduction reaction, so the standard cell potential can be calculated using the data in Appendix L.\n\n$$\n\\begin{array}{llrl}\n\\text { anode (oxidation): } & \\mathrm{Fe}(s) \\longrightarrow \\mathrm{Fe}^{2+}(a q)+2 \\mathrm{e}^{-} & E_{\\mathrm{Fe} 2+/ \\mathrm{Fe}}^{\\circ}=-0.447 \\mathrm{~V} \\\\\n\\text { cathode (reduction): } & 2 \\times\\left(\\mathrm{Ag}^{+}(a q)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Ag}(s)\\right) & E_{\\mathrm{Ag}^{+} / \\mathrm{Ag}}^{\\circ}=0.7996 \\mathrm{~V} \\\\\n& E_{\\text {cell }}^{\\circ}=E_{\\text {cathode }}^{\\circ}-E_{\\text {anode }}^{\\circ}= & E_{\\mathrm{Ag}^{+} / \\mathrm{Ag}}^{\\circ}-E_{\\mathrm{Fe}^{2+} / \\mathrm{Fe}}^{\\circ}=+1.247 \\mathrm{~V}\n\\end{array}\n$$\n\nWith $n=2$, the equilibrium constant is then\n\n$$\n\\begin{aligned}\nE_{\\mathrm{cell}}^{\\circ} & =\\frac{0.0592 \\mathrm{~V}}{n} \\log K \\\\\nK & =10^{n \\times E_{\\mathrm{cell}}^{\\circ} / 0.0592 \\mathrm{~V}} \\\\\nK & =10^{2 \\times 1.247 \\mathrm{~V} / 0.0592 \\mathrm{~V}} \\\\\nK & =10^{42.128} \\\\\nK & =1.3 \\times 10^{42}\n\\end{aligned}\n$$\n\nThe standard free energy is then"}
{"id": 4429, "contents": "1727. Solution - \nThe standard free energy is then\n\n$$\n\\begin{aligned}\n& \\Delta G^{\\circ}=-n F E_{\\mathrm{cell}}^{\\circ} \\\\\n& \\Delta G^{\\circ}=-2 \\times 96,485 \\frac{\\mathrm{C}}{\\mathrm{~mol}} \\times 1.247 \\frac{\\mathrm{~J}}{\\mathrm{C}}=-240.6 \\frac{\\mathrm{~kJ}}{\\mathrm{~mol}}\n\\end{aligned}\n$$\n\nThe reaction is spontaneous, as indicated by a negative free energy change and a positive cell potential. The $K$ value is very large, indicating the reaction proceeds to near completion to yield an equilibrium mixture containing mostly products."}
{"id": 4430, "contents": "1728. Check Your Learning - \nWhat is the standard free energy change and the equilibrium constant for the following reaction at room temperature? Is the reaction spontaneous?\n\n$$\n\\mathrm{Sn}(s)+2 \\mathrm{Cu}^{2+}(a q) \\rightleftharpoons \\mathrm{Sn}^{2+}(a q)+2 \\mathrm{Cu}^{+}(a q)\n$$\n\nAnswer:\nSpontaneous; $n=2 ; E_{\\text {cell }}^{\\circ}=+0.291 \\mathrm{~V} ; \\Delta G^{\\circ}=-56.2 \\frac{\\mathrm{~kJ}}{\\mathrm{~mol}} ; K=6.8 \\times 10^{9}$."}
{"id": 4431, "contents": "1729. Potentials at Nonstandard Conditions: The Nernst Equation - \nMost of the redox processes that interest science and society do not occur under standard state conditions, and so the potentials of these systems under nonstandard conditions are a property worthy of attention. Having established the relationship between potential and free energy change in this section, the previously discussed relation between free energy change and reaction mixture composition can be used for this purpose.\n\n$$\n\\Delta G=\\Delta G^{\\circ}+R T \\ln Q\n$$\n\nNotice the reaction quotient, $Q$, appears in this equation, making the free energy change dependent upon the composition of the reaction mixture. Substituting the equation relating free energy change to cell potential yields the Nernst equation:\n\n$$\n\\begin{gathered}\n-n F E_{\\mathrm{cell}}=-n F E_{\\mathrm{cell}}^{\\circ}+R T \\ln Q \\\\\nE_{\\mathrm{cell}}=E_{\\mathrm{cell}}^{\\circ}-\\frac{R T}{n F} \\ln Q\n\\end{gathered}\n$$\n\nThis equation describes how the potential of a redox system (such as a galvanic cell) varies from its standard state value, specifically, showing it to be a function of the number of electrons transferred, $n$, the temperature, $T$, and the reaction mixture composition as reflected in $Q$. A convenient form of the Nernst equation for most work is one in which values for the fundamental constants ( $R$ and $F$ ) and standard temperature (298) K), along with a factor converting from natural to base-10 logarithms, have been included:\n\n$$\nE_{\\mathrm{cell}}=E_{\\mathrm{cell}}^{\\circ}-\\frac{0.0592 \\mathrm{~V}}{n} \\log Q\n$$"}
{"id": 4432, "contents": "1731. Predicting Redox Spontaneity Under Nonstandard Conditions - \nUse the Nernst equation to predict the spontaneity of the redox reaction shown below.\n\n$$\n\\mathrm{Co}(s)+\\mathrm{Fe}^{2+}(a q, 1.94 M) \\longrightarrow \\mathrm{Co}^{2+}(a q, 0.15 M)+\\mathrm{Fe}(s)\n$$"}
{"id": 4433, "contents": "1732. Solution - \nCollecting information from Appendix L and the problem,\n\n$$\n\\begin{array}{llrl}\n\\text { Anode (oxidation): } & \\mathrm{Co}(s) \\longrightarrow \\mathrm{Co}^{2+}(a q)+2 \\mathrm{e}^{-} & E_{\\mathrm{Co}^{2+} / \\mathrm{Co}}^{\\circ}=-0.28 \\mathrm{~V} \\\\\n\\text { Cathode (reduction): } & \\mathrm{Fe}^{2+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Fe}(s) & E_{\\mathrm{Fe}^{2+} / \\mathrm{Fe}}^{\\circ}=-0.447 \\mathrm{~V} \\\\\n& E_{\\mathrm{cell}}^{\\circ}=E_{\\text {cathode }}^{\\circ}-E_{\\text {anode }}^{\\circ}=-0.447 \\mathrm{~V}-(-0.28 \\mathrm{~V})=-0.17 \\mathrm{~V}\n\\end{array}\n$$\n\nNotice the negative value of the standard cell potential indicates the process is not spontaneous under standard conditions. Substitution of the Nernst equation terms for the nonstandard conditions yields:\n\n$$\n\\begin{aligned}\nQ & =\\frac{\\left[\\mathrm{Co}^{2+}\\right]}{\\left[\\mathrm{Fe}^{2+}\\right]}=\\frac{0.15 M}{1.94 M}=0.077 \\\\\nE_{\\text {cell }} & =E_{\\text {cell }}^{\\circ}-\\frac{0.0592 \\mathrm{~V}}{n} \\log Q \\\\\nE_{\\text {cell }} & =-0.17 \\mathrm{~V}-\\frac{0.0592 \\mathrm{~V}}{2} \\log 0.077 \\\\\nE_{\\text {cell }} & =-0.17 \\mathrm{~V}+0.033 \\mathrm{~V}=-0.14 \\mathrm{~V}\n\\end{aligned}\n$$\n\nThe cell potential remains negative (slightly) under the specified conditions, and so the reaction remains nonspontaneous."}
{"id": 4434, "contents": "1733. Check Your Learning - \nFor the cell schematic below, identify values for $n$ and $Q$, and calculate the cell potential, $E_{\\text {cell }}$ -\n\n$$\n\\mathrm{Al}(s)\\left|\\mathrm{Al}^{3+}(a q, 0.15 M) \\| \\mathrm{Cu}^{2+}(a q, 0.025 M)\\right| \\mathrm{Cu}(s)\n$$"}
{"id": 4435, "contents": "1734. Answer: - \n$n=6 ; Q=1440 ; E_{\\text {cell }}=+1.97 \\mathrm{~V}$, spontaneous.\n\nA concentration cell is constructed by connecting two nearly identical half-cells, each based on the same halfreaction and using the same electrode, varying only in the concentration of one redox species. The potential of a concentration cell, therefore, is determined only by the difference in concentration of the chosen redox species. The example problem below illustrates the use of the Nernst equation in calculations involving concentration cells."}
{"id": 4436, "contents": "1736. Concentration Cells - \nWhat is the cell potential of the concentration cell described by\n\n$$\n\\mathrm{Zn}(s)\\left|\\mathrm{Zn}^{2+}(a q, 0.10 M) \\| \\mathrm{Zn}^{2+}(a q, 0.50 M)\\right| \\mathrm{Zn}(s)\n$$"}
{"id": 4437, "contents": "1737. Solution - \nFrom the information given:\n\n| Anode: | $\\mathrm{Zn}(s) \\longrightarrow \\mathrm{Zn}^{2+}(a q, 0.10 M)+2 \\mathrm{e}^{-}$ | $E_{\\text {anode }}^{\\circ}$ | $=-0.7618 \\mathrm{~V}$ |\n| :--- | :--- | :--- | :--- |\n| Cathode: | $\\mathrm{Zn}^{2+}(a q, 0.50 M)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Zn}(s)$ | $E_{\\text {cathode }}^{\\circ}$ | $=-0.7618 \\mathrm{~V}$ |\n| Overall: | $\\mathrm{Zn}^{2+}(a q, 0.50 M) \\longrightarrow \\mathrm{Zn}^{2+}(a q, 0.10 M)$ | $E_{\\text {cell }}^{\\circ}$ | $=0.000 \\mathrm{~V}$ |\n\nSubstituting into the Nernst equation,\n\n$$\nE_{\\text {cell }}=0.000 \\mathrm{~V}-\\frac{0.0592 \\mathrm{~V}}{2} \\log \\frac{0.10}{0.50}=+0.021 \\mathrm{~V}\n$$\n\nThe positive value for cell potential indicates the overall cell reaction (see above) is spontaneous. This spontaneous reaction is one in which the zinc ion concentration in the cathode falls (it is reduced to elemental zinc) while that in the anode rises (it is produced by oxidation of the zinc anode). A greater driving force for zinc reduction is present in the cathode, where the zinc(II) ion concentration is greater ( $E_{\\text {cathode }}>E_{\\text {anode }}$ )."}
{"id": 4438, "contents": "1738. Check Your Learning - \nThe concentration cell above was allowed to operate until the cell reaction reached equilibrium. What are the cell potential and the concentrations of zinc(II) in each half-cell for the cell now?"}
{"id": 4439, "contents": "1739. Answer: - \n$E_{\\text {cell }}=0.000 \\mathrm{~V} ;\\left[\\mathrm{Zn}^{2+}\\right]_{\\text {cathode }}=\\left[\\mathrm{Zn}^{2+}\\right]_{\\text {anode }}=0.30 \\mathrm{M}$"}
{"id": 4440, "contents": "1740. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the electrochemistry associated with several common batteries\n- Distinguish the operation of a fuel cell from that of a battery\n\nThere are many technological products associated with the past two centuries of electrochemistry research, none more immediately obvious than the battery. A battery is a galvanic cell that has been specially designed and constructed in a way that best suits its intended use a source of electrical power for specific applications. Among the first successful batteries was the Daniell cell, which relied on the spontaneous oxidation of zinc by copper(II) ions (Figure 16.8):\n\n$$\n\\mathrm{Zn}(s)+\\mathrm{Cu}^{2+}(a q) \\longrightarrow \\mathrm{Zn}^{2+}(a q)+\\mathrm{Cu}(s)\n$$\n\n\n\nFIGURE 16.8 Illustration of a Daniell cell taken from a 1904 journal publication (left) along with a simplified illustration depicting the electrochemistry of the cell (right). The 1904 design used a porous clay pot to both contain one of the half-cell's content and to serve as a salt bridge to the other half-cell.\n\nModern batteries exist in a multitude of forms to accommodate various applications, from tiny button batteries that provide the modest power needs of a wristwatch to the very large batteries used to supply backup energy to municipal power grids. Some batteries are designed for single-use applications and cannot be recharged\n(primary cells), while others are based on conveniently reversible cell reactions that allow recharging by an external power source (secondary cells). This section will provide a summary of the basic electrochemical aspects of several batteries familiar to most consumers, and will introduce a related electrochemical device called a fuel cell that can offer improved performance in certain applications."}
{"id": 4441, "contents": "1741. LINK TO LEARNING - \nVisit this site (http://openstax.org/l/16batteries) to learn more about batteries."}
{"id": 4442, "contents": "1742. Single-Use Batteries - \nA common primary battery is the dry cell, which uses a zinc can as both container and anode (\"-\" terminal) and a graphite rod as the cathode (\"+\" terminal). The Zn can is filled with an electrolyte paste containing manganese(IV) oxide, zinc(II) chloride, ammonium chloride, and water. A graphite rod is immersed in the electrolyte paste to complete the cell. The spontaneous cell reaction involves the oxidation of zinc:\n\n$$\n\\text { anode reaction: } \\mathrm{Zn}(s) \\longrightarrow \\mathrm{Zn}^{2+}(a q)+2 \\mathrm{e}^{-}\n$$\n\nand the reduction of manganese(IV)\nreduction reaction: $2 \\mathrm{MnO}_{2}(s)+2 \\mathrm{NH}_{4} \\mathrm{Cl}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Mn}_{2} \\mathrm{O}_{3}(s)+2 \\mathrm{NH}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{Cl}^{-}$\nwhich together yield the cell reaction:\ncell reaction: $2 \\mathrm{MnO}_{2}(s)+2 \\mathrm{NH}_{4} \\mathrm{Cl}(a q)+\\mathrm{Zn}(s) \\longrightarrow \\mathrm{Zn}^{2+}(a q)+\\mathrm{Mn}_{2} \\mathrm{O}_{3}(s)+2 \\mathrm{NH}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{Cl}^{-} E_{\\text {cell }} \\sim 1.5 \\mathrm{~V}$\nThe voltage (cell potential) of a dry cell is approximately 1.5 V . Dry cells are available in various sizes (e.g., D, C, AA, AAA). All sizes of dry cells comprise the same components, and so they exhibit the same voltage, but larger cells contain greater amounts of the redox reactants and therefore are capable of transferring correspondingly greater amounts of charge. Like other galvanic cells, dry cells may be connected in series to yield batteries with greater voltage outputs, if needed.\n\n\nFIGURE 16.9 A schematic diagram shows a typical dry cell."}
{"id": 4443, "contents": "1743. LINK TO LEARNING - \nVisit this site (http://openstax.org/l/16zinccarbon) to learn more about zinc-carbon batteries.\n\nAlkaline batteries (Figure 16.10) were developed in the 1950s to improve on the performance of the dry cell,\nand they were designed around the same redox couples. As their name suggests, these types of batteries use alkaline electrolytes, often potassium hydroxide. The reactions are\n\n| $\\mathrm{Zn}(s)+2 \\mathrm{OH}^{-}(a q)$ <br> anode: <br> cathode: $2 \\mathrm{MnO}_{2}(s)+\\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Mn}_{2} \\mathrm{O}_{3}(s)+\\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{OH}^{-}(a q)$ |\n| :--- |\n| cell: |\n\nAn alkaline battery can deliver about three to five times the energy of a zinc-carbon dry cell of similar size. Alkaline batteries are prone to leaking potassium hydroxide, so they should be removed from devices for longterm storage. While some alkaline batteries are rechargeable, most are not. Attempts to recharge an alkaline battery that is not rechargeable often leads to rupture of the battery and leakage of the potassium hydroxide electrolyte.\n\n\nFIGURE 16.10 Alkaline batteries were designed as improved replacements for zinc-carbon (dry cell) batteries."}
{"id": 4444, "contents": "1744. LINK TO LEARNING - \nVisit this site (http://openstax.org/l/16alkaline) to learn more about alkaline batteries."}
{"id": 4445, "contents": "1745. Rechargeable (Secondary) Batteries - \nNickel-cadmium, or NiCd, batteries (Figure 16.11) consist of a nickel-plated cathode, cadmium-plated anode, and a potassium hydroxide electrode. The positive and negative plates, which are prevented from shorting by the separator, are rolled together and put into the case. This is a \"jelly-roll\" design and allows the NiCd cell to deliver much more current than a similar-sized alkaline battery. The reactions are\n\n$$\n\\left.\\begin{array}{l}\n\\text { anode: } \\quad \\mathrm{Cd}(s)+2 \\mathrm{OH}^{-}(a q) \\longrightarrow \\mathrm{Cd}(\\mathrm{OH})_{2}(s)+2 \\mathrm{e}^{-} \\\\\n\\text {cathode: } \\mathrm{NiO}_{2}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Ni}(\\mathrm{OH})_{2}(s)+2 \\mathrm{OH}^{-}(a q)\n\\end{array}\\right]\n$$\n\nWhen properly treated, a NiCd battery can be recharged about 1000 times. Cadmium is a toxic heavy metal so NiCd batteries should never be ruptured or incinerated, and they should be disposed of in accordance with relevant toxic waste guidelines.\n\n\nFIGURE 16.11 NiCd batteries use a \"jelly-roll\" design that significantly increases the amount of current the battery can deliver as compared to a similar-sized alkaline battery."}
{"id": 4446, "contents": "1746. LINK TO LEARNING - \nVisit this site (http://openstax.org/l/16NiCdrecharge) for more information about nickel cadmium rechargeable batteries.\n\nLithium ion batteries (Figure 16.12) are among the most popular rechargeable batteries and are used in many portable electronic devices. The reactions are\n\n$$\n\\begin{array}{lll}\n\\text { anode: } & \\mathrm{LiCoO}_{2} \\rightleftharpoons \\mathrm{Li}_{1-x} \\mathrm{CoO}_{2}+x \\mathrm{Li}^{+}+x \\mathrm{e}^{-} \\\\\n\\text {cathode: } x \\mathrm{Li}^{+}+x \\mathrm{e}^{-}+x \\mathrm{C}_{6} \\rightleftharpoons x \\mathrm{LiC}_{6} & \\\\\n\\hline \\text { cell: } & \\mathrm{LiCoO}_{2}+x \\mathrm{C}_{6} \\rightleftharpoons \\mathrm{Li}_{1-x} \\mathrm{CoO}_{2}+x \\mathrm{LiC}_{6} & E_{\\text {cell }} \\sim 3.7 \\mathrm{~V}\n\\end{array}\n$$\n\nThe variable stoichiometry of the cell reaction leads to variation in cell voltages, but for typical conditions, $x$ is usually no more than 0.5 and the cell voltage is approximately 3.7 V . Lithium batteries are popular because they can provide a large amount current, are lighter than comparable batteries of other types, produce a nearly constant voltage as they discharge, and only slowly lose their charge when stored.\n\n\nFIGURE 16.12 In a lithium ion battery, charge flows as the lithium ions are transferred between the anode and cathode."}
{"id": 4447, "contents": "1747. LINK TO LEARNING - \nVisit this site (http://openstax.org/l/16lithiumion) for more information about lithium ion batteries.\n\nThe lead acid battery (Figure 16.13) is the type of secondary battery commonly used in automobiles. It is inexpensive and capable of producing the high current required by automobile starter motors. The reactions for a lead acid battery are\n\n$$\n\\begin{aligned}\n& \\text { anode: } \\mathrm{Pb}(s)+\\mathrm{HSO}_{4}^{-}(a q) \\longrightarrow \\mathrm{PbSO}_{4}(s)+\\mathrm{H}^{+}(a q)+2 \\mathrm{e}^{-} \\\\\n& \\text {cathode: } \\mathrm{PbO}_{2}(s)+\\mathrm{HSO}_{4}^{-}(a q)+3 \\mathrm{H}^{+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{PbSO}_{4}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\\\\n& \\hline \\text { cell: }\n\\end{aligned} \\quad \\mathrm{Pb}(s)+\\mathrm{PbO}_{2}(s)+2 \\mathrm{H}_{2} \\mathrm{SO}_{4}(a q) \\longrightarrow 2 \\mathrm{PbSO}_{4}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\quad E_{\\text {cell }} \\sim 2 \\mathrm{~V} \\text { lel }\n$$\n\nEach cell produces 2 V , so six cells are connected in series to produce a $12-\\mathrm{V}$ car battery. Lead acid batteries are heavy and contain a caustic liquid electrolyte, $\\mathrm{H}_{2} \\mathrm{SO}_{4}(\\mathrm{aq})$, but are often still the battery of choice because of their high current density. Since these batteries contain a significant amount of lead, they must always be disposed of properly.\n\n\nFIGURE 16.13 The lead acid battery in your automobile consists of six cells connected in series to give 12 V ."}
{"id": 4448, "contents": "1748. LINK TO LEARNING - \nVisit this site (http://openstax.org/1/16leadacid) for more information about lead acid batteries."}
{"id": 4449, "contents": "1749. Fuel Cells - \nA fuel cell is a galvanic cell that uses traditional combustive fuels, most often hydrogen or methane, that are continuously fed into the cell along with an oxidant. (An alternative, but not very popular, name for a fuel cell is a flow battery.) Within the cell, fuel and oxidant undergo the same redox chemistry as when they are combusted, but via a catalyzed electrochemical that is significantly more efficient. For example, a typical hydrogen fuel cell uses graphite electrodes embedded with platinum-based catalysts to accelerate the two halfcell reactions:\n\n\nFIGURE 16.14 In this hydrogen fuel cell, oxygen from the air reacts with hydrogen, producing water and electricity.\n\n$$\n\\begin{aligned}\n& \\text { Anode: } \\quad 2 \\mathrm{H}_{2}(g) \\longrightarrow 4 \\mathrm{H}^{+}(a q)+4 \\mathrm{e}^{-} \\\\\n& \\text {Cathode: } \\mathrm{O}_{2}(g)+4 \\mathrm{H}^{+}(a q)+4 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(g) \\\\\n& \\hline \\text { Cell: } \\quad 2 \\mathrm{H}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(g) \\quad \\\\\n& \\hline \\text { cell } \\sim 1.2 \\mathrm{~V}\n\\end{aligned}\n$$\n\nThese types of fuel cells generally produce voltages of approximately 1.2 V . Compared to an internal combustion engine, the energy efficiency of a fuel cell using the same redox reaction is typically more than double ( $\\sim 20 \\%-25 \\%$ for an engine versus $\\sim 50 \\%-75 \\%$ for a fuel cell). Hydrogen fuel cells are commonly used on extended space missions, and prototypes for personal vehicles have been developed, though the technology remains relatively immature."}
{"id": 4450, "contents": "1750. LINK TO LEARNING - \nCheck out this link (http://openstax.org/l/16fuelcells) to learn more about fuel cells."}
{"id": 4451, "contents": "1750. LINK TO LEARNING - 1750.1. Corrosion\nLEARNING OBJECTIVES\nBy the end of this section, you will be able to:\n\n- Define corrosion\n- List some of the methods used to prevent or slow corrosion\n\nCorrosion is usually defined as the degradation of metals by a naturally occurring electrochemical process. The formation of rust on iron, tarnish on silver, and the blue-green patina that develops on copper are all examples of corrosion. The total cost of corrosion remediation in the United States is significant, with estimates in excess of half a trillion dollars a year."}
{"id": 4452, "contents": "1752. Statue of Liberty: Changing Colors - \nThe Statue of Liberty is a landmark every American recognizes. The Statue of Liberty is easily identified by its height, stance, and unique blue-green color (Figure 16.15). When this statue was first delivered from France, its appearance was not green. It was brown, the color of its copper \"skin.\" So how did the Statue of Liberty change colors? The change in appearance was a direct result of corrosion. The copper that is the primary component of the statue slowly underwent oxidation from the air. The oxidation-reduction reactions of copper metal in the environment occur in several steps. Copper metal is oxidized to copper(I) oxide ( $\\mathrm{Cu}_{2} \\mathrm{O}$ ), which is red, and then to copper(II) oxide, which is black\n\n$$\n\\begin{aligned}\n2 \\mathrm{Cu}(s)+\\frac{1}{2} \\mathrm{O}_{2}(g) & \\longrightarrow \\mathrm{Cu}_{2} \\mathrm{O}(s) \\\\\n\\mathrm{Cu}_{2} \\mathrm{O}(s)+\\frac{1}{2} \\mathrm{O}_{2}(g) & \\longrightarrow 2 \\mathrm{CuO}(s)\n\\end{aligned}\n$$\n\nCoal, which was often high in sulfur, was burned extensively in the early part of the last century. As a result, atmospheric sulfur trioxide, carbon dioxide, and water all reacted with the CuO\n\n$$\n\\begin{gathered}\n2 \\mathrm{CuO}(s)+\\mathrm{CO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Cu}_{2} \\mathrm{CO}_{3}(\\mathrm{OH})_{2}(s) \\\\\n3 \\mathrm{CuO}(s)+2 \\mathrm{CO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Cu}_{2}\\left(\\mathrm{CO}_{3}\\right)_{2}(\\mathrm{OH})_{2}(s) \\\\\n4 \\mathrm{CuO}(s)+\\mathrm{SO}_{3}(g)+3 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Cu}_{4} \\mathrm{SO}_{4}(\\mathrm{OH})_{6}(s)\n\\end{gathered}\n$$"}
{"id": 4453, "contents": "1752. Statue of Liberty: Changing Colors - \nThese three compounds are responsible for the characteristic blue-green patina seen on the Statue of Liberty (and other outdoor copper structures). Fortunately, formation of patina creates a protective layer on the copper surface, preventing further corrosion of the underlying copper. The formation of the protective layer is called passivation, a phenomenon discussed further in another chapter of this text.\n\n\nFIGURE 16.15 (a) The Statue of Liberty is covered with a copper skin, and was originally brown, as shown in this painting. (b) Exposure to the elements has resulted in the formation of the blue-green patina seen today.\n\nPerhaps the most familiar example of corrosion is the formation of rust on iron. Iron will rust when it is exposed to oxygen and water. Rust formation involves the creation of a galvanic cell at an iron surface, as illustrated in Figure 16.15. The relevant redox reactions are described by the following equations:\n\n$$\n\\begin{array}{llr}\n\\text { anode: } & \\mathrm{Fe}(s) \\longrightarrow \\mathrm{Fe}^{2+}(a q)+2 \\mathrm{e}^{-} & E_{\\mathrm{Fe}^{2+} / \\mathrm{Fe}}^{\\circ}=-0.44 \\mathrm{~V} \\\\\n\\text { cathode: } & \\mathrm{O}_{2}(g)+4 \\mathrm{H}^{+}(a q)+4 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(l) & E_{\\mathrm{O}_{2} / \\mathrm{O}^{2}}^{\\circ}=+1.23 \\mathrm{~V} \\\\\n\\text { overall: } & 2 \\mathrm{Fe}(s)+\\mathrm{O}_{2}(g)+4 \\mathrm{H}^{+}(a q) \\longrightarrow 2 \\mathrm{Fe}^{2+}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l) & E_{\\mathrm{cell}}^{\\circ}=+1.67 \\mathrm{~V}\n\\end{array}\n$$\n\nFurther reaction of the iron(II) product in humid air results in the production of an iron(III) oxide hydrate known as rust:"}
{"id": 4454, "contents": "1752. Statue of Liberty: Changing Colors - \nFurther reaction of the iron(II) product in humid air results in the production of an iron(III) oxide hydrate known as rust:\n\n$$\n4 \\mathrm{Fe}^{2+}(a q)+\\mathrm{O}_{2}(g)+(4+2 x) \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{Fe}_{2} \\mathrm{O}_{3} \\cdot x \\mathrm{H}_{2} \\mathrm{O}(s)+8 \\mathrm{H}^{+}(a q)\n$$\n\nThe stoichiometry of the hydrate varies, as indicated by the use of $x$ in the compound formula. Unlike the patina on copper, the formation of rust does not create a protective layer and so corrosion of the iron continues as the rust flakes off and exposes fresh iron to the atmosphere.\n\n\nFIGURE 16.16 Corrosion can occur when a painted iron or steel surface is exposed to the environment by a scratch through the paint. A galvanic cell results that may be approximated by the simplified cell schematic $\\mathrm{Fe}(s) \\mid \\mathrm{Fe}^{2+}(\\mathrm{aq})$ $\\| \\mathrm{O}_{2}(\\mathrm{aq}), \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{D}) \\mid \\mathrm{Fe}(\\mathrm{s})$.\n\nOne way to keep iron from corroding is to keep it painted. The layer of paint prevents the water and oxygen necessary for rust formation from coming into contact with the iron. As long as the paint remains intact, the iron is protected from corrosion.\n\nOther strategies include alloying the iron with other metals. For example, stainless steel is an alloy of iron containing a small amount of chromium. The chromium tends to collect near the surface, where it corrodes and forms a passivating an oxide layer that protects the iron."}
{"id": 4455, "contents": "1752. Statue of Liberty: Changing Colors - \nIron and other metals may also be protected from corrosion by galvanization, a process in which the metal to be protected is coated with a layer of a more readily oxidized metal, usually zinc. When the zinc layer is intact, it prevents air from contacting the underlying iron and thus prevents corrosion. If the zinc layer is breached by either corrosion or mechanical abrasion, the iron may still be protected from corrosion by a cathodic protection process, which is described in the next paragraph.\n\nAnother important way to protect metal is to make it the cathode in a galvanic cell. This is cathodic protection and can be used for metals other than just iron. For example, the rusting of underground iron storage tanks and pipes can be prevented or greatly reduced by connecting them to a more active metal such as zinc or magnesium (Figure 16.17). This is also used to protect the metal parts in water heaters. The more active metals (lower reduction potential) are called sacrificial anodes because as they get used up as they corrode (oxidize) at the anode. The metal being protected serves as the cathode for the reduction of oxygen in air, and so it simply serves to conduct (not react with) the electrons being transferred. When the anodes are properly monitored and periodically replaced, the useful lifetime of the iron storage tank can be greatly extended.\n\n\nFIGURE 16.17 Cathodic protection is a useful approach to electrochemically preventing corrosion of underground storage tanks."}
{"id": 4456, "contents": "1753. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the process of electrolysis\n- Compare the operation of electrolytic cells with that of galvanic cells\n- Perform stoichiometric calculations for electrolytic processes\n\nElectrochemical cells in which spontaneous redox reactions take place (galvanic cells) have been the topic of discussion so far in this chapter. In these cells, electrical work is done by a redox system on its surroundings as electrons produced by the redox reaction are transferred through an external circuit. This final section of the chapter will address an alternative scenario in which an external circuit does work on a redox system by imposing a voltage sufficient to drive an otherwise nonspontaneous reaction, a process known as electrolysis. A familiar example of electrolysis is recharging a battery, which involves use of an external power source to drive the spontaneous (discharge) cell reaction in the reverse direction, restoring to some extent the composition of the half-cells and the voltage of the battery. Perhaps less familiar is the use of electrolysis in the refinement of metallic ores, the manufacture of commodity chemicals, and the electroplating of metallic coatings on various products (e.g., jewelry, utensils, auto parts). To illustrate the essential concepts of electrolysis, a few specific processes will be considered."}
{"id": 4457, "contents": "1754. The Electrolysis of Molten Sodium Chloride - \nMetallic sodium, Na , and chlorine gas, $\\mathrm{Cl}_{2}$, are used in numerous applications, and their industrial production relies on the large-scale electrolysis of molten sodium chloride, $\\mathrm{NaCl}(1)$. The industrial process typically uses a Downs cell similar to the simplified illustration shown in Figure 16.18. The reactions associated with this process are:\n\n| anode: | $2 \\mathrm{Cl}^{-}(l) \\longrightarrow \\mathrm{Cl}_{2}(g)+2 \\mathrm{e}^{-}$ |\n| :--- | :---: |\n| cathode: | $\\mathrm{Na}^{+}(l)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Na}(l)$ |\n| cell: | $2 \\mathrm{Na}^{+}(l)+2 \\mathrm{Cl}^{-}(l) \\longrightarrow 2 \\mathrm{Na}(l)+\\mathrm{Cl}_{2}(g)$ |\n\nThe cell potential for the above process is negative, indicating the reaction as written (decomposition of liquid $\\mathrm{NaCl})$ is not spontaneous. To force this reaction, a positive potential of magnitude greater than the negative cell potential must be applied to the cell.\n\n\nFIGURE 16.18 Cells of this sort (a cell for the electrolysis of molten sodium chloride) are used in the Downs process for production of sodium and chlorine, and they typically use iron cathodes and carbon anodes."}
{"id": 4458, "contents": "1755. The Electrolysis of Water - \nWater may be electrolytically decomposed in a cell similar to the one illustrated in Figure 16.19. To improve electrical conductivity without introducing a different redox species, the hydrogen ion concentration of the water is typically increased by addition of a strong acid. The redox processes associated with this cell are\n\n| anode: | $2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{O}_{2}(g)+4 \\mathrm{H}^{+}(a q)+4 \\mathrm{e}^{-}$ | $E_{\\text {anode }}^{\\circ}$ |\n| :--- | ---: | :--- |\n| cathode: $2 \\mathrm{H}^{+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{H}_{2}(g)$ | $E_{\\text {cathode }}^{\\circ}$ | $=0 \\mathrm{~V}$ |\n| cell: | $2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 229 \\mathrm{~V}$ |  |\n|  |  | $E_{\\text {cell }}^{\\circ}$ |\n\nAgain, the cell potential as written is negative, indicating a nonspontaneous cell reaction that must be driven by imposing a cell voltage greater than +1.229 V . Keep in mind that standard electrode potentials are used to inform thermodynamic predictions here, though the cell is not operating under standard state conditions. Therefore, at best, calculated cell potentials should be considered ballpark estimates.\n\n\nFIGURE 16.19 The electrolysis of water produces stoichiometric amounts of oxygen gas at the anode and hydrogen at the anode."}
{"id": 4459, "contents": "1756. The Electrolysis of Aqueous Sodium Chloride - \nWhen aqueous solutions of ionic compounds are electrolyzed, the anode and cathode half-reactions may involve the electrolysis of either water species $\\left(\\mathrm{H}_{2} \\mathrm{O}, \\mathrm{H}^{+}, \\mathrm{OH}^{-}\\right)$or solute species (the cations and anions of the compound). As an example, the electrolysis of aqueous sodium chloride could involve either of these two anode reactions:\n\n$$\n\\begin{array}{ll}\n\\text { (i) } 2 \\mathrm{Cl}^{-}(a q) \\longrightarrow \\mathrm{Cl}_{2}(g)+2 \\mathrm{e}^{-} & E_{\\text {anode }}^{\\circ}=+1.35827 \\mathrm{~V} \\\\\n\\text { (ii) } 2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{O}_{2}(g)+4 \\mathrm{H}^{+}(a q)+4 \\mathrm{e}^{-} & E_{\\text {anode }}^{\\circ}=+1.229 \\mathrm{~V}\n\\end{array}\n$$\n\nThe standard electrode (reduction) potentials of these two half-reactions indicate water may be oxidized at a less negative/more positive potential $(-1.229 \\mathrm{~V})$ than chloride ion $(-1.358 \\mathrm{~V})$. Thermodynamics thus predicts that water would be more readily oxidized, though in practice it is observed that both water and chloride ion are oxidized under typical conditions, producing a mixture of oxygen and chlorine gas.\n\nTurning attention to the cathode, the possibilities for reduction are:"}
{"id": 4460, "contents": "1756. The Electrolysis of Aqueous Sodium Chloride - \nTurning attention to the cathode, the possibilities for reduction are:\n\n$$\n\\begin{array}{ll}\n\\text { (iii) } 2 \\mathrm{H}^{+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{H}_{2}(g) & E_{\\text {cathode }}^{\\circ}=0 \\mathrm{~V} \\\\\n\\text { (iv) } 2 \\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{H}_{2}(g)+2 \\mathrm{OH}^{-}(a q) & E_{\\text {cathode }}^{\\circ}=-0.8277 \\mathrm{~V} \\\\\n\\text { (v) } \\mathrm{Na}^{+}(a q)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Na}(s) & E_{\\text {cathode }}^{\\circ}=-2.71 \\mathrm{~V}\n\\end{array}\n$$\n\nComparison of these standard half-reaction potentials suggests the reduction of hydrogen ion is\nthermodynamically favored. However, in a neutral aqueous sodium chloride solution, the concentration of hydrogen ion is far below the standard state value of 1 M (approximately $10^{-7} \\mathrm{M}$ ), and so the observed cathode reaction is actually reduction of water. The net cell reaction in this case is then\n\n$$\n\\text { cell: } 2 \\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{Cl}^{-}(a q) \\longrightarrow \\mathrm{H}_{2}(g)+\\mathrm{Cl}_{2}(g)+2 \\mathrm{OH}^{-}(a q) \\quad E_{\\mathrm{cell}}^{\\circ}=-2.186 \\mathrm{~V}\n$$\n\nThis electrolysis reaction is part of the chlor-alkali process used by industry to produce chlorine and sodium hydroxide (lye)."}
{"id": 4461, "contents": "1757. Chemistry in Everyday Life Electroplating - \nAn important use for electrolytic cells is in electroplating. Electroplating results in a thin coating of one metal on top of a conducting surface. Reasons for electroplating include making the object more corrosion resistant, strengthening the surface, producing a more attractive finish, or for purifying metal. The metals commonly used in electroplating include cadmium, chromium, copper, gold, nickel, silver, and tin. Common consumer products include silver-plated or gold-plated tableware, chrome-plated automobile parts, and jewelry. The silver plating of eating utensils is used here to illustrate the process. (Figure 16.20).\n\n\nFIGURE 16.20 This schematic shows an electrolytic cell for silver plating eating utensils.\nIn the figure, the anode consists of a silver electrode, shown on the left. The cathode is located on the right and is the spoon, which is made from inexpensive metal. Both electrodes are immersed in a solution of silver nitrate. Applying a sufficient potential results in the oxidation of the silver anode\n\n$$\n\\text { anode }: \\mathrm{Ag}(s) \\longrightarrow \\mathrm{Ag}^{+}(a q)+\\mathrm{e}^{-}\n$$\n\nand reduction of silver ion at the (spoon) cathode:\n\n$$\n\\text { cathode: } \\mathrm{Ag}^{+}(a q)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Ag}(s)\n$$\n\nThe net result is the transfer of silver metal from the anode to the cathode. Several experimental factors must be carefully controlled to obtain high-quality silver coatings, including the exact composition of the electrolyte solution, the cell voltage applied, and the rate of the electrolysis reaction (electrical current)."}
{"id": 4462, "contents": "1758. Quantitative Aspects of Electrolysis - \nElectrical current is defined as the rate of flow for any charged species. Most relevant to this discussion is the flow of electrons. Current is measured in a composite unit called an ampere, defined as one coulomb per second ( $\\mathrm{A}=1 \\mathrm{C} / \\mathrm{s}$ ). The charge transferred, $Q$, by passage of a constant current, $I$, over a specified time interval, $t$, is then given by the simple mathematical product\n\n$$\nQ=I t\n$$\n\nWhen electrons are transferred during a redox process, the stoichiometry of the reaction may be used to derive the total amount of (electronic) charge involved. For example, the generic reduction process\n\n$$\n\\mathrm{M}^{\\mathrm{n}+}(a q)+\\mathrm{ne}^{-} \\longrightarrow \\mathrm{M}(s)\n$$\n\ninvolves the transfer of $n$ mole of electrons. The charge transferred is, therefore,\n\n$$\nQ=n F\n$$\n\nwhere $F$ is Faraday's constant, the charge in coulombs for one mole of electrons. If the reaction takes place in an electrochemical cell, the current flow is conveniently measured, and it may be used to assist in stoichiometric calculations related to the cell reaction."}
{"id": 4463, "contents": "1760. Converting Current to Moles of Electrons - \nIn one process used for electroplating silver, a current of 10.23 A was passed through an electrolytic cell for exactly 1 hour. How many moles of electrons passed through the cell? What mass of silver was deposited at the cathode from the silver nitrate solution?"}
{"id": 4464, "contents": "1761. Solution - \nFaraday's constant can be used to convert the charge ( $Q$ ) into moles of electrons (n). The charge is the current (I) multiplied by the time\n\n$$\nn=\\frac{Q}{F}=\\frac{\\frac{10.23 \\mathrm{C}}{\\mathrm{~s}} \\times 1 \\mathrm{hr} \\times \\frac{60 \\mathrm{~min}}{\\mathrm{hr}} \\times \\frac{60 \\mathrm{~s}}{\\mathrm{~min}}}{96,485 \\mathrm{C} / \\mathrm{mol} \\mathrm{e}^{-}}=\\frac{36,830 \\mathrm{C}}{96,485 \\mathrm{C} / \\mathrm{mol} \\mathrm{e}^{-}}=0.3817 \\mathrm{~mol} \\mathrm{e}^{-}\n$$\n\nFrom the problem, the solution contains $\\mathrm{AgNO}_{3}$, so the reaction at the cathode involves 1 mole of electrons for each mole of silver\n\n$$\n\\text { cathode: } \\mathrm{Ag}^{+}(a q)+\\mathrm{e}^{-} \\longrightarrow \\operatorname{Ag}(s)\n$$\n\nThe atomic mass of silver is $107.9 \\mathrm{~g} / \\mathrm{mol}$, so\n\n$$\n\\text { mass } \\mathrm{Ag}=0.3817 \\mathrm{~mol} \\mathrm{e} ~-~ \\times \\frac{1 \\mathrm{~mol} \\mathrm{Ag}}{1 \\mathrm{~mol} \\mathrm{e}^{-}} \\times \\frac{107.9 \\mathrm{~g} \\mathrm{Ag}}{1 \\mathrm{~mol} \\mathrm{Ag}}=41.19 \\mathrm{~g} \\mathrm{Ag}\n$$"}
{"id": 4465, "contents": "1762. Check Your Learning - \nAluminum metal can be made from aluminum(III) ions by electrolysis. What is the half-reaction at the cathode? What mass of aluminum metal would be recovered if a current of 25.0 A passed through the solution for 15.0 minutes?\n\nAnswer:\n$\\mathrm{Al}^{3+}(a q)+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Al}(s) ; 0.0777 \\mathrm{~mol} \\mathrm{Al}=2.10 \\mathrm{~g} \\mathrm{Al}$."}
{"id": 4466, "contents": "1764. Time Required for Deposition - \nIn one application, a $0.010-\\mathrm{mm}$ layer of chromium must be deposited on a part with a total surface area of 3.3 $\\mathrm{m}^{2}$ from a solution of containing chromium(III) ions. How long would it take to deposit the layer of chromium if the current was 33.46 A? The density of chromium (metal) is $7.19 \\mathrm{~g} / \\mathrm{cm}^{3}$."}
{"id": 4467, "contents": "1765. Solution - \nFirst, compute the volume of chromium that must be produced (equal to the product of surface area and thickness):\n\n$$\n\\text { volume }=\\left(0.010 \\mathrm{~mm} \\times \\frac{1 \\mathrm{~cm}}{10 \\mathrm{~mm}}\\right) \\times\\left(3.3 \\mathrm{~m}^{2} \\times\\left(\\frac{10,000 \\mathrm{~cm}^{2}}{1 \\mathrm{~m}^{2}}\\right)\\right)=33 \\mathrm{~cm}^{3}\n$$\n\nUse the computed volume and the provided density to calculate the molar amount of chromium required:\n\n$$\n\\begin{gathered}\n\\text { mass }=\\text { volume } \\times \\text { density }=33 \\mathrm{~cm}^{3} \\times \\frac{7.19 \\mathrm{~g}}{\\mathrm{~cm}^{3}}=237 \\mathrm{~g} \\mathrm{Cr} \\\\\n\\\\\n\\mathrm{~mol} \\mathrm{Cr}=237 \\mathrm{~g} \\mathrm{Cr} \\times \\frac{1 \\mathrm{~mol} \\mathrm{Cr}}{52.00 \\mathrm{~g} \\mathrm{Cr}}=4.56 \\mathrm{~mol} \\mathrm{Cr}\n\\end{gathered}\n$$\n\nThe stoichiometry of the chromium(III) reduction process requires three moles of electrons for each mole of chromium(0) produced, and so the total charge required is:\n\n$$\nQ=4.56 \\mathrm{~mol} \\mathrm{Cr} \\times \\frac{3 \\mathrm{~mol} \\mathrm{e}^{-}}{1 \\mathrm{~mol} \\mathrm{Cr}} \\times \\frac{96485 \\mathrm{C}}{\\mathrm{~mol} \\mathrm{e}^{-}}=1.32 \\times 10^{6} \\mathrm{C}\n$$\n\nFinally, if this charge is passed at a rate of $33.46 \\mathrm{C} / \\mathrm{s}$, the required time is:"}
{"id": 4468, "contents": "1765. Solution - \nFinally, if this charge is passed at a rate of $33.46 \\mathrm{C} / \\mathrm{s}$, the required time is:\n\n$$\nt=\\frac{Q}{I}=\\frac{1.32 \\times 10^{6} \\mathrm{C}}{33.46 \\mathrm{C} / \\mathrm{s}}=3.95 \\times 10^{4} \\mathrm{~s}=11.0 \\mathrm{hr}\n$$"}
{"id": 4469, "contents": "1766. Check Your Learning - \nWhat mass of zinc is required to galvanize the top of a $3.00 \\mathrm{~m} \\times 5.50 \\mathrm{~m}$ sheet of iron to a thickness of 0.100 mm of zinc? If the zinc comes from a solution of $\\mathrm{Zn}\\left(\\mathrm{NO}_{3}\\right)_{2}$ and the current is 25.5 A , how long will it take to galvanize the top of the iron? The density of zinc is $7.140 \\mathrm{~g} / \\mathrm{cm}^{3}$."}
{"id": 4470, "contents": "1767. Answer: - \n11.8 kg Zn requires 382 hours."}
{"id": 4471, "contents": "1768. Key Terms - \nactive electrode electrode that participates as a reactant or product in the oxidation-reduction reaction of an electrochemical cell; the mass of an active electrode changes during the oxidationreduction reaction\nalkaline battery primary battery similar to a dry cell that uses an alkaline (often potassium hydroxide) electrolyte; designed to be an improved replacement for the dry cell, but with more energy storage and less electrolyte leakage than typical dry cell\nanode electrode in an electrochemical cell at which oxidation occurs\nbattery single or series of galvanic cells designed for use as a source of electrical power\ncathode electrode in an electrochemical cell at which reduction occurs\ncathodic protection approach to preventing corrosion of a metal object by connecting it to a sacrificial anode composed of a more readily oxidized metal\ncell notation (schematic) symbolic representation of the components and reactions in an electrochemical cell\ncell potential ( $\\boldsymbol{E}_{\\text {cell }}$ ) difference in potential of the cathode and anode half-cells\nconcentration cell galvanic cell comprising halfcells of identical composition but for the concentration of one redox reactant or product\ncorrosion degradation of metal via a natural electrochemical process\ndry cell primary battery, also called a zinc-carbon battery, based on the spontaneous oxidation of zinc by manganese(IV)\nelectrode potential ( $E_{\\mathrm{X}}$ ) the potential of a cell in which the half-cell of interest acts as a cathode when connected to the standard hydrogen electrode\nelectrolysis process using electrical energy to cause a nonspontaneous process to occur\nelectrolytic cell electrochemical cell in which an external source of electrical power is used to drive an otherwise nonspontaneous process\nFaraday's constant (F) charge on 1 mol of electrons; $F=96,485 \\mathrm{C} / \\mathrm{mol} \\mathrm{e}^{-}$\nfuel cell devices similar to galvanic cells that require a continuous feed of redox reactants; also called a flow battery\ngalvanic (voltaic) cell electrochemical cell in\nwhich a spontaneous redox reaction takes place; also called a voltaic cell\ngalvanization method of protecting iron or similar metals from corrosion by coating with a thin layer of more easily oxidized zinc."}
{"id": 4472, "contents": "1768. Key Terms - \nwhich a spontaneous redox reaction takes place; also called a voltaic cell\ngalvanization method of protecting iron or similar metals from corrosion by coating with a thin layer of more easily oxidized zinc.\nhalf cell component of a cell that contains the redox conjugate pair (\"couple\") of a single reactant\ninert electrode electrode that conducts electrons to and from the reactants in a half-cell but that is not itself oxidized or reduced\nlead acid battery rechargeable battery commonly used in automobiles; it typically comprises six galvanic cells based on Pb half-reactions in acidic solution\nlithium ion battery widely used rechargeable battery commonly used in portable electronic devices, based on lithium ion transfer between the anode and cathode\nNernst equation relating the potential of a redox system to its composition\nnickel-cadmium battery rechargeable battery based on Ni/Cd half-cells with applications similar to those of lithium ion batteries\nprimary cell nonrechargeable battery, suitable for single use only\nsacrificial anode electrode constructed from an easily oxidized metal, often magnesium or zinc, used to prevent corrosion of metal objects via cathodic protection\nsalt bridge tube filled with inert electrolyte solution\nsecondary cell battery designed to allow recharging\nstandard cell potential ( $E_{\\text {cell }}^{\\circ}$ ) the cell potential when all reactants and products are in their standard states (1 bar or 1 atm or gases; 1 M for solutes), usually at 298.15 K\nstandard electrode potential $\\left(\\left(E_{X}^{\\circ}\\right)\\right)$ electrode potential measured under standard conditions (1 bar or 1 atm for gases; 1 M for solutes) usually at 298.15 K\nstandard hydrogen electrode (SHE) half-cell based on hydrogen ion production, assigned a potential of exactly 0 V under standard state conditions, used as the universal reference for measuring electrode potential"}
{"id": 4473, "contents": "1769. Key Equations - \n$E_{\\text {cell }}^{\\circ}=E_{\\text {cathode }}^{\\circ}-E_{\\text {anode }}^{\\circ}$\n$E_{\\text {cell }}^{\\circ}=\\frac{R T}{n F} \\ln K$\n$E_{\\text {cell }}^{\\circ}=\\frac{0.0257 \\mathrm{~V}}{n} \\ln K=\\frac{0.0592 \\mathrm{~V}}{n} \\log K$\n$E_{\\text {cell }}=E_{\\text {cell }}^{\\circ}-\\frac{R T}{n F} \\ln Q \\quad$ (Nernst equation)\n$E_{\\text {cell }}=E^{\\circ}{ }_{\\text {cell }}-\\frac{0.0592 \\mathrm{~V}}{n} \\log Q$\n$\\Delta G=-n F E_{\\text {cell }}$\n$\\Delta G^{\\circ}=-n F E_{\\text {cell }}^{\\circ}$\n$w_{\\text {ele }}=w_{\\text {max }}=-n F E_{\\text {cell }}$\n$Q=I \\times t=n \\times F$"}
{"id": 4474, "contents": "1770. Summary - 1770.1. Review of Redox Chemistry\nRedox reactions are defined by changes in reactant oxidation numbers, and those most relevant to electrochemistry involve actual transfer of electrons. Aqueous phase redox processes often involve water or its characteristic ions, $\\mathrm{H}^{+}$and $\\mathrm{OH}^{-}$, as reactants in addition to the oxidant and reductant, and equations representing these reactions can be challenging to balance. The half-reaction method is a systematic approach to balancing such equations that involves separate treatment of the oxidation and reduction half-reactions."}
{"id": 4475, "contents": "1770. Summary - 1770.2. Galvanic Cells\nGalvanic cells are devices in which a spontaneous redox reaction occurs indirectly, with the oxidant and reductant redox couples contained in separate half-cells. Electrons are transferred from the reductant (in the anode half-cell) to the oxidant (in the cathode half-cell) through an external circuit, and inert solution phase ions are transferred between half-cells, through a salt bridge, to maintain charge neutrality. The construction and composition of a galvanic cell may be succinctly represented using chemical formulas and others symbols in the form of a cell schematic (cell notation)."}
{"id": 4476, "contents": "1770. Summary - 1770.3. Electrode and Cell Potentials\nThe property of potential, $E$, is the energy associated with the separation/transfer of charge. In electrochemistry, the potentials of cells and halfcells are thermodynamic quantities that reflect the driving force or the spontaneity of their redox processes. The cell potential of an electrochemical cell is the difference in between its cathode and anode. To permit easy sharing of half-cell potential data, the standard hydrogen electrode (SHE) is\nassigned a potential of exactly 0 V and used to define a single electrode potential for any given half-cell. The electrode potential of a half-cell, $E_{X}$, is the cell potential of said half-cell acting as a cathode when connected to a SHE acting as an anode. When the half-cell is operating under standard state conditions, its potential is the standard electrode potential, $E^{\\circ}{ }_{\\mathrm{X}}$. Standard electrode potentials reflect the relative oxidizing strength of the half-reaction's reactant, with stronger oxidants exhibiting larger (more positive) $E^{o}{ }_{X}$ values. Tabulations of standard electrode potentials may be used to compute standard cell potentials, $E^{o}$ cell, for many redox reactions. The arithmetic sign of a cell potential indicates the spontaneity of the cell reaction, with positive values for spontaneous reactions and negative values for nonspontaneous reactions (spontaneous in the reverse direction)."}
{"id": 4477, "contents": "1770. Summary - 1770.4. Potential, Free Energy, and Equilibrium\nPotential is a thermodynamic quantity reflecting the intrinsic driving force of a redox process, and it is directly related to the free energy change and equilibrium constant for the process. For redox processes taking place in electrochemical cells, the maximum (electrical) work done by the system is easily computed from the cell potential and the reaction stoichiometry and is equal to the free energy change for the process. The equilibrium constant for a redox reaction is logarithmically related to the reaction's cell potential, with larger (more positive) potentials indicating reactions with greater driving force that equilibrate when the reaction has proceeded far towards completion (large value of $K$ ). Finally, the potential of a redox process varies with the composition of the reaction mixture, being related to the reactions standard potential and the value of its reaction quotient, $Q$, as\ndescribed by the Nernst equation."}
{"id": 4478, "contents": "1770. Summary - 1770.5. Batteries and Fuel Cells\nGalvanic cells designed specifically to function as electrical power supplies are called batteries. A variety of both single-use batteries (primary cells) and rechargeable batteries (secondary cells) are commercially available to serve a variety of applications, with important specifications including voltage, size, and lifetime. Fuel cells, sometimes called flow batteries, are devices that harness the energy of spontaneous redox reactions normally associated with combustion processes. Like batteries, fuel cells enable the reaction's electron transfer via an external circuit, but they require continuous input of the redox reactants (fuel and oxidant) from an external reservoir. Fuel cells are typically much more efficient in converting the energy released by the reaction to useful work in comparison to internal combustion engines."}
{"id": 4479, "contents": "1770. Summary - 1770.6. Corrosion\nSpontaneous oxidation of metals by natural\nelectrochemical processes is called corrosion, familiar examples including the rusting of iron and the tarnishing of silver. Corrosion process involve the creation of a galvanic cell in which different sites on the metal object function as anode and cathode, with the corrosion taking place at the anodic site. Approaches to preventing corrosion of metals include use of a protective coating of zinc (galvanization) and the use of sacrificial anodes connected to the metal object (cathodic protection)."}
{"id": 4480, "contents": "1770. Summary - 1770.7. Electrolysis\nNonspontaneous redox processes may be forced to occur in electrochemical cells by the application of an appropriate potential using an external power source-a process known as electrolysis. Electrolysis is the basis for certain ore refining processes, the industrial production of many chemical commodities, and the electroplating of metal coatings on various products. Measurement of the current flow during electrolysis permits stoichiometric calculations."}
{"id": 4481, "contents": "1771. Exercises - 1771.1. Review of Redox Chemistry\n1. Identify each half-reaction below as either oxidation or reduction.\n(a) $\\mathrm{Fe}^{3+}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Fe}$\n(b) $\\mathrm{Cr} \\longrightarrow \\mathrm{Cr}^{3+}+3 \\mathrm{e}^{-}$\n(c) $\\mathrm{MnO}_{4}{ }^{2-} \\longrightarrow \\mathrm{MnO}_{4}^{-}+\\mathrm{e}^{-}$\n(d) $\\mathrm{Li}^{+}+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Li}$\n2. Identify each half-reaction below as either oxidation or reduction.\n(a) $\\mathrm{Cl}^{-} \\longrightarrow \\mathrm{Cl}_{2}$\n(b) $\\mathrm{Mn}^{2+} \\longrightarrow \\mathrm{MnO}_{2}$\n(c) $\\mathrm{H}_{2} \\longrightarrow \\mathrm{H}^{+}$\n(d) $\\mathrm{NO}_{3}{ }^{-} \\longrightarrow \\mathrm{NO}$\n3. Assuming each pair of half-reactions below takes place in an acidic solution, write a balanced equation for the overall reaction.\n(a) $\\mathrm{Ca} \\longrightarrow \\mathrm{Ca}^{2+}+2 \\mathrm{e}^{-}, \\mathrm{F}_{2}+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{~F}^{-}$\n(b) $\\mathrm{Li} \\longrightarrow \\mathrm{Li}^{+}+\\mathrm{e}^{-}, \\mathrm{Cl}_{2}+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Cl}^{-}$\n(c) $\\mathrm{Fe} \\longrightarrow \\mathrm{Fe}^{3+}+3 \\mathrm{e}^{-}, \\mathrm{Br}_{2}+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Br}^{-}$"}
{"id": 4482, "contents": "1771. Exercises - 1771.1. Review of Redox Chemistry\n(d) $\\mathrm{Ag} \\longrightarrow \\mathrm{Ag}^{+}+\\mathrm{e}^{-}, \\mathrm{MnO}_{4}^{-}+4 \\mathrm{H}^{+}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{MnO}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}$\n4. Balance the equations below assuming they occur in an acidic solution.\n(a) $\\mathrm{H}_{2} \\mathrm{O}_{2}+\\mathrm{Sn}^{2+} \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{Sn}^{4+}$\n(b) $\\mathrm{PbO}_{2}+\\mathrm{Hg} \\longrightarrow \\mathrm{Hg}_{2}{ }^{2+}+\\mathrm{Pb}^{2+}$\n(c) $\\mathrm{Al}+\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-} \\longrightarrow \\mathrm{Al}^{3+}+\\mathrm{Cr}^{3+}$\n5. Identify the oxidant and reductant of each reaction of the previous exercise.\n6. Balance the equations below assuming they occur in a basic solution.\n(a) $\\mathrm{SO}_{3}{ }^{2-}(a q)+\\mathrm{Cu}(\\mathrm{OH})_{2}(s) \\longrightarrow \\mathrm{SO}_{4}{ }^{2-}(a q)+\\mathrm{Cu}(\\mathrm{OH})(s)$\n(b) $\\mathrm{O}_{2}(g)+\\mathrm{Mn}(\\mathrm{OH})_{2}(s) \\longrightarrow \\mathrm{MnO}_{2}(s)$\n(c) $\\mathrm{NO}_{3}{ }^{-}(a q)+\\mathrm{H}_{2}(g) \\longrightarrow \\mathrm{NO}(g)$\n(d) $\\mathrm{Al}(s)+\\mathrm{CrO}_{4}{ }^{2-}(a q) \\longrightarrow \\mathrm{Al}(\\mathrm{OH})_{3}(s)+\\mathrm{Cr}(\\mathrm{OH})_{4}{ }^{-}(a q)$\n7. Identify the oxidant and reductant of each reaction of the previous exercise."}
{"id": 4483, "contents": "1771. Exercises - 1771.1. Review of Redox Chemistry\n7. Identify the oxidant and reductant of each reaction of the previous exercise.\n8. Why don't hydroxide ions appear in equations for half-reactions occurring in acidic solution?\n9. Why don't hydrogen ions appear in equations for half-reactions occurring in basic solution?\n10. Why must the charge balance in oxidation-reduction reactions?"}
{"id": 4484, "contents": "1771. Exercises - 1771.2. Galvanic Cells\n11. Write cell schematics for the following cell reactions, using platinum as an inert electrode as needed.\n(a) $\\mathrm{Mg}(s)+\\mathrm{Ni}^{2+}(a q) \\longrightarrow \\mathrm{Mg}^{2+}(a q)+\\mathrm{Ni}(s)$\n(b) $2 \\mathrm{Ag}^{+}(a q)+\\mathrm{Cu}(s) \\longrightarrow \\mathrm{Cu}^{2+}(a q)+2 \\mathrm{Ag}(s)$\n(c) $\\mathrm{Mn}(s)+\\mathrm{Sn}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q) \\longrightarrow \\mathrm{Mn}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+\\mathrm{Sn}(s)$\n(d) $3 \\mathrm{CuNO}_{3}(a q)+\\mathrm{Au}\\left(\\mathrm{NO}_{3}\\right)_{3}(a q) \\longrightarrow 3 \\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+\\mathrm{Au}(s)$\n12. Assuming the schematics below represent galvanic cells as written, identify the half-cell reactions occurring in each.\n(a) $\\mathrm{Mg}(s)\\left|\\mathrm{Mg}^{2+}{ }^{2+}(a q) \\| \\mathrm{Cu}^{2+}(a q)\\right| \\mathrm{Cu}(s)$\n(b) $\\mathrm{Ni}(s)\\left|\\mathrm{Ni}^{2+}(a q) \\| \\mathrm{Ag}^{+}(a q)\\right| \\operatorname{Ag}(s)$\n13. Write a balanced equation for the cell reaction of each cell in the previous exercise.\n14. Balance each reaction below, and write a cell schematic representing the reaction as it would occur in a galvanic cell.\n(a) $\\mathrm{Al}(s)+\\mathrm{Zr}^{4+}(a q) \\longrightarrow \\mathrm{Al}^{3+}(a q)+\\mathrm{Zr}(s)$"}
{"id": 4485, "contents": "1771. Exercises - 1771.2. Galvanic Cells\n(a) $\\mathrm{Al}(s)+\\mathrm{Zr}^{4+}(a q) \\longrightarrow \\mathrm{Al}^{3+}(a q)+\\mathrm{Zr}(s)$\n(b) $\\mathrm{Ag}^{+}(a q)+\\mathrm{NO}(g) \\longrightarrow \\mathrm{Ag}(s)+\\mathrm{NO}_{3}{ }^{-}(a q) \\quad$ (acidic solution)\n(c) $\\mathrm{SiO}_{3}{ }^{2-}(a q)+\\mathrm{Mg}(s) \\longrightarrow \\mathrm{Si}(s)+\\mathrm{Mg}(\\mathrm{OH})_{2}(s)$\n(basic solution)\n(d) $\\mathrm{ClO}_{3}^{-}(a q)+\\mathrm{MnO}_{2}(s) \\longrightarrow \\mathrm{Cl}^{-}(a q)+\\mathrm{MnO}_{4}^{-}(a q) \\quad$ (basic solution)\n15. Identify the oxidant and reductant in each reaction of the previous exercise.\n16. From the information provided, use cell notation to describe the following systems:\n(a) In one half-cell, a solution of $\\mathrm{Pt}\\left(\\mathrm{NO}_{3}\\right)_{2}$ forms Pt metal, while in the other half-cell, Cu metal goes into a\n$\\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}$ solution with all solute concentrations 1 M .\n(b) The cathode consists of a gold electrode in a $0.55 \\mathrm{M} \\mathrm{Au}\\left(\\mathrm{NO}_{3}\\right)_{3}$ solution and the anode is a magnesium electrode in $0.75 \\mathrm{M} \\mathrm{Mg}\\left(\\mathrm{NO}_{3}\\right)_{2}$ solution.\n(c) One half-cell consists of a silver electrode in a $1 \\mathrm{M} \\mathrm{AgNO}_{3}$ solution, and in the other half-cell, a copper electrode in $1 \\mathrm{MCu}\\left(\\mathrm{NO}_{3}\\right)_{2}$ is oxidized.\n17. Why is a salt bridge necessary in galvanic cells like the one in Figure 16.3?"}
{"id": 4486, "contents": "1771. Exercises - 1771.2. Galvanic Cells\n17. Why is a salt bridge necessary in galvanic cells like the one in Figure 16.3?\n18. An active (metal) electrode was found to gain mass as the oxidation-reduction reaction was allowed to proceed. Was the electrode an anode or a cathode? Explain.\n19. An active (metal) electrode was found to lose mass as the oxidation-reduction reaction was allowed to proceed. Was the electrode an anode or a cathode? Explain.\n20. The masses of three electrodes (A, B, and C), each from three different galvanic cells, were measured before and after the cells were allowed to pass current for a while. The mass of electrode A increased, that of electrode B was unchanged, and that of electrode C decreased. Identify each electrode as active or inert, and note (if possible) whether it functioned as anode or cathode."}
{"id": 4487, "contents": "1771. Exercises - 1771.3. Electrode and Cell Potentials\n21. Calculate the standard cell potential for each reaction below, and note whether the reaction is spontaneous under standard state conditions.\n(a) $\\mathrm{Mg}(s)+\\mathrm{Ni}^{2+}(a q) \\longrightarrow \\mathrm{Mg}^{2+}(a q)+\\mathrm{Ni}(s)$\n(b) $2 \\mathrm{Ag}^{+}(a q)+\\mathrm{Cu}(s) \\longrightarrow \\mathrm{Cu}^{2+}(a q)+2 \\mathrm{Ag}(s)$\n(c) $\\mathrm{Mn}(s)+\\mathrm{Sn}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q) \\longrightarrow \\mathrm{Mn}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+\\mathrm{Sn}(s)$\n(d) $3 \\mathrm{Fe}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+\\mathrm{Au}\\left(\\mathrm{NO}_{3}\\right)_{3}(a q) \\longrightarrow 3 \\mathrm{Fe}\\left(\\mathrm{NO}_{3}\\right)_{3}(a q)+\\mathrm{Au}(s)$\n22. Calculate the standard cell potential for each reaction below, and note whether the reaction is spontaneous under standard state conditions.\n(a) $\\mathrm{Mn}(s)+\\mathrm{Ni}^{2+}(a q) \\longrightarrow \\mathrm{Mn}^{2+}(a q)+\\mathrm{Ni}(s)$\n(b) $3 \\mathrm{Cu}^{2+}(a q)+2 \\mathrm{Al}(s) \\longrightarrow 2 \\mathrm{Al}^{3+}(a q)+3 \\mathrm{Cu}(s)$\n(c) $\\mathrm{Na}(s)+\\mathrm{LiNO}_{3}(a q) \\longrightarrow \\mathrm{NaNO}_{3}(a q)+\\mathrm{Li}(s)$\n(d) $\\mathrm{Ca}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+\\mathrm{Ba}(s) \\longrightarrow \\mathrm{Ba}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+\\mathrm{Ca}(s)$"}
{"id": 4488, "contents": "1771. Exercises - 1771.3. Electrode and Cell Potentials\n23. Write the balanced cell reaction for the cell schematic below, calculate the standard cell potential, and note whether the reaction is spontaneous under standard state conditions.\n$\\mathrm{Cu}(s)\\left|\\mathrm{Cu}^{2+}(a q) \\| \\mathrm{Au}^{3+}(a q)\\right| \\mathrm{Au}(s)$\n24. Determine the cell reaction and standard cell potential at $25^{\\circ} \\mathrm{C}$ for a cell made from a cathode half-cell consisting of a silver electrode in $1 M$ silver nitrate solution and an anode half-cell consisting of a zinc electrode in 1 M zinc nitrate. Is the reaction spontaneous at standard conditions?\n25. Determine the cell reaction and standard cell potential at $25^{\\circ} \\mathrm{C}$ for a cell made from an anode half-cell containing a cadmium electrode in 1 M cadmium nitrate and a cathode half-cell consisting of an aluminum electrode in 1 M aluminum nitrate solution. Is the reaction spontaneous at standard conditions?\n26. Write the balanced cell reaction for the cell schematic below, calculate the standard cell potential, and note whether the reaction is spontaneous under standard state conditions.\n$\\operatorname{Pt}(s)\\left|\\mathrm{H}_{2}(g)\\right| \\mathrm{H}^{+}(a q) \\| \\mathrm{Br}_{2}(a q), \\mathrm{Br}^{-}(a q) \\mid \\operatorname{Pt}(s)$"}
{"id": 4489, "contents": "1771. Exercises - 1771.4. Potential, Free Energy, and Equilibrium\n27. For each pair of standard cell potential and electron stoichiometry values below, calculate a corresponding standard free energy change (kJ).\n(a) $0.000 \\mathrm{~V}, \\mathrm{n}=2$\n(b) $+0.434 \\mathrm{~V}, \\mathrm{n}=2$\n(c) $-2.439 \\mathrm{~V}, \\mathrm{n}=1$\n28. For each pair of standard free energy change and electron stoichiometry values below, calculate a corresponding standard cell potential.\n(a) $12 \\mathrm{~kJ} / \\mathrm{mol}, \\mathrm{n}=3$\n(b) $-45 \\mathrm{~kJ} / \\mathrm{mol}, \\mathrm{n}=1$\n29. Determine the standard cell potential and the cell potential under the stated conditions for the electrochemical reactions described here. State whether each is spontaneous or nonspontaneous under each set of conditions at 298.15 K .\n(a) $\\mathrm{Hg}(l)+\\mathrm{S}^{2-}(a q, 0.10 M)+2 \\mathrm{Ag}^{+}(a q, 0.25 M) \\longrightarrow 2 \\mathrm{Ag}(s)+\\mathrm{HgS}(s)$\n(b) The cell made from an anode half-cell consisting of an aluminum electrode in 0.015 M aluminum nitrate solution and a cathode half-cell consisting of a nickel electrode in 0.25 M nickel(II) nitrate solution.\n(c) The cell comprised of a half-cell in which aqueous bromine $(1.0 \\mathrm{M})$ is being oxidized to bromide ion $(0.11 M)$ and a half-cell in which $\\mathrm{Al}^{3+}(0.023 M)$ is being reduced to aluminum metal.\n30. Determine $\\Delta G$ and $\\Delta G^{\\circ}$ for each of the reactions in the previous problem.\n31. Use the data in Appendix $L$ to calculate equilibrium constants for the following reactions. Assume 298.15 K if no temperature is given.\n(a) $\\mathrm{AgCl}(s) \\rightleftharpoons \\mathrm{Ag}^{+}(a q)+\\mathrm{Cl}^{-}(a q)$"}
{"id": 4490, "contents": "1771. Exercises - 1771.4. Potential, Free Energy, and Equilibrium\n(a) $\\mathrm{AgCl}(s) \\rightleftharpoons \\mathrm{Ag}^{+}(a q)+\\mathrm{Cl}^{-}(a q)$\n(b) $\\mathrm{CdS}(s) \\rightleftharpoons \\mathrm{Cd}^{2+}(a q)+\\mathrm{S}^{2-}(a q) \\quad$ at 377 K\n(c) $\\mathrm{Hg}^{2+}(a q)+4 \\mathrm{Br}^{-}(a q) \\rightleftharpoons\\left[\\mathrm{HgBr}_{4}\\right]^{2-}(a q)$\n(d) $\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}^{+}(a q)+\\mathrm{OH}^{-}(a q) \\quad$ at $25^{\\circ} \\mathrm{C}$"}
{"id": 4491, "contents": "1771. Exercises - 1771.5. Batteries and Fuel Cells\n32. Consider a battery made from one half-cell that consists of a copper electrode in $1 \\mathrm{MCuSO}_{4}$ solution and another half-cell that consists of a lead electrode in $1 \\mathrm{M} \\mathrm{Pb}\\left(\\mathrm{NO}_{3}\\right)_{2}$ solution.\n(a) What is the standard cell potential for the battery?\n(b) What are the reactions at the anode, cathode, and the overall reaction?\n(c) Most devices designed to use dry-cell batteries can operate between 1.0 and 1.5 V. Could this cell be used to make a battery that could replace a dry-cell battery? Why or why not.\n(d) Suppose sulfuric acid is added to the half-cell with the lead electrode and some $\\mathrm{PbSO}_{4}(s)$ forms. Would the cell potential increase, decrease, or remain the same?\n33. Consider a battery with the overall reaction: $\\mathrm{Cu}(s)+2 \\mathrm{Ag}^{+}(a q) \\longrightarrow 2 \\mathrm{Ag}(s)+\\mathrm{Cu}^{2+}(a q)$.\n(a) What is the reaction at the anode and cathode?\n(b) A battery is \"dead\" when its cell potential is zero. What is the value of $Q$ when this battery is dead?\n(c) If a particular dead battery was found to have $\\left[\\mathrm{Cu}^{2+}\\right]=0.11 \\mathrm{M}$, what was the concentration of silver ion?\n34. Why do batteries go dead, but fuel cells do not?\n35. Use the Nernst equation to explain the drop in voltage observed for some batteries as they discharge.\n36. Using the information thus far in this chapter, explain why battery-powered electronics perform poorly in low temperatures."}
{"id": 4492, "contents": "1771. Exercises - 1771.6. Corrosion\n37. Which member of each pair of metals is more likely to corrode (oxidize)?\n(a) Mg or Ca\n(b) Au or Hg\n(c) Fe or Zn\n(d) Ag or Pt\n38. Consider the following metals: $\\mathrm{Ag}, \\mathrm{Au}, \\mathrm{Mg}, \\mathrm{Ni}$, and Zn . Which of these metals could be used as a sacrificial anode in the cathodic protection of an underground steel storage tank? Steel is an alloy composed mostly of iron, so use -0.447 V as the standard reduction potential for steel.\n39. Aluminum $\\left(E_{\\mathrm{Al}^{3+} / \\mathrm{Al}}^{\\circ}=-2.07 \\mathrm{~V}\\right)$ is more easily oxidized than iron $\\left(E_{\\mathrm{Fe}^{3+} / \\mathrm{Fe}}^{\\circ}=-0.477 \\mathrm{~V}\\right)$, and yet when both are exposed to the environment, untreated aluminum has very good corrosion resistance while the corrosion resistance of untreated iron is poor. What might explain this observation?\n40. If a sample of iron and a sample of zinc come into contact, the zinc corrodes but the iron does not. If a sample of iron comes into contact with a sample of copper, the iron corrodes but the copper does not. Explain this phenomenon.\n41. Suppose you have three different metals, A, B, and C. When metals A and B come into contact, B corrodes and A does not corrode. When metals A and C come into contact, A corrodes and C does not corrode. Based on this information, which metal corrodes and which metal does not corrode when B and C come into contact?\n42. Why would a sacrificial anode made of lithium metal be a bad choice"}
{"id": 4493, "contents": "1771. Exercises - 1771.7. Electrolysis\n43. If a 2.5 A current flows through a circuit for 35 minutes, how many coulombs of charge moved through the circuit?\n44. For the scenario in the previous question, how many electrons moved through the circuit?\n45. Write the half-reactions and cell reaction occurring during electrolysis of each molten salt below.\n(a) $\\mathrm{CaCl}_{2}$\n(b) LiH\n(c) $\\mathrm{AlCl}_{3}$\n(d) $\\mathrm{CrBr}_{3}$\n46. What mass of each product is produced in each of the electrolytic cells of the previous problem if a total charge of $3.33 \\times 10^{5} \\mathrm{C}$ passes through each cell?\n47. How long would it take to reduce 1 mole of each of the following ions using the current indicated?\n(a) $\\mathrm{Al}^{3+}, 1.234 \\mathrm{~A}$\n(b) $\\mathrm{Ca}^{2+}, 22.2 \\mathrm{~A}$\n(c) $\\mathrm{Cr}^{5+}, 37.45 \\mathrm{~A}$\n(d) $\\mathrm{Au}^{3+}, 3.57 \\mathrm{~A}$\n48. A current of 2.345 A passes through the cell shown in Figure 16.19 for 45 minutes. What is the volume of the hydrogen collected at room temperature if the pressure is exactly 1 atm ? (Hint: Is hydrogen the only gas present above the water?)\n49. An irregularly shaped metal part made from a particular alloy was galvanized with zinc using a $\\mathrm{Zn}\\left(\\mathrm{NO}_{3}\\right)_{2}$ solution. When a current of 2.599 A was used, it took exactly 1 hour to deposit a 0.01123 -mm layer of zinc on the part. What was the total surface area of the part? The density of zinc is $7.140 \\mathrm{~g} / \\mathrm{cm}^{3}$.\n\n79816 \u2022 Exercises"}
{"id": 4494, "contents": "1772. CHAPTER 17 Kinetics - \nFigure 17.1 An agama lizard basks in the sun. As its body warms, the chemical reactions of its metabolism speed up."}
{"id": 4495, "contents": "1773. CHAPTER OUTLINE - 1773.1. Chemical Reaction Rates\n17.2 Factors Affecting Reaction Rates"}
{"id": 4496, "contents": "1773. CHAPTER OUTLINE - 1773.2. Rate Laws\n17.4 Integrated Rate Laws\n17.5 Collision Theory\n17.6 Reaction Mechanisms\n17.7 Catalysis\n\nINTRODUCTION The lizard in the photograph is not simply enjoying the sunshine or working on its tan. The heat from the sun's rays is critical to the lizard's survival. A warm lizard can move faster than a cold one because the chemical reactions that allow its muscles to move occur more rapidly at higher temperatures. A cold lizard is a slower lizard and an easier meal for predators.\n\nFrom baking a cake to determining the useful lifespan of a bridge, rates of chemical reactions play important roles in our understanding of processes that involve chemical changes. Two questions are typically posed when planning to carry out a chemical reaction. The first is: \"Will the reaction produce the desired products in useful quantities?\" The second question is: \"How rapidly will the reaction occur?\" A third question is often asked when investigating reactions in greater detail: \"What specific molecular-level processes take place as the reaction occurs?\" Knowing the answer to this question is of practical importance when the yield or rate of a reaction needs to be controlled.\n\nThe study of chemical kinetics concerns the second and third questions-that is, the rate at which a reaction yields products and the molecular-scale means by which a reaction occurs. This chapter examines the factors\nthat influence the rates of chemical reactions, the mechanisms by which reactions proceed, and the quantitative techniques used to describe the rates at which reactions occur."}
{"id": 4497, "contents": "1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates\nLEARNING OBJECTIVES\nBy the end of this section, you will be able to:\n\n- Define chemical reaction rate\n- Derive rate expressions from the balanced equation for a given chemical reaction\n- Calculate reaction rates from experimental data\n\nA rate is a measure of how some property varies with time. Speed is a familiar rate that expresses the distance traveled by an object in a given amount of time. Wage is a rate that represents the amount of money earned by a person working for a given amount of time. Likewise, the rate of a chemical reaction is a measure of how much reactant is consumed, or how much product is produced, by the reaction in a given amount of time.\n\nThe rate of reaction is the change in the amount of a reactant or product per unit time. Reaction rates are therefore determined by measuring the time dependence of some property that can be related to reactant or product amounts. Rates of reactions that consume or produce gaseous substances, for example, are conveniently determined by measuring changes in volume or pressure. For reactions involving one or more colored substances, rates may be monitored via measurements of light absorption. For reactions involving aqueous electrolytes, rates may be measured via changes in a solution's conductivity.\n\nFor reactants and products in solution, their relative amounts (concentrations) are conveniently used for purposes of expressing reaction rates. For example, the concentration of hydrogen peroxide, $\\mathrm{H}_{2} \\mathrm{O}_{2}$, in an aqueous solution changes slowly over time as it decomposes according to the equation:\n\n$$\n2 \\mathrm{H}_{2} \\mathrm{O}_{2}(a q) \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{O}_{2}(g)\n$$\n\nThe rate at which the hydrogen peroxide decomposes can be expressed in terms of the rate of change of its concentration, as shown here:"}
{"id": 4498, "contents": "1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates\nThe rate at which the hydrogen peroxide decomposes can be expressed in terms of the rate of change of its concentration, as shown here:\n\n$$\n\\text { rate of decomposition of } \\begin{aligned}\n\\mathrm{H}_{2} \\mathrm{O}_{2} & =-\\frac{\\text { change in concentration of reactant }}{\\text { time interval }} \\\\\n& =-\\frac{\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]_{t_{2}}-\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]_{t_{1}}}{t_{2}-t_{1}} \\\\\n& =-\\frac{\\Delta\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]}{\\Delta t}\n\\end{aligned}\n$$\n\nThis mathematical representation of the change in species concentration over time is the rate expression for the reaction. The brackets indicate molar concentrations, and the symbol delta ( $\\Delta$ ) indicates \"change in.\" Thus, $\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]_{t_{1}}$ represents the molar concentration of hydrogen peroxide at some time $t_{1}$; likewise, $\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]_{t_{2}}$ represents the molar concentration of hydrogen peroxide at a later time $t_{2}$; and $\\Delta\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]$ represents the change in molar concentration of hydrogen peroxide during the time interval $\\Delta t$ (that is, $t_{2}-t_{1}$ ). Since the reactant concentration decreases as the reaction proceeds, $\\Delta\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]$ is a negative quantity. Reaction rates are, by convention, positive quantities, and so this negative change in concentration is multiplied by -1 . Figure 17.2 provides an example of data collected during the decomposition of $\\mathrm{H}_{2} \\mathrm{O}_{2}$."}
{"id": 4499, "contents": "1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates\n| Time (h) | $\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]\\left(\\mathrm{mol} \\mathrm{L}^{-1}\\right)$ | $\\Delta\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]\\left(\\mathrm{mol} \\mathrm{L}^{-1}\\right)$ | $\\Delta t(\\mathrm{~h})$ | $\\begin{aligned} & \\text { Rate of Decomposition, } \\\\ & \\left(\\mathrm{mol} \\mathrm{~L}^{-1} \\mathrm{~h}^{-1}\\right) \\end{aligned}$ |\n| :---: | :---: | :---: | :---: | :---: |\n| 0.00 | 1.000 |  |  |  |\n|  |  | $\\geq-0.500$ | 6.00 | 0.0833 |\n| 6.00 | $0.500=$ |  |  |  |\n|  |  | $\\geq-0.250$ | 6.00 | 0.0417 |\n| 12.00 | $0.250=$ |  |  |  |\n|  | - | $\\geq-0.125$ | 6.00 | 0.0208 |\n| 18.00 | $0.125=$ |  |  |  |\n| 24.00 | - | $\\geq-0.062$ | 6.00 | 0.010 |\n\nFIGURE 17.2 The rate of decomposition of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ in an aqueous solution decreases as the concentration of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ decreases.\n\nTo obtain the tabulated results for this decomposition, the concentration of hydrogen peroxide was measured every 6 hours over the course of a day at a constant temperature of $40^{\\circ} \\mathrm{C}$. Reaction rates were computed for each time interval by dividing the change in concentration by the corresponding time increment, as shown here for the first 6-hour period:"}
{"id": 4500, "contents": "1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates\n$$\n\\frac{-\\Delta\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]}{\\Delta t}=\\frac{-(0.500 \\mathrm{~mol} / \\mathrm{L}-1.000 \\mathrm{~mol} / \\mathrm{L})}{(6.00 \\mathrm{~h}-0.00 \\mathrm{~h})}=0.0833 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~h}^{-1}\n$$\n\nNotice that the reaction rates vary with time, decreasing as the reaction proceeds. Results for the last 6-hour period yield a reaction rate of:\n\n$$\n\\frac{-\\Delta\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]}{\\Delta t}=\\frac{-(0.0625 \\mathrm{~mol} / \\mathrm{L}-0.125 \\mathrm{~mol} / \\mathrm{L})}{(24.00 \\mathrm{~h}-18.00 \\mathrm{~h})}=0.010 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~h}^{-1}\n$$"}
{"id": 4501, "contents": "1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates\nThis behavior indicates the reaction continually slows with time. Using the concentrations at the beginning and end of a time period over which the reaction rate is changing results in the calculation of an average rate for the reaction over this time interval. At any specific time, the rate at which a reaction is proceeding is known as its instantaneous rate. The instantaneous rate of a reaction at \"time zero,\" when the reaction commences, is its initial rate. Consider the analogy of a car slowing down as it approaches a stop sign. The vehicle's initial rate-analogous to the beginning of a chemical reaction-would be the speedometer reading at the moment the driver begins pressing the brakes ( $t_{0}$ ). A few moments later, the instantaneous rate at a specific moment-call it $t_{1}$-would be somewhat slower, as indicated by the speedometer reading at that point in time. As time passes, the instantaneous rate will continue to fall until it reaches zero, when the car (or reaction) stops. Unlike instantaneous speed, the car's average speed is not indicated by the speedometer; but it can be calculated as the ratio of the distance traveled to the time required to bring the vehicle to a complete stop ( $\\Delta t$ ). Like the decelerating car, the average rate of a chemical reaction will fall somewhere between its initial and final rates.\n\nThe instantaneous rate of a reaction may be determined one of two ways. If experimental conditions permit the measurement of concentration changes over very short time intervals, then average rates computed as described earlier provide reasonably good approximations of instantaneous rates. Alternatively, a graphical procedure may be used that, in effect, yields the results that would be obtained if short time interval measurements were possible. In a plot of the concentration of hydrogen peroxide against time, the instantaneous rate of decomposition of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ at any time $t$ is given by the slope of a straight line that is tangent to the curve at that time (Figure 17.3). These tangent line slopes may be evaluated using calculus, but the procedure for doing so is beyond the scope of this chapter."}
{"id": 4502, "contents": "1773. CHAPTER OUTLINE - 1773.3. Chemical Reaction Rates\nFIGURE 17.3 This graph shows a plot of concentration versus time for a 1.000 M solution of $\\mathrm{H}_{2} \\mathrm{O}_{2}$. The rate at any time is equal to the negative of the slope of a line tangent to the curve at that time. Tangents are shown at $t=0 \\mathrm{~h}$ (\"initial rate\") and at $t=12 \\mathrm{~h}$ (\"instantaneous rate\" at 12 h )."}
{"id": 4503, "contents": "1775. Reaction Rates in Analysis: Test Strips for Urinalysis - \nPhysicians often use disposable test strips to measure the amounts of various substances in a patient's urine (Figure 17.4). These test strips contain various chemical reagents, embedded in small pads at various locations along the strip, which undergo changes in color upon exposure to sufficient concentrations of specific substances. The usage instructions for test strips often stress that proper read time is critical for optimal results. This emphasis on read time suggests that kinetic aspects of the chemical reactions occurring on the test strip are important considerations.\n\nThe test for urinary glucose relies on a two-step process represented by the chemical equations shown here:\n\n$$\n\\begin{gathered}\n\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}+\\mathrm{O}_{2} \\xrightarrow{\\text { catalyst }} \\mathrm{C}_{6} \\mathrm{H}_{10} \\mathrm{O}_{6}+\\mathrm{H}_{2} \\mathrm{O}_{2} \\\\\n2 \\mathrm{H}_{2} \\mathrm{O}_{2}+2 \\mathrm{I}^{-} \\xrightarrow{\\text { catalyst }} \\mathrm{I}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{O}_{2}\n\\end{gathered}\n$$\n\nThe first equation depicts the oxidation of glucose in the urine to yield glucolactone and hydrogen peroxide. The hydrogen peroxide produced subsequently oxidizes colorless iodide ion to yield brown iodine, which may be visually detected. Some strips include an additional substance that reacts with iodine to produce a more distinct color change.\n\nThe two test reactions shown above are inherently very slow, but their rates are increased by special enzymes embedded in the test strip pad. This is an example of catalysis, a topic discussed later in this chapter. A typical glucose test strip for use with urine requires approximately 30 seconds for completion of the color-forming reactions. Reading the result too soon might lead one to conclude that the glucose concentration of the urine sample is lower than it actually is (a false-negative result). Waiting too long to assess the color change can lead to a false positive due to the slower (not catalyzed) oxidation of iodide ion by other substances found in urine."}
{"id": 4504, "contents": "1775. Reaction Rates in Analysis: Test Strips for Urinalysis - \nFIGURE 17.4 Test strips are commonly used to detect the presence of specific substances in a person's urine. Many test strips have several pads containing various reagents to permit the detection of multiple substances on a single strip. (credit: Iqbal Osman)"}
{"id": 4505, "contents": "1776. Relative Rates of Reaction - \nThe rate of a reaction may be expressed as the change in concentration of any reactant or product. For any given reaction, these rate expressions are all related simply to one another according to the reaction stoichiometry. The rate of the general reaction\n\n$$\n\\mathrm{aA} \\longrightarrow \\mathrm{bB}\n$$\n\ncan be expressed in terms of the decrease in the concentration of A or the increase in the concentration of B. These two rate expressions are related by the stoichiometry of the reaction:\n\n$$\n\\text { rate }=-\\left(\\frac{1}{\\mathrm{a}}\\right)\\left(\\frac{\\Delta \\mathrm{A}}{\\Delta t}\\right)=\\left(\\frac{1}{\\mathrm{~b}}\\right)\\left(\\frac{\\Delta \\mathrm{B}}{\\Delta t}\\right)\n$$\n\nConsider the reaction represented by the following equation:\n\n$$\n2 \\mathrm{NH}_{3}(g) \\longrightarrow \\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g)\n$$\n\nThe relation between the reaction rates expressed in terms of nitrogen production and ammonia consumption, for example, is:\n\n$$\n-\\frac{\\Delta \\mathrm{mol} \\mathrm{NH}_{3}}{\\Delta t} \\times \\frac{1 \\mathrm{~mol} \\mathrm{~N}_{2}}{2 \\mathrm{~mol} \\mathrm{NH}_{3}}=\\frac{\\Delta \\mathrm{mol} \\mathrm{~N}_{2}}{\\Delta t}\n$$\n\nThis may be represented in an abbreviated format by omitting the units of the stoichiometric factor:\n\n$$\n-\\frac{1}{2} \\frac{\\Delta \\mathrm{~mol} \\mathrm{NH}}{3} \\text { } \\quad \\Delta t \\quad=\\frac{\\Delta \\mathrm{mol} \\mathrm{~N}_{2}}{\\Delta t}\n$$\n\nNote that a negative sign has been included as a factor to account for the opposite signs of the two amount changes (the reactant amount is decreasing while the product amount is increasing). For homogeneous reactions, both the reactants and products are present in the same solution and thus occupy the same volume, so the molar amounts may be replaced with molar concentrations:"}
{"id": 4506, "contents": "1776. Relative Rates of Reaction - \n$$\n-\\frac{1}{2} \\frac{\\Delta\\left[\\mathrm{NH}_{3}\\right]}{\\Delta t}=\\frac{\\Delta\\left[\\mathrm{N}_{2}\\right]}{\\Delta t}\n$$\n\nSimilarly, the rate of formation of $\\mathrm{H}_{2}$ is three times the rate of formation of $\\mathrm{N}_{2}$ because three moles of $\\mathrm{H}_{2}$ are produced for each mole of $\\mathrm{N}_{2}$ produced.\n\n$$\n\\frac{1}{3} \\frac{\\Delta\\left[\\mathrm{H}_{2}\\right]}{\\Delta t}=\\frac{\\Delta\\left[\\mathrm{N}_{2}\\right]}{\\Delta t}\n$$\n\nFigure 17.5 illustrates the change in concentrations over time for the decomposition of ammonia into nitrogen\nand hydrogen at $1100^{\\circ} \\mathrm{C}$. Slopes of the tangent lines at $t=500 \\mathrm{~s}$ show that the instantaneous rates derived from all three species involved in the reaction are related by their stoichiometric factors. The rate of hydrogen production, for example, is observed to be three times greater than that for nitrogen production:\n\n$$\n\\frac{2.91 \\times 10^{-6} \\mathrm{M} / \\mathrm{s}}{9.70 \\times 10^{-7} \\mathrm{M} / \\mathrm{s}} \\approx 3\n$$\n\n\n\nFIGURE 17.5 Changes in concentrations of the reactant and products for the reaction $2 \\mathrm{NH}_{3} \\longrightarrow \\mathrm{~N}_{2}+3 \\mathrm{H}_{2}$. The rates of change of the three concentrations are related by the reaction stoichiometry, as shown by the different slopes of the tangents at $t=500 \\mathrm{~s}$.\n\nEXAMPLE 17.1"}
{"id": 4507, "contents": "1777. Expressions for Relative Reaction Rates - \nThe first step in the production of nitric acid is the combustion of ammonia:\n\n$$\n4 \\mathrm{NH}_{3}(g)+5 \\mathrm{O}_{2}(g) \\longrightarrow 4 \\mathrm{NO}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(g)\n$$\n\nWrite the equations that relate the rates of consumption of the reactants and the rates of formation of the products."}
{"id": 4508, "contents": "1778. Solution - \nConsidering the stoichiometry of this homogeneous reaction, the rates for the consumption of reactants and formation of products are:\n\n$$\n-\\frac{1}{4} \\frac{\\Delta\\left[\\mathrm{NH}_{3}\\right]}{\\Delta t}=-\\frac{1}{5} \\frac{\\Delta\\left[\\mathrm{O}_{2}\\right]}{\\Delta t}=\\frac{1}{4} \\frac{\\Delta[\\mathrm{NO}]}{\\Delta t}=\\frac{1}{6} \\frac{\\Delta\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]}{\\Delta t}\n$$\n\nCheck Your Learning\nThe rate of formation of $\\mathrm{Br}_{2}$ is $6.0 \\times 10^{-6} \\mathrm{~mol} / \\mathrm{L} / \\mathrm{s}$ in a reaction described by the following net ionic equation:\n\n$$\n5 \\mathrm{Br}^{-}+\\mathrm{BrO}_{3}^{-}+6 \\mathrm{H}^{+} \\longrightarrow 3 \\mathrm{Br}_{2}+3 \\mathrm{H}_{2} \\mathrm{O}\n$$\n\nWrite the equations that relate the rates of consumption of the reactants and the rates of formation of the products."}
{"id": 4509, "contents": "1779. Answer: - \n$-\\frac{1}{5} \\frac{\\Delta\\left[\\mathrm{Br}^{-}\\right]}{\\Delta t}=-\\frac{\\Delta\\left[\\mathrm{BrO}_{3}{ }^{-}\\right]}{\\Delta t}=-\\frac{1}{6} \\frac{\\Delta\\left[\\mathrm{H}^{+}\\right]}{\\Delta t}=\\frac{1}{3} \\frac{\\Delta\\left[\\mathrm{Br}_{2}\\right]}{\\Delta t}=\\frac{1}{3} \\frac{\\Delta\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]}{\\Delta t}$"}
{"id": 4510, "contents": "1781. Reaction Rate Expressions for Decomposition of $\\mathbf{H}_{\\mathbf{2}} \\mathbf{O}_{\\mathbf{2}}$ - \nThe graph in Figure 17.3 shows the rate of the decomposition of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ over time:\n\n$$\n2 \\mathrm{H}_{2} \\mathrm{O}_{2} \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{O}_{2}\n$$\n\nBased on these data, the instantaneous rate of decomposition of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ at $t=11.1 \\mathrm{~h}$ is determined to be $3.20 \\times 10^{-2} \\mathrm{~mol} / \\mathrm{L} / \\mathrm{h}$, that is:\n\n$$\n-\\frac{\\Delta\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]}{\\Delta t}=3.20 \\times 10^{-2} \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~h}^{-1}\n$$\n\nWhat is the instantaneous rate of production of $\\mathrm{H}_{2} \\mathrm{O}$ and $\\mathrm{O}_{2}$ ?"}
{"id": 4511, "contents": "1782. Solution - \nThe reaction stoichiometry shows that\n\n$$\n-\\frac{1}{2} \\frac{\\Delta\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]}{\\Delta t}=\\frac{1}{2} \\frac{\\Delta\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]}{\\Delta t}=\\frac{\\Delta\\left[\\mathrm{O}_{2}\\right]}{\\Delta t}\n$$\n\nTherefore:\n\n$$\n\\frac{1}{2} \\times 3.20 \\times 10^{-2} \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~h}^{-1}=\\frac{\\Delta\\left[\\mathrm{O}_{2}\\right]}{\\Delta t}\n$$\n\nand\n\n$$\n\\frac{\\Delta\\left[\\mathrm{O}_{2}\\right]}{\\Delta t}=1.60 \\times 10^{-2} \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~h}^{-1}\n$$"}
{"id": 4512, "contents": "1783. Check Your Learning - \nIf the rate of decomposition of ammonia, $\\mathrm{NH}_{3}$, at 1150 K is $2.10 \\times 10^{-6} \\mathrm{~mol} / \\mathrm{L} / \\mathrm{s}$, what is the rate of production of nitrogen and hydrogen?"}
{"id": 4513, "contents": "1784. Answer: - \n$1.05 \\times 10^{-6} \\mathrm{~mol} / \\mathrm{L} / \\mathrm{s}, \\mathrm{N}_{2}$ and $3.15 \\times 10^{-6} \\mathrm{~mol} / \\mathrm{L} / \\mathrm{s}, \\mathrm{H}_{2}$."}
{"id": 4514, "contents": "1785. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the effects of chemical nature, physical state, temperature, concentration, and catalysis on reaction rates\n\nThe rates at which reactants are consumed and products are formed during chemical reactions vary greatly. Five factors typically affecting the rates of chemical reactions will be explored in this section: the chemical nature of the reacting substances, the state of subdivision (one large lump versus many small particles) of the reactants, the temperature of the reactants, the concentration of the reactants, and the presence of a catalyst."}
{"id": 4515, "contents": "1786. The Chemical Nature of the Reacting Substances - \nThe rate of a reaction depends on the nature of the participating substances. Reactions that appear similar may have different rates under the same conditions, depending on the identity of the reactants. For example, when small pieces of the metals iron and sodium are exposed to air, the sodium reacts completely with air\novernight, whereas the iron is barely affected. The active metals calcium and sodium both react with water to form hydrogen gas and a base. Yet calcium reacts at a moderate rate, whereas sodium reacts so rapidly that the reaction is almost explosive."}
{"id": 4516, "contents": "1787. The Physical States of the Reactants - \nA chemical reaction between two or more substances requires intimate contact between the reactants. When reactants are in different physical states, or phases (solid, liquid, gaseous, dissolved), the reaction takes place only at the interface between the phases. Consider the heterogeneous reaction between a solid phase and either a liquid or gaseous phase. Compared with the reaction rate for large solid particles, the rate for smaller particles will be greater because the surface area in contact with the other reactant phase is greater. For example, large pieces of iron react more slowly with acids than they do with finely divided iron powder (Figure 17.6). Large pieces of wood smolder, smaller pieces burn rapidly, and saw dust burns explosively.\n\n(a)\n\n(b)\n\nFIGURE 17.6 (a) Iron powder reacts rapidly with dilute hydrochloric acid and produces bubbles of hydrogen gas: $2 \\mathrm{Fe}(s)+6 \\mathrm{HCl}(a q) \\longrightarrow 2 \\mathrm{FeCl}_{3}(a q)+3 \\mathrm{H}_{2}(g)$. (b) An iron nail reacts more slowly because the surface area exposed to the acid is much less."}
{"id": 4517, "contents": "1788. LINK TO LEARNING - \nWatch this video (http://openstax.org/l/16cesium) to see the reaction of cesium with water in slow motion and a discussion of how the state of reactants and particle size affect reaction rates."}
{"id": 4518, "contents": "1789. Temperature of the Reactants - \nChemical reactions typically occur faster at higher temperatures. Food can spoil quickly when left on the kitchen counter. However, the lower temperature inside of a refrigerator slows that process so that the same food remains fresh for days. Gas burners, hot plates, and ovens are often used in the laboratory to increase the speed of reactions that proceed slowly at ordinary temperatures. For many chemical processes, reaction rates are approximately doubled when the temperature is raised by $10^{\\circ} \\mathrm{C}$."}
{"id": 4519, "contents": "1790. Concentrations of the Reactants - \nThe rates of many reactions depend on the concentrations of the reactants. Rates usually increase when the concentration of one or more of the reactants increases. For example, calcium carbonate $\\left(\\mathrm{CaCO}_{3}\\right)$ deteriorates as a result of its reaction with the pollutant sulfur dioxide. The rate of this reaction depends on the amount of sulfur dioxide in the air (Figure 17.7). An acidic oxide, sulfur dioxide combines with water vapor in the air to produce sulfurous acid in the following reaction:\n\n$$\n\\mathrm{SO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(g) \\longrightarrow \\mathrm{H}_{2} \\mathrm{SO}_{3}(a q)\n$$\n\nCalcium carbonate reacts with sulfurous acid as follows:\n\n$$\n\\mathrm{CaCO}_{3}(s)+\\mathrm{H}_{2} \\mathrm{SO}_{3}(a q) \\longrightarrow \\mathrm{CaSO}_{3}(a q)+\\mathrm{CO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nIn a polluted atmosphere where the concentration of sulfur dioxide is high, calcium carbonate deteriorates more rapidly than in less polluted air. Similarly, phosphorus burns much more rapidly in an atmosphere of pure oxygen than in air, which is only about $20 \\%$ oxygen.\n\n\nFIGURE 17.7 Statues made from carbonate compounds such as limestone and marble typically weather slowly over time due to the actions of water, and thermal expansion and contraction. However, pollutants like sulfur dioxide can accelerate weathering. As the concentration of air pollutants increases, deterioration of limestone occurs more rapidly. (credit: James P Fisher III)"}
{"id": 4520, "contents": "1791. LINK TO LEARNING - \nPhosphorus burns rapidly in air, but it will burn even more rapidly if the concentration of oxygen is higher. Watch this video (http://openstax.org/l/16phosphor) to see an example."}
{"id": 4521, "contents": "1792. The Presence of a Catalyst - \nRelatively dilute aqueous solutions of hydrogen peroxide, $\\mathrm{H}_{2} \\mathrm{O}_{2}$, are commonly used as topical antiseptics. Hydrogen peroxide decomposes to yield water and oxygen gas according to the equation:\n\n$$\n2 \\mathrm{H}_{2} \\mathrm{O}_{2}(a q) \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{O}_{2}(\\mathrm{~g})\n$$\n\nUnder typical conditions, this decomposition occurs very slowly. When dilute $\\mathrm{H}_{2} \\mathrm{O}_{2}(\\mathrm{aq})$ is poured onto an open wound, however, the reaction occurs rapidly and the solution foams because of the vigorous production of oxygen gas. This dramatic difference is caused by the presence of substances within the wound's exposed tissues that accelerate the decomposition process. Substances that function to increase the rate of a reaction are called catalysts, a topic treated in greater detail later in this chapter."}
{"id": 4522, "contents": "1793. (C) LINK TO LEARNING - \nChemical reactions occur when molecules collide with each other and undergo a chemical transformation. Before physically performing a reaction in a laboratory, scientists can use molecular modeling simulations to predict how the parameters discussed earlier will influence the rate of a reaction. Use the PhET Reactions \\& Rates interactive (http://openstax.org/l/16PHETreaction) to explore how temperature, concentration, and the nature of the reactants affect reaction rates."}
{"id": 4523, "contents": "1794. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Explain the form and function of a rate law\n- Use rate laws to calculate reaction rates\n- Use rate and concentration data to identify reaction orders and derive rate laws\n\nAs described in the previous module, the rate of a reaction is often affected by the concentrations of reactants.\nRate laws (sometimes called differential rate laws) or rate equations are mathematical expressions that describe the relationship between the rate of a chemical reaction and the concentration of its reactants. As an example, consider the reaction described by the chemical equation\n\n$$\na A+b B \\longrightarrow \\text { products }\n$$\n\nwhere $a$ and $b$ are stoichiometric coefficients. The rate law for this reaction is written as:\n\n$$\n\\text { rate }=k[A]^{m}[B]^{n}\n$$\n\nin which $[A]$ and $[B]$ represent the molar concentrations of reactants, and $k$ is the rate constant, which is specific for a particular reaction at a particular temperature. The exponents $m$ and $n$ are the reaction orders and are typically positive integers, though they can be fractions, negative, or zero. The rate constant $k$ and the reaction orders $m$ and $n$ must be determined experimentally by observing how the rate of a reaction changes as the concentrations of the reactants are changed. The rate constant $k$ is independent of the reactant concentrations, but it does vary with temperature.\n\nThe reaction orders in a rate law describe the mathematical dependence of the rate on reactant concentrations. Referring to the generic rate law above, the reaction is $m$ order with respect to $A$ and $n$ order with respect to $B$. For example, if $m=1$ and $n=2$, the reaction is first order in $A$ and second order in $B$. The overall reaction order is simply the sum of orders for each reactant. For the example rate law here, the reaction is third order overall $(1+2=3)$. A few specific examples are shown below to further illustrate this concept.\n\nThe rate law:\n\n$$\n\\text { rate }=k\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]\n$$"}
{"id": 4524, "contents": "1794. LEARNING OBJECTIVES - \nThe rate law:\n\n$$\n\\text { rate }=k\\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]\n$$\n\ndescribes a reaction that is first order in hydrogen peroxide and first order overall. The rate law:\n\n$$\n\\text { rate }=k\\left[\\mathrm{C}_{4} \\mathrm{H}_{6}\\right]^{2}\n$$\n\ndescribes a reaction that is second order in $\\mathrm{C}_{4} \\mathrm{H}_{6}$ and second order overall. The rate law:\n\n$$\n\\text { rate }=k\\left[\\mathrm{H}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]\n$$\n\ndescribes a reaction that is first order in $\\mathrm{H}^{+}$, first order in $\\mathrm{OH}^{-}$, and second order overall."}
{"id": 4525, "contents": "1796. Writing Rate Laws from Reaction Orders - \nAn experiment shows that the reaction of nitrogen dioxide with carbon monoxide:\n\n$$\n\\mathrm{NO}_{2}(g)+\\mathrm{CO}(g) \\longrightarrow \\mathrm{NO}(g)+\\mathrm{CO}_{2}(g)\n$$\n\nis second order in $\\mathrm{NO}_{2}$ and zero order in CO at $100^{\\circ} \\mathrm{C}$. What is the rate law for the reaction?"}
{"id": 4526, "contents": "1797. Solution - \nThe reaction will have the form:\n\n$$\n\\text { rate }=k\\left[\\mathrm{NO}_{2}\\right]^{m}[\\mathrm{CO}]^{n}\n$$\n\nThe reaction is second order in $\\mathrm{NO}_{2}$; thus $m=2$. The reaction is zero order in CO; thus $n=0$. The rate law is:\n\n$$\n\\text { rate }=k\\left[\\mathrm{NO}_{2}\\right]^{2}[\\mathrm{CO}]^{0}=k\\left[\\mathrm{NO}_{2}\\right]^{2}\n$$\n\nRemember that a number raised to the zero power is equal to 1 , thus $[\\mathrm{CO}]^{0}=1$, which is why the CO concentration term may be omitted from the rate law: the rate of reaction is solely dependent on the concentration of $\\mathrm{NO}_{2}$. A later chapter section on reaction mechanisms will explain how a reactant's concentration can have no effect on a reaction rate despite being involved in the reaction."}
{"id": 4527, "contents": "1798. Check Your Learning - \nThe rate law for the reaction:\n\n$$\n\\mathrm{H}_{2}(g)+2 \\mathrm{NO}(g) \\longrightarrow \\mathrm{N}_{2} \\mathrm{O}(g)+\\mathrm{H}_{2} \\mathrm{O}(g)\n$$\n\nhas been determined to be rate $=k[N O]^{2}\\left[\\mathrm{H}_{2}\\right]$. What are the orders with respect to each reactant, and what is the\noverall order of the reaction?"}
{"id": 4528, "contents": "1799. Answer: - \norder in $\\mathrm{NO}=2$; order in $\\mathrm{H}_{2}=1$; overall order $=3$"}
{"id": 4529, "contents": "1800. Check Your Learning - \nIn a transesterification reaction, a triglyceride reacts with an alcohol to form an ester and glycerol. Many students learn about the reaction between methanol $\\left(\\mathrm{CH}_{3} \\mathrm{OH}\\right)$ and ethyl acetate $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OCOCH}_{3}\\right)$ as a sample reaction before studying the chemical reactions that produce biodiesel:\n\n$$\n\\mathrm{CH}_{3} \\mathrm{OH}+\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OCOCH}_{3} \\longrightarrow \\mathrm{CH}_{3} \\mathrm{OCOCH}_{3}+\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}\n$$\n\nThe rate law for the reaction between methanol and ethyl acetate is, under certain conditions, determined to be:\n\n$$\n\\text { rate }=k\\left[\\mathrm{CH}_{3} \\mathrm{OH}\\right]\n$$\n\nWhat is the order of reaction with respect to methanol and ethyl acetate, and what is the overall order of reaction?"}
{"id": 4530, "contents": "1801. Answer: - \norder in $\\mathrm{CH}_{3} \\mathrm{OH}=1$; order in $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OCOCH}_{3}=0$; overall order $=1$\n\nA common experimental approach to the determination of rate laws is the method of initial rates. This method involves measuring reaction rates for multiple experimental trials carried out using different initial reactant concentrations. Comparing the measured rates for these trials permits determination of the reaction orders and, subsequently, the rate constant, which together are used to formulate a rate law. This approach is illustrated in the next two example exercises."}
{"id": 4531, "contents": "1803. Determining a Rate Law from Initial Rates - \nOzone in the upper atmosphere is depleted when it reacts with nitrogen oxides. The rates of the reactions of nitrogen oxides with ozone are important factors in deciding how significant these reactions are in the formation of the ozone hole over Antarctica (Figure 17.8). One such reaction is the combination of nitric oxide, NO, with ozone, $\\mathrm{O}_{3}$ :\n\nFIGURE 17.8 A contour map showing stratospheric ozone concentration and the \"ozone hole\" that occurs over Antarctica during its spring months. (credit: modification of work by NASA)\n\n$$\n\\mathrm{NO}(\\mathrm{~g})+\\mathrm{O}_{3}(\\mathrm{~g}) \\longrightarrow \\mathrm{NO}_{2}(\\mathrm{~g})+\\mathrm{O}_{2}(\\mathrm{~g})\n$$\n\nThis reaction has been studied in the laboratory, and the following rate data were determined at $25^{\\circ} \\mathrm{C}$.\n\n| Trial | $\\mathrm{NO}$ | $\\left\\mathrm{O}_{3}\\right$ |\n| :--- | :--- | :--- |\n| 1 | $\\frac{\\Delta\\left[\\mathrm{NO}_{2}\\right]}{\\Delta t}\\left(\\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~s}^{-1}\\right)$ |  |\n| $1.00 \\times 10^{-6}$ | $3.00 \\times 10^{-6}$ | $6.60 \\times 10^{-5}$ |\n| 2 | $1.00 \\times 10^{-6}$ | $6.00 \\times 10^{-6}$ |\n| 3 | $1.00 \\times 10^{-6}$ | $9.00 \\times 10^{-6}$ |\n| 4 | $2.00 \\times 10^{-6}$ | $9.00 \\times 10^{-6}$ |\n| 5 | $3.00 \\times 10^{-6}$ | $9.00 \\times 10^{-6}$ |\n\nDetermine the rate law and the rate constant for the reaction at $25^{\\circ} \\mathrm{C}$."}
{"id": 4532, "contents": "1804. Solution - \nThe rate law will have the form:\n\n$$\n\\text { rate }=k[\\mathrm{NO}]^{m}\\left[\\mathrm{O}_{3}\\right]^{n}\n$$\n\nDetermine the values of $m, n$, and $k$ from the experimental data using the following three-part process:\nStep 1.\nDetermine the value of m from the data in which [ NO ] varies and $\\left[\\mathrm{O}_{3}\\right.$ ] is constant. In the last three experiments, [ NO ] varies while $\\left[\\mathrm{O}_{3}\\right]$ remains constant. When [NO] doubles from trial 3 to 4 , the rate doubles, and when [ NO ] triples from trial 3 to 5 , the rate also triples. Thus, the rate is also directly proportional to [NO], and $m$ in the rate law is equal to 1 .\n\nStep 2.\nDetermine the value of n from data in which $\\left[\\mathrm{O}_{3}\\right]$ varies and [NO] is constant. In the first three experiments, [ NO ] is constant and $\\left[\\mathrm{O}_{3}\\right]$ varies. The reaction rate changes in direct proportion to the change in [ $\\mathrm{O}_{3}$ ]. When $\\left[\\mathrm{O}_{3}\\right]$ doubles from trial 1 to 2, the rate doubles; when $\\left[\\mathrm{O}_{3}\\right]$ triples from trial 1 to 3 , the rate increases also triples. Thus, the rate is directly proportional to $\\left[\\mathrm{O}_{3}\\right]$, and $n$ is equal to 1.The rate law is thus:\n\n$$\n\\text { rate }=k[\\mathrm{NO}]^{1}\\left[\\mathrm{O}_{3}\\right]^{1}=k[\\mathrm{NO}]\\left[\\mathrm{O}_{3}\\right]\n$$\n\nStep 3.\nDetermine the value of k from one set of concentrations and the corresponding rate. The data from trial 1 are used below:"}
{"id": 4533, "contents": "1804. Solution - \nStep 3.\nDetermine the value of k from one set of concentrations and the corresponding rate. The data from trial 1 are used below:\n\n$$\n\\begin{aligned}\n& k=\\frac{\\text { rate }}{[\\mathrm{NO}]\\left[\\mathrm{O}_{3}\\right]} \\\\\n&=\\frac{6.60 \\times 10^{-5} \\overline{\\mathrm{molL}}^{-1}}{\\mathrm{~s}}-1 \\\\\n&\\left(1.00 \\times 10^{-6} \\mathrm{molL}^{-1}\\right)\\left(3.00 \\times 10^{-6} \\mathrm{~mol} \\mathrm{~L}^{-1}\\right)\n\\end{aligned}\n$$"}
{"id": 4534, "contents": "1805. Check Your Learning - \nAcetaldehyde decomposes when heated to yield methane and carbon monoxide according to the equation:\n\n$$\n\\mathrm{CH}_{3} \\mathrm{CHO}(\\mathrm{~g}) \\longrightarrow \\mathrm{CH}_{4}(\\mathrm{~g})+\\mathrm{CO}(\\mathrm{~g})\n$$\n\nDetermine the rate law and the rate constant for the reaction from the following experimental data:\n\n| Trial | $\\left\\mathrm{CH}_{3} \\mathrm{CHO}\\right$ | $-\\frac{\\Delta\\left[\\mathrm{CH}_{3} \\mathrm{CHO}\\right]}{\\Delta t}\\left(\\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~s}^{-1}\\right)$ |\n| :--- | :--- | :--- |\n| 1 | $1.75 \\times 10^{-3}$ | $2.06 \\times 10^{-11}$ |\n| 2 | $3.50 \\times 10^{-3}$ | $8.24 \\times 10^{-11}$ |\n| 3 | $7.00 \\times 10^{-3}$ | $3.30 \\times 10^{-10}$ |"}
{"id": 4535, "contents": "1806. Answer: - \nrate $=k\\left[\\mathrm{CH}_{3} \\mathrm{CHO}\\right]^{2}$ with $k=6.73 \\times 10^{-6} \\mathrm{~L} / \\mathrm{mol} / \\mathrm{s}$"}
{"id": 4536, "contents": "1808. Determining Rate Laws from Initial Rates - \nUsing the initial rates method and the experimental data, determine the rate law and the value of the rate constant for this reaction:\n\n$$\n2 \\mathrm{NO}(g)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{NOCl}(g)\n$$\n\n| Trial | $\\mathrm{NO}$ | $\\left\\mathrm{Cl}_{2}\\right$ | $-\\frac{\\Delta[\\mathrm{NO}]}{\\Delta t}\\left(\\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~s}^{-1}\\right)$ |\n| :--- | :--- | :--- | :--- |\n| 1 | 0.10 | 0.10 | 0.00300 |\n| 2 | 0.10 | 0.15 | 0.00450 |\n| 3 | 0.15 | 0.10 | 0.00675 |"}
{"id": 4537, "contents": "1809. Solution - \nThe rate law for this reaction will have the form:\n\n$$\n\\text { rate }=k[\\mathrm{NO}]^{m}\\left[\\mathrm{Cl}_{2}\\right]^{n}\n$$\n\nAs in Example 17.4, approach this problem in a stepwise fashion, determining the values of $m$ and $n$ from the experimental data and then using these values to determine the value of $k$. In this example, however, an explicit algebraic approach (vs. the implicit approach of the previous example) will be used to determine the values of $m$ and $n$ :"}
{"id": 4538, "contents": "1810. Step 1. - \nDetermine the value of m from the data in which $[\\mathrm{NO}]$ varies and $\\left[\\mathrm{Cl}_{2}\\right]$ is constant. Write the ratios with the subscripts $x$ and $y$ to indicate data from two different trials:\n\n$$\n\\frac{\\operatorname{rate}_{x}}{\\operatorname{rate}_{y}}=\\frac{k[\\mathrm{NO}]_{x}^{m}\\left[\\mathrm{Cl}_{2}\\right]_{x}^{n}}{k[\\mathrm{NO}]_{y}^{m}\\left[\\mathrm{Cl}_{2}\\right]_{y}^{n}}\n$$\n\nUsing the third trial and the first trial, in which $\\left[\\mathrm{Cl}_{2}\\right]$ does not vary, gives:\n\n$$\n\\frac{\\text { rate } 3}{\\text { rate } 1}=\\frac{0.00675}{0.00300}=\\frac{k(0.15)^{m}(0.10)^{n}}{k(0.10)^{m}(0.10)^{n}}\n$$\n\nCanceling equivalent terms in the numerator and denominator leaves:\n\n$$\n\\frac{0.00675}{0.00300}=\\frac{(0.15)^{m}}{(0.10)^{m}}\n$$\n\nwhich simplifies to:\n\n$$\n2.25=(1.5)^{m}\n$$\n\nUse logarithms to determine the value of the exponent $m$ :\n\n$$\n\\begin{aligned}\n\\ln (2.25) & =m \\ln (1.5) \\\\\n\\frac{\\ln (2.25)}{\\ln (1.5)} & =m \\\\\n2 & =m\n\\end{aligned}\n$$\n\nConfirm the result\n\n$$\n1.5^{2}=2.25\n$$\n\nStep 2.\nDetermine the value of n from data in which [ $\\mathrm{Cl}_{2}$ ] varies and [ NO ] is constant.\n\n$$\n\\frac{\\text { rate } 2}{\\text { rate } 1}=\\frac{0.00450}{0.00300}=\\frac{k(0.10)^{m}(0.15)^{n}}{k(0.10)^{m}(0.10)^{n}}\n$$"}
{"id": 4539, "contents": "1810. Step 1. - \nCancelation gives:\n\n$$\n\\frac{0.0045}{0.0030}=\\frac{(0.15)^{n}}{(0.10)^{n}}\n$$\n\nwhich simplifies to:\n\n$$\n1.5=(1.5)^{n}\n$$\n\nThus $n$ must be 1 , and the form of the rate law is:\n\n$$\n\\text { rate }=k[\\mathrm{NO}]^{m}\\left[\\mathrm{Cl}_{2}\\right]^{n}=k[\\mathrm{NO}]^{2}\\left[\\mathrm{Cl}_{2}\\right]\n$$\n\nStep 3.\nDetermine the numerical value of the rate constant k with appropriate units. The units for the rate of a reaction are $\\mathrm{mol} / \\mathrm{L} / \\mathrm{s}$. The units for $k$ are whatever is needed so that substituting into the rate law expression affords the appropriate units for the rate. In this example, the concentration units are $\\mathrm{mol}^{3} / \\mathrm{L}^{3}$. The units for $k$ should be $\\mathrm{mol}^{-2} \\mathrm{~L}^{2} / \\mathrm{s}$ so that the rate is in terms of $\\mathrm{mol} / \\mathrm{L} / \\mathrm{s}$.\n\nTo determine the value of $k$ once the rate law expression has been solved, simply plug in values from the first experimental trial and solve for $k$ :\n\n$$\n\\begin{aligned}\n0.00300 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~s}^{-1} & =k\\left(0.10 \\mathrm{~mol} \\mathrm{~L}^{-1}\\right)^{2}\\left(0.10 \\mathrm{~mol} \\mathrm{~L}^{-1}\\right)^{1} \\\\\nk & =3.0 \\mathrm{~mol}^{-2} \\mathrm{~L}^{2} \\mathrm{~s}^{-1}\n\\end{aligned}\n$$"}
{"id": 4540, "contents": "1811. Check Your Learning - \nUse the provided initial rate data to derive the rate law for the reaction whose equation is:\n\n| $\\mathrm{OCl}^{-}(a q)+\\mathrm{I}^{-}(a q) \\longrightarrow \\mathrm{OI}^{-}(a q)+\\mathrm{Cl}^{-}(a q)$ |  |  |  |\n| :--- | :--- | :--- | :--- |\n| Trial | $\\left\\mathrm{OCl}^{-}\\right$ | $\\left\\mathrm{I}^{-}\\right$ | Initial Rate (mol/L/s) |\n| 1 | 0.0040 | 0.0020 | 0.00184 |\n| 2 | 0.0020 | 0.0040 | 0.00092 |\n| 3 | 0.0020 | 0.0020 | 0.00046 |\n\nDetermine the rate law expression and the value of the rate constant $k$ with appropriate units for this reaction."}
{"id": 4541, "contents": "1812. Answer: - \n$$\n\\begin{aligned}\n& \\frac{\\text { rate } 2}{\\text { rate } 3}=\\frac{0.00092}{0.00046}=\\frac{k(0.0020)^{x}(0.0040)^{y}}{k(0.0020)^{x}(0.0020)^{y}} \\\\\n& 2.00=2.00^{y} \\\\\n& y=1 \\\\\n& \\frac{\\text { rate } 1}{\\text { rate } 2}=\\frac{0.00184}{0.00092}=\\frac{k(0.0040)^{x}(0.0020)^{y}}{k(0.0020)^{x}(0.0040)^{y}} \\\\\n& 2.00=\\frac{2^{x}}{2^{y}} \\\\\n& 2.00=\\frac{2^{x}}{2^{1}} \\\\\n& 4.00=2^{x} \\\\\n& x=2\n\\end{aligned}\n$$\n\nSubstituting the concentration data from trial 1 and solving for $k$ yields:\n\n$$\n\\begin{aligned}\n\\text { rate } & =k\\left[\\mathrm{OCl}^{-}\\right]^{2}\\left[\\mathrm{I}^{-}\\right]^{1} \\\\\n0.00184 & =k(0.0040)^{2}(0.0020)^{1} \\\\\nk & =5.75 \\times 10^{4} \\mathrm{~mol}^{-2} \\mathrm{~L}^{2} \\mathrm{~s}^{-1}\n\\end{aligned}\n$$"}
{"id": 4542, "contents": "1813. Reaction Order and Rate Constant Units - \nIn some of our examples, the reaction orders in the rate law happen to be the same as the coefficients in the\nchemical equation for the reaction. This is merely a coincidence and very often not the case.\nRate laws may exhibit fractional orders for some reactants, and negative reaction orders are sometimes observed when an increase in the concentration of one reactant causes a decrease in reaction rate. A few examples illustrating these points are provided:\n\n$$\n\\begin{array}{lc}\n\\mathrm{NO}_{2}+\\mathrm{CO} \\longrightarrow \\mathrm{NO}+\\mathrm{CO}_{2} & \\text { rate }=k\\left[\\mathrm{NO}_{2}\\right]^{2} \\\\\n\\mathrm{CH}_{3} \\mathrm{CHO} \\longrightarrow \\mathrm{CH}_{4}+\\mathrm{CO} & \\text { rate }=k\\left[\\mathrm{CH}_{3} \\mathrm{CHO}\\right]^{2} \\\\\n2 \\mathrm{~N}_{2} \\mathrm{O}_{5} \\longrightarrow \\mathrm{NO}_{2}+\\mathrm{O}_{2} & \\text { rate }=k\\left[\\mathrm{~N}_{2} \\mathrm{O}_{5}\\right] \\\\\n2 \\mathrm{NO}_{2}+\\mathrm{F}_{2} \\longrightarrow 2 \\mathrm{NO}_{2} \\mathrm{~F} & \\text { rate }=k\\left[\\mathrm{NO}_{2}\\right]\\left[\\mathrm{F}_{2}\\right] \\\\\n2 \\mathrm{NO}_{2} \\mathrm{Cl} \\longrightarrow 2 \\mathrm{NO}_{2}+\\mathrm{Cl}_{2} & \\text { rate }=k\\left[\\mathrm{NO}_{2} \\mathrm{Cl}\\right]\n\\end{array}\n$$\n\nIt is important to note that rate laws are determined by experiment only and are not reliably predicted by reaction stoichiometry."}
{"id": 4543, "contents": "1813. Reaction Order and Rate Constant Units - \nIt is important to note that rate laws are determined by experiment only and are not reliably predicted by reaction stoichiometry.\n\nThe units for a rate constant will vary as appropriate to accommodate the overall order of the reaction. The unit of the rate constant for the second-order reaction described in Example 17.4 was determined to be $\\mathrm{L} \\mathrm{mol}^{-1} \\mathrm{~s}^{-1}$. For the third-order reaction described in Example 17.5, the unit for $k$ was derived to be $\\mathrm{L}^{2} \\mathrm{~mol}^{-2} \\mathrm{~s}^{-1}$. Dimensional analysis requires the rate constant unit for a reaction whose overall order is $x$ to be $\\mathrm{L}^{x-1} \\mathrm{~mol}^{1-x} \\mathrm{~s}^{-1}$. Table 17.1 summarizes the rate constant units for common reaction orders.\n\nRate Constant Units for Common Reaction Orders\n\n| Overall Reaction Order $(\\boldsymbol{x})$ | Rate Constant Unit ( $\\mathrm{L}^{\\boldsymbol{x - 1}} \\mathrm{mol}^{1-x} \\mathrm{~s}^{-1}$ ) |\n| :--- | :--- |\n| 0 (zero) | $\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~s}^{-1}$ |\n| 1 (first) | $\\mathrm{s}^{-1}$ |\n| 2 (second) | $\\mathrm{L} \\mathrm{mol}^{-1} \\mathrm{~s}^{-1}$ |\n| 3 (third) | $\\mathrm{L}^{2} \\mathrm{~mol}^{-2} \\mathrm{~s}^{-1}$ |\n\nTABLE 17.1\n\nNote that the units in this table were derived using specific units for concentration ( $\\mathrm{mol} / \\mathrm{L}$ ) and time ( s ), though any valid units for these two properties may be used."}
{"id": 4544, "contents": "1813. Reaction Order and Rate Constant Units - 1813.1. Integrated Rate Laws\nLEARNING OBJECTIVES\nBy the end of this section, you will be able to:\n\n- Explain the form and function of an integrated rate law\n- Perform integrated rate law calculations for zero-, first-, and second-order reactions\n- Define half-life and carry out related calculations\n- Identify the order of a reaction from concentration/time data\n\nThe rate laws discussed thus far relate the rate and the concentrations of reactants. We can also determine a second form of each rate law that relates the concentrations of reactants and time. These are called integrated rate laws. We can use an integrated rate law to determine the amount of reactant or product present after a period of time or to estimate the time required for a reaction to proceed to a certain extent. For example, an integrated rate law is used to determine the length of time a radioactive material must be stored for its radioactivity to decay to a safe level.\n\nUsing calculus, the differential rate law for a chemical reaction can be integrated with respect to time to give an equation that relates the amount of reactant or product present in a reaction mixture to the elapsed time of the reaction. This process can either be very straightforward or very complex, depending on the complexity of the differential rate law. For purposes of discussion, we will focus on the resulting integrated rate laws for first, second-, and zero-order reactions."}
{"id": 4545, "contents": "1814. First-Order Reactions - \nIntegration of the rate law for a simple first-order reaction (rate $=k[A]$ ) results in an equation describing how the reactant concentration varies with time:\n\n$$\n[A]_{t}=[A]_{0} e^{-k t}\n$$\n\nwhere $[A] t$ is the concentration of $A$ at any time $t,[A]_{0}$ is the initial concentration of $A$, and $k$ is the first-order rate constant.\n\nFor mathematical convenience, this equation may be rearranged to other formats, including direct and indirect proportionalities:\n\n$$\n\\ln \\left(\\frac{[A]_{t}}{[A]_{0}}\\right)=-k t \\quad \\text { or } \\quad \\ln \\left(\\frac{[A]_{0}}{[A]_{t}}\\right)=k t\n$$\n\nand a format showing a linear dependence of concentration in time:\n\n$$\n\\ln [A]_{t}=\\ln [A]_{0}-k t\n$$"}
{"id": 4546, "contents": "1816. The Integrated Rate Law for a First-Order Reaction - \nThe rate constant for the first-order decomposition of cyclobutane, $\\mathrm{C}_{4} \\mathrm{H}_{8}$ at $500^{\\circ} \\mathrm{C}$ is $9.2 \\times 10^{-3} \\mathrm{~s}^{-1}$ :\n\n$$\n\\mathrm{C}_{4} \\mathrm{H}_{8} \\longrightarrow 2 \\mathrm{C}_{2} \\mathrm{H}_{4}\n$$\n\nHow long will it take for $80.0 \\%$ of a sample of $\\mathrm{C}_{4} \\mathrm{H}_{8}$ to decompose?"}
{"id": 4547, "contents": "1817. Solution - \nSince the relative change in reactant concentration is provided, a convenient format for the integrated rate law is:\n\n$$\n\\ln \\left(\\frac{[A]_{0}}{[A]_{t}}\\right)=k t\n$$\n\nThe initial concentration of $\\mathrm{C}_{4} \\mathrm{H}_{8},[A]_{0}$, is not provided, but the provision that $80.0 \\%$ of the sample has decomposed is enough information to solve this problem. Let $x$ be the initial concentration, in which case the concentration after $80.0 \\%$ decomposition is $20.0 \\%$ of $x$ or $0.200 x$. Rearranging the rate law to isolate $t$ and substituting the provided quantities yields:\n\n$$\n\\begin{aligned}\nt & =\\ln \\frac{[x]}{[0.200 x]} \\times \\frac{1}{k} \\\\\n& =\\ln 5 \\times \\frac{1}{9.2 \\times 10^{-3} \\mathrm{~s}^{-1}} \\\\\n& =1.609 \\times \\frac{1}{9.2 \\times 10^{-3} \\mathrm{~s}^{-1}} \\\\\n& =1.7 \\times 10^{2} \\mathrm{~s}\n\\end{aligned}\n$$"}
{"id": 4548, "contents": "1818. Check Your Learning - \nIodine-131 is a radioactive isotope that is used to diagnose and treat some forms of thyroid cancer. Iodine-131 decays to xenon-131 according to the equation:\n\n$$\n\\mathrm{I}-131 \\longrightarrow \\mathrm{Xe}-131+\\text { electron }\n$$\n\nThe decay is first-order with a rate constant of $0.138 \\mathrm{~d}^{-1}$. How many days will it take for $90 \\%$ of the iodine -131 in a 0.500 M solution of this substance to decay to Xe-131?"}
{"id": 4549, "contents": "1819. Answer: - \n16.7 days\n\nIn the next example exercise, a linear format for the integrated rate law will be convenient:\n\n$$\n\\begin{aligned}\n\\ln [A]_{t} & =(-k)(t)+\\ln [A]_{0} \\\\\ny & =m x+b\n\\end{aligned}\n$$\n\nA plot of $\\ln [A]_{t}$ versus $t$ for a first-order reaction is a straight line with a slope of $-k$ and a $y$-intercept of $\\ln [A]_{0}$. If a set of rate data are plotted in this fashion but do not result in a straight line, the reaction is not first order in A."}
{"id": 4550, "contents": "1821. Graphical Determination of Reaction Order and Rate Constant - \nShow that the data in Figure 17.2 can be represented by a first-order rate law by graphing $\\ln \\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]$ versus time. Determine the rate constant for the decomposition of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ from these data."}
{"id": 4551, "contents": "1822. Solution - \nThe data from Figure 17.2 are tabulated below, and a plot of $\\ln \\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]$ is shown in Figure 17.9.\n\n| Trial |  | Time (h) | $\\left\\mathrm{H}_{2} \\mathrm{O}_{2}\\right$ |\n| :--- | :--- | :--- | :--- |\n| 1 | 0.00 | 1.000 | 0.000 |\n| 2 | 6.00 | 0.500 | -0.693 |\n| 3 | 12.00 | 0.250 | -1.386 |\n| 4 | 18.00 | 0.125 | -2.079 |\n| 5 | 24.00 | 0.0625 | -2.772 |\n\n\n\nFIGURE 17.9 A linear relationship between $\\ln \\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]$ and time suggests the decomposition of hydrogen peroxide is a first-order reaction.\n\nThe plot of $\\ln \\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]$ versus time is linear, indicating that the reaction may be described by a first-order rate law.\n\nAccording to the linear format of the first-order integrated rate law, the rate constant is given by the negative of this plot's slope.\n\n$$\n\\text { slope }=\\frac{\\text { change in } y}{\\text { change in } x}=\\frac{\\Delta y}{\\Delta x}=\\frac{\\Delta \\ln \\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]}{\\Delta t}\n$$\n\nThe slope of this line may be derived from two values of $\\ln \\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]$ at different values of $t$ (one near each end of the line is preferable). For example, the value of $\\ln \\left[\\mathrm{H}_{2} \\mathrm{O}_{2}\\right]$ when $t$ is 0.00 h is 0.000 ; the value when $t=24.00 \\mathrm{~h}$ is $-2.772$"}
{"id": 4552, "contents": "1822. Solution - \n$$\n\\begin{aligned}\n\\text { slope } & =\\frac{-2.772-0.000}{24.00-0.00 \\mathrm{~h}} \\\\\n& =\\frac{-2.772}{24.00 \\mathrm{~h}} \\\\\n& =-0.116 \\mathrm{~h}^{-1} \\\\\nk & =- \\text { slope }=-\\left(-0.116 \\mathrm{~h}^{-1}\\right)=0.116 \\mathrm{~h}^{-1}\n\\end{aligned}\n$$"}
{"id": 4553, "contents": "1823. Check Your Learning - \nGraph the following data to determine whether the reaction $A \\longrightarrow B+C$ is first order.\n\n| Trial |  | Time (s) |\n| :--- | :--- | :--- |\n| $A]$ |  |  |\n| 1 | 4.0 | 0.220 |\n| 2 | 8.0 | 0.144 |\n| 3 | 12.0 | 0.110 |\n| 4 | 16.0 | 0.088 |\n| 5 | 20.0 | 0.074 |"}
{"id": 4554, "contents": "1824. Answer: - \nThe plot of $\\ln [A]_{t}$ vs. $t$ is not linear, indicating the reaction is not first order:"}
{"id": 4555, "contents": "1825. Second-Order Reactions - \nThe equations that relate the concentrations of reactants and the rate constant of second-order reactions can be fairly complicated. To illustrate the point with minimal complexity, only the simplest second-order reactions will be described here, namely, those whose rates depend on the concentration of just one reactant. For these types of reactions, the differential rate law is written as:\n\n$$\n\\text { rate }=k[A]^{2}\n$$\n\nFor these second-order reactions, the integrated rate law is:\n\n$$\n\\frac{1}{[A]_{t}}=k t+\\frac{1}{[A]_{0}}\n$$\n\nwhere the terms in the equation have their usual meanings as defined earlier."}
{"id": 4556, "contents": "1826. EXAMPLE 17.8 - \nThe Integrated Rate Law for a Second-Order Reaction\nThe reaction of butadiene gas $\\left(\\mathrm{C}_{4} \\mathrm{H}_{6}\\right)$ to yield $\\mathrm{C}_{8} \\mathrm{H}_{12}$ gas is described by the equation:\n\n$$\n2 \\mathrm{C}_{4} \\mathrm{H}_{6}(g) \\longrightarrow \\mathrm{C}_{8} \\mathrm{H}_{12}(g)\n$$\n\nThis \"dimerization\" reaction is second order with a rate constant equal to $5.76 \\times 10^{-2} \\mathrm{~L} \\mathrm{~mol}^{-1} \\mathrm{~min}^{-1}$ under certain conditions. If the initial concentration of butadiene is 0.200 M , what is the concentration after 10.0 min?"}
{"id": 4557, "contents": "1827. Solution - \nFor a second-order reaction, the integrated rate law is written\n\n$$\n\\frac{1}{[A]_{t}}=k t+\\frac{1}{[A]_{0}}\n$$\n\nWe know three variables in this equation: $[A]_{0}=0.200 \\mathrm{~mol} / \\mathrm{L}, k=5.76 \\times 10^{-2} \\mathrm{~L} / \\mathrm{mol} / \\mathrm{min}$, and $t=10.0 \\mathrm{~min}$. Therefore, we can solve for $[A]$, the fourth variable:\n\n$$\n\\begin{aligned}\n\\frac{1}{[A]_{t}} & =\\left(5.76 \\times 10^{-2} \\mathrm{~L} \\mathrm{~mol}^{-1} \\mathrm{~min}^{-1}\\right)(10 \\mathrm{~min})+\\frac{1}{0.200 \\mathrm{~mol}^{-1}} \\\\\n\\frac{1}{[A]_{t}} & =\\left(5.76 \\times 10^{-1} \\mathrm{~L} \\mathrm{~mol}^{-1}\\right)+5.00 \\mathrm{~L} \\mathrm{~mol}^{-1} \\\\\n\\frac{1}{[A]_{t}} & =5.58 \\mathrm{~L} \\mathrm{~mol}^{-1} \\\\\n{[A]_{t} } & =1.79 \\times 10^{-1} \\mathrm{~mol} \\mathrm{~L}^{-1}\n\\end{aligned}\n$$\n\nTherefore $0.179 \\mathrm{~mol} / \\mathrm{L}$ of butadiene remain at the end of 10.0 min , compared to the $0.200 \\mathrm{~mol} / \\mathrm{L}$ that was originally present."}
{"id": 4558, "contents": "1828. Check Your Learning - \nIf the initial concentration of butadiene is 0.0200 M , what is the concentration remaining after 20.0 min?"}
{"id": 4559, "contents": "1829. Answer: - \n$0.0195 \\mathrm{~mol} / \\mathrm{L}$\n\nThe integrated rate law for second-order reactions has the form of the equation of a straight line:\n\n$$\n\\begin{aligned}\n\\frac{1}{[A]_{t}} & =k t+\\frac{1}{[A]_{0}} \\\\\ny & =m x+b\n\\end{aligned}\n$$\n\nA plot of $\\frac{1}{[A]_{t}}$ versus $t$ for a second-order reaction is a straight line with a slope of $k$ and a $y$-intercept of $\\frac{1}{[A]_{0}}$. If the plot is not a straight line, then the reaction is not second order."}
{"id": 4560, "contents": "1831. Graphical Determination of Reaction Order and Rate Constant - \nThe data below are for the same reaction described in Example 17.8. Prepare and compare two appropriate data plots to identify the reaction as being either first or second order. After identifying the reaction order, estimate a value for the rate constant."}
{"id": 4561, "contents": "1832. Solution - \n| Trial | Time (s) | $\\left\\mathrm{C}_{4} \\mathrm{H}_{6}\\right$ |\n| :--- | :--- | :--- |\n| 1 | 0 | $1.00 \\times 10^{-2}$ |\n| 2 | 1600 | $5.04 \\times 10^{-3}$ |\n| 3 | 3200 | $3.37 \\times 10^{-3}$ |\n| 4 | 4800 | $2.53 \\times 10^{-3}$ |\n| 5 | 6200 | $2.08 \\times 10^{-3}$ |\n\nIn order to distinguish a first-order reaction from a second-order reaction, prepare a plot of $\\ln \\left[\\mathrm{C}_{4} \\mathrm{H}_{6}\\right]_{t}$ versus $t$ and compare it to a plot of $\\frac{1}{\\left[\\mathrm{C}_{4} \\mathrm{H}_{6}\\right]_{t}}$ versus $t$. The values needed for these plots follow.\n\n| Time (s) | $\\frac{1}{\\left[\\mathrm{C}_{4} \\mathrm{H}_{6}\\right]}\\left(M^{-1}\\right)$ | $\\ln \\left[\\mathrm{C}_{4} \\mathrm{H}_{6}\\right]$ |\n| :--- | :--- | :--- |\n| 0 | 100 | -4.605 |\n| 1600 | 198 | -5.289 |\n| 3200 | 296 | -5.692 |\n| 4800 | 395 | -5.978 |\n| 6200 | 481 | -6.175 |\n\nThe plots are shown in Figure 17.10, which clearly shows the plot of $\\ln \\left[\\mathrm{C}_{4} \\mathrm{H}_{6}\\right]_{t}$ versus $t$ is not linear, therefore the reaction is not first order. The plot of $\\frac{1}{\\left[\\mathrm{C}_{4} \\mathrm{H}_{6}\\right]_{t}}$ versus $t$ is linear, indicating that the reaction is second order."}
{"id": 4562, "contents": "1832. Solution - \nFIGURE 17.10 These two graphs show first- and second-order plots for the dimerization of $\\mathrm{C}_{4} \\mathrm{H}_{6}$. The linear trend in the second-order plot (right) indicates that the reaction follows second-order kinetics.\n\nAccording to the second-order integrated rate law, the rate constant is equal to the slope of the $\\frac{1}{[A]_{t}}$ versus $t$ plot. Using the data for $t=0 \\mathrm{~s}$ and $t=6200 \\mathrm{~s}$, the rate constant is estimated as follows:\n\n$$\nk=\\text { slope }=\\frac{\\left(481 M^{-1}-100 M^{-1}\\right)}{(6200 \\mathrm{~s}-0 \\mathrm{~s})}=0.0614 \\mathrm{M}^{-1} \\mathrm{~s}^{-1}\n$$"}
{"id": 4563, "contents": "1833. Check Your Learning - \nDo the following data fit a second-order rate law?\n\n| Trial | Time (s) | $A$ |\n| :--- | :--- | :--- |\n| 1 | 5 | 0.952 |\n| 2 | 10 | 0.625 |\n| 3 | 15 | 0.465 |\n| 4 | 20 | 0.370 |\n| 5 | 25 | 0.308 |\n| 6 | 35 | 0.230 |"}
{"id": 4564, "contents": "1834. Answer: - \nYes. The plot of $\\frac{1}{[A]_{t}}$ vs. $t$ is linear:"}
{"id": 4565, "contents": "1835. Zero-Order Reactions - \nFor zero-order reactions, the differential rate law is:\n\n$$\n\\text { rate }=k\n$$\n\nA zero-order reaction thus exhibits a constant reaction rate, regardless of the concentration of its reactant(s). This may seem counterintuitive, since the reaction rate certainly can't be finite when the reactant concentration is zero. For purposes of this introductory text, it will suffice to note that zero-order kinetics are observed for some reactions only under certain specific conditions. These same reactions exhibit different kinetic behaviors when the specific conditions aren't met, and for this reason the more prudent term pseudo-zero-order is sometimes used.\n\nThe integrated rate law for a zero-order reaction is a linear function:\n\n$$\n\\begin{aligned}\n{[A]_{t} } & =-k t+[A]_{0} \\\\\ny & =m x+b\n\\end{aligned}\n$$\n\nA plot of $[A]$ versus $t$ for a zero-order reaction is a straight line with a slope of $-k$ and a $y$-intercept of $[A]_{0}$. Figure 17.11 shows a plot of $\\left[\\mathrm{NH}_{3}\\right]$ versus $t$ for the thermal decomposition of ammonia at the surface of two different heated solids. The decomposition reaction exhibits first-order behavior at a quartz $\\left(\\mathrm{SiO}_{2}\\right)$ surface, as suggested by the exponentially decaying plot of concentration versus time. On a tungsten surface, however, the plot is linear, indicating zero-order kinetics."}
{"id": 4566, "contents": "1837. Graphical Determination of Zero-Order Rate Constant - \nUse the data plot in Figure 17.11 to graphically estimate the zero-order rate constant for ammonia decomposition at a tungsten surface."}
{"id": 4567, "contents": "1838. Solution - \nThe integrated rate law for zero-order kinetics describes a linear plot of reactant concentration, $[A]_{t}$, versus time, $t$, with a slope equal to the negative of the rate constant, $-k$. Following the mathematical approach of previous examples, the slope of the linear data plot (for decomposition on W ) is estimated from the graph. Using the ammonia concentrations at $t=0$ and $t=1000 \\mathrm{~s}$ :\n\n$$\nk=- \\text { slope }=-\\frac{\\left(0.0015 \\mathrm{~mol} \\mathrm{~L}^{-1}-0.0028 \\mathrm{~mol} \\mathrm{~L}^{-1}\\right)}{(1000 \\mathrm{~s}-0 \\mathrm{~s})}=1.3 \\times 10^{-6} \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~s}^{-1}\n$$"}
{"id": 4568, "contents": "1839. Check Your Learning - \nThe zero-order plot in Figure 17.11 shows an initial ammonia concentration of $0.0028 \\mathrm{~mol} \\mathrm{~L}^{-1}$ decreasing linearly with time for 1000 s . Assuming no change in this zero-order behavior, at what time (min) will the concentration reach $0.0001 \\mathrm{~mol} \\mathrm{~L}^{-1}$ ?"}
{"id": 4569, "contents": "1840. Answer: - \n35 min\n\n\nFIGURE 17.11 The decomposition of $\\mathrm{NH}_{3}$ on a tungsten $(\\mathrm{W})$ surface is a zero-order reaction, whereas on a quartz $\\left(\\mathrm{SiO}_{2}\\right)$ surface, the reaction is first order.\n\nThe Half-Life of a Reaction\nThe half-life of a reaction ( $\\boldsymbol{t}_{\\mathbf{1} / \\mathbf{2}}$ ) is the time required for one-half of a given amount of reactant to be consumed. In each succeeding half-life, half of the remaining concentration of the reactant is consumed. Using the decomposition of hydrogen peroxide (Figure 17.2) as an example, we find that during the first half-life (from 0.00 hours to 6.00 hours), the concentration of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ decreases from 1.000 M to 0.500 M . During the second half-life (from 6.00 hours to 12.00 hours), it decreases from 0.500 M to 0.250 M ; during the third half-life, it decreases from 0.250 M to 0.125 M . The concentration of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ decreases by half during each successive period of 6.00 hours. The decomposition of hydrogen peroxide is a first-order reaction, and, as can be shown, the half-life of a first-order reaction is independent of the concentration of the reactant. However, half-lives of reactions with other orders depend on the concentrations of the reactants."}
{"id": 4570, "contents": "1841. First-Order Reactions - \nAn equation relating the half-life of a first-order reaction to its rate constant may be derived from the integrated rate law as follows:\n\n$$\n\\begin{aligned}\n\\ln \\frac{[A]_{0}}{[A]_{t}} & =k t \\\\\nt & =\\ln \\frac{[A]_{0}}{[A]_{t}} \\times \\frac{1}{k}\n\\end{aligned}\n$$\n\nInvoking the definition of half-life, symbolized $t_{1 / 2}$, requires that the concentration of $A$ at this point is one-half its initial concentration: $t=t_{1 / 2},[A]_{t}=\\frac{1}{2}[A]_{0}$.\n\nSubstituting these terms into the rearranged integrated rate law and simplifying yields the equation for halflife:\n\n$$\n\\begin{aligned}\nt_{1 / 2} & =\\ln \\frac{[A]_{0}}{\\frac{1}{2}[A]_{0}} \\times \\frac{1}{k} \\\\\n& =\\ln 2 \\times \\frac{1}{k}=0.693 \\times \\frac{1}{k} \\\\\nt_{1 / 2} & =\\frac{0.693}{k}\n\\end{aligned}\n$$\n\nThis equation describes an expected inverse relation between the half-life of the reaction and its rate constant, k. Faster reactions exhibit larger rate constants and correspondingly shorter half-lives. Slower reactions exhibit smaller rate constants and longer half-lives."}
{"id": 4571, "contents": "1843. Calculation of a First-order Rate Constant using Half-Life - \nCalculate the rate constant for the first-order decomposition of hydrogen peroxide in water at $40^{\\circ} \\mathrm{C}$, using the data given in Figure 17.12.\n\n\nFIGURE 17.12 The decomposition of $\\mathrm{H}_{2} \\mathrm{O}_{2}\\left(2 \\mathrm{H}_{2} \\mathrm{O}_{2} \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{O}_{2}\\right)$ at $40^{\\circ} \\mathrm{C}$ is illustrated. The intensity of the color symbolizes the concentration of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ at the indicated times; $\\mathrm{H}_{2} \\mathrm{O}_{2}$ is actually colorless."}
{"id": 4572, "contents": "1844. Solution - \nInspecting the concentration/time data in Figure 17.12 shows the half-life for the decomposition of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ is $2.16 \\times 10^{4} \\mathrm{~s}$ :\n\n$$\n\\begin{aligned}\nt_{1 / 2} & =\\frac{0.693}{k} \\\\\nk & =\\frac{0.693}{t_{1 / 2}}=\\frac{0.693}{2.16 \\times 10^{4} \\mathrm{~s}}=3.21 \\times 10^{-5} \\mathrm{~s}^{-1}\n\\end{aligned}\n$$"}
{"id": 4573, "contents": "1845. Check Your Learning - \nThe first-order radioactive decay of iodine-131 exhibits a rate constant of $0.138 \\mathrm{~d}^{-1}$. What is the half-life for this decay?\n\nAnswer:\n5.02 d."}
{"id": 4574, "contents": "1846. Second-Order Reactions - \nFollowing the same approach as used for first-order reactions, an equation relating the half-life of a secondorder reaction to its rate constant and initial concentration may be derived from its integrated rate law:\n\n$$\n\\frac{1}{[A]_{t}}=k t+\\frac{1}{[A]_{0}}\n$$\n\nor\n\n$$\n\\frac{1}{[A]}-\\frac{1}{[A]_{0}}=k t\n$$\n\nRestrict $t$ to $t_{1 / 2}$\n\n$$\nt=t_{1 / 2}\n$$\n\ndefine $[A]_{t}$ as one-half $[A]_{0}$\n\n$$\n[A]_{t}=\\frac{1}{2}[A]_{0}\n$$\n\nand then substitute into the integrated rate law and simplify:\n\n$$\n\\begin{aligned}\n\\frac{1}{\\frac{1}{2}[A]_{0}}-\\frac{1}{[A]_{0}} & =k t_{1 / 2} \\\\\n\\frac{2}{[A]_{0}}-\\frac{1}{[A]_{0}} & =k t_{1 / 2} \\\\\n\\frac{1}{[A]_{0}} & =k t_{1 / 2} \\\\\nt_{1 / 2} & =\\frac{1}{k[A]_{0}}\n\\end{aligned}\n$$\n\nFor a second-order reaction, $t_{1 / 2}$ is inversely proportional to the concentration of the reactant, and the half-life increases as the reaction proceeds because the concentration of reactant decreases. Unlike with first-order reactions, the rate constant of a second-order reaction cannot be calculated directly from the half-life unless the initial concentration is known."}
{"id": 4575, "contents": "1847. Zero-Order Reactions - \nAs for other reaction orders, an equation for zero-order half-life may be derived from the integrated rate law:\n\n$$\n[A]=-k t+[A]_{0}\n$$\n\nRestricting the time and concentrations to those defined by half-life: $t=t_{1 / 2}$ and $[A]=\\frac{[A]_{0}}{2}$. Substituting these terms into the zero-order integrated rate law yields:\n\n$$\n\\begin{aligned}\n\\frac{[\\mathrm{A}]_{0}}{2} & =-k t_{1 / 2}+[\\mathrm{A}]_{0} \\\\\nk t_{1 / 2} & =\\frac{[\\mathrm{A}]_{0}}{2} \\\\\nt_{1 / 2} & =\\frac{[A]_{0}}{2 k}\n\\end{aligned}\n$$\n\nAs for all reaction orders, the half-life for a zero-order reaction is inversely proportional to its rate constant. However, the half-life of a zero-order reaction increases as the initial concentration increases.\n\nEquations for both differential and integrated rate laws and the corresponding half-lives for zero-, first-, and second-order reactions are summarized in Table 17.2.\n\nSummary of Rate Laws for Zero-, First-, and Second-Order Reactions\n\n| Zero-Order |  | First-Order | Second-Order |\n| :--- | :--- | :--- | :--- |\n| rate law | rate $=k$ | rate $=k[A]$ | rate $=k[A]^{2}$ |\n| units of rate constant | $M s^{-1}$ | $\\mathrm{~s}^{-1}$ | $M^{-1} \\mathrm{~s}^{-1}$ |\n| integrated rate law | $[A]=-k t+[A]_{0}$ | $\\ln [A]=-k t+\\ln [A]_{0}$ | $\\frac{1}{[A]}=k t+\\left(\\frac{1}{[A]_{0}}\\right)$ |\n\nTABLE 17.2"}
{"id": 4576, "contents": "1847. Zero-Order Reactions - \nTABLE 17.2\n\n|  | Zero-Order | First-Order | Second-Order |\n| :--- | :--- | :--- | :--- |\n| plot needed for linear fit of rate data | $[A]$ vs. $t$ | $\\ln [A]$ vs. $t$ | $\\frac{1}{[A]}$ vs. $t$ |\n| relationship between slope of linear <br> plot and rate constant | $k=-$ slope | $k=-$ slope | $k=$ slope |\n| half-life | $t_{1 / 2}=\\frac{[A]_{0}}{2 k}$ | $t_{1 / 2}=\\frac{0.693}{k}$ | $t_{1 / 2}=\\frac{1}{[A]_{0} k}$ |\n\nTABLE 17.2"}
{"id": 4577, "contents": "1849. Half-Life for Zero-Order and Second-Order Reactions - \nWhat is the half-life for the butadiene dimerization reaction described in Example 17.8?"}
{"id": 4578, "contents": "1850. Solution - \nThe reaction in question is second order, is initiated with a $0.200 \\mathrm{~mol} \\mathrm{~L}^{-1}$ reactant solution, and exhibits a rate constant of $0.0576 \\mathrm{~L} \\mathrm{~mol}^{-1} \\mathrm{~min}^{-1}$. Substituting these quantities into the second-order half-life equation:\n\n$$\nt_{1 / 2}=\\frac{1}{\\left[\\left(0.0576 \\mathrm{~L} \\mathrm{~mol}^{-1} \\mathrm{~min}^{-1}\\right)\\left(0.200 \\mathrm{~mol} \\mathrm{~L}^{-1}\\right)\\right]}=18 \\mathrm{~min}\n$$"}
{"id": 4579, "contents": "1851. Check Your Learning - \nWhat is the half-life (min) for the thermal decomposition of ammonia on tungsten (see Figure 17.11)?"}
{"id": 4580, "contents": "1852. Answer: - \n87 min"}
{"id": 4581, "contents": "1853. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Use the postulates of collision theory to explain the effects of physical state, temperature, and concentration on reaction rates\n- Define the concepts of activation energy and transition state\n- Use the Arrhenius equation in calculations relating rate constants to temperature\n\nWe should not be surprised that atoms, molecules, or ions must collide before they can react with each other. Atoms must be close together to form chemical bonds. This simple premise is the basis for a very powerful theory that explains many observations regarding chemical kinetics, including factors affecting reaction rates.\n\nCollision theory is based on the following postulates:\n\n1. The rate of a reaction is proportional to the rate of reactant collisions:\n\n$$\n\\text { reaction rate } \\propto \\frac{\\# \\text { collisions }}{\\text { time }}\n$$\n\n2. The reacting species must collide in an orientation that allows contact between the atoms that will become bonded together in the product.\n3. The collision must occur with adequate energy to permit mutual penetration of the reacting species' valence shells so that the electrons can rearrange and form new bonds (and new chemical species).\n\nWe can see the importance of the two physical factors noted in postulates 2 and 3 , the orientation and energy\nof collisions, when we consider the reaction of carbon monoxide with oxygen:\n\n$$\n2 \\mathrm{CO}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{CO}_{2}(g)\n$$\n\nCarbon monoxide is a pollutant produced by the combustion of hydrocarbon fuels. To reduce this pollutant, automobiles have catalytic converters that use a catalyst to carry out this reaction. It is also a side reaction of the combustion of gunpowder that results in muzzle flash for many firearms. If carbon monoxide and oxygen are present in sufficient amounts, the reaction will occur at high temperature and pressure.\n\nThe first step in the gas-phase reaction between carbon monoxide and oxygen is a collision between the two molecules:\n\n$$\n\\mathrm{CO}(g)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)+\\mathrm{O}(g)\n$$"}
{"id": 4582, "contents": "1853. LEARNING OBJECTIVES - \n$$\n\\mathrm{CO}(g)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)+\\mathrm{O}(g)\n$$\n\nAlthough there are many different possible orientations the two molecules can have relative to each other, consider the two presented in Figure 17.13. In the first case, the oxygen side of the carbon monoxide molecule collides with the oxygen molecule. In the second case, the carbon side of the carbon monoxide molecule collides with the oxygen molecule. The second case is clearly more likely to result in the formation of carbon dioxide, which has a central carbon atom bonded to two oxygen atoms $(\\mathrm{O}=\\mathrm{C}=\\mathrm{O})$. This is a rather simple example of how important the orientation of the collision is in terms of creating the desired product of the reaction.\n\n\nFIGURE 17.13 Illustrated are two collisions that might take place between carbon monoxide and oxygen molecules. The orientation of the colliding molecules partially determines whether a reaction between the two molecules will occur.\n\nIf the collision does take place with the correct orientation, there is still no guarantee that the reaction will proceed to form carbon dioxide. In addition to a proper orientation, the collision must also occur with sufficient energy to result in product formation. When reactant species collide with both proper orientation and adequate energy, they combine to form an unstable species called an activated complex or a transition state. These species are very short lived and usually undetectable by most analytical instruments. In some cases, sophisticated spectral measurements have been used to observe transition states.\n\nCollision theory explains why most reaction rates increase as concentrations increase. With an increase in the concentration of any reacting substance, the chances for collisions between molecules are increased because there are more molecules per unit of volume. More collisions mean a faster reaction rate, assuming the energy of the collisions is adequate."}
{"id": 4583, "contents": "1854. Activation Energy and the Arrhenius Equation - \nThe minimum energy necessary to form a product during a collision between reactants is called the activation energy $\\left(\\boldsymbol{E}_{\\mathrm{a}}\\right)$. How this energy compares to the kinetic energy provided by colliding reactant molecules is a primary factor affecting the rate of a chemical reaction. If the activation energy is much larger than the average kinetic energy of the molecules, the reaction will occur slowly since only a few fast-moving molecules will have enough energy to react. If the activation energy is much smaller than the average kinetic energy of the molecules, a large fraction of molecules will be adequately energetic and the reaction will proceed rapidly.\n\nFigure 17.14 shows how the energy of a chemical system changes as it undergoes a reaction converting reactants to products according to the equation\n\n$$\nA+B \\longrightarrow C+D\n$$\n\nThese reaction diagrams are widely used in chemical kinetics to illustrate various properties of the reaction of interest. Viewing the diagram from left to right, the system initially comprises reactants only, $A+B$. Reactant molecules with sufficient energy can collide to form a high-energy activated complex or transition state. The unstable transition state can then subsequently decay to yield stable products, $C+D$. The diagram depicts the reaction's activation energy, $E_{a}$, as the energy difference between the reactants and the transition state. Using a specific energy, the enthalpy (see chapter on thermochemistry), the enthalpy change of the reaction, $\\Delta H$, is estimated as the energy difference between the reactants and products. In this case, the reaction is exothermic $(\\Delta H<0)$ since it yields a decrease in system enthalpy.\n\n\nFIGURE 17.14 Reaction diagram for the exothermic reaction $A+B \\longrightarrow C+D$.\nThe Arrhenius equation relates the activation energy and the rate constant, $k$, for many chemical reactions:\n\n$$\nk=A e^{-E_{\\mathrm{a}} / R T}\n$$"}
{"id": 4584, "contents": "1854. Activation Energy and the Arrhenius Equation - \n$$\nk=A e^{-E_{\\mathrm{a}} / R T}\n$$\n\nIn this equation, $R$ is the ideal gas constant, which has a value $8.314 \\mathrm{~J} / \\mathrm{mol} / \\mathrm{K}, \\mathrm{T}$ is temperature on the Kelvin scale, $E_{\\mathrm{a}}$ is the activation energy in joules per mole, $e$ is the constant 2.7183 , and $A$ is a constant called the frequency factor, which is related to the frequency of collisions and the orientation of the reacting molecules.\n\nPostulates of collision theory are nicely accommodated by the Arrhenius equation. The frequency factor, $A$, reflects how well the reaction conditions favor properly oriented collisions between reactant molecules. An increased probability of effectively oriented collisions results in larger values for $A$ and faster reaction rates.\nThe exponential term, $e^{-E a / R T}$, describes the effect of activation energy on reaction rate. According to kinetic molecular theory (see chapter on gases), the temperature of matter is a measure of the average kinetic energy of its constituent atoms or molecules. The distribution of energies among the molecules composing a sample of matter at any given temperature is described by the plot shown in Figure 17.15(a). Two shaded areas under the curve represent the numbers of molecules possessing adequate energy $(R T)$ to overcome the activation barriers $\\left(E_{a}\\right)$. A lower activation energy results in a greater fraction of adequately energized molecules and a faster reaction.\n\nThe exponential term also describes the effect of temperature on reaction rate. A higher temperature represents a correspondingly greater fraction of molecules possessing sufficient energy ( $R T$ ) to overcome the activation barrier $\\left(E_{a}\\right)$, as shown in Figure $17.15(\\mathbf{b})$. This yields a greater value for the rate constant and a correspondingly faster reaction rate.\n\n\nFIGURE 17.15 Molecular energy distributions showing numbers of molecules with energies exceeding (a) two different activation energies at a given temperature, and (b) a given activation energy at two different temperatures.\n\nA convenient approach for determining $E_{\\mathrm{a}}$ for a reaction involves the measurement of $k$ at two or more different temperatures and using an alternate version of the Arrhenius equation that takes the form of a linear equation"}
{"id": 4585, "contents": "1854. Activation Energy and the Arrhenius Equation - \nA convenient approach for determining $E_{\\mathrm{a}}$ for a reaction involves the measurement of $k$ at two or more different temperatures and using an alternate version of the Arrhenius equation that takes the form of a linear equation\n\n$$\n\\begin{aligned}\n\\ln k & =\\left(\\frac{-E_{\\mathrm{a}}}{R}\\right)\\left(\\frac{1}{T}\\right)+\\ln A \\\\\ny & =m x+b\n\\end{aligned}\n$$\n\nA plot of $\\ln k$ versus $\\frac{1}{T}$ is linear with a slope equal to $\\frac{-E_{\\mathrm{a}}}{R}$ and a $y$-intercept equal to $\\ln A$."}
{"id": 4586, "contents": "1855. Determination of $E_{\\mathrm{a}}$ - \nThe variation of the rate constant with temperature for the decomposition of $\\mathrm{HI}(g)$ to $\\mathrm{H}_{2}(g)$ and $\\mathrm{I}_{2}(g)$ is given here. What is the activation energy for the reaction?\n\n$$\n2 \\mathrm{HI}(g) \\longrightarrow \\mathrm{H}_{2}(g)+\\mathrm{I}_{2}(g)\n$$\n\n| $\\boldsymbol{T}(\\mathrm{K})$ | $k(\\mathrm{~L} / \\mathrm{mol} / \\mathrm{s})$ |\n| :---: | :---: |\n| 555 | $3.52 \\times 10^{-7}$ |\n| 575 | $1.22 \\times 10^{-6}$ |\n| 645 | $8.59 \\times 10^{-5}$ |\n| 700 | $1.16 \\times 10^{-3}$ |\n| 781 | $3.95 \\times 10^{-2}$ |"}
{"id": 4587, "contents": "1856. Solution - \nUse the provided data to derive values of $\\frac{1}{T}$ and $\\ln k$ :\n\n| $\\frac{1}{\\mathrm{~T}}\\left(\\mathrm{~K}^{-1}\\right)$ | $\\ln \\boldsymbol{k}$ |\n| :--- | :--- |\n| $1.80 \\times 10^{-3}$ | -14.860 |\n| $1.74 \\times 10^{-3}$ | -13.617 |\n\n\n| $\\frac{1}{\\mathrm{~T}}\\left(\\mathrm{~K}^{-1}\\right)$ | $\\ln k$ |\n| :--- | :--- |\n| $1.55 \\times 10^{-3}$ | -9.362 |\n| $1.43 \\times 10^{-3}$ | -6.759 |\n| $1.28 \\times 10^{-3}$ | -3.231 |\n\nFigure 17.16 is a graph of $\\ln k$ versus $\\frac{1}{T}$. In practice, the equation of the line (slope and $y$-intercept) that best fits these plotted data points would be derived using a statistical process called regression. This is helpful for most experimental data because a perfect fit of each data point with the line is rarely encountered. For the data here, the fit is nearly perfect and the slope may be estimated using any two of the provided data pairs. Using the first and last data points permits estimation of the slope.\n\n\nFIGURE 17.16 This graph shows the linear relationship between $\\ln k$ and $\\frac{1}{T}$ for the reaction $2 \\mathrm{HI} \\longrightarrow \\mathrm{H}_{2}+\\mathrm{I}_{2}$ according to the Arrhenius equation."}
{"id": 4588, "contents": "1856. Solution - \n$$\n\\begin{aligned}\n\\text { Slope }= & \\frac{\\Delta(\\ln k)}{\\Delta\\left(\\frac{1}{T}\\right)} \\\\\n= & \\frac{(-14.860)-(-3.231)}{\\left(1.80 \\times 10^{-3} \\mathrm{~K}^{-1}\\right)-\\left(1.28 \\times 10^{-3} \\mathrm{~K}^{-1}\\right)} \\\\\n= & \\frac{-11.629}{0.52 \\times 10^{-3} \\mathrm{~K}^{-1}}=-2.2 \\times 10^{4} \\mathrm{~K} \\\\\n= & -\\frac{E_{\\mathrm{a}}}{R} \\\\\nE_{\\mathrm{a}}= & - \\text { slope } \\times R=-\\left(-2.2 \\times 10^{4} \\mathrm{~K} \\times 8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}\\right) \\\\\n& 1.8 \\times 10^{5} \\mathrm{~J} \\mathrm{~mol}^{-1} \\text { or } 180 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\n\\end{aligned}\n$$\n\nAlternative approach: A more expedient approach involves deriving activation energy from measurements of the rate constant at just two temperatures. In this approach, the Arrhenius equation is rearranged to a convenient two-point form:\n\n$$\n\\ln \\frac{k_{1}}{k_{2}}=\\frac{E_{\\mathrm{a}}}{R}\\left(\\frac{1}{T_{2}}-\\frac{1}{T_{1}}\\right)\n$$\n\nRearranging this equation to isolate activation energy yields:\n\n$$\nE_{\\mathrm{a}}=-R\\left(\\frac{\\ln k_{2}-\\ln k_{1}}{\\left(\\frac{1}{T_{2}}\\right)-\\left(\\frac{1}{T_{1}}\\right)}\\right)\n$$\n\nAny two data pairs may be substituted into this equation-for example, the first and last entries from the above data table:"}
{"id": 4589, "contents": "1856. Solution - \nAny two data pairs may be substituted into this equation-for example, the first and last entries from the above data table:\n\n$$\nE_{\\mathrm{a}}=-8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}\\left(\\frac{-3.231-(-14.860)}{1.28 \\times 10^{-3} \\mathrm{~K}^{-1}-1.80 \\times 10^{-3} \\mathrm{~K}^{-1}}\\right)\n$$\n\nand the result is $E_{\\mathrm{a}}=1.8 \\times 10^{5} \\mathrm{~J} \\mathrm{~mol}^{-1}$ or $180 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nThis approach yields the same result as the more rigorous graphical approach used above, as expected. In practice, the graphical approach typically provides more reliable results when working with actual experimental data."}
{"id": 4590, "contents": "1857. Check Your Learning - \nThe rate constant for the rate of decomposition of $\\mathrm{N}_{2} \\mathrm{O}_{5}$ to NO and $\\mathrm{O}_{2}$ in the gas phase is $1.66 \\mathrm{~L} / \\mathrm{mol} / \\mathrm{s}$ at 650 K and $7.39 \\mathrm{~L} / \\mathrm{mol} / \\mathrm{s}$ at 700 K :\n\n$$\n2 \\mathrm{~N}_{2} \\mathrm{O}_{5}(g) \\longrightarrow 4 \\mathrm{NO}(g)+3 \\mathrm{O}_{2}(g)\n$$\n\nAssuming the kinetics of this reaction are consistent with the Arrhenius equation, calculate the activation energy for this decomposition."}
{"id": 4591, "contents": "1858. Answer: - \n$1.1 \\times 10^{5} \\mathrm{~J} \\mathrm{~mol}^{-1}$ or $110 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$"}
{"id": 4592, "contents": "1859. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Distinguish net reactions from elementary reactions (steps)\n- Identify the molecularity of elementary reactions\n- Write a balanced chemical equation for a process given its reaction mechanism\n- Derive the rate law consistent with a given reaction mechanism\n\nChemical reactions very often occur in a step-wise fashion, involving two or more distinct reactions taking place in sequence. A balanced equation indicates what is reacting and what is produced, but it reveals no details about how the reaction actually takes place. The reaction mechanism (or reaction path) provides details regarding the precise, step-by-step process by which a reaction occurs.\n\nThe decomposition of ozone, for example, appears to follow a mechanism with two steps:\n\n$$\n\\begin{aligned}\n& \\mathrm{O}_{3}(g) \\longrightarrow \\mathrm{O}_{2}(g)+\\mathrm{O} \\\\\n& \\mathrm{O}+\\mathrm{O}_{3}(g) \\longrightarrow 2 \\mathrm{O}_{2}(g)\n\\end{aligned}\n$$\n\nEach of the steps in a reaction mechanism is an elementary reaction. These elementary reactions occur precisely as represented in the step equations, and they must sum to yield the balanced chemical equation representing the overall reaction:\n\n$$\n2 \\mathrm{O}_{3}(g) \\longrightarrow 3 \\mathrm{O}_{2}(g)\n$$\n\nNotice that the oxygen atom produced in the first step of this mechanism is consumed in the second step and therefore does not appear as a product in the overall reaction. Species that are produced in one step and consumed in a subsequent step are called intermediates.\n\nWhile the overall reaction equation for the decomposition of ozone indicates that two molecules of ozone react\nto give three molecules of oxygen, the mechanism of the reaction does not involve the direct collision and reaction of two ozone molecules. Instead, one $\\mathrm{O}_{3}$ decomposes to yield $\\mathrm{O}_{2}$ and an oxygen atom, and a second $\\mathrm{O}_{3}$ molecule subsequently reacts with the oxygen atom to yield two additional $\\mathrm{O}_{2}$ molecules."}
{"id": 4593, "contents": "1859. LEARNING OBJECTIVES - \nUnlike balanced equations representing an overall reaction, the equations for elementary reactions are explicit representations of the chemical change taking place. The reactant(s) in an elementary reaction's equation undergo only the bond-breaking and/or making events depicted to yield the product(s). For this reason, the rate law for an elementary reaction may be derived directly from the balanced chemical equation describing the reaction. This is not the case for typical chemical reactions, for which rate laws may be reliably determined only via experimentation."}
{"id": 4594, "contents": "1860. Unimolecular Elementary Reactions - \nThe molecularity of an elementary reaction is the number of reactant species (atoms, molecules, or ions). For example, a unimolecular reaction involves the reaction of a single reactant species to produce one or more molecules of product:\n\n$$\nA \\longrightarrow \\text { products }\n$$\n\nThe rate law for a unimolecular reaction is first order:\n\n$$\n\\text { rate }=k[A]\n$$\n\nA unimolecular reaction may be one of several elementary reactions in a complex mechanism. For example, the reaction:\n\n$$\n\\mathrm{O}_{3} \\longrightarrow \\mathrm{O}_{2}+\\mathrm{O}\n$$\n\nillustrates a unimolecular elementary reaction that occurs as one part of a two-step reaction mechanism as described above. However, some unimolecular reactions may be the only step of a single-step reaction mechanism. (In other words, an \"overall\" reaction may also be an elementary reaction in some cases.) For example, the gas-phase decomposition of cyclobutane, $\\mathrm{C}_{4} \\mathrm{H}_{8}$, to ethylene, $\\mathrm{C}_{2} \\mathrm{H}_{4}$, is represented by the following chemical equation:\n\n\nThis equation represents the overall reaction observed, and it might also represent a legitimate unimolecular elementary reaction. The rate law predicted from this equation, assuming it is an elementary reaction, turns out to be the same as the rate law derived experimentally for the overall reaction, namely, one showing firstorder behavior:\n\n$$\n\\text { rate }=-\\frac{\\Delta\\left[\\mathrm{C}_{4} \\mathrm{H}_{8}\\right]}{\\Delta t}=k\\left[\\mathrm{C}_{4} \\mathrm{H}_{8}\\right]\n$$\n\nThis agreement between observed and predicted rate laws is interpreted to mean that the proposed unimolecular, single-step process is a reasonable mechanism for the butadiene reaction."}
{"id": 4595, "contents": "1861. Bimolecular Elementary Reactions - \nA bimolecular reaction involves two reactant species, for example:\n\n$$\n\\begin{gathered}\nA+B \\longrightarrow \\text { products } \\\\\n\\text { and } \\\\\n2 A \\longrightarrow \\text { products }\n\\end{gathered}\n$$\n\nFor the first type, in which the two reactant molecules are different, the rate law is first-order in $A$ and first order in $B$ (second-order overall):\n\n$$\n\\text { rate }=kA\n$$\n\nFor the second type, in which two identical molecules collide and react, the rate law is second order in $A$ :\n\n$$\n\\text { rate }=kA=k[A]^{2}\n$$\n\nSome chemical reactions occur by mechanisms that consist of a single bimolecular elementary reaction. One example is the reaction of nitrogen dioxide with carbon monoxide:\n\n$$\n\\mathrm{NO}_{2}(g)+\\mathrm{CO}(g) \\longrightarrow \\mathrm{NO}(g)+\\mathrm{CO}_{2}(g)\n$$\n\n(see Figure 17.17)\n\n\nFIGURE 17.17 The probable mechanism for the reaction between $\\mathrm{NO}_{2}$ and CO to yield NO and $\\mathrm{CO}_{2}$.\nBimolecular elementary reactions may also be involved as steps in a multistep reaction mechanism. The reaction of atomic oxygen with ozone is the second step of the two-step ozone decomposition mechanism discussed earlier in this section:\n\n$$\n\\mathrm{O}(\\mathrm{~g})+\\mathrm{O}_{3}(\\mathrm{~g}) \\longrightarrow 2 \\mathrm{O}_{2}(\\mathrm{~g})\n$$"}
{"id": 4596, "contents": "1862. Termolecular Elementary Reactions - \nAn elementary termolecular reaction involves the simultaneous collision of three atoms, molecules, or ions. Termolecular elementary reactions are uncommon because the probability of three particles colliding simultaneously is less than one one-thousandth of the probability of two particles colliding. There are, however, a few established termolecular elementary reactions. The reaction of nitric oxide with oxygen appears to involve termolecular steps:\n\n$$\n\\begin{aligned}\n& 2 \\mathrm{NO}+\\mathrm{O}_{2} \\longrightarrow 2 \\mathrm{NO}_{2} \\\\\n& \\text { rate }=k[\\mathrm{NO}]^{2}\\left[\\mathrm{O}_{2}\\right]\n\\end{aligned}\n$$\n\nLikewise, the reaction of nitric oxide with chlorine appears to involve termolecular steps:\n\n$$\n\\begin{aligned}\n& 2 \\mathrm{NO}+\\mathrm{Cl}_{2} \\longrightarrow 2 \\mathrm{NOCl} \\\\\n& \\text { rate }=k[\\mathrm{NO}]^{2}\\left[\\mathrm{Cl}_{2}\\right]\n\\end{aligned}\n$$"}
{"id": 4597, "contents": "1863. Relating Reaction Mechanisms to Rate Laws - \nIt's often the case that one step in a multistep reaction mechanism is significantly slower than the others. Because a reaction cannot proceed faster than its slowest step, this step will limit the rate at which the overall reaction occurs. The slowest step is therefore called the rate-limiting step (or rate-determining step) of the reaction Figure 17.18.\n\n\nFIGURE 17.18 A cattle chute is a nonchemical example of a rate-determining step. Cattle can only be moved from one holding pen to another as quickly as one animal can make its way through the chute. (credit: Loren Kerns)\n\nAs described earlier, rate laws may be derived directly from the chemical equations for elementary reactions. This is not the case, however, for ordinary chemical reactions. The balanced equations most often encountered represent the overall change for some chemical system, and very often this is the result of some multistep reaction mechanisms. In every case, the rate law must be determined from experimental data and the reaction mechanism subsequently deduced from the rate law (and sometimes from other data). The reaction of $\\mathrm{NO}_{2}$ and CO provides an illustrative example:\n\n$$\n\\mathrm{NO}_{2}(g)+\\mathrm{CO}(g) \\longrightarrow \\mathrm{CO}_{2}(g)+\\mathrm{NO}(g)\n$$\n\nFor temperatures above $225^{\\circ} \\mathrm{C}$, the rate law has been found to be:\n\n$$\n\\text { rate }=k\\left\\mathrm{NO}_{2}\\right\n$$\n\nThe reaction is first order with respect to $\\mathrm{NO}_{2}$ and first-order with respect to CO . This is consistent with a single-step bimolecular mechanism and it is possible that this is the mechanism for this reaction at high temperatures.\n\nAt temperatures below $225^{\\circ} \\mathrm{C}$, the reaction is described by a rate law that is second order with respect to $\\mathrm{NO}_{2}$ :\n\n$$\n\\text { rate }=k\\left[\\mathrm{NO}_{2}\\right]^{2}\n$$\n\nThis rate law is not consistent with the single-step mechanism, but is consistent with the following two-step mechanism:"}
{"id": 4598, "contents": "1863. Relating Reaction Mechanisms to Rate Laws - \n$$\n\\text { rate }=k\\left[\\mathrm{NO}_{2}\\right]^{2}\n$$\n\nThis rate law is not consistent with the single-step mechanism, but is consistent with the following two-step mechanism:\n\n$$\n\\begin{aligned}\n& \\mathrm{NO}_{2}(g)+\\mathrm{NO}_{2}(g) \\longrightarrow \\mathrm{NO}_{3}(g)+\\mathrm{NO}(g) \\text { (slow) } \\\\\n& \\mathrm{NO}_{3}(g)+\\mathrm{CO}(g) \\longrightarrow \\mathrm{NO}_{2}(g)+\\mathrm{CO}_{2}(g) \\text { (fast) }\n\\end{aligned}\n$$\n\nThe rate-determining (slower) step gives a rate law showing second-order dependence on the $\\mathrm{NO}_{2}$ concentration, and the sum of the two equations gives the net overall reaction.\n\nIn general, when the rate-determining (slower) step is the first step in a mechanism, the rate law for the overall reaction is the same as the rate law for this step. However, when the rate-determining step is preceded by a step involving a rapidly reversible reaction the rate law for the overall reaction may be more difficult to derive.\n\nAs discussed in several chapters of this text, a reversible reaction is at equilibrium when the rates of the forward and reverse processes are equal. Consider the reversible elementary reaction in which NO dimerizes to yield an intermediate species $\\mathrm{N}_{2} \\mathrm{O}_{2}$. When this reaction is at equilibrium:\n\n$$\n\\begin{aligned}\n& \\mathrm{NO}+\\mathrm{NO} \\rightleftharpoons \\mathrm{~N}_{2} \\mathrm{O}_{2} \\\\\n& \\mathrm{rate}_{\\text {forward }}=\\text { rate }_{\\text {reverse }} \\\\\n& k_{1}[\\mathrm{NO}]^{2}=k_{-1}\\left[\\mathrm{~N}_{2} \\mathrm{O}_{2}\\right]\n\\end{aligned}\n$$\n\nThis expression may be rearranged to express the concentration of the intermediate in terms of the reactant NO:"}
{"id": 4599, "contents": "1863. Relating Reaction Mechanisms to Rate Laws - \nThis expression may be rearranged to express the concentration of the intermediate in terms of the reactant NO:\n\n$$\n\\left(\\frac{\\mathrm{k}_{1}[\\mathrm{NO}]^{2}}{\\mathrm{k}_{-1}}\\right)=\\left[\\mathrm{N}_{2} \\mathrm{O}_{2}\\right]\n$$\n\nSince intermediate species concentrations are not used in formulating rate laws for overall reactions, this approach is sometimes necessary, as illustrated in the following example exercise."}
{"id": 4600, "contents": "1865. Deriving a Rate Law from a Reaction Mechanism - \nThe two-step mechanism below has been proposed for a reaction between nitrogen monoxide and molecular chlorine:\n\n$$\n\\begin{array}{ll}\n\\text { Step 1: } \\mathrm{NO}(g)+\\mathrm{Cl}_{2}(g) \\rightleftharpoons \\mathrm{NOCl}_{2}(g) & \\text { fast } \\\\\n\\text { Step 2: } \\mathrm{NOCl}_{2}(g)+\\mathrm{NO}(g) \\longrightarrow 2 \\mathrm{NOCl}(g) & \\text { slow }\n\\end{array}\n$$\n\nUse this mechanism to derive the equation and predicted rate law for the overall reaction."}
{"id": 4601, "contents": "1866. Solution - \nThe equation for the overall reaction is obtained by adding the two elementary reactions:\n\n$$\n2 \\mathrm{NO}(g)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{NOCl}(g)\n$$\n\nTo derive a rate law from this mechanism, first write rates laws for each of the two steps.\n\n$$\n\\begin{aligned}\n& \\operatorname{rate}_{1}=k_{1}[\\mathrm{NO}]\\left[\\mathrm{Cl}_{2}\\right] \\text { for the forward reaction of step } 1 \\\\\n& \\text { rate }_{-1}=k_{-1}\\left[\\mathrm{NOCl}_{2}\\right] \\text { for the reverse reaction of step } 1 \\\\\n& \\operatorname{rate}_{2}=k_{2}\\left\\mathrm{NOCl}_{2}\\right \\text { for step } 2\n\\end{aligned}\n$$\n\nStep 2 is the rate-determining step, and so the rate law for the overall reaction should be the same as for this step. However, the step 2 rate law, as written, contains an intermediate species concentration, [ $\\mathrm{NOCl}_{2}$ ]. To remedy this, use the first step's rate laws to derive an expression for the intermediate concentration in terms of the reactant concentrations.\n\nAssuming step 1 is at equilibrium:\n\n$$\n\\begin{aligned}\n\\text { rate }_{1} & =\\text { rate }_{-1} \\\\\nk_{1}[\\mathrm{NO}]\\left[\\mathrm{Cl}_{2}\\right] & =k_{-1}\\left[\\mathrm{NOCl}_{2}\\right] \\\\\n{\\left[\\mathrm{NOCl}_{2}\\right] } & =\\left(\\frac{k_{1}}{k_{-1}}\\right)[\\mathrm{NO}]\\left[\\mathrm{Cl}_{2}\\right]\n\\end{aligned}\n$$\n\nSubstituting this expression into the rate law for step 2 yields:\n\n$$\n\\operatorname{rate}_{2}=\\operatorname{rate}_{\\text {overall }}=\\left(\\frac{k_{2} k_{1}}{k_{-1}}\\right)[\\mathrm{NO}]^{2}\\left[\\mathrm{Cl}_{2}\\right]\n$$"}
{"id": 4602, "contents": "1867. Check Your Learning - \nThe first step of a proposed multistep mechanism is:\n\n$$\n\\mathrm{F}_{2}(g) \\rightleftharpoons 2 \\mathrm{~F}(g) \\text { fast }\n$$\n\nDerive the equation relating atomic fluorine concentration to molecular fluorine concentration."}
{"id": 4603, "contents": "1868. Answer: - \n$[\\mathrm{F}]=\\left(\\frac{k_{1}\\left[\\mathrm{~F}_{2}\\right]}{k_{-1}}\\right)^{1 / 2}$"}
{"id": 4604, "contents": "1868. Answer: - 1868.1. Catalysis\nLEARNING OBJECTIVES\nBy the end of this section, you will be able to:\n\n- Explain the function of a catalyst in terms of reaction mechanisms and potential energy diagrams\n- List examples of catalysis in natural and industrial processes"}
{"id": 4605, "contents": "1869. Catalysts Do Not Affect Equilibrium - \nA catalyst can speed up the rate of a reaction. Though this increase in reaction rate may cause a system to reach equilibrium more quickly (by speeding up the forward and reverse reactions), a catalyst has no effect on the value of an equilibrium constant nor on equilibrium concentrations.\n\nThe interplay of changes in concentration or pressure, temperature, and the lack of an influence of a catalyst on a chemical equilibrium is illustrated in the industrial synthesis of ammonia from nitrogen and hydrogen according to the equation\n\n$$\n\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g)\n$$\n\nA large quantity of ammonia is manufactured by this reaction. Each year, ammonia is among the top 10 chemicals, by mass, manufactured in the world. About 2 billion pounds are manufactured in the United States each year.\n\nAmmonia plays a vital role in our global economy. It is used in the production of fertilizers and is, itself, an important fertilizer for the growth of corn, cotton, and other crops. Large quantities of ammonia are converted to nitric acid, which plays an important role in the production of fertilizers, explosives, plastics, dyes, and fibers, and is also used in the steel industry."}
{"id": 4606, "contents": "1871. Fritz Haber - \nIn the early 20th century, German chemist Fritz Haber (Figure 17.19) developed a practical process for converting diatomic nitrogen, which cannot be used by plants as a nutrient, to ammonia, a form of nitrogen that is easiest for plants to absorb.\n\n$$\n\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g)\n$$\n\nThe availability of nitrogen is a strong limiting factor to the growth of plants. Despite accounting for $78 \\%$ of air, diatomic nitrogen $\\left(\\mathrm{N}_{2}\\right)$ is nutritionally unavailable due the tremendous stability of the nitrogennitrogen triple bond. For plants to use atmospheric nitrogen, the nitrogen must be converted to a more bioavailable form (this conversion is called nitrogen fixation).\n\nHaber was born in Breslau, Prussia (presently Wroclaw, Poland) in December 1868. He went on to study chemistry and, while at the University of Karlsruhe, he developed what would later be known as the Haber process: the catalytic formation of ammonia from hydrogen and atmospheric nitrogen under high temperatures and pressures. For this work, Haber was awarded the 1918 Nobel Prize in Chemistry for synthesis of ammonia from its elements. The Haber process was a boon to agriculture, as it allowed the production of fertilizers to no longer be dependent on mined feed stocks such as sodium nitrate. Currently, the annual production of synthetic nitrogen fertilizers exceeds 100 million tons and synthetic fertilizer production has increased the number of humans that arable land can support from 1.9 persons per hectare in 1908 to 4.3 in 2008.\n\n\nFIGURE 17.19 The work of Nobel Prize recipient Fritz Haber revolutionized agricultural practices in the early 20th century. His work also affected wartime strategies, adding chemical weapons to the artillery."}
{"id": 4607, "contents": "1871. Fritz Haber - \nFIGURE 17.19 The work of Nobel Prize recipient Fritz Haber revolutionized agricultural practices in the early 20th century. His work also affected wartime strategies, adding chemical weapons to the artillery.\n\nIn addition to his work in ammonia production, Haber is also remembered by history as one of the fathers of chemical warfare. During World War I, he played a major role in the development of poisonous gases used for trench warfare. Regarding his role in these developments, Haber said, \"During peace time a scientist belongs to the World, but during war time he belongs to his country.\" ${ }^{1}$ Haber defended the use of gas warfare against accusations that it was inhumane, saying that death was death, by whatever means it was inflicted. He stands as an example of the ethical dilemmas that face scientists in times of war and the double-edged nature of the sword of science.\n\nLike Haber, the products made from ammonia can be multifaceted. In addition to their value for agriculture, nitrogen compounds can also be used to achieve destructive ends. Ammonium nitrate has also been used in explosives, including improvised explosive devices. Ammonium nitrate was one of the components of the bomb used in the attack on the Alfred P. Murrah Federal Building in downtown Oklahoma City on April 19, 1995.\n\nIt has long been known that nitrogen and hydrogen react to form ammonia. However, it became possible to manufacture ammonia in useful quantities by the reaction of nitrogen and hydrogen only in the early 20th century after the factors that influence its equilibrium were understood.\n\nTo be practical, an industrial process must give a large yield of product relatively quickly. One way to increase the yield of ammonia is to increase the pressure on the system in which $\\mathrm{N}_{2}, \\mathrm{H}_{2}$, and $\\mathrm{NH}_{3}$ are at equilibrium or are coming to equilibrium.\n\n$$\n\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g)\n$$\n\nThe formation of additional amounts of ammonia reduces the total pressure exerted by the system and somewhat reduces the stress of the increased pressure."}
{"id": 4608, "contents": "1871. Fritz Haber - \nThe formation of additional amounts of ammonia reduces the total pressure exerted by the system and somewhat reduces the stress of the increased pressure.\n\nAlthough increasing the pressure of a mixture of $\\mathrm{N}_{2}, \\mathrm{H}_{2}$, and $\\mathrm{NH}_{3}$ will increase the yield of ammonia, at low temperatures, the rate of formation of ammonia is slow. At room temperature, for example, the reaction is so slow that if we prepared a mixture of $\\mathrm{N}_{2}$ and $\\mathrm{H}_{2}$, no detectable amount of ammonia would form during our lifetime. The formation of ammonia from hydrogen and nitrogen is an exothermic process:\n\n$$\n\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\longrightarrow 2 \\mathrm{NH}_{3}(g) \\quad \\Delta H=-92.2 \\mathrm{~kJ}\n$$\n\nThus, increasing the temperature to increase the rate lowers the yield. If we lower the temperature to shift the equilibrium to favor the formation of more ammonia, equilibrium is reached more slowly because of the large decrease of reaction rate with decreasing temperature.\n\nPart of the rate of formation lost by operating at lower temperatures can be recovered by using a catalyst. The net effect of the catalyst on the reaction is to cause equilibrium to be reached more rapidly.\n\n[^10]In the commercial production of ammonia, conditions of about $500^{\\circ} \\mathrm{C}, 150-900 \\mathrm{~atm}$, and the presence of a catalyst are used to give the best compromise among rate, yield, and the cost of the equipment necessary to produce and contain high-pressure gases at high temperatures (Figure 17.20).\n\n\nFIGURE 17.20 Commercial production of ammonia requires heavy equipment to handle the high temperatures and pressures required. This schematic outlines the design of an ammonia plant.\n\nAmong the factors affecting chemical reaction rates discussed earlier in this chapter was the presence of a catalyst, a substance that can increase the reaction rate without being consumed in the reaction. The concepts introduced in the previous section on reaction mechanisms provide the basis for understanding how catalysts are able to accomplish this very important function."}
{"id": 4609, "contents": "1871. Fritz Haber - \nFigure 17.21 shows reaction diagrams for a chemical process in the absence and presence of a catalyst. Inspection of the diagrams reveals several traits of these reactions. Consistent with the fact that the two diagrams represent the same overall reaction, both curves begin and end at the same energies (in this case, because products are more energetic than reactants, the reaction is endothermic). The reaction mechanisms, however, are clearly different. The uncatalyzed reaction proceeds via a one-step mechanism (one transition state observed), whereas the catalyzed reaction follows a two-step mechanism (two transition states observed) with a notably lesser activation energy. This difference illustrates the means by which a catalyst functions to accelerate reactions, namely, by providing an alternative reaction mechanism with a lower activation energy. Although the catalyzed reaction mechanism for a reaction needn't necessarily involve a different number of steps than the uncatalyzed mechanism, it must provide a reaction path whose rate determining step is faster (lower $E_{\\mathrm{a}}$ ).\n\n\nFIGURE 17.21 Reaction diagrams for an endothermic process in the absence (red curve) and presence (blue curve) of a catalyst. The catalyzed pathway involves a two-step mechanism (note the presence of two transition states) and an intermediate species (represented by the valley between the two transitions states)."}
{"id": 4610, "contents": "1873. Reaction Diagrams for Catalyzed Reactions - \nThe two reaction diagrams here represent the same reaction: one without a catalyst and one with a catalyst. Estimate the activation energy for each process, and identify which one involves a catalyst.\n\n(a)\n\n(b)"}
{"id": 4611, "contents": "1874. Solution - \nActivation energies are calculated by subtracting the reactant energy from the transition state energy.\n\n$$\n\\begin{aligned}\n& \\operatorname{diagram}(\\mathrm{a}): E_{\\mathrm{a}}=32 \\mathrm{~kJ}-6 \\mathrm{~kJ}=26 \\mathrm{~kJ} \\\\\n& \\operatorname{diagram}(\\mathrm{~b}): E_{\\mathrm{a}}=20 \\mathrm{~kJ}-6 \\mathrm{~kJ}=14 \\mathrm{~kJ}\n\\end{aligned}\n$$\n\nThe catalyzed reaction is the one with lesser activation energy, in this case represented by diagram (b)."}
{"id": 4612, "contents": "1875. Check Your Learning - \nReaction diagrams for a chemical process with and without a catalyst are shown below. Both reactions involve a two-step mechanism with a rate-determining first step. Compute activation energies for the first step of each mechanism, and identify which corresponds to the catalyzed reaction. How do the second steps of these two mechanisms compare?"}
{"id": 4613, "contents": "1876. Answer: - \nFor the first step, $E_{\\mathrm{a}}=80 \\mathrm{~kJ}$ for (a) and 70 kJ for (b), so diagram (b) depicts the catalyzed reaction. Activation energies for the second steps of both mechanisms are the same, 20 kJ ."}
{"id": 4614, "contents": "1877. Homogeneous Catalysts - \nA homogeneous catalyst is present in the same phase as the reactants. It interacts with a reactant to form an intermediate substance, which then decomposes or reacts with another reactant in one or more steps to regenerate the original catalyst and form product.\n\nAs an important illustration of homogeneous catalysis, consider the earth's ozone layer. Ozone in the upper atmosphere, which protects the earth from ultraviolet radiation, is formed when oxygen molecules absorb ultraviolet light and undergo the reaction:\n\n$$\n3 \\mathrm{O}_{2}(\\mathrm{~g}) \\xrightarrow{h v} 2 \\mathrm{O}_{3}(\\mathrm{~g})\n$$\n\nOzone is a relatively unstable molecule that decomposes to yield diatomic oxygen by the reverse of this equation. This decomposition reaction is consistent with the following two-step mechanism:\n\n$$\n\\begin{aligned}\n& \\mathrm{O}_{3} \\longrightarrow \\mathrm{O}_{2}+\\mathrm{O} \\\\\n& \\mathrm{O}+\\mathrm{O}_{3} \\longrightarrow 2 \\mathrm{O}_{2}\n\\end{aligned}\n$$\n\nA number of substances can catalyze the decomposition of ozone. For example, the nitric oxide -catalyzed decomposition of ozone is believed to occur via the following three-step mechanism:\n\n$$\n\\begin{aligned}\n& \\mathrm{NO}(g)+\\mathrm{O}_{3}(g) \\longrightarrow \\mathrm{NO}_{2}(g)+\\mathrm{O}_{2}(g) \\\\\n& \\mathrm{O}_{3}(g) \\longrightarrow \\mathrm{O}_{2}(g)+\\mathrm{O}(g) \\\\\n& \\mathrm{NO}_{2}(g)+\\mathrm{O}(g) \\longrightarrow \\mathrm{NO}(g)+\\mathrm{O}_{2}(g)\n\\end{aligned}\n$$\n\nAs required, the overall reaction is the same for both the two-step uncatalyzed mechanism and the three-step NO-catalyzed mechanism:\n\n$$\n2 \\mathrm{O}_{3}(g) \\longrightarrow 3 \\mathrm{O}_{2}(g)\n$$\n\nNotice that NO is a reactant in the first step of the mechanism and a product in the last step. This is another characteristic trait of a catalyst: Though it participates in the chemical reaction, it is not consumed by the reaction."}
{"id": 4615, "contents": "1879. Mario J. Molina - \nThe 1995 Nobel Prize in Chemistry was shared by Paul J. Crutzen, Mario J. Molina (Figure 17.22), and F. Sherwood Rowland \"for their work in atmospheric chemistry, particularly concerning the formation and decomposition of ozone.\" ${ }^{2}$ Molina, a Mexican citizen, carried out the majority of his work at the Massachusetts Institute of Technology (MIT).\n\n\nFIGURE 17.22 (a) Mexican chemist Mario Molina (1943 -) shared the Nobel Prize in Chemistry in 1995 for his research on (b) the Antarctic ozone hole. (credit a: courtesy of Mario Molina; credit b: modification of work by NASA)\n\nIn 1974, Molina and Rowland published a paper in the journal Nature detailing the threat of chlorofluorocarbon gases to the stability of the ozone layer in earth's upper atmosphere. The ozone layer protects earth from solar radiation by absorbing ultraviolet light. As chemical reactions deplete the amount of ozone in the upper atmosphere, a measurable \"hole\" forms above Antarctica, and an increase in the amount of solar ultraviolet radiation - strongly linked to the prevalence of skin cancers-reaches earth's surface. The work of Molina and Rowland was instrumental in the adoption of the Montreal Protocol, an international treaty signed in 1987 that successfully began phasing out production of chemicals linked to ozone destruction.\n\nMolina and Rowland demonstrated that chlorine atoms from human-made chemicals can catalyze ozone destruction in a process similar to that by which NO accelerates the depletion of ozone. Chlorine atoms are generated when chlorocarbons or chlorofluorocarbons-once widely used as refrigerants and propellants-are photochemically decomposed by ultraviolet light or react with hydroxyl radicals. A sample mechanism is shown here using methyl chloride:\n\n$$\n\\mathrm{CH}_{3} \\mathrm{Cl}+\\mathrm{OH} \\longrightarrow \\mathrm{Cl}+\\text { other products }\n$$\n\nChlorine radicals break down ozone and are regenerated by the following catalytic cycle:"}
{"id": 4616, "contents": "1879. Mario J. Molina - \nChlorine radicals break down ozone and are regenerated by the following catalytic cycle:\n\n$$\n\\begin{aligned}\n& \\mathrm{Cl}+\\mathrm{O}_{3} \\longrightarrow \\mathrm{ClO}+\\mathrm{O}_{2} \\\\\n& \\mathrm{ClO}+\\mathrm{O} \\longrightarrow \\mathrm{Cl}+\\mathrm{O}_{2} \\\\\n& \\text { overall Reaction: } \\mathrm{O}_{3}+\\mathrm{O} \\longrightarrow 2 \\mathrm{O}_{2}\n\\end{aligned}\n$$\n\nA single monatomic chlorine can break down thousands of ozone molecules. Luckily, the majority of atmospheric chlorine exists as the catalytically inactive forms $\\mathrm{Cl}_{2}$ and $\\mathrm{ClONO}_{2}$.\n\n[^11]Since receiving his portion of the Nobel Prize, Molina has continued his work in atmospheric chemistry at MIT."}
{"id": 4617, "contents": "1881. Glucose-6-Phosphate Dehydrogenase Deficiency - \nEnzymes in the human body act as catalysts for important chemical reactions in cellular metabolism. As such, a deficiency of a particular enzyme can translate to a life-threatening disease. G6PD (glucose-6-phosphate dehydrogenase) deficiency, a genetic condition that results in a shortage of the enzyme glucose-6-phosphate dehydrogenase, is the most common enzyme deficiency in humans. This enzyme, shown in Figure 17.23, is the rate-limiting enzyme for the metabolic pathway that supplies NADPH to cells (Figure 17.24).\n\n\nFIGURE 17.23 Glucose-6-phosphate dehydrogenase is a rate-limiting enzyme for the metabolic pathway that supplies NADPH to cells.\n\nA disruption in this pathway can lead to reduced glutathione in red blood cells; once all glutathione is consumed, enzymes and other proteins such as hemoglobin are susceptible to damage. For example, hemoglobin can be metabolized to bilirubin, which leads to jaundice, a condition that can become severe. People who suffer from G6PD deficiency must avoid certain foods and medicines containing chemicals that can trigger damage their glutathione-deficient red blood cells.\n\n\nFIGURE 17.24 In the mechanism for the pentose phosphate pathway, G6PD catalyzes the reaction that regulates NADPH, a co-enzyme that regulates glutathione, an antioxidant that protects red blood cells and other cells from oxidative damage."}
{"id": 4618, "contents": "1882. Heterogeneous Catalysts - \nA heterogeneous catalyst is a catalyst that is present in a different phase (usually a solid) than the reactants. Such catalysts generally function by furnishing an active surface upon which a reaction can occur. Gas and liquid phase reactions catalyzed by heterogeneous catalysts occur on the surface of the catalyst rather than within the gas or liquid phase.\n\nHeterogeneous catalysis typically involves the following processes:\n\n1. Adsorption of the reactant(s) onto the surface of the catalyst\n2. Activation of the adsorbed reactant(s)\n3. Reaction of the adsorbed reactant(s)\n4. Desorption of product(s) from the surface of the catalyst\n\nFigure 17.25 illustrates the steps of a mechanism for the reaction of compounds containing a carbon-carbon double bond with hydrogen on a nickel catalyst. Nickel is the catalyst used in the hydrogenation of polyunsaturated fats and oils (which contain several carbon-carbon double bonds) to produce saturated fats and oils (which contain only carbon-carbon single bonds).\n\n\nFIGURE 17.25 Mechanism for the Ni-catalyzed reaction $\\mathrm{C}_{2} \\mathrm{H}_{4}+\\mathrm{H}_{2} \\longrightarrow \\mathrm{C}_{2} \\mathrm{H}_{6}$. (a) Hydrogen is adsorbed on the surface, breaking the $\\mathrm{H}-\\mathrm{H}$ bonds and forming $\\mathrm{Ni}-\\mathrm{H}$ bonds. (b) Ethylene is adsorbed on the surface, breaking the $\\mathrm{C}-\\mathrm{C}$ $\\pi$-bond and forming $\\mathrm{Ni}-\\mathrm{C}$ bonds. (c) Atoms diffuse across the surface and form new $\\mathrm{C}-\\mathrm{H}$ bonds when they collide.\n(d) $\\mathrm{C}_{2} \\mathrm{H}_{6}$ molecules desorb from the Ni surface.\n\nMany important chemical products are prepared via industrial processes that use heterogeneous catalysts, including ammonia, nitric acid, sulfuric acid, and methanol. Heterogeneous catalysts are also used in the catalytic converters found on most gasoline-powered automobiles (Figure 17.26)."}
{"id": 4619, "contents": "1884. Automobile Catalytic Converters - \nScientists developed catalytic converters to reduce the amount of toxic emissions produced by burning gasoline in internal combustion engines. By utilizing a carefully selected blend of catalytically active metals, it is possible to effect complete combustion of all carbon-containing compounds to carbon dioxide while also reducing the output of nitrogen oxides. This is particularly impressive when we consider that one step involves adding more oxygen to the molecule and the other involves removing the oxygen (Figure 17.26).\n\n\nFIGURE 17.26 A catalytic converter allows for the combustion of all carbon-containing compounds to carbon dioxide, while at the same time reducing the output of nitrogen oxide and other pollutants in emissions from gasoline-burning engines.\n\nMost modern, three-way catalytic converters possess a surface impregnated with a platinum-rhodium catalyst, which catalyzes the conversion of nitric oxide into dinitrogen and oxygen as well as the conversion of carbon monoxide and hydrocarbons such as octane into carbon dioxide and water vapor:\n\n$$\n\\begin{aligned}\n& 2 \\mathrm{NO}_{2}(g) \\longrightarrow \\mathrm{N}_{2}(g)+2 \\mathrm{O}_{2}(g) \\\\\n& 2 \\mathrm{CO}(g)+\\mathrm{O}_{2}(\\mathrm{~g}) \\longrightarrow 2 \\mathrm{CO}_{2}(\\mathrm{~g}) \\\\\n& 2 \\mathrm{C}_{8} \\mathrm{H}_{18}(g)+25 \\mathrm{O}_{2}(\\mathrm{~g}) \\longrightarrow 16 \\mathrm{CO}_{2}(\\mathrm{~g})+18 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{~g})\n\\end{aligned}\n$$\n\nIn order to be as efficient as possible, most catalytic converters are preheated by an electric heater. This ensures that the metals in the catalyst are fully active even before the automobile exhaust is hot enough to maintain appropriate reaction temperatures."}
{"id": 4620, "contents": "1885. LINK TO LEARNING - \nThe University of California at Davis' \"ChemWiki\" provides a thorough explanation (http://openstax.org/l/ 16 catconvert) of how catalytic converters work."}
{"id": 4621, "contents": "1887. Enzyme Structure and Function - \nThe study of enzymes is an important interconnection between biology and chemistry. Enzymes are usually proteins (polypeptides) that help to control the rate of chemical reactions between biologically important compounds, particularly those that are involved in cellular metabolism. Different classes of enzymes perform a variety of functions, as shown in Table 17.3.\n\nClasses of Enzymes and Their Functions\n\n| Class | Function |\n| :--- | :--- |\n| oxidoreductases | redox reactions |\n| transferases | transfer of functional groups |\n| hydrolases | hydrolysis reactions |\n| lyases | group elimination to form double bonds |\n| isomerases | isomerization |\n| ligases | bond formation with ATP hydrolysis |\n\nTABLE 17.3\n\nEnzyme molecules possess an active site, a part of the molecule with a shape that allows it to bond to a specific substrate (a reactant molecule), forming an enzyme-substrate complex as a reaction intermediate. There are two models that attempt to explain how this active site works. The most simplistic model is referred to as the lock-and-key hypothesis, which suggests that the molecular shapes of the active site and substrate are complementary, fitting together like a key in a lock. The induced fit hypothesis, on the other hand, suggests that the enzyme molecule is flexible and changes shape to accommodate a bond with the substrate. This is not to suggest that an enzyme's active site is completely malleable, however. Both the lock-and-key model and the induced fit model account for the fact that enzymes can only bind with specific substrates, since in general a particular enzyme only catalyzes a particular reaction (Figure 17.27).\n\n\nFIGURE 17.27 (a) According to the lock-and-key model, the shape of an enzyme's active site is a perfect fit for the substrate. (b) According to the induced fit model, the active site is somewhat flexible, and can change shape in order to bond with the substrate."}
{"id": 4622, "contents": "1888. LINK TO LEARNING - \nThe Royal Society of Chemistry (http://openstax.org/l/16enzymes) provides an excellent introduction to enzymes for students and teachers.\n\nThe connection between the rate of a reaction and its equilibrium constant is one we can easily determine with just a bit of algebraic substitution. For a reaction where substance A forms B (and the reverse)\n\n$$\nA \\rightleftharpoons B\n$$\n\nThe rate of the forward reaction is\n\n$$\n\\operatorname{Rate}(f)=k(f)[A]\n$$\n\nAnd the rate of the reverse reaction is\n\n$$\n\\operatorname{Rate}(r)=k(r)[B]\n$$\n\nOnce equilibrium is established, the rates of the forward and reverse reactions are equal:\n\n$$\n\\operatorname{Rate}(f)=\\operatorname{Rate}(r)=k(f)[A]=k(r)[B]\n$$\n\nRearranging a bit, we get\n\n$$\n\\operatorname{Rate}_{f}=\\operatorname{Rate}_{r} \\text { so } k(f)[A]=k(r)[B]\n$$\n\nAlso recall that the equilibrium constant is simply the ratio of product to reactant concentration at equilibrium:\n\n$$\n\\begin{gathered}\n\\frac{k(f)}{k(r)}=\\frac{[B]}{[A]} \\\\\nK=\\frac{[B]}{[A]}\n\\end{gathered}\n$$\n\nSo the equilibrium constant turns out to be the ratio of the forward to the reverse rate constants. This relationship also helps cement our understanding of the nature of a catalyst. That is, a catalyst does not change the fundamental equilibrium (or the underlying thermodynamics) of a reaction. Rather, what it does is alter the rate constant for the reaction - that is, both rate constants, forward and reverse, equally. In doing so, catalysts usually speed up the rate at which reactions attain equilibrium (though they can be used to slow down the rate of reaction as well!)."}
{"id": 4623, "contents": "1889. Key Terms - \nactivated complex (also, transition state) unstable combination of reactant species formed during a chemical reaction\nactivation energy $\\left(\\boldsymbol{E}_{\\mathrm{a}}\\right)$ minimum energy necessary in order for a reaction to take place\nArrhenius equation mathematical relationship between a reaction's rate constant, activation energy, and temperature\naverage rate rate of a chemical reaction computed as the ratio of a measured change in amount or concentration of substance to the time interval over which the change occurred\nbimolecular reaction elementary reaction involving two reactant species\ncatalyst substance that increases the rate of a reaction without itself being consumed by the reaction\ncollision theory model that emphasizes the energy and orientation of molecular collisions to explain and predict reaction kinetics\nelementary reaction reaction that takes place in a single step, precisely as depicted in its chemical equation\nfrequency factor (A) proportionality constant in the Arrhenius equation, related to the relative number of collisions having an orientation capable of leading to product formation\nhalf-life of a reaction ( $\\boldsymbol{t}_{\\mathbf{l} / \\mathbf{2}}$ ) time required for half of a given amount of reactant to be consumed\nheterogeneous catalyst catalyst present in a different phase from the reactants, furnishing a surface at which a reaction can occur\nhomogeneous catalyst catalyst present in the same phase as the reactants\ninitial rate instantaneous rate of a chemical reaction at $t=0 \\mathrm{~s}$ (immediately after the reaction has begun)\ninstantaneous rate rate of a chemical reaction at any instant in time, determined by the slope of the line tangential to a graph of concentration as a function of time\nintegrated rate law equation that relates the\nconcentration of a reactant to elapsed time of reaction\nintermediate species produced in one step of a reaction mechanism and consumed in a subsequent step\nmethod of initial rates common experimental approach to determining rate laws that involves measuring reaction rates at varying initial reactant concentrations\nmolecularity number of reactant species involved in an elementary reaction\noverall reaction order sum of the reaction orders for each substance represented in the rate law\nrate constant (k) proportionality constant in a rate law\nrate expression mathematical representation defining reaction rate as change in amount, concentration, or pressure of reactant or product species per unit time"}
{"id": 4624, "contents": "1889. Key Terms - \nrate constant (k) proportionality constant in a rate law\nrate expression mathematical representation defining reaction rate as change in amount, concentration, or pressure of reactant or product species per unit time\nrate law (also, rate equation) (also, differential rate laws) mathematical equation showing the dependence of reaction rate on the rate constant and the concentration of one or more reactants\nrate of reaction measure of the speed at which a chemical reaction takes place\nrate-determining step (also, rate-limiting step) slowest elementary reaction in a reaction mechanism; determines the rate of the overall reaction\nreaction diagram used in chemical kinetics to illustrate various properties of a reaction\nreaction mechanism stepwise sequence of elementary reactions by which a chemical change takes place\nreaction order value of an exponent in a rate law (for example, zero order for 0 , first order for 1 , second order for 2 , and so on)\ntermolecular reaction elementary reaction involving three reactant species\nunimolecular reaction elementary reaction involving a single reactant species"}
{"id": 4625, "contents": "1890. Key Equations - \nrelative reaction rates for $a \\mathrm{~A} \\longrightarrow b \\mathrm{~B}=-\\frac{1}{a} \\frac{\\Delta[\\mathrm{~A}]}{\\Delta t}=\\frac{1}{b} \\frac{\\Delta[\\mathrm{~B}]}{\\Delta t}$\nintegrated rate law for zero-order reactions: $[A]_{t}=-k t+[A]_{0}$,\nhalf-life for a zero-order reaction $t_{1 / 2}=\\frac{[A]_{0}}{2 k}$\nintegrated rate law for first-order reactions: $\\ln [A]_{t}=-k t+\\ln [A]_{0}$,\nhalf-life for a first-order reaction $t_{1 / 2}=\\frac{0.693}{k}$\nintegrated rate law for second-order reactions: $\\frac{1}{[A]_{t}}=k t+\\frac{1}{[A]_{0}}$,\nhalf-life for a second-order reaction $t_{1 / 2}=\\frac{1}{[A]_{0} k}$\n$k=A e^{-E_{\\mathrm{a}} / R T}$\n$\\ln k=\\left(\\frac{-E_{\\mathrm{a}}}{R}\\right)\\left(\\frac{1}{T}\\right)+\\ln A$\n$\\ln \\frac{k_{1}}{k_{2}}=\\frac{E_{\\mathrm{a}}}{R}\\left(\\frac{1}{T_{2}}-\\frac{1}{T_{1}}\\right)$"}
{"id": 4626, "contents": "1891. Summary - 1891.1. Chemical Reaction Rates\nThe rate of a reaction can be expressed either in terms of the decrease in the amount of a reactant or the increase in the amount of a product per unit time. Relations between different rate expressions for a given reaction are derived directly from the stoichiometric coefficients of the equation representing the reaction."}
{"id": 4627, "contents": "1891. Summary - 1891.2. Factors Affecting Reaction Rates\nThe rate of a chemical reaction is affected by several parameters. Reactions involving two phases proceed more rapidly when there is greater surface area contact. If temperature or reactant concentration is increased, the rate of a given reaction generally increases as well. A catalyst can increase the rate of a reaction by providing an alternative pathway with a lower activation energy."}
{"id": 4628, "contents": "1891. Summary - 1891.3. Rate Laws\nRate laws (differential rate laws) provide a mathematical description of how changes in the concentration of a substance affect the rate of a chemical reaction. Rate laws are determined experimentally and cannot be predicted by reaction stoichiometry. The order of reaction describes how much a change in the concentration of each substance affects the overall rate, and the overall order of a reaction is the sum of the orders for each substance present in the reaction. Reaction orders are typically first order, second order, or zero order, but fractional and even negative orders are possible."}
{"id": 4629, "contents": "1891. Summary - 1891.4. Integrated Rate Laws\nIntegrated rate laws are mathematically derived from differential rate laws, and they describe the time dependence of reactant and product concentrations.\n\nThe half-life of a reaction is the time required to decrease the amount of a given reactant by one-half. A reaction's half-life varies with rate constant and,\nfor some reaction orders, reactant concentration. The half-life of a zero-order reaction decreases as the initial concentration of the reactant in the reaction decreases. The half-life of a first-order reaction is independent of concentration, and the half-life of a second-order reaction decreases as the concentration increases."}
{"id": 4630, "contents": "1891. Summary - 1891.5. Collision Theory\nChemical reactions typically require collisions between reactant species. These reactant collisions must be of proper orientation and sufficient energy in order to result in product formation. Collision theory provides a simple but effective explanation for the effect of many experimental parameters on reaction rates. The Arrhenius equation describes the relation between a reaction's rate constant, activation energy, temperature, and dependence on collision orientation."}
{"id": 4631, "contents": "1891. Summary - 1891.6. Reaction Mechanisms\nThe sequence of individual steps, or elementary reactions, by which reactants are converted into products during the course of a reaction is called the reaction mechanism. The molecularity of an elementary reaction is the number of reactant species involved, typically one (unimolecular), two (bimolecular), or, less commonly, three (termolecular). The overall rate of a reaction is determined by the rate of the slowest in its mechanism, called the rate-determining step. Unimolecular elementary reactions have first-order rate laws, while bimolecular elementary reactions have second-order rate laws. By comparing the rate laws derived from a reaction mechanism to that determined experimentally, the mechanism may be deemed either incorrect or plausible."}
{"id": 4632, "contents": "1891. Summary - 1891.7. Catalysis\nCatalysts affect the rate of a chemical reaction by altering its mechanism to provide a lower activation energy, but they do not affect equilibrium. Catalysts\ncan be homogenous (in the same phase as the reactants) or heterogeneous (a different phase than\nthe reactants)."}
{"id": 4633, "contents": "1892. Exercises - 1892.1. Chemical Reaction Rates\n1. What is the difference between average rate, initial rate, and instantaneous rate?\n2. Ozone decomposes to oxygen according to the equation $2 \\mathrm{O}_{3}(g) \\longrightarrow 3 \\mathrm{O}_{2}(g)$. Write the equation that relates the rate expressions for this reaction in terms of the disappearance of $\\mathrm{O}_{3}$ and the formation of oxygen.\n3. In the nuclear industry, chlorine trifluoride is used to prepare uranium hexafluoride, a volatile compound of uranium used in the separation of uranium isotopes. Chlorine trifluoride is prepared by the reaction $\\mathrm{Cl}_{2}(g)+3 \\mathrm{~F}_{2}(g) \\longrightarrow 2 \\mathrm{ClF}_{3}(g)$. Write the equation that relates the rate expressions for this reaction in terms of the disappearance of $\\mathrm{Cl}_{2}$ and $\\mathrm{F}_{2}$ and the formation of $\\mathrm{ClF}_{3}$.\n4. A study of the rate of dimerization of $\\mathrm{C}_{4} \\mathrm{H}_{6}$ gave the data shown in the table:\n$2 \\mathrm{C}_{4} \\mathrm{H}_{6} \\longrightarrow \\mathrm{C}_{8} \\mathrm{H}_{12}$\n\n| Time (s) | 0 | 1600 | 3200 | 4800 | 6200 |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| $\\left\\mathrm{C}_{4} \\mathrm{H}_{6}\\right$ | $1.00 \\times 10^{-2}$ | $5.04 \\times 10^{-3}$ | $3.37 \\times 10^{-3}$ | $2.53 \\times 10^{-3}$ | $2.08 \\times 10^{-3}$ |"}
{"id": 4634, "contents": "1892. Exercises - 1892.1. Chemical Reaction Rates\n(a) Determine the average rate of dimerization between 0 s and 1600 s , and between 1600 s and 3200 s .\n(b) Estimate the instantaneous rate of dimerization at 3200 s from a graph of time versus $\\left[\\mathrm{C}_{4} \\mathrm{H}_{6}\\right]$. What are the units of this rate?\n(c) Determine the average rate of formation of $\\mathrm{C}_{8} \\mathrm{H}_{12}$ at 1600 s and the instantaneous rate of formation at 3200 $s$ from the rates found in parts (a) and (b).\n5. A study of the rate of the reaction represented as $2 A \\longrightarrow B$ gave the following data:\n\n| Time (s) | 0.0 | 5.0 | 10.0 | 15.0 | 20.0 | 25.0 | 35.0 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $A$ | 1.00 | 0.775 | 0.625 | 0.465 | 0.360 | 0.285 | 0.230 |"}
{"id": 4635, "contents": "1892. Exercises - 1892.1. Chemical Reaction Rates\n(a) Determine the average rate of disappearance of $A$ between 0.0 s and 10.0 s , and between 10.0 s and 20.0 s .\n(b) Estimate the instantaneous rate of disappearance of $A$ at 15.0 s from a graph of time versus [ $A$ ]. What are the units of this rate?\n(c) Use the rates found in parts (a) and (b) to determine the average rate of formation of $B$ between 0.00 s and 10.0 s , and the instantaneous rate of formation of $B$ at 15.0 s .\n6. Consider the following reaction in aqueous solution:\n$5 \\mathrm{Br}^{-}(a q)+\\mathrm{BrO}_{3}{ }^{-}(a q)+6 \\mathrm{H}^{+}(a q) \\longrightarrow 3 \\mathrm{Br}_{2}(a q)+3 \\mathrm{H}_{2} \\mathrm{O}(l)$\nIf the rate of disappearance of $\\mathrm{Br}^{-}(\\mathrm{aq})$ at a particular moment during the reaction is $3.5 \\times 10^{-4} \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~s}^{-1}$, what is the rate of appearance of $\\mathrm{Br}_{2}(a q)$ at that moment?"}
{"id": 4636, "contents": "1892. Exercises - 1892.2. Factors Affecting Reaction Rates\n7. Describe the effect of each of the following on the rate of the reaction of magnesium metal with a solution of hydrochloric acid: the molarity of the hydrochloric acid, the temperature of the solution, and the size of the pieces of magnesium.\n8. Explain why an egg cooks more slowly in boiling water in Denver than in New York City. (Hint: Consider the effect of temperature on reaction rate and the effect of pressure on boiling point.)\n9. Go to the PhET Reactions \\& Rates (http://openstax.org/l/16PHETreaction) interactive. Use the Single Collision tab to represent how the collision between monatomic oxygen ( O ) and carbon monoxide ( CO ) results in the breaking of one bond and the formation of another. Pull back on the red plunger to release the atom and observe the results. Then, click on \"Reload Launcher\" and change to \"Angled shot\" to see the difference.\n(a) What happens when the angle of the collision is changed?\n(b) Explain how this is relevant to rate of reaction.\n10. In the PhET Reactions \\& Rates (http://openstax.org/l/16PHETreaction) interactive, use the \"Many Collisions\" tab to observe how multiple atoms and molecules interact under varying conditions. Select a molecule to pump into the chamber. Set the initial temperature and select the current amounts of each reactant. Select \"Show bonds\" under Options. How is the rate of the reaction affected by concentration and temperature?\n11. In the PhET Reactions \\& Rates (http://openstax.org/l/16PHETreaction) interactive, on the Many Collisions tab, set up a simulation with 15 molecules of A and 10 molecules of BC. Select \"Show Bonds\" under Options.\n(a) Leave the Initial Temperature at the default setting. Observe the reaction. Is the rate of reaction fast or slow?\n(b) Click \"Pause\" and then \"Reset All,\" and then enter 15 molecules of A and 10 molecules of BC once again. Select \"Show Bonds\" under Options. This time, increase the initial temperature until, on the graph, the total average energy line is completely above the potential energy curve. Describe what happens to the reaction."}
{"id": 4637, "contents": "1892. Exercises - 1892.3. Rate Laws\n12. How do the rate of a reaction and its rate constant differ?\n13. Doubling the concentration of a reactant increases the rate of a reaction four times. With this knowledge, answer the following questions:\n(a) What is the order of the reaction with respect to that reactant?\n(b) Tripling the concentration of a different reactant increases the rate of a reaction three times. What is the order of the reaction with respect to that reactant?\n14. Tripling the concentration of a reactant increases the rate of a reaction nine-fold. With this knowledge, answer the following questions:\n(a) What is the order of the reaction with respect to that reactant?\n(b) Increasing the concentration of a reactant by a factor of four increases the rate of a reaction four-fold."}
{"id": 4638, "contents": "1892. Exercises - 1892.3. Rate Laws\nWhat is the order of the reaction with respect to that reactant?\n15. How will the rate of reaction change for the process: $\\mathrm{CO}(g)+\\mathrm{NO}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)+\\mathrm{NO}(g)$ if the rate law for the reaction is rate $=k\\left[\\mathrm{NO}_{2}\\right]^{2}$ ?\n(a) Decreasing the pressure of $\\mathrm{NO}_{2}$ from 0.50 atm to 0.250 atm .\n(b) Increasing the concentration of CO from 0.01 M to 0.03 M .\n16. How will each of the following affect the rate of the reaction: $\\mathrm{CO}(g)+\\mathrm{NO}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)+\\mathrm{NO}(g)$ if the rate law for the reaction is rate $=k\\left\\mathrm{NO}_{2}\\right$ ?\n(a) Increasing the pressure of $\\mathrm{NO}_{2}$ from 0.1 atm to 0.3 atm\n(b) Increasing the concentration of CO from 0.02 M to 0.06 M .\n17. Regular flights of supersonic aircraft in the stratosphere are of concern because such aircraft produce nitric oxide, NO, as a byproduct in the exhaust of their engines. Nitric oxide reacts with ozone, and it has been suggested that this could contribute to depletion of the ozone layer. The reaction $\\mathrm{NO}+\\mathrm{O}_{3} \\longrightarrow \\mathrm{NO}_{2}+\\mathrm{O}_{2}$ is first order with respect to both NO and $\\mathrm{O}_{3}$ with a rate constant of $2.20 \\times 10^{7}$ $\\mathrm{L} / \\mathrm{mol} / \\mathrm{s}$. What is the instantaneous rate of disappearance of NO when $[\\mathrm{NO}]=3.3 \\times 10^{-6} \\mathrm{M}$ and $\\left[\\mathrm{O}_{3}\\right]=5.9 \\times$ $10^{-7} \\mathrm{M}$ ?"}
{"id": 4639, "contents": "1892. Exercises - 1892.3. Rate Laws\n18. Radioactive phosphorus is used in the study of biochemical reaction mechanisms because phosphorus atoms are components of many biochemical molecules. The location of the phosphorus (and the location of the molecule it is bound in) can be detected from the electrons (beta particles) it produces:\n${ }_{15}^{32} \\mathrm{P} \\longrightarrow{ }_{16}^{32} \\mathrm{~S}+\\mathrm{e}^{-}$\nrate $=4.85 \\times 10^{-2}$ day $^{-1}\\left[{ }^{32} \\mathrm{P}\\right]$\nWhat is the instantaneous rate of production of electrons in a sample with a phosphorus concentration of 0.0033 M ?\n19. The rate constant for the radioactive decay of ${ }^{14} \\mathrm{C}$ is $1.21 \\times 10^{-4}$ year ${ }^{-1}$. The products of the decay are nitrogen atoms and electrons (beta particles):\n${ }_{6}^{14} \\mathrm{C} \\longrightarrow{ }_{7}^{14} \\mathrm{~N}+\\mathrm{e}^{-}$\nrate $=k\\left[{ }_{6}^{14} \\mathrm{C}\\right]$\nWhat is the instantaneous rate of production of N atoms in a sample with a carbon-14 content of $6.5 \\times$ $10^{-9} \\mathrm{M}$ ?\n20. The decomposition of acetaldehyde is a second order reaction with a rate constant of $4.71 \\times 10^{-8} \\mathrm{~L} \\mathrm{~mol}^{-1}$ $\\mathrm{s}^{-1}$. What is the instantaneous rate of decomposition of acetaldehyde in a solution with a concentration of $5.55 \\times 10^{-4} \\mathrm{M}$ ?\n21. Alcohol is removed from the bloodstream by a series of metabolic reactions. The first reaction produces acetaldehyde; then other products are formed. The following data have been determined for the rate at which alcohol is removed from the blood of an average male, although individual rates can vary by $25-30 \\%$. Women metabolize alcohol a little more slowly than men:"}
{"id": 4640, "contents": "1892. Exercises - 1892.3. Rate Laws\n| $\\left\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}\\right$ | $4.4 \\times 10^{-2}$ | $3.3 \\times 10^{-2}$ | $2.2 \\times 10^{-2}$ |\n| :---: | :---: | :---: | :---: |\n| Rate $\\left(\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~h}^{-1}\\right)$ | $2.0 \\times 10^{-2}$ | $2.0 \\times 10^{-2}$ | $2.0 \\times 10^{-2}$ |\n\nDetermine the rate law, the rate constant, and the overall order for this reaction.\n22. Under certain conditions the decomposition of ammonia on a metal surface gives the following data:\n\n| $\\left\\mathrm{NH}_{3}\\right$ | $1.0 \\times 10^{-3}$ | $2.0 \\times 10^{-3}$ | $3.0 \\times 10^{-3}$ |\n| :---: | :---: | :---: | :---: |\n| Rate $\\left(\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~h}^{-1}\\right)$ | $1.5 \\times 10^{-6}$ | $1.5 \\times 10^{-6}$ | $1.5 \\times 10^{-6}$ |\n\nDetermine the rate law, the rate constant, and the overall order for this reaction.\n23. Nitrosyl chloride, NOCl , decomposes to NO and $\\mathrm{Cl}_{2}$.\n$2 \\mathrm{NOCl}(g) \\longrightarrow 2 \\mathrm{NO}(g)+\\mathrm{Cl}_{2}(g)$\nDetermine the rate law, the rate constant, and the overall order for this reaction from the following data:"}
{"id": 4641, "contents": "1892. Exercises - 1892.3. Rate Laws\n| $\\mathrm{NOCl}$ | 0.10 | 0.20 | 0.30 |\n| :---: | :---: | :---: | :---: |\n| Rate $\\left(\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~h}^{-1}\\right)$ | $8.0 \\times 10^{-10}$ | $3.2 \\times 10^{-9}$ | $7.2 \\times 10^{-9}$ |\n\n24. From the following data, determine the rate law, the rate constant, and the order with respect to $A$ for the reaction $A \\longrightarrow 2 C$.\n\n| $A$ | $1.33 \\times 10^{-2}$ | $2.66 \\times 10^{-2}$ | $3.99 \\times 10^{-2}$ |\n| :---: | :---: | :---: | :---: |\n| Rate $\\left(\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~h}^{-1}\\right)$ | $3.80 \\times 10^{-7}$ | $1.52 \\times 10^{-6}$ | $3.42 \\times 10^{-6}$ |\n\n25. Nitrogen monoxide reacts with chlorine according to the equation:\n$2 \\mathrm{NO}(g)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{NOCl}(g)$\nThe following initial rates of reaction have been observed for certain reactant concentrations:\n\n| [NO] (mol/L) | $\\left\\mathrm{Cl}_{2}\\right$ |  |\n| :---: | :---: | :---: |\n| Rate $\\left(\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~h}^{-1}\\right)$ |  |  |\n| 0.50 | 0.50 | 1.14 |\n| 1.00 | 0.50 | 4.56 |\n| 1.00 | 1.00 | 9.12 |"}
{"id": 4642, "contents": "1892. Exercises - 1892.3. Rate Laws\nWhat is the rate law that describes the rate's dependence on the concentrations of NO and $\\mathrm{Cl}_{2}$ ? What is the rate constant? What are the orders with respect to each reactant?\n26. Hydrogen reacts with nitrogen monoxide to form dinitrogen monoxide (laughing gas) according to the equation: $\\mathrm{H}_{2}(g)+2 \\mathrm{NO}(g) \\longrightarrow \\mathrm{N}_{2} \\mathrm{O}(g)+\\mathrm{H}_{2} \\mathrm{O}(g)$\nDetermine the rate law, the rate constant, and the orders with respect to each reactant from the following data:\n\n| $\\mathrm{NO}$ | 0.30 | 0.60 | 0.60 |\n| :---: | :---: | :---: | :---: |\n| $\\left\\mathrm{H}_{2}\\right$ | 0.35 | 0.35 | 0.70 |\n| Rate $\\left(\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~s}^{-1}\\right)$ | $2.835 \\times 10^{-3}$ | $1.134 \\times 10^{-2}$ | $2.268 \\times 10^{-2}$ |\n\n27. For the reaction $A \\longrightarrow B+C$, the following data were obtained at $30^{\\circ} \\mathrm{C}$ :\n\n| $A$ | 0.230 | 0.356 | 0.557 |\n| :---: | :---: | :---: | :---: |\n| Rate $\\left(\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~s}^{-1}\\right)$ | $4.17 \\times 10^{-4}$ | $9.99 \\times 10^{-4}$ | $2.44 \\times 10^{-3}$ |\n\n(a) What is the order of the reaction with respect to [ $A$ ], and what is the rate law?\n(b) What is the rate constant?\n28. For the reaction $Q \\longrightarrow W+X$, the following data were obtained at $30^{\\circ} \\mathrm{C}$ :"}
{"id": 4643, "contents": "1892. Exercises - 1892.3. Rate Laws\n| $[Q]_{\\text {initial }}(M)$ | 0.170 | 0.212 | 0.357 |\n| :---: | :---: | :---: | :---: |\n| Rate $\\left(\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~s}^{-1}\\right)$ | $6.68 \\times 10^{-3}$ | $1.04 \\times 10^{-2}$ | $2.94 \\times 10^{-2}$ |\n\n(a) What is the order of the reaction with respect to [ $Q$ ], and what is the rate law?\n(b) What is the rate constant?\n29. The rate constant for the first-order decomposition at $45^{\\circ} \\mathrm{C}$ of dinitrogen pentoxide, $\\mathrm{N}_{2} \\mathrm{O}_{5}$, dissolved in chloroform, $\\mathrm{CHCl}_{3}$, is $6.2 \\times 10^{-4} \\mathrm{~min}^{-1}$.\n$2 \\mathrm{~N}_{2} \\mathrm{O}_{5} \\longrightarrow 4 \\mathrm{NO}_{2}+\\mathrm{O}_{2}$\nWhat is the rate of the reaction when $\\left[\\mathrm{N}_{2} \\mathrm{O}_{5}\\right]=0.40 \\mathrm{M}$ ?\n30. The annual production of $\\mathrm{HNO}_{3}$ in 2013 was 60 million metric tons Most of that was prepared by the following sequence of reactions, each run in a separate reaction vessel.\n(a) $4 \\mathrm{NH}_{3}(g)+5 \\mathrm{O}_{2}(g) \\longrightarrow 4 \\mathrm{NO}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(g)$\n(b) $2 \\mathrm{NO}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{NO}_{2}(g)$\n(c) $3 \\mathrm{NO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{HNO}_{3}(a q)+\\mathrm{NO}(g)$"}
{"id": 4644, "contents": "1892. Exercises - 1892.3. Rate Laws\nThe first reaction is run by burning ammonia in air over a platinum catalyst. This reaction is fast. The reaction in equation (c) is also fast. The second reaction limits the rate at which nitric acid can be prepared from ammonia. If equation (b) is second order in NO and first order in $\\mathrm{O}_{2}$, what is the rate of formation of $\\mathrm{NO}_{2}$ when the oxygen concentration is 0.50 M and the nitric oxide concentration is 0.75 M ? The rate constant for the reaction is $5.8 \\times 10^{-6} \\mathrm{~L}^{2} \\mathrm{~mol}^{-2} \\mathrm{~s}^{-1}$.\n31. The following data have been determined for the reaction:\n$\\mathrm{I}^{-}+\\mathrm{OCl}^{-} \\longrightarrow \\mathrm{IO}^{-}+\\mathrm{Cl}^{-}$"}
{"id": 4645, "contents": "1893. 1 - \n2\n3\n\n| $\\left[\\mathrm{I}^{-}\\right]_{\\text {initial }}(M)$ | 0.10 | 0.20 | 0.30 |\n| :---: | :---: | :---: | :---: |\n| $\\left[\\mathrm{OCl}^{-}\\right]_{\\text {initial }}(M)$ | 0.050 | 0.050 | 0.010 |\n| Rate $\\left(\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~s}^{-1}\\right)$ | $3.05 \\times 10^{-4}$ | $6.20 \\times 10^{-4}$ | $1.83 \\times 10^{-4}$ |\n\nDetermine the rate law and the rate constant for this reaction."}
{"id": 4646, "contents": "1893. 1 - 1893.1. Integrated Rate Laws\n32. Describe how graphical methods can be used to determine the order of a reaction and its rate constant from a series of data that includes the concentration of $A$ at varying times.\n33. Use the data provided to graphically determine the order and rate constant of the following reaction:\n$\\mathrm{SO}_{2} \\mathrm{Cl}_{2} \\longrightarrow \\mathrm{SO}_{2}+\\mathrm{Cl}_{2}$\n\n| Time (s) | 0 | $5.00 \\times 10^{3}$ | $1.00 \\times 10^{4}$ | $1.50 \\times 10^{4}$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $\\left\\mathrm{SO}_{2} \\mathrm{Cl}_{2}\\right$ | 0.100 | 0.0896 | 0.0802 | 0.0719 |\n| Time (s) | $2.50 \\times 10^{4}$ | $3.00 \\times 10^{4}$ | $4.00 \\times 10^{4}$ |  |\n| $\\left\\mathrm{SO}_{2} \\mathrm{Cl}_{2}\\right$ | 0.0577 | 0.0517 | 0.0415 |  |\n\n34. Pure ozone decomposes slowly to oxygen, $2 \\mathrm{O}_{3}(g) \\longrightarrow 3 \\mathrm{O}_{2}(g)$. Use the data provided in a graphical method and determine the order and rate constant of the reaction."}
{"id": 4647, "contents": "1893. 1 - 1893.1. Integrated Rate Laws\n| Time (h) | 0 | $2.0 \\times 10^{3}$ | $7.6 \\times 10^{3}$ | $1.00 \\times 10^{4}$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $\\left\\mathrm{O}_{3}\\right$ | $1.00 \\times 10^{-5}$ | $4.98 \\times 10^{-6}$ | $2.07 \\times 10^{-6}$ | $1.66 \\times 10^{-6}$ |\n| Time (h) | $1.23 \\times 10^{4}$ | $1.43 \\times 10^{4}$ | $1.70 \\times 10^{4}$ |  |\n| $\\left\\mathrm{O}_{3}\\right$ | $1.39 \\times 10^{-6}$ | $1.22 \\times 10^{-6}$ | $1.05 \\times 10^{-6}$ |  |\n\n35. From the given data, use a graphical method to determine the order and rate constant of the following reaction:\n$2 X \\longrightarrow Y+Z$\n\n| $2 X \\longrightarrow Y+$ Time (s) | 5.0 | 10.0 | 15.0 | 20.0 | 25.0 | 30.0 | 35.0 | 40.0 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $X$ | 0.0990 | 0.0497 | 0.0332 | 0.0249 | 0.0200 | 0.0166 | 0.0143 | 0.0125 |"}
{"id": 4648, "contents": "1893. 1 - 1893.1. Integrated Rate Laws\n36. What is the half-life for the first-order decay of phosphorus- 32 ? $\\left({ }_{15}^{32} \\mathrm{P} \\longrightarrow{ }_{16}^{32} \\mathrm{~S}+\\mathrm{e}^{-}\\right)$The rate constant for the decay is $4.85 \\times 10^{-2}$ day $^{-1}$.\n37. What is the half-life for the first-order decay of carbon-14? $\\left({ }_{6}^{14} \\mathrm{C} \\longrightarrow{ }_{7}^{14} \\mathrm{~N}+\\mathrm{e}^{-}\\right)$The rate constant for the decay is $1.21 \\times 10^{-4}$ year $^{-1}$.\n38. What is the half-life for the decomposition of NOCl when the concentration of NOCl is 0.15 M ? The rate constant for this second-order reaction is $8.0 \\times 10^{-8} \\mathrm{~L} \\mathrm{~mol}^{-1} \\mathrm{~s}^{-1}$.\n39. What is the half-life for the decomposition of $\\mathrm{O}_{3}$ when the concentration of $\\mathrm{O}_{3}$ is $2.35 \\times 10^{-6} \\mathrm{M}$ ? The rate constant for this second-order reaction is $50.4 \\mathrm{~L} \\mathrm{~mol}^{-1} \\mathrm{~h}^{-1}$.\n40. The reaction of compound $A$ to give compounds $C$ and $D$ was found to be second-order in $A$. The rate constant for the reaction was determined to be $2.42 \\mathrm{~L} \\mathrm{~mol}^{-1} \\mathrm{~s}^{-1}$. If the initial concentration is $0.500 \\mathrm{~mol} / \\mathrm{L}$, what is the value of $t_{1 / 2}$ ?"}
{"id": 4649, "contents": "1893. 1 - 1893.1. Integrated Rate Laws\n41. The half-life of a reaction of compound $A$ to give compounds $D$ and $E$ is 8.50 min when the initial concentration of $A$ is 0.150 M . How long will it take for the concentration to drop to 0.0300 M if the reaction is (a) first order with respect to $A$ or (b) second order with respect to $A$ ?\n42. Some bacteria are resistant to the antibiotic penicillin because they produce penicillinase, an enzyme with a molecular weight of $3 \\times 10^{4} \\mathrm{~g} / \\mathrm{mol}$ that converts penicillin into inactive molecules. Although the kinetics of enzyme-catalyzed reactions can be complex, at low concentrations this reaction can be described by a rate law that is first order in the catalyst (penicillinase) and that also involves the concentration of penicillin. From the following data: 1.0 L of a solution containing $0.15 \\mu \\mathrm{~g}\\left(0.15 \\times 10^{-6} \\mathrm{~g}\\right)$ of penicillinase, determine the order of the reaction with respect to penicillin and the value of the rate constant."}
{"id": 4650, "contents": "1893. 1 - 1893.1. Integrated Rate Laws\n| [Penicillin] (M) | Rate $\\left(\\mathrm{mol} \\mathrm{L}^{-1} \\mathbf{m i n}^{-1}\\right)$ |  |\n| :--- | :--- | :--- |\n| $2.0 \\times 10^{-6}$ | $1.0 \\times 10^{-10}$ |  |\n| $3.0 \\times 10^{-6}$ | $1.5 \\times 10^{-10}$ |  |\n| $4.0 \\times 10^{-6}$ | $2.0 \\times 10^{-10}$ |  |\n\n43. Both technetium-99 and thallium-201 are used to image heart muscle in patients with suspected heart problems. The half-lives are 6 h and 73 h , respectively. What percent of the radioactivity would remain for each of the isotopes after 2 days ( 48 h )?\n44. There are two molecules with the formula $\\mathrm{C}_{3} \\mathrm{H}_{6}$. Propene, $\\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{CH}_{2}$, is the monomer of the polymer polypropylene, which is used for indoor-outdoor carpets. Cyclopropane is used as an anesthetic:"}
{"id": 4651, "contents": "1893. 1 - 1893.1. Integrated Rate Laws\nWhen heated to $499^{\\circ} \\mathrm{C}$, cyclopropane rearranges (isomerizes) and forms propene with a rate constant of $5.95 \\times 10^{-4} \\mathrm{~s}^{-1}$. What is the half-life of this reaction? What fraction of the cyclopropane remains after 0.75 h at $499^{\\circ} \\mathrm{C}$ ?\n45. Fluorine-18 is a radioactive isotope that decays by positron emission to form oxygen-18 with a half-life of 109.7 min . (A positron is a particle with the mass of an electron and a single unit of positive charge; the equation is ${ }_{9}^{18} \\mathrm{~F} \\longrightarrow{ }_{8}^{18} \\mathrm{O}+{ }_{+1}^{0} \\mathrm{e}$ ) Physicians use ${ }^{18} \\mathrm{~F}$ to study the brain by injecting a quantity of fluorosubstituted glucose into the blood of a patient. The glucose accumulates in the regions where the brain is active and needs nourishment.\n(a) What is the rate constant for the decomposition of fluorine-18?\n(b) If a sample of glucose containing radioactive fluorine-18 is injected into the blood, what percent of the radioactivity will remain after 5.59 h ?\n(c) How long does it take for $99.99 \\%$ of the ${ }^{18} \\mathrm{~F}$ to decay?\n46. Suppose that the half-life of steroids taken by an athlete is 42 days. Assuming that the steroids biodegrade by a first-order process, how long would it take for $\\frac{1}{64}$ of the initial dose to remain in the athlete's body?\n47. Recently, the skeleton of King Richard III was found under a parking lot in England. If tissue samples from the skeleton contain about $93.79 \\%$ of the carbon-14 expected in living tissue, what year did King Richard III die? The half-life for carbon-14 is 5730 years."}
{"id": 4652, "contents": "1893. 1 - 1893.1. Integrated Rate Laws\n48. Nitroglycerine is an extremely sensitive explosive. In a series of carefully controlled experiments, samples of the explosive were heated to $160^{\\circ} \\mathrm{C}$ and their first-order decomposition studied. Determine the average rate constants for each experiment using the following data:"}
{"id": 4653, "contents": "1893. 1 - 1893.1. Integrated Rate Laws\n|  |  |  |  |  |  |  |  |  |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| Initial <br> $\\left[\\mathrm{C}_{3} \\mathrm{H}_{5} \\mathrm{~N}_{3} \\mathrm{O}_{9}\\right]$ <br> $(M)$ | 4.88 | 3.52 | 2.29 | 1.81 | 5.33 | 4.05 | 2.95 | 1.72 |\n| $t(\\mathrm{~s})$ | 300 | 300 | 300 | 300 | 180 | 180 | 180 | 180 |\n| $\\%$ <br> Decomposed | 52.0 | 52.9 | 53.2 | 53.9 | 34.6 | 35.9 | 36.0 | 35.4 |\n\n49. For the past 10 years, the unsaturated hydrocarbon 1,3-butadiene $\\left(\\mathrm{CH}_{2}=\\mathrm{CH}-\\mathrm{CH}=\\mathrm{CH}_{2}\\right)$ has ranked 38th among the top 50 industrial chemicals. It is used primarily for the manufacture of synthetic rubber. An isomer exists also as cyclobutene:\n\n\nThe isomerization of cyclobutene to butadiene is first-order and the rate constant has been measured as $2.0 \\times 10^{-4} \\mathrm{~s}^{-1}$ at $150^{\\circ} \\mathrm{C}$ in a 0.53 -L flask. Determine the partial pressure of cyclobutene and its concentration after 30.0 minutes if an isomerization reaction is carried out at $150^{\\circ} \\mathrm{C}$ with an initial pressure of 55 torr."}
{"id": 4654, "contents": "1893. 1 - 1893.2. Collision Theory\n50. Chemical reactions occur when reactants collide. What are two factors that may prevent a collision from producing a chemical reaction?\n51. When every collision between reactants leads to a reaction, what determines the rate at which the reaction occurs?\n52. What is the activation energy of a reaction, and how is this energy related to the activated complex of the reaction?\n53. Account for the relationship between the rate of a reaction and its activation energy.\n54. Describe how graphical methods can be used to determine the activation energy of a reaction from a series of data that includes the rate of reaction at varying temperatures.\n55. How does an increase in temperature affect rate of reaction? Explain this effect in terms of the collision theory of the reaction rate.\n56. The rate of a certain reaction doubles for every $10^{\\circ} \\mathrm{C}$ rise in temperature.\n(a) How much faster does the reaction proceed at $45^{\\circ} \\mathrm{C}$ than at $25^{\\circ} \\mathrm{C}$ ?\n(b) How much faster does the reaction proceed at $95^{\\circ} \\mathrm{C}$ than at $25^{\\circ} \\mathrm{C}$ ?\n57. In an experiment, a sample of $\\mathrm{NaClO}_{3}$ was $90 \\%$ decomposed in 48 min . Approximately how long would this decomposition have taken if the sample had been heated $20^{\\circ} \\mathrm{C}$ higher? (Hint: Assume the rate doubles for each $10^{\\circ} \\mathrm{C}$ rise in temperature.)\n58. The rate constant at $325^{\\circ} \\mathrm{C}$ for the decomposition reaction $\\mathrm{C}_{4} \\mathrm{H}_{8} \\longrightarrow 2 \\mathrm{C}_{2} \\mathrm{H}_{4}$ is $6.1 \\times 10^{-8} \\mathrm{~s}^{-1}$, and the activation energy is 261 kJ per mole of $\\mathrm{C}_{4} \\mathrm{H}_{8}$. Determine the frequency factor for the reaction."}
{"id": 4655, "contents": "1893. 1 - 1893.2. Collision Theory\n59. The rate constant for the decomposition of acetaldehyde, $\\mathrm{CH}_{3} \\mathrm{CHO}$, to methane, $\\mathrm{CH}_{4}$, and carbon monoxide, CO , in the gas phase is $1.1 \\times 10^{-2} \\mathrm{~L} \\mathrm{~mol}^{-1} \\mathrm{~s}^{-1}$ at 703 K and $4.95 \\mathrm{~L} \\mathrm{~mol}^{-1} \\mathrm{~s}^{-1}$ at 865 K . Determine the activation energy for this decomposition.\n60. An elevated level of the enzyme alkaline phosphatase (ALP) in human serum is an indication of possible liver or bone disorder. The level of serum ALP is so low that it is very difficult to measure directly. However, ALP catalyzes a number of reactions, and its relative concentration can be determined by measuring the rate of one of these reactions under controlled conditions. One such reaction is the conversion of p-nitrophenyl phosphate (PNPP) to p-nitrophenoxide ion (PNP) and phosphate ion. Control of temperature during the test is very important; the rate of the reaction increases 1.47 times if the temperature changes from $30^{\\circ} \\mathrm{C}$ to $37^{\\circ} \\mathrm{C}$. What is the activation energy for the ALP-catalyzed conversion of PNPP to PNP and phosphate?\n61. In terms of collision theory, to which of the following is the rate of a chemical reaction proportional?\n(a) the change in free energy per second\n(b) the change in temperature per second\n(c) the number of collisions per second\n(d) the number of product molecules\n62. Hydrogen iodide, HI , decomposes in the gas phase to produce hydrogen, $\\mathrm{H}_{2}$, and iodine, $\\mathrm{I}_{2}$. The value of the rate constant, $k$, for the reaction was measured at several different temperatures and the data are shown here:"}
{"id": 4656, "contents": "1893. 1 - 1893.2. Collision Theory\n| Temperature (K) | $\\boldsymbol{k}\\left(\\mathrm{L} \\mathrm{mol}^{-\\mathbf{1}} \\mathbf{s}^{\\mathbf{- 1}}\\right)$ |\n| :--- | :--- |\n| 555 |  |\n| 575 | $6.23 \\times 10^{-7}$ |\n| 645 | $2.42 \\times 10^{-6}$ |\n| 700 | $1.44 \\times 10^{-4}$ |\n\nWhat is the value of the activation energy (in $\\mathrm{kJ} / \\mathrm{mol}$ ) for this reaction?\n63. The element Co exists in two oxidation states, $\\mathrm{Co}(\\mathrm{II})$ and $\\mathrm{Co}(\\mathrm{III})$, and the ions form many complexes. The rate at which one of the complexes of Co (III) was reduced by Fe (II) in water was measured. Determine the activation energy of the reaction from the following data:\n\n| $\\boldsymbol{T}(\\mathrm{K})$ | $\\boldsymbol{k}\\left(\\mathbf{s}^{-1}\\right)$ |  |\n| :--- | :--- | :--- |\n| 293 |  | 0.054 |\n| 298 | 0.100 |  |"}
{"id": 4657, "contents": "1893. 1 - 1893.2. Collision Theory\n64. The hydrolysis of the sugar sucrose to the sugars glucose and fructose,\n$\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}+\\mathrm{H}_{2} \\mathrm{O} \\longrightarrow \\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}+\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$\nfollows a first-order rate law for the disappearance of sucrose: rate $=k\\left[\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}\\right]$ (The products of the reaction, glucose and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.)\n(a) In neutral solution, $k=2.1 \\times 10^{-11} \\mathrm{~s}^{-1}$ at $27^{\\circ} \\mathrm{C}$ and $8.5 \\times 10^{-11} \\mathrm{~s}^{-1}$ at $37^{\\circ} \\mathrm{C}$. Determine the activation energy, the frequency factor, and the rate constant for this equation at $47^{\\circ} \\mathrm{C}$ (assuming the kinetics remain consistent with the Arrhenius equation at this temperature).\n(b) When a solution of sucrose with an initial concentration of 0.150 M reaches equilibrium, the concentration of sucrose is $1.65 \\times 10^{-7} \\mathrm{M}$. How long will it take the solution to reach equilibrium at $27^{\\circ} \\mathrm{C}$ in the absence of a catalyst? Because the concentration of sucrose at equilibrium is so low, assume that the reaction is irreversible.\n(c) Why does assuming that the reaction is irreversible simplify the calculation in part (b)?"}
{"id": 4658, "contents": "1893. 1 - 1893.2. Collision Theory\n(c) Why does assuming that the reaction is irreversible simplify the calculation in part (b)?\n65. Use the PhET Reactions \\& Rates interactive simulation (http://openstax.org/l/16PHETreaction) to simulate a system. On the \"Single collision\" tab of the simulation applet, enable the \"Energy view\" by clicking the \" + \" icon. Select the first $A+B C \\longrightarrow A B+C$ reaction (A is yellow, B is purple, and C is navy blue). Using the \"straight shot\" default option, try launching the $A$ atom with varying amounts of energy. What changes when the Total Energy line at launch is below the transition state of the Potential Energy line? Why? What happens when it is above the transition state? Why?\n66. Use the PhET Reactions \\& Rates interactive simulation (http://openstax.org/l/16PHETreaction) to simulate a system. On the \"Single collision\" tab of the simulation applet, enable the \"Energy view\" by clicking the \" + \" icon. Select the first $A+B C \\longrightarrow A B+C$ reaction (A is yellow, $B$ is purple, and $C$ is navy blue). Using the \"angled shot\" option, try launching the $A$ atom with varying angles, but with more Total energy than the transition state. What happens when the $A$ atom hits the $B C$ molecule from different directions? Why?"}
{"id": 4659, "contents": "1893. 1 - 1893.3. Reaction Mechanisms\n67. Why are elementary reactions involving three or more reactants very uncommon?\n68. In general, can we predict the effect of doubling the concentration of $A$ on the rate of the overall reaction $A+B \\longrightarrow C$ ? Can we predict the effect if the reaction is known to be an elementary reaction?\n69. Define these terms:\n(a) unimolecular reaction\n(b) bimolecular reaction\n(c) elementary reaction\n(d) overall reaction\n70. What is the rate law for the elementary termolecular reaction $A+2 B \\longrightarrow$ products? For\n$3 A \\longrightarrow$ products?\n71. Given the following reactions and the corresponding rate laws, in which of the reactions might the elementary reaction and the overall reaction be the same?\n(a) $\\mathrm{Cl}_{2}+\\mathrm{CO} \\longrightarrow \\mathrm{Cl}_{2} \\mathrm{CO}$\nrate $=k\\left[\\mathrm{Cl}_{2}\\right]^{3 / 2}[\\mathrm{CO}]$\n(b) $\\mathrm{PCl}_{3}+\\mathrm{Cl}_{2} \\longrightarrow \\mathrm{PCl}_{5}$ rate $=k\\left[\\mathrm{PCl}_{3}\\right]\\left[\\mathrm{Cl}_{2}\\right]$\n(c) $2 \\mathrm{NO}+\\mathrm{H}_{2} \\longrightarrow \\mathrm{~N}_{2}+\\mathrm{H}_{2} \\mathrm{O}_{2}$\nrate $=k[\\mathrm{NO}]\\left[\\mathrm{H}_{2}\\right]$\n(d) $2 \\mathrm{NO}+\\mathrm{O}_{2} \\longrightarrow 2 \\mathrm{NO}_{2}$\nrate $=k[\\mathrm{NO}]^{2}\\left[\\mathrm{O}_{2}\\right]$\n(e) $\\mathrm{NO}+\\mathrm{O}_{3} \\longrightarrow \\mathrm{NO}_{2}+\\mathrm{O}_{2}$\nrate $=k[\\mathrm{NO}]\\left[\\mathrm{O}_{3}\\right]$\n72. Write the rate law for each of the following elementary reactions:\n(a) $\\mathrm{O}_{3} \\xrightarrow{\\text { sunlight }} \\mathrm{O}_{2}+\\mathrm{O}$"}
{"id": 4660, "contents": "1893. 1 - 1893.3. Reaction Mechanisms\n72. Write the rate law for each of the following elementary reactions:\n(a) $\\mathrm{O}_{3} \\xrightarrow{\\text { sunlight }} \\mathrm{O}_{2}+\\mathrm{O}$\n(b) $\\mathrm{O}_{3}+\\mathrm{Cl} \\longrightarrow \\mathrm{O}_{2}+\\mathrm{ClO}$\n(c) $\\mathrm{ClO}+\\mathrm{O} \\longrightarrow \\mathrm{Cl}+\\mathrm{O}_{2}$\n(d) $\\mathrm{O}_{3}+\\mathrm{NO} \\longrightarrow \\mathrm{NO}_{2}+\\mathrm{O}_{2}$\n(e) $\\mathrm{NO}_{2}+\\mathrm{O} \\longrightarrow \\mathrm{NO}+\\mathrm{O}_{2}$\n73. Nitrogen monoxide, NO , reacts with hydrogen, $\\mathrm{H}_{2}$, according to the following equation:\n$2 \\mathrm{NO}+2 \\mathrm{H}_{2} \\longrightarrow \\mathrm{~N}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}$\nWhat would the rate law be if the mechanism for this reaction were:\n$2 \\mathrm{NO}+\\mathrm{H}_{2} \\longrightarrow \\mathrm{~N}_{2}+\\mathrm{H}_{2} \\mathrm{O}_{2}$ (slow)\n$\\mathrm{H}_{2} \\mathrm{O}_{2}+\\mathrm{H}_{2} \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}$ (fast)\n74. Experiments were conducted to study the rate of the reaction represented by this equation. ${ }^{-3}$\n$2 \\mathrm{NO}(g)+2 \\mathrm{H}_{2}(g) \\longrightarrow \\mathrm{N}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(g)$\nInitial concentrations and rates of reaction are given here."}
{"id": 4661, "contents": "1893. 1 - 1893.3. Reaction Mechanisms\n| Experiment | Initial Concentration <br> $[\\mathrm{NO}]\\left(\\mathrm{mol} \\mathrm{L}^{-1}\\right)$ | Initial Concentration, <br> $\\left[\\mathrm{H}_{2}\\right]\\left(\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~min}^{-1}\\right)$ | Initial Rate of Formation <br> of $\\mathrm{N}_{2}\\left(\\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~min}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| 1 | 0.0060 | 0.0010 | $1.8 \\times 10^{-4}$ |\n| 2 | 0.0060 | 0.0020 | $3.6 \\times 10^{-4}$ |\n| 3 | 0.0010 | 0.0060 | $0.30 \\times 10^{-4}$ |\n| 4 | 0.0020 | 0.0060 | $1.2 \\times 10^{-4}$ |\n\nConsider the following questions:\n(a) Determine the order for each of the reactants, NO and $\\mathrm{H}_{2}$, from the data given and show your reasoning.\n(b) Write the overall rate law for the reaction.\n(c) Calculate the value of the rate constant, $k$, for the reaction. Include units.\n(d) For experiment 2, calculate the concentration of NO remaining when exactly one-half of the original amount of $\\mathrm{H}_{2}$ had been consumed.\n(e) The following sequence of elementary steps is a proposed mechanism for the reaction."}
{"id": 4662, "contents": "1893. 1 - 1893.3. Reaction Mechanisms\nStep 1: $\\mathrm{NO}+\\mathrm{NO} \\rightleftharpoons \\mathrm{N}_{2} \\mathrm{O}_{2}$\nStep 2: $\\mathrm{N}_{2} \\mathrm{O}_{2}+\\mathrm{H}_{2} \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{N}_{2} \\mathrm{O}$\nStep 3: $\\mathrm{N}_{2} \\mathrm{O}+\\mathrm{H}_{2} \\rightleftharpoons \\mathrm{~N}_{2}+\\mathrm{H}_{2} \\mathrm{O}$\nBased on the data presented, which of these is the rate determining step? Show that the mechanism is consistent with the observed rate law for the reaction and the overall stoichiometry of the reaction.\n75. The reaction of CO with $\\mathrm{Cl}_{2}$ gives phosgene $\\left(\\mathrm{COCl}_{2}\\right)$, a nerve gas that was used in World War I. Use the mechanism shown here to complete the following exercises:\n$\\mathrm{Cl}_{2}(g) \\rightleftharpoons 2 \\mathrm{Cl}(g)$ (fast, $\\mathrm{k}_{1}$ represents the forward rate constant, $k_{-1}$ the reverse rate constant)\n$\\mathrm{CO}(g)+\\mathrm{Cl}(g) \\longrightarrow \\mathrm{COCl}(g)$ (slow, $k_{2}$ the rate constant)\n$\\mathrm{COCl}(g)+\\mathrm{Cl}(g) \\longrightarrow \\mathrm{COCl}_{2}(g)$ (fast, $k_{3}$ the rate constant)\n(a) Write the overall reaction.\n(b) Identify all intermediates.\n(c) Write the rate law for each elementary reaction.\n(d) Write the overall rate law expression."}
{"id": 4663, "contents": "1893. 1 - 1893.4. Catalysis\n76. Account for the increase in reaction rate brought about by a catalyst.\n77. Compare the functions of homogeneous and heterogeneous catalysts.\n78. Consider this scenario and answer the following questions: Chlorine atoms resulting from decomposition of chlorofluoromethanes, such as $\\mathrm{CCl}_{2} \\mathrm{~F}_{2}$, catalyze the decomposition of ozone in the atmosphere. One simplified mechanism for the decomposition is:\n$\\mathrm{O}_{3} \\xrightarrow{\\text { sunlight }} \\mathrm{O}_{2}+\\mathrm{O}$\n$\\mathrm{O}_{3}+\\mathrm{Cl} \\longrightarrow \\mathrm{O}_{2}+\\mathrm{ClO}$\n$\\mathrm{ClO}+\\mathrm{O} \\longrightarrow \\mathrm{Cl}+\\mathrm{O}_{2}$\n(a) Explain why chlorine atoms are catalysts in the gas-phase transformation:\n$2 \\mathrm{O}_{3} \\longrightarrow 3 \\mathrm{O}_{2}$\n(b) Nitric oxide is also involved in the decomposition of ozone by the mechanism:\n$\\mathrm{O}_{3} \\xrightarrow{\\text { sunlight }} \\mathrm{O}_{2}+\\mathrm{O}$\n$\\mathrm{O}_{3}+\\mathrm{NO} \\longrightarrow \\mathrm{NO}_{2}+\\mathrm{O}_{2}$\n$\\mathrm{NO}_{2}+\\mathrm{O} \\longrightarrow \\mathrm{NO}+\\mathrm{O}_{2}$\nIs NO a catalyst for the decomposition? Explain your answer.\n79. Water gas is a 1:1 mixture of carbon monoxide and hydrogen gas and is called water gas because it is formed from steam and hot carbon in the following reaction: $\\mathrm{H}_{2} \\mathrm{O}(g)+\\mathrm{C}(s) \\rightleftharpoons \\mathrm{H}_{2}(g)+\\mathrm{CO}(g)$. Methanol, a liquid fuel that could possibly replace gasoline, can be prepared from water gas and hydrogen at high temperature and pressure in the presence of a suitable catalyst. What will happen to the concentrations of $\\mathrm{H}_{2}$, CO , and $\\mathrm{CH}_{3} \\mathrm{OH}$ at equilibrium if more catalyst is added?"}
{"id": 4664, "contents": "1893. 1 - 1893.4. Catalysis\n80. Nitrogen and oxygen react at high temperatures. What will happen to the concentrations of $\\mathrm{N}_{2}, \\mathrm{O}_{2}$, and NO at equilibrium if a catalyst is added?\n81. For each of the following pairs of reaction diagrams, identify which of the pair is catalyzed:\n(a)"}
{"id": 4665, "contents": "1893. 1 - 1893.4. Catalysis\n(a)\n(b)\n\n(a)\n\n(b)\n\n(b)\n82. For each of the following pairs of reaction diagrams, identify which of the pairs is catalyzed:\n(a)\n\n83. For each of the following reaction diagrams, estimate the activation energy $\\left(E_{\\mathrm{a}}\\right)$ of the reaction:\n(a)\n\n(b)\n\n84. For each of the following reaction diagrams, estimate the activation energy $\\left(E_{\\mathrm{a}}\\right)$ of the reaction:\n(a)\n\n(b)\n\n85. Assuming the diagrams in Exercise 17.83 represent different mechanisms for the same reaction, which of the reactions has the faster rate?\n86. Consider the similarities and differences in the two reaction diagrams shown in Exercise 17.84. Do these diagrams represent two different overall reactions, or do they represent the same overall reaction taking place by two different mechanisms? Explain your answer."}
{"id": 4666, "contents": "1894. CHAPTER 18 <br> Representative Metals, Metalloids, and Nonmetals - \nFigure 18.1 Purity is extremely important when preparing silicon wafers. Technicians in a cleanroom prepare silicon without impurities (left). The CEO of VLSI Research, Don Hutcheson, shows off a pure silicon wafer (center). A silicon wafer covered in Pentium chips is an enlarged version of the silicon wafers found in many electronics used today (right). (credit middle: modification of work by \"Intel Free Press\"/Flickr; credit right: modification of work by Naotake Murayama)"}
{"id": 4667, "contents": "1895. CHAPTER OUTLINE - 1895.1. Periodicity\n18.2 Occurrence and Preparation of the Representative Metals\n18.3 Structure and General Properties of the Metalloids\n18.4 Structure and General Properties of the Nonmetals\n18.5 Occurrence, Preparation, and Compounds of Hydrogen\n18.6 Occurrence, Preparation, and Properties of Carbonates\n18.7 Occurrence, Preparation, and Properties of Nitrogen\n18.8 Occurrence, Preparation, and Properties of Phosphorus\n18.9 Occurrence, Preparation, and Compounds of Oxygen\n18.10 Occurrence, Preparation, and Properties of Sulfur\n18.11 Occurrence, Preparation, and Properties of Halogens\n18.12 Occurrence, Preparation, and Properties of the Noble Gases\n\nINTRODUCTION The development of the periodic table in the mid-1800s came from observations that there was a periodic relationship between the properties of the elements. Chemists, who have an understanding of the variations of these properties, have been able to use this knowledge to solve a wide variety of technical challenges. For example, silicon and other semiconductors form the backbone of modern electronics because of our ability to fine-tune the electrical properties of these materials. This chapter explores important properties of representative metals, metalloids, and nonmetals in the periodic table."}
{"id": 4668, "contents": "1896. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Classify elements\n- Make predictions about the periodicity properties of the representative elements\n\nWe begin this section by examining the behaviors of representative metals in relation to their positions in the periodic table. The primary focus of this section will be the application of periodicity to the representative metals.\n\nIt is possible to divide elements into groups according to their electron configurations. The representative elements are elements where the $s$ and $p$ orbitals are filling. The transition elements are elements where the $d$ orbitals (groups 3-11 on the periodic table) are filling, and the inner transition metals are the elements where the $f$ orbitals are filling. The $d$ orbitals fill with the elements in group 11; therefore, the elements in group 12 qualify as representative elements because the last electron enters an $s$ orbital. Metals among the representative elements are the representative metals. Metallic character results from an element's ability to lose its outer valence electrons and results in high thermal and electrical conductivity, among other physical and chemical properties. There are 20 nonradioactive representative metals in groups $1,2,3,12,13,14$, and 15 of the periodic table (the elements shaded in yellow in Figure 18.2). The radioactive elements copernicium, flerovium, polonium, and livermorium are also metals but are beyond the scope of this chapter.\n\nIn addition to the representative metals, some of the representative elements are metalloids. A metalloid is an element that has properties that are between those of metals and nonmetals; these elements are typically semiconductors.\n\nThe remaining representative elements are nonmetals. Unlike metals, which typically form cations and ionic compounds (containing ionic bonds), nonmetals tend to form anions or molecular compounds. In general, the combination of a metal and a nonmetal produces a salt. A salt is an ionic compound consisting of cations and anions.\n\n\nFIGURE 18.2 The location of the representative metals is shown in the periodic table. Nonmetals are shown in green, metalloids in purple, and the transition metals and inner transition metals in blue."}
{"id": 4669, "contents": "1896. LEARNING OBJECTIVES - \nFIGURE 18.2 The location of the representative metals is shown in the periodic table. Nonmetals are shown in green, metalloids in purple, and the transition metals and inner transition metals in blue.\n\nMost of the representative metals do not occur naturally in an uncombined state because they readily react with water and oxygen in the air. However, it is possible to isolate elemental beryllium, magnesium, zinc, cadmium, mercury, aluminum, tin, and lead from their naturally occurring minerals and use them because they react very slowly with air. Part of the reason why these elements react slowly is that these elements react with air to form a protective coating. The formation of this protective coating is passivation. The coating is a nonreactive film of oxide or some other compound. Elemental magnesium, aluminum, zinc, and tin are important in the fabrication of many familiar items, including wire, cookware, foil, and many household and personal objects. Although beryllium, cadmium, mercury, and lead are readily available, there are limitations in their use because of their toxicity."}
{"id": 4670, "contents": "1897. Group 1: The Alkali Metals - \nThe alkali metals lithium, sodium, potassium, rubidium, cesium, and francium constitute group 1 of the periodic table. Although hydrogen is in group 1 (and also in group 17), it is a nonmetal and deserves separate consideration later in this chapter. The name alkali metal is in reference to the fact that these metals and their oxides react with water to form very basic (alkaline) solutions.\n\nThe properties of the alkali metals are similar to each other as expected for elements in the same family. The alkali metals have the largest atomic radii and the lowest first ionization energy in their periods. This combination makes it very easy to remove the single electron in the outermost (valence) shell of each. The easy loss of this valence electron means that these metals readily form stable cations with a charge of $1+$. Their\nreactivity increases with increasing atomic number due to the ease of losing the lone valence electron (decreasing ionization energy). Since oxidation is so easy, the reverse, reduction, is difficult, which explains why it is hard to isolate the elements. The solid alkali metals are very soft; lithium, shown in Figure 18.3, has the lowest density of any metal $\\left(0.5 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)$.\n\nThe alkali metals all react vigorously with water to form hydrogen gas and a basic solution of the metal hydroxide. This means they are easier to oxidize than is hydrogen. As an example, the reaction of lithium with water is:\n\n\nFIGURE 18.3 Lithium floats in paraffin oil because its density is less than the density of paraffin oil.\nAlkali metals react directly with all the nonmetals (except the noble gases) to yield binary ionic compounds containing $1+$ metal ions. These metals are so reactive that it is necessary to avoid contact with both moisture and oxygen in the air. Therefore, they are stored in sealed containers under mineral oil, as shown in Figure 18.4 , to prevent contact with air and moisture. The pure metals never exist free (uncombined) in nature due to their high reactivity. In addition, this high reactivity makes it necessary to prepare the metals by electrolysis of alkali metal compounds."}
{"id": 4671, "contents": "1897. Group 1: The Alkali Metals - \nFIGURE 18.4 To prevent contact with air and water, potassium for laboratory use comes as sticks or beads stored under kerosene or mineral oil, or in sealed containers. (credit: http://images-of-elements.com/potassium.php)\n\nUnlike many other metals, the reactivity and softness of the alkali metals make these metals unsuitable for structural applications. However, there are applications where the reactivity of the alkali metals is an advantage. For example, the production of metals such as titanium and zirconium relies, in part, on the ability of sodium to reduce compounds of these metals. The manufacture of many organic compounds, including\ncertain dyes, drugs, and perfumes, utilizes reduction by lithium or sodium.\nSodium and its compounds impart a bright yellow color to a flame, as seen in Figure 18.5. Passing an electrical discharge through sodium vapor also produces this color. In both cases, this is an example of an emission spectrum as discussed in the chapter on electronic structure. Streetlights sometime employ sodium vapor lights because the sodium vapor penetrates fog better than most other light. This is because the fog does not scatter yellow light as much as it scatters white light. The other alkali metals and their salts also impart color to a flame. Lithium creates a bright, crimson color, whereas the others create a pale, violet color.\n\n\nFIGURE 18.5 Dipping a wire into a solution of a sodium salt and then heating the wire causes emission of a bright yellow light, characteristic of sodium."}
{"id": 4672, "contents": "1898. LINK TO LEARNING - \nThis video (http://openstax.org/l/16alkalih2o) demonstrates the reactions of the alkali metals with water."}
{"id": 4673, "contents": "1899. Group 2: The Alkaline Earth Metals - \nThe alkaline earth metals (beryllium, magnesium, calcium, strontium, barium, and radium) constitute group 2 of the periodic table. The name alkaline metal comes from the fact that the oxides of the heavier members of the group react with water to form alkaline solutions. The nuclear charge increases when going from group 1 to group 2. Because of this charge increase, the atoms of the alkaline earth metals are smaller and have higher first ionization energies than the alkali metals within the same period. The higher ionization energy makes the alkaline earth metals less reactive than the alkali metals; however, they are still very reactive elements. Their reactivity increases, as expected, with increasing size and decreasing ionization energy. In chemical reactions, these metals readily lose both valence electrons to form compounds in which they exhibit an oxidation state of $2+$. Due to their high reactivity, it is common to produce the alkaline earth metals, like the alkali metals, by\nelectrolysis. Even though the ionization energies are low, the two metals with the highest ionization energies (beryllium and magnesium) do form compounds that exhibit some covalent characters. Like the alkali metals, the heavier alkaline earth metals impart color to a flame. As in the case of the alkali metals, this is part of the emission spectrum of these elements. Calcium and strontium produce shades of red, whereas barium produces a green color.\n\nMagnesium is a silver-white metal that is malleable and ductile at high temperatures. Passivation decreases the reactivity of magnesium metal. Upon exposure to air, a tightly adhering layer of magnesium oxycarbonate forms on the surface of the metal and inhibits further reaction. (The carbonate comes from the reaction of carbon dioxide in the atmosphere.) Magnesium is the lightest of the widely used structural metals, which is why most magnesium production is for lightweight alloys."}
{"id": 4674, "contents": "1899. Group 2: The Alkaline Earth Metals - \nMagnesium (shown in Figure 18.6), calcium, strontium, and barium react with water and air. At room temperature, barium shows the most vigorous reaction. The products of the reaction with water are hydrogen and the metal hydroxide. The formation of hydrogen gas indicates that the heavier alkaline earth metals are better reducing agents (more easily oxidized) than is hydrogen. As expected, these metals react with both acids and nonmetals to form ionic compounds. Unlike most salts of the alkali metals, many of the common salts of the alkaline earth metals are insoluble in water because of the high lattice energies of these compounds, containing a divalent metal ion.\n\n\nFIGURE 18.6 From left to right: $\\mathrm{Mg}(s)$, warm water at pH 7 , and the resulting solution with a pH greater than 7 , as indicated by the pink color of the phenolphthalein indicator. (credit: modification of work by Sahar Atwa)\n\nThe potent reducing power of hot magnesium is useful in preparing some metals from their oxides. Indeed, magnesium's affinity for oxygen is so great that burning magnesium reacts with carbon dioxide, producing elemental carbon:\n\n$$\n2 \\mathrm{Mg}(s)+\\mathrm{CO}_{2}(g) \\longrightarrow 2 \\mathrm{MgO}(s)+\\mathrm{C}(s)\n$$\n\nFor this reason, $\\mathrm{CO}_{2}$ fire extinguisher will not extinguish a magnesium fire. Additionally, the brilliant white light emitted by burning magnesium makes it useful in flares and fireworks."}
{"id": 4675, "contents": "1900. Group 12 - \nThe elements in group 12 are transition elements; however, the last electron added is not a $d$ electron, but an $s$ electron. Since the last electron added is an $s$ electron, these elements qualify as representative metals, or post-transition metals. The group 12 elements behave more like the alkaline earth metals than transition metals. Group 12 contains the four elements zinc, cadmium, mercury, and copernicium. Each of these elements has two electrons in its outer shell $\\left(n s^{2}\\right)$. When atoms of these metals form cations with a charge of $2+$, where the two outer electrons are lost, they have pseudo-noble gas electron configurations. Mercury is sometimes an exception because it also exhibits an oxidation state of $1+i n$ compounds that contain a diatomic $\\mathrm{Hg}_{2}{ }^{2+}$ ion. In their elemental forms and in compounds, cadmium and mercury are both toxic.\n\nZinc is the most reactive in group 12, and mercury is the least reactive. (This is the reverse of the reactivity\ntrend of the metals of groups 1 and 2, in which reactivity increases down a group. The increase in reactivity with increasing atomic number only occurs for the metals in groups 1 and 2.) The decreasing reactivity is due to the formation of ions with a pseudo-noble gas configuration and to other factors that are beyond the scope of this discussion. The chemical behaviors of zinc and cadmium are quite similar to each other but differ from that of mercury.\n\nZinc and cadmium have lower reduction potentials than hydrogen, and, like the alkali metals and alkaline earth metals, they will produce hydrogen gas when they react with acids. The reaction of zinc with hydrochloric acid, shown in Figure 18.7, is:\n\n\nFIGURE 18.7 Zinc is an active metal. It dissolves in hydrochloric acid, forming a solution of colorless $\\mathrm{Zn}^{2+}$ ions, $\\mathrm{Cl}^{-}$ ions, and hydrogen gas."}
{"id": 4676, "contents": "1900. Group 12 - \nZinc is a silvery metal that quickly tarnishes to a blue-gray appearance. This change in color is due to an adherent coating of a basic carbonate, $\\mathrm{Zn}_{2}(\\mathrm{OH})_{2} \\mathrm{CO}_{3}$, which passivates the metal to inhibit further corrosion. Dry cell and alkaline batteries contain a zinc anode. Brass ( Cu and Zn ) and some bronze ( Cu , Sn , and sometimes Zn ) are important zinc alloys. About half of zinc production serves to protect iron and other metals from corrosion. This protection may take the form of a sacrificial anode (also known as a galvanic anode, which is a means of providing cathodic protection for various metals) or as a thin coating on the protected metal. Galvanized steel is steel with a protective coating of zinc."}
{"id": 4677, "contents": "1902. Sacrificial Anodes - \nA sacrificial anode, or galvanic anode, is a means of providing cathodic protection of various metals. Cathodic protection refers to the prevention of corrosion by converting the corroding metal into a cathode. As a cathode, the metal resists corrosion, which is an oxidation process. Corrosion occurs at the sacrificial anode instead of at the cathode.\n\nThe construction of such a system begins with the attachment of a more active metal (more negative reduction potential) to the metal needing protection. Attachment may be direct or via a wire. To complete the circuit, a salt bridge is necessary. This salt bridge is often seawater or ground water. Once the circuit is complete, oxidation (corrosion) occurs at the anode and not the cathode.\n\nThe commonly used sacrificial anodes are magnesium, aluminum, and zinc. Magnesium has the most negative reduction potential of the three and serves best when the salt bridge is less efficient due to a low electrolyte concentration such as in freshwater. Zinc and aluminum work better in saltwater than does magnesium. Aluminum is lighter than zinc and has a higher capacity; however, an oxide coating may passivate the aluminum. In special cases, other materials are useful. For example, iron will protect copper.\n\nMercury is very different from zinc and cadmium. Mercury is the only metal that is liquid at $25^{\\circ} \\mathrm{C}$. Many metals dissolve in mercury, forming solutions called amalgams (see the feature on Amalgams), which are alloys of mercury with one or more other metals. Mercury, shown in Figure 18.8, is a nonreactive element that is more difficult to oxidize than hydrogen. Thus, it does not displace hydrogen from acids; however, it will react with strong oxidizing acids, such as nitric acid:\n\n$$\n\\begin{gathered}\n\\mathrm{Hg}(l)+\\mathrm{HCl}(a q) \\longrightarrow \\text { no reaction } \\\\\n3 \\mathrm{Hg}(l)+8 \\mathrm{HNO}_{3}(a q) \\longrightarrow 3 \\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+4 \\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{NO}(g)\n\\end{gathered}\n$$"}
{"id": 4678, "contents": "1902. Sacrificial Anodes - \nThe clear NO initially formed quickly undergoes further oxidation to the reddish brown $\\mathrm{NO}_{2}$.\n\n\nFIGURE 18.8 From left to right: $\\mathrm{Hg}\\left(\\mathrm{l}, \\mathrm{Hg}+\\right.$ concentrated $\\mathrm{HCl}, \\mathrm{Hg}+$ concentrated $\\mathrm{HNO}_{3}$. (credit: Sahar Atwa)\nMost mercury compounds decompose when heated. Most mercury compounds contain mercury with a 2+oxidation state. When there is a large excess of mercury, it is possible to form compounds containing the $\\mathrm{Hg}_{2}{ }^{2+}$ ion. All mercury compounds are toxic, and it is necessary to exercise great care in their synthesis."}
{"id": 4679, "contents": "1904. Amalgams - \nAn amalgam is an alloy of mercury with one or more other metals. This is similar to considering steel to be an alloy of iron with other metals. Most metals will form an amalgam with mercury, with the main exceptions being iron, platinum, tungsten, and tantalum.\n\nDue to toxicity issues with mercury, there has been a significant decrease in the use of amalgams. Historically, amalgams were important in electrolytic cells and in the extraction of gold. Amalgams of the alkali metals still find use because they are strong reducing agents and easier to handle than the pure alkali metals.\n\nProspectors had a problem when they found finely divided gold. They learned that adding mercury to their pans collected the gold into the mercury to form an amalgam for easier collection. Unfortunately, losses of small amounts of mercury over the years left many streams in California polluted with mercury.\n\nDentists use amalgams containing silver and other metals to fill cavities. There are several reasons to use an amalgam including low cost, ease of manipulation, and longevity compared to alternate materials. Dental amalgams are approximately $50 \\%$ mercury by weight, which, in recent years, has become a concern due to the toxicity of mercury.\n\nAfter reviewing the best available data, the Food and Drug Administration (FDA) considers amalgam-based fillings to be safe for adults and children over six years of age. Even with multiple fillings, the mercury levels in the patients remain far below the lowest levels associated with harm. Clinical studies have found no link between dental amalgams and health problems. Health issues may not be the same in cases of\nchildren under six or pregnant women. The FDA conclusions are in line with the opinions of the Environmental Protection Agency (EPA) and Centers for Disease Control (CDC). The only health consideration noted is that some people are allergic to the amalgam or one of its components."}
{"id": 4680, "contents": "1905. Group 13 - \nGroup 13 contains the metalloid boron and the metals aluminum, gallium, indium, and thallium. The lightest element, boron, is semiconducting, and its binary compounds tend to be covalent and not ionic. The remaining elements of the group are metals, but their oxides and hydroxides change characters. The oxides and hydroxides of aluminum and gallium exhibit both acidic and basic behaviors. A substance, such as these two, that will react with both acids and bases is amphoteric. This characteristic illustrates the combination of nonmetallic and metallic behaviors of these two elements. Indium and thallium oxides and hydroxides exhibit only basic behavior, in accordance with the clearly metallic character of these two elements. The melting point of gallium is unusually low (about $30^{\\circ} \\mathrm{C}$ ) and will melt in your hand.\n\nAluminum is amphoteric because it will react with both acids and bases. A typical reaction with an acid is:\n\n$$\n2 \\mathrm{Al}(s)+6 \\mathrm{HCl}(a q) \\longrightarrow 2 \\mathrm{AlCl}_{3}(a q)+3 \\mathrm{H}_{2}(g)\n$$\n\nThe products of the reaction of aluminum with a base depend upon the reaction conditions, with the following being one possibility:\n\n$$\n2 \\mathrm{Al}(s)+2 \\mathrm{NaOH}(a q)+6 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{Na}\\left\\mathrm{Al}(\\mathrm{OH})_{4}\\right+3 \\mathrm{H}_{2}(g)\n$$"}
{"id": 4681, "contents": "1905. Group 13 - \nWith both acids and bases, the reaction with aluminum generates hydrogen gas.\nThe group 13 elements have a valence shell electron configuration of $n s^{2} n p^{1}$. Aluminum normally uses all of its valence electrons when it reacts, giving compounds in which it has an oxidation state of $3+$. Although many of these compounds are covalent, others, such as $\\mathrm{AlF}_{3}$ and $\\mathrm{Al}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}$, are ionic. Aqueous solutions of aluminum salts contain the cation $\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{3+}$, abbreviated as $\\mathrm{Al}^{3+}(a q)$. Gallium, indium, and thallium also form ionic compounds containing $\\mathrm{M}^{3+}$ ions. These three elements exhibit not only the expected oxidation state of $3+$ from the three valence electrons but also an oxidation state (in this case, $1+$ ) that is two below the expected value. This phenomenon, the inert pair effect, refers to the formation of a stable ion with an oxidation state two lower than expected for the group. The pair of electrons is the valence $s$ orbital for those elements. In general, the inert pair effect is important for the lower $p$-block elements. In an aqueous solution, the $\\mathrm{Tl}^{+}(a q)$ ion is more stable than is $\\mathrm{Tl}^{3+}(a q)$. In general, these metals will react with air and water to form $3+\\mathrm{ions}$; however, thallium reacts to give thallium(I) derivatives. The metals of group 13 all react directly with nonmetals such as sulfur, phosphorus, and the halogens, forming binary compounds."}
{"id": 4682, "contents": "1905. Group 13 - \nThe metals of group 13 ( $\\mathrm{Al}, \\mathrm{Ga}, \\mathrm{In}$, and Tl ) are all reactive. However, passivation occurs as a tough, hard, thin film of the metal oxide forms upon exposure to air. Disruption of this film may counter the passivation, allowing the metal to react. One way to disrupt the film is to expose the passivated metal to mercury. Some of the metal dissolves in the mercury to form an amalgam, which sheds the protective oxide layer to expose the metal to further reaction. The formation of an amalgam allows the metal to react with air and water."}
{"id": 4683, "contents": "1906. LINK TO LEARNING - \nAlthough easily oxidized, the passivation of aluminum makes it very useful as a strong, lightweight building material. Because of the formation of an amalgam, mercury is corrosive to structural materials made of aluminum. This video (http://openstax.org/l/16aluminumhg) demonstrates how the integrity of an aluminum beam can be destroyed by the addition of a small amount of elemental mercury.\n\nThe most important uses of aluminum are in the construction and transportation industries, and in the manufacture of aluminum cans and aluminum foil. These uses depend on the lightness, toughness, and strength of the metal, as well as its resistance to corrosion. Because aluminum is an excellent conductor of heat and resists corrosion, it is useful in the manufacture of cooking utensils.\n\nAluminum is a very good reducing agent and may replace other reducing agents in the isolation of certain metals from their oxides. Although more expensive than reduction by carbon, aluminum is important in the isolation of $\\mathrm{Mo}, \\mathrm{W}$, and Cr from their oxides."}
{"id": 4684, "contents": "1907. Group 14 - \nThe metallic members of group 14 are tin, lead, and flerovium. Carbon is a typical nonmetal. The remaining elements of the group, silicon and germanium, are examples of semimetals or metalloids. Tin and lead form the stable divalent cations, $\\mathrm{Sn}^{2+}$ and $\\mathrm{Pb}^{2+}$, with oxidation states two below the group oxidation state of $4+$. The stability of this oxidation state is a consequence of the inert pair effect. Tin and lead also form covalent compounds with a formal $4+$-oxidation state. For example, $\\mathrm{SnCl}_{4}$ and $\\mathrm{PbCl}_{4}$ are low-boiling covalent liquids.\n\n\nFIGURE 18.9 (a) Tin(II) chloride is an ionic solid; (b) tin(IV) chloride is a covalent liquid.\nTin reacts readily with nonmetals and acids to form tin(II) compounds (indicating that it is more easily oxidized than hydrogen) and with nonmetals to form either tin(II) or tin(IV) compounds (shown in Figure 18.9), depending on the stoichiometry and reaction conditions. Lead is less reactive. It is only slightly easier to oxidize than hydrogen, and oxidation normally requires a hot concentrated acid."}
{"id": 4685, "contents": "1907. Group 14 - \nMany of these elements exist as allotropes. Allotropes are two or more forms of the same element in the same physical state with different chemical and physical properties. There are two common allotropes of tin. These allotropes are grey (brittle) tin and white tin. As with other allotropes, the difference between these forms of tin is in the arrangement of the atoms. White tin is stable above $13.2^{\\circ} \\mathrm{C}$ and is malleable like other metals. At low temperatures, gray tin is the more stable form. Gray tin is brittle and tends to break down to a powder. Consequently, articles made of tin will disintegrate in cold weather, particularly if the cold spell is lengthy. The change progresses slowly from the spot of origin, and the gray tin that is first formed catalyzes further change. In a way, this effect is similar to the spread of an infection in a plant or animal body, leading people to call this process tin disease or tin pest.\n\nThe principal use of tin is in the coating of steel to form tin plate-sheet iron, which constitutes the tin in tin cans. Important tin alloys are bronze ( Cu and Sn ) and solder ( Sn and Pb ). Lead is important in the lead storage batteries in automobiles."}
{"id": 4686, "contents": "1908. Group 15 - \nBismuth, the heaviest member of group 15, is a less reactive metal than the other representative metals. It readily gives up three of its five valence electrons to active nonmetals to form the tri-positive ion, $\\mathrm{Bi}^{3+}$. It forms compounds with the group oxidation state of $5+$ only when treated with strong oxidizing agents. The stability of the $3+$-oxidation state is another example of the inert pair effect."}
{"id": 4687, "contents": "1909. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Identify natural sources of representative metals\n- Describe electrolytic and chemical reduction processes used to prepare these elements from natural sources\n\nBecause of their reactivity, we do not find most representative metals as free elements in nature. However, compounds that contain ions of most representative metals are abundant. In this section, we will consider the two common techniques used to isolate the metals from these compounds-electrolysis and chemical reduction.\n\nThese metals primarily occur in minerals, with lithium found in silicate or phosphate minerals, and sodium and potassium found in salt deposits from evaporation of ancient seas and in silicates. The alkaline earth metals occur as silicates and, with the exception of beryllium, as carbonates and sulfates. Beryllium occurs as the mineral beryl, $\\mathrm{Be}_{3} \\mathrm{Al}_{2} \\mathrm{Si}_{6} \\mathrm{O}_{18}$, which, with certain impurities, may be either the gemstone emerald or aquamarine. Magnesium is in seawater and, along with the heavier alkaline earth metals, occurs as silicates, carbonates, and sulfates. Aluminum occurs abundantly in many types of clay and in bauxite, an impure aluminum oxide hydroxide. The principle tin ore is the oxide cassiterite, $\\mathrm{SnO}_{2}$, and the principle lead and thallium ores are the sulfides or the products of weathering of the sulfides. The remaining representative metals occur as impurities in zinc or aluminum ores."}
{"id": 4688, "contents": "1910. Electrolysis - \nIons of metals in of groups 1 and 2, along with aluminum, are very difficult to reduce; therefore, it is necessary to prepare these elements by electrolysis, an important process discussed in the chapter on electrochemistry. Briefly, electrolysis involves using electrical energy to drive unfavorable chemical reactions to completion; it is useful in the isolation of reactive metals in their pure forms. Sodium, aluminum, and magnesium are typical examples.\n\nThe Preparation of Sodium\nThe most important method for the production of sodium is the electrolysis of molten sodium chloride; the set-up is a Downs cell, shown in Figure 18.10. The reaction involved in this process is:\n\n$$\n2 \\mathrm{NaCl}(l) \\xrightarrow[600^{\\circ} \\mathrm{C}]{\\text { electrolysis }} 2 \\mathrm{Na}(l)+\\mathrm{Cl}_{2}(g)\n$$\n\nThe electrolysis cell contains molten sodium chloride (melting point $801^{\\circ} \\mathrm{C}$ ), to which calcium chloride has been added to lower the melting point to $600^{\\circ} \\mathrm{C}$ (a colligative effect). The passage of a direct current through the cell causes the sodium ions to migrate to the negatively charged cathode and pick up electrons, reducing the ions to sodium metal. Chloride ions migrate to the positively charged anode, lose electrons, and undergo oxidation to chlorine gas. The overall cell reaction comes from adding the following reactions:\n\n$$\n\\begin{aligned}\n& \\text { at the cathode: } 2 \\mathrm{Na}^{+}+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Na}(l) \\\\\n& \\text { at the anode: } 2 \\mathrm{Cl}^{-} \\longrightarrow \\mathrm{Cl}_{2}(g)+2 \\mathrm{e}^{-} \\\\\n& \\text {overall change: } 2 \\mathrm{Na}^{+}+2 \\mathrm{Cl}^{-} \\longrightarrow 2 \\mathrm{Na}(l)+\\mathrm{Cl}_{2}(g)\n\\end{aligned}\n$$"}
{"id": 4689, "contents": "1910. Electrolysis - \nSeparation of the molten sodium and chlorine prevents recombination. The liquid sodium, which is less dense than molten sodium chloride, floats to the surface and flows into a collector. The gaseous chlorine goes to storage tanks. Chlorine is also a valuable product.\n\n\nFIGURE 18.10 Pure sodium metal is isolated by electrolysis of molten sodium chloride using a Downs cell. It is not possible to isolate sodium by electrolysis of aqueous solutions of sodium salts because hydrogen ions are more easily reduced than are sodium ions; as a result, hydrogen gas forms at the cathode instead of the desired sodium metal. The high temperature required to melt NaCl means that liquid sodium metal forms.\n\nThe Preparation of Aluminum\nThe preparation of aluminum utilizes a process invented in 1886 by Charles M. Hall, who began to work on the problem while a student at Oberlin College in Ohio. Paul L. T. H\u00e9roult discovered the process independently a month or two later in France. In honor to the two inventors, this electrolysis cell is known as the Hall-H\u00e9roult cell. The Hall-H\u00e9roult cell is an electrolysis cell for the production of aluminum. Figure 18.11 illustrates the Hall-H\u00e9roult cell.\n\nThe production of aluminum begins with the purification of bauxite, the most common source of aluminum. The reaction of bauxite, $\\mathrm{AlO}(\\mathrm{OH})$, with hot sodium hydroxide forms soluble sodium aluminate, while clay and other impurities remain undissolved:\n\n$$\n\\mathrm{AlO}(\\mathrm{OH})(s)+\\mathrm{NaOH}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Na}\\left\\mathrm{Al}(\\mathrm{OH})_{4}\\right\n$$\n\nAfter the removal of the impurities by filtration, the addition of acid to the aluminate leads to the reprecipitation of aluminum hydroxide:"}
{"id": 4690, "contents": "1910. Electrolysis - \nAfter the removal of the impurities by filtration, the addition of acid to the aluminate leads to the reprecipitation of aluminum hydroxide:\n\n$$\n\\mathrm{Na}\\left\\mathrm{Al}(\\mathrm{OH})_{4}\\right+\\mathrm{H}_{3} \\mathrm{O}^{+}(a q) \\longrightarrow \\mathrm{Al}(\\mathrm{OH})_{3}(s)+\\mathrm{Na}^{+}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nThe next step is to remove the precipitated aluminum hydroxide by filtration. Heating the hydroxide produces aluminum oxide, $\\mathrm{Al}_{2} \\mathrm{O}_{3}$, which dissolves in a molten mixture of cryolite, $\\mathrm{Na}_{3} \\mathrm{AlF}_{6}$, and calcium fluoride, $\\mathrm{CaF}_{2}$. Electrolysis of this solution takes place in a cell like that shown in Figure 18.11. Reduction of aluminum ions to the metal occurs at the cathode, while oxygen, carbon monoxide, and carbon dioxide form at the anode.\n\n\nFIGURE 18.11 An electrolytic cell is used for the production of aluminum. The electrolysis of a solution of cryolite and calcium fluoride results in aluminum metal at the cathode, and oxygen, carbon monoxide, and carbon dioxide at the anode.\n\nThe Preparation of Magnesium\nMagnesium is the other metal that is isolated in large quantities by electrolysis. Seawater, which contains approximately $0.5 \\%$ magnesium chloride, serves as the major source of magnesium. Addition of calcium hydroxide to seawater precipitates magnesium hydroxide. The addition of hydrochloric acid to magnesium hydroxide, followed by evaporation of the resultant aqueous solution, leaves pure magnesium chloride. The electrolysis of molten magnesium chloride forms liquid magnesium and chlorine gas:"}
{"id": 4691, "contents": "1910. Electrolysis - \n$$\n\\begin{gathered}\n\\mathrm{MgCl}_{2}(a q)+\\mathrm{Ca}(\\mathrm{OH})_{2}(a q) \\longrightarrow \\mathrm{Mg}(\\mathrm{OH})_{2}(s)+\\mathrm{CaCl}_{2}(a q) \\\\\n\\mathrm{Mg}(\\mathrm{OH})_{2}(s)+2 \\mathrm{HCl}(a q) \\longrightarrow \\mathrm{MgCl}_{2}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\\\\n\\mathrm{MgCl}_{2}(l) \\longrightarrow \\mathrm{Mg}(l)+\\mathrm{Cl}_{2}(g)\n\\end{gathered}\n$$\n\nSome production facilities have moved away from electrolysis completely. In the next section, we will see how the Pidgeon process leads to the chemical reduction of magnesium."}
{"id": 4692, "contents": "1911. Chemical Reduction - \nIt is possible to isolate many of the representative metals by chemical reduction using other elements as reducing agents. In general, chemical reduction is much less expensive than electrolysis, and for this reason, chemical reduction is the method of choice for the isolation of these elements. For example, it is possible to produce potassium, rubidium, and cesium by chemical reduction, as it is possible to reduce the molten chlorides of these metals with sodium metal. This may be surprising given that these metals are more reactive than sodium; however, the metals formed are more volatile than sodium and can be distilled for collection. The removal of the metal vapor leads to a shift in the equilibrium to produce more metal (see how reactions can be driven in the discussions of Le Ch\u00e2telier's principle in the chapter on fundamental equilibrium concepts).\n\nThe production of magnesium, zinc, and tin provide additional examples of chemical reduction.\n\nThe Preparation of Magnesium\nThe Pidgeon process involves the reaction of magnesium oxide with elemental silicon at high temperatures to form pure magnesium:\n\n$$\n\\mathrm{Si}(s)+2 \\mathrm{MgO}(s) \\xrightarrow{\\Delta} \\mathrm{SiO}_{2}(s)+2 \\mathrm{Mg}(g)\n$$\n\nAlthough this reaction is unfavorable in terms of thermodynamics, the removal of the magnesium vapor produced takes advantage of Le Ch\u00e2telier's principle to continue the forward progress of the reaction. Over $75 \\%$ of the world's production of magnesium, primarily in China, comes from this process.\n\nThe Preparation of Zinc\nZinc ores usually contain zinc sulfide, zinc oxide, or zinc carbonate. After separation of these compounds from the ores, heating in air converts the ore to zinc oxide by one of the following reactions:\n\n$$\n\\begin{gathered}\n2 \\mathrm{ZnS}(s)+3 \\mathrm{O}_{2}(g) \\xrightarrow{\\Delta} 2 \\mathrm{ZnO}(s)+2 \\mathrm{SO}_{2}(g) \\\\\n\\mathrm{ZnCO}_{3}(s) \\xrightarrow{\\Delta} \\mathrm{ZnO}(s)+\\mathrm{CO}_{2}(g)\n\\end{gathered}\n$$\n\nCarbon, in the form of coal, reduces the zinc oxide to form zinc vapor:"}
{"id": 4693, "contents": "1911. Chemical Reduction - \nCarbon, in the form of coal, reduces the zinc oxide to form zinc vapor:\n\n$$\n\\mathrm{ZnO}(s)+\\mathrm{C}(s) \\longrightarrow \\mathrm{Zn}(g)+\\mathrm{CO}(g)\n$$\n\nThe zinc can be distilled (boiling point $907^{\\circ} \\mathrm{C}$ ) and condensed. This zinc contains impurities of cadmium ( 767 $\\left.{ }^{\\circ} \\mathrm{C}\\right)$, iron $\\left(2862^{\\circ} \\mathrm{C}\\right)$, lead $\\left(1750^{\\circ} \\mathrm{C}\\right)$, and arsenic $\\left(613^{\\circ} \\mathrm{C}\\right)$. Careful redistillation produces pure zinc. Arsenic and cadmium are distilled from the zinc because they have lower boiling points. At higher temperatures, the zinc is distilled from the other impurities, mainly lead and iron."}
{"id": 4694, "contents": "1912. The Preparation of Tin - \nThe ready reduction of tin(IV) oxide by the hot coals of a campfire accounts for the knowledge of tin in the ancient world. In the modern process, the roasting of tin ores containing $\\mathrm{SnO}_{2}$ removes contaminants such as arsenic and sulfur as volatile oxides. Treatment of the remaining material with hydrochloric acid removes the oxides of other metals. Heating the purified ore with carbon at temperature above $1000{ }^{\\circ} \\mathrm{C}$ produces tin:\n\n$$\n\\mathrm{SnO}_{2}(s)+2 \\mathrm{C}(s) \\xrightarrow{\\Delta} \\mathrm{Sn}(s)+2 \\mathrm{CO}(g)\n$$\n\nThe molten tin collects at the bottom of the furnace and is drawn off and cast into blocks."}
{"id": 4695, "contents": "1912. The Preparation of Tin - 1912.1. Structure and General Properties of the Metalloids\nLEARNING OBJECTIVES\nBy the end of this section, you will be able to:\n\n- Describe the general preparation, properties, and uses of the metalloids\n- Describe the preparation, properties, and compounds of boron and silicon\n\nA series of six elements called the metalloids separate the metals from the nonmetals in the periodic table. The metalloids are boron, silicon, germanium, arsenic, antimony, and tellurium. These elements look metallic; however, they do not conduct electricity as well as metals so they are semiconductors. They are semiconductors because their electrons are more tightly bound to their nuclei than are those of metallic conductors. Their chemical behavior falls between that of metals and nonmetals. For example, the pure metalloids form covalent crystals like the nonmetals, but like the metals, they generally do not form monatomic anions. This intermediate behavior is in part due to their intermediate electronegativity values. In this section, we will briefly discuss the chemical behavior of metalloids and deal with two of these elements-boron and silicon-in more detail.\n\nThe metalloid boron exhibits many similarities to its neighbor carbon and its diagonal neighbor silicon. All three elements form covalent compounds. However, boron has one distinct difference in that its $2 s^{2} 2 p^{1}$ outer electron structure gives it one less valence electron than it has valence orbitals. Although boron exhibits an oxidation state of $3+$ in most of its stable compounds, this electron deficiency provides boron with the ability to form other, sometimes fractional, oxidation states, which occur, for example, in the boron hydrides."}
{"id": 4696, "contents": "1912. The Preparation of Tin - 1912.1. Structure and General Properties of the Metalloids\nSilicon has the valence shell electron configuration $3 s^{2} 3 p^{2}$, and it commonly forms tetrahedral structures in which it is $s p^{3}$ hybridized with a formal oxidation state of $4+$. The major differences between the chemistry of carbon and silicon result from the relative strength of the carbon-carbon bond, carbon's ability to form stable bonds to itself, and the presence of the empty $3 d$ valence-shell orbitals in silicon. Silicon's empty $d$ orbitals and boron's empty $p$ orbital enable tetrahedral silicon compounds and trigonal planar boron compounds to act as Lewis acids. Carbon, on the other hand, has no available valence shell orbitals; tetrahedral carbon compounds cannot act as Lewis acids. Germanium is very similar to silicon in its chemical behavior.\n\nArsenic and antimony generally form compounds in which an oxidation state of $3+$ or $5+$ is exhibited; however, arsenic can form arsenides with an oxidation state of $3-$. These elements tarnish only slightly in dry air but readily oxidize when warmed.\n\nTellurium combines directly with most elements. The most stable tellurium compounds are the tellurides-salts of $\\mathrm{Te}^{2-}$ formed with active metals and lanthanides-and compounds with oxygen, fluorine, and chlorine, in which tellurium normally exhibits an oxidation state $2+$ or $4+$. Although tellurium(VI) compounds are known (for example, $\\mathrm{TeF}_{6}$ ), there is a marked resistance to oxidation to this maximum group oxidation state."}
{"id": 4697, "contents": "1913. Structures of the Metalloids - \nCovalent bonding is the key to the crystal structures of the metalloids. In this regard, these elements resemble nonmetals in their behavior.\n\nElemental silicon, germanium, arsenic, antimony, and tellurium are lustrous, metallic-looking solids. Silicon and germanium crystallize with a diamond structure. Each atom within the crystal has covalent bonds to four neighboring atoms at the corners of a regular tetrahedron. Single crystals of silicon and germanium are giant, three-dimensional molecules. There are several allotropes of arsenic with the most stable being layer like and containing puckered sheets of arsenic atoms. Each arsenic atom forms covalent bonds to three other atoms within the sheet. The crystal structure of antimony is similar to that of arsenic, both shown in Figure 18.12. The structures of arsenic and antimony are similar to the structure of graphite, covered later in this chapter. Tellurium forms crystals that contain infinite spiral chains of tellurium atoms. Each atom in the chain bonds to two other atoms."}
{"id": 4698, "contents": "1914. LINK TO LEARNING - \nExplore a cubic diamond (http://openstax.org/l/16crystal) crystal structure.\n\n\nFIGURE 18.12 (a) Arsenic and (b) antimony have a layered structure similar to that of (c) graphite, except that the layers are puckered rather than planar. (d) Elemental tellurium forms spiral chains.\n\nPure crystalline boron is transparent. The crystals consist of icosahedra, as shown in Figure 18.13, with a boron atom at each corner. In the most common form of boron, the icosahedra pack together in a manner similar to the cubic closest packing of spheres. All boron-boron bonds within each icosahedron are identical and are approximately 176 pm in length. In the different forms of boron, there are different arrangements and connections between the icosahedra.\n\n\nFIGURE 18.13 An icosahedron is a symmetrical, solid shape with 20 faces, each of which is an equilateral triangle. The faces meet at 12 corners.\n\nThe name silicon is derived from the Latin word for flint, silex. The metalloid silicon readily forms compounds containing Si-O-Si bonds, which are of prime importance in the mineral world. This bonding capability is in contrast to the nonmetal carbon, whose ability to form carbon-carbon bonds gives it prime importance in the plant and animal worlds.\n\nOccurrence, Preparation, and Compounds of Boron and Silicon\nBoron constitutes less than $0.001 \\%$ by weight of the earth's crust. In nature, it only occurs in compounds with oxygen. Boron is widely distributed in volcanic regions as boric acid, $\\mathrm{B}(\\mathrm{OH})_{3}$, and in dry lake regions, including the desert areas of California, as borates and salts of boron oxyacids, such as borax, $\\mathrm{Na}_{2} \\mathrm{~B}_{4} \\mathrm{O}_{7} \\cdot 10 \\mathrm{H}_{2} \\mathrm{O}$."}
{"id": 4699, "contents": "1914. LINK TO LEARNING - \nElemental boron is chemically inert at room temperature, reacting with only fluorine and oxygen to form boron trifluoride, $\\mathrm{BF}_{3}$, and boric oxide, $\\mathrm{B}_{2} \\mathrm{O}_{3}$, respectively. At higher temperatures, boron reacts with all nonmetals, except tellurium and the noble gases, and with nearly all metals; it oxidizes to $\\mathrm{B}_{2} \\mathrm{O}_{3}$ when heated with concentrated nitric or sulfuric acid. Boron does not react with nonoxidizing acids. Many boron compounds react readily with water to give boric acid, $\\mathrm{B}(\\mathrm{OH})_{3}$ (sometimes written as $\\mathrm{H}_{3} \\mathrm{BO}_{3}$ ).\n\nReduction of boric oxide with magnesium powder forms boron ( $95-98.5 \\%$ pure) as a brown, amorphous powder:\n\n$$\n\\mathrm{B}_{2} \\mathrm{O}_{3}(s)+3 \\mathrm{Mg}(s) \\longrightarrow 2 \\mathrm{~B}(s)+3 \\mathrm{MgO}(s)\n$$\n\nAn amorphous substance is a material that appears to be a solid, but does not have a long-range order like a true solid. Treatment with hydrochloric acid removes the magnesium oxide. Further purification of the boron begins with conversion of the impure boron into boron trichloride. The next step is to heat a mixture of boron trichloride and hydrogen:\n\n$$\n2 \\mathrm{BCl}_{3}(g)+3 \\mathrm{H}_{2}(g) \\xrightarrow{1500^{\\circ} \\mathrm{C}} 2 \\mathrm{~B}(s)+6 \\mathrm{HCl}(g) \\quad \\Delta H^{\\circ}=253.7 \\mathrm{~kJ}\n$$"}
{"id": 4700, "contents": "1914. LINK TO LEARNING - \nSilicon makes up nearly one-fourth of the mass of the earth's crust-second in abundance only to oxygen. The crust is composed almost entirely of minerals in which the silicon atoms are at the center of the silicon-oxygen tetrahedron, which connect in a variety of ways to produce, among other things, chains, layers, and threedimensional frameworks. These minerals constitute the bulk of most common rocks, soil, and clays. In addition, materials such as bricks, ceramics, and glasses contain silicon compounds.\n\nIt is possible to produce silicon by the high-temperature reduction of silicon dioxide with strong reducing agents, such as carbon and magnesium:\n\n$$\n\\begin{aligned}\n& \\mathrm{SiO}_{2}(s)+2 \\mathrm{C}(s) \\xrightarrow{\\Delta} \\mathrm{Si}(s)+2 \\mathrm{CO}(g) \\\\\n& \\mathrm{SiO}_{2}(s)+2 \\mathrm{Mg}(s) \\xrightarrow{\\Delta} \\mathrm{Si}(s)+2 \\mathrm{MgO}(s)\n\\end{aligned}\n$$\n\nExtremely pure silicon is necessary for the manufacture of semiconductor electronic devices. This process begins with the conversion of impure silicon into silicon tetrahalides, or silane ( $\\mathrm{SiH}_{4}$ ), followed by\ndecomposition at high temperatures. Zone refining, illustrated in Figure 18.14, completes the purification. In this method, a rod of silicon is heated at one end by a heat source that produces a thin cross-section of molten silicon. Slowly lowering the rod through the heat source moves the molten zone from one end of the rod to other. As this thin, molten region moves, impurities in the silicon dissolve in the liquid silicon and move with the molten region. Ultimately, the impurities move to one end of the rod, which is then cut off.\n\n\nFIGURE 18.14 A zone-refining apparatus used to purify silicon.\nThis highly purified silicon, containing no more than one part impurity per million parts of silicon, is the most important element in the computer industry. Pure silicon is necessary in semiconductor electronic devices such as transistors, computer chips, and solar cells."}
{"id": 4701, "contents": "1914. LINK TO LEARNING - \nLike some metals, passivation of silicon occurs due the formation of a very thin film of oxide (primarily silicon dioxide, $\\mathrm{SiO}_{2}$ ). Silicon dioxide is soluble in hot aqueous base; thus, strong bases destroy the passivation. Removal of the passivation layer allows the base to dissolve the silicon, forming hydrogen gas and silicate anions. For example:\n\n$$\n\\mathrm{Si}(s)+4 \\mathrm{OH}^{-}(a q) \\longrightarrow \\mathrm{SiO}_{4}{ }^{4-}(a q)+2 \\mathrm{H}_{2}(g)\n$$\n\nSilicon reacts with halogens at high temperatures, forming volatile tetrahalides, such as $\\operatorname{SiF}_{4}$.\nUnlike carbon, silicon does not readily form double or triple bonds. Silicon compounds of the general formula $\\mathrm{SiX}_{4}$, where X is a highly electronegative group, can act as Lewis acids to form six-coordinate silicon. For example, silicon tetrafluoride, $\\mathrm{SiF}_{4}$, reacts with sodium fluoride to yield $\\mathrm{Na}_{2}\\left[\\mathrm{SiF}_{6}\\right]$, which contains the octahedral $\\left[\\mathrm{SiF}_{6}\\right]^{2-}$ ion in which silicon is $s p^{3} d^{2}$ hybridized:\n\n$$\n2 \\mathrm{NaF}(s)+\\mathrm{SiF}_{4}(g) \\longrightarrow \\mathrm{Na}_{2} \\mathrm{SiF}_{6}(s)\n$$\n\nAntimony reacts readily with stoichiometric amounts of fluorine, chlorine, bromine, or iodine, yielding trihalides or, with excess fluorine or chlorine, forming the pentahalides $\\mathrm{SbF}_{5}$ and $\\mathrm{SbCl}_{5}$. Depending on the stoichiometry, it forms antimony(III) sulfide, $\\mathrm{Sb}_{2} \\mathrm{~S}_{3}$, or antimony(V) sulfide when heated with sulfur. As expected, the metallic nature of the element is greater than that of arsenic, which lies immediately above it in group 15."}
{"id": 4702, "contents": "1915. Boron and Silicon Halides - \nBoron trihalides $-\\mathrm{BF}_{3}, \\mathrm{BCl}_{3}, \\mathrm{BBr}_{3}$, and $\\mathrm{BI}_{3}-$ can be prepared by the direct reaction of the elements. These nonpolar molecules contain boron with $s p^{2}$ hybridization and a trigonal planar molecular geometry. The fluoride and chloride compounds are colorless gasses, the bromide is a liquid, and the iodide is a white crystalline solid.\n\nExcept for boron trifluoride, the boron trihalides readily hydrolyze in water to form boric acid and the corresponding hydrohalic acid. Boron trichloride reacts according to the equation:\n\n$$\n\\mathrm{BCl}_{3}(g)+3 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{B}(\\mathrm{OH})_{3}(a q)+3 \\mathrm{HCl}(a q)\n$$\n\nBoron trifluoride reacts with hydrofluoric acid, to yield a solution of fluoroboric acid, $\\mathrm{HBF}_{4}$ :\n\n$$\n\\mathrm{BF}_{3}(a q)+\\mathrm{HF}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{BF}_{4}^{-}(a q)\n$$\n\nIn this reaction, the $\\mathrm{BF}_{3}$ molecule acts as the Lewis acid (electron pair acceptor) and accepts a pair of electrons from a fluoride ion:\n\n\nAll the tetrahalides of silicon, $\\mathrm{SiX}_{4}$, have been prepared. Silicon tetrachloride can be prepared by direct chlorination at elevated temperatures or by heating silicon dioxide with chlorine and carbon:\n\n$$\n\\mathrm{SiO}_{2}(s)+2 \\mathrm{C}(s)+2 \\mathrm{Cl}_{2}(g) \\xrightarrow{\\Delta} \\mathrm{SiCl}_{4}(g)+2 \\mathrm{CO}(g)\n$$\n\nSilicon tetrachloride is a covalent tetrahedral molecule, which is a nonpolar, low-boiling $\\left(57^{\\circ} \\mathrm{C}\\right)$, colorless liquid."}
{"id": 4703, "contents": "1915. Boron and Silicon Halides - \nSilicon tetrachloride is a covalent tetrahedral molecule, which is a nonpolar, low-boiling $\\left(57^{\\circ} \\mathrm{C}\\right)$, colorless liquid.\n\nIt is possible to prepare silicon tetrafluoride by the reaction of silicon dioxide with hydrofluoric acid:\n\n$$\n\\mathrm{SiO}_{2}(s)+4 \\mathrm{HF}(g) \\longrightarrow \\mathrm{SiF}_{4}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\quad \\Delta H^{\\circ}=-191.2 \\mathrm{~kJ}\n$$\n\nHydrofluoric acid is the only common acid that will react with silicon dioxide or silicates. This reaction occurs because the silicon-fluorine bond is the only bond that silicon forms that is stronger than the silicon-oxygen bond. For this reason, it is possible to store all common acids, other than hydrofluoric acid, in glass containers.\n\nExcept for silicon tetrafluoride, silicon halides are extremely sensitive to water. Upon exposure to water, $\\mathrm{SiCl}_{4}$ reacts rapidly with hydroxide groups, replacing all four chlorine atoms to produce unstable orthosilicic acid, $\\mathrm{Si}(\\mathrm{OH})_{4}$ or $\\mathrm{H}_{4} \\mathrm{SiO}_{4}$, which slowly decomposes into $\\mathrm{SiO}_{2}$.\n\nBoron and Silicon Oxides and Derivatives\nBoron burns at $700^{\\circ} \\mathrm{C}$ in oxygen, forming boric oxide, $\\mathrm{B}_{2} \\mathrm{O}_{3}$. Boric oxide is necessary for the production of heat-resistant borosilicate glass, like that shown in Figure 18.15 and certain optical glasses. Boric oxide dissolves in hot water to form boric acid, $\\mathrm{B}(\\mathrm{OH})_{3}$ :\n\n$$\n\\mathrm{B}_{2} \\mathrm{O}_{3}(s)+3 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{~B}(\\mathrm{OH})_{3}(a q)\n$$"}
{"id": 4704, "contents": "1915. Boron and Silicon Halides - \nFIGURE 18.15 Laboratory glassware, such as Pyrex and Kimax, is made of borosilicate glass because it does not break when heated. The inclusion of borates in the glass helps to mediate the effects of thermal expansion and contraction. This reduces the likelihood of thermal shock, which causes silicate glass to crack upon rapid heating or cooling. (credit: \"Tweenk\"/Wikimedia Commons)\n\nThe boron atom in $\\mathrm{B}(\\mathrm{OH})_{3}$ is $s p^{2}$ hybridized and is located at the center of an equilateral triangle with oxygen atoms at the corners. In solid $\\mathrm{B}(\\mathrm{OH})_{3}$, hydrogen bonding holds these triangular units together. Boric acid, shown in Figure 18.16, is a very weak acid that does not act as a proton donor but rather as a Lewis acid, accepting an unshared pair of electrons from the Lewis base $\\mathrm{OH}^{-}$:\n\n$$\n\\mathrm{B}(\\mathrm{OH})_{3}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{B}(\\mathrm{OH})_{4}^{-}(a q)+\\mathrm{H}_{3} \\mathrm{O}^{+}(a q) \\quad K_{\\mathrm{a}}=5.8 \\times 10^{-10}\n$$\n\n\n\nFIGURE 18.16 Boric acid has a planar structure with three -OH groups spread out equally at $120^{\\circ}$ angles from each other.\n\nHeating boric acid to $100^{\\circ} \\mathrm{C}$ causes molecules of water to split out between pairs of adjacent -OH groups to form metaboric acid, $\\mathrm{HBO}_{2}$. At about $150^{\\circ} \\mathrm{C}$, additional B-O-B linkages form, connecting the $\\mathrm{BO}_{3}$ groups together with shared oxygen atoms to form tetraboric acid, $\\mathrm{H}_{2} \\mathrm{~B}_{4} \\mathrm{O}_{7}$. Complete water loss, at still higher temperatures, results in boric oxide."}
{"id": 4705, "contents": "1915. Boron and Silicon Halides - \nBorates are salts of the oxyacids of boron. Borates result from the reactions of a base with an oxyacid or from the fusion of boric acid or boric oxide with a metal oxide or hydroxide. Borate anions range from the simple trigonal planar $\\mathrm{BO}_{3}{ }^{3-}$ ion to complex species containing chains and rings of three- and four-coordinated boron atoms. The structures of the anions found in $\\mathrm{CaB}_{2} \\mathrm{O}_{4}, \\mathrm{~K}\\left[\\mathrm{~B}_{5} \\mathrm{O}_{6}(\\mathrm{OH})_{4}\\right] \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$ (commonly written $\\mathrm{KB}_{5} \\mathrm{O}_{8} \\cdot 4 \\mathrm{H}_{2} \\mathrm{O}$ ) and $\\mathrm{Na}_{2}\\left[\\mathrm{~B}_{4} \\mathrm{O}_{5}(\\mathrm{OH})_{4}\\right] \\cdot 8 \\mathrm{H}_{2} \\mathrm{O}$ (commonly written $\\mathrm{Na}_{2} \\mathrm{~B}_{4} \\mathrm{O}_{7} \\cdot 10 \\mathrm{H}_{2} \\mathrm{O}$ ) are shown in Figure 18.17. Commercially, the most important borate is borax, $\\mathrm{Na}_{2}\\left[\\mathrm{~B}_{4} \\mathrm{O}_{5}(\\mathrm{OH})_{4}\\right] \\cdot 8 \\mathrm{H}_{2} \\mathrm{O}$, which is an important component of some laundry detergents. Most of the supply of borax comes directly from dry lakes, such as Searles Lake in California, or is prepared from kernite, $\\mathrm{Na}_{2} \\mathrm{~B}_{4} \\mathrm{O}_{7} \\cdot 4 \\mathrm{H}_{2} \\mathrm{O}$.\n\n(a)\n\n(b)\n\n(c)"}
{"id": 4706, "contents": "1915. Boron and Silicon Halides - \n(a)\n\n(b)\n\n(c)\n\nFIGURE 18.17 The borate anions are (a) $\\mathrm{CaB}_{2} \\mathrm{O}_{4}$, (b) $\\mathrm{KB}_{5} \\mathrm{O}_{8} \\cdot 4 \\mathrm{H}_{2} \\mathrm{O}$, and (c) $\\mathrm{Na}_{2} \\mathrm{~B}_{4} \\mathrm{O}_{7} \\cdot 10 \\mathrm{H}_{2} \\mathrm{O}$. The anion in $\\mathrm{CaB}_{2} \\mathrm{O}_{4}$ is an \"infinite\" chain.\n\nSilicon dioxide, silica, occurs in both crystalline and amorphous forms. The usual crystalline form of silicon dioxide is quartz, a hard, brittle, clear, colorless solid. It is useful in many ways-for architectural decorations, semiprecious jewels, and frequency control in radio transmitters. Silica takes many crystalline forms, or polymorphs, in nature. Trace amounts of $\\mathrm{Fe}^{3+}$ in quartz give amethyst its characteristic purple color. The term quartz is also used for articles such as tubing and lenses that are manufactured from amorphous silica. Opal is a naturally occurring form of amorphous silica."}
{"id": 4707, "contents": "1915. Boron and Silicon Halides - \nThe contrast in structure and physical properties between silicon dioxide and carbon dioxide is interesting, as illustrated in Figure 18.18. Solid carbon dioxide (dry ice) contains single $\\mathrm{CO}_{2}$ molecules with each of the two oxygen atoms attached to the carbon atom by double bonds. Very weak intermolecular forces hold the molecules together in the crystal. The volatility of dry ice reflect these weak forces between molecules. In contrast, silicon dioxide is a covalent network solid. In silicon dioxide, each silicon atom links to four oxygen atoms by single bonds directed toward the corners of a regular tetrahedron, and $\\mathrm{SiO}_{4}$ tetrahedra share oxygen atoms. This arrangement gives a three dimensional, continuous, silicon-oxygen network. A quartz crystal is a macromolecule of silicon dioxide. The difference between these two compounds is the ability of the group 14 elements to form strong $\\pi$ bonds. Second-period elements, such as carbon, form very strong $\\pi$ bonds, which is why carbon dioxide forms small molecules with strong double bonds. Elements below the second period, such as silicon, do not form $\\pi$ bonds as readily as second-period elements, and when they do form, the $\\pi$ bonds are weaker than those formed by second-period elements. For this reason, silicon dioxide does not contain $\\pi$ bonds but only $\\sigma$ bonds.\n\n\nFIGURE 18.18 Because carbon tends to form double and triple bonds and silicon does not, (a) carbon dioxide is a discrete molecule with two $\\mathrm{C}=\\mathrm{O}$ double bonds and (b) silicon dioxide is an infinite network of oxygen atoms bridging between silicon atoms with each silicon atom possessing four $\\mathrm{Si}-\\mathrm{O}$ single bonds. (credit a photo: modification of work by Erica Gerdes; credit b photo: modification of work by Didier Descouens)"}
{"id": 4708, "contents": "1915. Boron and Silicon Halides - \nAt $1600^{\\circ} \\mathrm{C}$, quartz melts to yield a viscous liquid. When the liquid cools, it does not crystallize readily but usually supercools and forms a glass, also called silica. The $\\mathrm{SiO}_{4}$ tetrahedra in glassy silica have a random arrangement characteristic of supercooled liquids, and the glass has some very useful properties. Silica is highly transparent to both visible and ultraviolet light. For this reason, it is important in the manufacture of lamps that give radiation rich in ultraviolet light and in certain optical instruments that operate with ultraviolet light. The coefficient of expansion of silica glass is very low; therefore, rapid temperature changes do not cause it to fracture. CorningWare and other ceramic cookware contain amorphous silica.\n\nSilicates are salts containing anions composed of silicon and oxygen. In nearly all silicates, $s p^{3}$-hybridized silicon atoms occur at the centers of tetrahedra with oxygen at the corners. There is a variation in the silicon-to-oxygen ratio that occurs because silicon-oxygen tetrahedra may exist as discrete, independent units or may share oxygen atoms at corners in a variety of ways. In addition, the presence of a variety of cations gives rise to the large number of silicate minerals.\n\nMany ceramics are composed of silicates. By including small amounts of other compounds, it is possible to modify the physical properties of the silicate materials to produce ceramics with useful characteristics."}
{"id": 4709, "contents": "1916. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe structure and properties of nonmetals\n\nThe nonmetals are elements located in the upper right portion of the periodic table. Their properties and behavior are quite different from those of metals on the left side. Under normal conditions, more than half of the nonmetals are gases, one is a liquid, and the rest include some of the softest and hardest of solids. The nonmetals exhibit a rich variety of chemical behaviors. They include the most reactive and least reactive of elements, and they form many different ionic and covalent compounds. This section presents an overview of the properties and chemical behaviors of the nonmetals, as well as the chemistry of specific elements. Many of these nonmetals are important in biological systems.\n\nIn many cases, trends in electronegativity enable us to predict the type of bonding and the physical states in compounds involving the nonmetals. We know that electronegativity decreases as we move down a given group and increases as we move from left to right across a period. The nonmetals have higher electronegativities than do metals, and compounds formed between metals and nonmetals are generally ionic in nature because of the large differences in electronegativity between them. The metals form cations, the nonmetals form anions, and the resulting compounds are solids under normal conditions. On the other hand, compounds formed between two or more nonmetals have small differences in electronegativity between the atoms, and covalent bonding-sharing of electrons-results. These substances tend to be molecular in nature and are gases, liquids, or volatile solids at room temperature and pressure.\n\nIn normal chemical processes, nonmetals do not form monatomic positive ions (cations) because their ionization energies are too high. All monatomic nonmetal ions are anions; examples include the chloride ion, $\\mathrm{Cl}^{-}$, the nitride ion, $\\mathrm{N}^{3-}$, and the selenide ion, $\\mathrm{Se}^{2-}$."}
{"id": 4710, "contents": "1916. LEARNING OBJECTIVES - \nThe common oxidation states that the nonmetals exhibit in their ionic and covalent compounds are shown in Figure 18.19. Remember that an element exhibits a positive oxidation state when combined with a more electronegative element and that it exhibits a negative oxidation state when combined with a less electronegative element.\n\n\nFIGURE 18.19 Nonmetals exhibit these common oxidation states in ionic and covalent compounds.\nThe first member of each nonmetal group exhibits different behaviors, in many respects, from the other group members. The reasons for this include smaller size, greater ionization energy, and (most important) the fact that the first member of each group has only four valence orbitals (one $2 s$ and three $2 p$ ) available for bonding, whereas other group members have empty $d$ orbitals in their valence shells, making possible five, six, or even more bonds around the central atom. For example, nitrogen forms only $\\mathrm{NF}_{3}$, whereas phosphorus forms both $\\mathrm{PF}_{3}$ and $\\mathrm{PF}_{5}$.\n\nAnother difference between the first group member and subsequent members is the greater ability of the first member to form $\\pi$ bonds. This is primarily a function of the smaller size of the first member of each group, which allows better overlap of atomic orbitals. Nonmetals, other than the first member of each group, rarely form $\\pi$ bonds to nonmetals that are the first member of a group. For example, sulfur-oxygen $\\pi$ bonds are well known, whereas sulfur does not normally form stable $\\pi$ bonds to itself.\n\nThe variety of oxidation states displayed by most of the nonmetals means that many of their chemical reactions involve changes in oxidation state through oxidation-reduction reactions. There are five general aspects of the oxidation-reduction chemistry:\n\n1. Nonmetals oxidize most metals. The oxidation state of the metal becomes positive as it undergoes oxidation and that of the nonmetal becomes negative as it undergoes reduction. For example:\n\n$$\n\\begin{array}{cccc}\n4 \\mathrm{Fe}(s)+ & 3 \\mathrm{O}_{2}(g) & \\longrightarrow \\mathrm{Fe}_{2} \\mathrm{O}_{3}(s) \\\\\n0 & 0 & +3-2\n\\end{array}\n$$"}
{"id": 4711, "contents": "1916. LEARNING OBJECTIVES - \n2. With the exception of nitrogen and carbon, which are poor oxidizing agents, a more electronegative nonmetal oxidizes a less electronegative nonmetal or the anion of the nonmetal:\n\n$$\n\\begin{array}{cccc}\n\\begin{array}{llll}\n\\mathrm{S}(s)+ & \\mathrm{O}_{2}(g) & \\longrightarrow & 2 \\mathrm{SO}_{2}(s) \\\\\n0 & 0 & & +4-2 \\\\\n& & & \\\\\n\\mathrm{Cl}_{2}(g)+ & 2 \\mathrm{I}^{-}(a q) & \\longrightarrow & \\mathrm{I}_{2}(s)+2 \\mathrm{Cl}^{-}(a q) \\\\\n0 & & 0\n\\end{array}\n\\end{array}\n$$\n\n3. Fluorine and oxygen are the strongest oxidizing agents within their respective groups; each oxidizes all the elements that lie below it in the group. Within any period, the strongest oxidizing agent is in group 17. A nonmetal often oxidizes an element that lies to its left in the same period. For example:\n\n$$\n\\begin{array}{llll}\n2 \\operatorname{As}(s)+ & 3 \\mathrm{Br}_{2}(l) \\longrightarrow & 2 \\mathrm{AsBr}_{3}(s) \\\\\n0 & +3-1\n\\end{array}\n$$\n\n4. The stronger a nonmetal is as an oxidizing agent, the more difficult it is to oxidize the anion formed by the nonmetal. This means that the most stable negative ions are formed by elements at the top of the group or in group 17 of the period.\n5. Fluorine and oxygen are the strongest oxidizing elements known. Fluorine does not form compounds in which it exhibits positive oxidation states; oxygen exhibits a positive oxidation state only when combined with fluorine. For example:\n\n$$\n\\begin{array}{lll}\n2 \\mathrm{~F}_{2}(g)+2 \\mathrm{OH}^{-}(a q) \\longrightarrow & \\mathrm{OF}_{2}(g)+2 \\mathrm{~F}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\\\\n0 & +2 \\quad-1\n\\end{array}\n$$"}
{"id": 4712, "contents": "1916. LEARNING OBJECTIVES - \nWith the exception of most of the noble gases, all nonmetals form compounds with oxygen, yielding covalent oxides. Most of these oxides are acidic, that is, they react with water to form oxyacids. Recall from the acid-base chapter that an oxyacid is an acid consisting of hydrogen, oxygen, and some other element. Notable exceptions are carbon monoxide, CO , nitrous oxide, $\\mathrm{N}_{2} \\mathrm{O}$, and nitric oxide, NO . There are three characteristics of these acidic oxides:\n\n1. Oxides such as $\\mathrm{SO}_{2}$ and $\\mathrm{N}_{2} \\mathrm{O}_{5}$, in which the nonmetal exhibits one of its common oxidation states, are acid anhydrides and react with water to form acids with no change in oxidation state. The product is an oxyacid. For example:\n\n$$\n\\begin{gathered}\n\\mathrm{SO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{H}_{2} \\mathrm{SO}_{3}(a q) \\\\\n\\mathrm{N}_{2} \\mathrm{O}_{5}(s)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{HNO}_{3}(a q)\n\\end{gathered}\n$$\n\n2. Those oxides such as $\\mathrm{NO}_{2}$ and $\\mathrm{ClO}_{2}$, in which the nonmetal does not exhibit one of its common oxidation states, also react with water. In these reactions, the nonmetal is both oxidized and reduced. For example:\n\n$$\n\\begin{array}{ll}\n3 \\mathrm{NO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow & 2 \\mathrm{HNO}_{3}(a q)+ \\\\\n+4 & +5\n\\end{array}\n$$\n\nReactions in which the same element is both oxidized and reduced are called disproportionation reactions.\n3. The acid strength increases as the electronegativity of the central atom increases. To learn more, see the discussion in the chapter on acid-base chemistry."}
{"id": 4713, "contents": "1916. LEARNING OBJECTIVES - \nThe binary hydrogen compounds of the nonmetals also exhibit an acidic behavior in water, although only HCl , HBr , and HI are strong acids. The acid strength of the nonmetal hydrogen compounds increases from left to right across a period and down a group. For example, ammonia, $\\mathrm{NH}_{3}$, is a weaker acid than is water, $\\mathrm{H}_{2} \\mathrm{O}$, which is weaker than is hydrogen fluoride, HF. Water, $\\mathrm{H}_{2} \\mathrm{O}$, is also a weaker acid than is hydrogen sulfide, $\\mathrm{H}_{2} \\mathrm{~S}$, which is weaker than is hydrogen selenide, $\\mathrm{H}_{2} \\mathrm{Se}$. Weaker acidic character implies greater basic character."}
{"id": 4714, "contents": "1917. Structures of the Nonmetals - \nThe structures of the nonmetals differ dramatically from those of metals. Metals crystallize in closely packed arrays that do not contain molecules or covalent bonds. Nonmetal structures contain covalent bonds, and many nonmetals consist of individual molecules. The electrons in nonmetals are localized in covalent bonds,\nwhereas in a metal, there is delocalization of the electrons throughout the solid.\nThe noble gases are all monatomic, whereas the other nonmetal gases-hydrogen, nitrogen, oxygen, fluorine, and chlorine-normally exist as the diatomic molecules $\\mathrm{H}_{2}, \\mathrm{~N}_{2}, \\mathrm{O}_{2}, \\mathrm{~F}_{2}$, and $\\mathrm{Cl}_{2}$. The other halogens are also diatomic; $\\mathrm{Br}_{2}$ is a liquid and $\\mathrm{I}_{2}$ exists as a solid under normal conditions. The changes in state as one moves down the halogen family offer excellent examples of the increasing strength of intermolecular London forces with increasing molecular mass and increasing polarizability.\n\nOxygen has two allotropes: $\\mathrm{O}_{2}$, dioxygen, and $\\mathrm{O}_{3}$, ozone. Phosphorus has three common allotropes, commonly referred to by their colors: white, red, and black. Sulfur has several allotropes. There are also many carbon allotropes. Most people know of diamond, graphite, and charcoal, but fewer people know of the recent discovery of fullerenes, carbon nanotubes, and graphene.\n\nDescriptions of the physical properties of three nonmetals that are characteristic of molecular solids follow."}
{"id": 4715, "contents": "1918. Carbon - \nCarbon occurs in the uncombined (elemental) state in many forms, such as diamond, graphite, charcoal, coke, carbon black, graphene, and fullerene.\n\nDiamond, shown in Figure 18.20, is a very hard crystalline material that is colorless and transparent when pure. Each atom forms four single bonds to four other atoms at the corners of a tetrahedron ( $s p^{3}$ hybridization); this makes the diamond a giant molecule. Carbon-carbon single bonds are very strong, and, because they extend throughout the crystal to form a three-dimensional network, the crystals are very hard and have high melting points $\\left(\\sim 4400^{\\circ} \\mathrm{C}\\right)$.\n\n\nFIGURE 18.20 (a) Diamond and (b) graphite are two forms of carbon. (c) In the crystal structure of diamond, the covalent bonds form three-dimensional tetrahedrons. (d) In the crystal structure of graphite, each planar layer is composed of six-membered rings. (credit a: modification of work by \"Fancy Diamonds\"/Flickr; credit b: modification of work from http://images-of-elements.com/carbon.php)\n\nGraphite, also shown in Figure 18.20, is a soft, slippery, grayish-black solid that conducts electricity. These properties relate to its structure, which consists of layers of carbon atoms, with each atom surrounded by three other carbon atoms in a trigonal planar arrangement. Each carbon atom in graphite forms three $\\sigma$ bonds, one to each of its nearest neighbors, by means of $s p^{2}$-hybrid orbitals. The unhybridized $p$ orbital on each carbon atom will overlap unhybridized orbitals on adjacent carbon atoms in the same layer to form $\\pi$ bonds. Many resonance forms are necessary to describe the electronic structure of a graphite layer; Figure 18.21 illustrates two of these forms.\n\n(a)\n\n(b)\n\nFIGURE 18.21 (a) Carbon atoms in graphite have unhybridized $p$ orbitals. Each $p$ orbital is perpendicular to the plane of carbon atoms. (b) These are two of the many resonance forms of graphite necessary to describe its electronic structure as a resonance hybrid."}
{"id": 4716, "contents": "1918. Carbon - \nAtoms within a graphite layer are bonded together tightly by the $\\sigma$ and $\\pi$ bonds; however, the forces between layers are weak. London dispersion forces hold the layers together. To learn more, see the discussion of these weak forces in the chapter on liquids and solids. The weak forces between layers give graphite the soft, flaky character that makes it useful as the so-called \"lead\" in pencils and the slippery character that makes it useful as a lubricant. The loosely held electrons in the resonating $\\pi$ bonds can move throughout the solid and are responsible for the electrical conductivity of graphite.\n\nOther forms of elemental carbon include carbon black, charcoal, and coke. Carbon black is an amorphous form of carbon prepared by the incomplete combustion of natural gas, $\\mathrm{CH}_{4}$. It is possible to produce charcoal and coke by heating wood and coal, respectively, at high temperatures in the absence of air.\n\nRecently, new forms of elemental carbon molecules have been identified in the soot generated by a smoky flame and in the vapor produced when graphite is heated to very high temperatures in a vacuum or in helium. One of these new forms, first isolated by Professor Richard Smalley and coworkers at Rice University, consists of icosahedral (soccer-ball-shaped) molecules that contain 60 carbon atoms, $\\mathrm{C}_{60}$. This is buckminsterfullerene (often called bucky balls) after the architect Buckminster Fuller, who designed domed structures, which have a similar appearance (Figure 18.22).\n\n\nFIGURE 18.22 The molecular structure of $\\mathrm{C}_{60}$, buckminsterfullerene, is icosahedral."}
{"id": 4717, "contents": "1920. Nanotubes and Graphene - \nGraphene and carbon nanotubes are two recently discovered allotropes of carbon. Both of the forms bear some relationship to graphite. Graphene is a single layer of graphite (one atom thick), as illustrated in Figure 18.23, whereas carbon nanotubes roll the layer into a small tube, as illustrated in Figure 18.23.\n\n\nFIGURE 18.23 (a) Graphene and (b) carbon nanotubes are both allotropes of carbon.\nGraphene is a very strong, lightweight, and efficient conductor of heat and electricity discovered in 2003. As in graphite, the carbon atoms form a layer of six-membered rings with $s p^{2}$-hybridized carbon atoms at the corners. Resonance stabilizes the system and leads to its conductivity. Unlike graphite, there is no stacking of the layers to give a three-dimensional structure. Andre Geim and Kostya Novoselov at the University of Manchester won the 2010 Nobel Prize in Physics for their pioneering work characterizing graphene.\n\nThe simplest procedure for preparing graphene is to use a piece of adhesive tape to remove a single layer of graphene from the surface of a piece of graphite. This method works because there are only weak London dispersion forces between the layers in graphite. Alternative methods are to deposit a single layer of carbon atoms on the surface of some other material (ruthenium, iridium, or copper) or to synthesize it at the surface of silicon carbide via the sublimation of silicon.\n\nThere currently are no commercial applications of graphene. However, its unusual properties, such as high electron mobility and thermal conductivity, should make it suitable for the manufacture of many advanced electronic devices and for thermal management applications.\n\nCarbon nanotubes are carbon allotropes, which have a cylindrical structure. Like graphite and graphene, nanotubes consist of rings of $s p^{2}$-hybridized carbon atoms. Unlike graphite and graphene, which occur in layers, the layers wrap into a tube and bond together to produce a stable structure. The walls of the tube may be one atom or multiple atoms thick.\n\nCarbon nanotubes are extremely strong materials that are harder than diamond. Depending upon the shape of the nanotube, it may be a conductor or semiconductor. For some applications, the conducting form is preferable, whereas other applications utilize the semiconducting form."}
{"id": 4718, "contents": "1920. Nanotubes and Graphene - \nThe basis for the synthesis of carbon nanotubes is the generation of carbon atoms in a vacuum. It is possible to produce carbon atoms by an electrical discharge through graphite, vaporization of graphite\nwith a laser, and the decomposition of a carbon compound.\nThe strength of carbon nanotubes will eventually lead to some of their most exciting applications, as a thread produced from several nanotubes will support enormous weight. However, the current applications only employ bulk nanotubes. The addition of nanotubes to polymers improves the mechanical, thermal, and electrical properties of the bulk material. There are currently nanotubes in some bicycle parts, skis, baseball bats, fishing rods, and surfboards."}
{"id": 4719, "contents": "1921. Phosphorus - \nThe name phosphorus comes from the Greek words meaning light bringing. When phosphorus was first isolated, scientists noted that it glowed in the dark and burned when exposed to air. Phosphorus is the only member of its group that does not occur in the uncombined state in nature; it exists in many allotropic forms. We will consider two of those forms: white phosphorus and red phosphorus.\n\nWhite phosphorus is a white, waxy solid that melts at $44.2^{\\circ} \\mathrm{C}$ and boils at $280^{\\circ} \\mathrm{C}$. It is insoluble in water (in which it is stored-see Figure 18.24), is very soluble in carbon disulfide, and bursts into flame in air. As a solid, as a liquid, as a gas, and in solution, white phosphorus exists as $\\mathrm{P}_{4}$ molecules with four phosphorus atoms at the corners of a regular tetrahedron, as illustrated in Figure 18.24. Each phosphorus atom covalently bonds to the other three atoms in the molecule by single covalent bonds. White phosphorus is the most reactive allotrope and is very toxic.\n\n\nFIGURE 18.24 (a) Because white phosphorus bursts into flame in air, it is stored in water. (b) The structure of white phosphorus consists of $\\mathrm{P}_{4}$ molecules arranged in a tetrahedron. (c) Red phosphorus is much less reactive than is white phosphorus. (d) The structure of red phosphorus consists of networks of $\\mathrm{P}_{4}$ tetrahedra joined by P-P single bonds. (credit a: modification of work from http://images-of-elements.com/phosphorus.php)"}
{"id": 4720, "contents": "1921. Phosphorus - \nHeating white phosphorus to $270-300^{\\circ} \\mathrm{C}$ in the absence of air yields red phosphorus. Red phosphorus (shown in Figure 18.24 ) is denser, has a higher melting point $\\left(\\sim 600^{\\circ} \\mathrm{C}\\right)$, is much less reactive, is essentially nontoxic, and is easier and safer to handle than is white phosphorus. Its structure is highly polymeric and appears to contain three-dimensional networks of $\\mathrm{P}_{4}$ tetrahedra joined by P-P single bonds. Red phosphorus is insoluble in solvents that dissolve white phosphorus. When red phosphorus is heated, $\\mathrm{P}_{4}$ molecules sublime from the solid."}
{"id": 4721, "contents": "1922. Sulfur - \nThe allotropy of sulfur is far greater and more complex than that of any other element. Sulfur is the brimstone referred to in the Bible and other places, and references to sulfur occur throughout recorded history-right up to the relatively recent discovery that it is a component of the atmospheres of Venus and of Io, a moon of Jupiter. The most common and most stable allotrope of sulfur is yellow, rhombic sulfur, so named because of the shape of its crystals. Rhombic sulfur is the form to which all other allotropes revert at room temperature. Crystals of rhombic sulfur melt at $113^{\\circ} \\mathrm{C}$. Cooling this liquid gives long needles of monoclinic sulfur. This form is stable from $96^{\\circ} \\mathrm{C}$ to the melting point, $119^{\\circ} \\mathrm{C}$. At room temperature, it gradually reverts to the rhombic form.\n\nBoth rhombic sulfur and monoclinic sulfur contain $\\mathrm{S}_{8}$ molecules in which atoms form eight-membered, puckered rings that resemble crowns, as illustrated in Figure 18.25. Each sulfur atom is bonded to each of its\ntwo neighbors in the ring by covalent S-S single bonds.\n\n\nFIGURE 18.25 These four sulfur allotropes show eight-membered, puckered rings. Each sulfur atom bonds to each of its two neighbors in the ring by covalent $\\mathrm{S}-\\mathrm{S}$ single bonds. Here are (a) individual $\\mathrm{S}_{8}$ rings, (b) $\\mathrm{S}_{8}$ chains formed when the rings open, (c) longer chains formed by adding sulfur atoms to $\\mathrm{S}_{8}$ chains, and (d) part of the very long sulfur chains formed at higher temperatures."}
{"id": 4722, "contents": "1922. Sulfur - \nWhen rhombic sulfur melts, the straw-colored liquid is quite mobile; its viscosity is low because $\\mathrm{S}_{8}$ molecules are essentially spherical and offer relatively little resistance as they move past each other. As the temperature rises, S-S bonds in the rings break, and polymeric chains of sulfur atoms result. These chains combine end to end, forming still longer chains that tangle with one another. The liquid gradually darkens in color and becomes so viscous that finally (at about $230^{\\circ} \\mathrm{C}$ ) it does not pour easily. The dangling atoms at the ends of the chains of sulfur atoms are responsible for the dark red color because their electronic structure differs from those of sulfur atoms that have bonds to two adjacent sulfur atoms. This causes them to absorb light differently and results in a different visible color. Cooling the liquid rapidly produces a rubberlike amorphous mass, called plastic sulfur.\n\nSulfur boils at $445{ }^{\\circ} \\mathrm{C}$ and forms a vapor consisting of $\\mathrm{S}_{2}, \\mathrm{~S}_{6}$, and $\\mathrm{S}_{8}$ molecules; at about $1000^{\\circ} \\mathrm{C}$, the vapor density corresponds to the formula $\\mathrm{S}_{2}$, which is a paramagnetic molecule like $\\mathrm{O}_{2}$ with a similar electronic structure and a weak sulfur-sulfur double bond."}
{"id": 4723, "contents": "1922. Sulfur - \nAs seen in this discussion, an important feature of the structural behavior of the nonmetals is that the elements usually occur with eight electrons in their valence shells. If necessary, the elements form enough covalent bonds to supplement the electrons already present to possess an octet. For example, members of group 15 have five valence electrons and require only three additional electrons to fill their valence shells. These elements form three covalent bonds in their free state: triple bonds in the $\\mathrm{N}_{2}$ molecule or single bonds to three different atoms in arsenic and phosphorus. The elements of group 16 require only two additional electrons. Oxygen forms a double bond in the $\\mathrm{O}_{2}$ molecule, and sulfur, selenium, and tellurium form two single bonds in various rings and chains. The halogens form diatomic molecules in which each atom is involved in only one bond. This provides the electron required necessary to complete the octet on the halogen atom. The noble gases do not form covalent bonds to other noble gas atoms because they already have a filled outer shell."}
{"id": 4724, "contents": "1922. Sulfur - 1922.1. Occurrence, Preparation, and Compounds of Hydrogen\nLEARNING OBJECTIVES\nBy the end of this section, you will be able to:\n\n- Describe the properties, preparation, and compounds of hydrogen\n\nHydrogen is the most abundant element in the universe. The sun and other stars are composed largely of hydrogen. Astronomers estimate that $90 \\%$ of the atoms in the universe are hydrogen atoms. Hydrogen is a component of more compounds than any other element. Water is the most abundant compound of hydrogen found on earth. Hydrogen is an important part of petroleum, many minerals, cellulose and starch, sugar, fats, oils, alcohols, acids, and thousands of other substances.\n\nAt ordinary temperatures, hydrogen is a colorless, odorless, tasteless, and nonpoisonous gas consisting of the diatomic molecule $\\mathrm{H}_{2}$. Hydrogen is composed of three isotopes, and unlike other elements, these isotopes have different names and chemical symbols: protium, ${ }^{1} \\mathrm{H}$, deuterium, ${ }^{2} \\mathrm{H}$ (or \" D \"), and tritium ${ }^{3} \\mathrm{H}$ (or \" T \"). In a naturally occurring sample of hydrogen, there is one atom of deuterium for every 7000 H atoms and one atom of radioactive tritium for every $10^{18} \\mathrm{H}$ atoms. The chemical properties of the different isotopes are very similar because they have identical electron structures, but they differ in some physical properties because of their differing atomic masses. Elemental deuterium and tritium have lower vapor pressure than ordinary hydrogen. Consequently, when liquid hydrogen evaporates, the heavier isotopes are concentrated in the last portions to evaporate. Electrolysis of heavy water, $\\mathrm{D}_{2} \\mathrm{O}$, yields deuterium. Most tritium originates from nuclear reactions."}
{"id": 4725, "contents": "1923. Preparation of Hydrogen - \nElemental hydrogen must be prepared from compounds by breaking chemical bonds. The most common methods of preparing hydrogen follow.\n\nFrom Steam and Carbon or Hydrocarbons\nWater is the cheapest and most abundant source of hydrogen. Passing steam over coke (an impure form of elemental carbon) at $1000^{\\circ} \\mathrm{C}$ produces a mixture of carbon monoxide and hydrogen known as water gas:\n\n$$\n\\mathrm{C}(\\mathrm{~s})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{~g}) \\xrightarrow{1000^{\\circ} \\mathrm{C}} \\mathrm{CO}(g)+\\mathrm{H}_{2}(g)\n$$\n\nwater gas\nWater gas is as an industrial fuel. It is possible to produce additional hydrogen by mixing the water gas with steam in the presence of a catalyst to convert the CO to $\\mathrm{CO}_{2}$. This reaction is the water gas shift reaction.\n\nIt is also possible to prepare a mixture of hydrogen and carbon monoxide by passing hydrocarbons from natural gas or petroleum and steam over a nickel-based catalyst. Propane is an example of a hydrocarbon reactant:\n\n$$\n\\mathrm{C}_{3} \\mathrm{H}_{8}(\\mathrm{~g})+3 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{~g}) \\xrightarrow[\\text { catalyst }]{90{ }^{\\circ} \\mathrm{C}} 3 \\mathrm{CO}(\\mathrm{~g})+7 \\mathrm{H}_{2}(\\mathrm{~g})\n$$\n\nElectrolysis\nHydrogen forms when direct current electricity passes through water containing an electrolyte such as $\\mathrm{H}_{2} \\mathrm{SO}_{4}$, as illustrated in Figure 18.26. Bubbles of hydrogen form at the cathode, and oxygen evolves at the anode. The\nnet reaction is:\n\n$$\n2 \\mathrm{H}_{2} \\mathrm{O}(l)+\\text { electrical energy } \\longrightarrow 2 \\mathrm{H}_{2}(g)+\\mathrm{O}_{2}(g)\n$$"}
{"id": 4726, "contents": "1923. Preparation of Hydrogen - \n$$\n2 \\mathrm{H}_{2} \\mathrm{O}(l)+\\text { electrical energy } \\longrightarrow 2 \\mathrm{H}_{2}(g)+\\mathrm{O}_{2}(g)\n$$\n\n\n\nFIGURE 18.26 The electrolysis of water produces hydrogen and oxygen. Because there are twice as many hydrogen atoms as oxygen atoms and both elements are diatomic, there is twice the volume of hydrogen produced at the cathode as there is oxygen produced at the anode.\n\nReaction of Metals with Acids\nThis is the most convenient laboratory method of producing hydrogen. Metals with lower reduction potentials reduce the hydrogen ion in dilute acids to produce hydrogen gas and metal salts. For example, as shown in Figure 18.27, iron in dilute hydrochloric acid produces hydrogen gas and iron(II) chloride:\n\n$$\n\\mathrm{Fe}(s)+2 \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+2 \\mathrm{Cl}^{-}(a q) \\longrightarrow \\mathrm{Fe}^{2+}(a q)+2 \\mathrm{Cl}^{-}(a q)+\\mathrm{H}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\n\n\nFIGURE 18.27 The reaction of iron with an acid produces hydrogen. Here, iron reacts with hydrochloric acid. (credit: Mark Ott)"}
{"id": 4727, "contents": "1924. Reaction of Ionic Metal Hydrides with Water - \nIt is possible to produce hydrogen from the reaction of hydrides of the active metals, which contain the very strongly basic $\\mathrm{H}^{-}$anion, with water:\n\n$$\n\\mathrm{CaH}_{2}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Ca}^{2+}(a q)+2 \\mathrm{OH}^{-}(a q)+2 \\mathrm{H}_{2}(g)\n$$\n\nMetal hydrides are expensive but convenient sources of hydrogen, especially where space and weight are important factors. They are important in the inflation of life jackets, life rafts, and military balloons."}
{"id": 4728, "contents": "1925. Reactions - \nUnder normal conditions, hydrogen is relatively inactive chemically, but when heated, it enters into many chemical reactions.\n\nTwo thirds of the world's hydrogen production is devoted to the manufacture of ammonia, which is a fertilizer and used in the manufacture of nitric acid. Large quantities of hydrogen are also important in the process of hydrogenation, discussed in the chapter on organic chemistry.\n\nIt is possible to use hydrogen as a nonpolluting fuel. The reaction of hydrogen with oxygen is a very exothermic reaction, releasing 286 kJ of energy per mole of water formed. Hydrogen burns without explosion under controlled conditions. The oxygen-hydrogen torch, because of the high heat of combustion of hydrogen, can achieve temperatures up to $2800^{\\circ} \\mathrm{C}$. The hot flame of this torch is useful in cutting thick sheets of many metals. Liquid hydrogen is also an important rocket fuel (Figure 18.28).\n\n\nFIGURE 18.28 Before the fleet's retirement in 2011, liquid hydrogen and liquid oxygen were used in the three main engines of a space shuttle. Two compartments in the large tank held these liquids until the shuttle was launched. (credit: \"reynermedia\"/Flickr)\n\nAn uncombined hydrogen atom consists of a nucleus and one valence electron in the $1 s$ orbital. The $n=1$ valence shell has a capacity for two electrons, and hydrogen can rightfully occupy two locations in the periodic table. It is possible to consider hydrogen a group 1 element because hydrogen can lose an electron to form the cation, $\\mathrm{H}^{+}$. It is also possible to consider hydrogen to be a group 17 element because it needs only one electron to fill its valence orbital to form a hydride ion, $\\mathrm{H}^{-}$, or it can share an electron to form a single, covalent bond. In reality, hydrogen is a unique element that almost deserves its own location in the periodic table."}
{"id": 4729, "contents": "1926. Reactions with Elements - \nWhen heated, hydrogen reacts with the metals of group 1 and with $\\mathrm{Ca}, \\mathrm{Sr}$, and Ba (the more active metals in group 2). The compounds formed are crystalline, ionic hydrides that contain the hydride anion, $\\mathrm{H}^{-}$, a strong reducing agent and a strong base, which reacts vigorously with water and other acids to form hydrogen gas.\n\nThe reactions of hydrogen with nonmetals generally produce acidic hydrogen compounds with hydrogen in the 1+ oxidation state. The reactions become more exothermic and vigorous as the electronegativity of the nonmetal increases. Hydrogen reacts with nitrogen and sulfur only when heated, but it reacts explosively with fluorine (forming HF) and, under some conditions, with chlorine (forming HCl ). A mixture of hydrogen and oxygen explodes if ignited. Because of the explosive nature of the reaction, it is necessary to exercise caution when handling hydrogen (or any other combustible gas) to avoid the formation of an explosive mixture in a confined space. Although most hydrides of the nonmetals are acidic, ammonia and phosphine $\\left(\\mathrm{PH}_{3}\\right)$ are very, very weak acids and generally function as bases. There is a summary of these reactions of hydrogen with the elements in Table 18.1."}
{"id": 4730, "contents": "1926. Reactions with Elements - \n| General Equation | Comments |\n| :--- | :--- |\n| MH or $\\mathrm{MH}_{2} \\longrightarrow$ MOH or $\\mathrm{M}(\\mathrm{OH})_{2}+\\mathrm{H}_{2}$ | ionic hydrides with group 1 and Ca, Sr, and Ba |\n| $\\mathrm{H}_{2}+\\mathrm{C} \\longrightarrow$ (no reaction) |  |\n| $3 \\mathrm{H}_{2}+\\mathrm{N}_{2} \\longrightarrow 2 \\mathrm{NH}_{3}$ | requires high pressure and temperature; low yield |\n| $2 \\mathrm{H}_{2}+\\mathrm{O}_{2} \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}$ | exothermic and potentially explosive |\n| $\\mathrm{H}_{2}+\\mathrm{S} \\longrightarrow \\mathrm{H}_{2} \\mathrm{~S}$ | requires heating; low yield |\n| $\\mathrm{H}_{2}+\\mathrm{X}_{2} \\longrightarrow 2 \\mathrm{HX}$ | $\\mathrm{X}=\\mathrm{F}, \\mathrm{Cl}, \\mathrm{Br}$, and I; explosive with $\\mathrm{F}_{2}$; low yield with $\\mathrm{I}_{2}$ |\n\nTABLE 18.1"}
{"id": 4731, "contents": "1927. Reaction with Compounds - \nHydrogen reduces the heated oxides of many metals, with the formation of the metal and water vapor. For example, passing hydrogen over heated CuO forms copper and water.\n\nHydrogen may also reduce the metal ions in some metal oxides to lower oxidation states:\n\n$$\n\\mathrm{H}_{2}(g)+\\mathrm{MnO}_{2}(s) \\xrightarrow{\\Delta} \\mathrm{MnO}(s)+\\mathrm{H}_{2} \\mathrm{O}(g)\n$$"}
{"id": 4732, "contents": "1928. Hydrogen Compounds - \nOther than the noble gases, each of the nonmetals forms compounds with hydrogen. For brevity, we will discuss only a few hydrogen compounds of the nonmetals here."}
{"id": 4733, "contents": "1929. Nitrogen Hydrogen Compounds - \nAmmonia, $\\mathrm{NH}_{3}$, forms naturally when any nitrogen-containing organic material decomposes in the absence of air. The laboratory preparation of ammonia is by the reaction of an ammonium salt with a strong base such as sodium hydroxide. The acid-base reaction with the weakly acidic ammonium ion gives ammonia, illustrated in Figure 18.29. Ammonia also forms when ionic nitrides react with water. The nitride ion is a much stronger base than the hydroxide ion:\n\n$$\n\\mathrm{Mg}_{3} \\mathrm{~N}_{2}(s)+6 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 3 \\mathrm{Mg}(\\mathrm{OH})_{2}(s)+2 \\mathrm{NH}_{3}(g)\n$$\n\nThe commercial production of ammonia is by the direct combination of the elements in the Haber process:\n\n$$\n\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\stackrel{\\text { catalyst }}{\\rightleftharpoons} 2 \\mathrm{NH}_{3}(g) \\quad \\Delta H^{\\circ}=-92 \\mathrm{~kJ}\n$$\n\nFIGURE 18.29 The structure of ammonia is shown with a central nitrogen atom and three hydrogen atoms.\nAmmonia is a colorless gas with a sharp, pungent odor. Smelling salts utilize this powerful odor. Gaseous ammonia readily liquefies to give a colorless liquid that boils at $-33^{\\circ} \\mathrm{C}$. Due to intermolecular hydrogen bonding, the enthalpy of vaporization of liquid ammonia is higher than that of any other liquid except water, so ammonia is useful as a refrigerant. Ammonia is quite soluble in water ( 658 L at STP dissolves in $1 \\mathrm{~L} \\mathrm{H}_{2} \\mathrm{O}$ ).\n\nThe chemical properties of ammonia are as follows:"}
{"id": 4734, "contents": "1929. Nitrogen Hydrogen Compounds - \nThe chemical properties of ammonia are as follows:\n\n1. Ammonia acts as a Br\u00f8nsted base, as discussed in the chapter on acid-base chemistry. The ammonium ion is similar in size to the potassium ion; compounds of the two ions exhibit many similarities in their structures and solubilities.\n2. Ammonia can display acidic behavior, although it is a much weaker acid than water. Like other acids, ammonia reacts with metals, although it is so weak that high temperatures are necessary. Hydrogen and (depending on the stoichiometry) amides (salts of $\\mathrm{NH}_{2}{ }^{-}$), imides (salts of $\\mathrm{NH}^{2-}$ ), or nitrides (salts of $\\mathrm{N}^{3-}$ ) form.\n3. The nitrogen atom in ammonia has its lowest possible oxidation state (3-) and thus is not susceptible to reduction. However, it can be oxidized. Ammonia burns in air, giving NO and water. Hot ammonia and the ammonium ion are active reducing agents. Of particular interest are the oxidations of ammonium ion by nitrite ion, $\\mathrm{NO}_{2}{ }^{-}$, to yield pure nitrogen and by nitrate ion to yield nitrous oxide, $\\mathrm{N}_{2} \\mathrm{O}$.\n4. There are a number of compounds that we can consider derivatives of ammonia through the replacement of one or more hydrogen atoms with some other atom or group of atoms. Inorganic derivations include chloramine, $\\mathrm{NH}_{2} \\mathrm{Cl}$, and hydrazine, $\\mathrm{N}_{2} \\mathrm{H}_{4}$ :\n\nammonia\n(a)\n\nchloramine\n(b)\n\nhydrazine\n(c)\n\nChloramine, $\\mathrm{NH}_{2} \\mathrm{Cl}$, results from the reaction of sodium hypochlorite, NaOCl , with ammonia in basic solution. In the presence of a large excess of ammonia at low temperature, the chloramine reacts further to produce hydrazine, $\\mathrm{N}_{2} \\mathrm{H}_{4}$ :"}
{"id": 4735, "contents": "1929. Nitrogen Hydrogen Compounds - \n$$\n\\begin{gathered}\n\\mathrm{NH}_{3}(a q)+\\mathrm{OCl}^{-}(a q) \\longrightarrow \\mathrm{NH}_{2} \\mathrm{Cl}(a q)+\\mathrm{OH}^{-}(a q) \\\\\n\\mathrm{NH}_{2} \\mathrm{Cl}(a q)+\\mathrm{NH}_{3}(a q)+\\mathrm{OH}^{-}(a q) \\longrightarrow \\mathrm{N}_{2} \\mathrm{H}_{4}(a q)+\\mathrm{Cl}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)\n\\end{gathered}\n$$\n\nAnhydrous hydrazine is relatively stable in spite of its positive free energy of formation:\n\n$$\n\\mathrm{N}_{2}(g)+2 \\mathrm{H}_{2}(g) \\longrightarrow \\mathrm{N}_{2} \\mathrm{H}_{4}(l) \\quad \\Delta G_{\\mathrm{f}}^{\\circ}=149.2 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\n$$\n\nHydrazine is a fuming, colorless liquid that has some physical properties remarkably similar to those of $\\mathrm{H}_{2} \\mathrm{O}$ (it melts at $2{ }^{\\circ} \\mathrm{C}$, boils at $113.5^{\\circ} \\mathrm{C}$, and has a density at $25^{\\circ} \\mathrm{C}$ of $1.00 \\mathrm{~g} / \\mathrm{mL}$ ). It burns rapidly and completely in air with substantial evolution of heat:\n\n$$\n\\mathrm{N}_{2} \\mathrm{H}_{4}(l)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{N}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\quad \\Delta H^{\\circ}=-621.5 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\n$$"}
{"id": 4736, "contents": "1929. Nitrogen Hydrogen Compounds - \nLike ammonia, hydrazine is both a Br\u00f8nsted base and a Lewis base, although it is weaker than ammonia. It reacts with strong acids and forms two series of salts that contain the $\\mathrm{N}_{2} \\mathrm{H}_{5}{ }^{+}$and $\\mathrm{N}_{2} \\mathrm{H}_{6}{ }^{2+}$ ions, respectively. Some rockets use hydrazine as a fuel."}
{"id": 4737, "contents": "1930. Phosphorus Hydrogen Compounds - \nThe most important hydride of phosphorus is phosphine, $\\mathrm{PH}_{3}$, a gaseous analog of ammonia in terms of both formula and structure. Unlike ammonia, it is not possible to form phosphine by direct union of the elements. There are two methods for the preparation of phosphine. One method is by the action of an acid on an ionic phosphide. The other method is the disproportionation of white phosphorus with hot concentrated base to produce phosphine and the hydrogen phosphite ion:\n\n$$\n\\begin{gathered}\n\\mathrm{AlP}(s)+3 \\mathrm{H}_{3} \\mathrm{O}^{+}(a q) \\longrightarrow \\mathrm{PH}_{3}(g)+\\mathrm{Al}^{3+}(a q)+3 \\mathrm{H}_{2} \\mathrm{O}(l) \\\\\n\\mathrm{P}_{4}(s)+4 \\mathrm{OH}^{-}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{HPO}_{3}^{2-}(a q)+2 \\mathrm{PH}_{3}(g)\n\\end{gathered}\n$$"}
{"id": 4738, "contents": "1930. Phosphorus Hydrogen Compounds - \nPhosphine is a colorless, very poisonous gas, which has an odor like that of decaying fish. Heat easily decomposes phosphine $\\left(4 \\mathrm{PH}_{3} \\longrightarrow \\mathrm{P}_{4}+6 \\mathrm{H}_{2}\\right)$, and the compound burns in air. The major uses of phosphine are as a fumigant for grains and in semiconductor processing. Like ammonia, gaseous phosphine unites with gaseous hydrogen halides, forming phosphonium compounds like $\\mathrm{PH}_{4} \\mathrm{Cl}$ and $\\mathrm{PH}_{4} \\mathrm{I}$. Phosphine is a much weaker base than ammonia; therefore, these compounds decompose in water, and the insoluble $\\mathrm{PH}_{3}$ escapes\nfrom solution.\nSulfur Hydrogen Compounds\nHydrogen sulfide, $\\mathrm{H}_{2} \\mathrm{~S}$, is a colorless gas that is responsible for the offensive odor of rotten eggs and of many hot springs. Hydrogen sulfide is as toxic as hydrogen cyanide; therefore, it is necessary to exercise great care in handling it. Hydrogen sulfide is particularly deceptive because it paralyzes the olfactory nerves; after a short exposure, one does not smell it.\n\nThe production of hydrogen sulfide by the direct reaction of the elements $\\left(\\mathrm{H}_{2}+\\mathrm{S}\\right)$ is unsatisfactory because the yield is low. A more effective preparation method is the reaction of a metal sulfide with a dilute acid. For example:\n\n$$\n\\mathrm{FeS}(s)+2 \\mathrm{H}_{3} \\mathrm{O}^{+}(a q) \\longrightarrow \\mathrm{Fe}^{2+}(a q)+\\mathrm{H}_{2} \\mathrm{~S}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$"}
{"id": 4739, "contents": "1930. Phosphorus Hydrogen Compounds - \nIt is easy to oxidize the sulfur in metal sulfides and in hydrogen sulfide, making metal sulfides and $\\mathrm{H}_{2} \\mathrm{~S}$ good reducing agents. In acidic solutions, hydrogen sulfide reduces $\\mathrm{Fe}^{3+}$ to $\\mathrm{Fe}^{2+}, \\mathrm{MnO}_{4}{ }^{-}$to $\\mathrm{Mn}^{2+}, \\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}$ to $\\mathrm{Cr}^{3+}$, and $\\mathrm{HNO}_{3}$ to $\\mathrm{NO}_{2}$. The sulfur in $\\mathrm{H}_{2} \\mathrm{~S}$ usually oxidizes to elemental sulfur, unless a large excess of the oxidizing agent is present. In which case, the sulfide may oxidize to $\\mathrm{SO}_{3}{ }^{2-}$ or $\\mathrm{SO}_{4}{ }^{2-}$ (or to $\\mathrm{SO}_{2}$ or $\\mathrm{SO}_{3}$ in the absence of water):\n\n$$\n2 \\mathrm{H}_{2} \\mathrm{~S}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{~S}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nThis oxidation process leads to the removal of the hydrogen sulfide found in many sources of natural gas. The deposits of sulfur in volcanic regions may be the result of the oxidation of $\\mathrm{H}_{2} \\mathrm{~S}$ present in volcanic gases.\n\nHydrogen sulfide is a weak diprotic acid that dissolves in water to form hydrosulfuric acid. The acid ionizes in two stages, yielding hydrogen sulfide ions, $\\mathrm{HS}^{-}$, in the first stage and sulfide ions, $\\mathrm{S}^{2-}$, in the second. Since hydrogen sulfide is a weak acid, aqueous solutions of soluble sulfides and hydrogen sulfides are basic:"}
{"id": 4740, "contents": "1930. Phosphorus Hydrogen Compounds - \n$$\n\\begin{aligned}\n& \\mathrm{S}^{2-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{HS}^{-}(a q)+\\mathrm{OH}^{-}(a q) \\\\\n& \\mathrm{HS}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{~S}(g)+\\mathrm{OH}^{-}(a q)\n\\end{aligned}\n$$"}
{"id": 4741, "contents": "1931. Halogen Hydrogen Compounds - \nBinary compounds containing only hydrogen and a halogen are hydrogen halides. At room temperature, the pure hydrogen halides $\\mathrm{HF}, \\mathrm{HCl}, \\mathrm{HBr}$, and HI are gases.\n\nIn general, it is possible to prepare the halides by the general techniques used to prepare other acids. Fluorine, chlorine, and bromine react directly with hydrogen to form the respective hydrogen halide. This is a commercially important reaction for preparing hydrogen chloride and hydrogen bromide.\n\nThe acid-base reaction between a nonvolatile strong acid and a metal halide will yield a hydrogen halide. The escape of the gaseous hydrogen halide drives the reaction to completion. For example, the usual method of preparing hydrogen fluoride is by heating a mixture of calcium fluoride, $\\mathrm{CaF}_{2}$, and concentrated sulfuric acid:\n\n$$\n\\mathrm{CaF}_{2}(s)+\\mathrm{H}_{2} \\mathrm{SO}_{4}(a q) \\longrightarrow \\mathrm{CaSO}_{4}(s)+2 \\mathrm{HF}(g)\n$$\n\nGaseous hydrogen fluoride is also a by-product in the preparation of phosphate fertilizers by the reaction of fluoroapatite, $\\mathrm{Ca}_{5}\\left(\\mathrm{PO}_{4}\\right)_{3} \\mathrm{~F}$, with sulfuric acid. The reaction of concentrated sulfuric acid with a chloride salt produces hydrogen chloride both commercially and in the laboratory.\n\nIn most cases, sodium chloride is the chloride of choice because it is the least expensive chloride. Hydrogen bromide and hydrogen iodide cannot be prepared using sulfuric acid because this acid is an oxidizing agent capable of oxidizing both bromide and iodide. However, it is possible to prepare both hydrogen bromide and hydrogen iodide using an acid such as phosphoric acid because it is a weaker oxidizing agent. For example:\n\n$$\n\\mathrm{H}_{3} \\mathrm{PO}_{4}(l)+\\mathrm{Br}^{-}(a q) \\longrightarrow \\mathrm{HBr}(g)+\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}(a q)\n$$"}
{"id": 4742, "contents": "1931. Halogen Hydrogen Compounds - \nAll of the hydrogen halides are very soluble in water, forming hydrohalic acids. With the exception of hydrogen fluoride, which has a strong hydrogen-fluoride bond, they are strong acids. Reactions of hydrohalic acids with metals, metal hydroxides, oxides, or carbonates produce salts of the halides. Most chloride salts are soluble in water. $\\mathrm{AgCl}, \\mathrm{PbCl}_{2}$, and $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$ are the commonly encountered exceptions.\n\nThe halide ions give the substances the properties associated with $\\mathrm{X}^{-}(\\mathrm{aq})$. The heavier halide ions $\\left(\\mathrm{Cl}^{-}, \\mathrm{Br}^{-}\\right.$, and $\\mathrm{I}^{-}$) can act as reducing agents, and the lighter halogens or other oxidizing agents will oxidize them:\n\n$$\n\\begin{array}{rr}\n\\mathrm{Cl}_{2}(a q)+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Cl}^{-}(a q) & E^{\\circ}=1.36 \\mathrm{~V} \\\\\n\\mathrm{Br}_{2}(a q)+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Br}^{-}(a q) & E^{\\circ}=1.09 \\mathrm{~V} \\\\\n\\mathrm{I}_{2}(a q)+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{I}^{-}(a q) & E^{\\circ}=0.54 \\mathrm{~V}\n\\end{array}\n$$\n\nFor example, bromine oxidizes iodine:\n\n$$\n\\mathrm{Br}_{2}(a q)+2 \\mathrm{HI}(a q) \\longrightarrow 2 \\mathrm{HBr}(a q)+\\mathrm{I}_{2}(a q) \\quad E^{\\circ}=0.55 \\mathrm{~V}\n$$\n\nHydrofluoric acid is unique in its reactions with sand (silicon dioxide) and with glass, which is a mixture of silicates:"}
{"id": 4743, "contents": "1931. Halogen Hydrogen Compounds - \nHydrofluoric acid is unique in its reactions with sand (silicon dioxide) and with glass, which is a mixture of silicates:\n\n$$\n\\begin{gathered}\n\\mathrm{SiO}_{2}(s)+4 \\mathrm{HF}(a q) \\longrightarrow \\mathrm{SiF}_{4}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\\\\n\\mathrm{CaSiO}_{3}(s)+6 \\mathrm{HF}(a q) \\longrightarrow \\mathrm{CaF}_{2}(s)+\\mathrm{SiF}_{4}(g)+3 \\mathrm{H}_{2} \\mathrm{O}(l)\n\\end{gathered}\n$$\n\nThe volatile silicon tetrafluoride escapes from these reactions. Because hydrogen fluoride attacks glass, it can frost or etch glass and is used to etch markings on thermometers, burets, and other glassware.\n\nThe largest use for hydrogen fluoride is in production of hydrochlorofluorocarbons for refrigerants, in plastics, and in propellants. The second largest use is in the manufacture of cryolite, $\\mathrm{Na}_{3} \\mathrm{AlF}_{6}$, which is important in the production of aluminum. The acid is also important in the production of other inorganic fluorides (such as $\\mathrm{BF}_{3}$ ), which serve as catalysts in the industrial synthesis of certain organic compounds.\n\nHydrochloric acid is relatively inexpensive. It is an important and versatile acid in industry and is important for the manufacture of metal chlorides, dyes, glue, glucose, and various other chemicals. A considerable amount is also important for the activation of oil wells and as pickle liquor-an acid used to remove oxide coating from iron or steel that is to be galvanized, tinned, or enameled. The amounts of hydrobromic acid and hydroiodic acid used commercially are insignificant by comparison."}
{"id": 4744, "contents": "1932. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the preparation, properties, and uses of some representative metal carbonates\n\nThe chemistry of carbon is extensive; however, most of this chemistry is not relevant to this chapter. The other aspects of the chemistry of carbon will appear in the chapter covering organic chemistry. In this chapter, we will focus on the carbonate ion and related substances. The metals of groups 1 and 2, as well as zinc, cadmium, mercury, and lead(II), form ionic carbonates-compounds that contain the carbonate anions, $\\mathrm{CO}_{3}{ }^{2-}$. The metals of group 1, magnesium, calcium, strontium, and barium also form hydrogen carbonates-compounds that contain the hydrogen carbonate anion, $\\mathrm{HCO}_{3}{ }^{-}$, also known as the bicarbonate anion.\n\nWith the exception of magnesium carbonate, it is possible to prepare carbonates of the metals of groups 1 and 2 by the reaction of carbon dioxide with the respective oxide or hydroxide. Examples of such reactions include:\n\n$$\n\\begin{gathered}\n\\mathrm{Na}_{2} \\mathrm{O}(s)+\\mathrm{CO}_{2}(g) \\longrightarrow \\mathrm{Na}_{2} \\mathrm{CO}_{3}(s) \\\\\n\\mathrm{Ca}(\\mathrm{OH})_{2}(s)+\\mathrm{CO}_{2}(g) \\longrightarrow \\mathrm{CaCO}_{3}(s)+\\mathrm{H}_{2} \\mathrm{O}(l)\n\\end{gathered}\n$$\n\nThe carbonates of the alkaline earth metals of group 12 and lead(II) are not soluble. These carbonates precipitate upon mixing a solution of soluble alkali metal carbonate with a solution of soluble salts of these metals. Examples of net ionic equations for the reactions are:\n\n$$\n\\begin{aligned}\n& \\mathrm{Ca}^{2+}(a q)+\\mathrm{CO}_{3}^{2-}(a q) \\longrightarrow \\mathrm{CaCO}_{3}(s) \\\\\n& \\mathrm{Pb}^{2+}(a q)+\\mathrm{CO}_{3}^{2-}(a q) \\longrightarrow \\mathrm{PbCO}_{3}(s)\n\\end{aligned}\n$$"}
{"id": 4745, "contents": "1932. LEARNING OBJECTIVES - \nPearls and the shells of most mollusks are calcium carbonate. Tin(II) or one of the trivalent or tetravalent ions such as $\\mathrm{Al}^{3+}$ or $\\mathrm{Sn}^{4+}$ behave differently in this reaction as carbon dioxide and the corresponding oxide form\ninstead of the carbonate.\nAlkali metal hydrogen carbonates such as $\\mathrm{NaHCO}_{3}$ and $\\mathrm{CsHCO}_{3}$ form by saturating a solution of the hydroxides with carbon dioxide. The net ionic reaction involves hydroxide ion and carbon dioxide:\n\n$$\n\\mathrm{OH}^{-}(a q)+\\mathrm{CO}_{2}(a q) \\longrightarrow \\mathrm{HCO}_{3}^{-}(a q)\n$$\n\nIt is possible to isolate the solids by evaporation of the water from the solution.\nAlthough they are insoluble in pure water, alkaline earth carbonates dissolve readily in water containing carbon dioxide because hydrogen carbonate salts form. For example, caves and sinkholes form in limestone when $\\mathrm{CaCO}_{3}$ dissolves in water containing dissolved carbon dioxide:\n\n$$\n\\mathrm{CaCO}_{3}(s)+\\mathrm{CO}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Ca}^{2+}(a q)+2 \\mathrm{HCO}_{3}^{-}(a q)\n$$\n\nHydrogen carbonates of the alkaline earth metals remain stable only in solution; evaporation of the solution produces the carbonate. Stalactites and stalagmites, like those shown in Figure 18.30, form in caves when drops of water containing dissolved calcium hydrogen carbonate evaporate to leave a deposit of calcium carbonate.\n\n(a)\n\n(b)\n\nFIGURE 18.30 (a) Stalactites and (b) stalagmites are cave formations of calcium carbonate. (credit a: modification of work by Arvind Govindaraj; credit b: modification of work by the National Park Service.)"}
{"id": 4746, "contents": "1932. LEARNING OBJECTIVES - \nThe two carbonates used commercially in the largest quantities are sodium carbonate and calcium carbonate. In the United States, sodium carbonate is extracted from the mineral trona, $\\mathrm{Na}_{3}\\left(\\mathrm{CO}_{3}\\right)\\left(\\mathrm{HCO}_{3}\\right)\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{2}$. Following recrystallization to remove clay and other impurities, heating the recrystallized trona produces $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ :\n\n$$\n2 \\mathrm{Na}_{3}\\left(\\mathrm{CO}_{3}\\right)\\left(\\mathrm{HCO}_{3}\\right)\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{2}(s) \\longrightarrow 3 \\mathrm{Na}_{2} \\mathrm{CO}_{3}(s)+5 \\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{CO}_{2}(g)\n$$\n\nCarbonates are moderately strong bases. Aqueous solutions are basic because the carbonate ion accepts hydrogen ion from water in this reversible reaction:\n\n$$\n\\mathrm{CO}_{3}^{2-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\rightleftharpoons \\mathrm{HCO}_{3}^{-}(a q)+\\mathrm{OH}^{-}(a q)\n$$\n\nCarbonates react with acids to form salts of the metal, gaseous carbon dioxide, and water. The reaction of calcium carbonate, the active ingredient of the antacid Tums, with hydrochloric acid (stomach acid), as shown in Figure 18.31, illustrates the reaction:\n\n$$\n\\mathrm{CaCO}_{3}(s)+2 \\mathrm{HCl}(a q) \\longrightarrow \\mathrm{CaCl}_{2}(a q)+\\mathrm{CO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\n\n\nFIGURE 18.31 The reaction of calcium carbonate with hydrochloric acid is shown. (credit: Mark Ott)\nOther applications of carbonates include glass making-where carbonate ions serve as a source of oxide ions-and synthesis of oxides."}
{"id": 4747, "contents": "1932. LEARNING OBJECTIVES - \nHydrogen carbonates are amphoteric because they act as both weak acids and weak bases. Hydrogen carbonate ions act as acids and react with solutions of soluble hydroxides to form a carbonate and water:\n\n$$\n\\mathrm{KHCO}_{3}(a q)+\\mathrm{KOH}(a q) \\longrightarrow \\mathrm{K}_{2} \\mathrm{CO}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nWith acids, hydrogen carbonates form a salt, carbon dioxide, and water. Baking soda (bicarbonate of soda or sodium bicarbonate) is sodium hydrogen carbonate. Baking powder contains baking soda and a solid acid such as potassium hydrogen tartrate (cream of tartar), $\\mathrm{KHC}_{4} \\mathrm{H}_{4} \\mathrm{O}_{6}$. As long as the powder is dry, no reaction occurs; immediately after the addition of water, the acid reacts with the hydrogen carbonate ions to form carbon dioxide:\n\n$$\n\\mathrm{HC}_{4} \\mathrm{H}_{4} \\mathrm{O}_{6}{ }^{-}(a q)+\\mathrm{HCO}_{3}{ }^{-}(a q) \\longrightarrow \\mathrm{C}_{4} \\mathrm{H}_{4} \\mathrm{O}_{6}{ }^{2-}(a q)+\\mathrm{CO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nDough will trap the carbon dioxide, causing it to expand during baking, producing the characteristic texture of baked goods."}
{"id": 4748, "contents": "1933. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the properties, preparation, and uses of nitrogen\n\nMost pure nitrogen comes from the fractional distillation of liquid air. The atmosphere consists of 78\\% nitrogen by volume. This means there are more than 20 million tons of nitrogen over every square mile of the earth's surface. Nitrogen is a component of proteins and of the genetic material (DNA/RNA) of all plants and animals.\n\nUnder ordinary conditions, nitrogen is a colorless, odorless, and tasteless gas. It boils at 77 K and freezes at 63 K. Liquid nitrogen is a useful coolant because it is inexpensive and has a low boiling point. Nitrogen is very unreactive because of the very strong triple bond between the nitrogen atoms. The only common reactions at room temperature occur with lithium to form $\\mathrm{Li}_{3} \\mathrm{~N}$, with certain transition metal complexes, and with hydrogen or oxygen in nitrogen-fixing bacteria. The general lack of reactivity of nitrogen makes the remarkable ability of some bacteria to synthesize nitrogen compounds using atmospheric nitrogen gas as the source one of the most exciting chemical events on our planet. This process is one type of nitrogen fixation. In this case, nitrogen fixation is the process where organisms convert atmospheric nitrogen into biologically useful chemicals. Nitrogen fixation also occurs when lightning passes through air, causing molecular nitrogen to react with oxygen to form nitrogen oxides, which are then carried down to the soil."}
{"id": 4749, "contents": "1935. Nitrogen Fixation - \nAll living organisms require nitrogen compounds for survival. Unfortunately, most of these organisms cannot absorb nitrogen from its most abundant source-the atmosphere. Atmospheric nitrogen consists of $\\mathrm{N}_{2}$ molecules, which are very unreactive due to the strong nitrogen-nitrogen triple bond. However, a few organisms can overcome this problem through a process known as nitrogen fixation, illustrated in Figure 18.32.\n\n\nFIGURE 18.32 All living organisms require nitrogen. A few microorganisms are able to process atmospheric nitrogen using nitrogen fixation. (credit \"roots\": modification of work by the United States Department of Agriculture; credit \"root nodules\": modification of work by Louisa Howard)\n\nNitrogen fixation is the process where organisms convert atmospheric nitrogen into biologically useful chemicals. To date, the only known kind of biological organisms capable of nitrogen fixation are microorganisms. These organisms employ enzymes called nitrogenases, which contain iron and molybdenum. Many of these microorganisms live in a symbiotic relationship with plants, with the bestknown example being the presence of rhizobia in the root nodules of legumes.\n\nLarge volumes of atmospheric nitrogen are necessary for making ammonia-the principal starting material\nused for preparation of large quantities of other nitrogen-containing compounds. Most other uses for elemental nitrogen depend on its inactivity. It is helpful when a chemical process requires an inert atmosphere. Canned foods and luncheon meats cannot oxidize in a pure nitrogen atmosphere, so they retain a better flavor and color, and spoil less rapidly, when sealed in nitrogen instead of air. This technology allows fresh produce to be available year-round, regardless of growing season.\n\nThere are compounds with nitrogen in all of its oxidation states from 3- to $5+$. Much of the chemistry of nitrogen involves oxidation-reduction reactions. Some active metals (such as alkali metals and alkaline earth metals) can reduce nitrogen to form metal nitrides. In the remainder of this section, we will examine nitrogenoxygen chemistry."}
{"id": 4750, "contents": "1935. Nitrogen Fixation - \nThere are well-characterized nitrogen oxides in which nitrogen exhibits each of its positive oxidation numbers from $1+$ to $5+$. When ammonium nitrate is carefully heated, nitrous oxide (dinitrogen oxide) and water vapor form. Stronger heating generates nitrogen gas, oxygen gas, and water vapor. No one should ever attempt this reaction-it can be very explosive. In 1947, there was a major ammonium nitrate explosion in Texas City, Texas, and, in 2013, there was another major explosion in West, Texas. In the last 100 years, there were nearly 30 similar disasters worldwide, resulting in the loss of numerous lives. In this oxidation-reduction reaction, the nitrogen in the nitrate ion oxidizes the nitrogen in the ammonium ion. Nitrous oxide, shown in Figure 18.33, is a colorless gas possessing a mild, pleasing odor and a sweet taste. It finds application as an anesthetic for minor operations, especially in dentistry, under the name \"laughing gas.\"\n\n\n\nFIGURE 18.33 Nitrous oxide, $\\mathrm{N}_{2} \\mathrm{O}$, is an anesthetic that has these molecular (left) and resonance (right) structures.\nLow yields of nitric oxide, NO, form when heating nitrogen and oxygen together. NO also forms when lightning passes through air during thunderstorms. Burning ammonia is the commercial method of preparing nitric oxide. In the laboratory, the reduction of nitric acid is the best method for preparing nitric oxide. When copper reacts with dilute nitric acid, nitric oxide is the principal reduction product:\n\n$$\n3 \\mathrm{Cu}(s)+8 \\mathrm{HNO}_{3}(a q) \\longrightarrow 2 \\mathrm{NO}(g)+3 \\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+4 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nGaseous nitric oxide is the most thermally stable of the nitrogen oxides and is the simplest known thermally stable molecule with an unpaired electron. It is one of the air pollutants generated by internal combustion engines, resulting from the reaction of atmospheric nitrogen and oxygen during the combustion process."}
{"id": 4751, "contents": "1935. Nitrogen Fixation - \nAt room temperature, nitric oxide is a colorless gas consisting of diatomic molecules. As is often the case with molecules that contain an unpaired electron, two molecules combine to form a dimer by pairing their unpaired electrons to form a bond. Liquid and solid NO both contain $\\mathrm{N}_{2} \\mathrm{O}_{2}$ dimers, like that shown in Figure 18.34. Most substances with unpaired electrons exhibit color by absorbing visible light; however, NO is colorless because the absorption of light is not in the visible region of the spectrum.\n\n\nFIGURE 18.34 This shows the equilibrium between NO and $\\mathrm{N}_{2} \\mathrm{O}_{2}$. The molecule, $\\mathrm{N}_{2} \\mathrm{O}_{2}$, absorbs light.\nCooling a mixture of equal parts nitric oxide and nitrogen dioxide to $-21^{\\circ} \\mathrm{C}$ produces dinitrogen trioxide, a blue liquid consisting of $\\mathrm{N}_{2} \\mathrm{O}_{3}$ molecules (shown in Figure 18.35). Dinitrogen trioxide exists only in the liquid and solid states. When heated, it reverts to a mixture of NO and $\\mathrm{NO}_{2}$.\n\n\nFIGURE 18.35 Dinitrogen trioxide, $\\mathrm{N}_{2} \\mathrm{O}_{3}$, only exists in liquid or solid states and has these molecular (left) and resonance (right) structures.\n\nIt is possible to prepare nitrogen dioxide in the laboratory by heating the nitrate of a heavy metal, or by the reduction of concentrated nitric acid with copper metal, as shown in Figure 18.36. Commercially, it is possible to prepare nitrogen dioxide by oxidizing nitric oxide with air.\n\n\nFIGURE 18.36 The reaction of copper metal with concentrated $\\mathrm{HNO}_{3}$ produces a solution of $\\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}$ and brown fumes of $\\mathrm{NO}_{2}$. (credit: modification of work by Mark Ott)"}
{"id": 4752, "contents": "1935. Nitrogen Fixation - \nThe nitrogen dioxide molecule (illustrated in Figure 18.37) contains an unpaired electron, which is responsible for its color and paramagnetism. It is also responsible for the dimerization of $\\mathrm{NO}_{2}$. At low pressures or at high temperatures, nitrogen dioxide has a deep brown color that is due to the presence of the $\\mathrm{NO}_{2}$ molecule. At low temperatures, the color almost entirely disappears as dinitrogen tetraoxide, $\\mathrm{N}_{2} \\mathrm{O}_{4}$, forms. At room temperature, an equilibrium exists:\n\n\n$$\nK_{P}=6.86\n$$\n\n\n\n\nFIGURE 18.37 The molecular and resonance structures for nitrogen dioxide ( $\\mathrm{NO}_{2}$, left) and dinitrogen tetraoxide ( $\\mathrm{N}_{2} \\mathrm{O}_{4}$, right) are shown.\n\nDinitrogen pentaoxide, $\\mathrm{N}_{2} \\mathrm{O}_{5}$ (illustrated in Figure 18.38), is a white solid that is formed by the dehydration of nitric acid by phosphorus(V) oxide (tetraphosphorus decoxide):\n\n$$\n\\mathrm{P}_{4} \\mathrm{O}_{10}(s)+4 \\mathrm{HNO}_{3}(l) \\longrightarrow 4 \\mathrm{HPO}_{3}(s)+2 \\mathrm{~N}_{2} \\mathrm{O}_{5}(s)\n$$\n\nIt is unstable above room temperature, decomposing to $\\mathrm{N}_{2} \\mathrm{O}_{4}$ and $\\mathrm{O}_{2}$.\n\n\nFIGURE 18.38 This image shows the molecular structure and one resonance structure of a molecule of dinitrogen pentaoxide, $\\mathrm{N}_{2} \\mathrm{O}_{5}$."}
{"id": 4753, "contents": "1935. Nitrogen Fixation - \nFIGURE 18.38 This image shows the molecular structure and one resonance structure of a molecule of dinitrogen pentaoxide, $\\mathrm{N}_{2} \\mathrm{O}_{5}$.\n\nThe oxides of nitrogen(III), nitrogen(IV), and nitrogen(V) react with water and form nitrogen-containing oxyacids. Nitrogen(III) oxide, $\\mathrm{N}_{2} \\mathrm{O}_{3}$, is the anhydride of nitrous acid; $\\mathrm{HNO}_{2}$ forms when $\\mathrm{N}_{2} \\mathrm{O}_{3}$ reacts with water. There are no stable oxyacids containing nitrogen with an oxidation state of $4+$; therefore, nitrogen(IV) oxide, $\\mathrm{NO}_{2}$, disproportionates in one of two ways when it reacts with water. In cold water, a mixture of $\\mathrm{HNO}_{2}$ and $\\mathrm{HNO}_{3}$ forms. At higher temperatures, $\\mathrm{HNO}_{3}$ and NO will form. Nitrogen(V) oxide, $\\mathrm{N}_{2} \\mathrm{O}_{5}$, is the anhydride of nitric acid; $\\mathrm{HNO}_{3}$ is produced when $\\mathrm{N}_{2} \\mathrm{O}_{5}$ reacts with water:\n\n$$\n\\mathrm{N}_{2} \\mathrm{O}_{5}(s)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{HNO}_{3}(a q)\n$$\n\nThe nitrogen oxides exhibit extensive oxidation-reduction behavior. Nitrous oxide resembles oxygen in its behavior when heated with combustible substances. $\\mathrm{N}_{2} \\mathrm{O}$ is a strong oxidizing agent that decomposes when heated to form nitrogen and oxygen. Because one-third of the gas liberated is oxygen, nitrous oxide supports combustion better than air (one-fifth oxygen). A glowing splinter bursts into flame when thrust into a bottle of this gas. Nitric oxide acts both as an oxidizing agent and as a reducing agent. For example:"}
{"id": 4754, "contents": "1935. Nitrogen Fixation - \n$$\n\\begin{gathered}\n\\text { oxidizing agent: } \\mathrm{P}_{4}(s)+6 \\mathrm{NO}(g) \\longrightarrow \\mathrm{P}_{4} \\mathrm{O}_{6}(s)+3 \\mathrm{~N}_{2}(g) \\\\\n\\text { reducing agent: } \\mathrm{Cl}_{2}(g)+2 \\mathrm{NO}(g) \\longrightarrow 2 \\mathrm{ClNO}(g)\n\\end{gathered}\n$$\n\nNitrogen dioxide (or dinitrogen tetraoxide) is a good oxidizing agent. For example:\n\n$$\n\\begin{gathered}\n\\mathrm{NO}_{2}(g)+\\mathrm{CO}(g) \\longrightarrow \\mathrm{NO}(g)+\\mathrm{CO}_{2}(g) \\\\\n\\mathrm{NO}_{2}(g)+2 \\mathrm{HCl}(a q) \\longrightarrow \\mathrm{NO}(g)+\\mathrm{Cl}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l)\n\\end{gathered}\n$$"}
{"id": 4755, "contents": "1936. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the properties, preparation, and uses of phosphorus\n\nThe industrial preparation of phosphorus is by heating calcium phosphate, obtained from phosphate rock, with sand and coke:\n\n$$\n2 \\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}(s)+6 \\mathrm{SiO}_{2}(s)+10 \\mathrm{C}(s) \\xrightarrow{\\Delta} 6 \\mathrm{CaSiO}_{3}(l)+10 \\mathrm{CO}(g)+\\mathrm{P}_{4}(g)\n$$\n\nThe phosphorus distills out of the furnace and is condensed into a solid or burned to form $\\mathrm{P}_{4} \\mathrm{O}_{10}$. The preparation of many other phosphorus compounds begins with $\\mathrm{P}_{4} \\mathrm{O}_{10}$. The acids and phosphates are useful as fertilizers and in the chemical industry. Other uses are in the manufacture of special alloys such as ferrophosphorus and phosphor bronze. Phosphorus is important in making pesticides, matches, and some plastics. Phosphorus is an active nonmetal. In compounds, phosphorus usually occurs in oxidation states of $3-$, 3+, and 5+. Phosphorus exhibits oxidation numbers that are unusual for a group 15 element in compounds that contain phosphorus-phosphorus bonds; examples include diphosphorus tetrahydride, $\\mathrm{H}_{2} \\mathrm{P}-\\mathrm{PH}_{2}$, and tetraphosphorus trisulfide, $\\mathrm{P}_{4} \\mathrm{~S}_{3}$, illustrated in Figure 18.39.\n\n\nFIGURE 18.39 $\\mathrm{P}_{4} \\mathrm{~S}_{3}$ is a component of the heads of strike-anywhere matches."}
{"id": 4756, "contents": "1937. Phosphorus Oxygen Compounds - \nPhosphorus forms two common oxides, phosphorus(III) oxide (or tetraphosphorus hexaoxide), $\\mathrm{P}_{4} \\mathrm{O}_{6}$, and phosphorus(V) oxide (or tetraphosphorus decaoxide), $\\mathrm{P}_{4} \\mathrm{O}_{10}$, both shown in Figure 18.40. Phosphorus(III) oxide is a white crystalline solid with a garlic-like odor. Its vapor is very poisonous. It oxidizes slowly in air and inflames when heated to $70{ }^{\\circ} \\mathrm{C}$, forming $\\mathrm{P}_{4} \\mathrm{O}_{10}$. Phosphorus(III) oxide dissolves slowly in cold water to form phosphorous acid, $\\mathrm{H}_{3} \\mathrm{PO}_{3}$.\n\n\nFIGURE 18.40 This image shows the molecular structures of $\\mathrm{P}_{4} \\mathrm{O}_{6}$ (left) and $\\mathrm{P}_{4} \\mathrm{O}_{10}$ (right).\nPhosphorus $(\\mathrm{V})$ oxide, $\\mathrm{P}_{4} \\mathrm{O}_{10}$, is a white powder that is prepared by burning phosphorus in excess oxygen. Its enthalpy of formation is very high ( -2984 kJ ), and it is quite stable and a very poor oxidizing agent. Dropping $\\mathrm{P}_{4} \\mathrm{O}_{10}$ into water produces a hissing sound, heat, and orthophosphoric acid:\n\n$$\n\\mathrm{P}_{4} \\mathrm{O}_{10}(s)+6 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 4 \\mathrm{H}_{3} \\mathrm{PO}_{4}(a q)\n$$\n\nBecause of its great affinity for water, phosphorus(V) oxide is an excellent drying agent for gases and solvents, and for removing water from many compounds."}
{"id": 4757, "contents": "1938. Phosphorus Halogen Compounds - \nPhosphorus will react directly with the halogens, forming trihalides, $\\mathrm{PX}_{3}$, and pentahalides, $\\mathrm{PX}_{5}$. The trihalides are much more stable than the corresponding nitrogen trihalides; nitrogen pentahalides do not form because of nitrogen's inability to form more than four bonds.\n\nThe chlorides $\\mathrm{PCl}_{3}$ and $\\mathrm{PCl}_{5}$, both shown in Figure 18.41, are the most important halides of phosphorus. Phosphorus trichloride is a colorless liquid that is prepared by passing chlorine over molten phosphorus. Phosphorus pentachloride is an off-white solid that is prepared by oxidizing the trichloride with excess chlorine. The pentachloride sublimes when warmed and forms an equilibrium with the trichloride and chlorine when heated.\n\n\nFIGURE 18.41 This image shows the molecular structure of $\\mathrm{PCl}_{3}$ (left) and $\\mathrm{PCl}_{5}$ (right) in the gas phase.\nLike most other nonmetal halides, both phosphorus chlorides react with an excess of water and yield hydrogen chloride and an oxyacid: $\\mathrm{PCl}_{3}$ yields phosphorous acid $\\mathrm{H}_{3} \\mathrm{PO}_{3}$ and $\\mathrm{PCl}_{5}$ yields phosphoric acid, $\\mathrm{H}_{3} \\mathrm{PO}_{4}$."}
{"id": 4758, "contents": "1938. Phosphorus Halogen Compounds - \nThe pentahalides of phosphorus are Lewis acids because of the empty valence $d$ orbitals of phosphorus. These compounds readily react with halide ions (Lewis bases) to give the anion $\\mathrm{PX}_{6}{ }^{-}$. Whereas phosphorus pentafluoride is a molecular compound in all states, X -ray studies show that solid phosphorus pentachloride is an ionic compound, $\\left[\\mathrm{PCl}_{4}{ }^{+}\\right]\\left[\\mathrm{PCl}_{6}{ }^{-}\\right]$, as are phosphorus pentabromide, $\\left[\\mathrm{PBr}_{4}{ }^{+}\\right]\\left[\\mathrm{Br}^{-}\\right]$, and phosphorus pentaiodide, $\\left[\\mathrm{PI}_{4}{ }^{+}\\right]\\left[\\mathrm{I}^{-}\\right]$."}
{"id": 4759, "contents": "1939. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the properties, preparation, and compounds of oxygen\n- Describe the preparation, properties, and uses of some representative metal oxides, peroxides, and hydroxides\n\nOxygen is the most abundant element on the earth's crust. The earth's surface is composed of the crust, atmosphere, and hydrosphere. About $50 \\%$ of the mass of the earth's crust consists of oxygen (combined with other elements, principally silicon). Oxygen occurs as $\\mathrm{O}_{2}$ molecules and, to a limited extent, as $\\mathrm{O}_{3}$ (ozone) molecules in air. It forms about $20 \\%$ of the mass of the air. About $89 \\%$ of water by mass consists of combined oxygen. In combination with carbon, hydrogen, and nitrogen, oxygen is a large part of plants and animals.\n\nOxygen is a colorless, odorless, and tasteless gas at ordinary temperatures. It is slightly denser than air. Although it is only slightly soluble in water ( 49 mL of gas dissolves in 1 L at STP), oxygen's solubility is very important to aquatic life.\n\nMost of the oxygen isolated commercially comes from air and the remainder from the electrolysis of water. The separation of oxygen from air begins with cooling and compressing the air until it liquefies. As liquid air warms, oxygen with its higher boiling point ( 90 K ) separates from nitrogen, which has a lower boiling point (77 $K$ ). It is possible to separate the other components of air at the same time based on differences in their boiling points.\n\nOxygen is essential in combustion processes such as the burning of fuels. Plants and animals use the oxygen from the air in respiration. The administration of oxygen-enriched air is an important medical practice when a patient is receiving an inadequate supply of oxygen because of shock, pneumonia, or some other illness."}
{"id": 4760, "contents": "1939. LEARNING OBJECTIVES - \nThe chemical industry employs oxygen for oxidizing many substances. A significant amount of oxygen produced commercially is important in the removal of carbon from iron during steel production. Large quantities of pure oxygen are also necessary in metal fabrication and in the cutting and welding of metals with\noxyhydrogen and oxyacetylene torches.\nLiquid oxygen is important to the space industry. It is an oxidizing agent in rocket engines. It is also the source of gaseous oxygen for life support in space.\n\nAs we know, oxygen is very important to life. The energy required for the maintenance of normal body functions in human beings and in other organisms comes from the slow oxidation of chemical compounds. Oxygen is the final oxidizing agent in these reactions. In humans, oxygen passes from the lungs into the blood, where it combines with hemoglobin, producing oxyhemoglobin. In this form, blood transports the oxygen to tissues, where it is transferred to the tissues. The ultimate products are carbon dioxide and water. The blood carries the carbon dioxide through the veins to the lungs, where the blood releases the carbon dioxide and collects another supply of oxygen. Digestion and assimilation of food regenerate the materials consumed by oxidation in the body; the energy liberated is the same as if the food burned outside the body.\n\nGreen plants continually replenish the oxygen in the atmosphere by a process called photosynthesis. The products of photosynthesis may vary, but, in general, the process converts carbon dioxide and water into glucose (a sugar) and oxygen using the energy of light:\n\n\nThus, the oxygen that became carbon dioxide and water by the metabolic processes in plants and animals returns to the atmosphere by photosynthesis.\n\nWhen dry oxygen is passed between two electrically charged plates, ozone ( $\\mathrm{O}_{3}$, illustrated in Figure 18.42), an allotrope of oxygen possessing a distinctive odor, forms. The formation of ozone from oxygen is an endothermic reaction, in which the energy comes from an electrical discharge, heat, or ultraviolet light:\n\n$$\n3 \\mathrm{O}_{2}(g) \\xrightarrow{\\text { electric discharge }} 2 \\mathrm{O}_{3}(g) \\quad \\Delta H^{\\circ}=287 \\mathrm{~kJ}\n$$\n\nThe sharp odor associated with sparking electrical equipment is due, in part, to ozone."}
{"id": 4761, "contents": "1939. LEARNING OBJECTIVES - \nThe sharp odor associated with sparking electrical equipment is due, in part, to ozone.\n\n\n\nFIGURE 18.42 The image shows the bent ozone $\\left(\\mathrm{O}_{3}\\right)$ molecule and the resonance structures necessary to describe its bonding.\n\nOzone forms naturally in the upper atmosphere by the action of ultraviolet light from the sun on the oxygen there. Most atmospheric ozone occurs in the stratosphere, a layer of the atmosphere extending from about 10 to 50 kilometers above the earth's surface. This ozone acts as a barrier to harmful ultraviolet light from the sun by absorbing it via a chemical decomposition reaction:\n\n$$\n\\mathrm{O}_{3}(g) \\xrightarrow{\\text { ultraviolet light }} \\mathrm{O}(g)+\\mathrm{O}_{2}(g)\n$$\n\nThe reactive oxygen atoms recombine with molecular oxygen to complete the ozone cycle. The presence of stratospheric ozone decreases the frequency of skin cancer and other damaging effects of ultraviolet radiation. It has been clearly demonstrated that chlorofluorocarbons, CFCs (known commercially as Freons), which were\npresent as aerosol propellants in spray cans and as refrigerants, caused depletion of ozone in the stratosphere. This occurred because ultraviolet light also causes CFCs to decompose, producing atomic chlorine. The chlorine atoms react with ozone molecules, resulting in a net removal of $\\mathrm{O}_{3}$ molecules from stratosphere. This process is explored in detail in our coverage of chemical kinetics. There is a worldwide effort to reduce the amount of CFCs used commercially, and the ozone hole is already beginning to decrease in size as atmospheric concentrations of atomic chlorine decrease. While ozone in the stratosphere helps protect us, ozone in the troposphere is a problem. This ozone is a toxic component of photochemical smog.\n\nThe uses of ozone depend on its reactivity with other substances. It can be used as a bleaching agent for oils, waxes, fabrics, and starch: It oxidizes the colored compounds in these substances to colorless compounds. It is an alternative to chlorine as a disinfectant for water."}
{"id": 4762, "contents": "1940. Reactions - \nElemental oxygen is a strong oxidizing agent. It reacts with most other elements and many compounds."}
{"id": 4763, "contents": "1941. Reaction with Elements - \nOxygen reacts directly at room temperature or at elevated temperatures with all other elements except the noble gases, the halogens, and few second- and third-row transition metals of low reactivity (those with higher reduction potentials than copper). Rust is an example of the reaction of oxygen with iron. The more active metals form peroxides or superoxides. Less active metals and the nonmetals give oxides. Two examples of these reactions are:\n\n$$\n\\begin{aligned}\n& 2 \\mathrm{Mg}(s)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{MgO}(s) \\\\\n& \\mathrm{P}_{4}(s)+5 \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{P}_{4} \\mathrm{O}_{10}(s)\n\\end{aligned}\n$$\n\nThe oxides of halogens, at least one of the noble gases, and metals with higher reduction potentials than copper do not form by the direct action of the elements with oxygen."}
{"id": 4764, "contents": "1942. Reaction with Compounds - \nElemental oxygen also reacts with some compounds. If it is possible to oxidize any of the elements in a given compound, further oxidation by oxygen can occur. For example, hydrogen sulfide, $\\mathrm{H}_{2} \\mathrm{~S}$, contains sulfur with an oxidation state of $2-$. Because the sulfur does not exhibit its maximum oxidation state, we would expect $\\mathrm{H}_{2} \\mathrm{~S}$ to react with oxygen. It does, yielding water and sulfur dioxide. The reaction is:\n\n$$\n2 \\mathrm{H}_{2} \\mathrm{~S}(g)+3 \\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{SO}_{2}(g)\n$$\n\nIt is also possible to oxidize oxides such as CO and $\\mathrm{P}_{4} \\mathrm{O}_{6}$ that contain an element with a lower oxidation state. The ease with which elemental oxygen picks up electrons is mirrored by the difficulty of removing electrons from oxygen in most oxides. Of the elements, only the very reactive fluorine can oxidize oxides to form oxygen gas."}
{"id": 4765, "contents": "1943. Oxides, Peroxides, and Hydroxides - \nCompounds of the representative metals with oxygen fall into three categories: (1) oxides, containing oxide ions, $\\mathrm{O}^{2-}$; (2) peroxides, containing peroxides ions, $\\mathrm{O}_{2}{ }^{2-}$, with oxygen-oxygen covalent single bonds and a very limited number of superoxides, containing superoxide ions, $\\mathrm{O}_{2}{ }^{-}$, with oxygen-oxygen covalent bonds that have a bond order of $1 \\frac{1}{2}$, In addition, there are (3) hydroxides, containing hydroxide ions, $\\mathrm{OH}^{-}$. All representative metals form oxides. Some of the metals of group 2 also form peroxides, $\\mathrm{MO}_{2}$, and the metals of group 1 also form peroxides, $\\mathrm{M}_{2} \\mathrm{O}_{2}$, and superoxides, $\\mathrm{MO}_{2}$."}
{"id": 4766, "contents": "1944. Oxides - \nIt is possible to produce the oxides of most representative metals by heating the corresponding hydroxides (forming the oxide and gaseous water) or carbonates (forming the oxide and gaseous $\\mathrm{CO}_{2}$ ). Equations for example reactions are:\n\n$$\n\\begin{gathered}\n2 \\mathrm{Al}(\\mathrm{OH})_{3}(s) \\xrightarrow{\\Delta} \\mathrm{Al}_{2} \\mathrm{O}_{3}(s)+3 \\mathrm{H}_{2} \\mathrm{O}(g) \\\\\n\\mathrm{CaCO}_{3}(s) \\xrightarrow{\\Delta} \\mathrm{CaO}(s)+\\mathrm{CO}_{2}(\\mathrm{~g})\n\\end{gathered}\n$$\n\nHowever, alkali metal salts generally are very stable and do not decompose easily when heated. Alkali metal oxides result from the oxidation-reduction reactions created by heating nitrates or hydroxides with the metals. Equations for sample reactions are:\n\n$$\n\\begin{gathered}\n2 \\mathrm{KNO}_{3}(s)+10 \\mathrm{~K}(s) \\xrightarrow{\\Delta} 6 \\mathrm{~K}_{2} \\mathrm{O}(s)+\\mathrm{N}_{2}(g) \\\\\n2 \\mathrm{LiOH}(s)+2 \\mathrm{Li}(s) \\xrightarrow{\\Delta} 2 \\mathrm{Li}_{2} \\mathrm{O}(s)+\\mathrm{H}_{2}(g)\n\\end{gathered}\n$$"}
{"id": 4767, "contents": "1944. Oxides - \nWith the exception of mercury(II) oxide, it is possible to produce the oxides of the metals of groups $2-15$ by burning the corresponding metal in air. The heaviest member of each group, the member for which the inert pair effect is most pronounced, forms an oxide in which the oxidation state of the metal ion is two less than the group oxidation state (inert pair effect). Thus, $\\mathrm{Tl}_{2} \\mathrm{O}, \\mathrm{PbO}$, and $\\mathrm{Bi}_{2} \\mathrm{O}_{3}$ form when burning thallium, lead, and bismuth, respectively. The oxides of the lighter members of each group exhibit the group oxidation state. For example, $\\mathrm{SnO}_{2}$ forms from burning tin. Mercury(II) oxide, HgO , forms slowly when mercury is warmed below $500^{\\circ} \\mathrm{C}$; it decomposes at higher temperatures.\n\nBurning the members of groups 1 and 2 in air is not a suitable way to form the oxides of these elements. These metals are reactive enough to combine with nitrogen in the air, so they form mixtures of oxides and ionic nitrides. Several also form peroxides or superoxides when heated in air.\n\nIonic oxides all contain the oxide ion, a very powerful hydrogen ion acceptor. With the exception of the very insoluble aluminum oxide, $\\mathrm{Al}_{2} \\mathrm{O}_{3}$, $\\mathrm{tin}(\\mathrm{IV}), \\mathrm{SnO}_{2}$, and lead(IV), $\\mathrm{PbO}_{2}$, the oxides of the representative metals react with acids to form salts. Some equations for these reactions are:"}
{"id": 4768, "contents": "1944. Oxides - \n$$\n\\begin{gathered}\n\\mathrm{Na}_{2} \\mathrm{O}+2 \\mathrm{HNO}_{3}(a q) \\longrightarrow 2 \\mathrm{NaNO}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\\\\n\\mathrm{CaO}(s)+2 \\mathrm{HCL}(a q) \\longrightarrow \\mathrm{CaCl}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\\\\n\\mathrm{SnO}(s)+2 \\mathrm{HClO}_{4}(a q) \\longrightarrow{\\mathrm{Sn}\\left(\\mathrm{ClO}_{4}\\right)_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)}^{\\text {an }} \\text { ( }\n\\end{gathered}\n$$\n\nThe oxides of the metals of groups 1 and 2 and of thallium(I) oxide react with water and form hydroxides. Examples of such reactions are:\n\n$$\n\\begin{gathered}\n\\mathrm{Na}_{2} \\mathrm{O}(s)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{NaOH}(a q) \\\\\n\\mathrm{CaO}(s)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Ca}(\\mathrm{OH})_{2}(a q) \\\\\n\\mathrm{T}_{2} \\mathrm{O}(s)+\\mathrm{H}_{2} \\mathrm{O}(a q) \\longrightarrow 2 \\mathrm{TlOH}(a q)\n\\end{gathered}\n$$\n\nThe oxides of the alkali metals have little industrial utility, unlike magnesium oxide, calcium oxide, and aluminum oxide. Magnesium oxide is important in making firebrick, crucibles, furnace linings, and thermal insulation-applications that require chemical and thermal stability. Calcium oxide, sometimes called quicklime or lime in the industrial market, is very reactive, and its principal uses reflect its reactivity. Pure calcium oxide emits an intense white light when heated to a high temperature (as illustrated in Figure 18.43). Blocks of calcium oxide heated by gas flames were the stage lights in theaters before electricity was available. This is the source of the phrase \"in the limelight.\""}
{"id": 4769, "contents": "1944. Oxides - \nFIGURE 18.43 Calcium oxide has many industrial uses. When it is heated at high temperatures, it emits an intense white light.\n\nCalcium oxide and calcium hydroxide are inexpensive bases used extensively in chemical processing,\nalthough most of the useful products prepared from them do not contain calcium. Calcium oxide, CaO, is made by heating calcium carbonate, $\\mathrm{CaCO}_{3}$, which is widely and inexpensively available as limestone or oyster shells:\n\n$$\n\\mathrm{CaCO}_{3}(s) \\longrightarrow \\mathrm{CaO}(s)+\\mathrm{CO}_{2}(g)\n$$\n\nAlthough this decomposition reaction is reversible, it is possible to obtain a $100 \\%$ yield of CaO by allowing the $\\mathrm{CO}_{2}$ to escape. It is possible to prepare calcium hydroxide by the familiar acid-base reaction of a soluble metal oxide with water:\n\n$$\n\\mathrm{CaO}(s)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Ca}(\\mathrm{OH})_{2}(s)\n$$\n\nBoth CaO and $\\mathrm{Ca}(\\mathrm{OH})_{2}$ are useful as bases; they accept protons and neutralize acids.\nAlumina $\\left(\\mathrm{Al}_{2} \\mathrm{O}_{3}\\right)$ occurs in nature as the mineral corundum, a very hard substance used as an abrasive for grinding and polishing. Corundum is important to the jewelry trade as ruby and sapphire. The color of ruby is due to the presence of a small amount of chromium; other impurities produce the wide variety of colors possible for sapphires. Artificial rubies and sapphires are now manufactured by melting aluminum oxide (melting point $=2050^{\\circ} \\mathrm{C}$ ) with small amounts of oxides to produce the desired colors and cooling the melt in such a way as to produce large crystals. Ruby lasers use synthetic ruby crystals."}
{"id": 4770, "contents": "1944. Oxides - \nZinc oxide, ZnO, was a useful white paint pigment; however, pollutants tend to discolor the compound. The compound is also important in the manufacture of automobile tires and other rubber goods, and in the preparation of medicinal ointments. For example, zinc-oxide-based sunscreens, as shown in Figure 18.44, help prevent sunburn. The zinc oxide in these sunscreens is present in the form of very small grains known as nanoparticles. Lead dioxide is a constituent of charged lead storage batteries. Lead(IV) tends to revert to the more stable lead(II) ion by gaining two electrons, so lead dioxide is a powerful oxidizing agent.\n\n\nFIGURE 18.44 Zinc oxide protects exposed skin from sunburn. (credit: modification of work by \"osseous\"/Flickr)"}
{"id": 4771, "contents": "1945. Peroxides and Superoxides - \nPeroxides and superoxides are strong oxidizers and are important in chemical processes. Hydrogen peroxide, $\\mathrm{H}_{2} \\mathrm{O}_{2}$, prepared from metal peroxides, is an important bleach and disinfectant. Peroxides and superoxides form when the metal or metal oxides of groups 1 and 2 react with pure oxygen at elevated temperatures. Sodium peroxide and the peroxides of calcium, strontium, and barium form by heating the corresponding metal or metal oxide in pure oxygen:\n\n$$\n\\begin{gathered}\n2 \\mathrm{Na}(s)+\\mathrm{O}_{2}(g) \\xrightarrow{\\Delta} \\mathrm{Na}_{2} \\mathrm{O}_{2}(s) \\\\\n2 \\mathrm{Na}_{2} \\mathrm{O}(s)+\\mathrm{O}_{2}(g) \\xrightarrow{\\Delta} 2 \\mathrm{Na}_{2} \\mathrm{O}_{2}(s) \\\\\n2 \\mathrm{SrO}(s)+\\mathrm{O}_{2}(g) \\xrightarrow{\\Delta} 2 \\mathrm{SrO}_{2}(s)\n\\end{gathered}\n$$\n\nThe peroxides of potassium, rubidium, and cesium can be prepared by heating the metal or its oxide in a carefully controlled amount of oxygen:\n\n$$\n2 \\mathrm{~K}(s)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{K}_{2} \\mathrm{O}_{2}(s) \\quad\\left(2 \\mathrm{~mol} \\mathrm{~K} \\operatorname{per~mol~} \\mathrm{O}_{2}\\right)\n$$\n\nWith an excess of oxygen, the superoxides $\\mathrm{KO}_{2}, \\mathrm{RbO}_{2}$, and $\\mathrm{CsO}_{2}$ form. For example:\n\n$$\n\\mathrm{K}(s)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{KO}_{2}(s) \\quad\\left(1 \\mathrm{~mol} \\mathrm{~K} \\text { per } \\mathrm{mol} \\mathrm{O}_{2}\\right)\n$$"}
{"id": 4772, "contents": "1945. Peroxides and Superoxides - \nThe stability of the peroxides and superoxides of the alkali metals increases as the size of the cation increases."}
{"id": 4773, "contents": "1946. Hydroxides - \nHydroxides are compounds that contain the $\\mathrm{OH}^{-}$ion. It is possible to prepare these compounds by two general types of reactions. Soluble metal hydroxides can be produced by the reaction of the metal or metal oxide with water. Insoluble metal hydroxides form when a solution of a soluble salt of the metal combines with a solution containing hydroxide ions.\n\nWith the exception of beryllium and magnesium, the metals of groups 1 and 2 react with water to form hydroxides and hydrogen gas. Examples of such reactions include:\n\n$$\n\\begin{gathered}\n2 \\mathrm{Li}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{LiOH}(a q)+\\mathrm{H}_{2}(g) \\\\\n\\mathrm{Ca}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Ca}(\\mathrm{OH})_{2}(a q)+\\mathrm{H}_{2}(g)\n\\end{gathered}\n$$\n\nHowever, these reactions can be violent and dangerous; therefore, it is preferable to produce soluble metal hydroxides by the reaction of the respective oxide with water:\n\n$$\n\\begin{gathered}\n\\mathrm{Li}_{2} \\mathrm{O}(s)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{LiOH}(a q) \\\\\n\\mathrm{CaO}(s)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Ca}(\\mathrm{OH})_{2}(a q)\n\\end{gathered}\n$$\n\nMost metal oxides are base anhydrides. This is obvious for the soluble oxides because they form metal hydroxides. Most other metal oxides are insoluble and do not form hydroxides in water; however, they are still base anhydrides because they will react with acids.\n\nIt is possible to prepare the insoluble hydroxides of beryllium, magnesium, and other representative metals by the addition of sodium hydroxide to a solution of a salt of the respective metal. The net ionic equations for the reactions involving a magnesium salt, an aluminum salt, and a zinc salt are:"}
{"id": 4774, "contents": "1946. Hydroxides - \n$$\n\\begin{gathered}\n\\mathrm{Mg}^{2+}(a q)+2 \\mathrm{OH}^{-}(a q) \\longrightarrow \\mathrm{Mg}(\\mathrm{OH})_{2}(s) \\\\\n\\mathrm{Al}^{3+}(a q)+3 \\mathrm{OH}^{-}(a q) \\longrightarrow \\mathrm{Al}(\\mathrm{OH})_{3}(s) \\\\\n\\mathrm{Zn}^{2+}(a q)+2 \\mathrm{OH}^{-}(a q) \\longrightarrow \\mathrm{Zn}(\\mathrm{OH})_{2}(s)\n\\end{gathered}\n$$\n\nAn excess of hydroxide must be avoided when preparing aluminum, gallium, zinc, and tin(II) hydroxides, or the hydroxides will dissolve with the formation of the corresponding complex ions: $\\mathrm{Al}(\\mathrm{OH})_{4}{ }^{-}, \\mathrm{Ga}(\\mathrm{OH})_{4}{ }^{-}$, $\\mathrm{Zn}(\\mathrm{OH})_{4}{ }^{2-}$, and $\\mathrm{Sn}(\\mathrm{OH})_{3}{ }^{-}$(see Figure 18.45). The important aspect of complex ions for this chapter is that they form by a Lewis acid-base reaction with the metal being the Lewis acid.\n\n\nFIGURE 18.45 (a) Mixing solutions of NaOH and $\\mathrm{Zn}\\left(\\mathrm{NO}_{3}\\right)_{2}$ produces a white precipitate of $\\mathrm{Zn}(\\mathrm{OH})_{2}$. (b) Addition of an excess of NaOH results in dissolution of the precipitate. (credit: modification of work by Mark Ott)\n\nIndustry uses large quantities of sodium hydroxide as a cheap, strong base. Sodium chloride is the starting material for the production of NaOH because NaCl is a less expensive starting material than the oxide. Sodium hydroxide is among the top 10 chemicals in production in the United States, and this production was almost entirely by electrolysis of solutions of sodium chloride. This process is the chlor-alkali process, and it is the primary method for producing chlorine."}
{"id": 4775, "contents": "1946. Hydroxides - \nSodium hydroxide is an ionic compound and melts without decomposition. It is very soluble in water, giving off\na great deal of heat and forming very basic solutions: 40 grams of sodium hydroxide dissolves in only 60 grams of water at $25^{\\circ} \\mathrm{C}$. Sodium hydroxide is employed in the production of other sodium compounds and is used to neutralize acidic solutions during the production of other chemicals such as petrochemicals and polymers.\n\nMany of the applications of hydroxides are for the neutralization of acids (such as the antacid shown in Figure 18.46) and for the preparation of oxides by thermal decomposition. An aqueous suspension of magnesium hydroxide constitutes the antacid milk of magnesia. Because of its ready availability (from the reaction of water with calcium oxide prepared by the decomposition of limestone, $\\mathrm{CaCO}_{3}$ ), low cost, and activity, calcium hydroxide is used extensively in commercial applications needing a cheap, strong base. The reaction of hydroxides with appropriate acids is also used to prepare salts.\n\n\nFIGURE 18.46 Calcium carbonate, $\\mathrm{CaCO}_{3}$, can be consumed in the form of an antacid to neutralize the effects of acid in your stomach. (credit: \"Midnightcomm\"/Wikimedia Commons)"}
{"id": 4776, "contents": "1948. The Chlor-Alkali Process - \nAlthough they are very different chemically, there is a link between chlorine and sodium hydroxide because there is an important electrochemical process that produces the two chemicals simultaneously. The process known as the chlor-alkali process, utilizes sodium chloride, which occurs in large deposits in many parts of the world. This is an electrochemical process to oxidize chloride ion to chlorine and generate sodium hydroxide.\n\nPassing a direct current of electricity through a solution of NaCl causes the chloride ions to migrate to the positive electrode where oxidation to gaseous chlorine occurs when the ion gives up an electron to the electrode:\n\n$$\n2 \\mathrm{Cl}^{-}(a q) \\longrightarrow \\mathrm{Cl}_{2}(g)+2 \\mathrm{e}^{-} \\quad(\\text { at the positive electrode })\n$$\n\nThe electrons produced travel through the outside electrical circuit to the negative electrode. Although the positive sodium ions migrate toward this negative electrode, metallic sodium does not form because sodium ions are too difficult to reduce under the conditions used. (Recall that metallic sodium is active enough to react with water and hence, even if produced, would immediately react with water to produce sodium ions again.) Instead, water molecules pick up electrons from the electrode and undergo reduction to form hydrogen gas and hydroxide ions:\n\n$$\n2 \\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{e}^{-} \\text {(from the negative electrode) } \\longrightarrow \\mathrm{H}_{2}(g)+2 \\mathrm{OH}^{-}(a q)\n$$\n\nThe overall result is the conversion of the aqueous solution of NaCl to an aqueous solution of NaOH , gaseous $\\mathrm{Cl}_{2}$, and gaseous $\\mathrm{H}_{2}$ :\n\n$$\n2 \\mathrm{Na}^{+}(a q)+2 \\mathrm{Cl}^{-}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\xrightarrow{\\text { electrolysis }} 2 \\mathrm{Na}^{+}(a q)+2 \\mathrm{OH}^{-}(a q)+\\mathrm{Cl}_{2}(g)+\\mathrm{H}_{2}(g)\n$$"}
{"id": 4777, "contents": "1949. Nonmetal Oxygen Compounds - \nMost nonmetals react with oxygen to form nonmetal oxides. Depending on the available oxidation states for the element, a variety of oxides might form. Fluorine will combine with oxygen to form fluorides such as $\\mathrm{OF}_{2}$, where the oxygen has a $2+$-oxidation state.\n\nSulfur Oxygen Compounds\nThe two common oxides of sulfur are sulfur dioxide, $\\mathrm{SO}_{2}$, and sulfur trioxide, $\\mathrm{SO}_{3}$. The odor of burning sulfur comes from sulfur dioxide. Sulfur dioxide, shown in Figure 18.47, occurs in volcanic gases and in the atmosphere near industrial plants that burn fuel containing sulfur compounds.\n\n\nFIGURE 18.47 This image shows the molecular structure (left) and resonance forms (right) of sulfur dioxide.\nCommercial production of sulfur dioxide is from either burning sulfur or roasting sulfide ores such as ZnS , $\\mathrm{FeS}_{2}$, and $\\mathrm{Cu}_{2} \\mathrm{~S}$ in air. (Roasting, which forms the metal oxide, is the first step in the separation of many metals from their ores.) A convenient method for preparing sulfur dioxide in the laboratory is by the action of a strong acid on either sulfite salts containing the $\\mathrm{SO}_{3}{ }^{2-}$ ion or hydrogen sulfite salts containing $\\mathrm{HSO}_{3}{ }^{-}$. Sulfurous acid, $\\mathrm{H}_{2} \\mathrm{SO}_{3}$, forms first, but quickly decomposes into sulfur dioxide and water. Sulfur dioxide also forms when many reducing agents react with hot, concentrated sulfuric acid. Sulfur trioxide forms slowly when heating sulfur dioxide and oxygen together, and the reaction is exothermic:\n\n$$\n2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{SO}_{3}(g) \\quad \\Delta H^{\\circ}=-197.8 \\mathrm{~kJ}\n$$"}
{"id": 4778, "contents": "1949. Nonmetal Oxygen Compounds - \nSulfur dioxide is a gas at room temperature, and the $\\mathrm{SO}_{2}$ molecule is bent. Sulfur trioxide melts at $17^{\\circ} \\mathrm{C}$ and boils at $43^{\\circ} \\mathrm{C}$. In the vapor state, its molecules are single $\\mathrm{SO}_{3}$ units (shown in Figure 18.48), but in the solid state, $\\mathrm{SO}_{3}$ exists in several polymeric forms.\n\n\n\nFIGURE 18.48 This image shows the structure (top) of sulfur trioxide in the gas phase and its resonance forms (bottom).\n\nThe sulfur oxides react as Lewis acids with many oxides and hydroxides in Lewis acid-base reactions, with the formation of sulfites or hydrogen sulfites, and sulfates or hydrogen sulfates, respectively."}
{"id": 4779, "contents": "1950. Halogen Oxygen Compounds - \nThe halogens do not react directly with oxygen, but it is possible to prepare binary oxygen-halogen compounds by the reactions of the halogens with oxygen-containing compounds. Oxygen compounds with chlorine, bromine, and iodine are oxides because oxygen is the more electronegative element in these compounds. On the other hand, fluorine compounds with oxygen are fluorides because fluorine is the more electronegative element.\n\nAs a class, the oxides are extremely reactive and unstable, and their chemistry has little practical importance. Dichlorine oxide, formally called dichlorine monoxide, and chlorine dioxide, both shown in Figure 18.49, are the only commercially important compounds. They are important as bleaching agents (for use with pulp and flour) and for water treatment.\n\n\nFIGURE 18.49 This image shows the structures of the (a) $\\mathrm{Cl}_{2} \\mathrm{O}$ and (b) $\\mathrm{ClO}_{2}$ molecules."}
{"id": 4780, "contents": "1951. Nonmetal Oxyacids and Their Salts - \nNonmetal oxides form acids when allowed to react with water; these are acid anhydrides. The resulting oxyanions can form salts with various metal ions."}
{"id": 4781, "contents": "1952. Nitrogen Oxyacids and Salts - \nNitrogen pentaoxide, $\\mathrm{N}_{2} \\mathrm{O}_{5}$, and $\\mathrm{NO}_{2}$ react with water to form nitric acid, $\\mathrm{HNO}_{3}$. Alchemists, as early as the eighth century, knew nitric acid (shown in Figure 18.50) as aqua fortis (meaning \"strong water\"). The acid was useful in the separation of gold from silver because it dissolves silver but not gold. Traces of nitric acid occur in the atmosphere after thunderstorms, and its salts are widely distributed in nature. There are tremendous deposits of Chile saltpeter, $\\mathrm{NaNO}_{3}$, in the desert region near the boundary of Chile and Peru. Bengal saltpeter, $\\mathrm{KNO}_{3}$, occurs in India and in other countries of the Far East.\n\n\nFIGURE 18.50 This image shows the molecular structure (left) of nitric acid, $\\mathrm{HNO}_{3}$ and its resonance forms (right).\nIn the laboratory, it is possible to produce nitric acid by heating a nitrate salt (such as sodium or potassium nitrate) with concentrated sulfuric acid:\n\n$$\n\\mathrm{NaNO}_{3}(s)+\\mathrm{H}_{2} \\mathrm{SO}_{4}(l) \\xrightarrow{\\Delta} \\mathrm{NaHSO}_{4}(s)+\\mathrm{HNO}_{3}(g)\n$$\n\nThe Ostwald process is the commercial method for producing nitric acid. This process involves the oxidation of ammonia to nitric oxide, NO ; oxidation of nitric oxide to nitrogen dioxide, $\\mathrm{NO}_{2}$; and further oxidation and hydration of nitrogen dioxide to form nitric acid:\n\n$$\n4 \\mathrm{NH}_{3}(g)+5 \\mathrm{O}_{2}(g) \\longrightarrow 4 \\mathrm{NO}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(g)\n$$"}
{"id": 4782, "contents": "1952. Nitrogen Oxyacids and Salts - \n$$\n4 \\mathrm{NH}_{3}(g)+5 \\mathrm{O}_{2}(g) \\longrightarrow 4 \\mathrm{NO}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(g)\n$$\n\n$$\n\\begin{gathered}\n2 \\mathrm{NO}(g)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{NO}_{2}(g) \\\\\n3 \\mathrm{NO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{HNO}_{3}(a q)+\\mathrm{NO}(g)\n\\end{gathered}\n$$\n\nOr\n\n$$\n4 \\mathrm{NO}_{2}(g)+\\mathrm{O}_{2}(\\mathrm{~g})+2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{~g}) \\longrightarrow 4 \\mathrm{HNO}_{3}(\\mathrm{l})\n$$\n\nPure nitric acid is a colorless liquid. However, it is often yellow or brown in color because $\\mathrm{NO}_{2}$ forms as the acid decomposes. Nitric acid is stable in aqueous solution; solutions containing $68 \\%$ of the acid are commercially available concentrated nitric acid. It is both a strong oxidizing agent and a strong acid.\n\nThe action of nitric acid on a metal rarely produces $\\mathrm{H}_{2}$ (by reduction of $\\mathrm{H}^{+}$) in more than small amounts. Instead, the reduction of nitrogen occurs. The products formed depend on the concentration of the acid, the activity of the metal, and the temperature. Normally, a mixture of nitrates, nitrogen oxides, and various reduction products form. Less active metals such as copper, silver, and lead reduce concentrated nitric acid primarily to nitrogen dioxide. The reaction of dilute nitric acid with copper produces NO. In each case, the nitrate salts of the metals crystallize upon evaporation of the resultant solutions.\n\nNonmetallic elements, such as sulfur, carbon, iodine, and phosphorus, undergo oxidation by concentrated nitric acid to their oxides or oxyacids, with the formation of $\\mathrm{NO}_{2}$ :"}
{"id": 4783, "contents": "1952. Nitrogen Oxyacids and Salts - \nNonmetallic elements, such as sulfur, carbon, iodine, and phosphorus, undergo oxidation by concentrated nitric acid to their oxides or oxyacids, with the formation of $\\mathrm{NO}_{2}$ :\n\n$$\n\\begin{gathered}\n\\mathrm{S}(s)+6 \\mathrm{HNO}_{3}(a q) \\longrightarrow \\mathrm{H}_{2} \\mathrm{SO}_{4}(a q)+6 \\mathrm{NO}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\\\\n\\mathrm{C}(s)+4 \\mathrm{HNO}_{3}(a q) \\longrightarrow \\mathrm{CO}_{2}(g)+4 \\mathrm{NO}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)\n\\end{gathered}\n$$\n\nNitric acid oxidizes many compounds; for example, concentrated nitric acid readily oxidizes hydrochloric acid to chlorine and chlorine dioxide. A mixture of one part concentrated nitric acid and three parts concentrated hydrochloric acid (called aqua regia, which means royal water) reacts vigorously with metals. This mixture is particularly useful in dissolving gold, platinum, and other metals that are more difficult to oxidize than hydrogen. A simplified equation to represent the action of aqua regia on gold is:\n\n$$\n\\mathrm{Au}(s)+4 \\mathrm{HCl}(a q)+3 \\mathrm{HNO}_{3}(a q) \\longrightarrow \\mathrm{HAuCl}_{4}(a q)+3 \\mathrm{NO}_{2}(g)+3 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$"}
{"id": 4784, "contents": "1953. LINK TO LEARNING - \nAlthough gold is generally unreactive, you can watch a video (http://openstax.org/l/16gold) of the complex mixture of compounds present in aqua regia dissolving it into solution.\n\nNitrates, salts of nitric acid, form when metals, oxides, hydroxides, or carbonates react with nitric acid. Most nitrates are soluble in water; indeed, one of the significant uses of nitric acid is to prepare soluble metal nitrates.\n\nNitric acid finds extensive use in the laboratory and in chemical industries as a strong acid and strong oxidizing agent. It is important in the manufacture of explosives, dyes, plastics, and drugs. Salts of nitric acid (nitrates) are valuable as fertilizers. Gunpowder is a mixture of potassium nitrate, sulfur, and charcoal.\nThe reaction of $\\mathrm{N}_{2} \\mathrm{O}_{3}$ with water gives a pale blue solution of nitrous acid, $\\mathrm{HNO}_{2}$. However, $\\mathrm{HNO}_{2}$ (shown in Figure 18.51) is easier to prepare by the addition of an acid to a solution of nitrite; nitrous acid is a weak acid, so the nitrite ion is basic in aqueous solution:\n\n$$\n\\mathrm{NO}_{2}^{-}(a q)+\\mathrm{H}_{3} \\mathrm{O}^{+}(a q) \\longrightarrow \\mathrm{HNO}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nNitrous acid is very unstable and exists only in solution. It disproportionates slowly at room temperature (rapidly when heated) into nitric acid and nitric oxide. Nitrous acid is an active oxidizing agent with strong reducing agents, and strong oxidizing agents oxidize it to nitric acid."}
{"id": 4785, "contents": "1953. LINK TO LEARNING - \nFIGURE 18.51 This image shows the molecular structure of a molecule of nitrous acid, $\\mathrm{HNO}_{2}$.\nSodium nitrite, $\\mathrm{NaNO}_{2}$, is an additive to meats such as hot dogs and cold cuts. The nitrite ion has two functions. It limits the growth of bacteria that can cause food poisoning, and it prolongs the meat's retention of its red color. The addition of sodium nitrite to meat products is controversial because nitrous acid reacts with certain organic compounds to form a class of compounds known as nitrosamines. Nitrosamines produce cancer in laboratory animals. This has prompted the FDA to limit the amount of $\\mathrm{NaNO}_{2}$ in foods.\n\nThe nitrites are much more stable than the acid, but nitrites, like nitrates, can explode. Nitrites, like nitrates, are also soluble in water $\\left(\\mathrm{AgNO}_{2}\\right.$ is only slightly soluble)."}
{"id": 4786, "contents": "1954. Phosphorus Oxyacids and Salts - \nPure orthophosphoric acid, $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ (shown in Figure 18.52), forms colorless, deliquescent crystals that melt at $42^{\\circ} \\mathrm{C}$. The common name of this compound is phosphoric acid, and is commercially available as a viscous $82 \\%$ solution known as syrupy phosphoric acid. One use of phosphoric acid is as an additive to many soft drinks.\n\nOne commercial method of preparing orthophosphoric acid is to treat calcium phosphate rock with concentrated sulfuric acid:\n\n\nFIGURE 18.52 Orthophosphoric acid, $\\mathrm{H}_{3} \\mathrm{PO}_{4}$, is colorless when pure and has this molecular (left) and Lewis structure (right).\n\nDilution of the products with water, followed by filtration to remove calcium sulfate, gives a dilute acid solution contaminated with calcium dihydrogen phosphate, $\\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$, and other compounds associated with calcium phosphate rock. It is possible to prepare pure orthophosphoric acid by dissolving $\\mathrm{P}_{4} \\mathrm{O}_{10}$ in water.\n\nThe action of water on $\\mathrm{P}_{4} \\mathrm{O}_{6}, \\mathrm{PCl}_{3}, \\mathrm{PBr}_{3}$, or $\\mathrm{PI}_{3}$ forms phosphorous acid, $\\mathrm{H}_{3} \\mathrm{PO}_{3}$ (shown in Figure 18.53). The best method for preparing pure phosphorous acid is by hydrolyzing phosphorus trichloride:\n\n$$\n\\mathrm{PCl}_{3}(l)+3 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{H}_{3} \\mathrm{PO}_{3}(a q)+3 \\mathrm{HCl}(g)\n$$"}
{"id": 4787, "contents": "1954. Phosphorus Oxyacids and Salts - \nHeating the resulting solution expels the hydrogen chloride and leads to the evaporation of water. When sufficient water evaporates, white crystals of phosphorous acid will appear upon cooling. The crystals are deliquescent, very soluble in water, and have an odor like that of garlic. The solid melts at $70.1^{\\circ} \\mathrm{C}$ and decomposes at about $200^{\\circ} \\mathrm{C}$ by disproportionation into phosphine and orthophosphoric acid:\n\n$$\n4 \\mathrm{H}_{3} \\mathrm{PO}_{3}(l) \\longrightarrow \\mathrm{PH}_{3}(g)+3 \\mathrm{H}_{3} \\mathrm{PO}_{4}(l)\n$$\n\n\n\n\nFIGURE 18.53 In a molecule of phosphorous acid, $\\mathrm{H}_{3} \\mathrm{PO}_{3}$, only the two hydrogen atoms bonded to an oxygen atom are acidic.\n\nPhosphorous acid forms only two series of salts, which contain the dihydrogen phosphite ion, $\\mathrm{H}_{2} \\mathrm{PO}_{3}{ }^{-}$, or the hydrogen phosphate ion, $\\mathrm{HPO}_{3}{ }^{2-}$, respectively. It is not possible to replace the third atom of hydrogen because it is not very acidic, as it is not easy to ionize the $\\mathrm{P}-\\mathrm{H}$ bond."}
{"id": 4788, "contents": "1955. Sulfur Oxyacids and Salts - \nThe preparation of sulfuric acid, $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ (shown in Figure 18.54), begins with the oxidation of sulfur to sulfur trioxide and then converting the trioxide to sulfuric acid. Pure sulfuric acid is a colorless, oily liquid that freezes at $10.5^{\\circ} \\mathrm{C}$. It fumes when heated because the acid decomposes to water and sulfur trioxide. The heating process causes the loss of more sulfur trioxide than water, until reaching a concentration of $98.33 \\%$ acid. Acid of this concentration boils at $338^{\\circ} \\mathrm{C}$ without further change in concentration (a constant boiling solution) and is commercially concentrated $\\mathrm{H}_{2} \\mathrm{SO}_{4}$. The amount of sulfuric acid used in industry exceeds that of any other manufactured compound.\n\n\nFIGURE 18.54 Sulfuric acid has a tetrahedral molecular structure.\nThe strong affinity of concentrated sulfuric acid for water makes it a good dehydrating agent. It is possible to dry gases and immiscible liquids that do not react with the acid by passing them through the acid.\n\nSulfuric acid is a strong diprotic acid that ionizes in two stages. In aqueous solution, the first stage is essentially complete. The secondary ionization is not nearly so complete, and $\\mathrm{HSO}_{4}{ }^{-}$is a moderately strong acid (about $25 \\%$ ionized in solution of a $\\mathrm{HSO}_{4}{ }^{-}$salt: $K_{\\mathrm{a}}=1.2 \\times 10^{-2}$ ).\n\nBeing a diprotic acid, sulfuric acid forms both sulfates, such as $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$, and hydrogen sulfates, such as $\\mathrm{NaHSO}_{4}$. Most sulfates are soluble in water; however, the sulfates of barium, strontium, calcium, and lead are only slightly soluble in water."}
{"id": 4789, "contents": "1955. Sulfur Oxyacids and Salts - \nAmong the important sulfates are $\\mathrm{Na}_{2} \\mathrm{SO}_{4} \\cdot 10 \\mathrm{H}_{2} \\mathrm{O}$ and Epsom salts, $\\mathrm{MgSO}_{4} \\cdot 7 \\mathrm{H}_{2} \\mathrm{O}$. Because the $\\mathrm{HSO}_{4}{ }^{-}$ion is an acid, hydrogen sulfates, such as $\\mathrm{NaHSO}_{4}$, exhibit acidic behavior, and this compound is the primary ingredient in some household cleansers.\n\nHot, concentrated sulfuric acid is an oxidizing agent. Depending on its concentration, the temperature, and the strength of the reducing agent, sulfuric acid oxidizes many compounds and, in the process, undergoes reduction to $\\mathrm{SO}_{2}, \\mathrm{HSO}_{3}^{-}, \\mathrm{SO}_{3}{ }^{2-}, \\mathrm{S}, \\mathrm{H}_{2} \\mathrm{~S}$, or $\\mathrm{S}^{2-}$.\n\nSulfur dioxide dissolves in water to form a solution of sulfurous acid, as expected for the oxide of a nonmetal.\n\nSulfurous acid is unstable, and it is not possible to isolate anhydrous $\\mathrm{H}_{2} \\mathrm{SO}_{3}$. Heating a solution of sulfurous acid expels the sulfur dioxide. Like other diprotic acids, sulfurous acid ionizes in two steps: The hydrogen sulfite ion, $\\mathrm{HSO}_{3}{ }^{-}$, and the sulfite ion, $\\mathrm{SO}_{3}{ }^{2-}$, form. Sulfurous acid is a moderately strong acid. Ionization is about $25 \\%$ in the first stage, but it is much less in the second ( $K_{\\mathrm{a} 1}=1.2 \\times 10^{-2}$ and $K_{\\mathrm{a} 2}=6.2 \\times 10^{-8}$ )."}
{"id": 4790, "contents": "1955. Sulfur Oxyacids and Salts - \nIn order to prepare solid sulfite and hydrogen sulfite salts, it is necessary to add a stoichiometric amount of a base to a sulfurous acid solution and then evaporate the water. These salts also form from the reaction of $\\mathrm{SO}_{2}$ with oxides and hydroxides. Heating solid sodium hydrogen sulfite forms sodium sulfite, sulfur dioxide, and water:\n\n$$\n2 \\mathrm{NaHSO}_{3}(s) \\xrightarrow{\\Delta} \\mathrm{Na}_{2} \\mathrm{SO}_{3}(s)+\\mathrm{SO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nStrong oxidizing agents can oxidize sulfurous acid. Oxygen in the air oxidizes it slowly to the more stable sulfuric acid:\n\n$$\n2 \\mathrm{H}_{2} \\mathrm{SO}_{3}(a q)+\\mathrm{O}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\xrightarrow{\\Delta} 2 \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+2 \\mathrm{HSO}_{4}^{-}(a q)\n$$\n\nSolutions of sulfites are also very susceptible to air oxidation to produce sulfates. Thus, solutions of sulfites always contain sulfates after exposure to air."}
{"id": 4791, "contents": "1956. Halogen Oxyacids and Their Salts - \nThe compounds $\\mathrm{HXO}, \\mathrm{HXO}_{2}, \\mathrm{HXO}_{3}$, and $\\mathrm{HXO}_{4}$, where X represents $\\mathrm{Cl}, \\mathrm{Br}$, or I, are the hypohalous, halous, halic, and perhalic acids, respectively. The strengths of these acids increase from the hypohalous acids, which are very weak acids, to the perhalic acids, which are very strong. Table 18.2 lists the known acids, and, where known, their $\\mathrm{pK}_{\\mathrm{a}}$ values are given in parentheses.\n\nOxyacids of the Halogens\n\n| Name | Fluorine |  | Chlorine | Bromine |\n| :--- | :--- | :--- | :--- | :--- |\n| lodine |  |  |  |  |\n| hypohalous | HOF | $\\mathrm{HOCl}(7.5)$ | $\\mathrm{HOBr}(8.7)$ | $\\mathrm{HOI}(11)$ |\n| halous |  | $\\mathrm{HClO}_{2}(2.0)$ |  |  |\n| halic |  | $\\mathrm{HClO}_{3}$ | $\\mathrm{HBrO}_{3}$ | $\\mathrm{HIO}_{3}(0.8)$ |\n| perhalic |  | $\\mathrm{HClO}_{4}$ | $\\mathrm{HBrO}_{4}$ | $\\mathrm{HIO}_{4}(1.6)$ |\n| paraperhalic |  |  |  | $\\mathrm{H}_{5} \\mathrm{IO}_{6}(1.6)$ |\n\nTABLE 18.2\n\nThe only known oxyacid of fluorine is the very unstable hypofluorous acid, HOF, which is prepared by the reaction of gaseous fluorine with ice:\n\n$$\n\\mathrm{F}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(s) \\longrightarrow \\mathrm{HOF}(g)+\\mathrm{HF}(g)\n$$"}
{"id": 4792, "contents": "1956. Halogen Oxyacids and Their Salts - \n$$\n\\mathrm{F}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(s) \\longrightarrow \\mathrm{HOF}(g)+\\mathrm{HF}(g)\n$$\n\nThe compound is very unstable and decomposes above $-40^{\\circ} \\mathrm{C}$. This compound does not ionize in water, and there are no known salts. It is uncertain whether the name hypofluorous acid is even appropriate for HOF; a more appropriate name might be hydrogen hypofluorite.\n\nThe reactions of chlorine and bromine with water are analogous to that of fluorine with ice, but these reactions do not go to completion, and mixtures of the halogen and the respective hypohalous and hydrohalic acids result. Other than HOF, the hypohalous acids only exist in solution. The hypohalous acids are all very weak acids; however, HOCl is a stronger acid than HOBr , which, in turn, is stronger than HOI.\n\nThe addition of base to solutions of the hypohalous acids produces solutions of salts containing the basic\nhypohalite ions, $\\mathrm{OX}^{-}$. It is possible to isolate these salts as solids. All of the hypohalites are unstable with respect to disproportionation in solution, but the reaction is slow for hypochlorite. Hypobromite and hypoiodite disproportionate rapidly, even in the cold:\n\n$$\n3 \\mathrm{XO}^{-}(a q) \\longrightarrow 2 \\mathrm{X}^{-}(a q)+\\mathrm{XO}_{3}^{-}(a q)\n$$\n\nSodium hypochlorite is an inexpensive bleach (Clorox) and germicide. The commercial preparation involves the electrolysis of cold, dilute, aqueous sodium chloride solutions under conditions where the resulting chlorine and hydroxide ion can react. The net reaction is:\n\n$$\n\\mathrm{Cl}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\xrightarrow{\\text { electrical energy }} \\mathrm{ClO}^{-}(a q)+\\mathrm{H}_{2}(g)\n$$\n\nThe only definitely known halous acid is chlorous acid, $\\mathrm{HClO}_{2}$, obtained by the reaction of barium chlorite with dilute sulfuric acid:"}
{"id": 4793, "contents": "1956. Halogen Oxyacids and Their Salts - \nThe only definitely known halous acid is chlorous acid, $\\mathrm{HClO}_{2}$, obtained by the reaction of barium chlorite with dilute sulfuric acid:\n\n$$\n\\mathrm{Ba}\\left(\\mathrm{ClO}_{2}\\right)_{2}(a q)+\\mathrm{H}_{2} \\mathrm{SO}_{4}(a q) \\longrightarrow \\mathrm{BaSO}_{4}(s)+2 \\mathrm{HClO}_{2}(a q)\n$$\n\nFiltering the insoluble barium sulfate leaves a solution of $\\mathrm{HClO}_{2}$. Chlorous acid is not stable; it slowly decomposes in solution to yield chlorine dioxide, hydrochloric acid, and water. Chlorous acid reacts with bases to give salts containing the chlorite ion (shown in Figure 18.55). Sodium chlorite finds an extensive application in the bleaching of paper because it is a strong oxidizing agent and does not damage the paper.\n\n\nFIGURE 18.55 Chlorite ions, $\\mathrm{ClO}_{2}{ }^{-}$, are produced when chlorous acid reacts with bases.\nChloric acid, $\\mathrm{HClO}_{3}$, and bromic acid, $\\mathrm{HBrO}_{3}$, are stable only in solution. The reaction of iodine with concentrated nitric acid produces stable white iodic acid, $\\mathrm{HIO}_{3}$ :\n\n$$\n\\mathrm{I}_{2}(s)+10 \\mathrm{HNO}_{3}(a q) \\longrightarrow 2 \\mathrm{HIO}_{3}(s)+10 \\mathrm{NO}_{2}(g)+4 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nIt is possible to obtain the lighter halic acids from their barium salts by reaction with dilute sulfuric acid. The reaction is analogous to that used to prepare chlorous acid. All of the halic acids are strong acids and very active oxidizing agents. The acids react with bases to form salts containing chlorate ions (shown in Figure 18.56). Another preparative method is the electrochemical oxidation of a hot solution of a metal halide to form the appropriate metal chlorates. Sodium chlorate is a weed killer; potassium chlorate is used as an oxidizing agent."}
{"id": 4794, "contents": "1956. Halogen Oxyacids and Their Salts - \nFIGURE 18.56 Chlorate ions, $\\mathrm{ClO}_{3}{ }^{-}$, are produced when halic acids react with bases.\nPerchloric acid, $\\mathrm{HClO}_{4}$, forms when treating a perchlorate, such as potassium perchlorate, with sulfuric acid under reduced pressure. The $\\mathrm{HClO}_{4}$ can be distilled from the mixture:\n\n$$\n\\mathrm{KClO}_{4}(s)+\\mathrm{H}_{2} \\mathrm{SO}_{4}(a q) \\longrightarrow \\mathrm{HClO}_{4}(g)+\\mathrm{KHSO}_{4}(s)\n$$\n\nDilute aqueous solutions of perchloric acid are quite stable thermally, but concentrations above $60 \\%$ are unstable and dangerous. Perchloric acid and its salts are powerful oxidizing agents, as the very electronegative\nchlorine is more stable in a lower oxidation state than 7+. Serious explosions have occurred when heating concentrated solutions with easily oxidized substances. However, its reactions as an oxidizing agent are slow when perchloric acid is cold and dilute. The acid is among the strongest of all acids. Most salts containing the perchlorate ion (shown in Figure 18.57) are soluble. It is possible to prepare them from reactions of bases with perchloric acid and, commercially, by the electrolysis of hot solutions of their chlorides.\n\n\nFIGURE 18.57 Perchlorate ions, $\\mathrm{ClO}_{4}{ }^{-}$, can be produced when perchloric acid reacts with a base or by electrolysis of hot solutions of their chlorides.\n\nPerbromate salts are difficult to prepare, and the best syntheses currently involve the oxidation of bromates in basic solution with fluorine gas followed by acidification. There are few, if any, commercial uses of this acid or its salts.\n\nThere are several different acids containing iodine in the 7+-oxidation state; they include metaperiodic acid, $\\mathrm{HIO}_{4}$, and paraperiodic acid, $\\mathrm{H}_{5} \\mathrm{IO}_{6}$. These acids are strong oxidizing agents and react with bases to form the appropriate salts."}
{"id": 4795, "contents": "1957. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the properties, preparation, and uses of sulfur\n\nSulfur exists in nature as elemental deposits as well as sulfides of iron, zinc, lead, and copper, and sulfates of sodium, calcium, barium, and magnesium. Hydrogen sulfide is often a component of natural gas and occurs in many volcanic gases, like those shown in Figure 18.58. Sulfur is a constituent of many proteins and is essential for life.\n\n\nFIGURE 18.58 Volcanic gases contain hydrogen sulfide. (credit: Daniel Julie/Wikimedia Commons)\nThe Frasch process, illustrated in Figure 18.59, is important in the mining of free sulfur from enormous\nunderground deposits in Texas and Louisiana. Superheated water ( $170^{\\circ} \\mathrm{C}$ and 10 atm pressure) is forced down the outermost of three concentric pipes to the underground deposit. The hot water melts the sulfur. The innermost pipe conducts compressed air into the liquid sulfur. The air forces the liquid sulfur, mixed with air, to flow up through the outlet pipe. Transferring the mixture to large settling vats allows the solid sulfur to separate upon cooling. This sulfur is $99.5 \\%$ to $99.9 \\%$ pure and requires no purification for most uses.\n\n\nFIGURE 18.59 The Frasch process is used to mine sulfur from underground deposits.\nLarger amounts of sulfur also come from hydrogen sulfide recovered during the purification of natural gas.\nSulfur exists in several allotropic forms. The stable form at room temperature contains eight-membered rings, and so the true formula is $S_{8}$. However, chemists commonly use $S$ to simplify the coefficients in chemical equations; we will follow this practice in this book."}
{"id": 4796, "contents": "1957. LEARNING OBJECTIVES - \nLike oxygen, which is also a member of group 16, sulfur exhibits a distinctly nonmetallic behavior. It oxidizes metals, giving a variety of binary sulfides in which sulfur exhibits a negative oxidation state (2-). Elemental sulfur oxidizes less electronegative nonmetals, and more electronegative nonmetals, such as oxygen and the halogens, will oxidize it. Other strong oxidizing agents also oxidize sulfur. For example, concentrated nitric acid oxidizes sulfur to the sulfate ion, with the concurrent formation of nitrogen(IV) oxide:\n\n$$\n\\mathrm{S}(s)+6 \\mathrm{HNO}_{3}(a q) \\longrightarrow 2 \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q)+6 \\mathrm{NO}_{2}(g)\n$$\n\nThe chemistry of sulfur with an oxidation state of $2-$ is similar to that of oxygen. Unlike oxygen, however, sulfur forms many compounds in which it exhibits positive oxidation states."}
{"id": 4797, "contents": "1958. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the preparation, properties, and uses of halogens\n- Describe the properties, preparation, and uses of halogen compounds\n\nThe elements in group 17 are the halogens. These are the elements fluorine, chlorine, bromine, iodine, and astatine. These elements are too reactive to occur freely in nature, but their compounds are widely distributed. Chlorides are the most abundant; although fluorides, bromides, and iodides are less common, they are reasonably available. In this section, we will examine the occurrence, preparation, and properties of halogens. Next, we will examine halogen compounds with the representative metals followed by an examination of the interhalogens. This section will conclude with some applications of halogens."}
{"id": 4798, "contents": "1959. Occurrence and Preparation - \nAll of the halogens occur in seawater as halide ions. The concentration of the chloride ion is 0.54 M ; that of the other halides is less than $10^{-4} \\mathrm{M}$. Fluoride also occurs in minerals such as $\\mathrm{CaF}_{2}, \\mathrm{Ca}\\left(\\mathrm{PO}_{4}\\right)_{3} \\mathrm{~F}$, and $\\mathrm{Na}_{3} \\mathrm{AlF}_{6}$. Chloride also occurs in the Great Salt Lake and the Dead Sea, and in extensive salt beds that contain $\\mathrm{NaCl}, \\mathrm{KCl}$, or $\\mathrm{MgCl}_{2}$. Part of the chlorine in your body is present as hydrochloric acid, which is a component of stomach acid. Bromine compounds occur in the Dead Sea and underground brines. Iodine compounds are found in small quantities in Chile saltpeter, underground brines, and sea kelp. Iodine is essential to the function of the thyroid gland.\n\nThe best sources of halogens (except iodine) are halide salts. It is possible to oxidize the halide ions to free diatomic halogen molecules by various methods, depending on the ease of oxidation of the halide ion. Fluoride is the most difficult to oxidize, whereas iodide is the easiest.\n\nThe major method for preparing fluorine is electrolytic oxidation. The most common electrolysis procedure is to use a molten mixture of potassium hydrogen fluoride, $\\mathrm{KHF}_{2}$, and anhydrous hydrogen fluoride. Electrolysis causes HF to decompose, forming fluorine gas at the anode and hydrogen at the cathode. It is necessary to keep the two gases separated to prevent their explosive recombination to reform hydrogen fluoride."}
{"id": 4799, "contents": "1959. Occurrence and Preparation - \nMost commercial chlorine comes from the electrolysis of the chloride ion in aqueous solutions of sodium chloride; this is the chlor-alkali process discussed previously. Chlorine is also a product of the electrolytic production of metals such as sodium, calcium, and magnesium from their fused chlorides. It is also possible to prepare chlorine by the chemical oxidation of the chloride ion in acid solution with strong oxidizing agents such as manganese dioxide $\\left(\\mathrm{MnO}_{2}\\right)$ or sodium dichromate $\\left(\\mathrm{Na}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}\\right)$. The reaction with manganese dioxide is:\n\n$$\n\\mathrm{MnO}_{2}(s)+2 \\mathrm{Cl}^{-}(a q)+4 \\mathrm{H}_{3} \\mathrm{O}^{+}(a q) \\longrightarrow \\mathrm{Mn}^{2+}(a q)+\\mathrm{Cl}_{2}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nThe commercial preparation of bromine involves the oxidation of bromide ion by chlorine:\n\n$$\n2 \\mathrm{Br}^{-}(a q)+\\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{Br}_{2}(l)+2 \\mathrm{Cl}^{-}(a q)\n$$\n\nChlorine is a stronger oxidizing agent than bromine. This method is important for the production of essentially all domestic bromine.\n\nSome iodine comes from the oxidation of iodine chloride, ICl , or iodic acid, $\\mathrm{HlO}_{3}$. The commercial preparation of iodine utilizes the reduction of sodium iodate, $\\mathrm{NaIO}_{3}$, an impurity in deposits of Chile saltpeter, with sodium hydrogen sulfite:\n\n$$\n2 \\mathrm{IO}_{3}^{-}(a q)+5 \\mathrm{HSO}_{3}^{-}(a q) \\longrightarrow 3 \\mathrm{HSO}_{4}^{-}(a q)+2 \\mathrm{SO}_{4}{ }^{2-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{I}_{2}(s)\n$$"}
{"id": 4800, "contents": "1960. Properties of the Halogens - \nFluorine is a pale yellow gas, chlorine is a greenish-yellow gas, bromine is a deep reddish-brown liquid, and iodine is a grayish-black crystalline solid. Liquid bromine has a high vapor pressure, and the reddish vapor is readily visible in Figure 18.60. Iodine crystals have a noticeable vapor pressure. When gently heated, these crystals sublime and form a beautiful deep violet vapor.\n\n\nFIGURE 18.60 Chlorine is a pale yellow-green gas (left), gaseous bromine is deep orange (center), and gaseous iodine is purple (right). (Fluorine is so reactive that it is too dangerous to handle.) (credit: Sahar Atwa)\n\nBromine is only slightly soluble in water, but it is miscible in all proportions in less polar (or nonpolar) solvents such as chloroform, carbon tetrachloride, and carbon disulfide, forming solutions that vary from yellow to reddish-brown, depending on the concentration.\n\nIodine is soluble in chloroform, carbon tetrachloride, carbon disulfide, and many hydrocarbons, giving violet solutions of $\\mathrm{I}_{2}$ molecules. Iodine dissolves only slightly in water, giving brown solutions. It is quite soluble in aqueous solutions of iodides, with which it forms brown solutions. These brown solutions result because iodine molecules have empty valence $d$ orbitals and can act as weak Lewis acids towards the iodide ion. The equation for the reversible reaction of iodine (Lewis acid) with the iodide ion (Lewis base) to form triiodide ion, $\\mathrm{I}_{3}{ }^{-}$, is:\n\n$$\n\\mathrm{I}_{2}(s)+\\mathrm{I}^{-}(a q) \\longrightarrow \\mathrm{I}_{3}^{-}(a q)\n$$"}
{"id": 4801, "contents": "1960. Properties of the Halogens - \n$$\n\\mathrm{I}_{2}(s)+\\mathrm{I}^{-}(a q) \\longrightarrow \\mathrm{I}_{3}^{-}(a q)\n$$\n\nThe easier it is to oxidize the halide ion, the more difficult it is for the halogen to act as an oxidizing agent. Fluorine generally oxidizes an element to its highest oxidation state, whereas the heavier halogens may not. For example, when excess fluorine reacts with sulfur, $\\mathrm{SF}_{6}$ forms. Chlorine gives $\\mathrm{SCl}_{2}$ and bromine, $\\mathrm{S}_{2} \\mathrm{Br}_{2}$. Iodine does not react with sulfur.\n\nFluorine is the most powerful oxidizing agent of the known elements. It spontaneously oxidizes most other elements; therefore, the reverse reaction, the oxidation of fluorides, is very difficult to accomplish. Fluorine reacts directly and forms binary fluorides with all of the elements except the lighter noble gases ( $\\mathrm{He}, \\mathrm{Ne}$, and Ar ). Fluorine is such a strong oxidizing agent that many substances ignite on contact with it. Drops of water inflame in fluorine and form $\\mathrm{O}_{2}, \\mathrm{OF}_{2}, \\mathrm{H}_{2} \\mathrm{O}_{2}, \\mathrm{O}_{3}$, and HF . Wood and asbestos ignite and burn in fluorine gas. Most hot metals burn vigorously in fluorine. However, it is possible to handle fluorine in copper, iron, or nickel containers because an adherent film of the fluoride salt passivates their surfaces. Fluorine is the only element that reacts directly with the noble gas xenon."}
{"id": 4802, "contents": "1960. Properties of the Halogens - \nAlthough it is a strong oxidizing agent, chlorine is less active than fluorine. Mixing chlorine and hydrogen in the dark makes the reaction between them to be imperceptibly slow. Exposure of the mixture to light causes the two to react explosively. Chlorine is also less active towards metals than fluorine, and oxidation reactions usually require higher temperatures. Molten sodium ignites in chlorine. Chlorine attacks most nonmetals (C, $\\mathrm{N}_{2}$, and $\\mathrm{O}_{2}$ are notable exceptions), forming covalent molecular compounds. Chlorine generally reacts with compounds that contain only carbon and hydrogen (hydrocarbons) by adding to multiple bonds or by substitution.\n\nIn cold water, chlorine undergoes a disproportionation reaction:\n\n$$\n\\mathrm{Cl}_{2}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{HOCl}(a q)+\\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{Cl}^{-}(a q)\n$$\n\nHalf the chlorine atoms oxidize to the $1+$ oxidation state (hypochlorous acid), and the other half reduce to the $1-$ oxidation state (chloride ion). This disproportionation is incomplete, so chlorine water is an equilibrium mixture of chlorine molecules, hypochlorous acid molecules, hydronium ions, and chloride ions. When exposed to light, this solution undergoes a photochemical decomposition:\n\n$$\n2 \\mathrm{HOCl}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\xrightarrow{\\text { sunlight }} 2 \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+2 \\mathrm{Cl}^{-}(a q)+\\mathrm{O}_{2}(g)\n$$"}
{"id": 4803, "contents": "1960. Properties of the Halogens - \nThe nonmetal chlorine is more electronegative than any other element except fluorine, oxygen, and nitrogen. In general, very electronegative elements are good oxidizing agents; therefore, we would expect elemental chlorine to oxidize all of the other elements except for these three (and the nonreactive noble gases). Its oxidizing property, in fact, is responsible for its principal use. For example, phosphorus(V) chloride, an important intermediate in the preparation of insecticides and chemical weapons, is manufactured by oxidizing the phosphorus with chlorine:\n\n$$\n\\mathrm{P}_{4}(s)+10 \\mathrm{Cl}_{2}(g) \\longrightarrow 4 \\mathrm{PCl}_{5}(l)\n$$\n\nA great deal of chlorine is also used to oxidize, and thus to destroy, organic or biological materials in water purification and in bleaching.\n\nThe chemical properties of bromine are similar to those of chlorine, although bromine is the weaker oxidizing agent and its reactivity is less than that of chlorine.\n\nIodine is the least reactive of the halogens. It is the weakest oxidizing agent, and the iodide ion is the most easily oxidized halide ion. Iodine reacts with metals, but heating is often required. It does not oxidize other halide ions.\n\nCompared with the other halogens, iodine reacts only slightly with water. Traces of iodine in water react with a mixture of starch and iodide ion, forming a deep blue color. This reaction is a very sensitive test for the presence of iodine in water."}
{"id": 4804, "contents": "1961. Halides of the Representative Metals - \nThousands of salts of the representative metals have been prepared. The binary halides are an important subclass of salts. A salt is an ionic compound composed of cations and anions, other than hydroxide or oxide ions. In general, it is possible to prepare these salts from the metals or from oxides, hydroxides, or carbonates. We will illustrate the general types of reactions for preparing salts through reactions used to prepare binary halides.\n\nThe binary compounds of a metal with the halogens are the halides. Most binary halides are ionic. However, mercury, the elements of group 13 with oxidation states of $3+$, tin(IV), and lead(IV) form covalent binary halides.\n\nThe direct reaction of a metal and a halogen produce the halide of the metal. Examples of these oxidationreduction reactions include:\n\n$$\n\\begin{aligned}\n\\mathrm{Cd}(s)+\\mathrm{Cl}_{2}(g) & \\longrightarrow \\mathrm{CdCl}_{2}(s) \\\\\n2 \\mathrm{Ga}(l)+3 \\mathrm{Br}_{2}(l) & \\longrightarrow 2 \\mathrm{GaBr}_{3}(s)\n\\end{aligned}\n$$"}
{"id": 4805, "contents": "1962. LINK TO LEARNING - \nReactions of the alkali metals with elemental halogens are very exothermic and often quite violent. Under controlled conditions, they provide exciting demonstrations for budding students of chemistry. You can view the initial heating (http://openstax.org/l/16sodium) of the sodium that removes the coating of sodium hydroxide, sodium peroxide, and residual mineral oil to expose the reactive surface. The reaction with chlorine gas then proceeds very nicely.\n\nIf a metal can exhibit two oxidation states, it may be necessary to control the stoichiometry in order to obtain the halide with the lower oxidation state. For example, preparation of tin(II) chloride requires a 1:1 ratio of Sn to $\\mathrm{Cl}_{2}$, whereas preparation of $\\operatorname{tin}(\\mathrm{IV})$ chloride requires a 1:2 ratio:\n\n$$\n\\begin{gathered}\n\\mathrm{Sn}(s)+\\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{SnCl}_{2}(s) \\\\\n\\mathrm{Sn}(s)+2 \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{SnCl}_{4}(l)\n\\end{gathered}\n$$\n\nThe active representative metals-those that are easier to oxidize than hydrogen-react with gaseous hydrogen halides to produce metal halides and hydrogen. The reaction of zinc with hydrogen fluoride is:\n\n$$\n\\mathrm{Zn}(s)+2 \\mathrm{HF}(g) \\longrightarrow \\mathrm{ZnF}_{2}(s)+\\mathrm{H}_{2}(g)\n$$\n\nThe active representative metals also react with solutions of hydrogen halides to form hydrogen and solutions of the corresponding halides. Examples of such reactions include:\n\n$$\n\\begin{aligned}\n\\mathrm{Cd}(s)+2 \\mathrm{HBr}(a q) & \\longrightarrow \\mathrm{CdBr}_{2}(a q)+\\mathrm{H}_{2}(g) \\\\\n\\mathrm{Sn}(s)+2 \\mathrm{HI}(a q) & \\longrightarrow \\mathrm{SnI}_{2}(a q)+\\mathrm{H}_{2}(g)\n\\end{aligned}\n$$"}
{"id": 4806, "contents": "1962. LINK TO LEARNING - \nHydroxides, carbonates, and some oxides react with solutions of the hydrogen halides to form solutions of halide salts. It is possible to prepare additional salts by the reaction of these hydroxides, carbonates, and oxides with aqueous solution of other acids:\n\n$$\n\\begin{gathered}\n\\mathrm{CaCo}_{3}(s)+2 \\mathrm{HCl}(a q) \\longrightarrow \\mathrm{CaCl}_{2}(a q)+\\mathrm{CO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l) \\\\\n\\mathrm{TlOH}(a q)+\\mathrm{HF}(a q) \\longrightarrow \\mathrm{TlF}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)\n\\end{gathered}\n$$\n\nA few halides and many of the other salts of the representative metals are insoluble. It is possible to prepare these soluble salts by metathesis reactions that occur when solutions of soluble salts are mixed (see Figure 18.61). Metathesis reactions are examined in the chapter on the stoichiometry of chemical reactions.\n\n\nFIGURE 18.61 Solid $\\mathrm{HgI}_{2}$ forms when solutions of KI and $\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}$ are mixed. (credit: Sahar Atwa)\nSeveral halides occur in large quantities in nature. The ocean and underground brines contain many halides. For example, magnesium chloride in the ocean is the source of magnesium ions used in the production of magnesium. Large underground deposits of sodium chloride, like the salt mine shown in Figure 18.62, occur in many parts of the world. These deposits serve as the source of sodium and chlorine in almost all other compounds containing these elements. The chlor-alkali process is one example.\n\n\nFIGURE 18.62 Underground deposits of sodium chloride are found throughout the world and are often mined. This is a tunnel in the K\u0142odawa salt mine in Poland. (credit: Jarek Zok)"}
{"id": 4807, "contents": "1963. Interhalogens - \nCompounds formed from two or more different halogens are interhalogens. Interhalogen molecules consist of one atom of the heavier halogen bonded by single bonds to an odd number of atoms of the lighter halogen. The structures of $\\mathrm{IF}_{3}, \\mathrm{IF}_{5}$, and $\\mathrm{IF}_{7}$ are illustrated in Figure 18.63. Formulas for other interhalogens, each of which comes from the reaction of the respective halogens, are in Table 18.3.\n\n\nFIGURE 18.63 The structure of $\\mathrm{IF}_{3}$ is T -shaped (left), $\\mathrm{IF}_{5}$ is square pyramidal (center), and $\\mathrm{IF}_{7}$ is pentagonal bipyramidal (right).\n\nNote from Table 18.3 that fluorine is able to oxidize iodine to its maximum oxidation state, $7+$, whereas bromine and chlorine, which are more difficult to oxidize, achieve only the 5+-oxidation state. A 7+-oxidation state is the limit for the halogens. Because smaller halogens are grouped about a larger one, the maximum number of smaller atoms possible increases as the radius of the larger atom increases. Many of these compounds are unstable, and most are extremely reactive. The interhalogens react like their component halides; halogen fluorides, for example, are stronger oxidizing agents than are halogen chlorides.\n\nThe ionic polyhalides of the alkali metals, such as $\\mathrm{KI}_{3}, \\mathrm{KICl}_{2}, \\mathrm{KICl}_{4}, \\mathrm{CsIBr}_{2}$, and $\\mathrm{CsBrCl}_{2}$, which contain an anion composed of at least three halogen atoms, are closely related to the interhalogens. As seen previously, the formation of the polyhalide anion $\\mathrm{I}_{3}{ }^{-}$is responsible for the solubility of iodine in aqueous solutions\ncontaining an iodide ion.\n\nInterhalogens"}
{"id": 4808, "contents": "1963. Interhalogens - \nInterhalogens\n\n| YX | $\\mathrm{YX}_{3}$ | $\\mathrm{YX}_{5}$ | $\\mathrm{YX}_{7}$ |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{ClF}(g)$ | $\\mathrm{ClF}_{3}(g)$ | $\\mathrm{ClF}_{5}(g)$ |  |\n| $\\operatorname{BrF}(g)$ | $\\mathrm{BrF}_{3}(I)$ | $\\mathrm{BrF}_{5}(I)$ |  |\n| $\\operatorname{BrCl}(g)$ |  |  |  |\n| $\\mathrm{IF}(s)$ | $\\mathrm{IF}_{3}(s)$ | $\\mathrm{IF}_{5}(I)$ | $\\mathrm{IF}_{7}(g)$ |\n| $\\mathrm{ICl}(I)$ | $\\mathrm{ICl}_{3}(s)$ |  |  |\n| $\\mathrm{IBr}(s)$ |  |  |  |\n\nTABLE 18.3"}
{"id": 4809, "contents": "1964. Applications - \nThe fluoride ion and fluorine compounds have many important uses. Compounds of carbon, hydrogen, and fluorine are replacing Freons (compounds of carbon, chlorine, and fluorine) as refrigerants. Teflon is a polymer composed of $-\\mathrm{CF}_{2} \\mathrm{CF}_{2}-$ units. Fluoride ion is added to water supplies and to some toothpastes as $\\mathrm{SnF}_{2}$ or NaF to fight tooth decay. Fluoride partially converts teeth from $\\mathrm{Ca}_{5}\\left(\\mathrm{PO}_{4}\\right)_{3}(\\mathrm{OH})$ into $\\mathrm{Ca}_{5}\\left(\\mathrm{PO}_{4}\\right)_{3} \\mathrm{~F}$.\n\nChlorine is important to bleach wood pulp and cotton cloth. The chlorine reacts with water to form hypochlorous acid, which oxidizes colored substances to colorless ones. Large quantities of chlorine are important in chlorinating hydrocarbons (replacing hydrogen with chlorine) to produce compounds such as tetrachloride $\\left(\\mathrm{CCl}_{4}\\right)$, chloroform $\\left(\\mathrm{CHCl}_{3}\\right)$, and ethyl chloride $\\left(\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{Cl}\\right)$, and in the production of polyvinyl chloride (PVC) and other polymers. Chlorine is also important to kill the bacteria in community water supplies.\n\nBromine is important in the production of certain dyes, and sodium and potassium bromides are used as sedatives. At one time, light-sensitive silver bromide was a component of photographic film.\n\nIodine in alcohol solution with potassium iodide is an antiseptic (tincture of iodine). Iodide salts are essential for the proper functioning of the thyroid gland; an iodine deficiency may lead to the development of a goiter. Iodized table salt contains $0.023 \\%$ potassium iodide. Silver iodide is useful in the seeding of clouds to induce rain; it was important in the production of photographic film and iodoform, $\\mathrm{CHI}_{3}$, is an antiseptic."}
{"id": 4810, "contents": "1965. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the properties, preparation, and uses of the noble gases\n\nThe elements in group 18 are the noble gases (helium, neon, argon, krypton, xenon, and radon). They earned the name \"noble\" because they were assumed to be nonreactive since they have filled valence shells. In 1962, Dr. Neil Bartlett at the University of British Columbia proved this assumption to be false.\n\nThese elements are present in the atmosphere in small amounts. Some natural gas contains $1-2 \\%$ helium by mass. Helium is isolated from natural gas by liquefying the condensable components, leaving only helium as a gas. The United States possesses most of the world's commercial supply of this element in its helium-bearing gas fields. Argon, neon, krypton, and xenon come from the fractional distillation of liquid air. Radon comes from other radioactive elements. More recently, it was observed that this radioactive gas is present in very\nsmall amounts in soils and minerals. Its accumulation in well-insulated, tightly sealed buildings, however, constitutes a health hazard, primarily lung cancer.\n\nThe boiling points and melting points of the noble gases are extremely low relative to those of other substances of comparable atomic or molecular masses. This is because only weak London dispersion forces are present, and these forces can hold the atoms together only when molecular motion is very slight, as it is at very low temperatures. Helium is the only substance known that does not solidify on cooling at normal pressure. It remains liquid close to absolute zero $(0.001 \\mathrm{~K})$ at ordinary pressures, but it solidifies under elevated pressure.\n\nHelium is used for filling balloons and lighter-than-air craft because it does not burn, making it safer to use than hydrogen. Helium at high pressures is not a narcotic like nitrogen. Thus, mixtures of oxygen and helium are important for divers working under high pressures. Using a helium-oxygen mixture avoids the disoriented mental state known as nitrogen narcosis, the so-called rapture of the deep. Helium is important as an inert atmosphere for the melting and welding of easily oxidizable metals and for many chemical processes that are sensitive to air."}
{"id": 4811, "contents": "1965. LEARNING OBJECTIVES - \nLiquid helium (boiling point, 4.2 K ) is an important coolant to reach the low temperatures necessary for cryogenic research, and it is essential for achieving the low temperatures necessary to produce superconduction in traditional superconducting materials used in powerful magnets and other devices. This cooling ability is necessary for the magnets used for magnetic resonance imaging, a common medical diagnostic procedure. The other common coolant is liquid nitrogen (boiling point, 77 K ), which is significantly cheaper.\n\nNeon is a component of neon lamps and signs. Passing an electric spark through a tube containing neon at low pressure generates the familiar red glow of neon. It is possible to change the color of the light by mixing argon or mercury vapor with the neon or by utilizing glass tubes of a special color.\n\nArgon was useful in the manufacture of gas-filled electric light bulbs, where its lower heat conductivity and chemical inertness made it preferable to nitrogen for inhibiting the vaporization of the tungsten filament and prolonging the life of the bulb. Fluorescent tubes commonly contain a mixture of argon and mercury vapor. Argon is the third most abundant gas in dry air.\n\nKrypton-xenon flash tubes are used to take high-speed photographs. An electric discharge through such a tube gives a very intense light that lasts only $\\frac{1}{50,000}$ of a second. Krypton forms a difluoride, $\\mathrm{KrF}_{2}$, which is thermally unstable at room temperature.\n\nStable compounds of xenon form when xenon reacts with fluorine. Xenon difluoride, $\\mathrm{XeF}_{2}$, forms after heating an excess of xenon gas with fluorine gas and then cooling. The material forms colorless crystals, which are stable at room temperature in a dry atmosphere. Xenon tetrafluoride, $\\mathrm{XeF}_{4}$, and xenon hexafluoride, $\\mathrm{XeF}_{6}$, are prepared in an analogous manner, with a stoichiometric amount of fluorine and an excess of fluorine, respectively. Compounds with oxygen are prepared by replacing fluorine atoms in the xenon fluorides with oxygen."}
{"id": 4812, "contents": "1965. LEARNING OBJECTIVES - \nWhen $\\mathrm{XeF}_{6}$ reacts with water, a solution of $\\mathrm{XeO}_{3}$ results and the xenon remains in the $6+$-oxidation state:\n\n$$\n\\mathrm{XeF}_{6}(s)+3 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{XeO}_{3}(a q)+6 \\mathrm{HF}(a q)\n$$\n\nDry, solid xenon trioxide, $\\mathrm{XeO}_{3}$, is extremely explosive-it will spontaneously detonate. Both $\\mathrm{XeF}_{6}$ and $\\mathrm{XeO}_{3}$ disproportionate in basic solution, producing xenon, oxygen, and salts of the perxenate ion, $\\mathrm{XeO}_{6}{ }^{4-}$, in which xenon reaches its maximum oxidation sate of 8+.\n\nRadon apparently forms $\\mathrm{RnF}_{2}$-evidence of this compound comes from radiochemical tracer techniques.\nUnstable compounds of argon form at low temperatures, but stable compounds of helium and neon are not known."}
{"id": 4813, "contents": "1966. Key Terms - \nacid anhydride compound that reacts with water to form an acid or acidic solution\nalkaline earth metal any of the metals (beryllium, magnesium, calcium, strontium, barium, and radium) occupying group 2 of the periodic table; they are reactive, divalent metals that form basic oxides\nallotropes two or more forms of the same element, in the same physical state, with different chemical structures\namorphous solid material such as a glass that does not have a regular repeating component to its three-dimensional structure; a solid but not a crystal\nbase anhydride metal oxide that behaves as a base towards acids\nbicarbonate anion salt of the hydrogen carbonate ion, $\\mathrm{HCO}_{3}{ }^{-}$\nbismuth heaviest member of group 15; a less reactive metal than other representative metals\nborate compound containing boron-oxygen bonds, typically with clusters or chains as a part of the chemical structure\ncarbonate salt of the anion $\\mathrm{CO}_{3}{ }^{2-}$; often formed by the reaction of carbon dioxide with bases\nchemical reduction method of preparing a representative metal using a reducing agent\nchlor-alkali process electrolysis process for the synthesis of chlorine and sodium hydroxide\ndisproportionation reaction chemical reaction where a single reactant is simultaneously reduced and oxidized; it is both the reducing agent and the oxidizing agent\nDowns cell electrochemical cell used for the commercial preparation of metallic sodium (and chlorine) from molten sodium chloride\nFrasch process important in the mining of free sulfur from enormous underground deposits\nHaber process main industrial process used to produce ammonia from nitrogen and hydrogen; involves the use of an iron catalyst and elevated temperatures and pressures\nhalide compound containing an anion of a group 17 element in the 1-oxidation state (fluoride, $\\mathrm{F}^{-}$; chloride, $\\mathrm{Cl}^{-}$; bromide, $\\mathrm{Br}^{-}$; and iodide, $\\mathrm{I}^{-}$)\nHall-H\u00e9roult cell electrolysis apparatus used to isolate pure aluminum metal from a solution of alumina in molten cryolite"}
{"id": 4814, "contents": "1966. Key Terms - \nHall-H\u00e9roult cell electrolysis apparatus used to isolate pure aluminum metal from a solution of alumina in molten cryolite\nhydrogen carbonate salt of carbonic acid, $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ (containing the anion $\\mathrm{HCO}_{3}{ }^{-}$) in which one hydrogen atom has been replaced; an acid carbonate; also known as bicarbonate ion\nhydrogen halide binary compound formed between hydrogen and the halogens: $\\mathrm{HF}, \\mathrm{HCl}$, HBr , and HI\nhydrogen sulfate $\\mathrm{HSO}_{4}{ }^{-}$ion\nhydrogen sulfite $\\mathrm{HSO}_{3}{ }^{-}$ion\nhydrogenation addition of hydrogen $\\left(\\mathrm{H}_{2}\\right)$ to reduce a compound\nhydroxide compound of a metal with the hydroxide ion $\\mathrm{OH}^{-}$or the group -OH\ninterhalogen compound formed from two or more different halogens\nmetal (representative) atoms of the metallic elements of groups $1,2,12,13,14,15$, and 16 , which form ionic compounds by losing electrons from their outer $s$ or $p$ orbitals\nmetalloid element that has properties that are between those of metals and nonmetals; these elements are typically semiconductors\nnitrate $\\mathrm{NO}_{3}{ }^{-}$ion; salt of nitric acid\nnitrogen fixation formation of nitrogen compounds from molecular nitrogen\nOstwald process industrial process used to convert ammonia into nitric acid\noxide binary compound of oxygen with another element or group, typically containing $\\mathrm{O}^{2-}$ ions or the group -O - or $=0$\nozone allotrope of oxygen; $\\mathrm{O}_{3}$\npassivation metals with a protective nonreactive film of oxide or other compound that creates a barrier for chemical reactions; physical or chemical removal of the passivating film allows the metals to demonstrate their expected chemical reactivity\nperoxide molecule containing two oxygen atoms bonded together or as the anion, $\\mathrm{O}_{2}{ }^{2-}$\nphotosynthesis process whereby light energy promotes the reaction of water and carbon dioxide to form carbohydrates and oxygen; this allows photosynthetic organisms to store energy"}
{"id": 4815, "contents": "1966. Key Terms - \nphotosynthesis process whereby light energy promotes the reaction of water and carbon dioxide to form carbohydrates and oxygen; this allows photosynthetic organisms to store energy\nPidgeon process chemical reduction process used to produce magnesium through the thermal reaction of magnesium oxide with silicon\npolymorph variation in crystalline structure that results in different physical properties for the resulting compound\nrepresentative element element where the $s$ and $p$ orbitals are filling\nrepresentative metal metal among the representative elements\nsilicate compound containing silicon-oxygen bonds, with silicate tetrahedra connected in rings, sheets, or three-dimensional networks,\ndepending on the other elements involved in the formation of the compounds\nsulfate $\\mathrm{SO}_{4}{ }^{2-}$ ion"}
{"id": 4816, "contents": "1967. Summary - 1967.1. Periodicity\nThis section focuses on the periodicity of the representative elements. These are the elements where the electrons are entering the $s$ and $p$ orbitals. The representative elements occur in groups 1, 2, and 12-18. These elements are representative metals, metalloids, and nonmetals. The alkali metals (group 1) are very reactive, readily form ions with a charge of $1+$ to form ionic compounds that are usually soluble in water, and react vigorously with water to form hydrogen gas and a basic solution of the metal hydroxide. The outermost electrons of the alkaline earth metals (group 2) are more difficult to remove than the outer electron of the alkali metals, leading to the group 2 metals being less reactive than those in group 1. These elements easily form compounds in which the metals exhibit an oxidation state of $2+$. Zinc, cadmium, and mercury (group 12) commonly exhibit the group oxidation state of 2+ (although mercury also exhibits an oxidation state of $1+$ in compounds that contain $\\mathrm{Hg}_{2}{ }^{2+}$ ). Aluminum, gallium, indium, and thallium (group 13) are easier to oxidize than is hydrogen. Aluminum, gallium, and indium occur with an oxidation state 3+ (however, thallium also commonly occurs as the $\\mathrm{Tl}^{+}$ion). Tin and lead form stable divalent cations and covalent compounds in which the metals exhibit the 4+oxidation state."}
{"id": 4817, "contents": "1967. Summary - 1967.2. Occurrence and Preparation of the Representative Metals\nBecause of their chemical reactivity, it is necessary to produce the representative metals in their pure forms by reduction from naturally occurring compounds. Electrolysis is important in the production of sodium, potassium, and aluminum. Chemical reduction is the primary method for the isolation of magnesium, zinc, and tin. Similar procedures are important for the other representative metals."}
{"id": 4818, "contents": "1967. Summary - 1967.3. Structure and General Properties of the Metalloids\nThe elements boron, silicon, germanium, arsenic, antimony, and tellurium separate the metals from the nonmetals in the periodic table. These elements, called metalloids or sometimes semimetals, exhibit\nsulfite $\\mathrm{SO}_{3}{ }^{2-}$ ion\nsuperoxide oxide containing the anion $\\mathrm{O}_{2}{ }^{-}$\nproperties characteristic of both metals and nonmetals. The structures of these elements are similar in many ways to those of nonmetals, but the elements are electrical semiconductors."}
{"id": 4819, "contents": "1967. Summary - 1967.4. Structure and General Properties of the Nonmetals\nNonmetals have structures that are very different from those of the metals, primarily because they have greater electronegativity and electrons that are more tightly bound to individual atoms. Most nonmetal oxides are acid anhydrides, meaning that they react with water to form acidic solutions. Molecular structures are common for most of the nonmetals, and several have multiple allotropes with varying physical properties."}
{"id": 4820, "contents": "1967. Summary - 1967.5. Occurrence, Preparation, and Compounds of Hydrogen\nHydrogen is the most abundant element in the universe and its chemistry is truly unique. Although it has some chemical reactivity that is similar to that of the alkali metals, hydrogen has many of the same chemical properties of a nonmetal with a relatively low electronegativity. It forms ionic hydrides with active metals, covalent compounds in which it has an oxidation state of 1- with less electronegative elements, and covalent compounds in which it has an oxidation state of $1+$ with more electronegative nonmetals. It reacts explosively with oxygen, fluorine, and chlorine, less readily with bromine, and much less readily with iodine, sulfur, and nitrogen. Hydrogen reduces the oxides of metals with lower reduction potentials than chromium to form the metal and water. The hydrogen halides are all acidic when dissolved in water."}
{"id": 4821, "contents": "1967. Summary - 1967.6. Occurrence, Preparation, and Properties of Carbonates\nThe usual method for the preparation of the carbonates of the alkali and alkaline earth metals is by reaction of an oxide or hydroxide with carbon dioxide. Other carbonates form by precipitation. Metal carbonates or hydrogen carbonates such as limestone $\\left(\\mathrm{CaCO}_{3}\\right)$, the antacid Tums $\\left(\\mathrm{CaCO}_{3}\\right)$, and baking soda $\\left(\\mathrm{NaHCO}_{3}\\right)$ are common examples. Carbonates and hydrogen carbonates decompose in\nthe presence of acids and most decompose on heating."}
{"id": 4822, "contents": "1968. Properties of Nitrogen - \nNitrogen exhibits oxidation states ranging from 3to $5+$. Because of the stability of the $\\mathrm{N} \\equiv \\mathrm{N}$ triple bond, it requires a great deal of energy to make compounds from molecular nitrogen. Active metals such as the alkali metals and alkaline earth metals can reduce nitrogen to form metal nitrides. Nitrogen oxides and nitrogen hydrides are also important substances."}
{"id": 4823, "contents": "1968. Properties of Nitrogen - 1968.1. Occurrence, Preparation, and Properties of Phosphorus\nPhosphorus (group 15) commonly exhibits oxidation states of 3 - with active metals and of $3+$ and $5+$ with more electronegative nonmetals. The halogens and oxygen will oxidize phosphorus. The oxides are phosphorus(V) oxide, $\\mathrm{P}_{4} \\mathrm{O}_{10}$, and phosphorus(III) oxide, $\\mathrm{P}_{4} \\mathrm{O}_{6}$. The two common methods for preparing orthophosphoric acid, $\\mathrm{H}_{3} \\mathrm{PO}_{4}$, are either the reaction of a phosphate with sulfuric acid or the reaction of water with phosphorus(V) oxide. Orthophosphoric acid is a triprotic acid that forms three types of salts."}
{"id": 4824, "contents": "1968. Properties of Nitrogen - 1968.2. Occurrence, Preparation, and Compounds of Oxygen\nOxygen is one of the most reactive elements. This reactivity, coupled with its abundance, makes the chemistry of oxygen very rich and well understood.\n\nCompounds of the representative metals with oxygen exist in three categories (1) oxides, (2) peroxides and superoxides, and (3) hydroxides. Heating the corresponding hydroxides, nitrates, or carbonates is the most common method for producing oxides. Heating the metal or metal oxide in oxygen may lead to the formation of peroxides and superoxides. The soluble oxides dissolve in water to form solutions of hydroxides. Most metals oxides are base anhydrides and react with acids. The hydroxides of the representative metals react with acids in acid-base reactions to form salts and water.\n\nThe hydroxides have many commercial uses.\nAll nonmetals except fluorine form multiple oxides. Nearly all of the nonmetal oxides are acid anhydrides. The acidity of oxyacids requires that the hydrogen atoms bond to the oxygen atoms in the molecule rather than to the other nonmetal atom. Generally, the strength of the oxyacid increases with the number of oxygen atoms bonded to the nonmetal atom and not to a hydrogen."}
{"id": 4825, "contents": "1968. Properties of Nitrogen - 1968.3. Occurrence, Preparation, and Properties of Sulfur\nSulfur (group 16) reacts with almost all metals and readily forms the sulfide ion, $\\mathrm{S}^{2-}$, in which it has as oxidation state of $2-$. Sulfur reacts with most nonmetals."}
{"id": 4826, "contents": "1968. Properties of Nitrogen - 1968.4. Occurrence, Preparation, and Properties of Halogens\nThe halogens form halides with less electronegative elements. Halides of the metals vary from ionic to covalent; halides of nonmetals are covalent. Interhalogens form by the combination of two or more different halogens.\n\nAll of the representative metals react directly with elemental halogens or with solutions of the hydrohalic acids (HF, $\\mathrm{HCl}, \\mathrm{HBr}$, and HI ) to produce representative metal halides. Other laboratory preparations involve the addition of aqueous hydrohalic acids to compounds that contain such basic anions, such as hydroxides, oxides, or carbonates."}
{"id": 4827, "contents": "1968. Properties of Nitrogen - 1968.5. Occurrence, Preparation, and Properties of the Noble Gases\nThe most significant property of the noble gases (group 18) is their inactivity. They occur in low concentrations in the atmosphere. They find uses as inert atmospheres, neon signs, and as coolants. The three heaviest noble gases react with fluorine to form fluorides. The xenon fluorides are the best characterized as the starting materials for a few other noble gas compounds."}
{"id": 4828, "contents": "1969. Exercises - 1969.1. Periodicity\n1. How do alkali metals differ from alkaline earth metals in atomic structure and general properties?\n2. Why does the reactivity of the alkali metals decrease from cesium to lithium?\n3. Predict the formulas for the nine compounds that may form when each species in column 1 of the table reacts with each species in column 2.\n\n1 2\n\n| Na | I |\n| :---: | :---: |\n| Sr | Se |\n| Al | O |"}
{"id": 4829, "contents": "1969. Exercises - 1969.1. Periodicity\n4. Predict the best choice in each of the following. You may wish to review the chapter on electronic structure for relevant examples.\n(a) the most metallic of the elements $\\mathrm{Al}, \\mathrm{Be}$, and Ba\n(b) the most covalent of the compounds $\\mathrm{NaCl}, \\mathrm{CaCl}_{2}$, and $\\mathrm{BeCl}_{2}$\n(c) the lowest first ionization energy among the elements $\\mathrm{Rb}, \\mathrm{K}$, and Li\n(d) the smallest among $\\mathrm{Al}, \\mathrm{Al}^{+}$, and $\\mathrm{Al}^{3+}$\n(e) the largest among $\\mathrm{Cs}^{+}, \\mathrm{Ba}^{2+}$, and Xe\n5. Sodium chloride and strontium chloride are both white solids. How could you distinguish one from the other?\n6. The reaction of quicklime, CaO , with water produces slaked lime, $\\mathrm{Ca}(\\mathrm{OH})_{2}$, which is widely used in the construction industry to make mortar and plaster. The reaction of quicklime and water is highly exothermic:\n$\\mathrm{CaO}(s)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Ca}(\\mathrm{OH})_{2}(s) \\quad \\Delta H=-350 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n(a) What is the enthalpy of reaction per gram of quicklime that reacts?\n(b) How much heat, in kilojoules, is associated with the production of 1 ton of slaked lime?\n7. Write a balanced equation for the reaction of elemental strontium with each of the following:\n(a) oxygen\n(b) hydrogen bromide\n(c) hydrogen\n(d) phosphorus\n(e) water\n8. How many moles of ionic species are present in 1.0 L of a solution marked 1.0 M mercury(I) nitrate?"}
{"id": 4830, "contents": "1969. Exercises - 1969.1. Periodicity\n(c) hydrogen\n(d) phosphorus\n(e) water\n8. How many moles of ionic species are present in 1.0 L of a solution marked 1.0 M mercury(I) nitrate?\n9. What is the mass of fish, in kilograms, that one would have to consume to obtain a fatal dose of mercury, if the fish contains 30 parts per million of mercury by weight? (Assume that all the mercury from the fish ends up as mercury(II) chloride in the body and that a fatal dose is 0.20 g of $\\mathrm{HgCl}_{2}$.) How many pounds of fish is this?\n10. The elements sodium, aluminum, and chlorine are in the same period.\n(a) Which has the greatest electronegativity?\n(b) Which of the atoms is smallest?\n(c) Write the Lewis structure for the simplest covalent compound that can form between aluminum and chlorine.\n(d) Will the oxide of each element be acidic, basic, or amphoteric?\n11. Does metallic tin react with HCl ?\n12. What is tin pest, also known as tin disease?\n13. Compare the nature of the bonds in $\\mathrm{PbCl}_{2}$ to that of the bonds in $\\mathrm{PbCl}_{4}$.\n14. Is the reaction of rubidium with water more or less vigorous than that of sodium? How does the rate of reaction of magnesium compare?"}
{"id": 4831, "contents": "1969. Exercises - 1969.2. Occurrence and Preparation of the Representative Metals\n15. Write an equation for the reduction of cesium chloride by elemental calcium at high temperature.\n16. Why is it necessary to keep the chlorine and sodium, resulting from the electrolysis of sodium chloride, separate during the production of sodium metal?\n17. Give balanced equations for the overall reaction in the electrolysis of molten lithium chloride and for the reactions occurring at the electrodes. You may wish to review the chapter on electrochemistry for relevant examples.\n18. The electrolysis of molten sodium chloride or of aqueous sodium chloride produces chlorine. Calculate the mass of chlorine produced from 3.00 kg sodium chloride in each case. You may wish to review the chapter on electrochemistry for relevant examples.\n19. What mass, in grams, of hydrogen gas forms during the complete reaction of 10.01 g of calcium with water?\n20. How many grams of oxygen gas are necessary to react completely with $3.01 \\times 10^{21}$ atoms of magnesium to yield magnesium oxide?\n21. Magnesium is an active metal; it burns in the form of powder, ribbons, and filaments to provide flashes of brilliant light. Why is it possible to use magnesium in construction?\n22. Why is it possible for an active metal like aluminum to be useful as a structural metal?\n23. Describe the production of metallic aluminum by electrolytic reduction.\n24. What is the common ore of tin and how is tin separated from it?\n25. A chemist dissolves a 1.497 -g sample of a type of metal (an alloy of $\\mathrm{Sn}, \\mathrm{Pb}, \\mathrm{Sb}$, and Cu ) in nitric acid, and metastannic acid, $\\mathrm{H}_{2} \\mathrm{SnO}_{3}$, is precipitated. She heats the precipitate to drive off the water, which leaves 0.4909 g of $\\operatorname{tin}(\\mathrm{IV})$ oxide. What was the percentage of tin in the original sample?\n26. Consider the production of 100 kg of sodium metal using a current of $50,000 \\mathrm{~A}$, assuming a $100 \\%$ yield.\n(a) How long will it take to produce the 100 kg of sodium metal?"}
{"id": 4832, "contents": "1969. Exercises - 1969.2. Occurrence and Preparation of the Representative Metals\n(a) How long will it take to produce the 100 kg of sodium metal?\n(b) What volume of chlorine at $25^{\\circ} \\mathrm{C}$ and 1.00 atm forms?\n27. What mass of magnesium forms when $100,000 \\mathrm{~A}$ is passed through a $\\mathrm{MgCl}_{2}$ melt for 1.00 h if the yield of magnesium is $85 \\%$ of the theoretical yield?"}
{"id": 4833, "contents": "1969. Exercises - 1969.3. Structure and General Properties of the Metalloids\n28. Give the hybridization of the metalloid and the molecular geometry for each of the following compounds or ions. You may wish to review the chapters on chemical bonding and advanced covalent bonding for relevant examples.\n(a) $\\mathrm{GeH}_{4}$\n(b) $\\mathrm{SbF}_{3}$\n(c) $\\mathrm{Te}(\\mathrm{OH})_{6}$\n(d) $\\mathrm{H}_{2} \\mathrm{Te}$\n(e) $\\mathrm{GeF}_{2}$\n(f) $\\mathrm{TeCl}_{4}$\n(g) $\\mathrm{SiF}_{6}{ }^{2-}$\n(h) $\\mathrm{SbCl}_{5}$\n(i) $\\mathrm{TeF}_{6}$\n29. Write a Lewis structure for each of the following molecules or ions. You may wish to review the chapter on chemical bonding.\n(a) $\\mathrm{H}_{3} \\mathrm{BPH}_{3}$\n(b) $\\mathrm{BF}_{4}{ }^{-}$\n(c) $\\mathrm{BBr}_{3}$\n(d) $\\mathrm{B}\\left(\\mathrm{CH}_{3}\\right)_{3}$\n(e) $\\mathrm{B}(\\mathrm{OH})_{3}$\n30. Describe the hybridization of boron and the molecular structure about the boron in each of the following:\n(a) $\\mathrm{H}_{3} \\mathrm{BPH}_{3}$\n(b) $\\mathrm{BF}_{4}^{-}$\n(c) $\\mathrm{BBr}_{3}$\n(d) $\\mathrm{B}\\left(\\mathrm{CH}_{3}\\right)_{3}$\n(e) $\\mathrm{B}(\\mathrm{OH})_{3}$\n31. Using only the periodic table, write the complete electron configuration for silicon, including any empty orbitals in the valence shell. You may wish to review the chapter on electronic structure.\n32. Write a Lewis structure for each of the following molecules and ions:\n(a) $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{SiH}$\n(b) $\\mathrm{SiO}_{4}{ }^{4-}$\n(c) $\\mathrm{Si}_{2} \\mathrm{H}_{6}$\n(d) $\\mathrm{Si}(\\mathrm{OH})_{4}$"}
{"id": 4834, "contents": "1969. Exercises - 1969.3. Structure and General Properties of the Metalloids\n(b) $\\mathrm{SiO}_{4}{ }^{4-}$\n(c) $\\mathrm{Si}_{2} \\mathrm{H}_{6}$\n(d) $\\mathrm{Si}(\\mathrm{OH})_{4}$\n(e) $\\mathrm{SiF}_{6}{ }^{2-}$\n33. Describe the hybridization of silicon and the molecular structure of the following molecules and ions:\n(a) $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{SiH}$\n(b) $\\mathrm{SiO}_{4}{ }^{4-}$\n(c) $\\mathrm{Si}_{2} \\mathrm{H}_{6}$\n(d) $\\mathrm{Si}(\\mathrm{OH})_{4}$\n(e) $\\mathrm{SiF}_{6}{ }^{2-}$\n34. Describe the hybridization and the bonding of a silicon atom in elemental silicon.\n35. Classify each of the following molecules as polar or nonpolar. You may wish to review the chapter on chemical bonding.\n(a) $\\mathrm{SiH}_{4}$\n(b) $\\mathrm{Si}_{2} \\mathrm{H}_{6}$\n(c) $\\mathrm{SiCl}_{3} \\mathrm{H}$\n(d) $\\mathrm{SiF}_{4}$\n(e) $\\mathrm{SiCl}_{2} \\mathrm{~F}_{2}$\n36. Silicon reacts with sulfur at elevated temperatures. If 0.0923 g of silicon reacts with sulfur to give 0.3030 g of silicon sulfide, determine the empirical formula of silicon sulfide.\n37. Name each of the following compounds:\n(a) $\\mathrm{TeO}_{2}$\n(b) $\\mathrm{Sb}_{2} \\mathrm{~S}_{3}$\n(c) $\\mathrm{GeF}_{4}$\n(d) $\\mathrm{SiH}_{4}$\n(e) $\\mathrm{GeH}_{4}$\n38. Write a balanced equation for the reaction of elemental boron with each of the following (most of these reactions require high temperature):\n(a) $\\mathrm{F}_{2}$\n(b) $\\mathrm{O}_{2}$\n(c) S\n(d) Se\n(e) $\\mathrm{Br}_{2}$\n39. Why is boron limited to a maximum coordination number of four in its compounds?"}
{"id": 4835, "contents": "1969. Exercises - 1969.3. Structure and General Properties of the Metalloids\n(b) $\\mathrm{O}_{2}$\n(c) S\n(d) Se\n(e) $\\mathrm{Br}_{2}$\n39. Why is boron limited to a maximum coordination number of four in its compounds?\n40. Write a formula for each of the following compounds:\n(a) silicon dioxide\n(b) silicon tetraiodide\n(c) silane\n(d) silicon carbide\n(e) magnesium silicide\n41. From the data given in Appendix G, determine the standard enthalpy change and the standard free energy change for each of the following reactions:\n(a) $\\mathrm{BF}_{3}(g)+3 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{B}(\\mathrm{OH})_{3}(s)+3 \\mathrm{HF}(g)$\n(b) $\\mathrm{BCl}_{3}(g)+3 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{B}(\\mathrm{OH})_{3}(s)+3 \\mathrm{HCl}(g)$\n(c) $\\mathrm{B}_{2} \\mathrm{H}_{6}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{~B}(\\mathrm{OH})_{3}(s)+6 \\mathrm{H}_{2}(g)$\n42. A hydride of silicon prepared by the reaction of $\\mathrm{Mg}_{2} \\mathrm{Si}$ with acid exerted a pressure of 306 torr at $26^{\\circ} \\mathrm{C}$ in a bulb with a volume of 57.0 mL . If the mass of the hydride was 0.0861 g , what is its molecular mass? What is the molecular formula for the hydride?\n43. Suppose you discovered a diamond completely encased in a silicate rock. How would you chemically free the diamond without harming it?"}
{"id": 4836, "contents": "1969. Exercises - 1969.4. Structure and General Properties of the Nonmetals\n44. Carbon forms a number of allotropes, two of which are graphite and diamond. Silicon has a diamond structure. Why is there no allotrope of silicon with a graphite structure?\n45. Nitrogen in the atmosphere exists as very stable diatomic molecules. Why does phosphorus form less stable $\\mathrm{P}_{4}$ molecules instead of $\\mathrm{P}_{2}$ molecules?\n46. Write balanced chemical equations for the reaction of the following acid anhydrides with water:\n(a) $\\mathrm{SO}_{3}$\n(b) $\\mathrm{N}_{2} \\mathrm{O}_{3}$\n(c) $\\mathrm{Cl}_{2} \\mathrm{O}_{7}$\n(d) $\\mathrm{P}_{4} \\mathrm{O}_{10}$\n(e) $\\mathrm{NO}_{2}$\n47. Determine the oxidation number of each element in each of the following compounds:\n(a) HCN\n(b) $\\mathrm{OF}_{2}$\n(c) $\\mathrm{AsCl}_{3}$\n48. Determine the oxidation state of sulfur in each of the following:\n(a) $\\mathrm{SO}_{3}$\n(b) $\\mathrm{SO}_{2}$\n(c) $\\mathrm{SO}_{3}{ }^{2-}$\n49. Arrange the following in order of increasing electronegativity: $\\mathrm{F} ; \\mathrm{Cl} ; \\mathrm{O} ;$ and S .\n50. Why does white phosphorus consist of tetrahedral $P_{4}$ molecules while nitrogen consists of diatomic $N_{2}$ molecules?"}
{"id": 4837, "contents": "1969. Exercises - 1969.5. Occurrence, Preparation, and Compounds of Hydrogen\n51. Why does hydrogen not exhibit an oxidation state of 1 - when bonded to nonmetals?\n52. The reaction of calcium hydride, $\\mathrm{CaH}_{2}$, with water can be characterized as a Lewis acid-base reaction: $\\mathrm{CaH}_{2}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Ca}(\\mathrm{OH})_{2}(a q)+2 \\mathrm{H}_{2}(g)$\nIdentify the Lewis acid and the Lewis base among the reactants. The reaction is also an oxidationreduction reaction. Identify the oxidizing agent, the reducing agent, and the changes in oxidation number that occur in the reaction.\n53. In drawing Lewis structures, we learn that a hydrogen atom forms only one bond in a covalent compound. Why?\n54. What mass of $\\mathrm{CaH}_{2}$ is necessary to react with water to provide enough hydrogen gas to fill a balloon at 20 ${ }^{\\circ} \\mathrm{C}$ and 0.8 atm pressure with a volume of 4.5 L ? The balanced equation is:\n$\\mathrm{CaH}_{2}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Ca}(\\mathrm{OH})_{2}(a q)+2 \\mathrm{H}_{2}(g)$\n55. What mass of hydrogen gas results from the reaction of 8.5 g of KH with water?\n$\\mathrm{KH}+\\mathrm{H}_{2} \\mathrm{O} \\longrightarrow \\mathrm{KOH}+\\mathrm{H}_{2}$"}
{"id": 4838, "contents": "1969. Exercises - 1969.6. Occurrence, Preparation, and Properties of Carbonates\n56. Carbon forms the $\\mathrm{CO}_{3}{ }^{2-}$ ion, yet silicon does not form an analogous $\\mathrm{SiO}_{3}{ }^{2-}$ ion. Why?\n57. Complete and balance the following chemical equations:\n(a) hardening of plaster containing slaked lime\n$\\mathrm{Ca}(\\mathrm{OH})_{2}+\\mathrm{CO}_{2} \\longrightarrow$\n(b) removal of sulfur dioxide from the flue gas of power plants\n$\\mathrm{CaO}+\\mathrm{SO}_{2} \\longrightarrow$\n(c) the reaction of baking powder that produces carbon dioxide gas and causes bread to rise\n$\\mathrm{NaHCO}_{3}+\\mathrm{NaH}_{2} \\mathrm{PO}_{4} \\longrightarrow$\n58. Heating a sample of $\\mathrm{Na}_{2} \\mathrm{CO}_{3} \\cdot \\mathrm{xH}_{2} \\mathrm{O}$ weighing 4.640 g until the removal of the water of hydration leaves 1.720 g of anhydrous $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$. What is the formula of the hydrated compound?"}
{"id": 4839, "contents": "1969. Exercises - 1969.7. Occurrence, Preparation, and Properties of Nitrogen\n59. Write the Lewis structures for each of the following:\n(a) $\\mathrm{NH}^{2-}$\n(b) $\\mathrm{N}_{2} \\mathrm{~F}_{4}$\n(c) $\\mathrm{NH}_{2}{ }^{-}$\n(d) $\\mathrm{NF}_{3}$\n(e) $\\mathrm{N}_{3}{ }^{-}$\n60. For each of the following, indicate the hybridization of the nitrogen atom (for $\\mathrm{N}_{3}{ }^{-}$, the central nitrogen).\n(a) $\\mathrm{N}_{2} \\mathrm{~F}_{4}$\n(b) $\\mathrm{NH}_{2}{ }^{-}$\n(c) $\\mathrm{NF}_{3}$\n(d) $\\mathrm{N}_{3}{ }^{-}$\n61. Explain how ammonia can function both as a Br\u00f8nsted base and as a Lewis base.\n62. Determine the oxidation state of nitrogen in each of the following. You may wish to review the chapter on chemical bonding for relevant examples.\n(a) $\\mathrm{NCl}_{3}$\n(b) ClNO\n(c) $\\mathrm{N}_{2} \\mathrm{O}_{5}$\n(d) $\\mathrm{N}_{2} \\mathrm{O}_{3}$\n(e) $\\mathrm{NO}_{2}{ }^{-}$\n(f) $\\mathrm{N}_{2} \\mathrm{O}_{4}$\n(g) $\\mathrm{N}_{2} \\mathrm{O}$\n(h) $\\mathrm{NO}_{3}{ }^{-}$\n(i) $\\mathrm{HNO}_{2}$\n(j) $\\mathrm{HNO}_{3}$\n63. For each of the following, draw the Lewis structure, predict the ONO bond angle, and give the hybridization of the nitrogen. You may wish to review the chapters on chemical bonding and advanced theories of covalent bonding for relevant examples.\n(a) $\\mathrm{NO}_{2}$\n(b) $\\mathrm{NO}_{2}{ }^{-}$\n(c) $\\mathrm{NO}_{2}{ }^{+}$\n64. How many grams of gaseous ammonia will the reaction of 3.0 g hydrogen gas and 3.0 g of nitrogen gas produce?"}
{"id": 4840, "contents": "1969. Exercises - 1969.7. Occurrence, Preparation, and Properties of Nitrogen\n(c) $\\mathrm{NO}_{2}{ }^{+}$\n64. How many grams of gaseous ammonia will the reaction of 3.0 g hydrogen gas and 3.0 g of nitrogen gas produce?\n65. Although $\\mathrm{PF}_{5}$ and $\\mathrm{AsF}_{5}$ are stable, nitrogen does not form $\\mathrm{NF}_{5}$ molecules. Explain this difference among members of the same group.\n66. The equivalence point for the titration of a $25.00-\\mathrm{mL}$ sample of CsOH solution with $0.1062 \\mathrm{M} \\mathrm{HNO}_{3}$ is at 35.27 mL . What is the concentration of the CsOH solution?"}
{"id": 4841, "contents": "1969. Exercises - 1969.8. Occurrence, Preparation, and Properties of Phosphorus\n67. Write the Lewis structure for each of the following. You may wish to review the chapter on chemical bonding and molecular geometry.\n(a) $\\mathrm{PH}_{3}$\n(b) $\\mathrm{PH}_{4}{ }^{+}$\n(c) $\\mathrm{P}_{2} \\mathrm{H}_{4}$\n(d) $\\mathrm{PO}_{4}{ }^{3-}$\n(e) $\\mathrm{PF}_{5}$\n68. Describe the molecular structure of each of the following molecules or ions listed. You may wish to review the chapter on chemical bonding and molecular geometry.\n(a) $\\mathrm{PH}_{3}$\n(b) $\\mathrm{PH}_{4}{ }^{+}$\n(c) $\\mathrm{P}_{2} \\mathrm{H}_{4}$\n(d) $\\mathrm{PO}_{4}{ }^{3-}$\n69. Complete and balance each of the following chemical equations. (In some cases, there may be more than one correct answer.)\n(a) $\\mathrm{P}_{4}+\\mathrm{Al} \\longrightarrow$\n(b) $\\mathrm{P}_{4}+\\mathrm{Na} \\longrightarrow$\n(c) $\\mathrm{P}_{4}+\\mathrm{F}_{2} \\longrightarrow$\n(d) $\\mathrm{P}_{4}+\\mathrm{Cl}_{2} \\longrightarrow$\n(e) $\\mathrm{P}_{4}+\\mathrm{O}_{2} \\longrightarrow$\n(f) $\\mathrm{P}_{4} \\mathrm{O}_{6}+\\mathrm{O}_{2} \\longrightarrow$\n70. Describe the hybridization of phosphorus in each of the following compounds: $\\mathrm{P}_{4} \\mathrm{O}_{10}, \\mathrm{P}_{4} \\mathrm{O}_{6}, \\mathrm{PH}_{4} \\mathrm{I}$ (an ionic compound), $\\mathrm{PBr}_{3}, \\mathrm{H}_{3} \\mathrm{PO}_{4}, \\mathrm{H}_{3} \\mathrm{PO}_{3}, \\mathrm{PH}_{3}$, and $\\mathrm{P}_{2} \\mathrm{H}_{4}$. You may wish to review the chapter on advanced theories of covalent bonding."}
{"id": 4842, "contents": "1969. Exercises - 1969.8. Occurrence, Preparation, and Properties of Phosphorus\n71. What volume of 0.200 M NaOH is necessary to neutralize the solution produced by dissolving 2.00 g of $\\mathrm{PCl}_{3}$ is an excess of water? Note that when $\\mathrm{H}_{3} \\mathrm{PO}_{3}$ is titrated under these conditions, only one proton of the acid molecule reacts.\n72. How much $\\mathrm{POCl}_{3}$ can form from 25.0 g of $\\mathrm{PCl}_{5}$ and the appropriate amount of $\\mathrm{H}_{2} \\mathrm{O}$ ?\n73. How many tons of $\\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$ are necessary to prepare 5.0 tons of phosphorus if the yield is $90 \\%$ ?\n74. Write equations showing the stepwise ionization of phosphorous acid.\n75. Draw the Lewis structures and describe the geometry for the following:\n(a) $\\mathrm{PF}_{4}{ }^{+}$\n(b) $\\mathrm{PF}_{5}$\n(c) $\\mathrm{PF}_{6}{ }^{-}$\n(d) $\\mathrm{POF}_{3}$\n76. Why does phosphorous acid form only two series of salts, even though the molecule contains three hydrogen atoms?\n77. Assign an oxidation state to phosphorus in each of the following:\n(a) $\\mathrm{NaH}_{2} \\mathrm{PO}_{3}$\n(b) $\\mathrm{PF}_{5}$\n(c) $\\mathrm{P}_{4} \\mathrm{O}_{6}$\n(d) $\\mathrm{K}_{3} \\mathrm{PO}_{4}$\n(e) $\\mathrm{Na}_{3} \\mathrm{P}$\n(f) $\\mathrm{Na}_{4} \\mathrm{P}_{2} \\mathrm{O}_{7}$\n78. Phosphoric acid, one of the acids used in some cola drinks, is produced by the reaction of phosphorus(V) oxide, an acidic oxide, with water. Phosphorus(V) oxide is prepared by the combustion of phosphorus.\n(a) Write the empirical formula of phosphorus(V) oxide.\n(b) What is the molecular formula of phosphorus(V) oxide if the molar mass is about 280."}
{"id": 4843, "contents": "1969. Exercises - 1969.8. Occurrence, Preparation, and Properties of Phosphorus\n(a) Write the empirical formula of phosphorus(V) oxide.\n(b) What is the molecular formula of phosphorus(V) oxide if the molar mass is about 280.\n(c) Write balanced equations for the production of phosphorus(V) oxide and phosphoric acid.\n(d) Determine the mass of phosphorus required to make $1.00 \\times 10^{4} \\mathrm{~kg}$ of phosphoric acid, assuming a yield of $98.85 \\%$."}
{"id": 4844, "contents": "1969. Exercises - 1969.9. Occurrence, Preparation, and Compounds of Oxygen\n79. Predict the product of burning francium in air.\n80. Using equations, describe the reaction of water with potassium and with potassium oxide.\n81. Write balanced chemical equations for the following reactions:\n(a) zinc metal heated in a stream of oxygen gas\n(b) zinc carbonate heated until loss of mass stops\n(c) zinc carbonate added to a solution of acetic acid, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$\n(d) zinc added to a solution of hydrobromic acid\n82. Write balanced chemical equations for the following reactions:\n(a) cadmium burned in air\n(b) elemental cadmium added to a solution of hydrochloric acid\n(c) cadmium hydroxide added to a solution of acetic acid, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$\n83. Illustrate the amphoteric nature of aluminum hydroxide by citing suitable equations.\n84. Write balanced chemical equations for the following reactions:\n(a) metallic aluminum burned in air\n(b) elemental aluminum heated in an atmosphere of chlorine\n(c) aluminum heated in hydrogen bromide gas\n(d) aluminum hydroxide added to a solution of nitric acid\n85. Write balanced chemical equations for the following reactions:\n(a) sodium oxide added to water\n(b) cesium carbonate added to an excess of an aqueous solution of HF\n(c) aluminum oxide added to an aqueous solution of $\\mathrm{HClO}_{4}$\n(d) a solution of sodium carbonate added to solution of barium nitrate\n(e) titanium metal produced from the reaction of titanium tetrachloride with elemental sodium\n86. What volume of $0.250 \\mathrm{MH}_{2} \\mathrm{SO}_{4}$ solution is required to neutralize a solution that contains 5.00 g of $\\mathrm{CaCO}_{3}$ ?\n87. Which is the stronger acid, $\\mathrm{HClO}_{4}$ or $\\mathrm{HBrO}_{4}$ ? Why?\n88. Write a balanced chemical equation for the reaction of an excess of oxygen with each of the following. Remember that oxygen is a strong oxidizing agent and tends to oxidize an element to its maximum oxidation state.\n(a) Mg\n(b) Rb\n(c) Ga"}
{"id": 4845, "contents": "1969. Exercises - 1969.9. Occurrence, Preparation, and Compounds of Oxygen\n(a) Mg\n(b) Rb\n(c) Ga\n(d) $\\mathrm{C}_{2} \\mathrm{H}_{2}$\n(e) CO\n89. Which is the stronger acid, $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ or $\\mathrm{H}_{2} \\mathrm{SeO}_{4}$ ? Why? You may wish to review the chapter on acid-base equilibria."}
{"id": 4846, "contents": "1969. Exercises - 1969.10. Occurrence, Preparation, and Properties of Sulfur\n90. Explain why hydrogen sulfide is a gas at room temperature, whereas water, which has a lower molecular mass, is a liquid.\n91. Give the hybridization and oxidation state for sulfur in $\\mathrm{SO}_{2}$, in $\\mathrm{SO}_{3}$, and in $\\mathrm{H}_{2} \\mathrm{SO}_{4}$.\n92. Which is the stronger acid, $\\mathrm{NaHSO}_{3}$ or $\\mathrm{NaHSO}_{4}$ ?\n93. Determine the oxidation state of sulfur in $\\mathrm{SF}_{6}, \\mathrm{SO}_{2} \\mathrm{~F}_{2}$, and KHS.\n94. Which is a stronger acid, sulfurous acid or sulfuric acid? Why?\n95. Oxygen forms double bonds in $\\mathrm{O}_{2}$, but sulfur forms single bonds in $\\mathrm{S}_{8}$. Why?\n96. Give the Lewis structure of each of the following:\n(a) $\\mathrm{SF}_{4}$\n(b) $\\mathrm{K}_{2} \\mathrm{SO}_{4}$\n(c) $\\mathrm{SO}_{2} \\mathrm{Cl}_{2}$\n(d) $\\mathrm{H}_{2} \\mathrm{SO}_{3}$\n(e) $\\mathrm{SO}_{3}$\n97. Write two balanced chemical equations in which sulfuric acid acts as an oxidizing agent.\n98. Explain why sulfuric acid, $\\mathrm{H}_{2} \\mathrm{SO}_{4}$, which is a covalent molecule, dissolves in water and produces a solution that contains ions.\n99. How many grams of Epsom salts $\\left(\\mathrm{MgSO}_{4} \\cdot 7 \\mathrm{H}_{2} \\mathrm{O}\\right)$ will form from 5.0 kg of magnesium?"}
{"id": 4847, "contents": "1969. Exercises - 1969.11. Occurrence, Preparation, and Properties of Halogens\n100. What does it mean to say that mercury(II) halides are weak electrolytes?\n101. Why is $\\mathrm{SnCl}_{4}$ not classified as a salt?\n102. The following reactions are all similar to those of the industrial chemicals. Complete and balance the equations for these reactions:\n(a) reaction of a weak base and a strong acid\n$\\mathrm{NH}_{3}+\\mathrm{HClO}_{4} \\longrightarrow$\n(b) preparation of a soluble silver salt for silver plating\n$\\mathrm{Ag}_{2} \\mathrm{CO}_{3}+\\mathrm{HNO}_{3} \\longrightarrow$\n(c) preparation of strontium hydroxide by electrolysis of a solution of strontium chloride\n$\\mathrm{SrCl}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\xrightarrow{\\text { electrolysis }}$\n103. Which is the stronger acid, $\\mathrm{HClO}_{3}$ or $\\mathrm{HBrO}_{3}$ ? Why?\n104. What is the hybridization of iodine in $\\mathrm{IF}_{3}$ and $\\mathrm{IF}_{5}$ ?\n105. Predict the molecular geometries and draw Lewis structures for each of the following. You may wish to review the chapter on chemical bonding and molecular geometry.\n(a) $\\mathrm{IF}_{5}$\n(b) $\\mathrm{I}_{3}{ }^{-}$\n(c) $\\mathrm{PCl}_{5}$\n(d) $\\mathrm{SeF}_{4}$\n(e) $\\mathrm{ClF}_{3}$\n106. Which halogen has the highest ionization energy? Is this what you would predict based on what you have learned about periodic properties?\n107. Name each of the following compounds:\n(a) $\\mathrm{BrF}_{3}$\n(b) $\\mathrm{NaBrO}_{3}$\n(c) $\\mathrm{PBr}_{5}$\n(d) $\\mathrm{NaClO}_{4}$\n(e) KClO\n108. Explain why, at room temperature, fluorine and chlorine are gases, bromine is a liquid, and iodine is a solid.\n109. What is the oxidation state of the halogen in each of the following?\n(a) $\\mathrm{H}_{5} \\mathrm{IO}_{6}$"}
{"id": 4848, "contents": "1969. Exercises - 1969.11. Occurrence, Preparation, and Properties of Halogens\n109. What is the oxidation state of the halogen in each of the following?\n(a) $\\mathrm{H}_{5} \\mathrm{IO}_{6}$\n(b) $\\mathrm{IO}_{4}^{-}$\n(c) $\\mathrm{ClO}_{2}$\n(d) $\\mathrm{ICl}_{3}$\n(e) $\\mathrm{F}_{2}$\n110. Physiological saline concentration-that is, the sodium chloride concentration in our bodies-is approximately 0.16 M . A saline solution for contact lenses is prepared to match the physiological concentration. If you purchase 25 mL of contact lens saline solution, how many grams of sodium chloride have you bought?"}
{"id": 4849, "contents": "1969. Exercises - 1969.12. Occurrence, Preparation, and Properties of the Noble Gases\n111. Give the hybridization of xenon in each of the following. You may wish to review the chapter on the advanced theories of covalent bonding.\n(a) $\\mathrm{XeF}_{2}$\n(b) $\\mathrm{XeF}_{4}$\n(c) $\\mathrm{XeO}_{3}$\n(d) $\\mathrm{XeO}_{4}$\n(e) $\\mathrm{XeOF}_{4}$\n112. What is the molecular structure of each of the following molecules? You may wish to review the chapter on chemical bonding and molecular geometry.\n(a) $\\mathrm{XeF}_{2}$\n(b) $\\mathrm{XeF}_{4}$\n(c) $\\mathrm{XeO}_{3}$\n(d) $\\mathrm{XeO}_{4}$\n(e) $\\mathrm{XeOF}_{4}$\n113. Indicate whether each of the following molecules is polar or nonpolar. You may wish to review the chapter on chemical bonding and molecular geometry.\n(a) $\\mathrm{XeF}_{2}$\n(b) $\\mathrm{XeF}_{4}$\n(c) $\\mathrm{XeO}_{3}$\n(d) $\\mathrm{XeO}_{4}$\n(e) $\\mathrm{XeOF}_{4}$\n114. What is the oxidation state of the noble gas in each of the following? You may wish to review the chapter on chemical bonding and molecular geometry.\n(a) $\\mathrm{XeO}_{2} \\mathrm{~F}_{2}$\n(b) $\\mathrm{KrF}_{2}$\n(c) $\\mathrm{XeF}_{3}{ }^{+}$\n(d) $\\mathrm{XeO}_{6}{ }^{4-}$\n(e) $\\mathrm{XeO}_{3}$\n115. A mixture of xenon and fluorine was heated. A sample of the white solid that formed reacted with hydrogen to yield 81 mL of xenon (at STP) and hydrogen fluoride, which was collected in water, giving a solution of hydrofluoric acid. The hydrofluoric acid solution was titrated, and 68.43 mL of 0.3172 M sodium hydroxide was required to reach the equivalence point. Determine the empirical formula for the white solid and write balanced chemical equations for the reactions involving xenon."}
{"id": 4850, "contents": "1969. Exercises - 1969.12. Occurrence, Preparation, and Properties of the Noble Gases\n116. Basic solutions of $\\mathrm{Na}_{4} \\mathrm{XeO}_{6}$ are powerful oxidants. What mass of $\\mathrm{Mn}\\left(\\mathrm{NO}_{3}\\right)_{2} \\cdot 6 \\mathrm{H}_{2} \\mathrm{O}$ reacts with 125.0 mL of a 0.1717 M basic solution of $\\mathrm{Na}_{4} \\mathrm{XeO}_{6}$ that contains an excess of sodium hydroxide if the products include Xe and solution of sodium permanganate?"}
{"id": 4851, "contents": "1970. CHAPTER 19 Transition Metals and Coordination Chemistry - \nFigure 19.1 Transition metals often form vibrantly colored complexes. The minerals malachite (green), azurite (blue), and proustite (red) are some examples. (credit left: modification of work by James St. John; credit middle: modification of work by Stephanie Clifford; credit right: modification of work by Terry Wallace)"}
{"id": 4852, "contents": "1971. CHAPTER OUTLINE - 1971.1. Occurrence, Preparation, and Properties of Transition Metals and Their Compounds 19.2 Coordination Chemistry of Transition Metals 19.3 Spectroscopic and Magnetic Properties of Coordination Compounds\nINTRODUCTION We have daily contact with many transition metals. Iron occurs everywhere-from the rings in your spiral notebook and the cutlery in your kitchen to automobiles, ships, buildings, and in the hemoglobin in your blood. Titanium is useful in the manufacture of lightweight, durable products such as bicycle frames, artificial hips, and jewelry. Chromium is useful as a protective plating on plumbing fixtures and automotive detailing.\n\nIn addition to being used in their pure elemental forms, many compounds containing transition metals have numerous other applications. Silver nitrate is used to create mirrors, zirconium silicate provides friction in automotive brakes, and many important cancer-fighting agents, like the drug cisplatin and related species, are platinum compounds.\n\nThe variety of properties exhibited by transition metals is due to their complex valence shells. Unlike most main group metals where one oxidation state is normally observed, the valence shell structure of transition metals means that they usually occur in several different stable oxidation states. In addition, electron transitions in these elements can correspond with absorption of photons in the visible electromagnetic spectrum, leading to colored compounds. Because of these behaviors, transition metals exhibit a rich and fascinating chemistry."}
{"id": 4853, "contents": "1971. CHAPTER OUTLINE - 1971.2. Occurrence, Preparation, and Properties of Transition Metals and Their Compounds\nTransition metals are defined as those elements that have (or readily form) partially filled $d$ orbitals. As shown in Figure 19.2, the $\\boldsymbol{d}$-block elements in groups 3-11 are transition elements. The $\\boldsymbol{f}$-block elements, also called inner transition metals (the lanthanides and actinides), also meet this criterion because the $d$ orbital is partially occupied before the $f$ orbitals. The $d$ orbitals fill with the copper family (group 11); for this reason, the\nnext family (group 12) are technically not transition elements. However, the group 12 elements do display some of the same chemical properties and are commonly included in discussions of transition metals. Some chemists do treat the group 12 elements as transition metals.\n\n\nFIGURE 19.2 The transition metals are located in groups 3-11 of the periodic table. The inner transition metals are in the two rows below the body of the table.\n\nThe $d$-block elements are divided into the first transition series (the elements Sc through Cu ), the second transition series (the elements Y through Ag), and the third transition series (the element La and the elements Hf through Au). Actinium, Ac, is the first member of the fourth transition series, which also includes Rf through Rg.\n\nThe $f$-block elements are the elements Ce through Lu, which constitute the lanthanide series (or lanthanoid series), and the elements Th through Lr, which constitute the actinide series (or actinoid series). Because lanthanum behaves very much like the lanthanide elements, it is considered a lanthanide element, even though its electron configuration makes it the first member of the third transition series. Similarly, the behavior of actinium means it is part of the actinide series, although its electron configuration makes it the first member of the fourth transition series."}
{"id": 4854, "contents": "1973. Valence Electrons in Transition Metals - \nReview how to write electron configurations, covered in the chapter on electronic structure and periodic properties of elements. Recall that for the transition and inner transition metals, it is necessary to remove the $s$\nelectrons before the $d$ or $f$ electrons. Then, for each ion, give the electron configuration:\n(a) cerium(III)\n(b) lead(II)\n(c) $\\mathrm{Ti}^{2+}$\n(d) $\\mathrm{Am}^{3+}$\n(e) $\\mathrm{Pd}^{2+}$\n\nFor the examples that are transition metals, determine to which series they belong."}
{"id": 4855, "contents": "1974. Solution - \nFor ions, the $s$-valence electrons are lost prior to the $d$ or $f$ electrons.\n(a) $\\mathrm{Ce}^{3+}[\\mathrm{Xe}] 4 f^{1} ; \\mathrm{Ce}^{3+}$ is an inner transition element in the lanthanide series.\n(b) $\\mathrm{Pb}^{2+}[\\mathrm{Xe}] 6 s^{2} 5 d^{10} 4 f^{14}$; the electrons are lost from the $p$ orbital. This is a main group element.\n(c) titanium(II) $[\\mathrm{Ar}] 3 d^{2}$; first transition series\n(d) americium(III) $[\\mathrm{Rn}] 5 f^{6}$; actinide\n(e) palladium(II) $[\\mathrm{Kr}] 4 d^{8}$; second transition series"}
{"id": 4856, "contents": "1975. Check Your Learning - \nGive an example of an ion from the first transition series with no $d$ electrons."}
{"id": 4857, "contents": "1976. Answer: - \n$\\mathrm{V}^{5+}$ is one possibility. Other examples include $\\mathrm{Sc}^{3+}, \\mathrm{Ti}^{4+}, \\mathrm{Cr}^{6+}$, and $\\mathrm{Mn}^{7+}$."}
{"id": 4858, "contents": "1978. Uses of Lanthanides in Devices - \nLanthanides (elements 57-71) are fairly abundant in the earth's crust, despite their historic characterization as rare earth elements. Thulium, the rarest naturally occurring lanthanoid, is more common in the earth's crust than silver ( $4.5 \\times 10^{-5} \\%$ versus $0.79 \\times 10^{-5} \\%$ by mass). There are 17 rare earth elements, consisting of the 15 lanthanoids plus scandium and yttrium. They are called rare because they were once difficult to extract economically, so it was rare to have a pure sample; due to similar chemical properties, it is difficult to separate any one lanthanide from the others. However, newer separation methods, such as ion exchange resins similar to those found in home water softeners, make the separation of these elements easier and more economical. Most ores that contain these elements have low concentrations of all the rare earth elements mixed together.\n\nThe commercial applications of lanthanides are growing rapidly. For example, europium is important in flat screen displays found in computer monitors, cell phones, and televisions. Neodymium is useful in laptop hard drives and in the processes that convert crude oil into gasoline (Figure 19.3). Holmium is found in dental and medical equipment. In addition, many alternative energy technologies rely heavily on lanthanoids. Neodymium and dysprosium are key components of hybrid vehicle engines and the magnets used in wind turbines.\n\n\nFIGURE 19.3 (a) Europium is used in display screens for televisions, computer monitors, and cell phones. (b) Neodymium magnets are commonly found in computer hard drives. (credit b: modification of work by \"KUERT Datenrettung\"/Flickr)\n\nAs the demand for lanthanide materials has increased faster than supply, prices have also increased. In 2008 , dysprosium cost $\\$ 110 / \\mathrm{kg}$; by 2014 , the price had increased to $\\$ 470 / \\mathrm{kg}$. Increasing the supply of lanthanoid elements is one of the most significant challenges facing the industries that rely on the optical and magnetic properties of these materials."}
{"id": 4859, "contents": "1978. Uses of Lanthanides in Devices - \nThe transition elements have many properties in common with other metals. They are almost all hard, highmelting solids that conduct heat and electricity well. They readily form alloys and lose electrons to form stable cations. In addition, transition metals form a wide variety of stable coordination compounds, in which the central metal atom or ion acts as a Lewis acid and accepts one or more pairs of electrons. Many different molecules and ions can donate lone pairs to the metal center, serving as Lewis bases. In this chapter, we shall focus primarily on the chemical behavior of the elements of the first transition series."}
{"id": 4860, "contents": "1979. Properties of the Transition Elements - \nTransition metals demonstrate a wide range of chemical behaviors. As can be seen from their reduction potentials (see Appendix H), some transition metals are strong reducing agents, whereas others have very low reactivity. For example, the lanthanides all form stable 3+ aqueous cations. The driving force for such oxidations is similar to that of alkaline earth metals such as Be or Mg , forming $\\mathrm{Be}^{2+}$ and $\\mathrm{Mg}^{2+}$. On the other hand, materials like platinum and gold have much higher reduction potentials. Their ability to resist oxidation makes them useful materials for constructing circuits and jewelry.\n\nIons of the lighter $d$-block elements, such as $\\mathrm{Cr}^{3+}, \\mathrm{Fe}^{3+}$, and $\\mathrm{Co}^{2+}$, form colorful hydrated ions that are stable in water. However, ions in the period just below these $\\left(\\mathrm{Mo}^{3+}, \\mathrm{Ru}^{3+}\\right.$, and $\\mathrm{Ir}^{2+}$ ) are unstable and react readily with oxygen from the air. The majority of simple, water-stable ions formed by the heavier $d$-block elements are oxyanions such as $\\mathrm{MoO}_{4}{ }^{2-}$ and $\\mathrm{ReO}_{4}{ }^{-}$.\n\nRuthenium, osmium, rhodium, iridium, palladium, and platinum are the platinum metals. With difficulty, they form simple cations that are stable in water, and, unlike the earlier elements in the second and third transition series, they do not form stable oxyanions.\nBoth the $d$ - and $f$-block elements react with nonmetals to form binary compounds; heating is often required. These elements react with halogens to form a variety of halides ranging in oxidation state from $1+$ to $6+$. On heating, oxygen reacts with all of the transition elements except palladium, platinum, silver, and gold. The oxides of these latter metals can be formed using other reactants, but they decompose upon heating. The $f$-block elements, the elements of group 3, and the elements of the first transition series except copper react with aqueous solutions of acids, forming hydrogen gas and solutions of the corresponding salts."}
{"id": 4861, "contents": "1979. Properties of the Transition Elements - \nTransition metals can form compounds with a wide range of oxidation states. Some of the observed oxidation states of the elements of the first transition series are shown in Figure 19.4. As we move from left to right across the first transition series, we see that the number of common oxidation states increases at first to a maximum towards the middle of the table, then decreases. The values in the table are typical values; there are other known values, and it is possible to synthesize new additions. For example, in 2014, researchers were successful in synthesizing a new oxidation state of iridium (9+).\n\n| ${ }^{21} \\mathbf{S c}$ | ${ }^{22} \\mathrm{Ti}$ | ${ }^{23} \\mathrm{~V}$ | ${ }^{24} \\mathrm{Cr}$ | ${ }^{25} \\mathrm{Mn}$ | ${ }^{26} \\mathrm{Fe}$ | ${ }^{27} \\mathrm{Co}$ | ${ }^{28} \\mathrm{Ni}$ | ${ }^{29} \\mathrm{Cu}$ | ${ }^{30} \\mathrm{Zn}$ |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|  |  |  |  |  |  |  |  | $1+$ |  |\n|  |  | $2+$ | ${ }^{2+}$ | $2+$ | ${ }^{2+}$ | $2+$ | ${ }^{2+}$ | $2+$ | 2+ |\n| $3+$ | $3+$ | $3+$ | $3+$ | $3+$ | $3+$ | $3+$ | $3+$ | $3+$ |  |\n|  | $4+$ | $4+$ | $4+$ | $4+$ |  |  |  |  |  |\n|  |  | $5+$ |  |  |  |  |  |  |  |\n|  |  |  | 6+ | ${ }^{6+}$ | $6+$ |  |  |  |  |\n|  |  |  |  | $7+$ |  |  |  |  |  |"}
{"id": 4862, "contents": "1979. Properties of the Transition Elements - \nFIGURE 19.4 Transition metals of the first transition series can form compounds with varying oxidation states.\nFor the elements scandium through manganese (the first half of the first transition series), the highest oxidation state corresponds to the loss of all of the electrons in both the $s$ and $d$ orbitals of their valence shells. The titanium(IV) ion, for example, is formed when the titanium atom loses its two $3 d$ and two $4 s$ electrons. These highest oxidation states are the most stable forms of scandium, titanium, and vanadium. However, it is not possible to continue to remove all of the valence electrons from metals as we continue through the series. Iron is known to form oxidation states from $2+$ to $6+$, with iron(II) and iron(III) being the most common. Most of the elements of the first transition series form ions with a charge of $2+$ or $3+$ that are stable in water, although those of the early members of the series can be readily oxidized by air.\n\nThe elements of the second and third transition series generally are more stable in higher oxidation states than are the elements of the first series. In general, the atomic radius increases down a group, which leads to the ions of the second and third series being larger than are those in the first series. Removing electrons from orbitals that are located farther from the nucleus is easier than removing electrons close to the nucleus. For example, molybdenum and tungsten, members of group 6, are limited mostly to an oxidation state of 6+ in aqueous solution. Chromium, the lightest member of the group, forms stable $\\mathrm{Cr}^{3+}$ ions in water and, in the absence of air, less stable $\\mathrm{Cr}^{2+}$ ions. The sulfide with the highest oxidation state for chromium is $\\mathrm{Cr}_{2} \\mathrm{~S}_{3}$, which contains the $\\mathrm{Cr}^{3+}$ ion. Molybdenum and tungsten form sulfides in which the metals exhibit oxidation states of $4+$ and 6+."}
{"id": 4863, "contents": "1981. Activity of the Transition Metals - \nWhich is the strongest oxidizing agent in acidic solution: dichromate ion, which contains chromium(VI), permanganate ion, which contains manganese(VII), or titanium dioxide, which contains titanium(IV)?"}
{"id": 4864, "contents": "1982. Solution - \nFirst, we need to look up the reduction half reactions (in Appendix L) for each oxide in the specified oxidation state:\n\n$$\n\\begin{array}{cc}\n\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+14 \\mathrm{H}^{+}+6 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Cr}^{3+}+7 \\mathrm{H}_{2} \\mathrm{O} & +1.33 \\mathrm{~V} \\\\\n\\mathrm{MnO}_{4}^{-}+8 \\mathrm{H}^{+}+5 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Mn}^{2+}+\\mathrm{H}_{2} \\mathrm{O} & +1.51 \\mathrm{~V} \\\\\n\\mathrm{TiO}_{2}+4 \\mathrm{H}^{+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Ti}^{2+}+2 \\mathrm{H}_{2} \\mathrm{O} & -0.50 \\mathrm{~V}\n\\end{array}\n$$\n\nA larger reduction potential means that it is easier to reduce the reactant. Permanganate, with the largest reduction potential, is the strongest oxidizer under these conditions. Dichromate is next, followed by titanium dioxide as the weakest oxidizing agent (the hardest to reduce) of this set."}
{"id": 4865, "contents": "1983. Check Your Learning - \nPredict what reaction (if any) will occur between HCl and $\\mathrm{Co}(s)$, and between HBr and $\\mathrm{Pt}(s)$. You will need to use\nthe standard reduction potentials from Appendix L."}
{"id": 4866, "contents": "1984. Answer: - \n$\\mathrm{Co}(s)+2 \\mathrm{HCl} \\longrightarrow \\mathrm{H}_{2}+\\mathrm{CoCl}_{2}(a q) ;$ no reaction because $\\mathrm{Pt}(s)$ will not be oxidized by $\\mathrm{H}^{+}$"}
{"id": 4867, "contents": "1985. Preparation of the Transition Elements - \nAncient civilizations knew about iron, copper, silver, and gold. The time periods in human history known as the Bronze Age and Iron Age mark the advancements in which societies learned to isolate certain metals and use them to make tools and goods. Naturally occurring ores of copper, silver, and gold can contain high concentrations of these metals in elemental form (Figure 19.5). Iron, on the other hand, occurs on earth almost exclusively in oxidized forms, such as rust $\\left(\\mathrm{Fe}_{2} \\mathrm{O}_{3}\\right)$. The earliest known iron implements were made from iron meteorites. Surviving iron artifacts dating from approximately 4000 to 2500 BC are rare, but all known examples contain specific alloys of iron and nickel that occur only in extraterrestrial objects, not on earth. It took thousands of years of technological advances before civilizations developed iron smelting, the ability to extract a pure element from its naturally occurring ores and for iron tools to become common.\n\n(a)\n\n(b)\n\n(c)\n\nFIGURE 19.5 Transition metals occur in nature in various forms. Examples include (a) a nugget of copper, (b) a deposit of gold, and (c) an ore containing oxidized iron. (credit a: modification of work by http://images-of-elements.com/copper-2.jpg; credit c: modification of work by http://images-of-elements.com/iron-ore.jpg)\n\nGenerally, the transition elements are extracted from minerals found in a variety of ores. However, the ease of their recovery varies widely, depending on the concentration of the element in the ore, the identity of the other elements present, and the difficulty of reducing the element to the free metal.\n\nIn general, it is not difficult to reduce ions of the $d$-block elements to the free element. Carbon is a sufficiently strong reducing agent in most cases. However, like the ions of the more active main group metals, ions of the $f$-block elements must be isolated by electrolysis or by reduction with an active metal such as calcium.\n\nWe shall discuss the processes used for the isolation of iron, copper, and silver because these three processes illustrate the principal means of isolating most of the $d$-block metals. In general, each of these processes involves three principal steps: preliminary treatment, smelting, and refining."}
{"id": 4868, "contents": "1985. Preparation of the Transition Elements - \n1. Preliminary treatment. In general, there is an initial treatment of the ores to make them suitable for the extraction of the metals. This usually involves crushing or grinding the ore, concentrating the metalbearing components, and sometimes treating these substances chemically to convert them into compounds that are easier to reduce to the metal.\n2. Smelting. The next step is the extraction of the metal in the molten state, a process called smelting, which includes reduction of the metallic compound to the metal. Impurities may be removed by the addition of a compound that forms a slag-a substance with a low melting point that can be readily separated from the molten metal.\n3. Refining. The final step in the recovery of a metal is refining the metal. Low boiling metals such as zinc and mercury can be refined by distillation. When fused on an inclined table, low melting metals like tin flow away from higher-melting impurities. Electrolysis is another common method for refining metals."}
{"id": 4869, "contents": "1986. Isolation of Iron - \nThe early application of iron to the manufacture of tools and weapons was possible because of the wide distribution of iron ores and the ease with which iron compounds in the ores could be reduced by carbon. For a long time, charcoal was the form of carbon used in the reduction process. The production and use of iron became much more widespread about 1620, when coke was introduced as the reducing agent. Coke is a form of carbon formed by heating coal in the absence of air to remove impurities.\n\nThe first step in the metallurgy of iron is usually roasting the ore (heating the ore in air) to remove water, decomposing carbonates into oxides, and converting sulfides into oxides. The oxides are then reduced in a blast furnace that is $80-100$ feet high and about 25 feet in diameter (Figure 19.6) in which the roasted ore, coke, and limestone (impure $\\mathrm{CaCO}_{3}$ ) are introduced continuously into the top. Molten iron and slag are withdrawn at the bottom. The entire stock in a furnace may weigh several hundred tons.\n\n\nFIGURE 19.6 Within a blast furnace, different reactions occur in different temperature zones. Carbon monoxide is generated in the hotter bottom regions and rises upward to reduce the iron oxides to pure iron through a series of reactions that take place in the upper regions.\n\nNear the bottom of a furnace are nozzles through which preheated air is blown into the furnace. As soon as the air enters, the coke in the region of the nozzles is oxidized to carbon dioxide with the liberation of a great deal of heat. The hot carbon dioxide passes upward through the overlying layer of white-hot coke, where it is reduced to carbon monoxide:\n\n$$\n\\mathrm{CO}_{2}(g)+\\mathrm{C}(s) \\longrightarrow 2 \\mathrm{CO}(g)\n$$\n\nThe carbon monoxide serves as the reducing agent in the upper regions of the furnace. The individual reactions are indicated in Figure 19.6.\n\nThe iron oxides are reduced in the upper region of the furnace. In the middle region, limestone (calcium carbonate) decomposes, and the resulting calcium oxide combines with silica and silicates in the ore to form\nslag. The slag is mostly calcium silicate and contains most of the commercially unimportant components of the ore:"}
{"id": 4870, "contents": "1986. Isolation of Iron - \n$$\n\\mathrm{CaO}(s)+\\mathrm{SiO}_{2}(s) \\longrightarrow \\mathrm{CaSiO}_{3}(l)\n$$\n\nJust below the middle of the furnace, the temperature is high enough to melt both the iron and the slag. They collect in layers at the bottom of the furnace; the less dense slag floats on the iron and protects it from oxidation. Several times a day, the slag and molten iron are withdrawn from the furnace. The iron is transferred to casting machines or to a steelmaking plant (Figure 19.7).\n\n\nFIGURE 19.7 Molten iron is shown being cast as steel. (credit: Clint Budd)\nMuch of the iron produced is refined and converted into steel. Steel is made from iron by removing impurities and adding substances such as manganese, chromium, nickel, tungsten, molybdenum, and vanadium to produce alloys with properties that make the material suitable for specific uses. Most steels also contain small but definite percentages of carbon ( $0.04 \\%-2.5 \\%$ ). However, a large part of the carbon contained in iron must be removed in the manufacture of steel; otherwise, the excess carbon would make the iron brittle."}
{"id": 4871, "contents": "1987. LINK TO LEARNING - \nYou can watch an animation of steelmaking (http://openstax.org/l/16steelmaking) that walks you through the process."}
{"id": 4872, "contents": "1988. Isolation of Copper - \nThe most important ores of copper contain copper sulfides (such as covellite, CuS), although copper oxides (such as tenorite, CuO ) and copper hydroxycarbonates [such as malachite, $\\mathrm{Cu}_{2}(\\mathrm{OH})_{2} \\mathrm{CO}_{3}$ ] are sometimes found. In the production of copper metal, the concentrated sulfide ore is roasted to remove part of the sulfur as sulfur dioxide. The remaining mixture, which consists of $\\mathrm{Cu}_{2} \\mathrm{~S}, \\mathrm{FeS}, \\mathrm{FeO}$, and $\\mathrm{SiO}_{2}$, is mixed with limestone, which serves as a flux (a material that aids in the removal of impurities), and heated. Molten slag forms as the iron and silica are removed by Lewis acid-base reactions:\n\n$$\n\\begin{gathered}\n\\mathrm{CaCO}_{3}(s)+\\mathrm{SiO}_{2}(s) \\longrightarrow \\mathrm{CaSiO}_{3}(l)+\\mathrm{CO}_{2}(\\mathrm{~g}) \\\\\n\\mathrm{FeO}(s)+\\mathrm{SiO}_{2}(s) \\longrightarrow \\mathrm{FeSiO}_{3}(l)\n\\end{gathered}\n$$\n\nIn these reactions, the silicon dioxide behaves as a Lewis acid, which accepts a pair of electrons from the Lewis base (the oxide ion).\n\nReduction of the $\\mathrm{Cu}_{2} \\mathrm{~S}$ that remains after smelting is accomplished by blowing air through the molten material.\n\nThe air converts part of the $\\mathrm{Cu}_{2} \\mathrm{~S}$ into $\\mathrm{Cu}_{2} \\mathrm{O}$. As soon as copper(I) oxide is formed, it is reduced by the remaining copper(I) sulfide to metallic copper:"}
{"id": 4873, "contents": "1988. Isolation of Copper - \n$$\n\\begin{gathered}\n2 \\mathrm{Cu}_{2} \\mathrm{~S}(l)+3 \\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{Cu}_{2} \\mathrm{O}(l)+2 \\mathrm{SO}_{2}(g) \\\\\n2 \\mathrm{Cu}_{2} \\mathrm{O}(l)+\\mathrm{Cu}_{2} \\mathrm{~S}(l) \\longrightarrow 6 \\mathrm{Cu}(l)+\\mathrm{SO}_{2}(g)\n\\end{gathered}\n$$\n\nThe copper obtained in this way is called blister copper because of its characteristic appearance, which is due to the air blisters it contains (Figure 19.8). This impure copper is cast into large plates, which are used as anodes in the electrolytic refining of the metal (which is described in the chapter on electrochemistry).\n\n\nFIGURE 19.8 Blister copper is obtained during the conversion of copper-containing ore into pure copper. (credit: \"Tortie tude\"/Wikimedia Commons)"}
{"id": 4874, "contents": "1989. Isolation of Silver - \nSilver sometimes occurs in large nuggets (Figure 19.9) but more frequently in veins and related deposits. At one time, panning was an effective method of isolating both silver and gold nuggets. Due to their low reactivity, these metals, and a few others, occur in deposits as nuggets. The discovery of platinum was due to Spanish explorers in Central America mistaking platinum nuggets for silver. When the metal is not in the form of nuggets, it often useful to employ a process called hydrometallurgy to separate silver from its ores. Hydrology involves the separation of a metal from a mixture by first converting it into soluble ions and then extracting and reducing them to precipitate the pure metal. In the presence of air, alkali metal cyanides readily form the soluble dicyanoargentate(I) ion, $\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]^{-}$, from silver metal or silver-containing compounds such as $\\mathrm{Ag}_{2} \\mathrm{~S}$ and AgCl . Representative equations are:\n\n$$\n\\begin{gathered}\n4 \\mathrm{Ag}(s)+8 \\mathrm{CN}^{-}(a q)+\\mathrm{O}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 4\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]^{-}(a q)+4 \\mathrm{OH}^{-}(a q) \\\\\n2 \\mathrm{Ag}_{2} \\mathrm{~S}(s)+8 \\mathrm{CN}^{-}(a q)+\\mathrm{O}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 4\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]^{-}(a q)+2 \\mathrm{~S}(s)+4 \\mathrm{OH}^{-}(a q) \\\\\n\\mathrm{AgCl}(s)+2 \\mathrm{CN}^{-}(a q) \\longrightarrow\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]^{-}(a q)+\\mathrm{Cl}^{-}(a q)\n\\end{gathered}\n$$"}
{"id": 4875, "contents": "1989. Isolation of Silver - \nFIGURE 19.9 Naturally occurring free silver may be found as nuggets (a) or in veins (b). (credit a: modification of work by \"Teravolt\"/Wikimedia Commons; credit b: modification of work by James St. John)\n\nThe silver is precipitated from the cyanide solution by the addition of either zinc or iron(II) ions, which serves as the reducing agent:\n\n$$\n2\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]^{-}(a q)+\\mathrm{Zn}(s) \\longrightarrow 2 \\mathrm{Ag}(s)+\\left[\\mathrm{Zn}(\\mathrm{CN})_{4}\\right]^{2-}(a q)\n$$"}
{"id": 4876, "contents": "1991. Refining Redox - \nOne of the steps for refining silver involves converting silver into dicyanoargenate(I) ions:\n\n$$\n4 \\mathrm{Ag}(s)+8 \\mathrm{CN}^{-}(a q)+\\mathrm{O}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 4\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]^{-}(a q)+4 \\mathrm{OH}^{-}(a q)\n$$\n\nExplain why oxygen must be present to carry out the reaction. Why does the reaction not occur as:\n\n$$\n4 \\mathrm{Ag}(s)+8 \\mathrm{CN}^{-}(a q) \\longrightarrow 4\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]^{-}(a q) ?\n$$"}
{"id": 4877, "contents": "1992. Solution - \nThe charges, as well as the atoms, must balance in reactions. The silver atom is being oxidized from the 0 oxidation state to the $1+$ state. Whenever something loses electrons, something must also gain electrons (be reduced) to balance the equation. Oxygen is a good oxidizing agent for these reactions because it can gain electrons to go from the 0 oxidation state to the $2-$ state."}
{"id": 4878, "contents": "1993. Check Your Learning - \nDuring the refining of iron, carbon must be present in the blast furnace. Why is carbon necessary to convert iron oxide into iron?"}
{"id": 4879, "contents": "1994. Answer: - \nThe carbon is converted into CO, which is the reducing agent that accepts electrons so that iron(III) can be reduced to iron(0)."}
{"id": 4880, "contents": "1995. Transition Metal Compounds - \nThe bonding in the simple compounds of the transition elements ranges from ionic to covalent. In their lower oxidation states, the transition elements form ionic compounds; in their higher oxidation states, they form covalent compounds or polyatomic ions. The variation in oxidation states exhibited by the transition elements gives these compounds a metal-based, oxidation-reduction chemistry. The chemistry of several classes of compounds containing elements of the transition series follows."}
{"id": 4881, "contents": "1996. Halides - \nAnhydrous halides of each of the transition elements can be prepared by the direct reaction of the metal with halogens. For example:\n\n$$\n2 \\mathrm{Fe}(s)+3 \\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{FeCl}_{3}(s)\n$$\n\nHeating a metal halide with additional metal can be used to form a halide of the metal with a lower oxidation\nstate:\n\n$$\n\\mathrm{Fe}(s)+2 \\mathrm{FeCl}_{3}(s) \\longrightarrow 3 \\mathrm{FeCl}_{2}(s)\n$$\n\nThe stoichiometry of the metal halide that results from the reaction of the metal with a halogen is determined by the relative amounts of metal and halogen and by the strength of the halogen as an oxidizing agent. Generally, fluorine forms fluoride-containing metals in their highest oxidation states. The other halogens may not form analogous compounds.\n\nIn general, the preparation of stable water solutions of the halides of the metals of the first transition series is by the addition of a hydrohalic acid to carbonates, hydroxides, oxides, or other compounds that contain basic anions. Sample reactions are:\n\n$$\n\\begin{gathered}\n\\mathrm{NiCO}_{3}(s)+2 \\mathrm{HF}(a q) \\longrightarrow \\mathrm{NiF}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{CO}_{2}(g) \\\\\n\\mathrm{Co}(\\mathrm{OH})_{2}(s)+2 \\mathrm{HBr}(a q) \\longrightarrow \\mathrm{CoBr}_{2}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l)\n\\end{gathered}\n$$\n\nMost of the first transition series metals also dissolve in acids, forming a solution of the salt and hydrogen gas. For example:\n\n$$\n\\mathrm{Cr}(s)+2 \\mathrm{HCl}(a q) \\longrightarrow \\mathrm{CrCl}_{2}(a q)+\\mathrm{H}_{2}(g)\n$$"}
{"id": 4882, "contents": "1996. Halides - \n$$\n\\mathrm{Cr}(s)+2 \\mathrm{HCl}(a q) \\longrightarrow \\mathrm{CrCl}_{2}(a q)+\\mathrm{H}_{2}(g)\n$$\n\nThe polarity of bonds with transition metals varies based not only upon the electronegativities of the atoms involved but also upon the oxidation state of the transition metal. Remember that bond polarity is a continuous spectrum with electrons being shared evenly (covalent bonds) at one extreme and electrons being transferred completely (ionic bonds) at the other. No bond is ever 100\\% ionic, and the degree to which the electrons are evenly distributed determines many properties of the compound. Transition metal halides with low oxidation numbers form more ionic bonds. For example, titanium(II) chloride and titanium(III) chloride ( $\\mathrm{TiCl}_{2}$ and $\\mathrm{TiCl}_{3}$ ) have high melting points that are characteristic of ionic compounds, but titanium(IV) chloride ( $\\mathrm{TiCl}_{4}$ ) is a volatile liquid, consistent with having covalent titanium-chlorine bonds. All halides of the heavier $d$-block elements have significant covalent characteristics.\n\nThe covalent behavior of the transition metals with higher oxidation states is exemplified by the reaction of the metal tetrahalides with water. Like covalent silicon tetrachloride, both the titanium and vanadium tetrahalides react with water to give solutions containing the corresponding hydrohalic acids and the metal oxides:\n\n$$\n\\begin{aligned}\n& \\mathrm{SiCl}_{4}(l)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{SiO}_{2}(s)+4 \\mathrm{HCl}(a q) \\\\\n& \\mathrm{TiCl}_{4}(l)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{TiO}_{2}(s)+4 \\mathrm{HCl}(a q)\n\\end{aligned}\n$$"}
{"id": 4883, "contents": "1997. Oxides - \nAs with the halides, the nature of bonding in oxides of the transition elements is determined by the oxidation state of the metal. Oxides with low oxidation states tend to be more ionic, whereas those with higher oxidation states are more covalent. These variations in bonding are because the electronegativities of the elements are not fixed values. The electronegativity of an element increases with increasing oxidation state. Transition metals in low oxidation states have lower electronegativity values than oxygen; therefore, these metal oxides are ionic. Transition metals in very high oxidation states have electronegativity values close to that of oxygen, which leads to these oxides being covalent.\n\nThe oxides of the first transition series can be prepared by heating the metals in air. These oxides are $\\mathrm{Sc}_{2} \\mathrm{O}_{3}$, $\\mathrm{TiO}_{2}, \\mathrm{~V}_{2} \\mathrm{O}_{5}, \\mathrm{Cr}_{2} \\mathrm{O}_{3}, \\mathrm{Mn}_{3} \\mathrm{O}_{4}, \\mathrm{Fe}_{3} \\mathrm{O}_{4}, \\mathrm{Co}_{3} \\mathrm{O}_{4}$, NiO , and CuO .\n\nAlternatively, these oxides and other oxides (with the metals in different oxidation states) can be produced by heating the corresponding hydroxides, carbonates, or oxalates in an inert atmosphere. Iron(II) oxide can be prepared by heating iron(II) oxalate, and cobalt(II) oxide is produced by heating cobalt(II) hydroxide:\n\n$$\n\\begin{gathered}\n\\mathrm{FeC}_{2} \\mathrm{O}_{4}(s) \\longrightarrow \\mathrm{FeO}(s)+\\mathrm{CO}(g)+\\mathrm{CO}_{2}(g) \\\\\n\\mathrm{Co}(\\mathrm{OH})_{2}(s) \\longrightarrow \\mathrm{CoO}(s)+\\mathrm{H}_{2} \\mathrm{O}(g)\n\\end{gathered}\n$$"}
{"id": 4884, "contents": "1997. Oxides - \nWith the exception of $\\mathrm{CrO}_{3}$ and $\\mathrm{Mn}_{2} \\mathrm{O}_{7}$, transition metal oxides are not soluble in water. They can react with acids and, in a few cases, with bases. Overall, oxides of transition metals with the lowest oxidation states are basic (and react with acids), the intermediate ones are amphoteric, and the highest oxidation states are\nprimarily acidic. Basic metal oxides at a low oxidation state react with aqueous acids to form solutions of salts and water. Examples include the reaction of cobalt(II) oxide accepting protons from nitric acid, and scandium(III) oxide accepting protons from hydrochloric acid:\n\n$$\n\\begin{gathered}\n\\mathrm{CoO}(s)+2 \\mathrm{HNO}_{3}(a q) \\longrightarrow \\mathrm{Co}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\\\\n\\mathrm{Sc}_{2} \\mathrm{O}_{3}(s)+6 \\mathrm{HCl}(a q) \\longrightarrow 2 \\mathrm{ScCl}_{3}(a q)+3 \\mathrm{H}_{2} \\mathrm{O}(l)\n\\end{gathered}\n$$\n\nThe oxides of metals with oxidation states of 4+ are amphoteric, and most are not soluble in either acids or bases. Vanadium(V) oxide, chromium(VI) oxide, and manganese(VII) oxide are acidic. They react with solutions of hydroxides to form salts of the oxyanions $\\mathrm{VO}_{4}{ }^{3-}, \\mathrm{CrO}_{4}{ }^{2-}$, and $\\mathrm{MnO}_{4}{ }^{-}$. For example, the complete ionic equation for the reaction of chromium(VI) oxide with a strong base is given by:\n\n$$\n\\mathrm{CrO}_{3}(s)+2 \\mathrm{Na}^{+}(a q)+2 \\mathrm{OH}^{-}(a q) \\longrightarrow 2 \\mathrm{Na}^{+}(a q)+\\mathrm{CrO}_{4}{ }^{2-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)\n$$"}
{"id": 4885, "contents": "1997. Oxides - \nChromium(VI) oxide and manganese(VII) oxide react with water to form the acids $\\mathrm{H}_{2} \\mathrm{CrO}_{4}$ and $\\mathrm{HMnO}_{4}$, respectively."}
{"id": 4886, "contents": "1998. Hydroxides - \nWhen a soluble hydroxide is added to an aqueous solution of a salt of a transition metal of the first transition series, a gelatinous precipitate forms. For example, adding a solution of sodium hydroxide to a solution of cobalt sulfate produces a gelatinous pink or blue precipitate of cobalt(II) hydroxide. The net ionic equation is:\n\n$$\n\\mathrm{Co}^{2+}(a q)+2 \\mathrm{OH}^{-}(a q) \\longrightarrow \\mathrm{Co}(\\mathrm{OH})_{2}(s)\n$$\n\nIn this and many other cases, these precipitates are hydroxides containing the transition metal ion, hydroxide ions, and water coordinated to the transition metal. In other cases, the precipitates are hydrated oxides composed of the metal ion, oxide ions, and water of hydration:\n\n$$\n4 \\mathrm{Fe}^{3+}(a q)+6 \\mathrm{OH}^{-}(a q)+\\mathrm{nH}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{Fe}_{2} \\mathrm{O}_{3} \\cdot(\\mathrm{n}+3) \\mathrm{H}_{2} \\mathrm{O}(s)\n$$\n\nThese substances do not contain hydroxide ions. However, both the hydroxides and the hydrated oxides react with acids to form salts and water. When precipitating a metal from solution, it is necessary to avoid an excess of hydroxide ion, as this may lead to complex ion formation as discussed later in this chapter. The precipitated metal hydroxides can be separated for further processing or for waste disposal."}
{"id": 4887, "contents": "1999. Carbonates - \nMany of the elements of the first transition series form insoluble carbonates. It is possible to prepare these carbonates by the addition of a soluble carbonate salt to a solution of a transition metal salt. For example, nickel carbonate can be prepared from solutions of nickel nitrate and sodium carbonate according to the following net ionic equation:\n\n$$\n\\mathrm{Ni}^{2+}(a q)+\\mathrm{CO}_{3}{ }^{2-} \\longrightarrow \\mathrm{NiCO}_{3}(s)\n$$\n\nThe reactions of the transition metal carbonates are similar to those of the active metal carbonates. They react with acids to form metals salts, carbon dioxide, and water. Upon heating, they decompose, forming the transition metal oxides."}
{"id": 4888, "contents": "2000. Other Salts - \nIn many respects, the chemical behavior of the elements of the first transition series is very similar to that of the main group metals. In particular, the same types of reactions that are used to prepare salts of the main group metals can be used to prepare simple ionic salts of these elements.\n\nA variety of salts can be prepared from metals that are more active than hydrogen by reaction with the corresponding acids: Scandium metal reacts with hydrobromic acid to form a solution of scandium bromide:\n\n$$\n2 \\mathrm{Sc}(s)+6 \\mathrm{HBr}(a q) \\longrightarrow 2 \\mathrm{ScBr}_{3}(a q)+3 \\mathrm{H}_{2}(g)\n$$\n\nThe common compounds that we have just discussed can also be used to prepare salts. The reactions involved include the reactions of oxides, hydroxides, or carbonates with acids. For example:\n\n$$\n\\mathrm{Ni}(\\mathrm{OH})_{2}(s)+2 \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+2 \\mathrm{ClO}_{4}^{-}(a q) \\longrightarrow \\mathrm{Ni}^{2+}(a q)+2 \\mathrm{ClO}_{4}^{-}(a q)+4 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nSubstitution reactions involving soluble salts may be used to prepare insoluble salts. For example:\n\n$$\n\\mathrm{Ba}^{2+}(a q)+2 \\mathrm{Cl}^{-}(a q)+2 \\mathrm{~K}^{+}(a q)+\\mathrm{CrO}_{4}{ }^{2-}(a q) \\longrightarrow \\mathrm{BaCrO}_{4}(s)+2 \\mathrm{~K}^{+}(a q)+2 \\mathrm{Cl}^{-}(a q)\n$$\n\nIn our discussion of oxides in this section, we have seen that reactions of the covalent oxides of the transition elements with hydroxides form salts that contain oxyanions of the transition elements."}
{"id": 4889, "contents": "2002. High Temperature Superconductors - \nA superconductor is a substance that conducts electricity with no resistance. This lack of resistance means that there is no energy loss during the transmission of electricity. This would lead to a significant reduction in the cost of electricity.\n\nMost currently used, commercial superconducting materials, such as NbTi and $\\mathrm{Nb}_{3} \\mathrm{Sn}$, do not become superconducting until they are cooled below $23 \\mathrm{~K}\\left(-250^{\\circ} \\mathrm{C}\\right)$. This requires the use of liquid helium, which has a boiling temperature of 4 K and is expensive and difficult to handle. The cost of liquid helium has deterred the widespread application of superconductors.\n\nOne of the most exciting scientific discoveries of the 1980s was the characterization of compounds that exhibit superconductivity at temperatures above 90 K . (Compared to liquid helium, 90 K is a high temperature.) Typical among the high-temperature superconducting materials are oxides containing yttrium (or one of several rare earth elements), barium, and copper in a 1:2:3 ratio. The formula of the ionic yttrium compound is $\\mathrm{YBa}_{2} \\mathrm{Cu}_{3} \\mathrm{O}_{7}$.\n\nThe new materials become superconducting at temperatures close to 90 K (Figure 19.10), temperatures that can be reached by cooling with liquid nitrogen (boiling temperature of 77 K ). Not only are liquid nitrogencooled materials easier to handle, but the cooling costs are also about 1000 times lower than for liquid helium.\n\nFurther advances during the same period included materials that became superconducting at even higher temperatures and with a wider array of materials. The DuPont team led by Uma Chowdry and Arthur Sleight identified Bismouth-Strontium-Copper-Oxides that became superconducting at temperatures as high as 110 K and, importantly, did not contain rare earth elements. Advances continued through the subsequent decades until, in 2020, a team led by Ranga Dias at University of Rochester announced the development of a roomtemperature superconductor, opening doors to widespread applications. More research and development is needed to realize the potential of these materials, but the possibilities are very promising."}
{"id": 4890, "contents": "2002. High Temperature Superconductors - \nFIGURE 19.10 The resistance of the high-temperature superconductor $\\mathrm{YBa}_{2} \\mathrm{Cu}_{3} \\mathrm{O}_{7}$ varies with temperature. Note how the resistance falls to zero below 92 K , when the substance becomes superconducting.\n\nAlthough the brittle, fragile nature of these materials presently hampers their commercial applications, they have tremendous potential that researchers are hard at work improving their processes to help realize.\nSuperconducting transmission lines would carry current for hundreds of miles with no loss of power due to resistance in the wires. This could allow generating stations to be located in areas remote from population centers and near the natural resources necessary for power production. The first project demonstrating the viability of high-temperature superconductor power transmission was established in New York in 2008.\n\nResearchers are also working on using this technology to develop other applications, such as smaller and more powerful microchips. In addition, high-temperature superconductors can be used to generate magnetic fields for applications such as medical devices, magnetic levitation trains, and containment fields for nuclear fusion reactors (Figure 19.11).\n\n\nFIGURE 19.11 (a) This magnetic levitation train (or maglev) uses superconductor technology to move along its tracks. (b) A magnet can be levitated using a dish like this as a superconductor. (credit a: modification of work by Alex Needham; credit b: modification of work by Kevin Jarrett)"}
{"id": 4891, "contents": "2003. LINK TO LEARNING - \nWatch how a high-temperature superconductor (http://openstax.org/l/16supercond) levitates around a magnetic racetrack in the video."}
{"id": 4892, "contents": "2004. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- List the defining traits of coordination compounds\n- Describe the structures of complexes containing monodentate and polydentate ligands\n- Use standard nomenclature rules to name coordination compounds\n- Explain and provide examples of geometric and optical isomerism\n- Identify several natural and technological occurrences of coordination compounds\n\nThe hemoglobin in your blood, the chlorophyll in green plants, vitamin B-12, and the catalyst used in the manufacture of polyethylene all contain coordination compounds. Ions of the metals, especially the transition metals, are likely to form complexes. Many of these compounds are highly colored (Figure 19.12). In the remainder of this chapter, we will consider the structure and bonding of these remarkable compounds.\n\n\nFIGURE 19.12 Metal ions that contain partially filled $d$ subshell usually form colored complex ions; ions with empty $d$ subshell $\\left(d^{0}\\right)$ or with filled $d$ subshells $\\left(d^{10}\\right)$ usually form colorless complexes. This figure shows, from left to right, solutions containing $\\left[M\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{\\mathrm{n+}}$ ions with $M=\\mathrm{Sc}^{3+}\\left(d^{0}\\right), \\mathrm{Cr}^{3+}\\left(d^{3}\\right), \\mathrm{Co}^{2+}\\left(d^{7}\\right), \\mathrm{Ni}^{2+}\\left(d^{8}\\right), \\mathrm{Cu}^{2+}\\left(d^{9}\\right)$, and $\\mathrm{Zn}^{2+}\\left(d^{10}\\right)$. (credit: Sahar Atwa)"}
{"id": 4893, "contents": "2004. LEARNING OBJECTIVES - \nRemember that in most main group element compounds, the valence electrons of the isolated atoms combine to form chemical bonds that satisfy the octet rule. For instance, the four valence electrons of carbon overlap with electrons from four hydrogen atoms to form $\\mathrm{CH}_{4}$. The one valence electron leaves sodium and adds to the seven valence electrons of chlorine to form the ionic formula unit NaCl (Figure 19.13). Transition metals do not normally bond in this fashion. They primarily form coordinate covalent bonds, a form of the Lewis acid-base interaction in which both of the electrons in the bond are contributed by a donor (Lewis base) to an electron acceptor (Lewis acid). The Lewis acid in coordination complexes, often called a central metal ion (or atom), is often a transition metal or inner transition metal, although main group elements can also form coordination compounds. The Lewis base donors, called ligands, can be a wide variety of chemicals-atoms, molecules, or ions. The only requirement is that they have one or more electron pairs, which can be donated to the central metal. Most often, this involves a donor atom with a lone pair of electrons that can form a coordinate bond to the metal.\n\n(a)\n\n(b)\n\nFIGURE 19.13 (a) Covalent bonds involve the sharing of electrons, and ionic bonds involve the transferring of electrons associated with each bonding atom, as indicated by the colored electrons. (b) However, coordinate covalent bonds involve electrons from a Lewis base being donated to a metal center. The lone pairs from six water molecules form bonds to the scandium ion to form an octahedral complex. (Only the donated pairs are shown.)"}
{"id": 4894, "contents": "2004. LEARNING OBJECTIVES - \nThe coordination sphere consists of the central metal ion or atom plus its attached ligands. Brackets in a formula enclose the coordination sphere; species outside the brackets are not part of the coordination sphere. The coordination number of the central metal ion or atom is the number of donor atoms bonded to it. The coordination number for the silver ion in $\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}$is two (Figure 19.14). For the copper(II) ion in $\\left[\\mathrm{CuCl}_{4}\\right]^{2-}$, the coordination number is four, whereas for the cobalt(II) ion in $\\left[\\mathrm{Co}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}$ the coordination number is six. Each of these ligands is monodentate, from the Greek for \"one toothed,\" meaning that they connect with the central metal through only one atom. In this case, the number of ligands and the coordination number are equal.\n\n\n\n\nFIGURE 19.14 The complexes (a) $\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}$, (b) $\\left[\\mathrm{Cu}(\\mathrm{Cl})_{4}\\right]^{2-}$, and (c) $\\left[\\mathrm{Co}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}$ have coordination numbers of two, four, and six, respectively. The geometries of these complexes are the same as we have seen with VSEPR theory for main group elements: linear, tetrahedral, and octahedral."}
{"id": 4895, "contents": "2004. LEARNING OBJECTIVES - \nMany other ligands coordinate to the metal in more complex fashions. Bidentate ligands are those in which two atoms coordinate to the metal center. For example, ethylenediamine (en, $\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}$ ) contains two nitrogen atoms, each of which has a lone pair and can serve as a Lewis base (Figure 19.15). Both of the atoms can coordinate to a single metal center. In the complex $\\left[\\mathrm{Co}(\\mathrm{en})_{3}\\right]^{3+}$, there are three bidentate en ligands, and the coordination number of the cobalt(III) ion is six. The most common coordination numbers are two, four, and six, but examples of all coordination numbers from 1 to 15 are known.\n\n(a)\n\n(b)\n\nFIGURE 19.15 (a) The ethylenediamine (en) ligand contains two atoms with lone pairs that can coordinate to the metal center. (b) The cobalt(III) complex $\\left[\\mathrm{Co}(\\mathrm{en})_{3}\\right]^{3+}$ contains three of these ligands, each forming two bonds to the cobalt ion.\n\nAny ligand that bonds to a central metal ion by more than one donor atom is a polydentate ligand (or \"many teeth\") because it can bite into the metal center with more than one bond. The term chelate (pronounced \"KEY-late\") from the Greek for \"claw\" is also used to describe this type of interaction. Many polydentate ligands are chelating ligands, and a complex consisting of one or more of these ligands and a central metal is a chelate. A chelating ligand is also known as a chelating agent. A chelating ligand holds the metal ion rather like a crab's claw would hold a marble. Figure 19.15 showed one example of a chelate. The heme complex in hemoglobin is another important example (Figure 19.16). It contains a polydentate ligand with four donor atoms that coordinate to iron."}
{"id": 4896, "contents": "2004. LEARNING OBJECTIVES - \nFIGURE 19.16 The single ligand heme contains four nitrogen atoms that coordinate to iron in hemoglobin to form a\nchelate.\nPolydentate ligands are sometimes identified with prefixes that indicate the number of donor atoms in the ligand. As we have seen, ligands with one donor atom, such as $\\mathrm{NH}_{3}, \\mathrm{Cl}^{-}$, and $\\mathrm{H}_{2} \\mathrm{O}$, are monodentate ligands. Ligands with two donor groups are bidentate ligands. Ethylenediamine, $\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}$, and the anion of the acid glycine, $\\mathrm{NH}_{2} \\mathrm{CH}_{2} \\mathrm{CO}_{2}{ }^{-}$(Figure 19.17) are examples of bidentate ligands. Tridentate ligands, tetradentate ligands, pentadentate ligands, and hexadentate ligands contain three, four, five, and six donor atoms, respectively. The ligand in heme (Figure 19.16) is a tetradentate ligand.\n\n\nFIGURE 19.17 Each of the anionic ligands shown attaches in a bidentate fashion to platinum(II), with both a nitrogen and oxygen atom coordinating to the metal.\n\nThe Naming of Complexes\nThe nomenclature of the complexes is patterned after a system suggested by Alfred Werner, a Swiss chemist and Nobel laureate, whose outstanding work more than 100 years ago laid the foundation for a clearer understanding of these compounds. The following five rules are used for naming complexes:"}
{"id": 4897, "contents": "2004. LEARNING OBJECTIVES - \n1. If a coordination compound is ionic, name the cation first and the anion second, in accordance with the usual nomenclature.\n2. Name the ligands first, followed by the central metal. Name the ligands alphabetically. Negative ligands (anions) have names formed by adding -o to the stem name of the group. For examples, see Table 19.1. For most neutral ligands, the name of the molecule is used. The four common exceptions are aqua $\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)$, ammine $\\left(\\mathrm{NH}_{3}\\right)$, carbonyl (CO), and nitrosyl (NO). For example, name $\\left[\\mathrm{Pt}\\left(\\mathrm{NH}_{3}\\right)_{2} \\mathrm{Cl}_{4}\\right]$ as diamminetetrachloroplatinum(IV).\n\n| Examples of Anionic Ligands |  |\n| :--- | :--- |\n| Anionic Ligand | Name |\n| $\\mathrm{F}^{-}$ | fluoro |\n| $\\mathrm{Cl}^{-}$ | chloro |\n| $\\mathrm{Br}^{-}$ | bromo |\n| $\\mathrm{I}^{-}$ | iodo |\n| $\\mathrm{CN}^{-}$ | cyano |\n| $\\mathrm{NO}_{3}^{-}$ | hitrato |\n| $\\mathrm{OH}^{-}$ | oxo |\n| $\\mathrm{O}^{2-}$ | oxalato |\n| $\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}$ | carbonato |\n| $\\mathrm{CO}_{3}^{2-}$ |  |"}
{"id": 4898, "contents": "2004. LEARNING OBJECTIVES - \nTABLE 19.1\n3. If more than one ligand of a given type is present, the number is indicated by the prefixes di- (for two), tri(for three), tetra- (for four), penta- (for five), and hexa- (for six). Sometimes, the prefixes bis- (for two), tris(for three), and tetrakis- (for four) are used when the name of the ligand already includes di-, tri-, or tetra-, or when the ligand name begins with a vowel. For example, the ion bis(bipyridyl)osmium(II) uses bis- to signify that there are two ligands attached to Os, and each bipyridyl ligand contains two pyridine groups $\\left(\\mathrm{C}_{5} \\mathrm{H}_{4} \\mathrm{~N}\\right)$.\n\nWhen the complex is either a cation or a neutral molecule, the name of the central metal atom is spelled exactly like the name of the element and is followed by a Roman numeral in parentheses to indicate its oxidation state (Table 19.2 and Table 19.3). When the complex is an anion, the suffix -ate is added to the stem of the name of the metal, followed by the Roman numeral designation of its oxidation state (Table 19.4).\nSometimes, the Latin name of the metal is used when the English name is clumsy. For example, ferrate is used instead of ironate, plumbate instead leadate, and stannate instead of tinate. The oxidation state of the metal is determined based on the charges of each ligand and the overall charge of the coordination compound. For example, in $\\left[\\mathrm{Cr}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{4} \\mathrm{Cl}_{2}\\right] \\mathrm{Br}$, the coordination sphere (in brackets) has a charge of $1+$ to balance the bromide ion. The water ligands are neutral, and the chloride ligands are anionic with a charge of 1-each. To determine the oxidation state of the metal, we set the overall charge equal to the sum of the ligands and the metal: $+1=-2$ $+x$, so the oxidation state $(x)$ is equal to $3+$."}
{"id": 4899, "contents": "2004. LEARNING OBJECTIVES - \n| Examples in Which the Complex Is a Cation |  |\n| :--- | :--- |\n| $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]_{\\mathrm{Cl}}^{3}$ | hexaamminecobalt(III) chloride |\n| $\\left[\\mathrm{Pt}\\left(\\mathrm{NH}_{3}\\right)_{4} \\mathrm{Cl}_{2}\\right]^{2+}$ | tetraamminedichloroplatinum(IV) ion |\n| $\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}$ | diamminesilver(I) ion |\n| $\\left[\\mathrm{Cr}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{4} \\mathrm{Cl}_{2}\\right] \\mathrm{Cl}$ | tetraaquadichlorochromium(III) chloride |\n| $\\left[\\mathrm{Co}\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right)_{3}\\right]_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}$ | tris(ethylenediamine)cobalt(III) sulfate |\n\nTABLE 19.2\n\nExamples in Which the Complex Is Neutral\n\n| $\\left[\\mathrm{Pt}\\left(\\mathrm{NH}_{3}\\right)_{2} \\mathrm{Cl}_{4}\\right]$ | diamminetetrachloroplatinum(IV) |\n| :--- | :--- |\n| $\\left[\\mathrm{Ni}\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right)_{2} \\mathrm{Cl}_{2}\\right]$ | dichlorobis(ethylenediamine)nickel(II) |\n\nTABLE 19.3"}
{"id": 4900, "contents": "2004. LEARNING OBJECTIVES - \nTABLE 19.3\n\n| Examples in Which the Complex Is an Anion |  |\n| :--- | :--- |\n| $\\left[\\mathrm{PtCl}_{6}\\right]^{2-}$ | hexachloroplatinate(IV) ion |\n| $\\mathrm{Na}_{2}\\left[\\mathrm{SnCl}_{6}\\right]$ | sodium hexachlorostannate(IV) |\n\nTABLE 19.4"}
{"id": 4901, "contents": "2005. LINK TO LEARNING - \nDo you think you understand naming coordination complexes? You can look over more examples and test\nyourself with online quizzes (http://openstax.org/l/16namingcomps) at the University of Sydney's site."}
{"id": 4902, "contents": "2007. Coordination Numbers and Oxidation States - \nDetermine the name of the following complexes and give the coordination number of the central metal atom.\n(a) $\\mathrm{Na}_{2}\\left[\\mathrm{PtCl}_{6}\\right]$\n(b) $\\mathrm{K}_{3}\\left[\\mathrm{Fe}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}\\right)_{3}\\right]$\n(c) $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{5} \\mathrm{Cl}\\right] \\mathrm{Cl}_{2}$"}
{"id": 4903, "contents": "2008. Solution - \n(a) There are two $\\mathrm{Na}^{+}$ions, so the coordination sphere has a negative two charge: $\\left[\\mathrm{PtCl}_{6}\\right]^{2-}$. There are six anionic chloride ligands, so $-2=-6+x$, and the oxidation state of the platinum is $4+$. The name of the complex is sodium hexachloroplatinate(IV), and the coordination number is six. (b) The coordination sphere has a charge of 3 - (based on the potassium) and the oxalate ligands each have a charge of $2-$, so the metal oxidation state is given by $-3=-6+x$, and this is an iron(III) complex. The name is potassium trisoxalatoferrate(III) (note that tris is used instead of tri because the ligand name starts with a vowel). Because oxalate is a bidentate ligand, this complex has a coordination number of six. (c) In this example, the coordination sphere has a cationic charge of $2+$. The $\\mathrm{NH}_{3}$ ligand is neutral, but the chloro ligand has a charge of $1-$. The oxidation state is found by $+2=-1+x$ and is $3+$, so the complex is pentaamminechlorocobalt(III) chloride and the coordination number is six."}
{"id": 4904, "contents": "2009. Check Your Learning - \nThe complex potassium dicyanoargenate(I) is used to make antiseptic compounds. Give the formula and coordination number."}
{"id": 4905, "contents": "2010. Answer: - \n$\\mathrm{K}\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]$; coordination number two"}
{"id": 4906, "contents": "2011. The Structures of Complexes - \nThe most common structures of the complexes in coordination compounds are octahedral, tetrahedral, and square planar (see Figure 19.18). For transition metal complexes, the coordination number determines the geometry around the central metal ion. Table 19.5 compares coordination numbers to the molecular geometry:\n\n\nFIGURE 19.18 These are geometries of some complexes with coordination numbers of seven and eight.\n\nCoordination Numbers and Molecular Geometry\n\n| Coordination Number | Molecular Geometry | Example |\n| :--- | :--- | :--- |\n| 2 | linear | $\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}$ |\n| 3 | trigonal planar | $\\left[\\mathrm{Cu}(\\mathrm{CN})_{3}\\right]^{2-}$ |\n| 4 | tetrahedral $\\left(d^{0}\\right.$ or $\\left.d^{10}\\right)$, low oxidation states for M | $\\left[\\mathrm{Ni}(\\mathrm{CO})_{4}\\right]$ |\n| 4 | square planar $\\left(d^{8}\\right)$ | $\\left[\\mathrm{Ni}(\\mathrm{CN})_{4}\\right]^{2-}$ |\n| 5 | trigonal bipyramidal | $\\left[\\mathrm{CoCl}_{5}\\right]^{2-}$ |\n| 5 | square pyramidal | $\\left[\\mathrm{VO}_{3}(\\mathrm{CN})_{4}\\right]^{2-}$ |\n| 6 | octahedral | $\\left[\\mathrm{CoCl}_{6}\\right]^{3-}$ |\n| 7 | pentagonal bipyramid | $\\left[\\mathrm{ZrF}_{7}\\right]^{3-}$ |\n| 8 | square antiprism | $\\left[\\mathrm{ReF}_{8}\\right]^{2-}$ |\n| 8 | dodecahedron | $\\left[\\mathrm{Mo}^{2-}(\\mathrm{CN})_{8}\\right]^{4-}$ |\n| 9 and above | more complicated structures |  |\n\nTABLE 19.5"}
{"id": 4907, "contents": "2011. The Structures of Complexes - \nTABLE 19.5\n\nUnlike main group atoms in which both the bonding and nonbonding electrons determine the molecular shape, the nonbonding $d$-electrons do not change the arrangement of the ligands. Octahedral complexes have a coordination number of six, and the six donor atoms are arranged at the corners of an octahedron around the central metal ion. Examples are shown in Figure 19.19. The chloride and nitrate anions in $\\left[\\mathrm{Co}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right] \\mathrm{Cl}_{2}$ and $\\left[\\mathrm{Cr}(\\mathrm{en})_{3}\\right]\\left(\\mathrm{NO}_{3}\\right)_{3}$, and the potassium cations in $\\mathrm{K}_{2}\\left[\\mathrm{PtCl}_{6}\\right]$, are outside the brackets and are not bonded to the metal ion.\n\n$2 \\mathrm{Cl}^{-}$\n\n$3 \\mathrm{NO}_{3}{ }^{-}$\n\n$2 \\mathrm{~K}^{+}$\n\nFIGURE 19.19 Many transition metal complexes adopt octahedral geometries, with six donor atoms forming bond angles of $90^{\\circ}$ about the central atom with adjacent ligands. Note that only ligands within the coordination sphere affect the geometry around the metal center."}
{"id": 4908, "contents": "2011. The Structures of Complexes - \nFor transition metals with a coordination number of four, two different geometries are possible: tetrahedral or square planar. Unlike main group elements, where these geometries can be predicted from VSEPR theory, a more detailed discussion of transition metal orbitals (discussed in the section on Crystal Field Theory) is required to predict which complexes will be tetrahedral and which will be square planar. In tetrahedral complexes such as $\\left[\\mathrm{Zn}(\\mathrm{CN})_{4}\\right]^{2-}$ (Figure 19.20), each of the ligand pairs forms an angle of $109.5^{\\circ}$. In square\nplanar complexes, such as $\\left[\\mathrm{Pt}\\left(\\mathrm{NH}_{3}\\right)_{2} \\mathrm{Cl}_{2}\\right]$, each ligand has two other ligands at $90^{\\circ}$ angles (called the cis positions) and one additional ligand at an $180^{\\circ}$ angle, in the trans position.\n\n(a)\n\n(b)\n\nFIGURE 19.20 Transition metals with a coordination number of four can adopt a tetrahedral geometry (a) as in $\\mathrm{K}_{2}\\left[\\mathrm{Zn}(\\mathrm{CN})_{4}\\right]$ or a square planar geometry (b) as shown in $\\left[\\mathrm{Pt}\\left(\\mathrm{NH}_{3}\\right)_{2} \\mathrm{Cl}_{2}\\right]$."}
{"id": 4909, "contents": "2012. Isomerism in Complexes - \nIsomers are different chemical species that have the same chemical formula. Transition metal complexes often exist as geometric isomers, in which the same atoms are connected through the same types of bonds but with differences in their orientation in space. Coordination complexes with two different ligands in the cis and trans positions from a ligand of interest form isomers. For example, the octahedral $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{4} \\mathrm{Cl}_{2}\\right]^{+}$ion has two isomers. In the cis configuration, the two chloride ligands are adjacent to each other (Figure 19.21). The other isomer, the trans configuration, has the two chloride ligands directly across from one another.\n\nViolet, cis form\n\n\nGreen, trans form\n\n\nFIGURE 19.21 The cis and trans isomers of $\\left[\\mathrm{Co}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{4} \\mathrm{Cl}_{2}\\right]^{+}$contain the same ligands attached to the same metal ion, but the spatial arrangement causes these two compounds to have very different properties."}
{"id": 4910, "contents": "2012. Isomerism in Complexes - \nDifferent geometric isomers of a substance are different chemical compounds. They exhibit different properties, even though they have the same formula. For example, the two isomers of $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{4} \\mathrm{Cl}_{2}\\right] \\mathrm{NO}_{3}$ differ in color; the cis form is violet, and the trans form is green. Furthermore, these isomers have different dipole moments, solubilities, and reactivities. As an example of how the arrangement in space can influence the molecular properties, consider the polarity of the two $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{4} \\mathrm{Cl}_{2}\\right] \\mathrm{NO}_{3}$ isomers. Remember that the polarity of a molecule or ion is determined by the bond dipoles (which are due to the difference in electronegativity of the bonding atoms) and their arrangement in space. In one isomer, cis chloride ligands cause more electron density on one side of the molecule than on the other, making it polar. For the trans isomer, each ligand is directly across from an identical ligand, so the bond dipoles cancel out, and the molecule is nonpolar."}
{"id": 4911, "contents": "2014. Geometric Isomers - \nIdentify which geometric isomer of $\\left[\\mathrm{Pt}\\left(\\mathrm{NH}_{3}\\right)_{2} \\mathrm{Cl}_{2}\\right]$ is shown in Figure 19.20. Draw the other geometric isomer and give its full name."}
{"id": 4912, "contents": "2015. Solution - \nIn the Figure 19.20, the two chlorine ligands occupy cis positions. The other form is shown in Figure 19.22. When naming specific isomers, the descriptor is listed in front of the name. Therefore, this complex is trans-diamminedichloroplatinum(II).\n\n\nFIGURE 19.22 The trans isomer of $\\left[\\mathrm{Pt}\\left(\\mathrm{NH}_{3}\\right)_{2} \\mathrm{Cl}_{2}\\right]$ has each ligand directly across from an adjacent ligand."}
{"id": 4913, "contents": "2016. Check Your Learning - \nDraw the ion trans-diaqua-trans-dibromo-trans-dichlorocobalt(II)."}
{"id": 4914, "contents": "2017. Answer: - \nAnother important type of isomers are optical isomers, or enantiomers, in which two objects are exact mirror images of each other but cannot be lined up so that all parts match. This means that optical isomers are nonsuperimposable mirror images. A classic example of this is a pair of hands, in which the right and left hand are mirror images of one another but cannot be superimposed. Optical isomers are very important in organic and biochemistry because living systems often incorporate one specific optical isomer and not the other. Unlike geometric isomers, pairs of optical isomers have identical properties (boiling point, polarity, solubility, etc.). Optical isomers differ only in the way they affect polarized light and how they react with other optical isomers. For coordination complexes, many coordination compounds such as $\\left[\\mathrm{M}(\\mathrm{en})_{3}\\right]^{\\mathrm{n+}}\\left[\\right.$ in which $\\mathrm{M}^{\\mathrm{n}+}$ is a central metal ion such as iron(III) or cobalt(II)] form enantiomers, as shown in Figure 19.23. These two isomers will react differently with other optical isomers. For example, DNA helices are optical isomers, and the form that occurs in nature (right-handed DNA) will bind to only one isomer of $\\left[\\mathrm{M}(\\mathrm{en})_{3}\\right]^{\\mathrm{n}+}$ and not the other.\n\n\n\nFIGURE 19.23 The complex $\\left[M(e n)_{3}\\right]^{n+}\\left(M^{n+}=\\right.$ a metal ion, en $=$ ethylenediamine) has a nonsuperimposable mirror image.\n\nThe $\\left[\\mathrm{Co}(\\mathrm{en})_{2} \\mathrm{Cl}_{2}\\right]^{+}$ion exhibits geometric isomerism (cis/trans), and its cis isomer exists as a pair of optical isomers (Figure 19.24).\ncis form (optical isomers)"}
{"id": 4915, "contents": "2017. Answer: - \nFIGURE 19.24 Three isomeric forms of $\\left[\\mathrm{Co}(\\mathrm{en})_{2} \\mathrm{Cl}_{2}\\right]^{+}$exist. The trans isomer, formed when the chlorines are positioned at a $180^{\\circ}$ angle, has very different properties from the cis isomers. The mirror images of the cis isomer form a pair of optical isomers, which have identical behavior except when reacting with other enantiomers.\n\nLinkage isomers occur when the coordination compound contains a ligand that can bind to the transition metal center through two different atoms. For example, the CN ligand can bind through the carbon atom (cyano) or through the nitrogen atom (isocyano). Similarly, SCN- can be bound through the sulfur or nitrogen\natom, affording two distinct compounds $\\left(\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{5} \\mathrm{SCN}\\right]^{2+}\\right.$ or $\\left.\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{5} \\mathrm{NCS}\\right]^{2+}\\right)$.\nIonization isomers (or coordination isomers) occur when one anionic ligand in the inner coordination sphere is replaced with the counter ion from the outer coordination sphere. A simple example of two ionization isomers are $\\left\\mathrm{CoCl}_{6}\\right$ and $\\left\\mathrm{CoCl}_{5} \\mathrm{Br}\\right$."}
{"id": 4916, "contents": "2018. Coordination Complexes in Nature and Technology - \nChlorophyll, the green pigment in plants, is a complex that contains magnesium (Figure 19.25). This is an example of a main group element in a coordination complex. Plants appear green because chlorophyll absorbs red and purple light; the reflected light consequently appears green. The energy resulting from the absorption of light is used in photosynthesis.\n\n(a)\n\n(b)\n\nFIGURE 19.25 (a) Chlorophyll comes in several different forms, which all have the same basic structure around the magnesium center. (b) Copper phthalocyanine blue, a square planar copper complex, is present in some blue dyes."}
{"id": 4917, "contents": "2020. Transition Metal Catalysts - \nOne of the most important applications of transition metals is as industrial catalysts. As you recall from the chapter on kinetics, a catalyst increases the rate of reaction by lowering the activation energy and is regenerated in the catalytic cycle. Over $90 \\%$ of all manufactured products are made with the aid of one or more catalysts. The ability to bind ligands and change oxidation states makes transition metal catalysts well suited for catalytic applications. Vanadium oxide is used to produce $230,000,000$ tons of sulfuric acid worldwide each year, which in turn is used to make everything from fertilizers to cans for food. Plastics are made with the aid of transition metal catalysts, along with detergents, fertilizers, paints, and more (see Figure 19.26). Very complicated pharmaceuticals are manufactured with catalysts that are selective, reacting with one specific bond out of a large number of possibilities. Catalysts allow processes to be more economical and more environmentally friendly. Developing new catalysts and better understanding of existing systems are important areas of current research.\n\n\nFIGURE 19.26 (a) Detergents, (b) paints, and (c) fertilizers are all made using transition metal catalysts. (credit a: modification of work by \"Mr. Brian\"/Flickr; credit b: modification of work by Ewen Roberts; credit c: modification of work by \"osseous\"/Flickr)"}
{"id": 4918, "contents": "2022. Deanna D'Alessandro - \nDr. Deanna D'Alessandro develops new metal-containing materials that demonstrate unique electronic, optical, and magnetic properties. Her research combines the fields of fundamental inorganic and physical chemistry with materials engineering. She is working on many different projects that rely on transition metals. For example, one type of compound she is developing captures carbon dioxide waste from power plants and catalytically converts it into useful products (see Figure 19.27).\n\n\nFIGURE 19.27 Catalytic converters change carbon dioxide emissions from power plants into useful products, and, like the one shown here, are also found in cars.\n\nAnother project involves the development of porous, sponge-like materials that are \"photoactive.\" The absorption of light causes the pores of the sponge to change size, allowing gas diffusion to be controlled. This has many potential useful applications, from powering cars with hydrogen fuel cells to making better electronics components. Although not a complex, self-darkening sunglasses are an example of a photoactive substance.\n\nWatch this video (http://openstax.org/l/16DeannaD) to learn more about this research and listen to Dr. D'Alessandro (shown in Figure 19.28) describe what it is like being a research chemist.\n\n\nFIGURE 19.28 Dr. Deanna D'Alessandro is a functional materials researcher. Her work combines the inorganic and physical chemistry fields with engineering, working with transition metals to create new systems to power cars and convert energy (credit: image courtesy of Deanna D'Alessandro).\n\nMany other coordination complexes are also brightly colored. The square planar copper(II) complex phthalocyanine blue (from Figure 19.25) is one of many complexes used as pigments or dyes. This complex is used in blue ink, blue jeans, and certain blue paints."}
{"id": 4919, "contents": "2022. Deanna D'Alessandro - \nThe structure of heme (Figure 19.29), the iron-containing complex in hemoglobin, is very similar to that in chlorophyll. In hemoglobin, the red heme complex is bonded to a large protein molecule (globin) by the attachment of the protein to the heme ligand. Oxygen molecules are transported by hemoglobin in the blood by being bound to the iron center. When the hemoglobin loses its oxygen, the color changes to a bluish red. Hemoglobin will only transport oxygen if the iron is $\\mathrm{Fe}^{2+}$; oxidation of the iron to $\\mathrm{Fe}^{3+}$ prevents oxygen transport.\n\n\nFIGURE 19.29 Hemoglobin contains four protein subunits, each of which has an iron center attached to a heme ligand (shown in red), which is coordinated to a globin protein. Each subunit is shown in a different color.\n\nComplexing agents often are used for water softening because they tie up such ions as $\\mathrm{Ca}^{2+}, \\mathrm{Mg}^{2+}$, and $\\mathrm{Fe}^{2+}$, which make water hard. Many metal ions are also undesirable in food products because these ions can catalyze reactions that change the color of food. Coordination complexes are useful as preservatives. For example, the ligand EDTA, $\\left(\\mathrm{HO}_{2} \\mathrm{CCH}_{2}\\right)_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{~N}\\left(\\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}\\right)_{2}$, coordinates to metal ions through six donor atoms and prevents the metals from reacting (Figure 19.30). This ligand also is used to sequester metal ions in paper production, textiles, and detergents, and has pharmaceutical uses."}
{"id": 4920, "contents": "2022. Deanna D'Alessandro - \nFIGURE 19.30 The ligand EDTA binds tightly to a variety of metal ions by forming hexadentate complexes.\nComplexing agents that tie up metal ions are also used as drugs. British Anti-Lewisite (BAL), $\\mathrm{HSCH}_{2} \\mathrm{CH}(\\mathrm{SH}) \\mathrm{CH}_{2} \\mathrm{OH}$, is a drug developed during World War I as an antidote for the arsenic-based war gas Lewisite. BAL is now used to treat poisoning by heavy metals, such as arsenic, mercury, thallium, and chromium. The drug is a ligand and functions by making a water-soluble chelate of the metal; the kidneys eliminate this metal chelate (Figure 19.31). Another polydentate ligand, enterobactin, which is isolated from\ncertain bacteria, is used to form complexes of iron and thereby to control the severe iron buildup found in patients suffering from blood diseases such as Cooley's anemia, who require frequent transfusions. As the transfused blood breaks down, the usual metabolic processes that remove iron are overloaded, and excess iron can build up to fatal levels. Enterobactin forms a water-soluble complex with excess iron, and the body can safely eliminate this complex.\n\n(a)\n\n(b)\n\nFIGURE 19.31 Coordination complexes are used as drugs. (a) British Anti-Lewisite is used to treat heavy metal poisoning by coordinating metals (M), and enterobactin (b) allows excess iron in the blood to be removed."}
{"id": 4921, "contents": "2024. Chelation Therapy - \nLigands like BAL and enterobactin are important in medical treatments for heavy metal poisoning. However, chelation therapies can disrupt the normal concentration of ions in the body, leading to serious side effects, so researchers are searching for new chelation drugs. One drug that has been developed is dimercaptosuccinic acid (DMSA), shown in Figure 19.32. Identify which atoms in this molecule could act as donor atoms.\n\n\nFIGURE 19.32 Dimercaptosuccinic acid is used to treat heavy metal poisoning."}
{"id": 4922, "contents": "2025. Solution - \nAll of the oxygen and sulfur atoms have lone pairs of electrons that can be used to coordinate to a metal center, so there are six possible donor atoms. Geometrically, only two of these atoms can be coordinated to a metal at once. The most common binding mode involves the coordination of one sulfur atom and one oxygen atom, forming a five-member ring with the metal."}
{"id": 4923, "contents": "2026. Check Your Learning - \nSome alternative medicine practitioners recommend chelation treatments for ailments that are not clearly related to heavy metals, such as cancer and autism, although the practice is discouraged by many scientific organizations. ${ }^{\\frac{1}{~}}$ Identify at least two biologically important metals that could be disrupted by chelation therapy."}
{"id": 4924, "contents": "2027. Answer: - \n$\\mathrm{Ca}, \\mathrm{Fe}, \\mathrm{Zn}$, and Cu\n\nLigands are also used in the electroplating industry. When metal ions are reduced to produce thin metal coatings, metals can clump together to form clusters and nanoparticles. When metal coordination complexes are used, the ligands keep the metal atoms isolated from each other. It has been found that many metals plate out as a smoother, more uniform, better-looking, and more adherent surface when plated from a bath containing the metal as a complex ion. Thus, complexes such as $\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]^{-}$and $\\left[\\mathrm{Au}(\\mathrm{CN})_{2}\\right]^{-}$are used extensively in the electroplating industry.\n\nIn 1965, scientists at Michigan State University discovered that there was a platinum complex that inhibited cell division in certain microorganisms. Later work showed that the complex was cis-diamminedichloroplatinum(II), $\\left[\\mathrm{Pt}\\left(\\mathrm{NH}_{3}\\right)_{2}(\\mathrm{Cl})_{2}\\right]$, and that the trans isomer was not effective. The inhibition of cell division indicated that this square planar compound could be an anticancer agent. In 1978, the US Food and Drug Administration approved this compound, known as cisplatin, for use in the treatment of certain forms of cancer. Since that time, many similar platinum compounds have been developed for the treatment of cancer. In all cases, these are the cis isomers and never the trans isomers. The diammine $\\left(\\mathrm{NH}_{3}\\right)_{2}$ portion is retained with other groups, replacing the dichloro $\\left[(\\mathrm{Cl})_{2}\\right]$ portion. The newer drugs include carboplatin, oxaliplatin, and satraplatin."}
{"id": 4925, "contents": "2028. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Outline the basic premise of crystal field theory (CFT)\n- Identify molecular geometries associated with various d-orbital splitting patterns\n- Predict electron configurations of split d orbitals for selected transition metal atoms or ions\n- Explain spectral and magnetic properties in terms of CFT concepts\n\nThe behavior of coordination compounds cannot be adequately explained by the same theories used for main group element chemistry. The observed geometries of coordination complexes are not consistent with hybridized orbitals on the central metal overlapping with ligand orbitals, as would be predicted by valence bond theory. The observed colors indicate that the $d$ orbitals often occur at different energy levels rather than all being degenerate, that is, of equal energy, as are the three $p$ orbitals. To explain the stabilities, structures, colors, and magnetic properties of transition metal complexes, a different bonding model has been developed. Just as valence bond theory explains many aspects of bonding in main group chemistry, crystal field theory is useful in understanding and predicting the behavior of transition metal complexes."}
{"id": 4926, "contents": "2029. Crystal Field Theory - \nTo explain the observed behavior of transition metal complexes (such as how colors arise), a model involving electrostatic interactions between the electrons from the ligands and the electrons in the unhybridized $d$ orbitals of the central metal atom has been developed. This electrostatic model is crystal field theory (CFT). It allows us to understand, interpret, and predict the colors, magnetic behavior, and some structures of coordination compounds of transition metals.\n\nCFT focuses on the nonbonding electrons on the central metal ion in coordination complexes not on the metalligand bonds. Like valence bond theory, CFT tells only part of the story of the behavior of complexes. However, it tells the part that valence bond theory does not. In its pure form, CFT ignores any covalent bonding between ligands and metal ions. Both the ligand and the metal are treated as infinitesimally small point charges.\n\nAll electrons are negative, so the electrons donated from the ligands will repel the electrons of the central metal. Let us consider the behavior of the electrons in the unhybridized $d$ orbitals in an octahedral complex. The five $d$ orbitals consist of lobe-shaped regions and are arranged in space, as shown in Figure 19.33. In an octahedral complex, the six ligands coordinate along the axes.\n\n\nFIGURE 19.33 The directional characteristics of the five $d$ orbitals are shown here. The shaded portions indicate the phase of the orbitals. The ligands (L) coordinate along the axes. For clarity, the ligands have been omitted from the $d_{x^{2}-y^{2}}$ orbital so that the axis labels could be shown."}
{"id": 4927, "contents": "2029. Crystal Field Theory - \nIn an uncomplexed metal ion in the gas phase, the electrons are distributed among the five $d$ orbitals in accord with Hund's rule because the orbitals all have the same energy. However, when ligands coordinate to a metal ion, the energies of the $d$ orbitals are no longer the same.\nIn octahedral complexes, the lobes in two of the five $d$ orbitals, the $d_{z^{2}}$ and $d_{x^{2}-y^{2}}$ orbitals, point toward the ligands (Figure 19.33). These two orbitals are called the $\\boldsymbol{e}_{g}$ orbitals (the symbol actually refers to the symmetry of the orbitals, but we will use it as a convenient name for these two orbitals in an octahedral complex). The other three orbitals, the $d_{x y}, d_{x z}$, and $d_{y z}$ orbitals, have lobes that point between the ligands and are called the $\\boldsymbol{t}_{\\mathbf{2 g}}$ orbitals (again, the symbol really refers to the symmetry of the orbitals). As six ligands approach the metal ion along the axes of the octahedron, their point charges repel the electrons in the $d$ orbitals of the metal ion. However, the repulsions between the electrons in the $e_{g}$ orbitals (the $d_{z^{2}}$ and $d_{x^{2}-y^{2}}$ orbitals) and the ligands are greater than the repulsions between the electrons in the $t_{2 g}$ orbitals (the $d_{z y}, d_{x z}$, and $d_{y z}$ orbitals) and the ligands. This is because the lobes of the $e_{g}$ orbitals point directly at the ligands, whereas the lobes of the $t_{2 g}$ orbitals point between them. Thus, electrons in the $e_{g}$ orbitals of the metal ion in an octahedral complex have higher potential energies than those of electrons in the $t_{2 g}$ orbitals. The difference in energy may be represented as shown in Figure 19.34."}
{"id": 4928, "contents": "2029. Crystal Field Theory - \nFIGURE 19.34 In octahedral complexes, the $e_{g}$ orbitals are destabilized (higher in energy) compared to the $t_{2 g}$ orbitals because the ligands interact more strongly with the $d$ orbitals at which they are pointed directly.\n\nThe difference in energy between the $e_{g}$ and the $t_{2 g}$ orbitals is called the crystal field splitting and is symbolized by $\\boldsymbol{\\Delta}_{\\text {oct }}$, where oct stands for octahedral.\n\nThe magnitude of $\\Delta_{\\text {oct }}$ depends on many factors, including the nature of the six ligands located around the central metal ion, the charge on the metal, and whether the metal is using $3 d, 4 d$, or $5 d$ orbitals. Different ligands produce different crystal field splittings. The increasing crystal field splitting produced by ligands is expressed in the spectrochemical series, a short version of which is given here:\n\n$$\n\\mathrm{I}^{-}<\\mathrm{Br}^{-}<\\mathrm{Cl}^{-}<\\mathrm{F}^{-}<\\mathrm{H}_{2} \\mathrm{O}<\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}<\\mathrm{NH}_{3}<e n<\\mathrm{NO}_{2}-<\\mathrm{CN}^{-}\n$$\n\na few ligands of the spectrochemical series, in order of increasing field strength of the ligand\n\nIn this series, ligands on the left cause small crystal field splittings and are weak-field ligands, whereas those on the right cause larger splittings and are strong-field ligands. Thus, the $\\Delta_{\\text {oct }}$ value for an octahedral complex with iodide ligands $\\left(\\mathrm{I}^{-}\\right)$is much smaller than the $\\Delta_{\\text {oct }}$ value for the same metal with cyanide ligands $\\left(\\mathrm{CN}^{-}\\right)$."}
{"id": 4929, "contents": "2029. Crystal Field Theory - \nElectrons in the $d$ orbitals follow the aufbau (\"filling up\") principle, which says that the orbitals will be filled to give the lowest total energy, just as in main group chemistry. When two electrons occupy the same orbital, the like charges repel each other. The energy needed to pair up two electrons in a single orbital is called the pairing energy ( $\\mathbf{P}$ ). Electrons will always singly occupy each orbital in a degenerate set before pairing. $P$ is similar in magnitude to $\\Delta_{\\text {oct }}$. When electrons fill the $d$ orbitals, the relative magnitudes of $\\Delta_{\\text {oct }}$ and $P$ determine which orbitals will be occupied.\n\nIn $\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{4-}$, the strong field of six cyanide ligands produces a large $\\Delta_{\\text {oct. }}$. Under these conditions, the electrons require less energy to pair than they require to be excited to the $e_{g}$ orbitals ( $\\Delta_{\\text {oct }}>P$ ). The six $3 d$ electrons of the $\\mathrm{Fe}^{2+}$ ion pair in the three $t_{2 g}$ orbitals (Figure 19.35). Complexes in which the electrons are paired because of the large crystal field splitting are called low-spin complexes because the number of unpaired electrons (spins) is minimized.\n\n\nFIGURE 19.35 Iron(II) complexes have six electrons in the $5 d$ orbitals. In the absence of a crystal field, the orbitals are degenerate. For coordination complexes with strong-field ligands such as $\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{4-}, \\Delta_{\\text {oct }}$ is greater than P , and the electrons pair in the lower energy $t_{2 g}$ orbitals before occupying the eg orbitals. With weak-field ligands such as $\\mathrm{H}_{2} \\mathrm{O}$, the ligand field splitting is less than the pairing energy, $\\Delta_{\\text {oct }}$ less than P , so the electrons occupy all $d$ orbitals singly before any pairing occurs."}
{"id": 4930, "contents": "2029. Crystal Field Theory - \nIn $\\left[\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}$, on the other hand, the weak field of the water molecules produces only a small crystal field splitting ( $\\Delta_{\\text {oct }}<\\mathrm{P}$ ). Because it requires less energy for the electrons to occupy the $e_{g}$ orbitals than to pair together, there will be an electron in each of the five $3 d$ orbitals before pairing occurs. For the six $d$ electrons on the iron(II) center in $\\left[\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}$, there will be one pair of electrons and four unpaired electrons (Figure 19.35). Complexes such as the $\\left[\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}$ ion, in which the electrons are unpaired because the crystal field splitting is not large enough to cause them to pair, are called high-spin complexes because the number of unpaired electrons (spins) is maximized.\n\nA similar line of reasoning shows why the $\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{3-}$ ion is a low-spin complex with only one unpaired electron, whereas both the $\\left[\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{3+}$ and $\\left[\\mathrm{FeF}_{6}\\right]^{3-}$ ions are high-spin complexes with five unpaired electrons."}
{"id": 4931, "contents": "2031. High- and Low-Spin Complexes - \nPredict the number of unpaired electrons.\n(a) $\\mathrm{K}_{3}\\left[\\mathrm{CrI}_{6}\\right]$\n(b) $\\left[\\mathrm{Cu}(\\mathrm{en})_{2}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{2}\\right] \\mathrm{Cl}_{2}$\n(c) $\\mathrm{Na}_{3}\\left[\\mathrm{Co}\\left(\\mathrm{NO}_{2}\\right)_{6}\\right]$"}
{"id": 4932, "contents": "2032. Solution - \nThe complexes are octahedral.\n(a) $\\mathrm{Cr}^{3+}$ has a $d^{3}$ configuration. These electrons will all be unpaired.\n(b) $\\mathrm{Cu}^{2+}$ is $d^{9}$, so there will be one unpaired electron.\n(c) $\\mathrm{Co}^{3+}$ has $d^{6}$ valence electrons, so the crystal field splitting will determine how many are paired. Nitrite is a strong-field ligand, so the complex will be low spin. Six electrons will go in the $t_{2 g}$ orbitals, leaving 0 unpaired."}
{"id": 4933, "contents": "2033. Check Your Learning - \nThe size of the crystal field splitting only influences the arrangement of electrons when there is a choice between pairing electrons and filling the higher-energy orbitals. For which $d$-electron configurations will there be a difference between high- and low-spin configurations in octahedral complexes?"}
{"id": 4934, "contents": "2034. Answer: - \n$d^{4}, d^{5}, d^{6}$, and $d^{7}$"}
{"id": 4935, "contents": "2036. CFT for Other Geometries - \nCFT is applicable to molecules in geometries other than octahedral. In octahedral complexes, remember that the lobes of the $e_{g}$ set point directly at the ligands. For tetrahedral complexes, the $d$ orbitals remain in place, but now we have only four ligands located between the axes (Figure 19.36). None of the orbitals points directly at the tetrahedral ligands. However, the $e_{g}$ set (along the Cartesian axes) overlaps with the ligands less than does the $t_{2 g}$ set. By analogy with the octahedral case, predict the energy diagram for the $d$ orbitals in a tetrahedral crystal field. To avoid confusion, the octahedral $e_{g}$ set becomes a tetrahedral e set, and the octahedral $t_{2 g}$ set becomes a $t_{2}$ set.\n\n\nFIGURE 19.36 This diagram shows the orientation of the tetrahedral ligands with respect to the axis system for the orbitals."}
{"id": 4936, "contents": "2037. Solution - \nSince CFT is based on electrostatic repulsion, the orbitals closer to the ligands will be destabilized and raised in energy relative to the other set of orbitals. The splitting is less than for octahedral complexes because the overlap is less, so $\\Delta_{\\text {tet }}$ is usually small $\\left(\\Delta_{\\text {tet }}=\\frac{4}{9} \\Delta_{\\text {oct }}\\right)$ :\n\n\nCheck Your Learning\nExplain how many unpaired electrons a tetrahedral $d^{4}$ ion will have."}
{"id": 4937, "contents": "2038. Answer: - \n4; because $\\Delta_{\\text {tet }}$ is small, all tetrahedral complexes are high spin and the electrons go into the $t_{2}$ orbitals before pairing\n\nThe other common geometry is square planar. It is possible to consider a square planar geometry as an octahedral structure with a pair of trans ligands removed. The removed ligands are assumed to be on the $z$-axis. This changes the distribution of the $d$ orbitals, as orbitals on or near the $z$-axis become more stable, and those on or near the $x$ - or $y$-axes become less stable. This results in the octahedral $t_{2 g}$ and the $e_{g}$ sets splitting and gives a more complicated pattern, as depicted below:\n\n\nMagnetic Moments of Molecules and Ions\nExperimental evidence of magnetic measurements supports the theory of high- and low-spin complexes. Remember that molecules such as $\\mathrm{O}_{2}$ that contain unpaired electrons are paramagnetic. Paramagnetic substances are attracted to magnetic fields. Many transition metal complexes have unpaired electrons and hence are paramagnetic. Molecules such as $\\mathrm{N}_{2}$ and ions such as $\\mathrm{Na}^{+}$and $\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{4-}$ that contain no unpaired electrons are diamagnetic. Diamagnetic substances have a slight tendency to be repelled by magnetic fields.\n\nWhen an electron in an atom or ion is unpaired, the magnetic moment due to its spin makes the entire atom or ion paramagnetic. The size of the magnetic moment of a system containing unpaired electrons is related directly to the number of such electrons: the greater the number of unpaired electrons, the larger the magnetic moment. Therefore, the observed magnetic moment is used to determine the number of unpaired electrons present. The measured magnetic moment of low-spin $d^{6}\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{4-}$ confirms that iron is diamagnetic,\nwhereas high-spin $d^{6}\\left[\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}$ has four unpaired electrons with a magnetic moment that confirms this arrangement."}
{"id": 4938, "contents": "2039. Colors of Transition Metal Complexes - \nWhen atoms or molecules absorb light at the proper frequency, their electrons are excited to higher-energy orbitals. For many main group atoms and molecules, the absorbed photons are in the ultraviolet range of the electromagnetic spectrum, which cannot be detected by the human eye. For coordination compounds, the energy difference between the $d$ orbitals often allows photons in the visible range to be absorbed.\n\nThe human eye perceives a mixture of all the colors, in the proportions present in sunlight, as white light. Complementary colors, those located across from each other on a color wheel, are also used in color vision. The eye perceives a mixture of two complementary colors, in the proper proportions, as white light. Likewise, when a color is missing from white light, the eye sees its complement. For example, when red photons are absorbed from white light, the eyes see the color green. When violet photons are removed from white light, the eyes see lemon yellow. The blue color of the $\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{4}\\right]^{2+}$ ion results because this ion absorbs orange and red light, leaving the complementary colors of blue and green (Figure 19.37).\n\n\nFIGURE 19.37 (a) An object is black if it absorbs all colors of light. If it reflects all colors of light, it is white. An object has a color if it absorbs all colors except one, such as this yellow strip. The strip also appears yellow if it absorbs the complementary color from white light (in this case, indigo). (b) Complementary colors are located directly across from one another on the color wheel. (c) A solution of $\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{4}\\right]^{2+}$ ions absorbs red and orange light, so the transmitted light appears as the complementary color, blue."}
{"id": 4939, "contents": "2041. Colors of Complexes - \nThe octahedral complex $\\left[\\mathrm{Ti}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{3+}$ has a single $d$ electron. To excite this electron from the ground state $t_{2 g}$ orbital to the $e_{g}$ orbital, this complex absorbs light from 450 to 600 nm . The maximum absorbance corresponds to $\\Delta_{\\text {oct }}$ and occurs at 499 nm . Calculate the value of $\\Delta_{\\text {oct }}$ in Joules and predict what color the solution will appear."}
{"id": 4940, "contents": "2042. Solution - \nUsing Planck's equation (refer to the section on electromagnetic energy), we calculate:\n\n$$\n\\begin{gathered}\nv=\\frac{c}{\\lambda} \\text { so } \\frac{3.00 \\times 10^{8} \\mathrm{~m} / \\mathrm{s}}{\\frac{499 \\mathrm{~nm} \\times 1 \\mathrm{~m}}{10^{9} \\mathrm{~nm}}}=6.01 \\times 10^{14} \\mathrm{~Hz} \\\\\nE=h \\nu \\text { so } 6.63 \\times 10^{-34} \\mathrm{~J} \\cdot \\mathrm{~s} \\times 6.01 \\times 10^{14} \\mathrm{~Hz}=3.99 \\times 10^{-19} \\text { Joules/ion }\n\\end{gathered}\n$$\n\nBecause the complex absorbs 600 nm (orange) through 450 (blue), the indigo, violet, and red wavelengths will be transmitted, and the complex will appear purple."}
{"id": 4941, "contents": "2043. Check Your Learning - \nA complex that appears green, absorbs photons of what wavelengths?"}
{"id": 4942, "contents": "2044. Answer: - \nred, 620-800 nm\n\nSmall changes in the relative energies of the orbitals that electrons are transitioning between can lead to drastic shifts in the color of light absorbed. Therefore, the colors of coordination compounds depend on many factors. As shown in Figure 19.38, different aqueous metal ions can have different colors. In addition, different oxidation states of one metal can produce different colors, as shown for the vanadium complexes in the link below.\n\n\nFIGURE 19.38 The partially filled $d$ orbitals of the stable ions $\\mathrm{Cr}^{3+}(\\mathrm{aq}), \\mathrm{Fe}^{3+}(\\mathrm{aq})$, and $\\mathrm{Co}^{2+}(\\mathrm{aq})$ (left, center and right, respectively) give rise to various colors. (credit: Sahar Atwa)\n\nThe specific ligands coordinated to the metal center also influence the color of coordination complexes. For example, the iron(II) complex $\\left[\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right] \\mathrm{SO}_{4}$ appears blue-green because the high-spin complex absorbs photons in the red wavelengths (Figure 19.39). In contrast, the low-spin iron(II) complex $\\mathrm{K}_{4}\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]$ appears pale yellow because it absorbs higher-energy violet photons.\n\n\nFIGURE 19.39 Both (a) hexaaquairon(II) sulfate and (b) potassium hexacyanoferrate(II) contain $d^{6}$ iron(II) octahedral metal centers, but they absorb photons in different ranges of the visible spectrum."}
{"id": 4943, "contents": "2045. LINK TO LEARNING - \nWatch this video (http://openstax.org/l/16vanadium) of the reduction of vanadium complexes to observe the colorful effect of changing oxidation states.\n\nIn general, strong-field ligands cause a large split in the energies of $d$ orbitals of the central metal atom (large $\\left.\\Delta_{\\text {oct }}\\right)$. Transition metal coordination compounds with these ligands are yellow, orange, or red because they absorb higher-energy violet or blue light. On the other hand, coordination compounds of transition metals with weak-field ligands are often blue-green, blue, or indigo because they absorb lower-energy yellow, orange, or red light.\n\nA coordination compound of the $\\mathrm{Cu}^{+}$ion has a $d^{10}$ configuration, and all the $e_{g}$ orbitals are filled. To excite an electron to a higher level, such as the $4 p$ orbital, photons of very high energy are necessary. This energy corresponds to very short wavelengths in the ultraviolet region of the spectrum. No visible light is absorbed, so the eye sees no change, and the compound appears white or colorless. A solution containing $\\left[\\mathrm{Cu}(\\mathrm{CN})_{2}\\right]^{-}$, for example, is colorless. On the other hand, octahedral $\\mathrm{Cu}^{2+}$ complexes have a vacancy in the $e_{g}$ orbitals, and electrons can be excited to this level. The wavelength (energy) of the light absorbed corresponds to the visible part of the spectrum, and $\\mathrm{Cu}^{2+}$ complexes are almost always colored-blue, blue-green violet, or yellow (Figure 19.40). Although CFT successfully describes many properties of coordination complexes, molecular orbital explanations (beyond the introductory scope provided here) are required to understand fully the behavior of coordination complexes.\n\n(a)\n\n(b)\n\nFIGURE 19.40 (a) Copper(I) complexes with $d^{10}$ configurations such as CuI tend to be colorless, whereas (b) $d^{9}$ copper(II) complexes such as $\\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2} \\cdot 5 \\mathrm{H}_{2} \\mathrm{O}$ are brightly colored."}
{"id": 4944, "contents": "2046. Key Terms - \nactinide series (also, actinoid series) actinium and the elements in the second row or the $f$-block, atomic numbers 89-103\nbidentate ligand ligand that coordinates to one central metal through coordinate bonds from two different atoms\ncentral metal ion or atom to which one or more ligands is attached through coordinate covalent bonds\nchelate complex formed from a polydentate ligand attached to a central metal\nchelating ligand ligand that attaches to a central metal ion by bonds from two or more donor atoms\ncis configuration configuration of a geometrical isomer in which two similar groups are on the same side of an imaginary reference line on the molecule\ncoordination compound stable compound in which the central metal atom or ion acts as a Lewis acid and accepts one or more pairs of electrons\ncoordination compound substance consisting of atoms, molecules, or ions attached to a central atom through Lewis acid-base interactions\ncoordination number number of coordinate covalent bonds to the central metal atom in a complex or the number of closest contacts to an atom in a crystalline form\ncoordination sphere central metal atom or ion plus the attached ligands of a complex\ncrystal field splitting ( $\\boldsymbol{\\Delta}_{\\mathbf{o c t}}$ ) difference in energy between the $t_{2 g}$ and $e_{g}$ sets or $t$ and e sets of orbitals\ncrystal field theory model that explains the energies of the orbitals in transition metals in terms of electrostatic interactions with the ligands but does not include metal ligand bonding\n$\\boldsymbol{d}$-block element one of the elements in groups $3-11$ with valence electrons in $d$ orbitals\ndonor atom atom in a ligand with a lone pair of electrons that forms a coordinate covalent bond to a central metal\n$\\boldsymbol{e}_{\\boldsymbol{g}}$ orbitals set of two $d$ orbitals that are oriented on the Cartesian axes for coordination complexes; in octahedral complexes, they are higher in energy than the $t_{2 g}$ orbitals\n$\\boldsymbol{f}$-block element (also, inner transition element) one of the elements with atomic numbers 58-71 or 90-103 that have valence electrons in $f$ orbitals; they are frequently shown offset below\nthe periodic table"}
{"id": 4945, "contents": "2046. Key Terms - \nthe periodic table\nfirst transition series transition elements in the fourth period of the periodic table (first row of the $d$-block), atomic numbers 21-29\nfourth transition series transition elements in the seventh period of the periodic table (fourth row of the $d$-block), atomic numbers 89 and 104-111\ngeometric isomers isomers that differ in the way in which atoms are oriented in space relative to each other, leading to different physical and chemical properties\nhigh-spin complex complex in which the electrons maximize the total electron spin by singly populating all of the orbitals before pairing two electrons into the lower-energy orbitals\nhydrometallurgy process in which a metal is separated from a mixture by first converting it into soluble ions, extracting the ions, and then reducing the ions to precipitate the pure metal\nionization isomer (or coordination isomer) isomer in which an anionic ligand is replaced by the counter ion in the inner coordination sphere\nlanthanide series (also, lanthanoid series) lanthanum and the elements in the first row or the $f$-block, atomic numbers 57-71\nligand ion or neutral molecule attached to the central metal ion in a coordination compound\nlinkage isomer coordination compound that possesses a ligand that can bind to the transition metal in two different ways ( $\\mathrm{CN}^{-}$vs. $\\mathrm{NC}^{-}$)\nlow-spin complex complex in which the electrons minimize the total electron spin by pairing in the lower-energy orbitals before populating the higher-energy orbitals\nmonodentate ligand that attaches to a central metal through just one coordinate covalent bond\noptical isomer (also, enantiomer) molecule that is a nonsuperimposable mirror image with identical chemical and physical properties, except when it reacts with other optical isomers\npairing energy ( $\\mathbf{P}$ ) energy required to place two electrons with opposite spins into a single orbital\nplatinum metals group of six transition metals consisting of ruthenium, osmium, rhodium, iridium, palladium, and platinum that tend to occur in the same minerals and demonstrate similar chemical properties"}
{"id": 4946, "contents": "2046. Key Terms - \nplatinum metals group of six transition metals consisting of ruthenium, osmium, rhodium, iridium, palladium, and platinum that tend to occur in the same minerals and demonstrate similar chemical properties\npolydentate ligand ligand that is attached to a central metal ion by bonds from two or more donor atoms, named with prefixes specifying how many donors are present (e.g., hexadentate = six coordinate bonds formed)\nrare earth element collection of 17 elements including the lanthanides, scandium, and yttrium that often occur together and have similar chemical properties, making separation difficult\nsecond transition series transition elements in the fifth period of the periodic table (second row of the $d$-block), atomic numbers 39-47\nsmelting process of extracting a pure metal from a molten ore\nspectrochemical series ranking of ligands according to the magnitude of the crystal field splitting they induce\nsteel material made from iron by removing impurities in the iron and adding substances that produce alloys with properties suitable for specific uses\nstrong-field ligand ligand that causes larger"}
{"id": 4947, "contents": "2047. Summary - 2047.1. Occurrence, Preparation, and Properties of Transition Metals and Their Compounds\nThe transition metals are elements with partially filled $d$ orbitals, located in the $d$-block of the periodic table. The reactivity of the transition elements varies widely from very active metals such as scandium and iron to almost inert elements, such as the platinum metals. The type of chemistry used in the isolation of the elements from their ores depends upon the concentration of the element in its ore and the difficulty of reducing ions of the elements to the metals. Metals that are more active are more difficult to reduce.\n\nTransition metals exhibit chemical behavior typical of metals. For example, they oxidize in air upon heating and react with elemental halogens to form halides. Those elements that lie above hydrogen in the activity series react with acids, producing salts and hydrogen gas. Oxides, hydroxides, and carbonates of transition metal compounds in low oxidation states are basic. Halides and other salts are generally stable in water, although oxygen must be excluded in some cases. Most transition metals form a variety of stable oxidation states, allowing them to demonstrate a wide range of chemical reactivity."}
{"id": 4948, "contents": "2047. Summary - 2047.2. Coordination Chemistry of Transition Metals\nThe transition elements and main group elements can form coordination compounds, or complexes, in\ncrystal field splittings\nsuperconductor material that conducts electricity with no resistance\n$\\boldsymbol{t}_{\\mathbf{2 g}}$ orbitals set of three $d$ orbitals aligned between the Cartesian axes for coordination complexes; in octahedral complexes, they are lowered in energy compared to the $e_{g}$ orbitals according to CFT\nthird transition series transition elements in the sixth period of the periodic table (third row of the $d$-block), atomic numbers 57 and 72-79\ntrans configuration configuration of a geometrical isomer in which two similar groups are on opposite sides of an imaginary reference line on the molecule\nweak-field ligand ligand that causes small crystal field splittings\nwhich a central metal atom or ion is bonded to one or more ligands by coordinate covalent bonds. Ligands with more than one donor atom are called polydentate ligands and form chelates. The common geometries found in complexes are tetrahedral and square planar (both with a coordination number of four) and octahedral (with a coordination number of six). Cis and trans configurations are possible in some octahedral and square planar complexes. In addition to these geometrical isomers, optical isomers (molecules or ions that are mirror images but not superimposable) are possible in certain octahedral complexes. Coordination complexes have a wide variety of uses including oxygen transport in blood, water purification, and pharmaceutical use."}
{"id": 4949, "contents": "2047. Summary - 2047.3. Spectroscopic and Magnetic Properties of Coordination Compounds\nCrystal field theory treats interactions between the electrons on the metal and the ligands as a simple electrostatic effect. The presence of the ligands near the metal ion changes the energies of the metal $d$ orbitals relative to their energies in the free ion. Both the color and the magnetic properties of a complex can be attributed to this crystal field splitting. The magnitude of the splitting ( $\\Delta_{\\text {oct }}$ ) depends on the nature of the ligands bonded to the metal. Strongfield ligands produce large splitting and favor lowspin complexes, in which the $t_{2 g}$ orbitals are completely filled before any electrons occupy the $e_{g}$ orbitals. Weak-field ligands favor formation of highspin complexes. The $t_{2 g}$ and the $e_{g}$ orbitals are singly occupied before any are doubly occupied."}
{"id": 4950, "contents": "2048. Exercises - 2048.1. Occurrence, Preparation, and Properties of Transition Metals and Their Compounds\n1. Write the electron configurations for each of the following elements:\n(a) Sc\n(b) Ti\n(c) Cr\n(d) Fe\n(e) Ru\n2. Write the electron configurations for each of the following elements and its ions:\n(a) Ti\n(b) $\\mathrm{Ti}^{2+}$\n(c) $\\mathrm{Ti}^{3+}$\n(d) $\\mathrm{Ti}^{4+}$\n3. Write the electron configurations for each of the following elements and its $3+$ ions:\n(a) La\n(b) Sm\n(c) Lu\n4. Why are the lanthanoid elements not found in nature in their elemental forms?\n5. Which of the following elements is most likely to be used to prepare La by the reduction of $\\mathrm{La}_{2} \\mathrm{O}_{3}$ : $\\mathrm{Al}, \\mathrm{C}$, or Fe? Why?\n6. Which of the following is the strongest oxidizing agent: $\\mathrm{VO}_{4}{ }^{3}, \\mathrm{CrO}_{4}{ }^{2-}$, or $\\mathrm{MnO}_{4}{ }^{-}$?\n7. Which of the following elements is most likely to form an oxide with the formula $\\mathrm{MO}_{3}: \\mathrm{Zr}, \\mathrm{Nb}$, or Mo ?\n8. The following reactions all occur in a blast furnace. Which of these are redox reactions?\n(a) $3 \\mathrm{Fe}_{2} \\mathrm{O}_{3}(s)+\\mathrm{CO}(g) \\longrightarrow 2 \\mathrm{Fe}_{3} \\mathrm{O}_{4}(s)+\\mathrm{CO}_{2}(g)$\n(b) $\\mathrm{Fe}_{3} \\mathrm{O}_{4}(s)+\\mathrm{CO}(g) \\longrightarrow 3 \\mathrm{FeO}(s)+\\mathrm{CO}_{2}(g)$\n(c) $\\mathrm{FeO}(s)+\\mathrm{CO}(g) \\longrightarrow \\mathrm{Fe}(l)+\\mathrm{CO}_{2}(g)$\n(d) $\\mathrm{C}(s)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)$"}
{"id": 4951, "contents": "2048. Exercises - 2048.1. Occurrence, Preparation, and Properties of Transition Metals and Their Compounds\n(d) $\\mathrm{C}(s)+\\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)$\n(e) $\\mathrm{C}(\\mathrm{s})+\\mathrm{CO}_{2}(g) \\longrightarrow 2 \\mathrm{CO}(g)$\n(f) $\\mathrm{CaCO}_{3}(s) \\longrightarrow \\mathrm{CaO}(s)+\\mathrm{CO}_{2}(g)$\n(g) $\\mathrm{CaO}(s)+\\mathrm{SiO}_{2}(s) \\longrightarrow \\mathrm{CaSiO}_{3}(l)$\n9. Why is the formation of slag useful during the smelting of iron?\n10. Would you expect an aqueous manganese(VII) oxide solution to have a pH greater or less than 7.0 ? Justify your answer.\n11. Iron(II) can be oxidized to iron(III) by dichromate ion, which is reduced to chromium(III) in acid solution. A $2.5000-\\mathrm{g}$ sample of iron ore is dissolved and the iron converted into iron(II). Exactly 19.17 mL of 0.0100 $M \\mathrm{Na}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$ is required in the titration. What percentage of the ore sample was iron?\n12. How many cubic feet of air at a pressure of 760 torr and $0^{\\circ} \\mathrm{C}$ is required per ton of $\\mathrm{Fe}_{2} \\mathrm{O}_{3}$ to convert that $\\mathrm{Fe}_{2} \\mathrm{O}_{3}$ into iron in a blast furnace? For this exercise, assume air is $19 \\%$ oxygen by volume.\n13. Find the potentials of the following electrochemical cell:\n$\\mathrm{Cd}\\left|\\mathrm{Cd}^{2+}, M=0.10 \\| \\mathrm{Ni}^{2+}, M=0.50\\right| \\mathrm{Ni}$"}
{"id": 4952, "contents": "2048. Exercises - 2048.1. Occurrence, Preparation, and Properties of Transition Metals and Their Compounds\n$\\mathrm{Cd}\\left|\\mathrm{Cd}^{2+}, M=0.10 \\| \\mathrm{Ni}^{2+}, M=0.50\\right| \\mathrm{Ni}$\n14. A $2.5624-\\mathrm{g}$ sample of a pure solid alkali metal chloride is dissolved in water and treated with excess silver nitrate. The resulting precipitate, filtered and dried, weighs 3.03707 g . What was the percent by mass of chloride ion in the original compound? What is the identity of the salt?\n15. The standard reduction potential for the reaction $\\left[\\mathrm{Co}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{3+}(a q)+\\mathrm{e}^{-} \\longrightarrow\\left[\\mathrm{Co}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}(a q)$ is about 1.8 V . The reduction potential for the reaction $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]^{3+}(a q)+\\mathrm{e}^{-} \\longrightarrow\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]^{2+}(a q)$ is +0.1 V . Calculate the cell potentials to show whether the complex ions, $\\left[\\mathrm{Co}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}$ and/or $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]^{2+}$, can be oxidized to the corresponding cobalt(III) complex by oxygen.\n16. Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.)\n(a) $\\mathrm{MnCO}_{3}(s)+\\mathrm{HI}(a q) \\longrightarrow$\n(b) $\\mathrm{CoO}(s)+\\mathrm{O}_{2}(g)$\n(c) $\\mathrm{La}(s)+\\mathrm{O}_{2}(g) \\longrightarrow$"}
{"id": 4953, "contents": "2048. Exercises - 2048.1. Occurrence, Preparation, and Properties of Transition Metals and Their Compounds\n(b) $\\mathrm{CoO}(s)+\\mathrm{O}_{2}(g)$\n(c) $\\mathrm{La}(s)+\\mathrm{O}_{2}(g) \\longrightarrow$\n(d) $\\mathrm{V}(s)+\\mathrm{VCl}_{4}(s) \\longrightarrow$\n(e) $\\mathrm{Co}(s)+x s \\mathrm{~F}_{2}(g) \\longrightarrow$\n(f) $\\mathrm{CrO}_{3}(s)+\\mathrm{CsOH}(a q) \\longrightarrow$\n17. Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.)\n(a) $\\mathrm{Fe}(s)+\\mathrm{H}_{2} \\mathrm{SO}_{4}(a q) \\longrightarrow$\n(b) $\\mathrm{FeCl}_{3}(a q)+\\mathrm{NaOH}(a q) \\longrightarrow$\n(c) $\\mathrm{Mn}(\\mathrm{OH})_{2}(s)+\\mathrm{HBr}(a q) \\longrightarrow$\n(d) $\\mathrm{Cr}(s)+\\mathrm{O}_{2}(g) \\longrightarrow$\n(e) $\\mathrm{Mn}_{2} \\mathrm{O}_{3}(s)+\\mathrm{HCl}(a q) \\longrightarrow$\n(f) $\\mathrm{Ti}(s)+x s \\mathrm{~F}_{2}(g) \\longrightarrow$\n18. Describe the electrolytic process for refining copper.\n19. Predict the products of the following reactions and balance the equations.\n(a) Zn is added to a solution of $\\mathrm{Cr}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}$ in acid.\n(b) $\\mathrm{FeCl}_{2}$ is added to a solution containing an excess of $\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}$ in hydrochloric acid.\n(c) $\\mathrm{Cr}^{2+}$ is added to $\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}$ in acid solution.\n(d) Mn is heated with $\\mathrm{CrO}_{3}$."}
{"id": 4954, "contents": "2048. Exercises - 2048.1. Occurrence, Preparation, and Properties of Transition Metals and Their Compounds\n(d) Mn is heated with $\\mathrm{CrO}_{3}$.\n(e) CrO is added to $2 \\mathrm{HNO}_{3}$ in water.\n(f) $\\mathrm{FeCl}_{3}$ is added to an aqueous solution of NaOH .\n20. What is the gas produced when iron(II) sulfide is treated with a nonoxidizing acid?\n21. Predict the products of each of the following reactions and then balance the chemical equations.\n(a) Fe is heated in an atmosphere of steam.\n(b) NaOH is added to a solution of $\\mathrm{Fe}\\left(\\mathrm{NO}_{3}\\right)_{3}$.\n(c) $\\mathrm{FeSO}_{4}$ is added to an acidic solution of $\\mathrm{KMnO}_{4}$.\n(d) Fe is added to a dilute solution of $\\mathrm{H}_{2} \\mathrm{SO}_{4}$.\n(e) A solution of $\\mathrm{Fe}\\left(\\mathrm{NO}_{3}\\right)_{2}$ and $\\mathrm{HNO}_{3}$ is allowed to stand in air.\n(f) $\\mathrm{FeCO}_{3}$ is added to a solution of $\\mathrm{HClO}_{4}$.\n(g) Fe is heated in air.\n22. Balance the following equations by oxidation-reduction methods; note that three elements change oxidation state.\n$\\mathrm{Co}\\left(\\mathrm{NO}_{3}\\right)_{2}(s) \\longrightarrow \\mathrm{Co}_{2} \\mathrm{O}_{3}(s)+\\mathrm{NO}_{2}(g)+\\mathrm{O}_{2}(g)$\n23. Dilute sodium cyanide solution is slowly dripped into a slowly stirred silver nitrate solution. A white precipitate forms temporarily but dissolves as the addition of sodium cyanide continues. Use chemical equations to explain this observation. Silver cyanide is similar to silver chloride in its solubility.\n24. Predict which will be more stable, $\\left[\\mathrm{CrO}_{4}\\right]^{2-}$ or $\\left[\\mathrm{WO}_{4}\\right]^{2-}$, and explain."}
{"id": 4955, "contents": "2048. Exercises - 2048.1. Occurrence, Preparation, and Properties of Transition Metals and Their Compounds\n24. Predict which will be more stable, $\\left[\\mathrm{CrO}_{4}\\right]^{2-}$ or $\\left[\\mathrm{WO}_{4}\\right]^{2-}$, and explain.\n25. Give the oxidation state of the metal for each of the following oxides of the first transition series. (Hint: Oxides of formula $\\mathrm{M}_{3} \\mathrm{O}_{4}$ are examples of mixed valence compounds in which the metal ion is present in more than one oxidation state. It is possible to write these compound formulas in the equivalent format $\\mathrm{MO} \\cdot \\mathrm{M}_{2} \\mathrm{O}_{3}$, to permit estimation of the metal's two oxidation states.)\n(a) $\\mathrm{Sc}_{2} \\mathrm{O}_{3}$\n(b) $\\mathrm{TiO}_{2}$\n(c) $\\mathrm{V}_{2} \\mathrm{O}_{5}$\n(d) $\\mathrm{CrO}_{3}$\n(e) $\\mathrm{MnO}_{2}$\n(f) $\\mathrm{Fe}_{3} \\mathrm{O}_{4}$\n(g) $\\mathrm{Co}_{3} \\mathrm{O}_{4}$\n(h) NiO\n(i) $\\mathrm{Cu}_{2} \\mathrm{O}$"}
{"id": 4956, "contents": "2048. Exercises - 2048.2. Coordination Chemistry of Transition Metals\n26. Indicate the coordination number for the central metal atom in each of the following coordination compounds:\n(a) $\\left[\\mathrm{Pt}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{2} \\mathrm{Br}_{2}\\right]$\n(b) $\\left[\\mathrm{Pt}\\left(\\mathrm{NH}_{3}\\right)(\\mathrm{py})(\\mathrm{Cl})(\\mathrm{Br})\\right]\\left(\\mathrm{py}=\\right.$ pyridine, $\\left.\\mathrm{C}_{5} \\mathrm{H}_{5} \\mathrm{~N}\\right)$\n(c) $\\left[\\mathrm{Zn}\\left(\\mathrm{NH}_{3}\\right)_{2} \\mathrm{Cl}_{2}\\right]$\n(d) $\\left[\\mathrm{Zn}\\left(\\mathrm{NH}_{3}\\right)(\\mathrm{py})(\\mathrm{Cl})(\\mathrm{Br})\\right]$\n(e) $\\left[\\mathrm{Ni}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{4} \\mathrm{Cl}_{2}\\right]$\n(f) $\\left[\\mathrm{Fe}(\\mathrm{en})_{2}(\\mathrm{CN})_{2}\\right]^{+}\\left(\\mathrm{en}=\\right.$ ethylenediamine, $\\left.\\mathrm{C}_{2} \\mathrm{H}_{8} \\mathrm{~N}_{2}\\right)$\n27. Give the coordination numbers and write the formulas for each of the following, including all isomers where appropriate:\n(a) tetrahydroxozincate(II) ion (tetrahedral)\n(b) hexacyanopalladate(IV) ion\n(c) dichloroaurate(I) ion (note that aurum is Latin for \"gold\")\n(d) diamminedichloroplatinum(II)\n(e) potassium diamminetetrachlorochromate(III)\n(f) hexaamminecobalt(III) hexacyanochromate(III)\n(g) dibromobis(ethylenediamine) cobalt(III) nitrate\n28. Give the coordination number for each metal ion in the following compounds:"}
{"id": 4957, "contents": "2048. Exercises - 2048.2. Coordination Chemistry of Transition Metals\n(g) dibromobis(ethylenediamine) cobalt(III) nitrate\n28. Give the coordination number for each metal ion in the following compounds:\n(a) $\\left[\\mathrm{Co}\\left(\\mathrm{CO}_{3}\\right)_{3}\\right]^{3-}$ (note that $\\mathrm{CO}_{3}{ }^{2-}$ is bidentate in this complex)\n(b) $\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{4}\\right]^{2+}$\n(c) $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{4} \\mathrm{Br}_{2}\\right]_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}$\n(d) $\\left[\\mathrm{Pt}\\left(\\mathrm{NH}_{3}\\right)_{4}\\right]\\left[\\mathrm{PtCl}_{4}\\right]$\n(e) $\\left[\\mathrm{Cr}(\\mathrm{en})_{3}\\right]\\left(\\mathrm{NO}_{3}\\right)_{3}$\n(f) $\\left[\\mathrm{Pd}\\left(\\mathrm{NH}_{3}\\right)_{2} \\mathrm{Br}_{2}\\right]$ (square planar)\n(g) $\\mathrm{K}_{3}\\left[\\mathrm{Cu}(\\mathrm{Cl})_{5}\\right]$\n(h) $\\left[\\mathrm{Zn}\\left(\\mathrm{NH}_{3}\\right)_{2} \\mathrm{Cl}_{2}\\right]$\n29. Sketch the structures of the following complexes. Indicate any cis, trans, and optical isomers.\n(a) $\\left[\\mathrm{Pt}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{2} \\mathrm{Br}_{2}\\right]$ (square planar)\n(b) $\\left[\\mathrm{Pt}\\left(\\mathrm{NH}_{3}\\right)(\\mathrm{py})(\\mathrm{Cl})(\\mathrm{Br})\\right]$ (square planar, py $=$ pyridine, $\\mathrm{C}_{5} \\mathrm{H}_{5} \\mathrm{~N}$ )"}
{"id": 4958, "contents": "2048. Exercises - 2048.2. Coordination Chemistry of Transition Metals\n(c) $\\left[\\mathrm{Zn}\\left(\\mathrm{NH}_{3}\\right)_{3} \\mathrm{Cl}\\right]^{+}$(tetrahedral)\n(d) $\\left[\\mathrm{Pt}\\left(\\mathrm{NH}_{3}\\right)_{3} \\mathrm{Cl}\\right]^{+}$(square planar)\n(e) $\\left[\\mathrm{Ni}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{4} \\mathrm{Cl}_{2}\\right]$\n(f) $\\left[\\mathrm{Co}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}\\right)_{2} \\mathrm{Cl}_{2}\\right]^{3-}$ (note that $\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}$ is the bidentate oxalate ion, $-\\mathrm{O}_{2} \\mathrm{CCO}_{2}{ }^{-}$)\n30. Draw diagrams for any cis, trans, and optical isomers that could exist for the following (en is ethylenediamine):\n(a) $\\left[\\mathrm{Co}(\\mathrm{en})_{2}\\left(\\mathrm{NO}_{2}\\right) \\mathrm{Cl}\\right]^{+}$\n(b) $\\left[\\mathrm{Co}(\\mathrm{en})_{2} \\mathrm{Cl}_{2}\\right]^{+}$\n(c) $\\left[\\mathrm{Pt}\\left(\\mathrm{NH}_{3}\\right)_{2} \\mathrm{Cl}_{4}\\right]$\n(d) $\\left[\\mathrm{Cr}(\\mathrm{en})_{3}\\right]^{3+}$\n(e) $\\left[\\mathrm{Pt}\\left(\\mathrm{NH}_{3}\\right)_{2} \\mathrm{Cl}_{2}\\right]$\n31. Name each of the compounds or ions given in Exercise 19.28 , including the oxidation state of the metal.\n32. Name each of the compounds or ions given in Exercise 19.30.\n33. Specify whether the following complexes have isomers.\n(a) tetrahedral $\\left[\\mathrm{Ni}(\\mathrm{CO})_{2}(\\mathrm{Cl})_{2}\\right]$"}
{"id": 4959, "contents": "2048. Exercises - 2048.2. Coordination Chemistry of Transition Metals\n33. Specify whether the following complexes have isomers.\n(a) tetrahedral $\\left[\\mathrm{Ni}(\\mathrm{CO})_{2}(\\mathrm{Cl})_{2}\\right]$\n(b) trigonal bipyramidal $\\left[\\mathrm{Mn}(\\mathrm{CO})_{4} \\mathrm{NO}\\right]$\n(c) $\\left[\\mathrm{Pt}(\\mathrm{en}){ }_{2} \\mathrm{Cl}_{2}\\right] \\mathrm{Cl}_{2}$\n34. Predict whether the carbonate ligand $\\mathrm{CO}_{3}{ }^{2-}$ will coordinate to a metal center as a monodentate, bidentate, or tridentate ligand.\n35. Draw the geometric, linkage, and ionization isomers for $\\left\\mathrm{CoCl}_{5} \\mathrm{CN}\\right$."}
{"id": 4960, "contents": "2048. Exercises - 2048.3. Spectroscopic and Magnetic Properties of Coordination Compounds\n36. Determine the number of unpaired electrons expected for $\\left[\\mathrm{Fe}\\left(\\mathrm{NO}_{2}\\right)_{6}\\right]^{3-}$ and for $\\left[\\mathrm{FeF} \\mathrm{F}_{6}\\right]^{3-}$ in terms of crystal field theory.\n37. Draw the crystal field diagrams for $\\left[\\mathrm{Fe}\\left(\\mathrm{NO}_{2}\\right)_{6}\\right]^{4-}$ and $\\left[\\mathrm{FeF}_{6}\\right]^{3-}$. State whether each complex is high spin or low spin, paramagnetic or diamagnetic, and compare $\\Delta_{\\text {oct }}$ to $P$ for each complex.\n38. Give the oxidation state of the metal, number of $d$ electrons, and the number of unpaired electrons predicted for $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right] \\mathrm{Cl}_{3}$.\n39. The solid anhydrous solid $\\mathrm{CoCl}_{2}$ is blue in color. Because it readily absorbs water from the air, it is used as a humidity indicator to monitor if equipment (such as a cell phone) has been exposed to excessive levels of moisture. Predict what product is formed by this reaction, and how many unpaired electrons this complex will have.\n40. Is it possible for a complex of a metal in the transition series to have six unpaired electrons? Explain.\n41. How many unpaired electrons are present in each of the following?\n(a) $\\left[\\mathrm{CoF}_{6}\\right]^{3-}$ (high spin)\n(b) $\\left[\\mathrm{Mn}(\\mathrm{CN})_{6}\\right]^{3-}$ (low spin)\n(c) $\\left[\\mathrm{Mn}(\\mathrm{CN})_{6}\\right]^{4-}$ (low spin)\n(d) $\\left[\\mathrm{MnCl}_{6}\\right]^{4-}$ (high spin)\n(e) $\\left[\\mathrm{RhCl}_{6}\\right]^{3-}$ (low spin)"}
{"id": 4961, "contents": "2048. Exercises - 2048.3. Spectroscopic and Magnetic Properties of Coordination Compounds\n(d) $\\left[\\mathrm{MnCl}_{6}\\right]^{4-}$ (high spin)\n(e) $\\left[\\mathrm{RhCl}_{6}\\right]^{3-}$ (low spin)\n42. Explain how the diphosphate ion, $\\left[\\mathrm{O}_{3} \\mathrm{P}-\\mathrm{O}-\\mathrm{PO}_{3}\\right]^{4-}$, can function as a water softener that prevents the precipitation of $\\mathrm{Fe}^{2+}$ as an insoluble iron salt.\n43. For complexes of the same metal ion with no change in oxidation number, the stability increases as the number of electrons in the $t_{2 g}$ orbitals increases. Which complex in each of the following pairs of complexes is more stable?\n(a) $\\left[\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}$ or $\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{4-}$\n(b) $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]^{3+}$ or $\\left[\\mathrm{CoF}_{6}\\right]^{3-}$\n(c) $\\left[\\mathrm{Mn}(\\mathrm{CN})_{6}\\right]^{4-}$ or $\\left[\\mathrm{MnCl}_{6}\\right]^{4-}$\n44. Trimethylphosphine, $\\mathrm{P}\\left(\\mathrm{CH}_{3}\\right)_{3}$, can act as a ligand by donating the lone pair of electrons on the phosphorus atom. If trimethylphosphine is added to a solution of nickel(II) chloride in acetone, a blue compound that has a molecular mass of approximately 270 g and contains $21.5 \\% \\mathrm{Ni}, 26.0 \\% \\mathrm{Cl}$, and $52.5 \\% \\mathrm{P}\\left(\\mathrm{CH}_{3}\\right)_{3}$ can be isolated. This blue compound does not have any isomeric forms. What are the geometry and molecular formula of the blue compound?"}
{"id": 4962, "contents": "2048. Exercises - 2048.3. Spectroscopic and Magnetic Properties of Coordination Compounds\n45. Would you expect the complex $\\left[\\mathrm{Co}(\\mathrm{en})_{3}\\right]_{\\mathrm{Cl}}^{3}$ to have any unpaired electrons? Any isomers?\n46. Would you expect the $\\mathrm{Mg}_{3}\\left[\\mathrm{Cr}(\\mathrm{CN})_{6}\\right]_{2}$ to be diamagnetic or paramagnetic? Explain your reasoning.\n47. Would you expect salts of the gold(I) ion, $\\mathrm{Au}^{+}$, to be colored? Explain.\n48. $\\left[\\mathrm{CuCl}_{4}\\right]^{2-}$ is green. $\\left[\\mathrm{Cu}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}$ is blue. Which absorbs higher-energy photons? Which is predicted to have a larger crystal field splitting?"}
{"id": 4963, "contents": "2049. CHAPTER 20 <br> Nuclear Chemistry - \nFigure 20.1 Nuclear chemistry provides the basis for many useful diagnostic and therapeutic methods in medicine, such as these positron emission tomography (PET) scans. The PET/computed tomography scan on the left shows muscle activity. The brain scans in the center show chemical differences in dopamine signaling in the brains of addicts and nonaddicts. The images on the right show an oncological application of PET scans to identify lymph node metastasis."}
{"id": 4964, "contents": "2050. CHAPTER OUTLINE - \n20.1 Nuclear Structure and Stability\n20.2 Nuclear Equations\n20.3 Radioactive Decay\n20.4 Transmutation and Nuclear Energy\n20.5 Uses of Radioisotopes\n20.6 Biological Effects of Radiation\n\nINTRODUCTION The chemical reactions that we have considered in previous chapters involve changes in the electronic structure of the species involved, that is, the arrangement of the electrons around atoms, ions, or molecules. Nuclear structure, the numbers of protons and neutrons within the nuclei of the atoms involved, remains unchanged during chemical reactions.\n\nThis chapter will introduce the topic of nuclear chemistry, which began with the discovery of radioactivity in 1896 by French physicist Antoine Becquerel and has become increasingly important during the twentieth and twenty-first centuries, providing the basis for various technologies related to energy, medicine, geology, and many other areas."}
{"id": 4965, "contents": "2051. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe nuclear structure in terms of protons, neutrons, and electrons\n- Calculate mass defect and binding energy for nuclei\n- Explain trends in the relative stability of nuclei\n\nNuclear chemistry is the study of reactions that involve changes in nuclear structure. The chapter on atoms, molecules, and ions introduced the basic idea of nuclear structure, that the nucleus of an atom is composed of protons and, with the exception of ${ }_{1}^{1} \\mathrm{H}$, neutrons. Recall that the number of protons in the nucleus is called the atomic number ( Z ) of the element, and the sum of the number of protons and the number of neutrons is the mass number (A). Atoms with the same atomic number but different mass numbers are isotopes of the same element. When referring to a single type of nucleus, we often use the term nuclide and identify it by the notation ${ }_{Z}^{A} \\mathrm{X}$, where X is the symbol for the element, A is the mass number, and Z is the atomic number (for example, ${ }_{6}^{14} \\mathrm{C}$ ). Often a nuclide is referenced by the name of the element followed by a hyphen and the mass number. For example, ${ }_{6}^{14} \\mathrm{C}$ is called \"carbon-14.\""}
{"id": 4966, "contents": "2051. LEARNING OBJECTIVES - \nProtons and neutrons, collectively called nucleons, are packed together tightly in a nucleus. With a radius of about $10^{-15}$ meters, a nucleus is quite small compared to the radius of the entire atom, which is about $10^{-10}$ meters. Nuclei are extremely dense compared to bulk matter, averaging $1.8 \\times 10^{14}$ grams per cubic centimeter. For example, water has a density of 1 gram per cubic centimeter, and iridium, one of the densest elements known, has a density of $22.6 \\mathrm{~g} / \\mathrm{cm}^{3}$. If the earth's density were equal to the average nuclear density, the earth's radius would be only about 200 meters (earth's actual radius is approximately $6.4 \\times 10^{6}$ meters, 30,000 times larger). Example 20.1 demonstrates just how great nuclear densities can be in the natural world."}
{"id": 4967, "contents": "2053. Density of a Neutron Star - \nNeutron stars form when the core of a very massive star undergoes gravitational collapse, causing the star's outer layers to explode in a supernova. Composed almost completely of neutrons, they are the densest-known stars in the universe, with densities comparable to the average density of an atomic nucleus. A neutron star in a faraway galaxy has a mass equal to 2.4 solar masses ( 1 solar mass $=M_{\\odot}=$ mass of the sun $=1.99 \\times 10^{30} \\mathrm{~kg}$ ) and a diameter of 26 km .\n(a) What is the density of this neutron star?\n(b) How does this neutron star's density compare to the density of a uranium nucleus, which has a diameter of about $15 \\mathrm{fm}\\left(1 \\mathrm{fm}=10^{-15} \\mathrm{~m}\\right)$ ?"}
{"id": 4968, "contents": "2054. Solution - \nWe can treat both the neutron star and the U-235 nucleus as spheres. Then the density for both is given by:\n\n$$\nd=\\frac{m}{V} \\quad \\text { with } \\quad V=\\frac{4}{3} \\pi r^{3}\n$$\n\n(a) The radius of the neutron star is $\\frac{1}{2} \\times 26 \\mathrm{~km}=\\frac{1}{2} \\times 2.6 \\times 10^{4} \\mathrm{~m}=1.3 \\times 10^{4} \\mathrm{~m}$, so the density of the neutron star is:\n\n$$\nd=\\frac{m}{V}=\\frac{m}{\\frac{4}{3} \\pi r^{3}}=\\frac{2.4\\left(1.99 \\times 10^{30} \\mathrm{~kg}\\right)}{\\frac{4}{3} \\pi\\left(1.3 \\times 10^{4} \\mathrm{~m}\\right)^{3}}=5.2 \\times 10^{17} \\mathrm{~kg} / \\mathrm{m}^{3}\n$$\n\n(b) The radius of the $\\mathrm{U}-235$ nucleus is $\\frac{1}{2} \\times 15 \\times 10^{-15} \\mathrm{~m}=7.5 \\times 10^{-15} \\mathrm{~m}$, so the density of the U-235 nucleus is:\n\n$$\nd=\\frac{m}{V}=\\frac{m}{\\frac{4}{3} \\pi r^{3}}=\\frac{235 \\mathrm{amu}\\left(\\frac{1.66 \\times 10^{-27} \\mathrm{~kg}}{1 \\mathrm{amu}}\\right)}{\\frac{4}{3} \\pi\\left(7.5 \\times 10^{-15} \\mathrm{~m}\\right)^{3}}=2.2 \\times 10^{17} \\mathrm{~kg} / \\mathrm{m}^{3}\n$$\n\nThese values are fairly similar (same order of magnitude), but the neutron star is more than twice as dense as the U-235 nucleus."}
{"id": 4969, "contents": "2055. Check Your Learning - \nFind the density of a neutron star with a mass of 1.97 solar masses and a diameter of 13 km , and compare it to the density of a hydrogen nucleus, which has a diameter of $1.75 \\mathrm{fm}\\left(1 \\mathrm{fm}=1 \\times 10^{-15} \\mathrm{~m}\\right)$."}
{"id": 4970, "contents": "2056. Answer: - \nThe density of the neutron star is $3.4 \\times 10^{18} \\mathrm{~kg} / \\mathrm{m}^{3}$. The density of a hydrogen nucleus is $6.0 \\times 10^{17} \\mathrm{~kg} / \\mathrm{m}^{3}$. The neutron star is 5.7 times denser than the hydrogen nucleus.\n\nTo hold positively charged protons together in the very small volume of a nucleus requires very strong attractive forces because the positively charged protons repel one another strongly at such short distances. The force of attraction that holds the nucleus together is the strong nuclear force. (The strong force is one of the four fundamental forces that are known to exist. The others are the electromagnetic force, the gravitational force, and the nuclear weak force.) This force acts between protons, between neutrons, and between protons and neutrons. It is very different from the electrostatic force that holds negatively charged electrons around a positively charged nucleus (the attraction between opposite charges). Over distances less than $10^{-15}$ meters and within the nucleus, the strong nuclear force is much stronger than electrostatic repulsions between protons; over larger distances and outside the nucleus, it is essentially nonexistent."}
{"id": 4971, "contents": "2057. LINK TO LEARNING - \nVisit this website (http://openstax.org/1/16fourfund) for more information about the four fundamental forces."}
{"id": 4972, "contents": "2058. Nuclear Binding Energy - \nAs a simple example of the energy associated with the strong nuclear force, consider the helium atom composed of two protons, two neutrons, and two electrons. The total mass of these six subatomic particles may be calculated as:\n\n$$\n(2 \\times \\underset{\\text { protons }}{1.0073 \\mathrm{amu}})+\\underset{\\text { neutrons }}{(2 \\times 1.0087 \\mathrm{amu})}+\\underset{\\text { electrons }}{1 \\times 0.00055 \\mathrm{amu})}=4.0 \\begin{aligned}\n& 4.0331 \\mathrm{amu} \\\\\n& (2 \\times 0\n\\end{aligned}\n$$\n\nHowever, mass spectrometric measurements reveal that the mass of an ${ }_{2}^{4} \\mathrm{He}$ atom is 4.0026 amu , less than the combined masses of its six constituent subatomic particles. This difference between the calculated and experimentally measured masses is known as the mass defect of the atom. In the case of helium, the mass defect indicates a \"loss\" in mass of $4.0331 \\mathrm{amu}-4.0026 \\mathrm{amu}=0.0305 \\mathrm{amu}$. The loss in mass accompanying the formation of an atom from protons, neutrons, and electrons is due to the conversion of that mass into energy that is evolved as the atom forms. The nuclear binding energy is the energy produced when the atoms' nucleons are bound together; this is also the energy needed to break a nucleus into its constituent protons and neutrons. In comparison to chemical bond energies, nuclear binding energies are vastly greater, as we will learn in this section. Consequently, the energy changes associated with nuclear reactions are vastly greater than are those for chemical reactions.\n\nThe conversion between mass and energy is most identifiably represented by the mass-energy equivalence equation as stated by Albert Einstein:\n\n$$\nE=m c^{2}\n$$"}
{"id": 4973, "contents": "2058. Nuclear Binding Energy - \nThe conversion between mass and energy is most identifiably represented by the mass-energy equivalence equation as stated by Albert Einstein:\n\n$$\nE=m c^{2}\n$$\n\nwhere $E$ is energy, $m$ is mass of the matter being converted, and $c$ is the speed of light in a vacuum. This equation can be used to find the amount of energy that results when matter is converted into energy. Using this\nmass-energy equivalence equation, the nuclear binding energy of a nucleus may be calculated from its mass defect, as demonstrated in Example 20.2. A variety of units are commonly used for nuclear binding energies, including electron volts (eV), with 1 eV equaling the amount of energy necessary to the move the charge of an electron across an electric potential difference of 1 volt, making $1 \\mathrm{eV}=1.602 \\times 10^{-19} \\mathrm{~J}$."}
{"id": 4974, "contents": "2060. Calculation of Nuclear Binding Energy - \nDetermine the binding energy for the nuclide ${ }_{2}^{4} \\mathrm{He}$ in:\n(a) joules per mole of nuclei\n(b) joules per nucleus\n(c) MeV per nucleus"}
{"id": 4975, "contents": "2061. Solution - \nThe mass defect for a ${ }_{2}^{4} \\mathrm{He}$ nucleus is 0.0305 amu , as shown previously. Determine the binding energy in joules per nuclide using the mass-energy equivalence equation. To accommodate the requested energy units, the mass defect must be expressed in kilograms (recall that $1 \\mathrm{~J}=1 \\mathrm{~kg} \\mathrm{~m}^{2} / \\mathrm{s}^{2}$ ).\n(a) First, express the mass defect in $\\mathrm{g} / \\mathrm{mol}$. This is easily done considering the numerical equivalence of atomic mass (amu) and molar mass ( $\\mathrm{g} / \\mathrm{mol}$ ) that results from the definitions of the amu and mole units (refer to the previous discussion in the chapter on atoms, molecules, and ions if needed). The mass defect is therefore $0.0305 \\mathrm{~g} / \\mathrm{mol}$. To accommodate the units of the other terms in the mass-energy equation, the mass must be expressed in kg , since $1 \\mathrm{~J}=1 \\mathrm{~kg} \\mathrm{~m}^{2} / \\mathrm{s}^{2}$. Converting grams into kilograms yields a mass defect of $3.05 \\times 10^{-5} \\mathrm{~kg} /$ mol. Substituting this quantity into the mass-energy equivalence equation yields:\n\n$$\n\\begin{aligned}\n& E=m c^{2}=\\frac{3.05 \\times 10^{-5} \\mathrm{~kg}}{\\mathrm{~mol}} \\times\\left(\\frac{2.998 \\times 10^{8} \\mathrm{~m}}{\\mathrm{~s}}\\right)^{2}=2.74 \\times 10^{12} \\mathrm{~kg} \\mathrm{~m}^{2} \\mathrm{~s}^{-2} \\mathrm{~mol}^{-1} \\\\\n& =2.74 \\times 10^{12} \\mathrm{~J} \\mathrm{~mol}^{-1}=2.74 \\mathrm{TJ} \\mathrm{~mol}^{-1}\n\\end{aligned}\n$$"}
{"id": 4976, "contents": "2061. Solution - \nNote that this tremendous amount of energy is associated with the conversion of a very small amount of matter (about 30 mg , roughly the mass of typical drop of water).\n(b) The binding energy for a single nucleus is computed from the molar binding energy using Avogadro's number:\n\n$$\nE=2.74 \\times 10^{12} \\mathrm{~J} \\mathrm{~mol}^{-1} \\times \\frac{1 \\mathrm{~mol}}{6.022 \\times 10^{23} \\text { nuclei }}=4.55 \\times 10^{-12} \\mathrm{~J}=4.55 \\mathrm{pJ}\n$$\n\n(c) Recall that $1 \\mathrm{eV}=1.602 \\times 10^{-19} \\mathrm{~J}$. Using the binding energy computed in part (b):\n\n$$\nE=4.55 \\times 10^{-12} \\mathrm{~J} \\times \\frac{1 \\mathrm{eV}}{1.602 \\times 10^{-19} \\mathrm{~J}}=2.84 \\times 10^{7} \\mathrm{eV}=28.4 \\mathrm{MeV}\n$$"}
{"id": 4977, "contents": "2062. Check Your Learning - \nWhat is the binding energy for the nuclide ${ }_{9}^{19} \\mathrm{~F}$ (atomic mass: 18.9984 amu ) in MeV per nucleus?"}
{"id": 4978, "contents": "2063. Answer: - \n148.4 MeV\n\nBecause the energy changes for breaking and forming bonds are so small compared to the energy changes for breaking or forming nuclei, the changes in mass during all ordinary chemical reactions are virtually undetectable. As described in the chapter on thermochemistry, the most energetic chemical reactions exhibit\nenthalpies on the order of thousands of $\\mathrm{kJ} / \\mathrm{mol}$, which is equivalent to mass differences in the nanogram range $\\left(10^{-9} \\mathrm{~g}\\right)$. On the other hand, nuclear binding energies are typically on the order of billions of $\\mathrm{kJ} / \\mathrm{mol}$, corresponding to mass differences in the milligram range $\\left(10^{-3} \\mathrm{~g}\\right)$.\nNuclear Stability\nA nucleus is stable if it cannot be transformed into another configuration without adding energy from the outside. Of the thousands of nuclides that exist, about 250 are stable. A plot of the number of neutrons versus the number of protons for stable nuclei reveals that the stable isotopes fall into a narrow band. This region is known as the band of stability (also called the belt, zone, or valley of stability). The straight line in Figure 20.2 represents nuclei that have a 1:1 ratio of protons to neutrons ( $\\mathrm{n}: \\mathrm{p}$ ratio). Note that the lighter stable nuclei, in general, have equal numbers of protons and neutrons. For example, nitrogen-14 has seven protons and seven neutrons. Heavier stable nuclei, however, have increasingly more neutrons than protons. For example: iron-56 has 30 neutrons and 26 protons, an n:p ratio of 1.15, whereas the stable nuclide lead- 207 has 125 neutrons and 82 protons, an n:p ratio equal to 1.52. This is because larger nuclei have more proton-proton repulsions, and require larger numbers of neutrons to provide compensating strong forces to overcome these electrostatic repulsions and hold the nucleus together."}
{"id": 4979, "contents": "2063. Answer: - \nFIGURE 20.2 This plot shows the nuclides that are known to exist and those that are stable. The stable nuclides are indicated in blue, and the unstable nuclides are indicated in green. Note that all isotopes of elements with atomic numbers greater than 83 are unstable. The solid line is the line where $\\mathrm{n}=\\mathrm{Z}$.\n\nThe nuclei that are to the left or to the right of the band of stability are unstable and exhibit radioactivity. They change spontaneously (decay) into other nuclei that are either in, or closer to, the band of stability. These nuclear decay reactions convert one unstable isotope (or radioisotope) into another, more stable, isotope. We will discuss the nature and products of this radioactive decay in subsequent sections of this chapter.\n\nSeveral observations may be made regarding the relationship between the stability of a nucleus and its structure. Nuclei with even numbers of protons, neutrons, or both are more likely to be stable (see Table 20.1). Nuclei with certain numbers of nucleons, known as magic numbers, are stable against nuclear decay. These numbers of protons or neutrons $(2,8,20,28,50,82$, and 126 ) make complete shells in the nucleus. These are similar in concept to the stable electron shells observed for the noble gases. Nuclei that have magic numbers of\nboth protons and neutrons, such as ${ }_{2}^{4} \\mathrm{He},{ }_{8}^{16} \\mathrm{O},{ }_{20}^{40} \\mathrm{Ca}$, and ${ }_{82}^{208} \\mathrm{~Pb}$, are called \"double magic\" and are particularly stable. These trends in nuclear stability may be rationalized by considering a quantum mechanical model of nuclear energy states analogous to that used to describe electronic states earlier in this textbook. The details of this model are beyond the scope of this chapter.\n\nStable Nuclear Isotopes\n\n| Number of Stable Isotopes | Proton Number | Neutron Number |\n| :--- | :--- | :--- |\n| 157 | even | even |\n| 53 | even | odd |\n| 50 | odd | even |\n| 5 | odd | odd |\n\nTABLE 20.1"}
{"id": 4980, "contents": "2063. Answer: - \nTABLE 20.1\n\nThe relative stability of a nucleus is correlated with its binding energy per nucleon, the total binding energy for the nucleus divided by the number or nucleons in the nucleus. For instance, we saw in Example 20.2 that the binding energy for a ${ }_{2}^{4} \\mathrm{He}$ nucleus is 28.4 MeV . The binding energy per nucleon for a ${ }_{2}^{4} \\mathrm{He}$ nucleus is therefore:\n\n$$\n\\frac{28.4 \\mathrm{MeV}}{4 \\text { nucleons }}=7.10 \\mathrm{MeV} / \\text { nucleon }\n$$\n\nIn Example 20.3, we learn how to calculate the binding energy per nucleon of a nuclide on the curve shown in Figure 20.3.\n\n\nFIGURE 20.3 The binding energy per nucleon is largest for nuclides with mass number of approximately 56."}
{"id": 4981, "contents": "2065. Calculation of Binding Energy per Nucleon - \nThe iron nuclide ${ }_{26}^{56} \\mathrm{Fe}$ lies near the top of the binding energy curve (Figure 20.3) and is one of the most stable nuclides. What is the binding energy per nucleon (in MeV) for the nuclide ${ }_{26}^{56} \\mathrm{Fe}$ (atomic mass of 55.9349 amu )?"}
{"id": 4982, "contents": "2066. Solution - \nAs in Example 20.2, we first determine the mass defect of the nuclide, which is the difference between the mass of 26 protons, 30 neutrons, and 26 electrons, and the observed mass of an ${ }_{26}^{56} \\mathrm{Fe}$ atom:\n\n$$\n\\begin{aligned}\n& \\text { Mass defect }=[(26 \\times 1.0073 \\mathrm{amu})+(30 \\times 1.0087 \\mathrm{amu})+(26 \\times 0.00055 \\mathrm{amu})]-55.9349 \\mathrm{amu} \\\\\n& =56.4651 \\mathrm{amu}-55.9349 \\mathrm{amu} \\\\\n& =0.5302 \\mathrm{amu}\n\\end{aligned}\n$$\n\nWe next calculate the binding energy for one nucleus from the mass defect using the mass-energy equivalence equation:\n\n$$\n\\begin{aligned}\n& E=m c^{2}=0.5302 \\mathrm{amu} \\times \\frac{1.6605 \\times 10^{-27} \\mathrm{~kg}}{1 \\mathrm{amu}} \\times\\left(2.998 \\times 10^{8} \\mathrm{~m} / \\mathrm{s}\\right)^{2} \\\\\n& =7.913 \\times 10^{-11} \\mathrm{~kg} \\cdot \\mathrm{~m} / \\mathrm{s}^{2} \\\\\n& =7.913 \\times 10^{-11} \\mathrm{~J}\n\\end{aligned}\n$$\n\nWe then convert the binding energy in joules per nucleus into units of MeV per nuclide:\n\n$$\n7.913 \\times 10^{-11} \\mathrm{~J} \\times \\frac{1 \\mathrm{MeV}}{1.602 \\times 10^{-13} \\mathrm{~J}}=493.9 \\mathrm{MeV}\n$$\n\nFinally, we determine the binding energy per nucleon by dividing the total nuclear binding energy by the number of nucleons in the atom:\n\n$$\n\\text { Binding energy per nucleon }=\\frac{493.9 \\mathrm{MeV}}{56}=8.820 \\mathrm{MeV} / \\text { nucleon }\n$$"}
{"id": 4983, "contents": "2066. Solution - \n$$\n\\text { Binding energy per nucleon }=\\frac{493.9 \\mathrm{MeV}}{56}=8.820 \\mathrm{MeV} / \\text { nucleon }\n$$\n\nNote that this is almost $25 \\%$ larger than the binding energy per nucleon for ${ }_{2}^{4} \\mathrm{He}$.\n(Note also that this is the same process as in Example 20.1, but with the additional step of dividing the total nuclear binding energy by the number of nucleons.)"}
{"id": 4984, "contents": "2067. Check Your Learning - \nWhat is the binding energy per nucleon in ${ }_{9}^{19} \\mathrm{~F}$ (atomic mass, 18.9984 amu )?"}
{"id": 4985, "contents": "2068. Answer: - \n7.810 MeV/nucleon"}
{"id": 4986, "contents": "2069. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Identify common particles and energies involved in nuclear reactions\n- Write and balance nuclear equations\n\nChanges of nuclei that result in changes in their atomic numbers, mass numbers, or energy states are nuclear reactions. To describe a nuclear reaction, we use an equation that identifies the nuclides involved in the reaction, their mass numbers and atomic numbers, and the other particles involved in the reaction."}
{"id": 4987, "contents": "2070. Types of Particles in Nuclear Reactions - \nMany entities can be involved in nuclear reactions. The most common are protons, neutrons, alpha particles, beta particles, positrons, and gamma rays, as shown in Figure 20.4. Protons ( ${ }_{1}^{1} \\mathrm{p}$, also represented by the symbol $\\left.{ }_{1}^{1} \\mathrm{H}\\right)$ and neutrons $\\left({ }_{0}^{1} \\mathrm{n}\\right)$ are the constituents of atomic nuclei, and have been described previously. Alpha particles $\\left({ }_{2}^{4} \\mathrm{He}\\right.$, also represented by the symbol $\\left.{ }_{2}^{4} \\alpha\\right)$ are high-energy helium nuclei. Beta particles $\\left({ }_{-1}^{0} \\beta\\right.$, also represented by the symbol $\\left.{ }_{-1}^{0} \\mathrm{e}\\right)$ are high-energy electrons, and gamma rays are photons of very high-energy electromagnetic radiation. Positrons $\\left({ }_{+1}^{0} \\mathrm{e}\\right.$, also represented by the symbol $\\left.{ }_{+1}^{0} \\beta\\right)$ are positively charged electrons (\"anti-electrons\"). The subscripts and superscripts are necessary for balancing nuclear equations, but are usually optional in other circumstances. For example, an alpha particle is a helium nucleus (He) with a charge of +2 and a mass number of 4 , so it is symbolized ${ }_{2}^{4} \\mathrm{He}$. This works because, in general, the ion charge is not important in the balancing of nuclear equations."}
{"id": 4988, "contents": "2070. Types of Particles in Nuclear Reactions - \n| Name | Symbol(s) | Representation | Description |\n| :---: | :---: | :---: | :---: |\n| Alpha particle | ${ }_{2}^{4} \\mathrm{He}$ or ${ }_{2}^{4} \\alpha$ |  | (High-energy) helium nuclei consisting of two protons and two neutrons |\n| Beta particle | ${ }_{-1}^{0} \\mathrm{e}$ or ${ }_{-1}^{0} \\beta$ | $\\bigcirc$ | (High-energy) electrons |\n| Positron | ${ }_{+1}^{0} \\mathrm{e}$ or ${ }_{+1}^{0} \\beta$ | ( | Particles with the same mass as an electron but with 1 unit of positive charge |\n| Proton | ${ }_{1}^{1} \\mathrm{H}$ or ${ }_{1}^{1} \\mathrm{p}$ | $\\pm$ | Nuclei of hydrogen atoms |\n| Neutron | ${ }_{0}^{1} n$ |  | Particles with a mass approximately equal to that of a proton but with no charge |\n| Gamma ray | $\\gamma$ | $\\sim \\sim \\sim \\gamma$ | Very high-energy electromagnetic radiation |\n\nFIGURE 20.4 Although many species are encountered in nuclear reactions, this table summarizes the names, symbols, representations, and descriptions of the most common of these.\n\nNote that positrons are exactly like electrons, except they have the opposite charge. They are the most common example of antimatter, particles with the same mass but the opposite state of another property (for example, charge) than ordinary matter. When antimatter encounters ordinary matter, both are annihilated and their mass is converted into energy in the form of gamma rays $(\\boldsymbol{\\gamma})$-and other much smaller subnuclear particles, which are beyond the scope of this chapter-according to the mass-energy equivalence equation $E=m c^{2}$, seen in the preceding section. For example, when a positron and an electron collide, both are annihilated and two gamma ray photons are created:\n\n$$\n{ }_{-1}^{0} \\mathrm{e}+{ }_{+1}^{0} \\mathrm{e} \\longrightarrow \\gamma+\\gamma\n$$"}
{"id": 4989, "contents": "2070. Types of Particles in Nuclear Reactions - \n$$\n{ }_{-1}^{0} \\mathrm{e}+{ }_{+1}^{0} \\mathrm{e} \\longrightarrow \\gamma+\\gamma\n$$\n\nAs seen in the chapter discussing light and electromagnetic radiation, gamma rays compose short wavelength, high-energy electromagnetic radiation and are (much) more energetic than better-known X-rays that can behave as particles in the wave-particle duality sense. Gamma rays are a type of high energy electromagnetic radiation produced when a nucleus undergoes a transition from a higher to a lower energy state, similar to how a photon is produced by an electronic transition from a higher to a lower energy level. Due to the much larger energy differences between nuclear energy shells, gamma rays emanating from a nucleus have energies that are typically millions of times larger than electromagnetic radiation emanating from electronic transitions."}
{"id": 4990, "contents": "2071. Balancing Nuclear Reactions - \nA balanced chemical reaction equation reflects the fact that during a chemical reaction, bonds break and form, and atoms are rearranged, but the total numbers of atoms of each element are conserved and do not change. A balanced nuclear reaction equation indicates that there is a rearrangement during a nuclear reaction, but of nucleons (subatomic particles within the atoms' nuclei) rather than atoms. Nuclear reactions also follow conservation laws, and they are balanced in two ways:\n\n1. The sum of the mass numbers of the reactants equals the sum of the mass numbers of the products.\n2. The sum of the charges of the reactants equals the sum of the charges of the products.\n\nIf the atomic number and the mass number of all but one of the particles in a nuclear reaction are known, we can identify the particle by balancing the reaction. For instance, we could determine that ${ }_{8}^{17} \\mathrm{O}$ is a product of the nuclear reaction of ${ }_{7}^{14} \\mathrm{~N}$ and ${ }_{2}^{4} \\mathrm{He}$ if we knew that a proton, ${ }_{1}^{1} \\mathrm{H}$, was one of the two products. Example 20.4 shows how we can identify a nuclide by balancing the nuclear reaction."}
{"id": 4991, "contents": "2073. Balancing Equations for Nuclear Reactions - \nThe reaction of an $\\alpha$ particle with magnesium- $25\\left({ }_{12}^{25} \\mathrm{Mg}\\right)$ produces a proton and a nuclide of another element. Identify the new nuclide produced."}
{"id": 4992, "contents": "2074. Solution - \nThe nuclear reaction can be written as:\n\n$$\n{ }_{12}^{25} \\mathrm{Mg}+{ }_{2}^{4} \\mathrm{He} \\longrightarrow{ }_{1}^{1} \\mathrm{H}+{ }_{\\mathrm{Z}}^{\\mathrm{A}} \\mathrm{X}\n$$\n\nwhere $A$ is the mass number and $Z$ is the atomic number of the new nuclide, $X$. Because the sum of the mass numbers of the reactants must equal the sum of the mass numbers of the products:\n\n$$\n25+4=\\mathrm{A}+1, \\text { or } \\mathrm{A}=28\n$$\n\nSimilarly, the charges must balance, so:\n\n$$\n12+2=\\mathrm{Z}+1, \\text { and } \\mathrm{Z}=13\n$$\n\nCheck the periodic table: The element with nuclear charge $=+13$ is aluminum. Thus, the product is ${ }_{13}^{28} \\mathrm{Al}$."}
{"id": 4993, "contents": "2075. Check Your Learning - \nThe nuclide ${ }_{53}^{125}$ I combines with an electron and produces a new nucleus and no other massive particles. What is the equation for this reaction?"}
{"id": 4994, "contents": "2076. Answer: - \n${ }_{53}^{125} \\mathrm{I}+{ }_{-1}^{0} \\mathrm{e} \\longrightarrow{ }_{52}^{125} \\mathrm{Te}$\n\nFollowing are the equations of several nuclear reactions that have important roles in the history of nuclear chemistry:\n\n- The first naturally occurring unstable element that was isolated, polonium, was discovered by the Polish scientist Marie Curie and her husband Pierre in 1898. It decays, emitting $\\alpha$ particles:\n\n$$\n{ }_{84}^{212} \\mathrm{Po} \\longrightarrow{ }_{82}^{208} \\mathrm{~Pb}+{ }_{2}^{4} \\mathrm{He}\n$$\n\n- The first nuclide to be prepared by artificial means was an isotope of oxygen, ${ }^{17} \\mathrm{O}$. It was made by Ernest Rutherford in 1919 by bombarding nitrogen atoms with $\\alpha$ particles:\n\n$$\n{ }_{7}^{14} \\mathrm{~N}+{ }_{2}^{4} \\mathrm{He} \\longrightarrow{ }_{8}^{17} \\mathrm{O}+{ }_{1}^{1} \\mathrm{H}\n$$\n\n- James Chadwick discovered the neutron in 1932, as a previously unknown neutral particle produced along with ${ }^{12} \\mathrm{C}$ by the nuclear reaction between ${ }^{9} \\mathrm{Be}$ and ${ }^{4} \\mathrm{He}$ :\n\n$$\n{ }_{4}^{9} \\mathrm{Be}+{ }_{2}^{4} \\mathrm{He} \\longrightarrow{ }_{6}^{12} \\mathrm{C}+{ }_{0}^{1} \\mathrm{n}\n$$\n\n- The first element to be prepared that does not occur naturally on the earth, technetium, was created by bombardment of molybdenum by deuterons (heavy hydrogen, ${ }_{1}^{2} \\mathrm{H}$ ), by Emilio Segre and Carlo Perrier in 1937:"}
{"id": 4995, "contents": "2076. Answer: - \n$$\n{ }_{1}^{2} \\mathrm{H}+{ }_{42}^{97} \\mathrm{Mo} \\longrightarrow 2{ }_{0}^{1} \\mathrm{n}+{ }_{43}^{97} \\mathrm{Tc}\n$$\n\n- The first controlled nuclear chain reaction was carried out in a reactor at the University of Chicago in 1942. One of the many reactions involved was:\n\n$$\n{ }_{92}^{235} \\mathrm{U}+{ }_{0}^{1} \\mathrm{n} \\longrightarrow{ }_{35}^{87} \\mathrm{Br}+{ }_{57}^{146} \\mathrm{La}+3{ }_{0}^{1} \\mathrm{n}\n$$"}
{"id": 4996, "contents": "2077. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Recognize common modes of radioactive decay\n- Identify common particles and energies involved in nuclear decay reactions\n- Write and balance nuclear decay equations\n- Calculate kinetic parameters for decay processes, including half-life\n- Describe common radiometric dating techniques\n\nFollowing the somewhat serendipitous discovery of radioactivity by Becquerel, many prominent scientists began to investigate this new, intriguing phenomenon. Among them were Marie Curie (the first woman to win a Nobel Prize, and the only person to win two Nobel Prizes in different sciences-chemistry and physics), who was the first to coin the term \"radioactivity,\" and Ernest Rutherford (of gold foil experiment fame), who investigated and named three of the most common types of radiation. During the beginning of the twentieth century, many radioactive substances were discovered, the properties of radiation were investigated and quantified, and a solid understanding of radiation and nuclear decay was developed.\n\nThe spontaneous change of an unstable nuclide into another is radioactive decay. The unstable nuclide is called the parent nuclide; the nuclide that results from the decay is known as the daughter nuclide. The daughter nuclide may be stable, or it may decay itself. The radiation produced during radioactive decay is such that the daughter nuclide lies closer to the band of stability than the parent nuclide, so the location of a nuclide relative to the band of stability can serve as a guide to the kind of decay it will undergo (Figure 20.5).\n\n\nFIGURE 20.5 A nucleus of uranium-238 (the parent nuclide) undergoes $\\alpha$ decay to form thorium-234 (the daughter nuclide). The alpha particle removes two protons (green) and two neutrons (gray) from the uranium-238 nucleus."}
{"id": 4997, "contents": "2078. LINK TO LEARNING - \nAlthough the radioactive decay of a nucleus is too small to see with the naked eye, we can indirectly view radioactive decay in an environment called a cloud chamber. Click here (http://openstax.org/l/16cloudchamb) to learn about cloud chambers and to view an interesting Cloud Chamber Demonstration from the Jefferson Lab."}
{"id": 4998, "contents": "2079. Types of Radioactive Decay - \nErnest Rutherford's experiments involving the interaction of radiation with a magnetic or electric field (Figure 20.6) helped him determine that one type of radiation consisted of positively charged and relatively massive $\\alpha$ particles; a second type was made up of negatively charged and much less massive $\\beta$ particles; and a third was uncharged electromagnetic waves, $\\gamma$ rays. We now know that $\\alpha$ particles are high-energy helium nuclei, $\\beta$ particles are high-energy electrons, and $\\gamma$ radiation compose high-energy electromagnetic radiation. We classify different types of radioactive decay by the radiation produced.\n\n\nFIGURE 20.6 Alpha particles, which are attracted to the negative plate and deflected by a relatively small amount, must be positively charged and relatively massive. Beta particles, which are attracted to the positive plate and deflected a relatively large amount, must be negatively charged and relatively light. Gamma rays, which are unaffected by the electric field, must be uncharged.\n\nAlpha ( $\\alpha$ ) decay is the emission of an $\\alpha$ particle from the nucleus. For example, polonium-210 undergoes $\\alpha$ decay:\n\n$$\n{ }_{84}^{210} \\mathrm{Po} \\longrightarrow{ }_{2}^{4} \\mathrm{He}+{ }_{82}^{206} \\mathrm{~Pb} \\quad \\text { or } \\quad{ }_{84}^{210} \\mathrm{Po} \\longrightarrow{ }_{2}^{4} \\alpha+{ }_{82}^{206} \\mathrm{~Pb}\n$$\n\nAlpha decay occurs primarily in heavy nuclei ( $\\mathrm{A}>200, \\mathrm{Z}>83$ ). Because the loss of an $\\alpha$ particle gives a daughter nuclide with a mass number four units smaller and an atomic number two units smaller than those of the parent nuclide, the daughter nuclide has a larger n:p ratio than the parent nuclide. If the parent nuclide undergoing $\\alpha$ decay lies below the band of stability (refer to Figure 20.2), the daughter nuclide will lie closer to the band.\n\nBeta ( $\\boldsymbol{\\beta}$ ) decay is the emission of an electron from a nucleus. Iodine-131 is an example of a nuclide that undergoes $\\beta$ decay:"}
{"id": 4999, "contents": "2079. Types of Radioactive Decay - \nBeta ( $\\boldsymbol{\\beta}$ ) decay is the emission of an electron from a nucleus. Iodine-131 is an example of a nuclide that undergoes $\\beta$ decay:\n\n$$\n{ }_{53}^{131} \\mathrm{I} \\longrightarrow{ }_{-1}^{0} \\mathrm{e}+{ }_{54}^{131} \\mathrm{Xe} \\quad \\text { or } \\quad{ }_{53}^{131} \\mathrm{I} \\longrightarrow{ }_{-1}^{0} \\beta+{ }_{54}^{131} \\mathrm{Xe}\n$$\n\nBeta decay, which can be thought of as the conversion of a neutron into a proton and a $\\beta$ particle, is observed in nuclides with a large n:p ratio. The beta particle (electron) emitted is from the atomic nucleus and is not one of the electrons surrounding the nucleus. Such nuclei lie above the band of stability. Emission of an electron does not change the mass number of the nuclide but does increase the number of its protons and decrease the number of its neutrons. Consequently, the n:p ratio is decreased, and the daughter nuclide lies closer to the band of stability than did the parent nuclide.\n\nGamma emission ( $\\boldsymbol{\\gamma}$ emission) is observed when a nuclide is formed in an excited state and then decays to its ground state with the emission of a $\\gamma$ ray, a quantum of high-energy electromagnetic radiation. The presence of a nucleus in an excited state is often indicated by an asterisk (*). Cobalt-60 emits $\\gamma$ radiation and is used in many applications including cancer treatment:\n\n$$\n{ }_{27}^{60} \\mathrm{Co}^{*} \\longrightarrow{ }_{0}^{0} \\gamma+{ }_{27}^{60} \\mathrm{Co}\n$$\n\nThere is no change in mass number or atomic number during the emission of a $\\gamma$ ray unless the $\\gamma$ emission accompanies one of the other modes of decay.\n\nPositron emission ( $\\boldsymbol{\\beta}^{+}$decay) is the emission of a positron from the nucleus. Oxygen-15 is an example of a nuclide that undergoes positron emission:"}
{"id": 5000, "contents": "2079. Types of Radioactive Decay - \nPositron emission ( $\\boldsymbol{\\beta}^{+}$decay) is the emission of a positron from the nucleus. Oxygen-15 is an example of a nuclide that undergoes positron emission:\n\n$$\n{ }_{8}^{15} \\mathrm{O} \\longrightarrow{ }_{+1}^{0} \\mathrm{e}+{ }_{7}^{15} \\mathrm{~N} \\quad \\text { or } \\quad{ }_{8}^{15} \\mathrm{O} \\longrightarrow{ }_{+1}^{0} \\beta+{ }_{7}^{15} \\mathrm{~N}\n$$\n\nPositron emission is observed for nuclides in which the n:p ratio is low. These nuclides lie below the band of stability. Positron decay is the conversion of a proton into a neutron with the emission of a positron. The n:p ratio increases, and the daughter nuclide lies closer to the band of stability than did the parent nuclide.\n\nElectron capture occurs when one of the inner electrons in an atom is captured by the atom's nucleus. For example, potassium-40 undergoes electron capture:\n\n$$\n{ }_{19}^{40} \\mathrm{~K}+{ }_{-1}^{0} \\mathrm{e} \\longrightarrow{ }_{18}^{40} \\mathrm{Ar}\n$$\n\nElectron capture occurs when an inner shell electron combines with a proton and is converted into a neutron. The loss of an inner shell electron leaves a vacancy that will be filled by one of the outer electrons. As the outer electron drops into the vacancy, it will emit energy. In most cases, the energy emitted will be in the form of an X-ray. Like positron emission, electron capture occurs for \"proton-rich\" nuclei that lie below the band of stability. Electron capture has the same effect on the nucleus as does positron emission: The atomic number is decreased by one and the mass number does not change. This increases the n:p ratio, and the daughter nuclide lies closer to the band of stability than did the parent nuclide. Whether electron capture or positron emission occurs is difficult to predict. The choice is primarily due to kinetic factors, with the one requiring the smaller activation energy being the one more likely to occur."}
{"id": 5001, "contents": "2079. Types of Radioactive Decay - \nFigure 20.7 summarizes these types of decay, along with their equations and changes in atomic and mass numbers.\n\n| Type | Nuclear equation | Representation | Change in mass/atomic numbers |\n| :---: | :---: | :---: | :---: |\n| Alpha decay | ${ }_{\\mathrm{Z}}^{\\mathrm{A}} \\mathrm{X} \\rightarrow{ }_{2}^{4} \\mathrm{He}+{ }_{\\mathrm{Z}-2}^{\\mathrm{A}-4} \\mathrm{Y}$ |  | A: decrease by 4 <br> Z: decrease by 2 |\n| Beta decay | ${ }_{\\mathrm{Z}} \\mathrm{A} X \\rightarrow{ }_{-1}^{0} \\mathrm{e}+{ }_{\\mathrm{Z}+1}{ }^{\\text {a }} \\mathrm{Y}$ |  | A: unchanged <br> Z: increase by 1 |\n| Gamma decay | ${ }_{Z}^{A} X \\rightarrow{ }_{0}^{0} \\gamma+{ }_{Z}^{A} Y$ | Excited nuclear state | A: unchanged <br> Z: unchanged |\n| Positron emission | ${ }_{\\mathrm{Z}} \\mathrm{X} \\mathrm{X} \\rightarrow{ }_{+1}^{0} \\mathrm{e}+{ }_{Y-1}{ }_{1}^{\\text {P }} \\mathrm{Y}$ | $8 \\bigcirc \\underset{\\oplus}{\\downarrow} \\longrightarrow$ | A: unchanged <br> Z: decrease by 1 |\n| Electron capture | ${ }_{\\mathrm{Z}}{ }^{\\text {P }} \\rightarrow{ }_{-1}^{0} \\mathrm{e}+{ }_{\\mathrm{Y}-1}{ }_{1} \\mathrm{Y}$ |  | A: unchanged <br> Z: decrease by 1 |\n\nFIGURE 20.7 This table summarizes the type, nuclear equation, representation, and any changes in the mass or atomic numbers for various types of decay."}
{"id": 5002, "contents": "2081. PET Scan - \nPositron emission tomography (PET) scans use radiation to diagnose and track health conditions and monitor medical treatments by revealing how parts of a patient's body function (Figure 20.8). To perform a PET scan, a positron-emitting radioisotope is produced in a cyclotron and then attached to a substance that is used by the part of the body being investigated. This \"tagged\" compound, or radiotracer, is then put into the patient (injected via IV or breathed in as a gas), and how it is used by the tissue reveals how that organ or other area of the body functions.\n\n(a)\n\n(b)\n\n(c)"}
{"id": 5003, "contents": "2081. PET Scan - \n(a)\n\n(b)\n\n(c)\n\nFIGURE 20.8 A PET scanner (a) uses radiation to provide an image of how part of a patient's body functions. The scans it produces can be used to image a healthy brain (b) or can be used for diagnosing medical conditions such as Alzheimer's disease (c). (credit a: modification of work by Jens Maus)\nFor example, $\\mathrm{F}-18$ is produced by proton bombardment of ${ }^{18} \\mathrm{O}\\left({ }_{8}^{18} \\mathrm{O}+{ }_{1}^{1} \\mathrm{p} \\longrightarrow{ }_{9}^{18} \\mathrm{~F}+{ }_{0}^{1} \\mathrm{n}\\right)$ and incorporated into a glucose analog called fludeoxyglucose (FDG). How FDG is used by the body provides critical diagnostic information; for example, since cancers use glucose differently than normal tissues, FDG can reveal cancers. The ${ }^{18} \\mathrm{~F}$ emits positrons that interact with nearby electrons, producing a burst of gamma radiation. This energy is detected by the scanner and converted into a detailed, three-dimensional, color image that shows how that part of the patient's body functions. Different levels of gamma radiation produce different amounts of brightness and colors in the image, which can then be interpreted by a radiologist to reveal what is going on. PET scans can detect heart damage and heart disease, help diagnose Alzheimer's disease, indicate the part of a brain that is affected by epilepsy, reveal cancer, show what stage it is, and how much it has spread, and whether treatments are effective. Unlike magnetic resonance imaging and Xrays, which only show how something looks, the big advantage of PET scans is that they show how something functions. PET scans are now usually performed in conjunction with a computed tomography scan."}
{"id": 5004, "contents": "2082. Radioactive Decay Series - \nThe naturally occurring radioactive isotopes of the heaviest elements fall into chains of successive disintegrations, or decays, and all the species in one chain constitute a radioactive family, or radioactive decay series. Three of these series include most of the naturally radioactive elements of the periodic table. They are the uranium series, the actinide series, and the thorium series. The neptunium series is a fourth series, which is no longer significant on the earth because of the short half-lives of the species involved. Each series is characterized by a parent (first member) that has a long half-life and a series of daughter nuclides that ultimately lead to a stable end-product-that is, a nuclide on the band of stability (Figure 20.9). In all three series, the end-product is a stable isotope of lead. The neptunium series, previously thought to terminate with bismuth-209, terminates with thallium-205.\n\n\nFIGURE 20.9 Uranium-238 undergoes a radioactive decay series consisting of 14 separate steps before producing stable lead-206. This series consists of eight $\\alpha$ decays and $\\operatorname{six} \\beta$ decays."}
{"id": 5005, "contents": "2083. Radioactive Half-Lives - \nRadioactive decay follows first-order kinetics. Since first-order reactions have already been covered in detail in the kinetics chapter, we will now apply those concepts to nuclear decay reactions. Each radioactive nuclide has a characteristic, constant half-life $\\left(t_{1 / 2}\\right)$, the time required for half of the atoms in a sample to decay. An isotope's half-life allows us to determine how long a sample of a useful isotope will be available, and how long a sample of an undesirable or dangerous isotope must be stored before it decays to a low-enough radiation level that is no longer a problem.\n\nFor example, cobalt-60, an isotope that emits gamma rays used to treat cancer, has a half-life of 5.27 years (Figure 20.10). In a given cobalt-60 source, since half of the ${ }_{27}^{60}$ Co nuclei decay every 5.27 years, both the amount of material and the intensity of the radiation emitted is cut in half every 5.27 years. (Note that for a given substance, the intensity of radiation that it produces is directly proportional to the rate of decay of the substance and the amount of the substance.) This is as expected for a process following first-order kinetics. Thus, a cobalt-60 source that is used for cancer treatment must be replaced regularly to continue to be effective.\n\n\nFIGURE 20.10 For cobalt-60, which has a half-life of 5.27 years, $50 \\%$ remains after 5.27 years (one half-life), 25\\% remains after 10.54 years (two half-lives), $12.5 \\%$ remains after 15.81 years (three half-lives), and so on."}
{"id": 5006, "contents": "2083. Radioactive Half-Lives - \nSince nuclear decay follows first-order kinetics, we can adapt the mathematical relationships used for firstorder chemical reactions. We generally substitute the number of nuclei, $N$, for the concentration. If the rate is stated in nuclear decays per second, we refer to it as the activity of the radioactive sample. The rate for radioactive decay is:\ndecay rate $=\\lambda N$ with $\\lambda=$ the decay constant for the particular radioisotope\nThe decay constant, $\\lambda$, which is the same as a rate constant discussed in the kinetics chapter. It is possible to express the decay constant in terms of the half-life, $t_{1 / 2}$ :\n\n$$\n\\lambda=\\frac{\\ln 2}{t_{1 / 2}}=\\frac{0.693}{t_{1 / 2}} \\quad \\text { or } \\quad t_{1 / 2}=\\frac{\\ln 2}{\\lambda}=\\frac{0.693}{\\lambda}\n$$\n\nThe first-order equations relating amount, $N$, and time are:\n\n$$\nN_{t}=N_{0} e^{-\\lambda t} \\quad \\text { or } \\quad t=-\\frac{1}{\\lambda} \\ln \\left(\\frac{N_{t}}{N_{0}}\\right)\n$$\n\nwhere $N_{0}$ is the initial number of nuclei or moles of the isotope, and $N_{t}$ is the number of nuclei/moles remaining at time $t$. Example 20.5 applies these calculations to find the rates of radioactive decay for specific nuclides."}
{"id": 5007, "contents": "2085. Rates of Radioactive Decay - \n${ }_{27}^{60} \\mathrm{Co}$ decays with a half-life of 5.27 years to produce ${ }_{28}^{60} \\mathrm{Ni}$.\n(a) What is the decay constant for the radioactive disintegration of cobalt-60?\n(b) Calculate the fraction of a sample of the ${ }_{27}^{60}$ Co isotope that will remain after 15 years.\n(c) How long does it take for a sample of ${ }_{27}^{60} \\mathrm{Co}$ to disintegrate to the extent that only $2.0 \\%$ of the original amount remains?"}
{"id": 5008, "contents": "2086. Solution - \n(a) The value of the rate constant is given by:\n\n$$\n\\lambda=\\frac{\\ln 2}{t_{1 / 2}}=\\frac{0.693}{5.27 \\mathrm{y}}=0.132 \\mathrm{y}^{-1}\n$$\n\n(b) The fraction of ${ }_{27}^{60} \\mathrm{Co}$ that is left after time $t$ is given by $\\frac{N_{t}}{N_{0}}$. Rearranging the first-order relationship $N_{t}=$ $N_{0} \\mathrm{e}^{-\\lambda t}$ to solve for this ratio yields:\n\n$$\n\\frac{N_{t}}{N_{0}}=e^{-\\lambda t}=e^{-(0.132 / \\mathrm{y})(15 \\times \\mathrm{y})}=0.138\n$$\n\nThe fraction of ${ }_{27}^{60}$ Co that will remain after 15.0 years is 0.138 . Or put another way, $13.8 \\%$ of the ${ }_{27}^{60} \\mathrm{Co}$ originally present will remain after 15 years.\n(c) $2.00 \\%$ of the original amount of ${ }_{27}^{60} \\mathrm{Co}$ is equal to $0.0200 \\times N_{0}$. Substituting this into the equation for time for first-order kinetics, we have:\n\n$$\nt=-\\frac{1}{\\lambda} \\ln \\left(\\frac{N_{t}}{N_{0}}\\right)=-\\frac{1}{0.132 \\mathrm{y}^{-1}} \\ln \\left(\\frac{0.0200 \\times N_{0}}{N_{0}}\\right)=29.6 \\mathrm{y}\n$$"}
{"id": 5009, "contents": "2087. Check Your Learning - \nRadon-222, ${ }_{86}^{222} \\mathrm{Rn}$, has a half-life of 3.823 days. How long will it take a sample of radon- 222 with a mass of 0.750 g to decay into other elements, leaving only 0.100 g of radon-222?"}
{"id": 5010, "contents": "2088. Answer: - \n11.1 days\n\nBecause each nuclide has a specific number of nucleons, a particular balance of repulsion and attraction, and its own degree of stability, the half-lives of radioactive nuclides vary widely. For example: the half-life of ${ }_{83}^{209} \\mathrm{Bi}$ is $1.9 \\times 10^{19}$ years; ${ }_{94}^{239} \\mathrm{Ra}$ is 24,000 years; ${ }_{86}^{222} \\mathrm{Rn}$ is 3.82 days; and element- 111 ( Rg for roentgenium) is $1.5 \\times$ $10^{-3}$ seconds. The half-lives of a number of radioactive isotopes important to medicine are shown in Table 20.2, and others are listed in Appendix M.\n\nHalf-lives of Radioactive Isotopes Important to Medicine\n\n| Type $^{\\mathbf{1}}$ | Decay Mode | Half-Life | Uses |\n| :--- | :--- | :--- | :--- |\n| F-18 | $\\beta^{+}$decay | 110. minutes | PET scans |\n| Co-60 | $\\beta$ decay, $\\gamma$ decay | 5.27 years | cancer treatment |\n| Tc-99m | $\\gamma$ decay | 8.01 hours | scans of brain, lung, heart, bone |\n| I-131 | $\\beta$ decay | 8.02 days | thyroid scans and treatment |\n| Tl-201 | electron capture | 73 hours | heart and arteries scans; cardiac stress tests |\n\nTABLE 20.2"}
{"id": 5011, "contents": "2089. Radiometric Dating - \nSeveral radioisotopes have half-lives and other properties that make them useful for purposes of \"dating\" the origin of objects such as archaeological artifacts, formerly living organisms, or geological formations. This process is radiometric dating and has been responsible for many breakthrough scientific discoveries about\n\n[^12]the geological history of the earth, the evolution of life, and the history of human civilization. We will explore some of the most common types of radioactive dating and how the particular isotopes work for each type.\n\nRadioactive Dating Using Carbon-14\nThe radioactivity of carbon-14 provides a method for dating objects that were a part of a living organism. This method of radiometric dating, which is also called radiocarbon dating or carbon-14 dating, is accurate for dating carbon-containing substances that are up to about 30,000 years old, and can provide reasonably accurate dates up to a maximum of about 50,000 years old.\nNaturally occurring carbon consists of three isotopes: ${ }_{6}^{12} \\mathrm{C}$, which constitutes about $99 \\%$ of the carbon on earth; ${ }_{6}^{13} \\mathrm{C}$, about $1 \\%$ of the total; and trace amounts of ${ }_{6}^{14} \\mathrm{C}$. Carbon- 14 forms in the upper atmosphere by the reaction of nitrogen atoms with neutrons from cosmic rays in space:\n\n$$\n{ }_{7}^{14} \\mathrm{~N}+{ }_{0}^{1} \\mathrm{n} \\longrightarrow{ }_{6}^{14} \\mathrm{C}+{ }_{1}^{1} \\mathrm{H}\n$$"}
{"id": 5012, "contents": "2089. Radiometric Dating - \nAll isotopes of carbon react with oxygen to produce $\\mathrm{CO}_{2}$ molecules. The ratio of ${ }_{6}^{14} \\mathrm{CO}_{2}$ to ${ }_{6}^{12} \\mathrm{CO}_{2}$ depends on the ratio of ${ }_{6}^{14} \\mathrm{CO}$ to ${ }_{6}^{12} \\mathrm{CO}$ in the atmosphere. The natural abundance of ${ }_{6}^{14} \\mathrm{CO}$ in the atmosphere is approximately 1 part per trillion; until recently, this has generally been constant over time, as seen is gas samples found trapped in ice. The incorporation of ${ }_{6}^{14} \\mathrm{C}_{6}^{14} \\mathrm{CO}_{2}$ and ${ }_{6}^{12} \\mathrm{CO}_{2}$ into plants is a regular part of the photosynthesis process, which means that the ${ }_{6}^{14} \\mathrm{C}:{ }_{6}^{12} \\mathrm{C}$ ratio found in a living plant is the same as the ${ }_{6}^{14} \\mathrm{C}$ : ${ }_{6}^{12} \\mathrm{C}$ ratio in the atmosphere. But when the plant dies, it no longer traps carbon through photosynthesis. Because ${ }_{6}^{12} \\mathrm{C}$ is a stable isotope and does not undergo radioactive decay, its concentration in the plant does not change. However, carbon-14 decays by $\\beta$ emission with a half-life of 5730 years:\n\n$$\n{ }_{6}^{14} \\mathrm{C} \\longrightarrow{ }_{7}^{14} \\mathrm{~N}+{ }_{-1}^{0} \\mathrm{e}\n$$\n\nThus, the ${ }_{6}^{14} \\mathrm{C}:{ }_{6}^{12} \\mathrm{C}$ ratio gradually decreases after the plant dies. The decrease in the ratio with time provides a measure of the time that has elapsed since the death of the plant (or other organism that ate the plant). Figure 20.11 visually depicts this process."}
{"id": 5013, "contents": "2089. Radiometric Dating - \nFIGURE 20.11 Along with stable carbon-12, radioactive carbon-14 is taken in by plants and animals, and remains at a constant level within them while they are alive. After death, the $\\mathrm{C}-14$ decays and the $\\mathrm{C}-14: \\mathrm{C}-12$ ratio in the remains decreases. Comparing this ratio to the $\\mathrm{C}-14: \\mathrm{C}-12$ ratio in living organisms allows us to determine how long ago the organism lived (and died).\nFor example, with the half-life of ${ }_{6}^{14} \\mathrm{C}$ being 5730 years, if the ${ }_{6}^{14} \\mathrm{C}:{ }_{6}^{12} \\mathrm{C}$ ratio in a wooden object found in an archaeological dig is half what it is in a living tree, this indicates that the wooden object is 5730 years old. Highly accurate determinations of ${ }_{6}^{14} \\mathrm{C}:{ }_{6}^{12} \\mathrm{C}$ ratios can be obtained from very small samples (as little as a milligram) by the use of a mass spectrometer."}
{"id": 5014, "contents": "2090. LINK TO LEARNING - \nVisit this website (http://openstax.org/l/16phetradiom) to perform simulations of radiometric dating."}
{"id": 5015, "contents": "2092. Radiocarbon Dating - \nA tiny piece of paper (produced from formerly living plant matter) taken from the Dead Sea Scrolls has an activity of 10.8 disintegrations per minute per gram of carbon. If the initial C-14 activity was 13.6 disintegrations $/ \\mathrm{min} / \\mathrm{g}$ of C, estimate the age of the Dead Sea Scrolls."}
{"id": 5016, "contents": "2093. Solution - \nThe rate of decay (number of disintegrations/minute/gram of carbon) is proportional to the amount of radioactive $\\mathrm{C}-14$ left in the paper, so we can substitute the rates for the amounts, $N$, in the relationship:\n\n$$\nt=-\\frac{1}{\\lambda} \\ln \\left(\\frac{N_{t}}{N_{0}}\\right) \\rightarrow t=-\\frac{1}{\\lambda} \\ln \\left(\\frac{\\text { Rate }_{t}}{\\text { Rate }_{0}}\\right)\n$$\n\nwhere the subscript 0 represents the time when the plants were cut to make the paper, and the subscript $t$ represents the current time.\n\nThe decay constant can be determined from the half-life of C-14, 5730 years:\n\n$$\n\\lambda=\\frac{\\ln 2}{t_{1 / 2}}=\\frac{0.693}{5730 \\mathrm{y}}=1.21 \\times 10^{-4} \\mathrm{y}^{-1}\n$$\n\nSubstituting and solving, we have:\n\n$$\nt=-\\frac{1}{\\lambda} \\ln \\left(\\frac{\\text { Rate }_{t}}{\\text { Rate }_{0}}\\right)=-\\frac{1}{1.21 \\times 10^{-4} \\mathrm{y}^{-1}} \\ln \\left(\\frac{10.8 \\mathrm{dis} / \\mathrm{min} / \\mathrm{g} \\mathrm{C}}{13.6 \\mathrm{dis} / \\mathrm{min} / \\mathrm{g} \\mathrm{C}}\\right)=1910 \\mathrm{y}\n$$\n\nTherefore, the Dead Sea Scrolls are approximately 1900 years old (Figure 20.12).\n\n\nFIGURE 20.12 Carbon-14 dating has shown that these pages from the Dead Sea Scrolls were written or copied on paper made from plants that died between 100 BC and AD 50."}
{"id": 5017, "contents": "2094. Check Your Learning - \nMore accurate dates of the reigns of ancient Egyptian pharaohs have been determined recently using plants that were preserved in their tombs. Samples of seeds and plant matter from King Tutankhamun's tomb have a C-14 decay rate of 9.07 disintegrations $/ \\mathrm{min} / \\mathrm{g}$ of C . How long ago did King Tut's reign come to an end?"}
{"id": 5018, "contents": "2095. Answer: - \nabout 3350 years ago, or approximately 1340 BC\n\nThere have been some significant, well-documented changes to the ${ }_{6}^{14} \\mathrm{C}:{ }_{6}^{12} \\mathrm{C}$ ratio. The accuracy of a straightforward application of this technique depends on the ${ }_{6}^{14} \\mathrm{C}:{ }_{6}^{12} \\mathrm{C}$ ratio in a living plant being the same now as it was in an earlier era, but this is not always valid. Due to the increasing accumulation of $\\mathrm{CO}_{2}$ molecules (largely ${ }_{6}^{12} \\mathrm{CO}_{2}$ ) in the atmosphere caused by combustion of fossil fuels (in which essentially all of the ${ }_{6}^{14} \\mathrm{C}$ has decayed), the ratio of ${ }_{6}^{14} \\mathrm{C}:{ }_{6}^{12} \\mathrm{C}$ in the atmosphere may be changing. This manmade increase in ${ }_{6}^{12} \\mathrm{CO}_{2}$ in the atmosphere causes the ${ }_{6}^{14} \\mathrm{C}:{ }_{6}^{12} \\mathrm{C}$ ratio to decrease, and this in turn affects the ratio in currently living organisms on the earth. Fortunately, however, we can use other data, such as tree dating via examination of annual growth rings, to calculate correction factors. With these correction factors, accurate dates can be determined. In general, radioactive dating only works for about 10 half-lives; therefore, the limit for carbon-14 dating is about 57,000 years."}
{"id": 5019, "contents": "2096. Radioactive Dating Using Nuclides Other than Carbon-14 - \nRadioactive dating can also use other radioactive nuclides with longer half-lives to date older events. For example, uranium-238 (which decays in a series of steps into lead-206) can be used for establishing the age of rocks (and the approximate age of the oldest rocks on earth). Since U-238 has a half-life of 4.5 billion years, it takes that amount of time for half of the original $\\mathrm{U}-238$ to decay into $\\mathrm{Pb}-206$. In a sample of rock that does not contain appreciable amounts of $\\mathrm{Pb}-208$, the most abundant isotope of lead, we can assume that lead was not present when the rock was formed. Therefore, by measuring and analyzing the ratio of U-238:Pb-206, we can determine the age of the rock. This assumes that all of the lead-206 present came from the decay of uranium-238. If there is additional lead-206 present, which is indicated by the presence of other lead isotopes in the sample, it is necessary to make an adjustment. Potassium-argon dating uses a similar method. K-40 decays by positron emission and electron capture to form $\\mathrm{Ar}-40$ with a half-life of 1.25 billion years. If a rock sample is crushed and the amount of Ar-40 gas that escapes is measured, determination of the Ar-40:K-40\nratio yields the age of the rock. Other methods, such as rubidium-strontium dating ( $\\mathrm{Rb}-87$ decays into $\\mathrm{Sr}-87$ with a half-life of 48.8 billion years), operate on the same principle. To estimate the lower limit for the earth's age, scientists determine the age of various rocks and minerals, making the assumption that the earth is older than the oldest rocks and minerals in its crust. As of 2014, the oldest known rocks on earth are the Jack Hills zircons from Australia, found by uranium-lead dating to be almost 4.4 billion years old."}
{"id": 5020, "contents": "2098. Radioactive Dating of Rocks - \nAn igneous rock contains $9.58 \\times 10^{-5} \\mathrm{~g}$ of $\\mathrm{U}-238$ and $2.51 \\times 10^{-5} \\mathrm{~g}$ of $\\mathrm{Pb}-206$, and much, much smaller amounts of $\\mathrm{Pb}-208$. Determine the approximate time at which the rock formed."}
{"id": 5021, "contents": "2099. Solution - \nThe sample of rock contains very little $\\mathrm{Pb}-208$, the most common isotope of lead, so we can safely assume that all the $\\mathrm{Pb}-206$ in the rock was produced by the radioactive decay of $\\mathrm{U}-238$. When the rock formed, it contained all of the U-238 currently in it, plus some U-238 that has since undergone radioactive decay.\n\nThe amount of $\\mathrm{U}-238$ currently in the rock is:\n\n$$\n9.58 \\times 10^{-5} \\frac{\\mathrm{~g} U}{\\delta} \\times\\left(\\frac{1 \\mathrm{~mol} \\mathrm{U}}{238 \\mathrm{~g} U}\\right)=4.03 \\times 10^{-7} \\mathrm{~mol} \\mathrm{U}\n$$\n\nBecause when one mole of U-238 decays, it produces one mole of $\\mathrm{Pb}-206$, the amount of $\\mathrm{U}-238$ that has undergone radioactive decay since the rock was formed is:\n\n$$\n2.51 \\times 10^{-5} \\underset{\\S}{\\mathrm{~g} P b} \\times\\left(\\frac{1 \\mathrm{molPb}}{206 \\frac{\\mathrm{~g} P \\mathrm{~Pb}}{}}\\right) \\times\\left(\\frac{1 \\mathrm{~mol} \\mathrm{U}}{1 \\mathrm{molPb}}\\right)=1.22 \\times 10^{-7} \\mathrm{~mol} \\mathrm{U}\n$$\n\nThe total amount of U-238 originally present in the rock is therefore:\n\n$$\n4.03 \\times 10^{-7} \\mathrm{~mol}+1.22 \\times 10^{-7} \\mathrm{~mol}=5.25 \\times 10^{-7} \\mathrm{~mol} \\mathrm{U}\n$$\n\nThe amount of time that has passed since the formation of the rock is given by:\n\n$$\nt=-\\frac{1}{\\lambda} \\ln \\left(\\frac{N_{t}}{N_{0}}\\right)\n$$"}
{"id": 5022, "contents": "2099. Solution - \n$$\nt=-\\frac{1}{\\lambda} \\ln \\left(\\frac{N_{t}}{N_{0}}\\right)\n$$\n\nwith $N_{0}$ representing the original amount of $\\mathrm{U}-238$ and $N_{t}$ representing the present amount of $\\mathrm{U}-238$.\n$\\mathrm{U}-238$ decays into $\\mathrm{Pb}-206$ with a half-life of $4.5 \\times 10^{9} \\mathrm{y}$, so the decay constant $\\lambda$ is:\n\n$$\n\\lambda=\\frac{\\ln 2}{t_{1 / 2}}=\\frac{0.693}{4.5 \\times 10^{9} \\mathrm{y}}=1.54 \\times 10^{-10} \\mathrm{y}^{-1}\n$$\n\nSubstituting and solving, we have:\n\n$$\nt=-\\frac{1}{1.54 \\times 10^{-10} \\mathrm{y}^{-1}} \\ln \\left(\\frac{4.03 \\times 10^{-7} \\mathrm{~mol} U}{5.25 \\times 10^{-7} \\mathrm{molU}}\\right)=1.7 \\times 10^{9} \\mathrm{y}\n$$\n\nTherefore, the rock is approximately 1.7 billion years old."}
{"id": 5023, "contents": "2100. Check Your Learning - \nA sample of rock contains $6.14 \\times 10^{-4} \\mathrm{~g}$ of $\\mathrm{Rb}-87$ and $3.51 \\times 10^{-5} \\mathrm{~g}$ of $\\mathrm{Sr}-87$. Calculate the age of the rock. (The half-life of the $\\beta$ decay of $\\mathrm{Rb}-87$ is $4.7 \\times 10^{10} \\mathrm{y}$.)"}
{"id": 5024, "contents": "2101. Answer: - \n$3.7 \\times 10^{9} \\mathrm{y}$"}
{"id": 5025, "contents": "2102. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the synthesis of transuranium nuclides\n- Explain nuclear fission and fusion processes\n- Relate the concepts of critical mass and nuclear chain reactions\n- Summarize basic requirements for nuclear fission and fusion reactors\n\nAfter the discovery of radioactivity, the field of nuclear chemistry was created and developed rapidly during the early twentieth century. A slew of new discoveries in the 1930s and 1940s, along with World War II, combined to usher in the Nuclear Age in the mid-twentieth century. Scientists learned how to create new substances, and certain isotopes of certain elements were found to possess the capacity to produce unprecedented amounts of energy, with the potential to cause tremendous damage during war, as well as produce enormous amounts of power for society's needs during peace."}
{"id": 5026, "contents": "2103. Synthesis of Nuclides - \nNuclear transmutation is the conversion of one nuclide into another. It can occur by the radioactive decay of a nucleus, or the reaction of a nucleus with another particle. The first manmade nucleus was produced in Ernest Rutherford's laboratory in 1919 by a transmutation reaction, the bombardment of one type of nuclei with other nuclei or with neutrons. Rutherford bombarded nitrogen atoms with high-speed $\\alpha$ particles from a natural radioactive isotope of radium and observed protons resulting from the reaction:\n\n$$\n{ }_{7}^{14} \\mathrm{~N}+{ }_{2}^{4} \\mathrm{He} \\longrightarrow{ }_{8}^{17} \\mathrm{O}+{ }_{1}^{1} \\mathrm{H}\n$$\n\nThe ${ }_{8}^{17} \\mathrm{O}$ and ${ }_{1}^{1} \\mathrm{H}$ nuclei that are produced are stable, so no further (nuclear) changes occur.\nTo reach the kinetic energies necessary to produce transmutation reactions, devices called particle accelerators are used. These devices use magnetic and electric fields to increase the speeds of nuclear particles. In all accelerators, the particles move in a vacuum to avoid collisions with gas molecules. When neutrons are required for transmutation reactions, they are usually obtained from radioactive decay reactions or from various nuclear reactions occurring in nuclear reactors. The Chemistry in Everyday Life feature that follows discusses a famous particle accelerator that made worldwide news."}
{"id": 5027, "contents": "2105. CERN Particle Accelerator - \nLocated near Geneva, the CERN (\"Conseil Europ\u00e9en pour la Recherche Nucl\u00e9aire,\" or European Council for Nuclear Research) Laboratory is the world's premier center for the investigations of the fundamental particles that make up matter. It contains the 27-kilometer ( 17 mile ) long, circular Large Hadron Collider (LHC), the largest particle accelerator in the world (Figure 20.13). In the LHC, particles are boosted to high energies and are then made to collide with each other or with stationary targets at nearly the speed of light. Superconducting electromagnets are used to produce a strong magnetic field that guides the particles around the ring. Specialized, purpose-built detectors observe and record the results of these collisions, which are then analyzed by CERN scientists using powerful computers.\n\n\nFIGURE 20.13 A small section of the LHC is shown with workers traveling along it. (credit: Christophe Delaere)\nIn 2012, CERN announced that experiments at the LHC showed the first observations of the Higgs boson, an elementary particle that helps explain the origin of mass in fundamental particles. This longanticipated discovery made worldwide news and resulted in the awarding of the 2013 Nobel Prize in Physics to Fran\u00e7ois Englert and Peter Higgs, who had predicted the existence of this particle almost 50 years previously."}
{"id": 5028, "contents": "2106. LINK TO LEARNING - \nFamous physicist Brian Cox talks about his work on the Large Hadron Collider at CERN, providing an entertaining and engaging tour (http://openstax.org/l/16tedCERN) of this massive project and the physics behind it.\nView a short video (http://openstax.org/l/16CERNvideo) from CERN, describing the basics of how its particle accelerators work.\n\nPrior to 1940, the heaviest-known element was uranium, whose atomic number is 92 . Now, many artificial elements have been synthesized and isolated, including several on such a large scale that they have had a profound effect on society. One of these-element 93, neptunium ( Np ) -was first made in 1940 by McMillan and Abelson by bombarding uranium- 238 with neutrons. The reaction creates unstable uranium-239, with a half-life of 23.5 minutes, which then decays into neptunium-239. Neptunium-239 is also radioactive, with a half-life of 2.36 days, and it decays into plutonium-239. The nuclear reactions are:\n\n$$\n\\begin{aligned}\n& { }_{92}^{238} \\mathrm{U}+{ }_{0}^{1} \\mathrm{n} \\longrightarrow{ }_{92}^{239} \\mathrm{U} \\\\\n& \\begin{array}{r}\n239 \\\\\n92 \\\\\n239 \\\\\n\\end{array} \\longrightarrow{ }_{93}^{239} \\mathrm{~Np}+{ }_{-1}^{0} \\mathrm{e} \\\\\n& \\text { half-life }=23.5 \\mathrm{~min} \\\\\n& { }_{93}^{239} \\mathrm{~Np} \\longrightarrow{ }_{94}^{239} \\mathrm{Pu}+{ }_{-1}^{0} \\mathrm{e} \\\\\n& \\text { half-life }=2.36 \\text { days }\n\\end{aligned}\n$$"}
{"id": 5029, "contents": "2106. LINK TO LEARNING - \nPlutonium is now mostly formed in nuclear reactors as a byproduct during the fission of $\\mathrm{U}-235$. Additional neutrons are released during this fission process (see the next section), some of which combine with U-238 nuclei to form uranium-239; this undergoes $\\beta$ decay to form neptunium- 239 , which in turn undergoes $\\beta$ decay to form plutonium-239 as illustrated in the preceding three equations. These processes are summarized in the equation:\n\n$$\n{ }_{92}^{238} \\mathrm{U}+{ }_{0}^{1} \\mathrm{n} \\longrightarrow{ }_{92}^{239} \\mathrm{U} \\xrightarrow{\\beta^{-}}{ }_{93}^{239} \\mathrm{~Np} \\xrightarrow{\\beta^{-}}{ }_{94}^{239} \\mathrm{Pu}\n$$\n\nHeavier isotopes of plutonium-Pu-240, $\\mathrm{Pu}-241$, and $\\mathrm{Pu}-242-$ are also produced when lighter plutonium\nnuclei capture neutrons. Some of this highly radioactive plutonium is used to produce military weapons, and the rest presents a serious storage problem because they have half-lives from thousands to hundreds of thousands of years.\n\nAlthough they have not been prepared in the same quantity as plutonium, many other synthetic nuclei have been produced. Nuclear medicine has developed from the ability to convert atoms of one type into other types of atoms. Radioactive isotopes of several dozen elements are currently used for medical applications. The radiation produced by their decay is used to image or treat various organs or portions of the body, among other uses.\n\nThe elements beyond element 92 (uranium) are called transuranium elements. As of this writing, 22 transuranium elements have been produced and officially recognized by IUPAC; several other elements have formation claims that are waiting for approval. Some of these elements are shown in Table 20.3.\n\nPreparation of Some of the Transuranium Elements"}
{"id": 5030, "contents": "2106. LINK TO LEARNING - \n| Name | Symbol | Atomic Number | Reaction |\n| :---: | :---: | :---: | :---: |\n| americium | Am | 95 | ${ }_{94}^{239} \\mathrm{Pu}+{ }_{0}^{1} \\mathrm{n} \\longrightarrow{ }_{95}^{240} \\mathrm{Am}+{ }_{-1}^{0} \\mathrm{e}$ |\n| curium | Cm | 96 | ${ }_{94}^{239} \\mathrm{Pu}+{ }_{2}^{4} \\mathrm{He} \\longrightarrow{ }_{96}^{242} \\mathrm{Cm}+{ }_{0}^{1} \\mathrm{n}$ |\n| californium | Cf | 98 | ${ }_{96}^{242} \\mathrm{Cm}+{ }_{2}^{4} \\mathrm{He} \\longrightarrow{ }_{98}^{245} \\mathrm{Cf}+{ }_{0}^{1} \\mathrm{n}$ |\n| einsteinium | Es | 99 | ${ }_{92}^{238} \\mathrm{U}+15_{0}^{1} \\mathrm{n} \\longrightarrow{ }_{99}^{253} \\mathrm{Es}+7{ }_{-1}^{0} \\mathrm{e}$ |\n| mendelevium | Md | 101 | ${ }_{99}^{253} \\mathrm{Es}+{ }_{2}^{4} \\mathrm{He} \\longrightarrow{ }_{101}^{256} \\mathrm{Md}+{ }_{0}^{1} \\mathrm{n}$ |\n| nobelium | No | 102 | ${ }_{96}^{246} \\mathrm{Cm}+{ }_{6}^{12} \\mathrm{C} \\longrightarrow{ }_{102}^{254} \\mathrm{No}+4{ }_{0}^{1} \\mathrm{n}$ |"}
{"id": 5031, "contents": "2106. LINK TO LEARNING - \n| rutherfordium | Rf | 104 | ${ }_{98}^{249} \\mathrm{Cf}+{ }_{6}^{12} \\mathrm{C} \\longrightarrow{ }_{104}^{257} \\mathrm{Rf}+4{ }_{0}^{1} \\mathrm{n}$ |\n| seaborgium | Sg | 106 | $\\begin{array}{r} { }_{82}^{206} \\mathrm{~Pb}+{ }_{24}^{54} \\mathrm{Cr} \\longrightarrow{ }_{106}^{257} \\mathrm{Sg}+3{ }_{0}^{1} \\mathrm{n} \\\\ { }_{98}^{249} \\mathrm{Cf}+{ }_{8}^{18} \\mathrm{O} \\longrightarrow{ }_{106}^{263} \\mathrm{Sg}+4{ }_{0}^{1} \\mathrm{n} \\end{array}$ |\n| meitnerium | Mt | 107 | ${ }_{83}^{209} \\mathrm{Bi}+{ }_{26}^{58} \\mathrm{Fe} \\longrightarrow{ }_{109}^{266} \\mathrm{Mt}+{ }_{0}^{1} \\mathrm{n}$ |"}
{"id": 5032, "contents": "2106. LINK TO LEARNING - \nTABLE 20.3"}
{"id": 5033, "contents": "2107. Nuclear Fission - \nMany heavier elements with smaller binding energies per nucleon can decompose into more stable elements that have intermediate mass numbers and larger binding energies per nucleon-that is, mass numbers and binding energies per nucleon that are closer to the \"peak\" of the binding energy graph near 56 (see Figure 20.3). Sometimes neutrons are also produced. This decomposition is called fission, the breaking of a large nucleus into smaller pieces. The breaking is rather random with the formation of a large number of different products. Fission usually does not occur naturally, but is induced by bombardment with neutrons. The first reported nuclear fission occurred in 1939 when three German scientists, Lise Meitner, Otto Hahn, and Fritz Strassman, bombarded uranium-235 atoms with slow-moving neutrons that split the U-238 nuclei into smaller fragments that consisted of several neutrons and elements near the middle of the periodic table. Since then, fission has been observed in many other isotopes, including most actinide isotopes that have an odd number of neutrons. A typical nuclear fission reaction is shown in Figure 20.14.\n\n\nFIGURE 20.14 When a slow neutron hits a fissionable U-235 nucleus, it is absorbed and forms an unstable U-236 nucleus. The U-236 nucleus then rapidly breaks apart into two smaller nuclei (in this case, Ba-141 and Kr-92) along with several neutrons (usually two or three), and releases a very large amount of energy.\n\nAmong the products of Meitner, Hahn, and Strassman's fission reaction were barium, krypton, lanthanum, and cerium, all of which have nuclei that are more stable than uranium-235. Since then, hundreds of different isotopes have been observed among the products of fissionable substances. A few of the many reactions that occur for $\\mathrm{U}-235$, and a graph showing the distribution of its fission products and their yields, are shown in Figure 20.15. Similar fission reactions have been observed with other uranium isotopes, as well as with a variety of other isotopes such as those of plutonium."}
{"id": 5034, "contents": "2107. Nuclear Fission - \n$$\n\\begin{aligned}\n& { }_{92}^{235} U+{ }_{0}^{1} n \\longrightarrow{ }_{92}^{236} U \\longrightarrow{ }_{98}^{90} \\mathrm{Sr}+{ }_{54}^{144} \\mathrm{Xe}+2{ }_{0}^{1} \\mathrm{n} \\\\\n& { }_{92}^{235} U+{ }_{0}^{1} n \\longrightarrow{ }_{92}^{236} U \\longrightarrow{ }_{35}^{87} \\mathrm{Br}+{ }_{57}^{146} \\mathrm{La}+3{ }_{0}^{1} \\mathrm{n} \\\\\n& { }_{92}^{235} U+{ }_{0}^{1} n \\longrightarrow{ }_{92}^{236} U \\longrightarrow{ }_{37}^{96} R b+{ }_{55}^{137} \\mathrm{Cs}+3{ }_{0}^{1} \\mathrm{n} \\\\\n& { }_{92}^{235} U+{ }_{0}^{1} n \\longrightarrow{ }_{92}^{236} U \\longrightarrow{ }_{52}^{137} \\mathrm{Te}+{ }_{40}^{97} \\mathrm{Zr}+2{ }_{0}^{1} n \\\\\n& { }_{92}^{235} U+{ }_{0}^{1} n \\longrightarrow{ }_{92}^{236} U \\longrightarrow{ }_{56}^{141} \\mathrm{Ba}+{ }_{36}^{92} \\mathrm{Kr}+3{ }_{0}^{1} \\mathrm{n}\n\\end{aligned}\n$$\n\n(a)\n\n(b)\n\nFIGURE 20.15 (a) Nuclear fission of U-235 produces a range of fission products. (b) The larger fission products of $\\mathrm{U}-235$ are typically one isotope with a mass number around $85-105$, and another isotope with a mass number that is about $50 \\%$ larger, that is, about 130-150."}
{"id": 5035, "contents": "2108. LINK TO LEARNING - \nView this link (http://openstax.org/l/16fission) to see a simulation of nuclear fission.\n$\\mathrm{U}-235$ undergoes fission, the products weigh about 0.2 grams less than the reactants; this \"lost\" mass is converted into a very large amount of energy, about $1.8 \\times 10^{10} \\mathrm{~kJ}$ per mole of $\\mathrm{U}-235$. Nuclear fission reactions produce incredibly large amounts of energy compared to chemical reactions. The fission of 1 kilogram of uranium-235, for example, produces about 2.5 million times as much energy as is produced by burning 1 kilogram of coal.\n\nAs described earlier, when undergoing fission U-235 produces two \"medium-sized\" nuclei, and two or three neutrons. These neutrons may then cause the fission of other uranium-235 atoms, which in turn provide more neutrons that can cause fission of even more nuclei, and so on. If this occurs, we have a nuclear chain reaction (see Figure 20.16). On the other hand, if too many neutrons escape the bulk material without interacting with a nucleus, then no chain reaction will occur.\n\n\nFIGURE 20.16 The fission of a large nucleus, such as U-235, produces two or three neutrons, each of which is capable of causing fission of another nucleus by the reactions shown. If this process continues, a nuclear chain reaction occurs."}
{"id": 5036, "contents": "2108. LINK TO LEARNING - \nMaterial that can sustain a nuclear fission chain reaction is said to be fissile or fissionable. (Technically, fissile material can undergo fission with neutrons of any energy, whereas fissionable material requires high-energy neutrons.) Nuclear fission becomes self-sustaining when the number of neutrons produced by fission equals or exceeds the number of neutrons absorbed by splitting nuclei plus the number that escape into the surroundings. The amount of a fissionable material that will support a self-sustaining chain reaction is a critical mass. An amount of fissionable material that cannot sustain a chain reaction is a subcritical mass. An amount of material in which there is an increasing rate of fission is known as a supercritical mass. The critical mass depends on the type of material: its purity, the temperature, the shape of the sample, and how the neutron reactions are controlled (Figure 20.17).\n\nSub-critical mass\n\n(a)\n\nCritical mass\n\n(b)\n\nFIGURE 20.17 (a) In a subcritical mass, the fissile material is too small and allows too many neutrons to escape the material, so a chain reaction does not occur. (b) In a critical mass, a large enough number of neutrons in the fissile material induce fission to create a chain reaction.\n\nAn atomic bomb (Figure 20.18) contains several pounds of fissionable material, ${ }_{92}^{235} \\mathrm{U}$ or ${ }_{94}^{239} \\mathrm{Pu}$, a source of neutrons, and an explosive device for compressing it quickly into a small volume. When fissionable material is in small pieces, the proportion of neutrons that escape through the relatively large surface area is great, and a chain reaction does not take place. When the small pieces of fissionable material are brought together quickly to form a body with a mass larger than the critical mass, the relative number of escaping neutrons decreases, and a chain reaction and explosion result."}
{"id": 5037, "contents": "2108. LINK TO LEARNING - \nFIGURE 20.18 (a) The nuclear fission bomb that destroyed Hiroshima on August 6, 1945, consisted of two subcritical masses of U-235, where conventional explosives were used to fire one of the subcritical masses into the other, creating the critical mass for the nuclear explosion. (b) The plutonium bomb that destroyed Nagasaki on August 9,1945 , consisted of a hollow sphere of plutonium that was rapidly compressed by conventional explosives. This led to a concentration of plutonium in the center that was greater than the critical mass necessary for the nuclear explosion."}
{"id": 5038, "contents": "2109. Fission Reactors - \nChain reactions of fissionable materials can be controlled and sustained without an explosion in a nuclear reactor (Figure 20.19). Any nuclear reactor that produces power via the fission of uranium or plutonium by bombardment with neutrons must have at least five components: nuclear fuel consisting of fissionable material, a nuclear moderator, reactor coolant, control rods, and a shield and containment system. We will discuss these components in greater detail later in the section. The reactor works by separating the fissionable nuclear material such that a critical mass cannot be formed, controlling both the flux and absorption of neutrons to allow shutting down the fission reactions. In a nuclear reactor used for the production of electricity, the energy released by fission reactions is trapped as thermal energy and used to boil water and produce steam. The steam is used to turn a turbine, which powers a generator for the production of electricity.\n\n\nFIGURE 20.19 (a) The Diablo Canyon Nuclear Power Plant near San Luis Obispo is the only nuclear power plant currently in operation in California. The domes are the containment structures for the nuclear reactors, and the brown building houses the turbine where electricity is generated. Ocean water is used for cooling. (b) The Diablo Canyon uses a pressurized water reactor, one of a few different fission reactor designs in use around the world, to produce electricity. Energy from the nuclear fission reactions in the core heats water in a closed, pressurized system. Heat from this system produces steam that drives a turbine, which in turn produces electricity. (credit a: modification of work by \"Mike\" Michael L. Baird; credit b: modification of work by the Nuclear Regulatory Commission)"}
{"id": 5039, "contents": "2110. Nuclear Fuels - \nNuclear fuel consists of a fissionable isotope, such as uranium-235, which must be present in sufficient quantity to provide a self-sustaining chain reaction. In the United States, uranium ores contain from $0.05-0.3 \\%$ of the uranium oxide $\\mathrm{U}_{3} \\mathrm{O}_{8}$; the uranium in the ore is about $99.3 \\%$ nonfissionable $\\mathrm{U}-238$ with only $0.7 \\%$ fissionable $U-235$. Nuclear reactors require a fuel with a higher concentration of $U-235$ than is found in nature; it is normally enriched to have about $5 \\%$ of uranium mass as $\\mathrm{U}-235$. At this concentration, it is not possible to achieve the supercritical mass necessary for a nuclear explosion. Uranium can be enriched by gaseous diffusion (the only method currently used in the US), using a gas centrifuge, or by laser separation.\n\nIn the gaseous diffusion enrichment plant where U - 235 fuel is prepared, $\\mathrm{UF}_{6}$ (uranium hexafluoride) gas at low pressure moves through barriers that have holes just barely large enough for $\\mathrm{UF}_{6}$ to pass through. The slightly lighter ${ }^{235} \\mathrm{UF}_{6}$ molecules diffuse through the barrier slightly faster than the heavier ${ }^{238} \\mathrm{UF}_{6}$ molecules. This process is repeated through hundreds of barriers, gradually increasing the concentration of ${ }^{235} \\mathrm{UF}_{6}$ to the level needed by the nuclear reactor. The basis for this process, Graham's law, is described in the chapter on gases. The enriched $\\mathrm{UF}_{6}$ gas is collected, cooled until it solidifies, and then taken to a fabrication facility where it is made into fuel assemblies. Each fuel assembly consists of fuel rods that contain many thimble-sized, ceramicencased, enriched uranium (usually $\\mathrm{UO}_{2}$ ) fuel pellets. Modern nuclear reactors may contain as many as 10 million fuel pellets. The amount of energy in each of these pellets is equal to that in almost a ton of coal or 150 gallons of oil."}
{"id": 5040, "contents": "2111. Nuclear Moderators - \nNeutrons produced by nuclear reactions move too fast to cause fission (refer back to Figure 20.17). They must first be slowed to be absorbed by the fuel and produce additional nuclear reactions. A nuclear moderator is a substance that slows the neutrons to a speed that is low enough to cause fission. Early reactors used highpurity graphite as a moderator. Modern reactors in the US exclusively use heavy water $\\left({ }_{1}^{2} \\mathrm{H}_{2} \\mathrm{O}\\right)$ or light water (ordinary $\\mathrm{H}_{2} \\mathrm{O}$ ), whereas some reactors in other countries use other materials, such as carbon dioxide, beryllium, or graphite."}
{"id": 5041, "contents": "2112. Reactor Coolants - \nA nuclear reactor coolant is used to carry the heat produced by the fission reaction to an external boiler and turbine, where it is transformed into electricity. Two overlapping coolant loops are often used; this counteracts the transfer of radioactivity from the reactor to the primary coolant loop. All nuclear power plants in the US use water as a coolant. Other coolants include molten sodium, lead, a lead-bismuth mixture, or molten salts."}
{"id": 5042, "contents": "2113. Control Rods - \nNuclear reactors use control rods (Figure 20.20) to control the fission rate of the nuclear fuel by adjusting the number of slow neutrons present to keep the rate of the chain reaction at a safe level. Control rods are made of boron, cadmium, hafnium, or other elements that are able to absorb neutrons. Boron-10, for example, absorbs neutrons by a reaction that produces lithium-7 and alpha particles:\n\n$$\n{ }_{5}^{10} \\mathrm{~B}+{ }_{0}^{1} \\mathrm{n} \\longrightarrow{ }_{3}^{7} \\mathrm{Li}+{ }_{2}^{4} \\mathrm{He}\n$$\n\nWhen control rod assemblies are inserted into the fuel element in the reactor core, they absorb a larger fraction of the slow neutrons, thereby slowing the rate of the fission reaction and decreasing the power produced. Conversely, if the control rods are removed, fewer neutrons are absorbed, and the fission rate and energy production increase. In an emergency, the chain reaction can be shut down by fully inserting all of the control rods into the nuclear core between the fuel rods.\n\n\nFIGURE 20.20 The nuclear reactor core shown in (a) contains the fuel and control rod assembly shown in (b).\n(credit: modification of work by E. Generalic, http://glossary.periodni.com/glossary.php?en=control+rod)"}
{"id": 5043, "contents": "2114. Shield and Containment System - \nDuring its operation, a nuclear reactor produces neutrons and other radiation. Even when shut down, the decay products are radioactive. In addition, an operating reactor is thermally very hot, and high pressures result from the circulation of water or another coolant through it. Thus, a reactor must withstand high temperatures and pressures, and must protect operating personnel from the radiation. Reactors are equipped with a containment system (or shield) that consists of three parts:\n\n1. The reactor vessel, a steel shell that is 3-20-centimeters thick and, with the moderator, absorbs much of the radiation produced by the reactor\n2. A main shield of $1-3$ meters of high-density concrete\n3. A personnel shield of lighter materials that protects operators from $\\gamma$ rays and X-rays\n\nIn addition, reactors are often covered with a steel or concrete dome that is designed to contain any radioactive materials might be released by a reactor accident."}
{"id": 5044, "contents": "2115. LINK TO LEARNING - \nClick here to watch a 3-minute video (http://openstax.org/l/16nucreactors) from the Nuclear Energy Institute on how nuclear reactors work.\n\nNuclear power plants are designed in such a way that they cannot form a supercritical mass of fissionable material and therefore cannot create a nuclear explosion. But as history has shown, failures of systems and safeguards can cause catastrophic accidents, including chemical explosions and nuclear meltdowns (damage to the reactor core from overheating). The following Chemistry in Everyday Life feature explores three infamous meltdown incidents."}
{"id": 5045, "contents": "2117. Nuclear Accidents - \nThe importance of cooling and containment are amply illustrated by three major accidents that occurred with the nuclear reactors at nuclear power generating stations in the United States (Three Mile Island), the former Soviet Union (Chernobyl), and Japan (Fukushima).\n\nIn March 1979, the cooling system of the Unit 2 reactor at Three Mile Island Nuclear Generating Station in Pennsylvania failed, and the cooling water spilled from the reactor onto the floor of the containment building. After the pumps stopped, the reactors overheated due to the high radioactive decay heat produced in the first few days after the nuclear reactor shut down. The temperature of the core climbed to at least $2200^{\\circ} \\mathrm{C}$, and the upper portion of the core began to melt. In addition, the zirconium alloy cladding of the fuel rods began to react with steam and produced hydrogen:\n\n$$\n\\mathrm{Zr}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(g) \\longrightarrow \\mathrm{ZrO}_{2}(s)+2 \\mathrm{H}_{2}(g)\n$$\n\nThe hydrogen accumulated in the confinement building, and it was feared that there was danger of an explosion of the mixture of hydrogen and air in the building. Consequently, hydrogen gas and radioactive gases (primarily krypton and xenon) were vented from the building. Within a week, cooling water circulation was restored and the core began to cool. The plant was closed for nearly 10 years during the cleanup process.\n\nAlthough zero discharge of radioactive material is desirable, the discharge of radioactive krypton and xenon, such as occurred at the Three Mile Island plant, is among the most tolerable. These gases readily disperse in the atmosphere and thus do not produce highly radioactive areas. Moreover, they are noble gases and are not incorporated into plant and animal matter in the food chain. Effectively none of the heavy elements of the core of the reactor were released into the environment, and no cleanup of the area outside of the containment building was necessary (Figure 20.21)."}
{"id": 5046, "contents": "2117. Nuclear Accidents - \nFIGURE 20.21 (a) In this 2010 photo of Three Mile Island, the remaining structures from the damaged Unit 2 reactor are seen on the left, whereas the separate Unit 1 reactor, unaffected by the accident, continues generating power to this day (right). (b) President Jimmy Carter visited the Unit 2 control room a few days after the accident in 1979.\n\nAnother major nuclear accident involving a reactor occurred in April 1986, at the Chernobyl Nuclear Power Plant in Ukraine, which was still a part of the former Soviet Union. While operating at low power during an unauthorized experiment with some of its safety devices shut off, one of the reactors at the plant became unstable. Its chain reaction became uncontrollable and increased to a level far beyond what the reactor was designed for. The steam pressure in the reactor rose to between 100 and 500 times the full power pressure and ruptured the reactor. Because the reactor was not enclosed in a containment building, a large amount of radioactive material spewed out, and additional fission products were released, as the graphite (carbon) moderator of the core ignited and burned. The fire was controlled, but over 200 plant workers and firefighters developed acute radiation sickness and at least 32 soon died from the effects of the radiation. It is predicted that about 4000 more deaths will occur among emergency workers and former Chernobyl residents from radiation-induced cancer and leukemia. The reactor has since been encapsulated in steel and concrete, a now-decaying structure known as the sarcophagus. Almost 30 years later, significant radiation problems still persist in the area, and Chernobyl largely remains a wasteland.\n\nIn 2011, the Fukushima Daiichi Nuclear Power Plant in Japan was badly damaged by a 9.0-magnitude earthquake and resulting tsunami. Three reactors up and running at the time were shut down automatically, and emergency generators came online to power electronics and coolant systems. However, the tsunami quickly flooded the emergency generators and cut power to the pumps that circulated coolant water through the reactors. High-temperature steam in the reactors reacted with zirconium alloy to produce hydrogen gas. The gas escaped into the containment building, and the mixture of hydrogen and air exploded. Radioactive material was released from the containment vessels as the result of deliberate venting to reduce the hydrogen pressure, deliberate discharge of coolant water into the sea, and accidental or uncontrolled events."}
{"id": 5047, "contents": "2117. Nuclear Accidents - \nAn evacuation zone around the damaged plant extended over 12.4 miles away, and an estimated 200,000 people were evacuated from the area. All 48 of Japan's nuclear power plants were subsequently shut down, remaining shuttered as of December 2014. Since the disaster, public opinion has shifted from largely favoring to largely opposing increasing the use of nuclear power plants, and a restart of Japan's atomic energy program is still stalled (Figure 20.22).\n\n\nFIGURE 20.22 (a) After the accident, contaminated waste had to be removed, and (b) an evacuation zone was set up around the plant in areas that received heavy doses of radioactive fallout. (credit a: modification of work by \"Live Action Hero\"/Flickr)\n\nThe energy produced by a reactor fueled with enriched uranium results from the fission of uranium as well as from the fission of plutonium produced as the reactor operates. As discussed previously, the plutonium forms from the combination of neutrons and the uranium in the fuel. In any nuclear reactor, only about $0.1 \\%$ of the mass of the fuel is converted into energy. The other $99.9 \\%$ remains in the fuel rods as fission products and unused fuel. All of the fission products absorb neutrons, and after a period of several months to a few years, depending on the reactor, the fission products must be removed by changing the fuel rods. Otherwise, the concentration of these fission products would increase and absorb more neutrons until the reactor could no longer operate.\n\nSpent fuel rods contain a variety of products, consisting of unstable nuclei ranging in atomic number from 25 to 60 , some transuranium elements, including plutonium and americium, and unreacted uranium isotopes. The unstable nuclei and the transuranium isotopes give the spent fuel a dangerously high level of radioactivity. The long-lived isotopes require thousands of years to decay to a safe level. The ultimate fate of the nuclear reactor as a significant source of energy in the United States probably rests on whether or not a politically and scientifically satisfactory technique for processing and storing the components of spent fuel rods can be developed."}
{"id": 5048, "contents": "2118. LINK TO LEARNING - \nExplore the information in this link (http://openstax.org/l/16wastemgmt) to learn about the approaches to nuclear waste management."}
{"id": 5049, "contents": "2119. Nuclear Fusion and Fusion Reactors - \nThe process of converting very light nuclei into heavier nuclei is also accompanied by the conversion of mass into large amounts of energy, a process called fusion. The principal source of energy in the sun is a net fusion reaction in which four hydrogen nuclei fuse and produce one helium nucleus and two positrons. This is a net reaction of a more complicated series of events:\n\n$$\n4{ }_{1}^{1} \\mathrm{H} \\longrightarrow{ }_{2}^{4} \\mathrm{He}+2{ }_{+1}^{0} \\mathrm{e}^{+}\n$$\n\nA helium nucleus has a mass that is $0.7 \\%$ less than that of four hydrogen nuclei; this lost mass is converted into energy during the fusion. This reaction produces about $3.6 \\times 10^{11} \\mathrm{~kJ}$ of energy per mole of ${ }_{2}^{4} \\mathrm{He}$ produced. This is somewhat larger than the energy produced by the nuclear fission of one mole of U-235 ( $1.8 \\times 10^{10} \\mathrm{~kJ}$ ), and over 3 million times larger than the energy produced by the (chemical) combustion of one mole of octane ( 5471 kJ ).\n\nIt has been determined that the nuclei of the heavy isotopes of hydrogen, a deuteron, ${ }_{1}^{2} \\mathrm{H}$ and a triton, ${ }_{1}^{3} \\mathrm{H}$, undergo fusion at extremely high temperatures (thermonuclear fusion). They form a helium nucleus and a neutron:\n\n$$\n{ }_{1}^{2} \\mathrm{H}+{ }_{1}^{3} \\mathrm{H} \\longrightarrow{ }_{2}^{4} \\mathrm{He}+{ }_{0}^{1} \\mathrm{n}\n$$\n\nThis change proceeds with a mass loss of 0.0188 amu , corresponding to the release of $1.69 \\times 10^{9}$ kilojoules per mole of ${ }_{2}^{4} \\mathrm{He}$ formed. The very high temperature is necessary to give the nuclei enough kinetic energy to overcome the very strong repulsive forces resulting from the positive charges on their nuclei so they can collide."}
{"id": 5050, "contents": "2119. Nuclear Fusion and Fusion Reactors - \nUseful fusion reactions require very high temperatures for their initiation-about 15,000,000 K or more. At these temperatures, all molecules dissociate into atoms, and the atoms ionize, forming plasma. These conditions occur in an extremely large number of locations throughout the universe-stars are powered by fusion. Humans have already figured out how to create temperatures high enough to achieve fusion on a large scale in thermonuclear weapons. A thermonuclear weapon such as a hydrogen bomb contains a nuclear fission bomb that, when exploded, gives off enough energy to produce the extremely high temperatures necessary for fusion to occur.\n\nAnother much more beneficial way to create fusion reactions is in a fusion reactor, a nuclear reactor in which fusion reactions of light nuclei are controlled. Because no solid materials are stable at such high temperatures, mechanical devices cannot contain the plasma in which fusion reactions occur. Two techniques to contain plasma at the density and temperature necessary for a fusion reaction are currently the focus of intensive research efforts: containment by a magnetic field and by the use of focused laser beams (Figure 20.23). A number of large projects are working to attain one of the biggest goals in science: getting hydrogen fuel to ignite and produce more energy than the amount supplied to achieve the extremely high temperatures and pressures that are required for fusion. At the time of this writing, there are no self-sustaining fusion reactors operating in the world, although small-scale controlled fusion reactions have been run for very brief periods.\n\n(a)\n\n(b)"}
{"id": 5051, "contents": "2119. Nuclear Fusion and Fusion Reactors - \n(a)\n\n(b)\n\nFIGURE 20.23 (a) This model is of the International Thermonuclear Experimental Reactor (ITER) reactor. Currently under construction in the south of France with an expected completion date of 2027, the ITER will be the world's largest experimental Tokamak nuclear fusion reactor with a goal of achieving large-scale sustained energy production. (b) In 2012, the National Ignition Facility at Lawrence Livermore National Laboratory briefly produced over 500,000,000,000 watts ( 500 terawatts, or 500 TW ) of peak power and delivered 1,850,000 joules (1.85 MJ) of energy, the largest laser energy ever produced and 1000 times the power usage of the entire United States in any given moment. Although lasting only a few billionths of a second, the 192 lasers attained the conditions needed for nuclear fusion ignition. This image shows the target prior to the laser shot. (credit a: modification of work by Stephan Mosel)"}
{"id": 5052, "contents": "2120. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- List common applications of radioactive isotopes\n\nRadioactive isotopes have the same chemical properties as stable isotopes of the same element, but they emit radiation, which can be detected. If we replace one (or more) atom(s) with radioisotope(s) in a compound, we can track them by monitoring their radioactive emissions. This type of compound is called a radioactive tracer (or radioactive label). Radioisotopes are used to follow the paths of biochemical reactions or to determine how a substance is distributed within an organism. Radioactive tracers are also used in many medical applications, including both diagnosis and treatment. They are used to measure engine wear, analyze the geological formation around oil wells, and much more.\n\nRadioimmunossays (RIA), for example, rely on radioisotopes to detect the presence and/or concentration of certain antigens. Developed by Rosalyn Sussman Yalow and Solomon Berson in the 1950s, the technique is known for extreme sensitivity, meaning that it can detect and measure very small quantities of a substance. Prior to its discovery, most similar detection relied on large enough quantities to produce visible outcomes. RIA revolutionized and expanded entire fields of study, most notably endocrinology, and is commonly used in narcotics detection, blood bank screening, early cancer screening, hormone measurement, and allergy diagnosis. Based on her significant contribution to medicine, Yalow received a Nobel Prize, making her the second woman to be awarded the prize for medicine."}
{"id": 5053, "contents": "2120. LEARNING OBJECTIVES - \nRadioisotopes have revolutionized medical practice (see Appendix M), where they are used extensively. Over 10 million nuclear medicine procedures and more than 100 million nuclear medicine tests are performed annually in the United States. Four typical examples of radioactive tracers used in medicine are technetium-99 $\\left({ }_{43}^{99} \\mathrm{Tc}\\right)$, thallium-201 $\\left({ }_{81}^{201} \\mathrm{Tl}\\right)$, iodine-131 ( $\\left.{ }_{53}^{131} \\mathrm{I}\\right)$, and sodium-24 $\\left({ }_{11}^{24} \\mathrm{Na}\\right)$. Damaged tissues in the heart, liver, and lungs absorb certain compounds of technetium-99 preferentially. After it is injected, the location of the technetium compound, and hence the damaged tissue, can be determined by detecting the $\\gamma$ rays emitted by the Tc-99 isotope. Thallium-201 (Figure 20.24) becomes concentrated in healthy heart tissue, so the two isotopes, Tc-99 and Tl-201, are used together to study heart tissue. Iodine-131 concentrates in the thyroid gland, the liver, and some parts of the brain. It can therefore be used to monitor goiter and treat thyroid conditions, such as Grave's disease, as well as liver and brain tumors. Salt solutions containing compounds of sodium-24 are injected into the bloodstream to help locate obstructions to the flow of blood.\n\n\nFIGURE 20.24 Administering thallium-201 to a patient and subsequently performing a stress test offer medical professionals an opportunity to visually analyze heart function and blood flow. (credit: modification of work by \"BlueOctane\"/Wikimedia Commons)"}
{"id": 5054, "contents": "2120. LEARNING OBJECTIVES - \nRadioisotopes used in medicine typically have short half-lives-for example, the ubiquitous Tc-99m has a halflife of 6.01 hours. This makes Tc-99m essentially impossible to store and prohibitively expensive to transport, so it is made on-site instead. Hospitals and other medical facilities use Mo-99 (which is primarily extracted from U-235 fission products) to generate Tc-99. Mo-99 undergoes $\\beta$ decay with a half-life of 66 hours, and the $\\mathrm{Tc}-99$ is then chemically extracted (Figure 20.25). The parent nuclide Mo-99 is part of a molybdate ion, $\\mathrm{MoO}_{4}{ }^{2-}$; when it decays, it forms the pertechnetate ion, $\\mathrm{TcO}_{4}{ }^{-}$. These two water-soluble ions are separated by column chromatography, with the higher charge molybdate ion adsorbing onto the alumina in the column, and the lower charge pertechnetate ion passing through the column in the solution. A few micrograms of Mo-99 can produce enough Tc-99 to perform as many as 10,000 tests.\n\n\nFIGURE 20.25 (a) The first Tc-99m generator (circa 1958) is used to separate Tc-99 from Mo-99. The $\\mathrm{MoO}_{4}{ }^{2-}$ is retained by the matrix in the column, whereas the $\\mathrm{TcO}_{4}{ }^{-}$passes through and is collected. (b) Tc-99 was used in this scan of the neck of a patient with Grave's disease. The scan shows the location of high concentrations of Tc-99. (credit a: modification of work by the Department of Energy; credit b: modification of work by \"MBq\"/Wikimedia Commons)"}
{"id": 5055, "contents": "2120. LEARNING OBJECTIVES - \nRadioisotopes can also be used, typically in higher doses than as a tracer, as treatment. Radiation therapy is the use of high-energy radiation to damage the DNA of cancer cells, which kills them or keeps them from dividing (Figure 20.26). A cancer patient may receive external beam radiation therapy delivered by a machine outside the body, or internal radiation therapy (brachytherapy) from a radioactive substance that\nhas been introduced into the body. Note that chemotherapy is similar to internal radiation therapy in that the cancer treatment is injected into the body, but differs in that chemotherapy uses chemical rather than radioactive substances to kill the cancer cells.\n\n\nFIGURE 20.26 The cartoon in (a) shows a cobalt-60 machine used in the treatment of cancer. The diagram in (b) shows how the gantry of the Co-60 machine swings through an arc, focusing radiation on the targeted region (tumor) and minimizing the amount of radiation that passes through nearby regions.\n\nCobalt-60 is a synthetic radioisotope produced by the neutron activation of Co-59, which then undergoes $\\beta$ decay to form Ni-60, along with the emission of $\\gamma$ radiation. The overall process is:\n\n$$\n{ }_{27}^{59} \\mathrm{Co}+{ }_{0}^{1} \\mathrm{n} \\longrightarrow{ }_{27}^{60} \\mathrm{Co} \\longrightarrow{ }_{28}^{60} \\mathrm{Ni}+{ }_{-1}^{0} \\beta+2{ }_{0}^{0} \\gamma\n$$\n\nThe overall decay scheme for this is shown graphically in Figure 20.27.\n\n\nFIGURE 20.27 Co-60 undergoes a series of radioactive decays. The $\\gamma$ emissions are used for radiation therapy.\n\nRadioisotopes are used in diverse ways to study the mechanisms of chemical reactions in plants and animals. These include labeling fertilizers in studies of nutrient uptake by plants and crop growth, investigations of digestive and milk-producing processes in cows, and studies on the growth and metabolism of animals and plants.\n\nFor example, the radioisotope C-14 was used to elucidate the details of how photosynthesis occurs. The overall reaction is:"}
{"id": 5056, "contents": "2120. LEARNING OBJECTIVES - \nFor example, the radioisotope C-14 was used to elucidate the details of how photosynthesis occurs. The overall reaction is:\n\n$$\n6 \\mathrm{CO}_{2}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}(s)+6 \\mathrm{O}_{2}(g)\n$$\n\nbut the process is much more complex, proceeding through a series of steps in which various organic compounds are produced. In studies of the pathway of this reaction, plants were exposed to $\\mathrm{CO}_{2}$ containing a high concentration of ${ }_{6}^{14} \\mathrm{C}$. At regular intervals, the plants were analyzed to determine which organic compounds contained carbon-14 and how much of each compound was present. From the time sequence in which the compounds appeared and the amount of each present at given time intervals, scientists learned more about the pathway of the reaction.\n\nCommercial applications of radioactive materials are equally diverse (Figure 20.28). They include determining the thickness of films and thin metal sheets by exploiting the penetration power of various types of radiation. Flaws in metals used for structural purposes can be detected using high-energy gamma rays from cobalt-60 in a fashion similar to the way X-rays are used to examine the human body. In one form of pest control, flies are controlled by sterilizing male flies with $\\gamma$ radiation so that females breeding with them do not produce offspring. Many foods are preserved by radiation that kills microorganisms that cause the foods to spoil.\n\n\nFIGURE 20.28 Common commercial uses of radiation include (a) X-ray examination of luggage at an airport and (b) preservation of food. (credit a: modification of work by the Department of the Navy; credit b: modification of work by the US Department of Agriculture)"}
{"id": 5057, "contents": "2120. LEARNING OBJECTIVES - \nAmericium-241, an $\\alpha$ emitter with a half-life of 458 years, is used in tiny amounts in ionization-type smoke detectors (Figure 20.29). The $\\alpha$ emissions from Am-241 ionize the air between two electrode plates in the ionizing chamber. A battery supplies a potential that causes movement of the ions, thus creating a small electric current. When smoke enters the chamber, the movement of the ions is impeded, reducing the conductivity of the air. This causes a marked drop in the current, triggering an alarm.\n\n\nFIGURE 20.29 Inside a smoke detector, Am-241 emits $\\alpha$ particles that ionize the air, creating a small electric current. During a fire, smoke particles impede the flow of ions, reducing the current and triggering an alarm. (credit a: modification of work by \"Muffet\"/Wikimedia Commons)"}
{"id": 5058, "contents": "2121. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the biological impact of ionizing radiation\n- Define units for measuring radiation exposure\n- Explain the operation of common tools for detecting radioactivity\n- List common sources of radiation exposure in the US\n\nThe increased use of radioisotopes has led to increased concerns over the effects of these materials on biological systems (such as humans). All radioactive nuclides emit high-energy particles or electromagnetic waves. When this radiation encounters living cells, it can cause heating, break chemical bonds, or ionize molecules. The most serious biological damage results when these radioactive emissions fragment or ionize molecules. For example, alpha and beta particles emitted from nuclear decay reactions possess much higher energies than ordinary chemical bond energies. When these particles strike and penetrate matter, they produce ions and molecular fragments that are extremely reactive. The damage this does to biomolecules in living organisms can cause serious malfunctions in normal cell processes, taxing the organism's repair mechanisms and possibly causing illness or even death (Figure 20.30).\n\n\nFIGURE 20.30 Radiation can harm biological systems by damaging the DNA of cells. If this damage is not properly repaired, the cells may divide in an uncontrolled manner and cause cancer."}
{"id": 5059, "contents": "2122. Ionizing and Nonionizing Radiation - \nThere is a large difference in the magnitude of the biological effects of nonionizing radiation (for example, light and microwaves) and ionizing radiation, emissions energetic enough to knock electrons out of molecules (for example, $\\alpha$ and $\\beta$ particles, $\\gamma$ rays, X-rays, and high-energy ultraviolet radiation) (Figure 20.31).\n\n\nFIGURE 20.31 Lower frequency, lower-energy electromagnetic radiation is nonionizing, and higher frequency, higher-energy electromagnetic radiation is ionizing.\n\nEnergy absorbed from nonionizing radiation speeds up the movement of atoms and molecules, which is equivalent to heating the sample. Although biological systems are sensitive to heat (as we might know from touching a hot stove or spending a day at the beach in the sun), a large amount of nonionizing radiation is necessary before dangerous levels are reached. Ionizing radiation, however, may cause much more severe damage by breaking bonds or removing electrons in biological molecules, disrupting their structure and function. The damage can also be done indirectly, by first ionizing $\\mathrm{H}_{2} \\mathrm{O}$ (the most abundant molecule in living organisms), which forms a $\\mathrm{H}_{2} \\mathrm{O}^{+}$ion that reacts with water, forming a hydronium ion and a hydroxyl radical:\n\n\nBecause the hydroxyl radical has an unpaired electron, it is highly reactive. (This is true of any substance with unpaired electrons, known as a free radical.) This hydroxyl radical can react with all kinds of biological molecules (DNA, proteins, enzymes, and so on), causing damage to the molecules and disrupting physiological processes. Examples of direct and indirect damage are shown in Figure 20.32.\n\n\nFIGURE 20.32 Ionizing radiation can (a) directly damage a biomolecule by ionizing it or breaking its bonds, or (b) create an $\\mathrm{H}_{2} \\mathrm{O}^{+}$ion, which reacts with $\\mathrm{H}_{2} \\mathrm{O}$ to form a hydroxyl radical, which in turn reacts with the biomolecule,\ncausing damage indirectly."}
{"id": 5060, "contents": "2123. Biological Effects of Exposure to Radiation - \nRadiation can harm either the whole body (somatic damage) or eggs and sperm (genetic damage). Its effects are more pronounced in cells that reproduce rapidly, such as the stomach lining, hair follicles, bone marrow, and embryos. This is why patients undergoing radiation therapy often feel nauseous or sick to their stomach, lose hair, have bone aches, and so on, and why particular care must be taken when undergoing radiation therapy during pregnancy.\n\nDifferent types of radiation have differing abilities to pass through material (Figure 20.33). A very thin barrier, such as a sheet or two of paper, or the top layer of skin cells, usually stops alpha particles. Because of this, alpha particle sources are usually not dangerous if outside the body, but are quite hazardous if ingested or inhaled (see the Chemistry in Everyday Life feature on Radon Exposure). Beta particles will pass through a hand, or a thin layer of material like paper or wood, but are stopped by a thin layer of metal. Gamma radiation is very penetrating and can pass through a thick layer of most materials. Some high-energy gamma radiation is able to pass through a few feet of concrete. Certain dense, high atomic number elements (such as lead) can effectively attenuate gamma radiation with thinner material and are used for shielding. The ability of various kinds of emissions to cause ionization varies greatly, and some particles have almost no tendency to produce ionization. Alpha particles have about twice the ionizing power of fast-moving neutrons, about 10 times that of $\\beta$ particles, and about 20 times that of $\\gamma$ rays and X-rays.\n\n\nFIGURE 20.33 The ability of different types of radiation to pass through material is shown. From least to most penetrating, they are alpha < beta < neutron < gamma."}
{"id": 5061, "contents": "2125. Radon Exposure - \nFor many people, one of the largest sources of exposure to radiation is from radon gas (Rn-222). Radon-222 is an $\\alpha$ emitter with a half-life of 3.82 days. It is one of the products of the radioactive decay series of U-238 (Figure 20.9), which is found in trace amounts in soil and rocks. The radon gas that is produced slowly escapes from the ground and gradually seeps into homes and other structures above. Since it is about eight times more dense than air, radon gas accumulates in basements and lower floors, and slowly diffuses throughout buildings (Figure 20.34).\n\n\nFIGURE 20.34 Radon-222 seeps into houses and other buildings from rocks that contain uranium-238, a radon emitter. The radon enters through cracks in concrete foundations and basement floors, stone or porous cinderblock foundations, and openings for water and gas pipes.\n\nRadon is found in buildings across the country, with amounts depending on where you live. The average concentration of radon inside houses in the US ( $1.25 \\mathrm{pCi} / \\mathrm{L}$ ) is about three times the levels found in outside air, and about one in six houses have radon levels high enough that remediation efforts to reduce the radon concentration are recommended. Exposure to radon increases one's risk of getting cancer (especially lung cancer), and high radon levels can be as bad for health as smoking a carton of cigarettes a day. Radon is the number one cause of lung cancer in nonsmokers and the second leading cause of lung cancer overall. Radon exposure is believed to cause over 20,000 deaths in the US per year."}
{"id": 5062, "contents": "2126. Measuring Radiation Exposure - \nSeveral different devices are used to detect and measure radiation, including Geiger counters, scintillation counters (scintillators), and radiation dosimeters (Figure 20.35). Probably the best-known radiation instrument, the Geiger counter (also called the Geiger-M\u00fcller counter) detects and measures radiation. Radiation causes the ionization of the gas in a Geiger-M\u00fcller tube. The rate of ionization is proportional to the amount of radiation. A scintillation counter contains a scintillator-a material that emits light (luminesces) when excited by ionizing radiation-and a sensor that converts the light into an electric signal. Radiation dosimeters also measure ionizing radiation and are often used to determine personal radiation exposure. Commonly used types are electronic, film badge, thermoluminescent, and quartz fiber dosimeters.\n\n\nFIGURE 20.35 Devices such as (a) Geiger counters, (b) scintillators, and (c) dosimeters can be used to measure radiation. (credit c: modification of work by \"osaMu\"/Wikimedia commons)"}
{"id": 5063, "contents": "2126. Measuring Radiation Exposure - \nA variety of units are used to measure various aspects of radiation (Figure 20.36). The SI unit for rate of radioactive decay is the becquerel ( $\\mathbf{B q}$ ), with $1 \\mathrm{~Bq}=1$ disintegration per second. The curie ( $\\mathbf{C i}$ ) and millicurie $(\\mathbf{m C i})$ are much larger units and are frequently used in medicine ( 1 curie $=1 \\mathrm{Ci}=3.7 \\times 10^{10}$ disintegrations per second). The SI unit for measuring radiation dose is the gray (Gy), with $1 \\mathrm{~Gy}=1 \\mathrm{~J}$ of energy absorbed per kilogram of tissue. In medical applications, the radiation absorbed dose (rad) is more often used ( $1 \\mathrm{rad}=0.01$ Gy; 1 rad results in the absorption of $0.01 \\mathrm{~J} / \\mathrm{kg}$ of tissue). The SI unit measuring tissue damage caused by radiation is the sievert (Sv). This takes into account both the energy and the biological effects of the type of radiation involved in the radiation dose. The roentgen equivalent for man (rem) is the unit for radiation damage that is used most frequently in medicine ( $100 \\mathrm{rem}=1 \\mathrm{~Sv}$ ). Note that the tissue damage units (rem or Sv ) includes the energy of the radiation dose (rad or Gy) along with a biological factor referred to as the RBE (for relative biological effectiveness) that is an approximate measure of the relative damage done by the radiation. These are related by:\n\n$$\n\\text { number of rems }=\\mathrm{RBE} \\times \\text { number of rads }\n$$\n\nwith RBE approximately 10 for $\\alpha$ radiation, $2(+)$ for protons and neutrons, and 1 for $\\beta$ and $\\gamma$ radiation.\n\n\nFIGURE 20.36 Different units are used to measure the rate of emission from a radioactive source, the energy that is absorbed from the source, and the amount of damage the absorbed radiation does."}
{"id": 5064, "contents": "2127. Units of Radiation Measurement - \nTable 20.4 summarizes the units used for measuring radiation.\n\n| Measurement Purpose | Unit | Quantity Measured | Description |\n| :---: | :---: | :---: | :---: |\n| activity of source | becquerel (Bq) | radioactive decays or emissions | amount of sample that undergoes 1 decay/second |\n|  | curie (Ci) |  | amount of sample that undergoes 3.7 $\\times 10^{10}$ decays/second |\n| absorbed dose | gray (Gy) | energy absorbed per kg of tissue | $1 \\mathrm{~Gy}=1 \\mathrm{~J} / \\mathrm{kg}$ tissue |\n|  | radiation absorbed dose (rad) |  | $1 \\mathrm{rad}=0.01 \\mathrm{~J} / \\mathrm{kg}$ tissue |\n| biologically effective dose | sievert (Sv) | tissue damage | $\\mathrm{Sv}=\\mathrm{RBE} \\times \\mathrm{Gy}$ |\n|  | roentgen equivalent for man (rem) |  | $\\mathrm{Rem}=\\mathrm{RBE} \\times \\mathrm{rad}$ |\n\nTABLE 20.4"}
{"id": 5065, "contents": "2129. Amount of Radiation - \nCobalt-60 ( $t_{1 / 2}=5.26 \\mathrm{y}$ ) is used in cancer therapy since the $\\gamma$ rays it emits can be focused in small areas where the cancer is located. A 5.00-g sample of Co-60 is available for cancer treatment.\n(a) What is its activity in Bq?\n(b) What is its activity in Ci?"}
{"id": 5066, "contents": "2130. Solution - \nThe activity is given by:\n\n$$\n\\text { Activity }=\\lambda N=\\left(\\frac{\\ln 2}{t_{1 / 2}}\\right) N=\\left(\\frac{\\ln 2}{5.26 \\mathrm{y}}\\right) \\times 5.00 \\mathrm{~g}=0.659 \\frac{\\mathrm{~g}}{\\mathrm{y}} \\text { of } \\mathrm{Co}-60 \\text { that decay }\n$$\n\nAnd to convert this to decays per second:\n\n$$\n\\begin{aligned}\n& 0.659 \\frac{\\mathrm{~g}}{\\mathrm{y}} \\times \\frac{1 \\mathrm{y}}{365 \\mathrm{~d}} \\times \\frac{1 \\mathrm{~d}}{24 \\mathrm{~h}} \\times \\frac{1 \\mathrm{~h}}{3600 \\mathrm{~s}} \\times \\frac{1 \\mathrm{~mol}}{59.9 \\mathrm{~g}} \\times \\frac{6.02 \\times 10^{23} \\mathrm{atoms}}{1 \\mathrm{~mol}} \\times \\frac{1 \\text { decay }}{1 \\text { atom }} \\\\\n& =2.10 \\times 10^{14 \\frac{\\text { decay }}{\\mathrm{s}}}\n\\end{aligned}\n$$\n\n(a) Since $1 \\mathrm{~Bq}=\\frac{1 \\text { decay }}{\\mathrm{s}}$, the activity in Becquerel $(\\mathrm{Bq})$ is:\n\n$$\n2.10 \\times 10^{14} \\frac{\\text { decay }}{s} \\times\\left(\\frac{1 \\mathrm{~Bq}}{1 \\frac{\\text { decay }}{\\mathrm{s}}}\\right)=2.10 \\times 10^{14} \\mathrm{~Bq}\n$$\n\n(b) Since $1 \\mathrm{Ci}=\\frac{3.7 \\times 10^{11} \\text { decay }}{\\mathrm{s}}$, the activity in curie (Ci) is:"}
{"id": 5067, "contents": "2130. Solution - \n(b) Since $1 \\mathrm{Ci}=\\frac{3.7 \\times 10^{11} \\text { decay }}{\\mathrm{s}}$, the activity in curie (Ci) is:\n\n$$\n2.10 \\times 10^{14} \\frac{\\text { decay }}{s} \\times\\left(\\frac{1 \\mathrm{Ci}}{\\frac{3.7 \\times 10^{11} \\text { decay }}{\\mathrm{s}}}\\right)=5.7 \\times 10^{2} \\mathrm{Ci}\n$$"}
{"id": 5068, "contents": "2131. Check Your Learning - \nTritium is a radioactive isotope of hydrogen ( $t_{1 / 2}=12.32 \\mathrm{y}$ ) that has several uses, including self-powered lighting, in which electrons emitted in tritium radioactive decay cause phosphorus to glow. Its nucleus contains one proton and two neutrons, and the atomic mass of tritium is 3.016 amu . What is the activity of a sample containing 1.00 mg of tritium (a) in Bq and (b) in Ci ?"}
{"id": 5069, "contents": "2132. Answer: - \n(a) $3.56 \\times 10^{11} \\mathrm{~Bq}$; (b) 0.962 Ci"}
{"id": 5070, "contents": "2133. Effects of Long-term Radiation Exposure on the Human Body - \nThe effects of radiation depend on the type, energy, and location of the radiation source, and the length of exposure. As shown in Figure 20.37, the average person is exposed to background radiation, including cosmic rays from the sun and radon from uranium in the ground (see the Chemistry in Everyday Life feature on Radon Exposure); radiation from medical exposure, including CAT scans, radioisotope tests, X-rays, and so on; and small amounts of radiation from other human activities, such as airplane flights (which are bombarded by increased numbers of cosmic rays in the upper atmosphere), radioactivity from consumer products, and a variety of radionuclides that enter our bodies when we breathe (for example, carbon-14) or through the food chain (for example, potassium-40, strontium-90, and iodine-131).\n\n\nFIGURE 20.37 The total annual radiation exposure for a person in the US is about 620 mrem . The various sources and their relative amounts are shown in this bar graph. (source: U.S. Nuclear Regulatory Commission)\n\nA short-term, sudden dose of a large amount of radiation can cause a wide range of health effects, from\nchanges in blood chemistry to death. Short-term exposure to tens of rems of radiation will likely cause very noticeable symptoms or illness; a dose of about 500 rems is estimated to have a $50 \\%$ probability of causing the death of the victim within 30 days of exposure. Exposure to radioactive emissions has a cumulative effect on the body during a person's lifetime, which is another reason why it is important to avoid any unnecessary exposure to radiation. Health effects of short-term exposure to radiation are shown in Table 20.5.\n\nHealth Effects of Radiation ${ }^{\\underline{2}}$"}
{"id": 5071, "contents": "2133. Effects of Long-term Radiation Exposure on the Human Body - \nHealth Effects of Radiation ${ }^{\\underline{2}}$\n\n| Exposure (rem) | Health Effect | Time to Onset (without treatment) |\n| :--- | :--- | :--- |\n| $5-10$ | changes in blood chemistry | - |\n| 50 | nausea | hours |\n| 55 | fatigue | - |\n| 70 | vomiting | - |\n| 75 | hair loss | $2-3$ weeks |\n| 90 | hemorrhage | - |\n| 100 | possible death | - |\n| 400 | destruction of intestinal lining | - |\n|  | internal bleeding | - |\n|  | death | within 2 months |\n| 1000 | damage to central nervous system | - |\n|  | loss of consciousness; | minutes |\n\nTABLE 20.5\n\nIt is impossible to avoid some exposure to ionizing radiation. We are constantly exposed to background radiation from a variety of natural sources, including cosmic radiation, rocks, medical procedures, consumer products, and even our own atoms. We can minimize our exposure by blocking or shielding the radiation, moving farther from the source, and limiting the time of exposure."}
{"id": 5072, "contents": "2134. Key Terms - \nalpha ( $\\alpha$ ) decay loss of an alpha particle during radioactive decay\nalpha particle ( $\\alpha$ or $\\mathbf{2}_{\\mathbf{2}}^{\\mathbf{H}} \\mathbf{H e}$ or ${ }_{\\mathbf{2}}^{\\mathbf{4}} \\boldsymbol{\\alpha}$ ) high-energy helium nucleus; a helium atom that has lost two electrons and contains two protons and two neutrons\nantimatter particles with the same mass but opposite properties (such as charge) of ordinary particles\nband of stability (also, belt of stability, zone of stability, or valley of stability) region of graph of number of protons versus number of neutrons containing stable (nonradioactive) nuclides\nbecquerel (Bq) SI unit for rate of radioactive decay; $1 \\mathrm{~Bq}=1$ disintegration/s\nbeta ( $\\boldsymbol{\\beta}$ ) decay breakdown of a neutron into a proton, which remains in the nucleus, and an electron, which is emitted as a beta particle\nbeta particle $\\left(\\boldsymbol{\\beta}\\right.$ or ${ }_{-\\mathbf{1}}^{\\mathbf{0}} \\mathbf{e}$ or $\\left.\\underset{-\\mathbf{1}}{\\mathbf{0}} \\boldsymbol{\\beta}\\right)$ high-energy electron\nbinding energy per nucleon total binding energy for the nucleus divided by the number of nucleons in the nucleus\nchain reaction repeated fission caused when the neutrons released in fission bombard other atoms\nchemotherapy similar to internal radiation therapy, but chemical rather than radioactive substances are introduced into the body to kill cancer cells\ncontainment system (also, shield) a three-part structure of materials that protects the exterior of a nuclear fission reactor and operating personnel from the high temperatures, pressures, and radiation levels inside the reactor\ncontrol rod material inserted into the fuel assembly that absorbs neutrons and can be raised or lowered to adjust the rate of a fission reaction\ncritical mass amount of fissionable material that will support a self-sustaining (nuclear fission) chain reaction\ncurie ( $\\mathbf{C i}$ ) larger unit for rate of radioactive decay frequently used in medicine; $1 \\mathrm{Ci}=3.7 \\times 10^{10}$ disintegrations/s"}
{"id": 5073, "contents": "2134. Key Terms - \ncurie ( $\\mathbf{C i}$ ) larger unit for rate of radioactive decay frequently used in medicine; $1 \\mathrm{Ci}=3.7 \\times 10^{10}$ disintegrations/s\ndaughter nuclide nuclide produced by the radioactive decay of another nuclide; may be stable or may decay further\nelectron capture combination of a core electron with a proton to yield a neutron within the nucleus\nelectron volt (eV) measurement unit of nuclear binding energies, with 1 eV equaling the amount energy due to the moving an electron across an electric potential difference of 1 volt\nexternal beam radiation therapy radiation delivered by a machine outside the body\nfissile (or fissionable) when a material is capable of sustaining a nuclear fission reaction\nfission splitting of a heavier nucleus into two or more lighter nuclei, usually accompanied by the conversion of mass into large amounts of energy\nfusion combination of very light nuclei into heavier nuclei, accompanied by the conversion of mass into large amounts of energy\nfusion reactor nuclear reactor in which fusion reactions of light nuclei are controlled\ngamma ( $\\boldsymbol{\\gamma}$ ) emission decay of an excited-state nuclide accompanied by emission of a gamma ray\ngamma ray ( $\\boldsymbol{\\gamma}$ or $\\mathbf{0}_{\\mathbf{0}}^{\\boldsymbol{0}} \\boldsymbol{\\gamma}$ ) short wavelength, highenergy electromagnetic radiation that exhibits wave-particle duality\nGeiger counter instrument that detects and measures radiation via the ionization produced in a Geiger-M\u00fcller tube\ngray (Gy) SI unit for measuring radiation dose; 1 Gy = 1 J absorbed $/ \\mathrm{kg}$ tissue\nhalf-life ( $\\boldsymbol{t}_{\\mathbf{1} / \\mathbf{2}}$ ) time required for half of the atoms in a radioactive sample to decay\ninternal radiation therapy (also, brachytherapy) radiation from a radioactive substance introduced into the body to kill cancer cells\nionizing radiation radiation that can cause a molecule to lose an electron and form an ion\nmagic number nuclei with specific numbers of nucleons that are within the band of stability\nmass defect difference between the mass of an atom and the summed mass of its constituent subatomic particles (or the mass \"lost\" when nucleons are brought together to form a nucleus)"}
{"id": 5074, "contents": "2134. Key Terms - \nmass defect difference between the mass of an atom and the summed mass of its constituent subatomic particles (or the mass \"lost\" when nucleons are brought together to form a nucleus)\nmass-energy equivalence equation Albert Einstein's relationship showing that mass and energy are equivalent\nmillicurie ( $\\mathbf{m C i}$ ) larger unit for rate of radioactive decay frequently used in medicine; $1 \\mathrm{Ci}=3.7 \\times$ $10^{10}$ disintegrations/s\nnonionizing radiation radiation that speeds up the movement of atoms and molecules; it is equivalent to heating a sample, but is not energetic enough to cause the ionization of molecules\nnuclear binding energy energy lost when an\natom's nucleons are bound together (or the energy needed to break a nucleus into its constituent protons and neutrons)\nnuclear chemistry study of the structure of atomic nuclei and processes that change nuclear structure\nnuclear fuel fissionable isotope present in sufficient quantities to provide a self-sustaining chain reaction in a nuclear reactor\nnuclear moderator substance that slows neutrons to a speed low enough to cause fission\nnuclear reaction change to a nucleus resulting in changes in the atomic number, mass number, or energy state\nnuclear reactor environment that produces energy via nuclear fission in which the chain reaction is controlled and sustained without explosion\nnuclear transmutation conversion of one nuclide into another nuclide\nnucleon collective term for protons and neutrons in a nucleus\nnuclide nucleus of a particular isotope\nparent nuclide unstable nuclide that changes spontaneously into another (daughter) nuclide\nparticle accelerator device that uses electric and magnetic fields to increase the kinetic energy of nuclei used in transmutation reactions\npositron $\\left({ }_{+1}^{0} \\beta\\right.$ or $\\left.{ }_{+1}^{0} \\mathrm{e}\\right)$ antiparticle to the electron; it has identical properties to an electron, except for having the opposite (positive) charge\npositron emission (also, $\\beta^{+}$decay) conversion of a proton into a neutron, which remains in the nucleus, and a positron, which is emitted\nradiation absorbed dose (rad) SI unit for measuring radiation dose, frequently used in medical applications; $1 \\mathrm{rad}=0.01$ Gy"}
{"id": 5075, "contents": "2134. Key Terms - \nradiation absorbed dose (rad) SI unit for measuring radiation dose, frequently used in medical applications; $1 \\mathrm{rad}=0.01$ Gy\nradiation dosimeter device that measures ionizing radiation and is used to determine personal radiation exposure\nradiation therapy use of high-energy radiation to damage the DNA of cancer cells, which kills them or keeps them from dividing\nradioactive decay spontaneous decay of an unstable nuclide into another nuclide\nradioactive decay series chains of successive disintegrations (radioactive decays) that ultimately lead to a stable end-product\nradioactive tracer (also, radioactive label)\nradioisotope used to track or follow a substance by monitoring its radioactive emissions\nradioactivity phenomenon exhibited by an unstable nucleon that spontaneously undergoes change into a nucleon that is more stable; an unstable nucleon is said to be radioactive\nradiocarbon dating highly accurate means of dating objects 30,000-50,000 years old that were derived from once-living matter; achieved by calculating the ratio of ${ }_{6}^{14} \\mathrm{C}:{ }_{6}^{12} \\mathrm{C}$ in the object vs. the ratio of ${ }_{6}^{14} \\mathrm{C}$ : ${ }_{6}^{12} \\mathrm{C}$ in the present-day atmosphere\nradioisotope isotope that is unstable and undergoes conversion into a different, more stable isotope\nradiometric dating use of radioisotopes and their properties to date the formation of objects such as archeological artifacts, formerly living organisms, or geological formations\nreactor coolant assembly used to carry the heat produced by fission in a reactor to an external boiler and turbine where it is transformed into electricity\nrelative biological effectiveness (RBE) measure of the relative damage done by radiation\nroentgen equivalent man (rem) unit for radiation damage, frequently used in medicine; 100 rem = 1 Sv\nscintillation counter instrument that uses a scintillator-a material that emits light when excited by ionizing radiation-to detect and measure radiation\nsievert (Sv) SI unit measuring tissue damage caused by radiation; takes into account energy and biological effects of radiation\nstrong nuclear force force of attraction between nucleons that holds a nucleus together\nsubcritical mass amount of fissionable material that cannot sustain a chain reaction; less than a critical mass"}
{"id": 5076, "contents": "2134. Key Terms - \nstrong nuclear force force of attraction between nucleons that holds a nucleus together\nsubcritical mass amount of fissionable material that cannot sustain a chain reaction; less than a critical mass\nsupercritical mass amount of material in which there is an increasing rate of fission\ntransmutation reaction bombardment of one type of nuclei with other nuclei or neutrons\ntransuranium element element with an atomic number greater than 92; these elements do not occur in nature"}
{"id": 5077, "contents": "2135. Key Equations - \n$E=m c^{2}$\ndecay rate $=\\lambda N$\n$t_{1 / 2}=\\frac{\\ln 2}{\\lambda}=\\frac{0.693}{\\lambda}$\nrem $=\\mathrm{RBE} \\times \\mathrm{rad}$\n$\\mathrm{Sv}=\\mathrm{RBE} \\times \\mathrm{Gy}$"}
{"id": 5078, "contents": "2136. Summary - 2136.1. Nuclear Structure and Stability\nAn atomic nucleus consists of protons and neutrons, collectively called nucleons. Although protons repel each other, the nucleus is held tightly together by a short-range, but very strong, force called the strong nuclear force. A nucleus has less mass than the total mass of its constituent nucleons. This \"missing\" mass is the mass defect, which has been converted into the binding energy that holds the nucleus together according to Einstein's mass-energy equivalence equation, $E=m c^{2}$. Of the many nuclides that exist, only a small number are stable. Nuclides with even numbers of protons or neutrons, or those with magic numbers of nucleons, are especially likely to be stable. These stable nuclides occupy a narrow band of stability on a graph of number of protons versus number of neutrons. The binding energy per nucleon is largest for the elements with mass numbers near 56 ; these are the most stable nuclei."}
{"id": 5079, "contents": "2136. Summary - 2136.2. Nuclear Equations\nNuclei can undergo reactions that change their number of protons, number of neutrons, or energy state. Many different particles can be involved in nuclear reactions. The most common are protons, neutrons, positrons (which are positively charged electrons), alpha ( $\\alpha$ ) particles (which are highenergy helium nuclei), beta ( $\\beta$ ) particles (which are high-energy electrons), and gamma ( $\\gamma$ ) rays (which compose high-energy electromagnetic radiation). As with chemical reactions, nuclear reactions are always balanced. When a nuclear reaction occurs, the total mass (number) and the total charge remain unchanged."}
{"id": 5080, "contents": "2136. Summary - 2136.3. Radioactive Decay\nNuclei that have unstable n:p ratios undergo spontaneous radioactive decay. The most common types of radioactivity are $\\alpha$ decay, $\\beta$ decay, $\\gamma$ emission, positron emission, and electron capture. Nuclear reactions also often involve $\\gamma$ rays, and some nuclei decay by electron capture. Each of these modes of decay leads to the formation of a new nucleus with a more stable n:p ratio. Some\nsubstances undergo radioactive decay series, proceeding through multiple decays before ending in a stable isotope. All nuclear decay processes follow first-order kinetics, and each radioisotope has its own characteristic half-life, the time that is required for half of its atoms to decay. Because of the large differences in stability among nuclides, there is a very wide range of half-lives of radioactive substances. Many of these substances have found useful applications in medical diagnosis and treatment, determining the age of archaeological and geological objects, and more."}
{"id": 5081, "contents": "2136. Summary - 2136.4. Transmutation and Nuclear Energy\nIt is possible to produce new atoms by bombarding other atoms with nuclei or high-speed particles. The products of these transmutation reactions can be stable or radioactive. A number of artificial elements, including technetium, astatine, and the transuranium elements, have been produced in this way.\n\nNuclear power as well as nuclear weapon detonations can be generated through fission (reactions in which a heavy nucleus is split into two or more lighter nuclei and several neutrons). Because the neutrons may induce additional fission reactions when they combine with other heavy nuclei, a chain reaction can result. Useful power is obtained if the fission process is carried out in a nuclear reactor. The conversion of light nuclei into heavier nuclei (fusion) also produces energy. At present, this energy has not been contained adequately and is too expensive to be feasible for commercial energy production."}
{"id": 5082, "contents": "2136. Summary - 2136.5. Uses of Radioisotopes\nCompounds known as radioactive tracers can be used to follow reactions, track the distribution of a substance, diagnose and treat medical conditions, and much more. Other radioactive substances are helpful for controlling pests, visualizing structures, providing fire warnings, and for many other applications. Hundreds of millions of nuclear medicine tests and procedures, using a wide variety of radioisotopes with relatively short half-lives, are\nperformed every year in the US. Most of these radioisotopes have relatively short half-lives; some are short enough that the radioisotope must be made on-site at medical facilities. Radiation therapy uses high-energy radiation to kill cancer cells by damaging their DNA. The radiation used for this treatment may be delivered externally or internally."}
{"id": 5083, "contents": "2136. Summary - 2136.6. Biological Effects of Radiation\nWe are constantly exposed to radiation from a variety of naturally occurring and human-produced sources. This radiation can affect living organisms. Ionizing radiation is the most harmful because it can ionize molecules or break chemical bonds, which damages the molecule and causes malfunctions in cell processes. It can also create reactive hydroxyl radicals that damage biological molecules and disrupt physiological processes. Radiation can cause somatic or genetic damage, and is most harmful to\nrapidly reproducing cells. Types of radiation differ in their ability to penetrate material and damage tissue, with alpha particles the least penetrating but potentially most damaging and gamma rays the most penetrating.\n\nVarious devices, including Geiger counters, scintillators, and dosimeters, are used to detect and measure radiation, and monitor radiation exposure. We use several units to measure radiation: becquerels or curies for rates of radioactive decay; gray or rads for energy absorbed; and rems or sieverts for biological effects of radiation. Exposure to radiation can cause a wide range of health effects, from minor to severe, and including death. We can minimize the effects of radiation by shielding with dense materials such as lead, moving away from the source, and limiting time of exposure."}
{"id": 5084, "contents": "2137. Exercises - 2137.1. Nuclear Structure and Stability\n1. Write the following isotopes in hyphenated form (e.g., \"carbon-14\")\n(a) ${ }_{11}^{24} \\mathrm{Na}$\n(b) ${ }_{13}^{29} \\mathrm{Al}$\n(c) ${ }_{36}^{73} \\mathrm{Kr}$\n(d) ${ }_{77}^{194} \\mathrm{Ir}$\n2. Write the following isotopes in nuclide notation (e.g., \" ${ }_{6}^{14} \\mathrm{C}$ \")\n(a) oxygen-14\n(b) copper-70\n(c) tantalum-175\n(d) francium-217\n3. For the following isotopes that have missing information, fill in the missing information to complete the notation\n(a) ${ }_{14}^{34} \\mathrm{X}$\n(b) ${ }_{X}^{36} \\mathrm{P}$\n(c) ${ }_{\\mathrm{X}}{ }^{57} \\mathrm{Mn}$\n(d) ${ }_{56}^{121} \\mathrm{X}$\n4. For each of the isotopes in Exercise 20.1, determine the numbers of protons, neutrons, and electrons in a neutral atom of the isotope.\n5. Write the nuclide notation, including charge if applicable, for atoms with the following characteristics:\n(a) 25 protons, 20 neutrons, 24 electrons\n(b) 45 protons, 24 neutrons, 43 electrons\n(c) 53 protons, 89 neutrons, 54 electrons\n(d) 97 protons, 146 neutrons, 97 electrons\n6. Calculate the density of the ${ }_{12}^{24} \\mathrm{Mg}$ nucleus in $\\mathrm{g} / \\mathrm{mL}$, assuming that it has the typical nuclear diameter of $1 \\times$ $10^{-13} \\mathrm{~cm}$ and is spherical in shape.\n7. What are the two principal differences between nuclear reactions and ordinary chemical changes?\n8. The mass of the atom ${ }_{11}^{23} \\mathrm{Na}$ is 22.9898 amu .\n(a) Calculate its binding energy per atom in millions of electron volts.\n(b) Calculate its binding energy per nucleon."}
{"id": 5085, "contents": "2137. Exercises - 2137.1. Nuclear Structure and Stability\n(a) Calculate its binding energy per atom in millions of electron volts.\n(b) Calculate its binding energy per nucleon.\n9. Which of the following nuclei lie within the band of stability shown in Figure 20.2?\n(a) chlorine-37\n(b) calcium-40\n(c) ${ }^{204} \\mathrm{Bi}$\n(d) ${ }^{56} \\mathrm{Fe}$\n(e) ${ }^{206} \\mathrm{~Pb}$\n(f) ${ }^{211} \\mathrm{~Pb}$\n(g) ${ }^{222} \\mathrm{Rn}$\n(h) carbon-14\n10. Which of the following nuclei lie within the band of stability shown in Figure 20.2?\n(a) argon-40\n(b) oxygen-16\n(c) ${ }^{122} \\mathrm{Ba}$\n(d) ${ }^{58} \\mathrm{Ni}$\n(e) ${ }^{205} \\mathrm{Tl}$\n(f) ${ }^{210} \\mathrm{Tl}$\n(g) ${ }^{226} \\mathrm{Ra}$\n(h) magnesium-24"}
{"id": 5086, "contents": "2137. Exercises - 2137.2. Nuclear Equations\n11. Write a brief description or definition of each of the following:\n(a) nucleon\n(b) $\\alpha$ particle\n(c) $\\beta$ particle\n(d) positron\n(e) $\\gamma$ ray\n(f) nuclide\n(g) mass number\n(h) atomic number\n12. Which of the various particles ( $\\alpha$ particles, $\\beta$ particles, and so on) that may be produced in a nuclear reaction are actually nuclei?\n13. Complete each of the following equations by adding the missing species:\n(a) ${ }_{13}^{27} \\mathrm{Al}+{ }_{2}^{4} \\mathrm{He} \\longrightarrow$ ? $+{ }_{0}^{1} \\mathrm{n}$\n(b) ${ }_{94}^{239} \\mathrm{Pu}+$ ? $\\longrightarrow{ }_{96}^{242} \\mathrm{Cm}+{ }_{0}^{1} \\mathrm{n}$\n(c) ${ }_{7}^{14} \\mathrm{~N}+{ }_{2}^{4} \\mathrm{He} \\longrightarrow$ ? $+{ }_{1}^{1} \\mathrm{H}$\n(d) ${ }_{92}^{235} \\mathrm{U} \\longrightarrow ?+{ }_{55}^{135} \\mathrm{Cs}+4{ }_{0}^{1} \\mathrm{n}$\n14. Complete each of the following equations:\n(a) ${ }_{3}^{7} \\mathrm{Li}+? \\longrightarrow 2{ }_{2}^{4} \\mathrm{He}$\n(b) ${ }_{6}^{14} \\mathrm{C} \\longrightarrow{ }_{7}^{14} \\mathrm{~N}+$ ?\n(c) ${ }_{13}^{27} \\mathrm{Al}+{ }_{2}^{4} \\mathrm{He} \\longrightarrow ?+{ }_{0}^{1} \\mathrm{n}$\n(d) ${ }_{96}^{250} \\mathrm{Cm} \\longrightarrow ?+{ }_{38}^{98} \\mathrm{Sr}+4{ }_{0}^{1} \\mathrm{n}$\n15. Write a balanced equation for each of the following nuclear reactions:"}
{"id": 5087, "contents": "2137. Exercises - 2137.2. Nuclear Equations\n15. Write a balanced equation for each of the following nuclear reactions:\n(a) the production of ${ }^{17} \\mathrm{O}$ from ${ }^{14} \\mathrm{~N}$ by $\\alpha$ particle bombardment\n(b) the production of ${ }^{14} \\mathrm{C}$ from ${ }^{14} \\mathrm{~N}$ by neutron bombardment\n(c) the production of ${ }^{233} \\mathrm{Th}$ from ${ }^{232} \\mathrm{Th}$ by neutron bombardment\n(d) the production of ${ }^{239} \\mathrm{U}$ from ${ }^{238} \\mathrm{U}$ by ${ }_{1}^{2} \\mathrm{H}$ bombardment\n16. Technetium-99 is prepared from ${ }^{98} \\mathrm{Mo}$. Molybdenum- 98 combines with a neutron to give molybdenum-99, an unstable isotope that emits a $\\beta$ particle to yield an excited form of technetium-99, represented as ${ }^{99} \\mathrm{Tc}$ \". This excited nucleus relaxes to the ground state, represented as ${ }^{99} \\mathrm{Tc}$, by emitting a $\\gamma$ ray. The ground state of ${ }^{99} \\mathrm{Tc}$ then emits a $\\beta$ particle. Write the equations for each of these nuclear reactions.\n17. The mass of the atom ${ }_{9}^{19} \\mathrm{~F}$ is 18.99840 amu .\n(a) Calculate its binding energy per atom in millions of electron volts.\n(b) Calculate its binding energy per nucleon.\n18. For the reaction ${ }_{6}^{14} \\mathrm{C} \\longrightarrow{ }_{7}^{14} \\mathrm{~N}+$ ?, if 100.0 g of carbon reacts, what volume of nitrogen gas $\\left(\\mathrm{N}_{2}\\right)$ is produced at 273 K and 1 atm ?"}
{"id": 5088, "contents": "2137. Exercises - 2137.3. Radioactive Decay\n19. What are the types of radiation emitted by the nuclei of radioactive elements?\n20. What changes occur to the atomic number and mass of a nucleus during each of the following decay scenarios?\n(a) an $\\alpha$ particle is emitted\n(b) a $\\beta$ particle is emitted\n(c) $\\gamma$ radiation is emitted\n(d) a positron is emitted\n(e) an electron is captured\n21. What is the change in the nucleus that results from the following decay scenarios?\n(a) emission of a $\\beta$ particle\n(b) emission of a $\\beta^{+}$particle\n(c) capture of an electron\n22. Many nuclides with atomic numbers greater than 83 decay by processes such as electron emission. Explain the observation that the emissions from these unstable nuclides also normally include $\\alpha$ particles.\n23. Why is electron capture accompanied by the emission of an X-ray?\n24. Explain, in terms of Figure 20.2, how unstable heavy nuclides (atomic number $>83$ ) may decompose to form nuclides of greater stability (a) if they are below the band of stability and (b) if they are above the band of stability.\n25. Which of the following nuclei is most likely to decay by positron emission? Explain your choice.\n(a) chromium-53\n(b) manganese-51\n(c) iron-59\n26. The following nuclei do not lie in the band of stability. How would they be expected to decay? Explain your answer.\n(a) ${ }_{15}^{34} \\mathrm{P}$\n(b) ${ }_{92}^{239} \\mathrm{U}$\n(c) ${ }_{20}^{38} \\mathrm{Ca}$\n(d) ${ }_{1}^{3} \\mathrm{H}$\n(e) ${ }_{94}^{245} \\mathrm{Pu}$\n27. The following nuclei do not lie in the band of stability. How would they be expected to decay?\n(a) ${ }_{15}^{28} \\mathrm{P}$\n(b) ${ }_{92}^{235} \\mathrm{U}$\n(c) ${ }_{20}^{37} \\mathrm{Ca}$\n(d) ${ }_{3}^{9} \\mathrm{Li}$"}
{"id": 5089, "contents": "2137. Exercises - 2137.3. Radioactive Decay\n(b) ${ }_{92}^{235} \\mathrm{U}$\n(c) ${ }_{20}^{37} \\mathrm{Ca}$\n(d) ${ }_{3}^{9} \\mathrm{Li}$\n(e) ${ }_{96}^{245} \\mathrm{Cm}$\n28. Predict by what mode(s) of spontaneous radioactive decay each of the following unstable isotopes might proceed:\n(a) ${ }_{2}^{6} \\mathrm{He}$\n(b) ${ }_{30}^{60} \\mathrm{Zn}$\n(c) ${ }_{91}^{235} \\mathrm{~Pa}$\n(d) ${ }_{94}^{241} \\mathrm{~Np}$\n(e) ${ }^{18} \\mathrm{~F}$\n(f) ${ }^{129} \\mathrm{Ba}$\n(g) ${ }^{237} \\mathrm{Pu}$\n29. Write a nuclear reaction for each step in the formation of ${ }_{84}^{218} \\operatorname{Po}$ from ${ }_{98}^{238} \\mathrm{U}$, which proceeds by a series of decay reactions involving the step-wise emission of $\\alpha, \\beta, \\beta, \\alpha, \\alpha, \\alpha$ particles, in that order.\n30. Write a nuclear reaction for each step in the formation of ${ }_{82}^{208} \\mathrm{~Pb}$ from ${ }_{90}^{228} \\mathrm{Th}$, which proceeds by a series of decay reactions involving the step-wise emission of $\\alpha, \\alpha, \\alpha, \\alpha, \\beta, \\beta, \\alpha$ particles, in that order.\n31. Define the term half-life and illustrate it with an example.\n32. A $1.00 \\times 10^{-6}$-g sample of nobelium, ${ }_{102}^{254} \\mathrm{No}$, has a half-life of 55 seconds after it is formed. What is the percentage of ${ }_{102}^{254}$ No remaining at the following times?\n(a) 5.0 min after it forms\n(b) 1.0 h after it forms"}
{"id": 5090, "contents": "2137. Exercises - 2137.3. Radioactive Decay\n(a) 5.0 min after it forms\n(b) 1.0 h after it forms\n33. ${ }^{239} \\mathrm{Pu}$ is a nuclear waste byproduct with a half-life of $24,000 \\mathrm{y}$. What fraction of the ${ }^{239} \\mathrm{Pu}$ present today will be present in 1000 y ?\n34. The isotope ${ }^{208} \\mathrm{Tl}$ undergoes $\\beta$ decay with a half-life of 3.1 min .\n(a) What isotope is produced by the decay?\n(b) How long will it take for $99.0 \\%$ of a sample of pure ${ }^{208} \\mathrm{Tl}$ to decay?\n(c) What percentage of a sample of pure ${ }^{208} \\mathrm{Tl}$ remains un-decayed after 1.0 h ?\n35. If 1.000 g of ${ }_{88}^{226} \\mathrm{Ra}$ produces 0.0001 mL of the gas ${ }_{86}^{222} \\mathrm{Rn}$ at STP (standard temperature and pressure) in 24 h , what is the half-life of ${ }^{226} \\mathrm{Ra}$ in years?\n36. The isotope ${ }_{38}^{90} \\mathrm{Sr}$ is one of the extremely hazardous species in the residues from nuclear power generation. The strontium in a $0.500-\\mathrm{g}$ sample diminishes to 0.393 g in 10.0 y . Calculate the half-life.\n37. Technetium-99 is often used for assessing heart, liver, and lung damage because certain technetium compounds are absorbed by damaged tissues. It has a half-life of 6.0 h . Calculate the rate constant for the decay of ${ }_{43}^{99} \\mathrm{Tc}$.\n38. What is the age of mummified primate skin that contains $8.25 \\%$ of the original quantity of ${ }^{14} \\mathrm{C}$ ?\n39. A sample of rock was found to contain 8.23 mg of rubidium -87 and 0.47 mg of strontium- 87 ."}
{"id": 5091, "contents": "2137. Exercises - 2137.3. Radioactive Decay\n39. A sample of rock was found to contain 8.23 mg of rubidium -87 and 0.47 mg of strontium- 87 .\n(a) Calculate the age of the rock if the half-life of the decay of rubidium by $\\beta$ emission is $4.7 \\times 10^{10} \\mathrm{y}$.\n(b) If some ${ }_{38}^{87} \\mathrm{Sr}$ was initially present in the rock, would the rock be younger, older, or the same age as the age calculated in (a)? Explain your answer.\n40. A laboratory investigation shows that a sample of uranium ore contains 5.37 mg of ${ }_{92}^{238} \\mathrm{U}$ and 2.52 mg of ${ }_{82}^{206} \\mathrm{~Pb}$. Calculate the age of the ore. The half-life of ${ }_{92}^{238} \\mathrm{U}$ is $4.5 \\times 10^{9} \\mathrm{yr}$.\n41. Plutonium was detected in trace amounts in natural uranium deposits by Glenn Seaborg and his associates in 1941. They proposed that the source of this ${ }^{239} \\mathrm{Pu}$ was the capture of neutrons by ${ }^{238} \\mathrm{U}$ nuclei. Why is this plutonium not likely to have been trapped at the time the solar system formed $4.7 \\times 10^{9}$ years ago?\n42. $\\mathrm{A}{ }_{4}^{7} \\mathrm{Be}$ atom (mass $=7.0169 \\mathrm{amu}$ ) decays into a ${ }_{3}^{7} \\mathrm{Li}$ atom (mass $=7.0160 \\mathrm{amu}$ ) by electron capture. How much energy (in millions of electron volts, MeV ) is produced by this reaction?"}
{"id": 5092, "contents": "2137. Exercises - 2137.3. Radioactive Decay\n43. $\\mathrm{A}{ }_{5}^{8} \\mathrm{~B}$ atom (mass $=8.0246 \\mathrm{amu}$ ) decays into a ${ }_{4}^{8} \\mathrm{~B}$ atom (mass $=8.0053 \\mathrm{amu}$ ) by loss of a $\\beta^{+}$particle (mass $=0.00055 \\mathrm{amu}$ ) or by electron capture. How much energy (in millions of electron volts) is produced by this reaction?\n44. Isotopes such as ${ }^{26} \\mathrm{Al}$ (half-life: $7.2 \\times 10^{5}$ years) are believed to have been present in our solar system as it formed, but have since decayed and are now called extinct nuclides.\n(a) ${ }^{26} \\mathrm{Al}$ decays by $\\beta^{+}$emission or electron capture. Write the equations for these two nuclear transformations.\n(b) The earth was formed about $4.7 \\times 10^{9}$ ( 4.7 billion) years ago. How old was the earth when $99.999999 \\%$ of the ${ }^{26} \\mathrm{Al}$ originally present had decayed?\n45. Write a balanced equation for each of the following nuclear reactions:\n(a) bismuth-212 decays into polonium-212\n(b) beryllium-8 and a positron are produced by the decay of an unstable nucleus\n(c) neptunium-239 forms from the reaction of uranium-238 with a neutron and then spontaneously converts into plutonium-239\n(d) strontium-90 decays into yttrium-90\n46. Write a balanced equation for each of the following nuclear reactions:\n(a) mercury-180 decays into platinum-176\n(b) zirconium-90 and an electron are produced by the decay of an unstable nucleus\n(c) thorium-232 decays and produces an alpha particle and a radium-228 nucleus, which decays into actinium- 228 by beta decay\n(d) neon-19 decays into fluorine-19"}
{"id": 5093, "contents": "2137. Exercises - 2137.4. Transmutation and Nuclear Energy\n47. Write the balanced nuclear equation for the production of the following transuranium elements:\n(a) berkelium-244, made by the reaction of Am-241 and $\\mathrm{He}-4$\n(b) fermium-254, made by the reaction of $\\mathrm{Pu}-239$ with a large number of neutrons\n(c) lawrencium-257, made by the reaction of Cf-250 and B-11\n(d) dubnium-260, made by the reaction of Cf-249 and N-15\n48. How does nuclear fission differ from nuclear fusion? Why are both of these processes exothermic?\n49. Both fusion and fission are nuclear reactions. Why is a very high temperature required for fusion, but not for fission?\n50. Cite the conditions necessary for a nuclear chain reaction to take place. Explain how it can be controlled to produce energy, but not produce an explosion.\n51. Describe the components of a nuclear reactor.\n52. In usual practice, both a moderator and control rods are necessary to operate a nuclear chain reaction safely for the purpose of energy production. Cite the function of each and explain why both are necessary.\n53. Describe how the potential energy of uranium is converted into electrical energy in a nuclear power plant.\n54. The mass of a hydrogen atom $\\left({ }_{1}^{1} \\mathrm{H}\\right)$ is 1.007825 amu ; that of a tritium atom $\\left({ }_{1}^{3} \\mathrm{H}\\right)$ is 3.01605 amu ; and that of an $\\alpha$ particle is 4.00150 amu . How much energy in kilojoules per mole of ${ }_{2}^{4} \\mathrm{He}$ produced is released by the following fusion reaction: ${ }_{1}^{1} \\mathrm{H}+{ }_{1}^{3} \\mathrm{H} \\longrightarrow{ }_{2}^{4} \\mathrm{He}$."}
{"id": 5094, "contents": "2137. Exercises - 2137.5. Uses of Radioisotopes\n55. How can a radioactive nuclide be used to show that the equilibrium: $\\mathrm{AgCl}(s) \\rightleftharpoons \\mathrm{Ag}^{+}(a q)+\\mathrm{Cl}^{-}(a q)$ is a dynamic equilibrium?\n56. Technetium-99m has a half-life of 6.01 hours. If a patient injected with technetium- 99 m is safe to leave the hospital once $75 \\%$ of the dose has decayed, when is the patient allowed to leave?\n57. Iodine that enters the body is stored in the thyroid gland from which it is released to control growth and metabolism. The thyroid can be imaged if iodine-131 is injected into the body. In larger doses, I-133 is also used as a means of treating cancer of the thyroid. I- 131 has a half-life of 8.70 days and decays by $\\beta^{-}$ emission.\n(a) Write an equation for the decay.\n(b) How long will it take for $95.0 \\%$ of a dose of I-131 to decay?"}
{"id": 5095, "contents": "2137. Exercises - 2137.6. Biological Effects of Radiation\n58. If a hospital were storing radioisotopes, what is the minimum containment needed to protect against:\n(a) cobalt-60 (a strong $\\gamma$ emitter used for irradiation)\n(b) molybdenum-99 (a beta emitter used to produce technetium-99 for imaging)\n59. Based on what is known about Radon- 222 's primary decay method, why is inhalation so dangerous?\n60. Given specimens uranium-232 ( $t_{1 / 2}=68.9 \\mathrm{y}$ ) and uranium-233 ( $\\left.t_{1 / 2}=159,200 \\mathrm{y}\\right)$ of equal mass, which one would have greater activity and why?\n61. A scientist is studying a 2.234 g sample of thorium-229 ( $\\left.t_{1 / 2}=7340 \\mathrm{y}\\right)$ in a laboratory.\n(a) What is its activity in Bq?\n(b) What is its activity in Ci ?\n62. Given specimens neon-24 ( $t_{1 / 2}=3.38 \\mathrm{~min}$ ) and bismuth-211 ( $\\left.t_{1 / 2}=2.14 \\mathrm{~min}\\right)$ of equal mass, which one would have greater activity and why?\n\n\nFigure 21.1 All organic compounds contain carbon and most are formed by living things, although they are also formed by geological and artificial processes. (credit left: modification of work by Jon Sullivan; credit left middle: modification of work by Deb Tremper; credit right middle: modification of work by \"annszyp\"/Wikimedia Commons; credit right: modification of work by George Shuklin)"}
{"id": 5096, "contents": "2138. CHAPTER OUTLINE - 2138.1. Hydrocarbons\n21.2 Alcohols and Ethers\n21.3 Aldehydes, Ketones, Carboxylic Acids, and Esters\n21.4 Amines and Amides\n\nINTRODUCTION All living things on earth are formed mostly of carbon compounds. The prevalence of carbon compounds in living things has led to the epithet \"carbon-based\" life. The truth is we know of no other kind of life. Early chemists regarded substances isolated from organisms (plants and animals) as a different type of matter that could not be synthesized artificially, and these substances were thus known as organic compounds. The widespread belief called vitalism held that organic compounds were formed by a vital force present only in living organisms. The German chemist Friedrich Wohler was one of the early chemists to refute this aspect of vitalism, when, in 1828, he reported the synthesis of urea, a component of many body fluids, from nonliving materials. Since then, it has been recognized that organic molecules obey the same natural laws as inorganic substances, and the category of organic compounds has evolved to include both natural and synthetic compounds that contain carbon. Some carbon-containing compounds are not classified as organic, for example, carbonates and cyanides, and simple oxides, such as CO and $\\mathrm{CO}_{2}$. Although a single, precise definition has yet to be identified by the chemistry community, most agree that a defining trait of organic molecules is the presence of carbon as the principal element, bonded to hydrogen and other carbon atoms.\n\nToday, organic compounds are key components of plastics, soaps, perfumes, sweeteners, fabrics, pharmaceuticals, and many other substances that we use every day. The value to us of organic compounds ensures that organic chemistry is an important discipline within the general field of chemistry. In this chapter, we discuss why the element carbon gives rise to a vast number and variety of compounds, how those compounds are classified, and the role of organic compounds in representative biological and industrial settings."}
{"id": 5097, "contents": "2139. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Explain the importance of hydrocarbons and the reason for their diversity\n- Name saturated and unsaturated hydrocarbons, and molecules derived from them\n- Describe the reactions characteristic of saturated and unsaturated hydrocarbons\n- Identify structural and geometric isomers of hydrocarbons\n\nThe largest database ${ }^{\\frac{1}{1}}$ of organic compounds lists about 10 million substances, which include compounds originating from living organisms and those synthesized by chemists. The number of potential organic compounds has been estimated ${ }^{2}$ at $10^{60}$-an astronomically high number. The existence of so many organic molecules is a consequence of the ability of carbon atoms to form up to four strong bonds to other carbon atoms, resulting in chains and rings of many different sizes, shapes, and complexities.\n\nThe simplest organic compounds contain only the elements carbon and hydrogen, and are called hydrocarbons. Even though they are composed of only two types of atoms, there is a wide variety of hydrocarbons because they may consist of varying lengths of chains, branched chains, and rings of carbon atoms, or combinations of these structures. In addition, hydrocarbons may differ in the types of carbon-carbon bonds present in their molecules. Many hydrocarbons are found in plants, animals, and their fossils; other hydrocarbons have been prepared in the laboratory. We use hydrocarbons every day, mainly as fuels, such as natural gas, acetylene, propane, butane, and the principal components of gasoline, diesel fuel, and heating oil. The familiar plastics polyethylene, polypropylene, and polystyrene are also hydrocarbons. We can distinguish several types of hydrocarbons by differences in the bonding between carbon atoms. This leads to differences in geometries and in the hybridization of the carbon orbitals."}
{"id": 5098, "contents": "2140. Alkanes - \nAlkanes, or saturated hydrocarbons, contain only single covalent bonds between carbon atoms. Each of the carbon atoms in an alkane has $s p^{3}$ hybrid orbitals and is bonded to four other atoms, each of which is either carbon or hydrogen. The Lewis structures and models of methane, ethane, and pentane are illustrated in Figure 21.2. Carbon chains are usually drawn as straight lines in Lewis structures, but one has to remember that Lewis structures are not intended to indicate the geometry of molecules. Notice that the carbon atoms in the structural models (the ball-and-stick and space-filling models) of the pentane molecule do not lie in a straight line. Because of the $s p^{3}$ hybridization, the bond angles in carbon chains are close to $109.5^{\\circ}$, giving such chains in an alkane a zigzag shape.\n\nThe structures of alkanes and other organic molecules may also be represented in a less detailed manner by condensed structural formulas (or simply, condensed formulas). Instead of the usual format for chemical formulas in which each element symbol appears just once, a condensed formula is written to suggest the bonding in the molecule. These formulas have the appearance of a Lewis structure from which most or all of the bond symbols have been removed. Condensed structural formulas for ethane and pentane are shown at the bottom of Figure 21.2, and several additional examples are provided in the exercises at the end of this chapter.\n\n[^13]\n\n\n\n\n\nmethane\n$\\mathrm{CH}_{4}$\n\nethane\n$\\mathrm{CH}_{3} \\mathrm{CH}_{3}$ or $\\mathrm{C}_{2} \\mathrm{H}_{6}$\n\npentane\n$\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$ or $\\mathrm{C}_{5} \\mathrm{H}_{12}$\n\nFIGURE 21.2 Pictured are the Lewis structures, ball-and-stick models, and space-filling models for molecules of methane, ethane, and pentane."}
{"id": 5099, "contents": "2140. Alkanes - \nFIGURE 21.2 Pictured are the Lewis structures, ball-and-stick models, and space-filling models for molecules of methane, ethane, and pentane.\n\nA common method used by organic chemists to simplify the drawings of larger molecules is to use a skeletal structure (also called a line-angle structure). In this type of structure, carbon atoms are not symbolized with a C, but represented by each end of a line or bend in a line. Hydrogen atoms are not drawn if they are attached to a carbon. Other atoms besides carbon and hydrogen are represented by their elemental symbols. Figure 21.3 shows three different ways to draw the same structure.\n\n\nExpanded formula\n\n\n\nCondensed formula\nSkeletal structure\nFIGURE 21.3 The same structure can be represented three different ways: an expanded formula, a condensed formula, and a skeletal structure."}
{"id": 5100, "contents": "2142. Drawing Skeletal Structures - \nDraw the skeletal structures for these two molecules:\n\n(a)\n\n(b)"}
{"id": 5101, "contents": "2143. Solution - \nEach carbon atom is converted into the end of a line or the place where lines intersect. All hydrogen atoms attached to the carbon atoms are left out of the structure (although we still need to recognize they are there):\n\n(a)\n\n(b)"}
{"id": 5102, "contents": "2144. Check Your Learning - \nDraw the skeletal structures for these two molecules:\n\n(a)\n\n$$\n\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}\n$$\n\n(b)\n\nAnswer:\n\n(a)\n\n(b)"}
{"id": 5103, "contents": "2146. Interpreting Skeletal Structures - \nIdentify the chemical formula of the molecule represented here:"}
{"id": 5104, "contents": "2147. Solution - \nThere are eight places where lines intersect or end, meaning that there are eight carbon atoms in the molecule. Since we know that carbon atoms tend to make four bonds, each carbon atom will have the number of hydrogen atoms that are required for four bonds. This compound contains 16 hydrogen atoms for a molecular formula of $\\mathrm{C}_{8} \\mathrm{H}_{16}$.\n\nLocation of the hydrogen atoms:\n\n\nCheck Your Learning\nIdentify the chemical formula of the molecule represented here:"}
{"id": 5105, "contents": "2148. Answer: - \n$\\mathrm{C}_{9} \\mathrm{H}_{20}$\n\nAll alkanes are composed of carbon and hydrogen atoms, and have similar bonds, structures, and formulas; noncyclic alkanes all have a formula of $\\mathrm{C}_{\\mathrm{n}} \\mathrm{H}_{2 \\mathrm{n}+2}$. The number of carbon atoms present in an alkane has no limit. Greater numbers of atoms in the molecules will lead to stronger intermolecular attractions (dispersion forces) and correspondingly different physical properties of the molecules. Properties such as melting point and boiling point (Table 21.1) usually change smoothly and predictably as the number of carbon and hydrogen atoms in the molecules change."}
{"id": 5106, "contents": "2149. Properties of Some Alkanes ${ }^{\\text {\u00b3 }}$ - \n| Alkane | Molecular <br> Formula | Melting Point <br> $\\left({ }^{\\circ} \\mathrm{C}\\right)$ | Boiling Point <br> $\\left({ }^{\\circ} \\mathrm{C}\\right)$ | Phase at <br> STP | Number of Structural <br> Isomers |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n| methane | $\\mathrm{CH}_{4}$ | -182.5 | -161.5 | gas | 1 |\n| ethane | $\\mathrm{C}_{2} \\mathrm{H}_{6}$ | -183.3 | -88.6 | gas | 1 |\n| propane | $\\mathrm{C}_{3} \\mathrm{H}_{8}$ | -187.7 | -42.1 | gas | 1 |\n| putane | $\\mathrm{C}_{4} \\mathrm{H}_{10}$ | -138.3 | -0.5 | gas | 2 |\n| pentane | $\\mathrm{C}_{5} \\mathrm{H}_{12}$ | -129.7 | 36.1 | liquid | 3 |\n| hepane | $\\mathrm{C}_{6} \\mathrm{H}_{14}$ | -95.3 | 68.7 | liquid | 5 |\n| octane | $\\mathrm{C}_{7} \\mathrm{H}_{16} \\mathrm{H}_{18}$ | -90.6 | 98.4 | liquid | 9 |\n| nonane | $\\mathrm{C}_{9} \\mathrm{H}_{20}$ | -56.8 | 125.7 | liquid | 18 |\n| decane | $\\mathrm{C}_{10} \\mathrm{H}_{22}$ | -53.6 | -29.7 | 150.8 | liquid |\n| tetradecane | $\\mathrm{C}_{14} \\mathrm{H}_{30}$ | 5.9 | liquid | 75 |  |\n| octadecane | $\\mathrm{C}_{18} \\mathrm{H}_{38}$ | 28.2 | solid | 1858 |  |"}
{"id": 5107, "contents": "2149. Properties of Some Alkanes ${ }^{\\text {\u00b3 }}$ - \n[^14]Hydrocarbons with the same formula, including alkanes, can have different structures. For example, two alkanes have the formula $\\mathrm{C}_{4} \\mathrm{H}_{10}$ : They are called $n$-butane and 2-methylpropane (or isobutane), and have the following Lewis structures:\n\n\n$n$-butane\n\n\n\n2-methylpropane\n\nThe compounds $n$-butane and 2-methylpropane are structural isomers (the term constitutional isomers is also commonly used). Constitutional isomers have the same molecular formula but different spatial arrangements of the atoms in their molecules. The $n$-butane molecule contains an unbranched chain, meaning that no carbon atom is bonded to more than two other carbon atoms. We use the term normal, or the prefix $n$, to refer to a chain of carbon atoms without branching. The compound 2 -methylpropane has a branched chain (the carbon atom in the center of the Lewis structure is bonded to three other carbon atoms)\n\nIdentifying isomers from Lewis structures is not as easy as it looks. Lewis structures that look different may actually represent the same isomers. For example, the three structures in Figure 21.4 all represent the same molecule, $n$-butane, and hence are not different isomers. They are identical because each contains an unbranched chain of four carbon atoms.\n\n[^15]\n\n\n\n\n\n\n\nFIGURE 21.4 These three representations of the structure of $n$-butane are not isomers because they all contain the same arrangement of atoms and bonds."}
{"id": 5108, "contents": "2150. The Basics of Organic Nomenclature: Naming Alkanes - \nThe International Union of Pure and Applied Chemistry (IUPAC) has devised a system of nomenclature that begins with the names of the alkanes and can be adjusted from there to account for more complicated structures. The nomenclature for alkanes is based on two rules:\n\n1. To name an alkane, first identify the longest chain of carbon atoms in its structure. A two-carbon chain is called ethane; a three-carbon chain, propane; and a four-carbon chain, butane. Longer chains are named as follows: pentane (five-carbon chain), hexane (6), heptane (7), octane (8), nonane (9), and decane (10). These prefixes can be seen in the names of the alkanes described in Table 21.1.\n2. Add prefixes to the name of the longest chain to indicate the positions and names of substituents. Substituents are branches or functional groups that replace hydrogen atoms on a chain. The position of a substituent or branch is identified by the number of the carbon atom it is bonded to in the chain. We number the carbon atoms in the chain by counting from the end of the chain nearest the substituents. Multiple substituents are named individually and placed in alphabetical order at the front of the name.\n\npropane\n\n\n2-chloropropane\n\n\n2-methylpropane\n\n\n2,4-difluorohexane\n\n\n1-bromo-3-chlorohexane\n\nWhen more than one substituent is present, either on the same carbon atom or on different carbon atoms, the substituents are listed alphabetically. Because the carbon atom numbering begins at the end closest to a substituent, the longest chain of carbon atoms is numbered in such a way as to produce the lowest number for the substituents. The ending -o replaces -ide at the end of the name of an electronegative substituent (in ionic compounds, the negatively charged ion ends with -ide like chloride; in organic compounds, such atoms are treated as substituents and the -o ending is used). The number of substituents of the same type is indicated by the prefixes di-(two), tri- (three), tetra-(four), and so on (for example, difluoro- indicates two fluoride substituents)."}
{"id": 5109, "contents": "2152. Naming Halogen-substituted Alkanes - \nName the molecule whose structure is shown here:"}
{"id": 5110, "contents": "2153. Solution - \nThe four-carbon chain is numbered from the end with the chlorine atom. This puts the substituents on positions 1 and 2 (numbering from the other end would put the substituents on positions 3 and 4). Four carbon atoms means that the base name of this compound will be butane. The bromine at position 2 will be described by adding 2 -bromo-; this will come at the beginning of the name, since bromo- comes before chloroalphabetically. The chlorine at position 1 will be described by adding 1-chloro-, resulting in the name of the molecule being 2-bromo-1-chlorobutane."}
{"id": 5111, "contents": "2154. Check Your Learning - \nName the following molecule:"}
{"id": 5112, "contents": "2155. Answer: - \n3,3-dibromo-2-iodopentane\n\nWe call a substituent that contains one less hydrogen than the corresponding alkane an alkyl group. The name of an alkyl group is obtained by dropping the suffix -ane of the alkane name and adding -yl:\n\nmethane\n\nmethyl group\n\nethane\n\nethyl group\n\nThe open bonds in the methyl and ethyl groups indicate that these alkyl groups are bonded to another atom."}
{"id": 5113, "contents": "2157. Naming Substituted Alkanes - \nName the molecule whose structure is shown here:"}
{"id": 5114, "contents": "2158. Solution - \nThe longest carbon chain runs horizontally across the page and contains six carbon atoms (this makes the base of the name hexane, but we will also need to incorporate the name of the branch). In this case, we want to number from right to left (as shown by the blue numbers) so the branch is connected to carbon 3 (imagine the numbers from left to right-this would put the branch on carbon 4, violating our rules). The branch attached to position 3 of our chain contains two carbon atoms (numbered in red)-so we take our name for two carbons eth- and attach $-y l$ at the end to signify we are describing a branch. Putting all the pieces together, this molecule is 3-ethylhexane."}
{"id": 5115, "contents": "2159. Check Your Learning - \nName the following molecule:"}
{"id": 5116, "contents": "2160. Answer: - \n4-propyloctane\n\nSome hydrocarbons can form more than one type of alkyl group when the hydrogen atoms that would be removed have different \"environments\" in the molecule. This diversity of possible alkyl groups can be identified in the following way: The four hydrogen atoms in a methane molecule are equivalent; they all have the same environment. They are equivalent because each is bonded to a carbon atom (the same carbon atom) that is bonded to three hydrogen atoms. (It may be easier to see the equivalency in the ball and stick models in Figure 21.2. Removal of any one of the four hydrogen atoms from methane forms a methyl group. Likewise, the six hydrogen atoms in ethane are equivalent (Figure 21.2) and removing any one of these hydrogen atoms produces an ethyl group. Each of the six hydrogen atoms is bonded to a carbon atom that is bonded to two other hydrogen atoms and a carbon atom. However, in both propane and $2-$ methylpropane, there are hydrogen atoms in two different environments, distinguished by the adjacent atoms or groups of atoms:\n\npropane\n\n\n2-methylpropane\n\nEach of the six equivalent hydrogen atoms of the first type in propane and each of the nine equivalent hydrogen atoms of that type in 2-methylpropane (all shown in black) are bonded to a carbon atom that is bonded to only one other carbon atom. The two purple hydrogen atoms in propane are of a second type. They differ from the six hydrogen atoms of the first type in that they are bonded to a carbon atom bonded to two other carbon atoms. The green hydrogen atom in 2-methylpropane differs from the other nine hydrogen atoms in that molecule and from the purple hydrogen atoms in propane. The green hydrogen atom in 2 -methylpropane is bonded to a carbon atom bonded to three other carbon atoms. Two different alkyl groups can be formed from each of these molecules, depending on which hydrogen atom is removed. The names and structures of these and several other alkyl groups are listed in Figure 21.5."}
{"id": 5117, "contents": "2160. Answer: - \n| Alkyl Group | Structure |\n| :---: | :---: |\n| methyl | $\\mathrm{CH}_{3}$ - |\n| ethyl | $\\mathrm{CH}_{3} \\mathrm{CH}_{2}$ - |\n| $n$-propyl | $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2}$ - |\n| isopropyl | <smiles>CC(C)I</smiles> |\n| $n$-butyl | $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2}$ - |\n| sec-butyl | <smiles>CCC(C)I</smiles> |\n| isobutyl | <smiles>CCCCC</smiles> |\n| tert-butyl | <smiles>CC(C)(C)C</smiles> |\n\nFIGURE 21.5 This listing gives the names and formulas for various alkyl groups formed by the removal of hydrogen atoms from different locations.\n\nNote that alkyl groups do not exist as stable independent entities. They are always a part of some larger molecule. The location of an alkyl group on a hydrocarbon chain is indicated in the same way as any other substituent:\n\n\nAlkanes are relatively stable molecules, but heat or light will activate reactions that involve the breaking of $\\mathrm{C}-\\mathrm{H}$ or $\\mathrm{C}-\\mathrm{C}$ single bonds. Combustion is one such reaction:\n\n$$\n\\mathrm{CH}_{4}(g)+2 \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{CO}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{~g})\n$$"}
{"id": 5118, "contents": "2160. Answer: - \nAlkanes burn in the presence of oxygen, a highly exothermic oxidation-reduction reaction that produces carbon dioxide and water. As a consequence, alkanes are excellent fuels. For example, methane, $\\mathrm{CH}_{4}$, is the principal component of natural gas. Butane, $\\mathrm{C}_{4} \\mathrm{H}_{10}$, used in camping stoves and lighters is an alkane. Gasoline is a liquid mixture of continuous- and branched-chain alkanes, each containing from five to nine carbon atoms, plus various additives to improve its performance as a fuel. Kerosene, diesel oil, and fuel oil are primarily mixtures of alkanes with higher molecular masses. The main source of these liquid alkane fuels is crude oil, a complex mixture that is separated by fractional distillation. Fractional distillation takes advantage of differences in the boiling points of the components of the mixture (see Figure 21.6). You may recall that boiling point is a function of intermolecular interactions, which was discussed in the chapter on solutions and colloids.\n\n\nSmall molecules:\n\n- Low boiling point\n- Very volatile\n- Flows easily\n- Ignites easily\n\n\nLarge molecules:\n\n- High boiling point\n- Not very volatile\n- Does not flow easily\n- Does not ignite easily\n\nFIGURE 21.6 In a column for the fractional distillation of crude oil, oil heated to about $425^{\\circ} \\mathrm{C}$ in the furnace vaporizes when it enters the base of the tower. The vapors rise through bubble caps in a series of trays in the tower. As the vapors gradually cool, fractions of higher, then of lower, boiling points condense to liquids and are drawn off. (credit left: modification of work by Luigi Chiesa)\n\nIn a substitution reaction, another typical reaction of alkanes, one or more of the alkane's hydrogen atoms is replaced with a different atom or group of atoms. No carbon-carbon bonds are broken in these reactions, and\nthe hybridization of the carbon atoms does not change. For example, the reaction between ethane and molecular chlorine depicted here is a substitution reaction:"}
{"id": 5119, "contents": "2160. Answer: - \nThe $\\mathrm{C}-\\mathrm{Cl}$ portion of the chloroethane molecule is an example of a functional group, the part or moiety of a molecule that imparts a specific chemical reactivity. The types of functional groups present in an organic molecule are major determinants of its chemical properties and are used as a means of classifying organic compounds as detailed in the remaining sections of this chapter."}
{"id": 5120, "contents": "2161. (C) LINK TO LEARNING - \nWant more practice naming alkanes? Watch this brief video tutorial (http://openstax.org/l/16alkanes) to review the nomenclature process."}
{"id": 5121, "contents": "2162. Alkenes - \nOrganic compounds that contain one or more double or triple bonds between carbon atoms are described as unsaturated. You have likely heard of unsaturated fats. These are complex organic molecules with long chains of carbon atoms, which contain at least one double bond between carbon atoms. Unsaturated hydrocarbon molecules that contain one or more double bonds are called alkenes. Carbon atoms linked by a double bond are bound together by two bonds, one $\\sigma$ bond and one $\\pi$ bond. Double and triple bonds give rise to a different geometry around the carbon atom that participates in them, leading to important differences in molecular shape and properties. The differing geometries are responsible for the different properties of unsaturated versus saturated fats.\n\nEthene, $\\mathrm{C}_{2} \\mathrm{H}_{4}$, is the simplest alkene. Each carbon atom in ethene, commonly called ethylene, has a trigonal planar structure. The second member of the series is propene (propylene) (Figure 21.7); the butene isomers follow in the series. Four carbon atoms in the chain of butene allows for the formation of isomers based on the position of the double bond, as well as a new form of isomerism.\n\n\nFIGURE 21.7 Expanded structures, ball-and-stick structures, and space-filling models for the alkenes ethene, propene, and 1-butene are shown.\n\nEthylene (the common industrial name for ethene) is a basic raw material in the production of polyethylene and other important compounds. Over 135 million tons of ethylene were produced worldwide in 2010 for use in the polymer, petrochemical, and plastic industries. Ethylene is produced industrially in a process called cracking, in which the long hydrocarbon chains in a petroleum mixture are broken into smaller molecules."}
{"id": 5122, "contents": "2164. Recycling Plastics - \nPolymers (from Greek words poly meaning \"many\" and mer meaning \"parts\") are large molecules made up of repeating units, referred to as monomers. Polymers can be natural (starch is a polymer of sugar residues and proteins are polymers of amino acids) or synthetic [like polyethylene, polyvinyl chloride (PVC), and polystyrene]. The variety of structures of polymers translates into a broad range of properties and uses that make them integral parts of our everyday lives. Adding functional groups to the structure of a polymer can result in significantly different properties (see the discussion about Kevlar later in this chapter).\n\nAn example of a polymerization reaction is shown in Figure 21.8. The monomer ethylene $\\left(\\mathrm{C}_{2} \\mathrm{H}_{4}\\right)$ is a gas at room temperature, but when polymerized, using a transition metal catalyst, it is transformed into a solid material made up of long chains of $-\\mathrm{CH}_{2}$ - units called polyethylene. Polyethylene is a commodity plastic used primarily for packaging (bags and films).\n\n\nFIGURE 21.8 The reaction for the polymerization of ethylene to polyethylene is shown.\nPolyethylene is a member of one subset of synthetic polymers classified as plastics. Plastics are synthetic organic solids that can be molded; they are typically organic polymers with high molecular masses. Most of the monomers that go into common plastics (ethylene, propylene, vinyl chloride, styrene, and ethylene terephthalate) are derived from petrochemicals and are not very biodegradable, making them candidate materials for recycling. Recycling plastics helps minimize the need for using more of the petrochemical supplies and also minimizes the environmental damage caused by throwing away these nonbiodegradable materials."}
{"id": 5123, "contents": "2164. Recycling Plastics - \nPlastic recycling is the process of recovering waste, scrap, or used plastics, and reprocessing the material into useful products. For example, polyethylene terephthalate (soft drink bottles) can be melted down and used for plastic furniture, in carpets, or for other applications. Other plastics, like polyethylene (bags) and polypropylene (cups, plastic food containers), can be recycled or reprocessed to be used again. Many areas of the country have recycling programs that focus on one or more of the commodity plastics that have been assigned a recycling code (see Figure 21.9). These operations have been in effect since the 1970s and have made the production of some plastics among the most efficient industrial operations today.\n\n| polyethylene terephthalate (PETE) | Soda bottles and oven-ready food trays |\n| :---: | :---: |\n| high-density polyethylene (HDPE) | Bottles for milk and dishwashing liquids |\n| polyvinyl chloride (PVC) <br> V | Food trays, plastic wrap, bottles for mineral water and shampoo |\n| low density polyethylene (LDPE) | Shopping bags and garbage bags |\n| polypropylene (PP) <br> PP | Margarine tubs, microwaveable food trays |\n| polystyrene (PS) | Yogurt tubs, foam meat trays, egg cartons, vending cups, plastic cutlery, packaging for electronics and toys |\n| any other plastics (OTHER) | Plastics that do not fall into any of the above categories One example is melamine resin (plastic plates, plastic cups) |\n\nFIGURE 21.9 Each type of recyclable plastic is imprinted with a code for easy identification.\n\nThe name of an alkene is derived from the name of the alkane with the same number of carbon atoms. The presence of the double bond is signified by replacing the suffix -ane with the suffix -ene. The location of the double bond is identified by naming the smaller of the numbers of the carbon atoms participating in the double bond:\n\nethene (ethylene)\n\npropene (propylene)\n\n\n1-butene\n\n\n2-butene"}
{"id": 5124, "contents": "2165. Isomers of Alkenes - \nMolecules of 1-butene and 2-butene are structural isomers; the arrangement of the atoms in these two molecules differs. As an example of arrangement differences, the first carbon atom in 1-butene is bonded to two hydrogen atoms; the first carbon atom in 2-butene is bonded to three hydrogen atoms.\n\nThe compound 2-butene and some other alkenes also form a second type of isomer called a geometric isomer. In a set of geometric isomers, the same types of atoms are attached to each other in the same order, but the geometries of the two molecules differ. Geometric isomers of alkenes differ in the orientation of the groups on either side of a $\\mathbf{C}=\\mathbf{C}$ bond.\n\nCarbon atoms are free to rotate around a single bond but not around a double bond; a double bond is rigid.\n\nThis makes it possible to have two isomers of 2-butene, one with both methyl groups on the same side of the double bond and one with the methyl groups on opposite sides. When structures of butene are drawn with $120^{\\circ}$ bond angles around the $s p^{2}$-hybridized carbon atoms participating in the double bond, the isomers are apparent. The 2-butene isomer in which the two methyl groups are on the same side is called a cis-isomer; the one in which the two methyl groups are on opposite sides is called a trans-isomer (Figure 21.10). The different geometries produce different physical properties, such as boiling point, that may make separation of the isomers possible:\n\n\n\n\n1-butene\n\n\ncis isomer\n\n\n\ntrans isomer"}
{"id": 5125, "contents": "2165. Isomers of Alkenes - \n1-butene\n\n\ncis isomer\n\n\n\ntrans isomer\n\nFIGURE 21.10 These molecular models show the structural and geometric isomers of butene.\nAlkenes are much more reactive than alkanes because the $\\mathrm{C}=\\mathrm{C}$ moiety is a reactive functional group. $\\mathrm{A} \\pi$ bond, being a weaker bond, is disrupted much more easily than a $\\sigma$ bond. Thus, alkenes undergo a characteristic reaction in which the $\\pi$ bond is broken and replaced by two $\\sigma$ bonds. This reaction is called an addition reaction. The hybridization of the carbon atoms in the double bond in an alkene changes from $s p^{2}$ to $s p^{3}$ during an addition reaction. For example, halogens add to the double bond in an alkene instead of replacing hydrogen, as occurs in an alkane:"}
{"id": 5126, "contents": "2167. Alkene Reactivity and Naming - \nProvide the IUPAC names for the reactant and product of the halogenation reaction shown here:"}
{"id": 5127, "contents": "2168. Solution - \nThe reactant is a five-carbon chain that contains a carbon-carbon double bond, so the base name will be\npentene. We begin counting at the end of the chain closest to the double bond-in this case, from the left-the double bond spans carbons 2 and 3, so the name becomes 2-pentene. Since there are two carbon-containing groups attached to the two carbon atoms in the double bond-and they are on the same side of the double bond-this molecule is the cis-isomer, making the name of the starting alkene cis-2-pentene. The product of the halogenation reaction will have two chlorine atoms attached to the carbon atoms that were a part of the carbon-carbon double bond:\n\n\nThis molecule is now a substituted alkane and will be named as such. The base of the name will be pentane. We will count from the end that numbers the carbon atoms where the chlorine atoms are attached as 2 and 3, making the name of the product 2,3-dichloropentane."}
{"id": 5128, "contents": "2169. Check Your Learning - \nProvide names for the reactant and product of the reaction shown:"}
{"id": 5129, "contents": "2170. Answer: - \nreactant: cis-3-hexene product: 3,4-dichlorohexane"}
{"id": 5130, "contents": "2171. Alkynes - \nHydrocarbon molecules with one or more triple bonds are called alkynes; they make up another series of unsaturated hydrocarbons. Two carbon atoms joined by a triple bond are bound together by one $\\sigma$ bond and two $\\pi$ bonds. The $s p$-hybridized carbons involved in the triple bond have bond angles of $180^{\\circ}$, giving these types of bonds a linear, rod-like shape.\n\nThe simplest member of the alkyne series is ethyne, $\\mathrm{C}_{2} \\mathrm{H}_{2}$, commonly called acetylene. The Lewis structure for ethyne, a linear molecule, is:\n\n```\nH\u2014C\u4e09C\u2014H\nethyne (acetylene)\n```\n\nThe IUPAC nomenclature for alkynes is similar to that for alkenes except that the suffix -yne is used to indicate a triple bond in the chain. For example, $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{C} \\equiv \\mathrm{CH}$ is called 1-butyne."}
{"id": 5131, "contents": "2173. Structure of Alkynes - \nDescribe the geometry and hybridization of the carbon atoms in the following molecule:"}
{"id": 5132, "contents": "2174. Solution - \nCarbon atoms 1 and 4 have four single bonds and are thus tetrahedral with $s p^{3}$ hybridization. Carbon atoms 2 and 3 are involved in the triple bond, so they have linear geometries and would be classified as $s p$ hybrids."}
{"id": 5133, "contents": "2175. Check Your Learning - \nIdentify the hybridization and bond angles at the carbon atoms in the molecule shown:"}
{"id": 5134, "contents": "2176. Answer: - \ncarbon 1: $s p, 180^{\\circ}$; carbon 2: $s p, 180^{\\circ}$; carbon 3: $s p^{2}, 120^{\\circ}$; carbon 4: $s p^{2}, 120^{\\circ}$; carbon 5: $s p^{3}, 109.5^{\\circ}$\n\nChemically, the alkynes are similar to the alkenes. Since the $\\mathrm{C} \\equiv \\mathrm{C}$ functional group has two $\\pi$ bonds, alkynes typically react even more readily, and react with twice as much reagent in addition reactions. The reaction of acetylene with bromine is a typical example:\n\n\nAcetylene and the other alkynes also burn readily. An acetylene torch takes advantage of the high heat of combustion for acetylene."}
{"id": 5135, "contents": "2177. Aromatic Hydrocarbons - \nBenzene, $\\mathrm{C}_{6} \\mathrm{H}_{6}$, is the simplest member of a large family of hydrocarbons, called aromatic hydrocarbons. These compounds contain ring structures and exhibit bonding that must be described using the resonance hybrid concept of valence bond theory or the delocalization concept of molecular orbital theory. (To review these concepts, refer to the earlier chapters on chemical bonding). The resonance structures for benzene, $\\mathrm{C}_{6} \\mathrm{H}_{6}$, are:\n\n\nValence bond theory describes the benzene molecule and other planar aromatic hydrocarbon molecules as hexagonal rings of $s p^{2}$-hybridized carbon atoms with the unhybridized $p$ orbital of each carbon atom perpendicular to the plane of the ring. Three valence electrons in the $s p^{2}$ hybrid orbitals of each carbon atom and the valence electron of each hydrogen atom form the framework of $\\sigma$ bonds in the benzene molecule. The fourth valence electron of each carbon atom is shared with an adjacent carbon atom in their unhybridized $p$ orbitals to yield the $\\pi$ bonds. Benzene does not, however, exhibit the characteristics typical of an alkene. Each of the six bonds between its carbon atoms is equivalent and exhibits properties that are intermediate between those of a $\\mathrm{C}-\\mathrm{C}$ single bond and a $\\mathrm{C}=\\mathrm{C}$ double bond. To represent this unique bonding, structural formulas for benzene and its derivatives are typically drawn with single bonds between the carbon atoms and a circle within the ring as shown in Figure 21.11.\n\n\nFIGURE 21.11 This condensed formula shows the unique bonding structure of benzene.\nThere are many derivatives of benzene. The hydrogen atoms can be replaced by many different substituents. Aromatic compounds more readily undergo substitution reactions than addition reactions; replacement of one\nof the hydrogen atoms with another substituent will leave the delocalized double bonds intact. The following are typical examples of substituted benzene derivatives:\n\ntoluene\n\nxylene\n\nstyrene\n\nToluene and xylene are important solvents and raw materials in the chemical industry. Styrene is used to produce the polymer polystyrene."}
{"id": 5136, "contents": "2179. Structure of Aromatic Hydrocarbons - \nOne possible isomer created by a substitution reaction that replaces a hydrogen atom attached to the aromatic ring of toluene with a chlorine atom is shown here. Draw two other possible isomers in which the chlorine atom replaces a different hydrogen atom attached to the aromatic ring:"}
{"id": 5137, "contents": "2180. Solution - \nSince the six-carbon ring with alternating double bonds is necessary for the molecule to be classified as aromatic, appropriate isomers can be produced only by changing the positions of the chloro-substituent relative to the methyl-substituent:\n\n\nCheck Your Learning\nDraw three isomers of a six-membered aromatic ring compound substituted with two bromines.\n\nAnswer:"}
{"id": 5138, "contents": "2181. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the structure and properties of alcohols\n- Describe the structure and properties of ethers\n- Name and draw structures for alcohols and ethers\n\nIn this section, we will learn about alcohols and ethers."}
{"id": 5139, "contents": "2182. Alcohols - \nIncorporation of an oxygen atom into carbon- and hydrogen-containing molecules leads to new functional groups and new families of compounds. When the oxygen atom is attached by single bonds, the molecule is either an alcohol or ether.\n\nAlcohols are derivatives of hydrocarbons in which an - OH group has replaced a hydrogen atom. Although all alcohols have one or more hydroxyl (-OH) functional groups, they do not behave like bases such as NaOH and $\\mathrm{KOH} . \\mathrm{NaOH}$ and KOH are ionic compounds that contain $\\mathrm{OH}^{-}$ions. Alcohols are covalent molecules; the - OH group in an alcohol molecule is attached to a carbon atom by a covalent bond.\n\nEthanol, $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$, also called ethyl alcohol, is a particularly important alcohol for human use. Ethanol is the alcohol produced by some species of yeast that is found in wine, beer, and distilled drinks. It has long been prepared by humans harnessing the metabolic efforts of yeasts in fermenting various sugars:\n\n\nLarge quantities of ethanol are synthesized from the addition reaction of water with ethylene using an acid as a catalyst:\n\n\nAlcohols containing two or more hydroxyl groups can be made. Examples include 1,2-ethanediol (ethylene glycol, used in antifreeze) and 1,2,3-propanetriol (glycerine, used as a solvent for cosmetics and medicines):\n\n\n1,2-ethanediol\n\n\n1,2,3-propanetriol\n\nNaming Alcohols\nThe name of an alcohol comes from the hydrocarbon from which it was derived. The final -e in the name of the hydrocarbon is replaced by -ol, and the carbon atom to which the -OH group is bonded is indicated by a number placed before the name. ${ }^{-5}$"}
{"id": 5140, "contents": "2184. Naming Alcohols - \nConsider the following example. How should it be named?"}
{"id": 5141, "contents": "2185. Solution - \nThe carbon chain contains five carbon atoms. If the hydroxyl group was not present, we would have named this molecule pentane. To address the fact that the hydroxyl group is present, we change the ending of the name to -ol. In this case, since the - OH is attached to carbon 2 in the chain, we would name this molecule 2-pentanol."}
{"id": 5142, "contents": "2186. Check Your Learning - \nName the following molecule:\n\n\nAnswer:\n2-methyl-2-pentanol"}
{"id": 5143, "contents": "2187. Ethers - \nEthers are compounds that contain the functional group -O-. Ethers do not have a designated suffix like the other types of molecules we have named so far. In the IUPAC system, the oxygen atom and the smaller carbon branch are named as an alkoxy substituent and the remainder of the molecule as the base chain, as in alkanes. As shown in the following compound, the red symbols represent the smaller alkyl group and the oxygen atom, which would be named \"methoxy.\" The larger carbon branch would be ethane, making the molecule methoxyethane. Many ethers are referred to with common names instead of the IUPAC system names. For common names, the two branches connected to the oxygen atom are named separately and followed by \"ether.\" The common name for the compound shown in Example 21.9 is ethylmethyl ether:\n\n[^16]\n\nEXAMPLE 21.9"}
{"id": 5144, "contents": "2188. Naming Ethers - \nProvide the IUPAC and common name for the ether shown here:"}
{"id": 5145, "contents": "2189. Solution - \nIUPAC: The molecule is made up of an ethoxy group attached to an ethane chain, so the IUPAC name would be ethoxyethane.\n\nCommon: The groups attached to the oxygen atom are both ethyl groups, so the common name would be diethyl ether."}
{"id": 5146, "contents": "2190. Check Your Learning - \nProvide the IUPAC and common name for the ether shown:"}
{"id": 5147, "contents": "2191. Answer: - \nIUPAC: 2-methoxypropane; common: isopropylmethyl ether\n\nEthers can be obtained from alcohols by the elimination of a molecule of water from two molecules of the alcohol. For example, when ethanol is treated with a limited amount of sulfuric acid and heated to $140^{\\circ} \\mathrm{C}$, diethyl ether and water are formed:\n\n\nIn the general formula for ethers, $R-\\mathbf{O}-\\mathrm{R}$, the hydrocarbon groups ( R ) may be the same or different. Diethyl ether, the most widely used compound of this class, is a colorless, volatile liquid that is highly flammable. It was first used in 1846 as an anesthetic, but better anesthetics have now largely taken its place. Diethyl ether and other ethers are presently used primarily as solvents for gums, fats, waxes, and resins. Tertiary-butyl methyl ether, $\\mathrm{C}_{4} \\mathrm{H}_{9} \\mathrm{OCH}_{3}$ (abbreviated MTBE-italicized portions of names are not counted when ranking the groups alphabetically-so butyl comes before methyl in the common name), is used as an additive for gasoline. MTBE belongs to a group of chemicals known as oxygenates due to their capacity to increase the oxygen content of gasoline."}
{"id": 5148, "contents": "2192. LINK TO LEARNING - \nWant more practice naming ethers? This brief video review (http://openstax.org/l/16ethers) summarizes the nomenclature for ethers."}
{"id": 5149, "contents": "2194. Carbohydrates and Diabetes - \nCarbohydrates are large biomolecules made up of carbon, hydrogen, and oxygen. The dietary forms of carbohydrates are foods rich in these types of molecules, like pastas, bread, and candy. The name \"carbohydrate\" comes from the formula of the molecules, which can be described by the general formula $\\mathrm{C}_{\\mathrm{m}}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{\\mathrm{n}}$, which shows that they are in a sense \"carbon and water\" or \"hydrates of carbon.\" In many cases, $m$ and $n$ have the same value, but they can be different. The smaller carbohydrates are generally referred to as \"sugars,\" the biochemical term for this group of molecules is \"saccharide\" from the Greek word for sugar (Figure 21.12). Depending on the number of sugar units joined together, they may be classified as monosaccharides (one sugar unit), disaccharides (two sugar units), oligosaccharides (a few sugars), or polysaccharides (the polymeric version of sugars-polymers were described in the feature box earlier in this chapter on recycling plastics). The scientific names of sugars can be recognized by the suffix -ose at the end of the name (for instance, fruit sugar is a monosaccharide called \"fructose\" and milk sugar is a disaccharide called lactose composed of two monosaccharides, glucose and galactose, connected together). Sugars contain some of the functional groups we have discussed: Note the alcohol groups present in the structures and how monosaccharide units are linked to form a disaccharide by formation of an ether.\n\n\nfructos\n\n\nFIGURE 21.12 The illustrations show the molecular structures of fructose, a five-carbon monosaccharide, and of lactose, a disaccharide composed of two isomeric, six-carbon sugars."}
{"id": 5150, "contents": "2194. Carbohydrates and Diabetes - \nFIGURE 21.12 The illustrations show the molecular structures of fructose, a five-carbon monosaccharide, and of lactose, a disaccharide composed of two isomeric, six-carbon sugars.\n\nOrganisms use carbohydrates for a variety of functions. Carbohydrates can store energy, such as the polysaccharides glycogen in animals or starch in plants. They also provide structural support, such as the polysaccharide cellulose in plants and the modified polysaccharide chitin in fungi and animals. The sugars ribose and deoxyribose are components of the backbones of RNA and DNA, respectively. Other sugars play key roles in the function of the immune system, in cell-cell recognition, and in many other biological roles.\n\nDiabetes is a group of metabolic diseases in which a person has a high sugar concentration in their blood (Figure 21.13). Diabetes may be caused by insufficient insulin production by the pancreas or by the body's cells not responding properly to the insulin that is produced. In a healthy person, insulin is produced when it is needed and functions to transport glucose from the blood into the cells where it can be used for energy.\n\nThe long-term complications of diabetes can include loss of eyesight, heart disease, and kidney failure.\nIn 2013, it was estimated that approximately $3.3 \\%$ of the world's population ( $\\sim 380$ million people) suffered from diabetes, resulting in over a million deaths annually. Prevention involves eating a healthy diet, getting plenty of exercise, and maintaining a normal body weight. Treatment involves all of these lifestyle practices and may require injections of insulin."}
{"id": 5151, "contents": "2194. Carbohydrates and Diabetes - \nEven after treatment protocols were introduced, the need to continually monitor their glucose levels posed a challenge for people with diabetes. The first tests required a doctor or lab, and therefore limited access and frequency. Eventually, researchers developed small tablets that would react to the presence of glucose in urine, but these still required a relatively complex process. Chemist Helen Free, who was working on improvements to the tablets, conceived a simpler device: a small test strip. With her husband and research partner, Alfred Free, she produced the first such product for measuring glucose; soon after, she expanded the technology to provide test strips for other compounds and conditions. While very recent advances (such as breath tests, discussed earlier in the text) have shown promise in replacing test strips, they have been widely used for decades and remain a primary method today.\n\n\nFIGURE 21.13 Diabetes is a disease characterized by high concentrations of glucose in the blood. Treating diabetes involves making lifestyle changes, monitoring blood-sugar levels, and sometimes insulin injections. (credit: \"Blausen Medical Communications\"/Wikimedia Commons)"}
{"id": 5152, "contents": "2195. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the structure and properties of aldehydes, ketones, carboxylic acids and esters\n\nAnother class of organic molecules contains a carbon atom connected to an oxygen atom by a double bond, commonly called a carbonyl group. The trigonal planar carbon in the carbonyl group can attach to two other substituents leading to several subfamilies (aldehydes, ketones, carboxylic acids and esters) described in this section."}
{"id": 5153, "contents": "2196. Aldehydes and Ketones - \nBoth aldehydes and ketones contain a carbonyl group, a functional group with a carbon-oxygen double bond. The names for aldehyde and ketone compounds are derived using similar nomenclature rules as for alkanes and alcohols, and include the class-identifying suffixes -al and -one, respectively:\n\n\nIn an aldehyde, the carbonyl group is bonded to at least one hydrogen atom. In a ketone, the carbonyl group is bonded to two carbon atoms:\n\n\nFunctional group of an aldehyde\n\n\n\nAn aldehyde ethanal (acetaldehyde)\n\n\nFunctional group of a ketone\n\n\n\nA ketone\nbutanone\n\nAs text, an aldehyde group is represented as -CHO ; a ketone is represented as $-\\mathrm{C}(\\mathrm{O})-$ or $-\\mathrm{CO}-$.\nIn both aldehydes and ketones, the geometry around the carbon atom in the carbonyl group is trigonal planar; the carbon atom exhibits $s p^{2}$ hybridization. Two of the $s p^{2}$ orbitals on the carbon atom in the carbonyl group are used to form $\\sigma$ bonds to the other carbon or hydrogen atoms in a molecule. The remaining $s p^{2}$ hybrid orbital forms a $\\sigma$ bond to the oxygen atom. The unhybridized $p$ orbital on the carbon atom in the carbonyl group overlaps a $p$ orbital on the oxygen atom to form the $\\pi$ bond in the double bond.\n\nLike the $\\mathrm{C}=\\mathrm{O}$ bond in carbon dioxide, the $\\mathrm{C}=\\mathrm{O}$ bond of a carbonyl group is polar (recall that oxygen is significantly more electronegative than carbon, and the shared electrons are pulled toward the oxygen atom and away from the carbon atom). Many of the reactions of aldehydes and ketones start with the reaction between a Lewis base and the carbon atom at the positive end of the polar $\\mathrm{C}=\\mathrm{O}$ bond to yield an unstable intermediate that subsequently undergoes one or more structural rearrangements to form the final product (Figure 21.14).\n\n\nFIGURE 21.14 The carbonyl group is polar, and the geometry of the bonds around the central carbon is trigonal planar."}
{"id": 5154, "contents": "2196. Aldehydes and Ketones - \nFIGURE 21.14 The carbonyl group is polar, and the geometry of the bonds around the central carbon is trigonal planar.\n\nThe importance of molecular structure in the reactivity of organic compounds is illustrated by the reactions that produce aldehydes and ketones. We can prepare a carbonyl group by oxidation of an alcohol-for organic molecules, oxidation of a carbon atom is said to occur when a carbon-hydrogen bond is replaced by a carbonoxygen bond. The reverse reaction-replacing a carbon-oxygen bond by a carbon-hydrogen bond-is a reduction of that carbon atom. Recall that oxygen is generally assigned a -2 oxidation number unless it is elemental or attached to a fluorine. Hydrogen is generally assigned an oxidation number of +1 unless it is attached to a metal. Since carbon does not have a specific rule, its oxidation number is determined algebraically by factoring the atoms it is attached to and the overall charge of the molecule or ion. In general, a carbon atom attached to an oxygen atom will have a more positive oxidation number and a carbon atom attached to a hydrogen atom will have a more negative oxidation number. This should fit nicely with your understanding of the polarity of $\\mathrm{C}-\\mathrm{O}$ and $\\mathrm{C}-\\mathrm{H}$ bonds. The other reagents and possible products of these reactions are beyond the scope of this chapter, so we will focus only on the changes to the carbon atoms:\n\n\nEXAMPLE 21.10"}
{"id": 5155, "contents": "2197. Oxidation and Reduction in Organic Chemistry - \nMethane represents the completely reduced form of an organic molecule that contains one carbon atom. Sequentially replacing each of the carbon-hydrogen bonds with a carbon-oxygen bond would lead to an alcohol, then an aldehyde, then a carboxylic acid (discussed later), and, finally, carbon dioxide:\n\n$$\n\\mathrm{CH}_{4} \\longrightarrow \\mathrm{CH}_{3} \\mathrm{OH} \\longrightarrow \\mathrm{CH}_{2} \\mathrm{O} \\longrightarrow \\mathrm{HCO}_{2} \\mathrm{H} \\longrightarrow \\mathrm{CO}_{2}\n$$\n\nWhat are the oxidation numbers for the carbon atoms in the molecules shown here?"}
{"id": 5156, "contents": "2198. Solution - \nIn this example, we can calculate the oxidation number (review the chapter on oxidation-reduction reactions if necessary) for the carbon atom in each case (note how this would become difficult for larger molecules with additional carbon atoms and hydrogen atoms, which is why organic chemists use the definition dealing with replacing $\\mathrm{C}-\\mathrm{H}$ bonds with $\\mathrm{C}-\\mathrm{O}$ bonds described). For $\\mathrm{CH}_{4}$, the carbon atom carries a -4 oxidation number (the hydrogen atoms are assigned oxidation numbers of +1 and the carbon atom balances that by having an oxidation number of -4 ). For the alcohol (in this case, methanol), the carbon atom has an oxidation number of -2 (the oxygen atom is assigned -2 , the four hydrogen atoms each are assigned +1 , and the carbon atom balances the sum by having an oxidation number of -2 ; note that compared to the carbon atom in $\\mathrm{CH}_{4}$, this carbon atom has lost two electrons so it was oxidized); for the aldehyde, the carbon atom's oxidation number is 0 ( -2 for the oxygen atom and +1 for each hydrogen atom already balances to 0 , so the oxidation number for the carbon atom is 0 ); for the carboxylic acid, the carbon atom's oxidation number is +2 (two oxygen atoms each at -2 and two hydrogen atoms at +1 ); and for carbon dioxide, the carbon atom's oxidation number is +4 (here, the carbon atom needs to balance the -4 sum from the two oxygen atoms)."}
{"id": 5157, "contents": "2199. Check Your Learning - \nIndicate whether the marked carbon atoms in the three molecules here are oxidized or reduced relative to the marked carbon atom in ethanol:\n\n\nThere is no need to calculate oxidation states in this case; instead, just compare the types of atoms bonded to the marked carbon atoms:\n\n(a)\n\n(b)\n\n(c)"}
{"id": 5158, "contents": "2200. Answer: - \n(a) reduced (bond to oxygen atom replaced by bond to hydrogen atom); (b) oxidized (one bond to hydrogen atom replaced by one bond to oxygen atom); (c) oxidized ( 2 bonds to hydrogen atoms have been replaced by bonds to an oxygen atom)\n\nAldehydes are commonly prepared by the oxidation of alcohols whose - OH functional group is located on the carbon atom at the end of the chain of carbon atoms in the alcohol:\n\n\nAlcohols that have their -OH groups in the middle of the chain are necessary to synthesize a ketone, which requires the carbonyl group to be bonded to two other carbon atoms:\n\n\nAn alcohol with its -OH group bonded to a carbon atom that is bonded to no or one other carbon atom will form an aldehyde. An alcohol with its -OH group attached to two other carbon atoms will form a ketone. If three carbons are attached to the carbon bonded to the -OH , the molecule will not have a $\\mathrm{C}-\\mathrm{H}$ bond to be replaced, so it will not be susceptible to oxidation.\n\nFormaldehyde, an aldehyde with the formula HCHO , is a colorless gas with a pungent and irritating odor. It is sold in an aqueous solution called formalin, which contains about $37 \\%$ formaldehyde by weight. Formaldehyde causes coagulation of proteins, so it kills bacteria (and any other living organism) and stops many of the biological processes that cause tissue to decay. Thus, formaldehyde is used for preserving tissue specimens and embalming bodies. It is also used to sterilize soil or other materials. Formaldehyde is used in the manufacture of Bakelite, a hard plastic having high chemical and electrical resistance.\n\nDimethyl ketone, $\\mathrm{CH}_{3} \\mathrm{COCH}_{3}$, commonly called acetone, is the simplest ketone. It is made commercially by fermenting corn or molasses, or by oxidation of 2-propanol. Acetone is a colorless liquid. Among its many uses are as a solvent for lacquer (including fingernail polish), cellulose acetate, cellulose nitrate, acetylene, plastics, and varnishes; as a paint and varnish remover; and as a solvent in the manufacture of pharmaceuticals and chemicals."}
{"id": 5159, "contents": "2201. Carboxylic Acids and Esters - \nThe odor of vinegar is caused by the presence of acetic acid, a carboxylic acid, in the vinegar. The odor of ripe bananas and many other fruits is due to the presence of esters, compounds that can be prepared by the reaction of a carboxylic acid with an alcohol. Because esters do not have hydrogen bonds between molecules, they have lower vapor pressures than the alcohols and carboxylic acids from which they are derived (see Figure 21.15).\n\n\n\n\n\n\n\n\n\nWintergreen methyl salicylate\n\n\nHoney methyl phenylacetate\n\n\nStrawberry ethyl methylphenylglycidate\n\nFIGURE 21.15 Esters are responsible for the odors associated with various plants and their fruits.\nBoth carboxylic acids and esters contain a carbonyl group with a second oxygen atom bonded to the carbon atom in the carbonyl group by a single bond. In a carboxylic acid, the second oxygen atom also bonds to a hydrogen atom. In an ester, the second oxygen atom bonds to another carbon atom. The names for carboxylic acids and esters include prefixes that denote the lengths of the carbon chains in the molecules and are derived following nomenclature rules similar to those for inorganic acids and salts (see these examples):\n\nethanoic acid\n(acetic acid)\n\nmethyl ethanoate\n(methyl acetate)\n\nThe functional groups for an acid and for an ester are shown in red in these formulas.\nThe hydrogen atom in the functional group of a carboxylic acid will react with a base to form an ionic salt:\n\n\nCarboxylic acids are weak acids (see the chapter on acids and bases), meaning they are not 100\\% ionized in water. Generally only about $1 \\%$ of the molecules of a carboxylic acid dissolved in water are ionized at any given time. The remaining molecules are undissociated in solution.\n\nWe prepare carboxylic acids by the oxidation of aldehydes or alcohols whose - OH functional group is located on the carbon atom at the end of the chain of carbon atoms in the alcohol:"}
{"id": 5160, "contents": "2201. Carboxylic Acids and Esters - \nWe prepare carboxylic acids by the oxidation of aldehydes or alcohols whose - OH functional group is located on the carbon atom at the end of the chain of carbon atoms in the alcohol:\n\n\nEsters are produced by the reaction of acids with alcohols. For example, the ester ethyl acetate, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$, is formed when acetic acid reacts with ethanol:\n\n\nThe simplest carboxylic acid is formic acid, $\\mathrm{HCO}_{2} \\mathrm{H}$, known since 1670. Its name comes from the Latin word formicus, which means \"ant\"; it was first isolated by the distillation of red ants. It is partially responsible for the pain and irritation of ant and wasp stings, and is responsible for a characteristic odor of ants that can be sometimes detected in their nests.\n\nAcetic acid, $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$, constitutes $3-6 \\%$ vinegar. Cider vinegar is produced by allowing apple juice to ferment without oxygen present. Yeast cells present in the juice carry out the fermentation reactions. The fermentation reactions change the sugar present in the juice to ethanol, then to acetic acid. Pure acetic acid has a penetrating odor and produces painful burns. It is an excellent solvent for many organic and some inorganic compounds, and it is essential in the production of cellulose acetate, a component of many synthetic fibers such as rayon."}
{"id": 5161, "contents": "2201. Carboxylic Acids and Esters - \nThe distinctive and attractive odors and flavors of many flowers, perfumes, and ripe fruits are due to the presence of one or more esters (Figure 21.16). Among the most important of the natural esters are fats (such as lard, tallow, and butter) and oils (such as linseed, cottonseed, and olive oils), which are esters of the trihydroxyl alcohol glycerine, $\\mathrm{C}_{3} \\mathrm{H}_{5}(\\mathrm{OH})_{3}$, with large carboxylic acids, such as palmitic acid, $\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{14} \\mathrm{CO}_{2} \\mathrm{H}$, stearic acid, $\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{16} \\mathrm{CO}_{2} \\mathrm{H}$, and oleic acid, $\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{7} \\mathrm{CH}=\\mathrm{CH}\\left(\\mathrm{CH}_{2}\\right)_{7} \\mathrm{CO}_{2} \\mathrm{H}$. Oleic acid is an unsaturated acid; it contains a $\\mathrm{C}=\\mathrm{C}$ double bond. Palmitic and stearic acids are saturated acids that contain no double or triple bonds.\n\n\nFIGURE 21.16 Over 350 different volatile molecules (many members of the ester family) have been identified in strawberries. (credit: Rebecca Siegel)"}
{"id": 5162, "contents": "2202. LEARNING OBJECTIVES - \nBy the end of this section, you will be able to:\n\n- Describe the structure and properties of an amine\n- Describe the structure and properties of an amide\n\nAmines are molecules that contain carbon-nitrogen bonds. The nitrogen atom in an amine has a lone pair of electrons and three bonds to other atoms, either carbon or hydrogen. Various nomenclatures are used to derive names for amines, but all involve the class-identifying suffix -ine as illustrated here for a few simple examples:\n\n\nIn some amines, the nitrogen atom replaces a carbon atom in an aromatic hydrocarbon. Pyridine (Figure 21.17) is one such heterocyclic amine. A heterocyclic compound contains atoms of two or more different elements in its ring structure.\n\n\nFIGURE 21.17 The illustration shows one of the resonance structures of pyridine."}
{"id": 5163, "contents": "2204. DNA in Forensics and Paternity - \nThe genetic material for all living things is a polymer of four different molecules, which are themselves a combination of three subunits. The genetic information, the code for developing an organism, is contained in the specific sequence of the four molecules, similar to the way the letters of the alphabet can be sequenced to form words that convey information. The information in a DNA sequence is used to form two other types of polymers, one of which are proteins. The proteins interact to form a specific type of organism with individual characteristics.\n\nA genetic molecule is called DNA, which stands for deoxyribonucleic acid. The four molecules that make up DNA are called nucleotides. Each nucleotide consists of a single- or double-ringed molecule containing nitrogen, carbon, oxygen, and hydrogen called a nitrogenous base. Each base is bonded to a five-carbon sugar called deoxyribose. The sugar is in turn bonded to a phosphate group ( $-\\mathrm{PO}_{4}{ }^{3-}$ ) When new DNA is made, a polymerization reaction occurs that binds the phosphate group of one nucleotide to the sugar group of a second nucleotide. The nitrogenous bases of each nucleotide stick out from this sugar-phosphate backbone. DNA is actually formed from two such polymers coiled around each other and held together by hydrogen bonds between the nitrogenous bases. Thus, the two backbones are on the outside of the coiled pair of strands, and the bases are on the inside. The shape of the two strands wound around each other is called a double helix (see Figure 21.18).\n\nIt probably makes sense that the sequence of nucleotides in the DNA of a cat differs from those of a dog. But it is also true that the sequences of the DNA in the cells of two individual pugs differ. Likewise, the sequences of DNA in you and a sibling differ (unless your sibling is an identical twin), as do those between you and an unrelated individual. However, the DNA sequences of two related individuals are more similar than the sequences of two unrelated individuals, and these similarities in sequence can be observed in various ways.\n\nThis is the principle behind DNA fingerprinting, which is a method used to determine whether two DNA samples came from related (or the same) individuals or unrelated individuals.\n\n(a)\n\n(b)\n\n(c)"}
{"id": 5164, "contents": "2204. DNA in Forensics and Paternity - \nThis is the principle behind DNA fingerprinting, which is a method used to determine whether two DNA samples came from related (or the same) individuals or unrelated individuals.\n\n(a)\n\n(b)\n\n(c)\n\nFIGURE 21.18 DNA is an organic molecule and the genetic material for all living organisms. (a) DNA is a double helix consisting of two single DNA strands hydrogen bonded together at each nitrogenous base. (b) This detail shows the hydrogen bonding (dotted lines) between nitrogenous bases on each DNA strand and the way in which each nucleotide is joined to the next, forming a backbone of sugars and phosphate groups along each strand. (c) This detail shows the structure of one of the four nucleotides that makes up the DNA polymer. Each nucleotide consists of a nitrogenous base (a double-ring molecule, in this case), a five-carbon sugar (deoxyribose), and a phosphate group.\n\nUsing similarities in sequences, technicians can determine whether a man is the father of a child (the identity of the mother is rarely in doubt, except in the case of an adopted child and a potential birth mother). Likewise, forensic geneticists can determine whether a crime scene sample of human tissue, such as blood or skin cells, contains DNA that matches exactly the DNA of a suspect."}
{"id": 5165, "contents": "2205. LINK TO LEARNING - \nWatch this video animation (http://openstax.org/l/16dnapackaging) of how DNA is packaged for a visual lesson\nin its structure.\n\nLike ammonia, amines are weak bases due to the lone pair of electrons on their nitrogen atoms:\n\n\nThe basicity of an amine's nitrogen atom plays an important role in much of the compound's chemistry. Amine functional groups are found in a wide variety of compounds, including natural and synthetic dyes, polymers, vitamins, and medications such as penicillin and codeine. They are also found in many molecules essential to life, such as amino acids, hormones, neurotransmitters, and DNA."}
{"id": 5166, "contents": "2207. Addictive Alkaloids - \nSince ancient times, plants have been used for medicinal purposes. One class of substances, called alkaloids, found in many of these plants has been isolated and found to contain cyclic molecules with an amine functional group. These amines are bases. They can react with $\\mathrm{H}_{3} \\mathrm{O}^{+}$in a dilute acid to form an ammonium salt, and this property is used to extract them from the plant:\n\n$$\n\\mathrm{R}_{3} \\mathrm{~N}+\\mathrm{H}_{3} \\mathrm{O}^{+}+\\mathrm{Cl}^{-} \\longrightarrow\\left[\\mathrm{R}_{3} \\mathrm{NH}^{+}\\right] \\mathrm{Cl}^{-}+\\mathrm{H}_{2} \\mathrm{O}\n$$\n\nThe name alkaloid means \"like an alkali.\" Thus, an alkaloid reacts with acid. The free compound can be recovered after extraction by reaction with a base:\n\n$$\n\\left[\\mathrm{R}_{3} \\mathrm{NH}^{+}\\right] \\mathrm{Cl}^{-}+\\mathrm{OH}^{-} \\longrightarrow \\mathrm{R}_{3} \\mathrm{~N}+\\mathrm{H}_{2} \\mathrm{O}+\\mathrm{Cl}^{-}\n$$\n\nThe structures of many naturally occurring alkaloids have profound physiological and psychotropic effects in humans. Examples of these drugs include nicotine, morphine, codeine, and heroin. The plant produces these substances, collectively called secondary plant compounds, as chemical defenses against the numerous pests that attempt to feed on the plant:\n\nnicotine\n\ncodeine\n\nmorphine\n\nheroin"}
{"id": 5167, "contents": "2207. Addictive Alkaloids - \nnicotine\n\ncodeine\n\nmorphine\n\nheroin\n\nIn these diagrams, as is common in representing structures of large organic compounds, carbon atoms in the rings and the hydrogen atoms bonded to them have been omitted for clarity. The solid wedges indicate bonds that extend out of the page. The dashed wedges indicate bonds that extend into the page. Notice that small changes to a part of the molecule change the properties of morphine, codeine, and heroin. Morphine, a strong narcotic used to relieve pain, contains two hydroxyl functional groups, located at the bottom of the molecule in this structural formula. Changing one of these hydroxyl groups to a methyl ether group forms codeine, a less potent drug used as a local anesthetic. If both hydroxyl groups are converted to esters of acetic acid, the powerfully addictive drug heroin results (Figure 21.19).\n\n\nFIGURE 21.19 Poppies can be used in the production of opium, a plant latex that contains morphine from which other opiates, such as heroin, can be synthesized. (credit: Karen Roe)\n\nAmides are molecules that contain nitrogen atoms connected to the carbon atom of a carbonyl group. Like amines, various nomenclature rules may be used to name amides, but all include use of the class-specific\nsuffix -amide:\n\n\nacetamide\n\n\nAmides can be produced when carboxylic acids react with amines or ammonia in a process called amidation. A water molecule is eliminated from the reaction, and the amide is formed from the remaining pieces of the carboxylic acid and the amine (note the similarity to formation of an ester from a carboxylic acid and an alcohol discussed in the previous section):\n\n\nThe reaction between amines and carboxylic acids to form amides is biologically important. It is through this reaction that amino acids (molecules containing both amine and carboxylic acid substituents) link together in a polymer to form proteins."}
{"id": 5168, "contents": "2209. Proteins and Enzymes - \nProteins are large biological molecules made up of long chains of smaller molecules called amino acids. Organisms rely on proteins for a variety of functions-proteins transport molecules across cell membranes, replicate DNA, and catalyze metabolic reactions, to name only a few of their functions. The properties of proteins are functions of the combination of amino acids that compose them and can vary greatly. Interactions between amino acid sequences in the chains of proteins result in the folding of the chain into specific, threedimensional structures that determine the protein's activity.\n\nAmino acids are organic molecules that contain an amine functional group ( $-\\mathrm{NH}_{2}$ ), a carboxylic acid functional group ( -COOH ), and a side chain (that is specific to each individual amino acid). Most living things build proteins from the same 20 different amino acids. Amino acids connect by the formation of a peptide bond, which is a covalent bond formed between two amino acids when the carboxylic acid group of one amino acid reacts with the amine group of the other amino acid. The formation of the bond results in the production of a molecule of water (in general, reactions that result in the production of water when two other molecules combine are referred to as condensation reactions). The resulting bond-between the carbonyl group carbon atom and the amine nitrogen atom is called a peptide link or peptide bond. Since each of the original amino acids has an unreacted group (one has an unreacted amine and the other an unreacted carboxylic acid), more peptide bonds can form to other amino acids, extending the structure. (Figure 21.20) A chain of connected amino acids is called a polypeptide. Proteins contain at least one long polypeptide chain.\n\n\nFIGURE 21.20 This condensation reaction forms a dipeptide from two amino acids and leads to the formation of water."}
{"id": 5169, "contents": "2209. Proteins and Enzymes - \nFIGURE 21.20 This condensation reaction forms a dipeptide from two amino acids and leads to the formation of water.\n\nEnzymes are large biological molecules, mostly composed of proteins, which are responsible for the thousands of metabolic processes that occur in living organisms. Enzymes are highly specific catalysts; they speed up the rates of certain reactions. Enzymes function by lowering the activation energy of the reaction they are catalyzing, which can dramatically increase the rate of the reaction. Most reactions catalyzed by enzymes have rates that are millions of times faster than the noncatalyzed version. Like all catalysts, enzymes are not consumed during the reactions that they catalyze. Enzymes do differ from other catalysts in how specific they are for their substrates (the molecules that an enzyme will convert into a different product). Each enzyme is only capable of speeding up one or a few very specific reactions or types of reactions. Since the function of enzymes is so specific, the lack or malfunctioning of an enzyme can lead to serious health consequences. One disease that is the result of an enzyme malfunction is phenylketonuria. In this disease, the enzyme that catalyzes the first step in the degradation of the amino acid phenylalanine is not functional (Figure 21.21). Untreated, this can lead to an accumulation of phenylalanine, which can lead to intellectual disabilities.\n\n\nFIGURE 21.21 A computer rendering shows the three-dimensional structure of the enzyme phenylalanine hydroxylase. In the disease phenylketonuria, a defect in the shape of phenylalanine hydroxylase causes it to lose its function in breaking down phenylalanine."}
{"id": 5170, "contents": "2211. Kevlar - \nKevlar (Figure 21.22) is a synthetic polymer made from two monomers 1,4-phenylene-diamine and terephthaloyl chloride (Kevlar is a registered trademark of DuPont). The material was developed by Susan Kwolek while she worked to find a replacement for steel in tires. Kwolek's work involved synthesizing polyamides and dissolving them in solvents, then spinning the resulting solution into fibers. One of her solutions proved to be quite different in initial appearance and structure. And once spun, the resulting fibers were particularly strong. From this initial discovery, Kevlar was created. The material has a high tensile strength-to-weight ratio (it is about 5 times stronger than an equal weight of steel), making it useful for many applications from bicycle tires to sails to body armor.\n\n\nFIGURE 21.22 This illustration shows the formula for polymeric Kevlar.\nThe material owes much of its strength to hydrogen bonds between polymer chains (refer back to the chapter on intermolecular interactions). These bonds form between the carbonyl group oxygen atom (which has a partial negative charge due to oxygen's electronegativity) on one monomer and the partially positively charged hydrogen atom in the $\\mathrm{N}-\\mathrm{H}$ bond of an adjacent monomer in the polymer structure (see dashed line in Figure 21.23). There is additional strength derived from the interaction between the unhybridized $p$ orbitals in the six-membered rings, called aromatic stacking.\n\n\nFIGURE 21.23 The diagram shows the polymer structure of Kevlar, with hydrogen bonds between polymer chains represented by dotted lines."}
{"id": 5171, "contents": "2211. Kevlar - \nFIGURE 21.23 The diagram shows the polymer structure of Kevlar, with hydrogen bonds between polymer chains represented by dotted lines.\n\nKevlar may be best known as a component of body armor, combat helmets, and face masks. Since the 1980s, the US military has used Kevlar as a component of the PASGT (personal armor system for ground troops) helmet and vest. Kevlar is also used to protect armored fighting vehicles and aircraft carriers. Civilian applications include protective gear for emergency service personnel such as body armor for police officers and heat-resistant clothing for fire fighters. Kevlar based clothing is considerably lighter and thinner than equivalent gear made from other materials (Figure 21.24). Beyond Kevlar, Susan Kwolek was instrumental in the development of Nomex, a fireproof material, and was also involved in the creation of Lycra. She became just the fourth woman inducted into the National Inventors Hall of Fame, and received a number of other awards for her significant contributions to science and society.\n\n\nFIGURE 21.24 (a) These soldiers are sorting through pieces of a Kevlar helmet that helped absorb a grenade blast. Kevlar is also used to make (b) canoes and (c) marine mooring lines. (credit a: modification of work by \"Cla68\"/Wikimedia Commons; credit b: modification of work by \"OakleyOriginals\"/Flickr; credit c: modification of work by Casey H. Kyhl)\n\nIn addition to its better-known uses, Kevlar is also often used in cryogenics for its very low thermal conductivity (along with its high strength). Kevlar maintains its high strength when cooled to the temperature of liquid nitrogen $\\left(-196^{\\circ} \\mathrm{C}\\right)$.\n\nThe table here summarizes the structures discussed in this chapter:"}
{"id": 5172, "contents": "2211. Kevlar - \nThe table here summarizes the structures discussed in this chapter:\n\n| Compound Name | Structure of Compound and Functional Group (red) | Example |  |\n| :---: | :---: | :---: | :---: |\n|  |  | Formula | Name |\n| alkene | $\\mathrm{C}=\\mathrm{C}$ | $\\mathrm{C}_{2} \\mathrm{H}_{4}$ | ethene |\n| alkyne | $\\mathrm{C} \\equiv \\mathrm{C}$ | $\\mathrm{C}_{2} \\mathrm{H}_{2}$ E | ethyne |\n| alcohol | $R-\\ddot{0}-\\mathrm{H}$ | $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$ | ethanol |\n| ether | $R-\\ddot{O}-R^{\\prime}$ | $\\left(\\mathrm{C}_{2} \\mathrm{H}_{5}\\right)_{2} \\mathrm{O}$ | diethyl ether |\n| aldehyde | <smiles>2H=[OH+]</smiles> | $\\mathrm{CH}_{3} \\mathrm{CHO}$ | ethanal |\n| ketone | <smiles>[R]C([R])=[Os]</smiles> | $\\mathrm{CH}_{3} \\mathrm{COCH}_{2} \\mathrm{CH}_{3}$ | methyl ethyl ketone |\n| carboxylic <br> acid | <smiles></smiles> | $\\mathrm{CH}_{3} \\mathrm{COOH}$ | acetic acid |\n| ester | <smiles>[R]OC-[O-]</smiles> | $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$ | ethyl acetate |\n| amine | <smiles>[R]NN[R]</smiles> | $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{NH}_{2}$ | ethylamine |\n| amide | <smiles></smiles> | $\\mathrm{CH}_{3} \\mathrm{CONH}_{2}$ | acetamide |"}
{"id": 5173, "contents": "2212. Key Terms - \naddition reaction reaction in which a double carbon-carbon bond forms a single carboncarbon bond by the addition of a reactant. Typical reaction for an alkene.\nalcohol organic compound with a hydroxyl group $(-\\mathrm{OH})$ bonded to a carbon atom\naldehyde organic compound containing a carbonyl group bonded to two hydrogen atoms or a hydrogen atom and a carbon substituent\nalkane molecule consisting of only carbon and hydrogen atoms connected by single ( $\\sigma$ ) bonds\nalkene molecule consisting of carbon and hydrogen containing at least one carbon-carbon double bond\nalkyl group substituent, consisting of an alkane missing one hydrogen atom, attached to a larger structure\nalkyne molecule consisting of carbon and hydrogen containing at least one carbon-carbon triple bond\namide organic molecule that features a nitrogen atom connected to the carbon atom in a carbonyl group\namine organic molecule in which a nitrogen atom is bonded to one or more alkyl group\naromatic hydrocarbon cyclic molecule consisting of carbon and hydrogen with delocalized alternating carbon-carbon single and double bonds, resulting in enhanced stability\ncarbonyl group carbon atom double bonded to an"}
{"id": 5174, "contents": "2213. Summary - 2213.1. Hydrocarbons\nStrong, stable bonds between carbon atoms produce complex molecules containing chains, branches, and rings. The chemistry of these compounds is called organic chemistry. Hydrocarbons are organic compounds composed of only carbon and hydrogen. The alkanes are saturated hydrocarbons-that is, hydrocarbons that contain only single bonds. Alkenes contain one or more carbon-carbon double bonds. Alkynes contain one or more carbon-carbon triple bonds. Aromatic hydrocarbons contain ring structures with delocalized $\\pi$ electron systems."}
{"id": 5175, "contents": "2213. Summary - 2213.2. Alcohols and Ethers\nMany organic compounds that are not hydrocarbons can be thought of as derivatives of hydrocarbons. A hydrocarbon derivative can be formed by replacing one or more hydrogen atoms of a hydrocarbon by a\noxygen atom\ncarboxylic acid organic compound containing a carbonyl group with an attached hydroxyl group\nester organic compound containing a carbonyl group with an attached oxygen atom that is bonded to a carbon substituent\nether organic compound with an oxygen atom that is bonded to two carbon atoms\nfunctional group part of an organic molecule that imparts a specific chemical reactivity to the molecule\nketone organic compound containing a carbonyl group with two carbon substituents attached to it\norganic compound natural or synthetic compound that contains carbon\nsaturated hydrocarbon molecule containing carbon and hydrogen that has only single bonds between carbon atoms\nskeletal structure shorthand method of drawing organic molecules in which carbon atoms are represented by the ends of lines and bends in between lines, and hydrogen atoms attached to the carbon atoms are not shown (but are understood to be present by the context of the structure)\nsubstituent branch or functional group that replaces hydrogen atoms in a larger hydrocarbon chain\nsubstitution reaction reaction in which one atom replaces another in a molecule\nfunctional group, which contains at least one atom of an element other than carbon or hydrogen. The properties of hydrocarbon derivatives are determined largely by the functional group. The - OH group is the functional group of an alcohol. The $-\\mathrm{R}-\\mathrm{O}-\\mathrm{R}-$ group is the functional group of an ether."}
{"id": 5176, "contents": "2213. Summary - 2213.3. Aldehydes, Ketones, Carboxylic Acids, and Esters\nFunctional groups related to the carbonyl group include the - CHO group of an aldehyde, the - $\\mathrm{CO}-$ group of a ketone, the $-\\mathrm{CO}_{2} \\mathrm{H}$ group of a carboxylic acid, and the $-\\mathrm{CO}_{2} \\mathrm{R}$ group of an ester. The carbonyl group, a carbon-oxygen double bond, is the key structure in these classes of organic molecules: Aldehydes contain at least one hydrogen atom attached to the carbonyl carbon atom, ketones contain two carbon groups attached to the carbonyl carbon atom, carboxylic acids contain a hydroxyl\ngroup attached to the carbonyl carbon atom, and esters contain an oxygen atom attached to another carbon group connected to the carbonyl carbon atom. All of these compounds contain oxidized carbon atoms relative to the carbon atom of an alcohol group."}
{"id": 5177, "contents": "2213. Summary - 2213.4. Amines and Amides\nThe addition of nitrogen into an organic framework\nleads to two families of molecules. Compounds containing a nitrogen atom bonded in a hydrocarbon framework are classified as amines. Compounds that have a nitrogen atom bonded to one side of a carbonyl group are classified as amides. Amines are a basic functional group. Amines and carboxylic acids can combine in a condensation reaction to form amides."}
{"id": 5178, "contents": "2214. Exercises - 2214.1. Hydrocarbons\n1. Write the chemical formula and Lewis structure of the following, each of which contains five carbon atoms:\n(a) an alkane\n(b) an alkene\n(c) an alkyne\n2. What is the difference between the hybridization of carbon atoms' valence orbitals in saturated and unsaturated hydrocarbons?\n3. On a microscopic level, how does the reaction of bromine with a saturated hydrocarbon differ from its reaction with an unsaturated hydrocarbon? How are they similar?\n4. On a microscopic level, how does the reaction of bromine with an alkene differ from its reaction with an alkyne? How are they similar?\n5. Explain why unbranched alkenes can form geometric isomers while unbranched alkanes cannot. Does this explanation involve the macroscopic domain or the microscopic domain?\n6. Explain why these two molecules are not isomers:\n\n\n7. Explain why these two molecules are not isomers:\n\n\n8. How does the carbon-atom hybridization change when polyethylene is prepared from ethylene?\n9. Write the Lewis structure and molecular formula for each of the following hydrocarbons:\n(a) hexane\n(b) 3-methylpentane\n(c) cis-3-hexene\n(d) 4-methyl-1-pentene\n(e) 3-hexyne\n(f) 4-methyl-2-pentyne\n10. Write the chemical formula, condensed formula, and Lewis structure for each of the following hydrocarbons:\n(a) heptane\n(b) 3-methylhexane\n(c) trans-3-heptene\n(d) 4-methyl-1-hexene\n(e) 2-heptyne\n(f) 3,4-dimethyl-1-pentyne\n11. Give the complete IUPAC name for each of the following compounds:\n(a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CBr}_{2} \\mathrm{CH}_{3}$\n(b) $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CCl}$\n(c)\n\n(d)\ne)\n\n(f)"}
{"id": 5179, "contents": "2214. Exercises - 2214.1. Hydrocarbons\n(d)\ne)\n\n(f)\n\n(g) $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CHCH}_{2} \\mathrm{CH}=\\mathrm{CH}_{2}$\n12. Give the complete IUPAC name for each of the following compounds:\n(a) $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CHF}$\n(b) $\\mathrm{CH}_{3} \\mathrm{CHClCHClCH}_{3}$\n(c)\n\n(d) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}=\\mathrm{CHCH}_{3}$\n(e)\n\n(f) $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CCH}_{2} \\mathrm{C} \\equiv \\mathrm{CH}$\n13. Butane is used as a fuel in disposable lighters. Write the Lewis structure for each isomer of butane.\n14. Write Lewis structures and name the five structural isomers of hexane.\n15. Write Lewis structures for the cis-trans isomers of $\\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{CHCl}$.\n16. Write structures for the three isomers of the aromatic hydrocarbon xylene, $\\mathrm{C}_{6} \\mathrm{H}_{4}\\left(\\mathrm{CH}_{3}\\right)_{2}$.\n17. Isooctane is the common name of the isomer of $\\mathrm{C}_{8} \\mathrm{H}_{18}$ used as the standard of 100 for the gasoline octane rating:"}
{"id": 5180, "contents": "2214. Exercises - 2214.1. Hydrocarbons\n(a) What is the IUPAC name for the compound?\n(b) Name the other isomers that contain a five-carbon chain with three methyl substituents.\n18. Write Lewis structures and IUPAC names for the alkyne isomers of $\\mathrm{C}_{4} \\mathrm{H}_{6}$.\n19. Write Lewis structures and IUPAC names for all isomers of $\\mathrm{C}_{4} \\mathrm{H}_{9} \\mathrm{Cl}$.\n20. Name and write the structures of all isomers of the propyl and butyl alkyl groups.\n21. Write the structures for all the isomers of the $-\\mathrm{C}_{5} \\mathrm{H}_{11}$ alkyl group.\n22. Write Lewis structures and describe the molecular geometry at each carbon atom in the following compounds:\n(a) cis-3-hexene\n(b) cis-1-chloro-2-bromoethene\n(c) 2-pentyne\n(d) trans-6-ethyl-7-methyl-2-octene\n23. Benzene is one of the compounds used as an octane enhancer in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene:\n$3 \\mathrm{C}_{2} \\mathrm{H}_{2} \\longrightarrow \\mathrm{C}_{6} \\mathrm{H}_{6}$\nDraw Lewis structures for these compounds, with resonance structures as appropriate, and determine the hybridization of the carbon atoms in each.\n24. Teflon is prepared by the polymerization of tetrafluoroethylene. Write the equation that describes the polymerization using Lewis symbols.\n25. Write two complete, balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures.\n(a) 1 mol of 1-butyne reacts with 2 mol of iodine.\n(b) Pentane is burned in air.\n26. Write two complete, balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures.\n(a) 2-butene reacts with chlorine.\n(b) benzene burns in air.\n27. What mass of 2 -bromopropane could be prepared from 25.5 g of propene? Assume a $100 \\%$ yield of product."}
{"id": 5181, "contents": "2214. Exercises - 2214.1. Hydrocarbons\n(b) benzene burns in air.\n27. What mass of 2 -bromopropane could be prepared from 25.5 g of propene? Assume a $100 \\%$ yield of product.\n28. Acetylene is a very weak acid; however, it will react with moist silver(I) oxide and form water and a compound composed of silver and carbon. Addition of a solution of HCl to a $0.2352-\\mathrm{g}$ sample of the compound of silver and carbon produced acetylene and 0.2822 g of AgCl .\n(a) What is the empirical formula of the compound of silver and carbon?\n(b) The production of acetylene on addition of HCl to the compound of silver and carbon suggests that the carbon is present as the acetylide ion, $\\mathrm{C}_{2}{ }^{2-}$. Write the formula of the compound showing the acetylide ion.\n29. Ethylene can be produced by the pyrolysis of ethane:\n$\\mathrm{C}_{2} \\mathrm{H}_{6} \\longrightarrow \\mathrm{C}_{2} \\mathrm{H}_{4}+\\mathrm{H}_{2}$\nHow many kilograms of ethylene is produced by the pyrolysis of $1.000 \\times 10^{3} \\mathrm{~kg}$ of ethane, assuming a $100.0 \\%$ yield?"}
{"id": 5182, "contents": "2214. Exercises - 2214.2. Alcohols and Ethers\n30. Why do the compounds hexane, hexanol, and hexene have such similar names?\n31. Write condensed formulas and provide IUPAC names for the following compounds:\n(a) ethyl alcohol (in beverages)\n(b) methyl alcohol (used as a solvent, for example, in shellac)\n(c) ethylene glycol (antifreeze)\n(d) isopropyl alcohol (used in rubbing alcohol)\n(e) glycerine\n32. Give the complete IUPAC name for each of the following compounds:\n(a)\n\n(b)\n\n(c)\n\n33. Give the complete IUPAC name and the common name for each of the following compounds:\n(a)\n\n(b)\n\n(c)"}
{"id": 5183, "contents": "2214. Exercises - 2214.2. Alcohols and Ethers\n(b)\n\n(c)\n\n33. Give the complete IUPAC name and the common name for each of the following compounds:\n(a)\n\n(b)\n\n(c)\n\n34. Write the condensed structures of both isomers with the formula $\\mathrm{C}_{2} \\mathrm{H}_{6} \\mathrm{O}$. Label the functional group of each isomer.\n35. Write the condensed structures of all isomers with the formula $\\mathrm{C}_{2} \\mathrm{H}_{6} \\mathrm{O}_{2}$. Label the functional group (or groups) of each isomer.\n36. Draw the condensed formulas for each of the following compounds:\n(a) dipropyl ether\n(b) 2,2-dimethyl-3-hexanol\n(c) 2-ethoxybutane\n37. MTBE, Methyl tert-butyl ether, $\\mathrm{CH}_{3} \\mathrm{OC}\\left(\\mathrm{CH}_{3}\\right)_{3}$, is used as an oxygen source in oxygenated gasolines. MTBE is manufactured by reacting 2 -methylpropene with methanol.\n(a) Using Lewis structures, write the chemical equation representing the reaction.\n(b) What volume of methanol, density $0.7915 \\mathrm{~g} / \\mathrm{mL}$, is required to produce exactly 1000 kg of MTBE, assuming a $100 \\%$ yield?\n38. Write two complete balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures.\n(a) propanol is converted to dipropyl ether\n(b) propene is treated with water in dilute acid.\n39. Write two complete balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures.\n(a) 2-butene is treated with water in dilute acid\n(b) ethanol is dehydrated to yield ethene"}
{"id": 5184, "contents": "2214. Exercises - 2214.3. Aldehydes, Ketones, Carboxylic Acids, and Esters\n40. Order the following molecules from least to most oxidized, based on the marked carbon atom:\n\n(a)\n\n(b)\n\n(c)\n41. Predict the products of oxidizing the molecules shown in this problem. In each case, identify the product that will result from the minimal increase in oxidation state for the highlighted carbon atom:\n(a)\n\n(b)\n\n(c)\n\n42. Predict the products of reducing the following molecules. In each case, identify the product that will result from the minimal decrease in oxidation state for the highlighted carbon atom:\n(a)\n\n(b)\n\n(c)"}
{"id": 5185, "contents": "2214. Exercises - 2214.3. Aldehydes, Ketones, Carboxylic Acids, and Esters\n43. Explain why it is not possible to prepare a ketone that contains only two carbon atoms.\n44. How does hybridization of the substituted carbon atom change when an alcohol is converted into an aldehyde? An aldehyde to a carboxylic acid?\n45. Fatty acids are carboxylic acids that have long hydrocarbon chains attached to a carboxylate group. How does a saturated fatty acid differ from an unsaturated fatty acid? How are they similar?\n46. Write a condensed structural formula, such as $\\mathrm{CH}_{3} \\mathrm{CH}_{3}$, and describe the molecular geometry at each carbon atom.\n(a) propene\n(b) 1-butanol\n(c) ethyl propyl ether\n(d) cis-4-bromo-2-heptene\n(e) 2,2,3-trimethylhexane\n(f) formaldehyde\n47. Write a condensed structural formula, such as $\\mathrm{CH}_{3} \\mathrm{CH}_{3}$, and describe the molecular geometry at each carbon atom.\n(a) 2-propanol\n(b) acetone\n(c) dimethyl ether\n(d) acetic acid\n(e) 3-methyl-1-hexene\n48. The foul odor of rancid butter is caused by butyric acid, $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CO}_{2} \\mathrm{H}$.\n(a) Draw the Lewis structure and determine the oxidation number and hybridization for each carbon atom in the molecule.\n(b) The esters formed from butyric acid are pleasant-smelling compounds found in fruits and used in perfumes. Draw the Lewis structure for the ester formed from the reaction of butyric acid with 2-propanol.\n49. Write the two-resonance structures for the acetate ion.\n50. Write two complete, balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures:\n(a) ethanol reacts with propionic acid\n(b) benzoic acid, $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CO}_{2} \\mathrm{H}$, is added to a solution of sodium hydroxide"}
{"id": 5186, "contents": "2214. Exercises - 2214.3. Aldehydes, Ketones, Carboxylic Acids, and Esters\n(b) benzoic acid, $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CO}_{2} \\mathrm{H}$, is added to a solution of sodium hydroxide\n51. Write two complete balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures.\n(a) 1-butanol reacts with acetic acid\n(b) propionic acid is poured onto solid calcium carbonate\n52. Yields in organic reactions are sometimes low. What is the percent yield of a process that produces 13.0 g of ethyl acetate from 10.0 g of $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$ ?\n53. Alcohols A, B, and $C$ all have the composition $\\mathrm{C}_{4} \\mathrm{H}_{10} \\mathrm{O}$. Molecules of alcohol A contain a branched carbon chain and can be oxidized to an aldehyde; molecules of alcohol B contain a linear carbon chain and can be oxidized to a ketone; and molecules of alcohol C can be oxidized to neither an aldehyde nor a ketone. Write the Lewis structures of these molecules."}
{"id": 5187, "contents": "2214. Exercises - 2214.4. Amines and Amides\n54. Write the Lewis structures of both isomers with the formula $\\mathrm{C}_{2} \\mathrm{H}_{7} \\mathrm{~N}$.\n55. What is the molecular structure about the nitrogen atom in trimethyl amine and in the trimethyl ammonium ion, $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{NH}^{+}$? What is the hybridization of the nitrogen atom in trimethyl amine and in the trimethyl ammonium ion?\n56. Write the two resonance structures for the pyridinium ion, $\\mathrm{C}_{5} \\mathrm{H}_{5} \\mathrm{NH}^{+}$.\n57. Draw Lewis structures for pyridine and its conjugate acid, the pyridinium ion, $\\mathrm{C}_{5} \\mathrm{H}_{5} \\mathrm{NH}^{+}$. What are the hybridizations, electron domain geometries, and molecular geometries about the nitrogen atoms in pyridine and in the pyridinium ion?\n58. Write the Lewis structures of all isomers with the formula $\\mathrm{C}_{3} \\mathrm{H}_{7} \\mathrm{ON}$ that contain an amide linkage.\n59. Write two complete balanced equations for the following reaction, one using condensed formulas and one using Lewis structures. Methyl amine is added to a solution of HCl .\n60. Write two complete, balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures. Ethylammonium chloride is added to a solution of sodium hydroxide.\n61. Identify any carbon atoms that change hybridization and the change in hybridization during the reactions in Exercise 21.26.\n62. Identify any carbon atoms that change hybridization and the change in hybridization during the reactions in Exercise 21.39.\n63. Identify any carbon atoms that change hybridization and the change in hybridization during the reactions in Exercise 21.51."}
{"id": 5188, "contents": "2216. The Periodic Table - \nPeriodic Table of the Elements"}
{"id": 5189, "contents": "2216. The Periodic Table - \n|  |  |  |  |  |  | 13 | 14 | 15 | 16 | 17 | $\\begin{gathered} 18 \\\\ \\qquad \\begin{array}{c} 2 \\mathrm{He} \\\\ \\begin{array}{c} \\text { 4.03 } \\\\ \\text { nelum } \\end{array} \\\\ \\hline \\end{array} . \\end{gathered}$ |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|  |  |  |  |  |  | $\\stackrel{5}{5}$ |  | $\\left\\lvert\\, \\begin{gathered} 74.01 \\\\ \\text { nitrogen } \\\\ \\mathbf{N} \\\\ \\hline \\end{gathered}\\right.$ | $\\underbrace{}_{\\substack{80 \\\\ \\text { 16.jo } \\\\ \\text { oxygen }}}$ |  | $\\begin{array}{\\|c} 10 \\\\ \\mathrm{Ne} \\\\ 20.18 \\\\ \\text { neon } \\\\ \\hline \\end{array}$ |\n| 7 | 8 | 9 | 10 | 11 | 12 |  |  | $\\begin{aligned} & 15 \\mathbf{P} \\\\ & \\text { 30.97 } \\\\ & \\text { phoshous } \\end{aligned}$ |  | ${ }^{17} \\mathrm{Cl}$ <br> 35.45 <br> chine | $\\stackrel{\\substack{18 \\\\ \\text { 3r } \\\\ \\text { 3agan } \\\\ \\text { arg }}}{ }$ |"}
{"id": 5190, "contents": "2216. The Periodic Table - \n|  | $\\left\\lvert\\, \\begin{gathered} \\begin{array}{c} 26 \\\\ \\mathrm{Fe} \\\\ 5 \\text { Fe. } \\\\ \\text { 1ron } \\end{array} \\\\ \\hline \\end{gathered}\\right.$ | $\\underset{\\substack{27 \\\\ \\hline 50.83 \\\\ \\text { cobat }}}{27}$ | $\\left\\lvert\\, \\begin{gathered} 28 \\\\ \\begin{array}{c} \\text { Ni } \\\\ 58.69 \\\\ \\text { nokel } \\end{array} \\\\ \\hline \\end{gathered}\\right.$ |  | $\\left\\lvert\\, \\begin{gathered} 30 \\\\ \\mathrm{Zn} \\\\ \\text { En } \\\\ \\text { zinc } \\\\ \\hline \\end{gathered}\\right.$ |  | $\\begin{gathered} 32 \\mathbf{G e} \\\\ 72.63 \\\\ \\text { gemanum } \\end{gathered}$ | $\\underbrace{}_{\\substack{33 \\\\ \\text { As } \\\\ \\text { atsenc } \\\\ \\text { anc }}}$ | $\\underset{\\substack{34 \\\\ \\text { seen } \\\\ \\text { senenum }}}{ }$ | $\\begin{gathered} 35 \\mathrm{Br} \\\\ \\text { Bran } \\\\ \\text { bomine } \\end{gathered}$ | $\\underset{\\substack{36 \\\\ \\hline 8.3, ~ \\\\ \\text { krpipon }}}{ }$ |"}
{"id": 5191, "contents": "2216. The Periodic Table - \n| ${ }^{43} \\mathbf{T C}$ |  |  |  |  |  | $\\begin{gathered} \\ln ^{114.8} \\\\ 1104 \\\\ \\hline \\text { nourum } \\end{gathered}$ | ${\\underset{S}{50}}_{\\substack{518.7 \\\\ \\text { in }}}$ | $\\underset{\\substack{\\text { Sxili.8 } \\\\ \\text { and } \\\\ \\text { and }}}{ }$ |  | $\\left\\lvert\\, \\begin{gathered} 53 \\\\ \\text { I } \\\\ \\text { 126.9 } \\\\ \\text { iodine } \\end{gathered}\\right.$ | $\\left\\lvert\\, \\begin{gathered} 54 \\\\ \\mathrm{Xe}^{131.3} \\\\ \\text { xenon } \\end{gathered}\\right.$ |"}
{"id": 5192, "contents": "2216. The Periodic Table - \n| $\\left\\begin{array}{c}75 \\\\ \\mathbf{R e} \\\\ \\text { 186.2. } \\\\ \\text { thenium }\\end{array}\\right.$ | ${ }^{76} \\mathrm{Os}$ <br> 190. | $\\left\\lvert\\, \\begin{array}{c\\|l} 77 \\\\ \\text { Ir } \\\\ \\text { 1922. } \\\\ \\text { indium } \\end{array}\\right.$ | ![ | $\\begin{array}{\\|c} 79 \\\\ \\text { Au } \\\\ \\text { 197.0 } \\\\ \\text { gold } \\end{array}$ | $\\begin{array}{\\|c} 80 \\\\ \\begin{array}{c} 80 \\\\ 200.6 \\\\ \\text { mercury } \\end{array} \\\\ \\hline \\end{array}$ | $\\begin{gathered} 81 \\\\ \\mathrm{TI} \\\\ \\text { 204.4. } \\\\ \\text { Raluium } \\end{gathered}$ | 82 <br> $\\mathbf{P b}$ <br> 207.2 <br> lead <br> lead | $\\begin{gathered} 83 \\\\ \\begin{array}{c} 8 i \\\\ \\text { 2ision } \\\\ \\text { biswnt } \\end{array} \\end{gathered}$ | $\\begin{array}{\\|c} 84 \\\\ \\mathbf{P O}_{0} \\\\ \\text { [209] } \\\\ \\text { polonium } \\end{array}$ |  |  |\n|  | $\\stackrel{108}{\\mathrm{Hs}}$ [277] |  | ${ }^{110} \\mathrm{DS}_{2}$ |  |  | ${\\underset{c}{123}}_{\\substack{\\text { N285] } \\\\ \\text { nhtonim }}}$ |  |  |  | $\\left.\\right\\|_{117} ^{11} \\text { TS }$ | ${ }^{118} \\mathbf{O g}$ |"}
{"id": 5193, "contents": "2216. The Periodic Table - \n|  | ${ }^{58} \\mathrm{Ce}$ 140.1 |  | ${ }^{60} \\mathrm{Nd}$ |  | $\\sqrt[\\substack{62 \\\\ \\text { Sm } \\\\ \\text { singati }}]{ }$ | ${ }^{63} \\mathrm{Eu}$ | ${ }^{64} \\mathbf{G d} \\text { Gd. }$ | ${ }^{65} \\mathrm{~Tb}$ 158.9 |  | $\\left\\lvert\\, \\begin{gathered} 67 \\mathrm{Ho} \\\\ \\text { He4.9 } \\\\ \\text { nomium } \\end{gathered}\\right.$ | ${ }^{68}$ Er | ${ }_{\\substack{\\text { a }}}^{\\substack{\\text { Tm } \\\\ \\text { 168.9 } \\\\ \\text { thuium }}}$ |  | ${ }^{71} \\mathrm{Lu}$ |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|  | $\\begin{gathered} 90 \\\\ \\hline 2320 \\\\ \\hline \\end{gathered}$ |  | ${ }_{\\substack{92 \\\\ \\hline \\\\ 238.0}}$ |  | ${ }^{94} \\mathrm{Pu}$ |  |  |  | ${ }^{98} \\mathbf{C f}$ | ${ }^{99} \\mathrm{Es}$ | $\\underset{\\substack{100 \\\\ \\text { Fm } \\\\ \\text { fetemiun }}}{ }$ |  | $\\stackrel{1}{102}_{\\substack{\\text { No } \\\\ \\text { nosediun }}}$ | $\\int_{103}^{\\mathrm{Lr}_{1}}$ |\n\nColor Code\n\n\nFIGURE A1\n\n1082 A \u2022 The Periodic Table"}
{"id": 5194, "contents": "2219. Exponential Arithmetic - \nExponential notation is used to express very large and very small numbers as a product of two numbers. The first number of the product, the digit term, is usually a number not less than 1 and not equal to or greater than 10. The second number of the product, the exponential term, is written as 10 with an exponent. Some examples of exponential notation are:\n\n$$\n\\begin{aligned}\n1000 & =1 \\times 10^{3} \\\\\n100 & =1 \\times 10^{2} \\\\\n10 & =1 \\times 10^{1} \\\\\n1 & =1 \\times 10^{0} \\\\\n0.1 & =1 \\times 10^{-1} \\\\\n0.001 & =1 \\times 10^{-3} \\\\\n2386 & =2.386 \\times 1000=2.386 \\times 10^{3} \\\\\n0.123 & =1.23 \\times 0.1=1.23 \\times 10^{-1}\n\\end{aligned}\n$$\n\nThe power (exponent) of 10 is equal to the number of places the decimal is shifted to give the digit number. The exponential method is particularly useful notation for very large and very small numbers. For example, $1,230,000,000=1.23 \\times 10^{9}$, and $0.00000000036=3.6 \\times 10^{-10}$."}
{"id": 5195, "contents": "2220. Addition of Exponentials - \nConvert all numbers to the same power of 10, add the digit terms of the numbers, and if appropriate, convert the digit term back to a number between 1 and 10 by adjusting the exponential term."}
{"id": 5196, "contents": "2222. Adding Exponentials - \nAdd $5.00 \\times 10^{-5}$ and $3.00 \\times 10^{-3}$."}
{"id": 5197, "contents": "2223. Solution - \n$$\n\\begin{aligned}\n3.00 \\times 10^{-3} & =300 \\times 10^{-5} \\\\\n\\left(5.00 \\times 10^{-5}\\right)+\\left(300 \\times 10^{-5}\\right) & =305 \\times 10^{-5}=3.05 \\times 10^{-3}\n\\end{aligned}\n$$"}
{"id": 5198, "contents": "2224. Subtraction of Exponentials - \nConvert all numbers to the same power of 10, take the difference of the digit terms, and if appropriate, convert the digit term back to a number between 1 and 10 by adjusting the exponential term."}
{"id": 5199, "contents": "2225. Subtracting Exponentials - \nSubtract $4.0 \\times 10^{-7}$ from $5.0 \\times 10^{-6}$."}
{"id": 5200, "contents": "2226. Solution - \n$$\n\\begin{aligned}\n& 4.0 \\times 10^{-7}=0.40 \\times 10^{-6} \\\\\n& \\left(5.0 \\times 10^{-6}\\right)-\\left(0.40 \\times 10^{-6}\\right)=4.6 \\times 10^{-6}\n\\end{aligned}\n$$"}
{"id": 5201, "contents": "2227. Multiplication of Exponentials - \nMultiply the digit terms in the usual way and add the exponents of the exponential terms."}
{"id": 5202, "contents": "2229. Multiplying Exponentials - \nMultiply $4.2 \\times 10^{-8}$ by $2.0 \\times 10^{3}$."}
{"id": 5203, "contents": "2230. Solution - \n$$\n\\left(4.2 \\times 10^{-8}\\right) \\times\\left(2.0 \\times 10^{3}\\right)=(4.2 \\times 2.0) \\times 10^{(-8)+(+3)}=8.4 \\times 10^{-5}\n$$"}
{"id": 5204, "contents": "2231. Division of Exponentials - \nDivide the digit term of the numerator by the digit term of the denominator and subtract the exponents of the exponential terms."}
{"id": 5205, "contents": "2233. Dividing Exponentials - \nDivide $3.6 \\times 10^{-5}$ by $6.0 \\times 10^{-4}$."}
{"id": 5206, "contents": "2234. Solution - \n$$\n\\frac{3.6 \\times 10^{-5}}{6.0 \\times 10^{-4}}=\\left(\\frac{3.6}{6.0}\\right) \\times 10^{(-5)-(-4)}=0.60 \\times 10^{-1}=6.0 \\times 10^{-2}\n$$"}
{"id": 5207, "contents": "2235. Squaring of Exponentials - \nSquare the digit term in the usual way and multiply the exponent of the exponential term by 2."}
{"id": 5208, "contents": "2237. Squaring Exponentials - \nSquare the number $4.0 \\times 10^{-6}$."}
{"id": 5209, "contents": "2238. Solution - \n$$\n\\left(4.0 \\times 10^{-6}\\right)^{2}=4 \\times 4 \\times 10^{2 \\times(-6)}=16 \\times 10^{-12}=1.6 \\times 10^{-11}\n$$"}
{"id": 5210, "contents": "2239. Cubing of Exponentials - \nCube the digit term in the usual way and multiply the exponent of the exponential term by 3."}
{"id": 5211, "contents": "2240. Cubing Exponentials - \nCube the number $2 \\times 10^{4}$."}
{"id": 5212, "contents": "2241. Solution - \n$\\left(2 \\times 10^{4}\\right)^{3}=2 \\times 2 \\times 2 \\times 10^{3 \\times 4}=8 \\times 10^{12}$"}
{"id": 5213, "contents": "2242. Taking Square Roots of Exponentials - \nIf necessary, decrease or increase the exponential term so that the power of 10 is evenly divisible by 2 . Extract the square root of the digit term and divide the exponential term by 2."}
{"id": 5214, "contents": "2244. Finding the Square Root of Exponentials - \nFind the square root of $1.6 \\times 10^{-7}$."}
{"id": 5215, "contents": "2245. Solution - \n$$\n\\begin{aligned}\n1.6 \\times 10^{-7} & =16 \\times 10^{-8} \\\\\n\\sqrt{16 \\times 10^{-8}}=\\sqrt{16} \\times \\sqrt{10^{-8}} & =\\sqrt{16} \\times 10^{-\\frac{8}{2}}=4.0 \\times 10^{-4}\n\\end{aligned}\n$$"}
{"id": 5216, "contents": "2246. Significant Figures - \nA beekeeper reports that he has 525,341 bees. The last three figures of the number are obviously inaccurate, for during the time the keeper was counting the bees, some of them died and others hatched; this makes it quite difficult to determine the exact number of bees. It would have been more reasonable if the beekeeper had reported the number 525,000. In other words, the last three figures are not significant, except to set the position of the decimal point. Their exact values have no useful meaning in this situation. When reporting quantities, use only as many significant figures as the accuracy of the measurement warrants.\n\nThe importance of significant figures lies in their application to fundamental computation. In addition and subtraction, the sum or difference should contain as many digits to the right of the decimal as that in the least certain of the numbers used in the computation (indicated by underscoring in the following example).\n\nEXAMPLE B8"}
{"id": 5217, "contents": "2247. Addition and Subtraction with Significant Figures - \nAdd 4.383 g and 0.0023 g ."}
{"id": 5218, "contents": "2248. Solution - \n$$\n\\begin{gathered}\n4.38 \\underline{\\mathrm{~g}} \\\\\n\\frac{0.002 \\underline{\\mathrm{3}} \\mathrm{~g}}{4.38 \\mathrm{~g}}\n\\end{gathered}\n$$\n\nIn multiplication and division, the product or quotient should contain no more digits than that in the factor containing the least number of significant figures."}
{"id": 5219, "contents": "2249. Multiplication and Division with Significant Figures - \nMultiply 0.6238 by 6.6 ."}
{"id": 5220, "contents": "2250. Solution - \n$0.623 \\underline{8} \\times 6 . \\underline{6}=4 . \\underline{1}$\n\nWhen rounding numbers, increase the retained digit by 1 if it is followed by a number larger than 5 (\"round up\"). Do not change the retained digit if the digits that follow are less than 5 (\"round down\"). If the retained digit is followed by 5 , round up if the retained digit is odd, or round down if it is even (after rounding, the retained digit will thus always be even)."}
{"id": 5221, "contents": "2251. The Use of Logarithms and Exponential Numbers - \nThe common logarithm of a number (log) is the power to which 10 must be raised to equal that number. For example, the common logarithm of 100 is 2 , because 10 must be raised to the second power to equal 100 . Additional examples follow.\n\nLogarithms and Exponential Numbers\n\n| Number | Number Expressed Exponentially | Common Logarithm |\n| :--- | :--- | :--- |\n| 1000 | $10^{3}$ | 3 |\n| 10 | $10^{1}$ | 1 |\n| 1 | $10^{0}$ | 0 |\n| 0.1 | $10^{-1}$ | -1 |\n| 0.001 | $10^{-3}$ | -3 |\n\nTABLE B1\n\nWhat is the common logarithm of 60 ? Because 60 lies between 10 and 100 , which have logarithms of 1 and 2, respectively, the logarithm of 60 is 1.7782 ; that is,\n\n$$\n60=10^{1.7782}\n$$\n\nThe common logarithm of a number less than 1 has a negative value. The logarithm of 0.03918 is -1.4069 , or\n\n$$\n0.03918=10^{-1.4069}=\\frac{1}{10^{1.4069}}\n$$\n\nTo obtain the common logarithm of a number, use the log button on your calculator. To calculate a number from its logarithm, take the inverse log of the logarithm, or calculate $10^{x}$ (where $x$ is the logarithm of the number).\n\nThe natural logarithm of a number ( $\\ln$ ) is the power to which e must be raised to equal the number; $e$ is the constant 2.7182818 . For example, the natural logarithm of 10 is 2.303 ; that is,\n\n$$\n10=e^{2.303}=2.7182818^{2.303}\n$$\n\nTo obtain the natural logarithm of a number, use the $\\ln$ button on your calculator. To calculate a number from its natural logarithm, enter the natural logarithm and take the inverse ln of the natural logarithm, or calculate $e^{x}$ (where $x$ is the natural logarithm of the number)."}
{"id": 5222, "contents": "2251. The Use of Logarithms and Exponential Numbers - \nLogarithms are exponents; thus, operations involving logarithms follow the same rules as operations involving\nexponents.\n\n1. The logarithm of a product of two numbers is the sum of the logarithms of the two numbers.\n\n$$\n\\log x y=\\log x+\\log y, \\text { and } \\ln x y=\\ln x+\\ln y\n$$\n\n2. The logarithm of the number resulting from the division of two numbers is the difference between the logarithms of the two numbers.\n\n$$\n\\log \\frac{x}{y}=\\log x-\\log y, \\text { and } \\ln \\frac{x}{y}=\\ln x-\\ln y\n$$\n\n3. The logarithm of a number raised to an exponent is the product of the exponent and the logarithm of the number.\n\n$$\n\\log x^{n}=n \\log x \\text { and } \\ln x^{n}=n \\ln x\n$$"}
{"id": 5223, "contents": "2252. The Solution of Quadratic Equations - \nMathematical functions of this form are known as second-order polynomials or, more commonly, quadratic functions.\n\n$$\na x^{2}+b x+c=0\n$$\n\nThe solution or roots for any quadratic equation can be calculated using the following formula:\n\n$$\nx=\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}\n$$"}
{"id": 5224, "contents": "2254. Solving Quadratic Equations - \nSolve the quadratic equation $3 x^{2}+13 x-10=0$."}
{"id": 5225, "contents": "2255. Solution - \nSubstituting the values $a=3, b=13, c=-10$ in the formula, we obtain\n\n$$\n\\begin{gathered}\nx=\\frac{-13 \\pm \\sqrt{(13)^{2}-4 \\times 3 \\times(-10)}}{2 \\times 3} \\\\\nx=\\frac{-13 \\pm \\sqrt{169+120}}{6}=\\frac{-13 \\pm \\sqrt{289}}{6}=\\frac{-13 \\pm 17}{6}\n\\end{gathered}\n$$\n\nThe two roots are therefore\n\n$$\nx=\\frac{-13+17}{6}=\\frac{2}{3} \\text { and } x=\\frac{-13-17}{6}=-5\n$$\n\nQuadratic equations constructed on physical data always have real roots, and of these real roots, often only those having positive values are of any significance."}
{"id": 5226, "contents": "2256. Two-Dimensional ( $x-y$ ) Graphing - \nThe relationship between any two properties of a system can be represented graphically by a two-dimensional data plot. Such a graph has two axes: a horizontal one corresponding to the independent variable, or the variable whose value is being controlled ( $x$ ), and a vertical axis corresponding to the dependent variable, or the variable whose value is being observed or measured ( $y$ ).\n\nWhen the value of $y$ is changing as a function of $x$ (that is, different values of $x$ correspond to different values of $y$ ), a graph of this change can be plotted or sketched. The graph can be produced by using specific values for $(x, y)$ data pairs."}
{"id": 5227, "contents": "2257. Graphing the Dependence of $y$ on $x$ - \nThis table contains the following points: $(1,5),(2,10),(3,7)$, and $(4,14)$. Each of these points can be plotted on a graph and connected to produce a graphical representation of the dependence of $y$ on $x$.\n\n\nIf the function that describes the dependence of $y$ on $x$ is known, it may be used to compute $x, y$ data pairs that may subsequently be plotted."}
{"id": 5228, "contents": "2258. Plotting Data Pairs - \nIf we know that $y=x^{2}+2$, we can produce a table of a few $(x, y)$ values and then plot the line based on the data shown here.\n\n\n| $\\boldsymbol{x}$ | $\\boldsymbol{y}=\\boldsymbol{x}^{2}+2$ |\n| :--- | :--- |\n| 3 | 11 |\n| 4 | 18 |"}
{"id": 5229, "contents": "2260. Units and Conversion Factors - \n| Units of Length |  |\n| :--- | :--- |\n| meter (m) | $=39.37$ inches (in.) |\n| $=1.094$ yards (yd) |  |$]$| centimeter (cm) | $=0.01 \\mathrm{~m}$ (exact, definition) |\n| :--- | :--- |\n| millimeter (mm) | $=0.001 \\mathrm{~m}$ (exact, definition) |\n| kilometer (km) | $=1000 \\mathrm{~m}$ (exact, definition) |\n| angstrom (\u00c5) | $=10^{-8} \\mathrm{~cm}$ (exact, definition) |\n| yard (yd) | $=0.9144 \\mathrm{~m}$ |\n| inch (in.) | $=2.54 \\mathrm{~cm}$ (exact, definition) |\n| mile (US) | $=1.60934 \\mathrm{~km}$ |\n\nTABLE C1\n\\$\\left.\\begin{array}{l|l} \\& Units of Volume <br>\nliter (L) \\& =0.001 \\mathrm{~m}^{3} (exact, definition) <br>\n=1000 \\mathrm{~cm}^{3} (exact, definition) <br>\n\n=1.057 (US) quarts\\end{array}\\right]\\$\\begin{tabular}{ll}\n\nmilliliter (mL) \\& | $=0.001 \\mathrm{~L}$ (exact, definition) |\n| :--- |\n| $=1 \\mathrm{~cm}^{3}$ (exact, definition) | <br>\n\n\\hline microliter ( $\\mu \\mathrm{L}$ ) \\& | $=10^{-6} \\mathrm{~L}$ (exact, definition) |\n| :--- |\n| $=10^{-3} \\mathrm{~cm}^{3}$ (exact, definition) | <br>\n\n\n\\hline liquid quart (US) \\& | $=32(\\mathrm{US})$ liquid ounces (exact, definition) |\n| :--- |\n| $=0.25$ (US) gallon (exact, definition) |\n| $=0.9463 \\mathrm{~L}$ | <br>"}
{"id": 5230, "contents": "2260. Units and Conversion Factors - \n\\hline dry quart \\& $=1.1012 \\mathrm{~L}$ <br>\n\\hline cubic foot (US) \\& $=28.316 \\mathrm{~L}$ <br>\n\\hline\n\\end{tabular}\n\nTABLE C2\n\n| gram (g) | $=0.001 \\mathrm{~kg}$ (exact, definition) |\n| :--- | :--- |\n| milligram (mg) | $=0.001 \\mathrm{~g}$ (exact, definition) |\n| kilogram (kg) | $=1000 \\mathrm{~g}$ (exact, definition) <br> $=2.205 \\mathrm{lb}$ |\n| ton (metric) | $=1000 \\mathrm{~kg}$ (exact, definition) <br> $=2204.62 \\mathrm{lb}$ |\n| ounce (oz) | $=28.35 \\mathrm{~g}$ |\n| pound (lb) | $=0.4535924 \\mathrm{~kg}$ |\n| ton (short) | $=2000 \\mathrm{lb}$ (exact, definition) <br> $=907.185 \\mathrm{~kg}$ |\n| ton (long) | $=2240 \\mathrm{lb}$ (exact, definition) <br> $=1.016$ metric ton |\n\nTABLE C3\n\nUnits of Energy"}
{"id": 5231, "contents": "2260. Units and Conversion Factors - \nTABLE C3\n\nUnits of Energy\n\n| 4.184 joule (J) | $=1$ thermochemical calorie (cal) |\n| :--- | :--- |\n| 1 thermochemical calorie (cal) | $=4.184 \\times 10^{7} \\mathrm{erg}$ |\n| erg | $=10^{-7} \\mathrm{~J}($ exact, definition) |\n| electron-volt (eV) | $=1.60218 \\times 10^{-19} \\mathrm{~J}=23.061 \\mathrm{kcal} \\mathrm{mol}{ }^{-1}$ |\n| liter $\\bullet$ atmosphere | $=24.217 \\mathrm{cal}=101.325 \\mathrm{~J}$ (exact, definition) |\n| nutritional calorie (Cal) | $=1000 \\mathrm{cal}($ exact, definition) $=4184 \\mathrm{~J}$ |\n| British thermal unit (BTU) | $=1054.804 \\mathrm{~J}^{\\underline{1}}$ |\n\nTABLE C4\n\n| Units of Pressure |  |\n| :--- | :--- |\n| torr | $=1 \\mathrm{~mm} \\mathrm{Hg}$ (exact, definition) |\n\nTABLE C5\n\n1 BTU is the amount of energy needed to heat one pound of water by one degree Fahrenheit. Therefore, the exact relationship of BTU to joules and other energy units depends on the temperature at which BTU is measured. $59^{\\circ} \\mathrm{F}\\left(15^{\\circ} \\mathrm{C}\\right)$ is the most widely used reference temperature for BTU definition in the United States. At this temperature, the conversion factor is the one provided in this table."}
{"id": 5232, "contents": "2260. Units and Conversion Factors - \n| pascal (Pa) | $=\\mathrm{N} \\mathrm{m}^{-2}$ (exact, definition) <br> $=\\mathrm{kg} \\mathrm{m}^{-1} \\mathrm{~s}^{-2}$ (exact, definition) |\n| :--- | :--- |\n| atmosphere (atm) $=760 \\mathrm{~mm} \\mathrm{Hg}$ (exact, definition) <br> $=760$ torr (exact, definition)  <br> $=101,325 \\mathrm{~N} \\mathrm{~m}$  <br> -2  <br>  $=101,325 \\mathrm{~Pa}$ (exact, definition) |  |\n| bar | $=10^{5} \\mathrm{~Pa}$ (exact, definition)  <br>  $=10^{5} \\mathrm{~kg} \\mathrm{~m}^{-1} \\mathrm{~s}^{-2}$ (exact, definition) |\n\nTABLE C5\n\n1094 C\u2022 Units and Conversion Factors"}
{"id": 5233, "contents": "2262. Fundamental Physical Constants - \nFundamental Physical Constants"}
{"id": 5234, "contents": "2262. Fundamental Physical Constants - \n| Name and Symbol | Value |\n| :---: | :---: |\n| atomic mass unit (amu) | $1.6605402 \\times 10^{-27} \\mathrm{~kg}$ |\n| Avogadro's number | $6.02214076 \\times 10^{23} \\mathrm{~mol}^{-1}$ |\n| Boltzmann's constant (k) | $1.380649 \\times 10^{-23} \\mathrm{~J} \\mathrm{~K}^{-1}$ |\n| charge-to-mass ratio for electron $\\left(e / m_{e}\\right)$ | $1.75881962 \\times 10^{11} \\mathrm{C} \\mathrm{kg}^{-1}$ |\n| fundamental unit of charge (e) | $1.602176634 \\times 10^{-19} \\mathrm{C}$ |\n| electron rest mass ( $m_{e}$ ) | $9.1093897 \\times 10^{-31} \\mathrm{~kg}$ |\n| Faraday's constant (F) | $9.6485309 \\times 10^{4} \\mathrm{C} \\mathrm{mol}^{-1}$ |\n| gas constant ( $R$ ) | $8.205784 \\times 10^{-2} \\mathrm{~L} \\mathrm{~atm} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}=8.314510 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}$ |\n| molar volume of an ideal gas, $1 \\mathrm{~atm}, 0^{\\circ} \\mathrm{C}$ | $22.41409 \\mathrm{~L} \\mathrm{~mol}^{-1}$ |\n| molar volume of an ideal gas, $1 \\mathrm{bar}, 0^{\\circ} \\mathrm{C}$ | $22.71108 \\mathrm{~L} \\mathrm{~mol}^{-1}$ |\n| neutron rest mass ( $m_{n}$ ) | $1.6749274 \\times 10^{-27} \\mathrm{~kg}$ |"}
{"id": 5235, "contents": "2262. Fundamental Physical Constants - \n| neutron rest mass ( $m_{n}$ ) | $1.6749274 \\times 10^{-27} \\mathrm{~kg}$ |\n| Planck's constant (h) | $6.62607015 \\times 10^{-34} \\mathrm{~J} \\mathrm{~S}$ |\n| proton rest mass ( $m_{p}$ ) | $1.6726231 \\times 10^{-27} \\mathrm{~kg}$ |\n| Rydberg constant (R) | $1.0973731534 \\times 10^{7} \\mathrm{~m}^{-1}=2.1798736 \\times 10^{-18} \\mathrm{~J}$ |\n| speed of light (in vacuum) (c) | $2.99792458 \\times 10^{8} \\mathrm{~m} \\mathrm{~s}^{-1}$ |"}
{"id": 5236, "contents": "2262. Fundamental Physical Constants - \nTABLE D1"}
{"id": 5237, "contents": "2264. Water Properties - \n| Temperature | Density ( $\\mathrm{g} / \\mathrm{mL}$ ) |\n| :---: | :---: |\n| 0 | 0.9998395 |\n| 4 | 0.9999720 (density maximum) |\n| 10 | 0.9997026 |\n| 15 | 0.9991026 |\n| 20 | 0.9982071 |\n| 22 | 0.9977735 |\n| 25 | 0.9970479 |\n| 30 | 0.9956502 |\n| 40 | 0.9922 |\n| 60 | 0.9832 |\n| 80 | 0.9718 |\n| 100 | 0.9584 |\n\nTABLE E1\n\nDensity of Water as a Function of Temperature\n\n\n| Temperature | Vapor Pressure (torr) | Vapor Pressure (Pa) |\n| :---: | :---: | :---: |\n| 0 | 4.6 | 613.2812 |\n| 4 | 6.1 | 813.2642 |\n| 10 | 9.2 | 1226.562 |\n| 15 | 12.8 | 1706.522 |\n| 20 | 17.5 | 2333.135 |\n| 22 | 19.8 | 2639.776 |\n| 25 | 23.8 | 3173.064 |\n| 30 | 31.8 | 4239.64 |\n| 35 | 42.2 | 5626.188 |\n| 40 | 55.3 | 7372.707 |\n| 45 | 71.9 | 9585.852 |\n| 50 | 92.5 | 12332.29 |\n| 55 | 118.0 | 15732 |\n\nTABLE E2"}
{"id": 5238, "contents": "2264. Water Properties - \nTABLE E2\n\n| Temperature |  | Vapor Pressure (torr) |  | Vapor Pressure (Pa) |\n| :--- | :--- | :--- | :---: | :---: |\n| 60 | 149.4 | 19918.31 |  |  |\n| 65 | 187.5 | 24997.88 |  |  |\n| 70 | 233.7 | 31157.35 |  |  |\n| 75 | 289.1 | 38543.39 |  |  |\n| 80 | 435.1 | 47342.64 |  |  |\n| 85 | 525.8 | 57808.42 |  |  |\n| 90 | 760.0 | 70100.71 |  |  |\n| 95 |  | 84512.82 |  |  |\n| 100 |  |  |  |  |\n| TABLE E2 |  |  |  |  |\n\n\n\nWater $\\mathrm{K}_{\\mathrm{w}}$ and $\\mathrm{pK}_{\\mathrm{W}}$ at Different Temperatures $\\left({ }^{\\circ} \\mathrm{C}\\right)$\n\n| Temperature | $\\mathrm{K}_{\\mathrm{w}} \\mathbf{1 0}^{\\mathbf{- 1 4}}$ | $\\mathrm{pK}_{\\mathrm{w}}{ }^{\\mathbf{1}}$ |\n| :--- | :--- | :--- |\n| 0 | 0.112 | 14.95 |\n\nTABLE E3"}
{"id": 5239, "contents": "2264. Water Properties - \nTABLE E3\n\n| Temperature | $\\mathrm{K}_{\\mathrm{w}} 10^{-14}$ | $\\mathrm{pK}_{\\mathrm{w}}{ }^{1}$ |\n| :---: | :---: | :---: |\n| 5 | 0.182 | 14.74 |\n| 10 | 0.288 | 14.54 |\n| 15 | 0.465 | 14.33 |\n| 20 | 0.671 | 14.17 |\n| 25 | 0.991 | 14.00 |\n| 30 | 1.432 | 13.84 |\n| 35 | 2.042 | 13.69 |\n| 40 | 2.851 | 13.55 |\n| 45 | 3.917 | 13.41 |\n| 50 | 5.297 | 13.28 |\n| 55 | 7.080 | 13.15 |\n| 60 | 9.311 | 13.03 |\n| 75 | 19.95 | 12.70 |\n| 100 | 56.23 | 12.25 |\n\nTABLE E3\n\n\nStandard Water Melting and Boiling Temperatures and Enthalpies of the Transitions\n\n|  | Temperature (K) | $\\Delta H(\\mathrm{~kJ} / \\mathrm{mol})$ |\n| :--- | :--- | :--- |\n| melting | 273.15 | 6.088 |\n| boiling | 373.15 | $40.656(44.016$ at 298 K$)$ |\n\nTABLE E5\n\nWater Cryoscopic (Freezing Point Depression) and Ebullioscopic (Boiling Point Elevation) Constants\n$\\mathrm{K}_{\\mathrm{f}}=1.86^{\\circ} \\mathrm{C} \\cdot \\mathrm{kg} \\cdot \\mathrm{mol}^{-1}($ cryoscopic constant $)$\n$\\mathrm{K}_{\\mathrm{b}}=0.51^{\\circ} \\mathrm{C} \\cdot \\mathrm{kg} \\cdot \\mathrm{mol}^{-1}($ ebullioscopic constant)\n\nTABLE E6"}
{"id": 5240, "contents": "2264. Water Properties - \nTABLE E6\n\n\nFIGURE E1 The plot shows the extent of light absorption versus wavelength for water. Absorption is reported in reciprocal meters and corresponds to the inverse of the distance light may travel through water before its intensity is diminished by $1 / e$ ( $\\sim 37 \\%$ )."}
{"id": 5241, "contents": "2266. Composition of Commercial Acids and Bases - \nComposition of Commercial Acids and Bases\n\n| Acid or Base $^{\\mathbf{1}}$ | Density (g/mL) | Percentage by Mass | Molarity |\n| :--- | :--- | :--- | :--- |\n| acetic acid, glacial | 1.05 | $99.5 \\%$ | 17.4 |\n| aqueous ammonia ${ }^{\\frac{3}{2}}$ | 0.90 | $28 \\%$ | 14.8 |\n| hydrochloric acid | 1.18 | $36 \\%$ | 11.6 |\n| nitric acid | 1.42 | $71 \\%$ | 16.0 |\n| perchloric acid | 1.67 | 1.70 | $50 \\%$ |\n| phosphoric acid | 1.53 | $96 \\%$ | 11.65 |\n| sodium hydroxide | 1.84 | 19.7 |  |\n| sulfuric acid |  |  | 18.0 |\n\nTABLE F1\n\n[^17]"}
{"id": 5242, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \nStandard Thermodynamic Properties for Selected Substances"}
{"id": 5243, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| aluminum |  |  |  |\n| $\\mathrm{Al}(\\mathrm{s})$ | 0 | 0 | 28.3 |\n| $\\mathrm{Al}(\\mathrm{g})$ | 324.4 | 285.7 | 164.54 |\n| $\\mathrm{Al}^{3+}(\\mathrm{aq})$ | -531 | -485 | -321.7 |\n| $\\mathrm{Al}_{2} \\mathrm{O}_{3}(\\mathrm{~s})$ | -1676 | -1582 | 50.92 |\n| $\\mathrm{AlF}_{3}(s)$ | -1510.4 | -1425 | 66.5 |\n| $\\mathrm{AlCl}_{3}(S)$ | -704.2 | -628.8 | 110.67 |\n| $\\mathrm{AlCl}_{3} \\cdot 6 \\mathrm{H}_{2} \\mathrm{O}(s)$ | -2691.57 | -2269.40 | 376.56 |\n| $\\mathrm{Al}_{2} \\mathrm{~S}_{3}(\\mathrm{~S})$ | -724.0 | -492.4 | 116.9 |\n| $\\mathrm{Al}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}(\\mathrm{~s})$ | -3445.06 | -3506.61 | 239.32 |\n| antimony |  |  |  |\n| $\\mathrm{Sb}(\\mathrm{s})$ | 0 | 0 | 45.69 |"}
{"id": 5244, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| antimony |  |  |  |\n| $\\mathrm{Sb}(\\mathrm{s})$ | 0 | 0 | 45.69 |\n| $\\mathrm{Sb}(\\mathrm{g})$ | 262.34 | 222.17 | 180.16 |\n| $\\mathrm{Sb}_{4} \\mathrm{O}_{6}(s)$ | -1440.55 | -1268.17 | 220.92 |\n| $\\mathrm{SbCl}_{3}(\\mathrm{~g})$ | -313.8 | -301.2 | 337.80 |\n| $\\mathrm{SbCl}_{5}(\\mathrm{~g})$ | -394.34 | -334.29 | 401.94 |\n| $\\mathrm{Sb}_{2} \\mathrm{~S}_{3}(\\mathrm{~s})$ | -174.89 | -173.64 | 182.00 |\n| $\\mathrm{SbCl}_{3}(\\mathrm{~s})$ | -382.17 | -323.72 | 184.10 |\n| SbOCl(s) | -374.0 | - | - |"}
{"id": 5245, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \nTABLE G1\n\n| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :--- | :--- | :--- | :--- |\n\narsenic\n\n| $\\mathrm{As}(\\mathrm{s})$ | 0 | 0 | 35.1 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{As}(\\mathrm{g})$ | 302.5 | 261.0 | 174.21 |\n| $\\mathrm{As}_{4}(\\mathrm{~g})$ | 143.9 | 92.4 | 314 |\n| $\\mathrm{As}_{4} \\mathrm{O}_{6}(s)$ | -1313.94 | -1152.52 | 214.22 |\n| $\\mathrm{As}_{2} \\mathrm{O}_{5}(\\mathrm{~s})$ | -924.87 | -782.41 | 105.44 |\n| $\\mathrm{AsCl}_{3}(\\mathrm{~g})$ | -261.50 | -248.95 | 327.06 |\n| $\\mathrm{As}_{2} \\mathrm{~S}_{3}(\\mathrm{~s})$ | -169.03 | -168.62 | 163.59 |\n| $\\mathrm{AsH}_{3}(\\mathrm{~g})$ | 66.44 | 68.93 | 222.78 |\n| $\\mathrm{H}_{3} \\mathrm{AsO}_{4}(s)$ | -906.3 | - | - |\n\nbarium"}
{"id": 5246, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \nbarium\n\n| $\\mathrm{Ba}(s)$ | 0 | 0 | 62.5 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{Ba}(g)$ | 180 | 146 | 170.24 |\n| $\\mathrm{Ba}^{2+}(a q)$ | -537.6 | -560.8 | 9.6 |\n| $\\mathrm{BaO}(s)$ | -548.0 | -520.3 | 72.1 |\n| $\\mathrm{BaCl}_{2}(s)$ | -855.0 | -806.7 | 123.7 |\n| $\\mathrm{BaSO}_{4}(s)$ | -1473.2 | -1362.3 | 132.2 |\n\nberyllium\n\n| $\\operatorname{Be}(s)$ | 0 | 0 | 9.50 |\n| :--- | :--- | :--- | :--- |\n| $\\operatorname{Be}(g)$ | 324.3 | 286.6 | 136.27 |\n| $\\operatorname{BeO}(s)$ | -609.4 | -580.1 | 13.8 |\n\nbismuth\n\n| $\\operatorname{Bi}(s)$ | 0 | 0 | 56.74 |\n| :--- | :--- | :--- | :--- |\n| $\\operatorname{Bi}(g)$ | 207.1 | 168.2 | 187.00 |\n\nTABLE G1"}
{"id": 5247, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{Bi}_{2} \\mathrm{O}_{3}(\\mathrm{~s})$ | -573.88 | -493.7 | 151.5 |\n| $\\mathrm{BiCl}_{3}(s)$ | -379.07 | -315.06 | 176.98 |\n| $\\mathrm{Bi}_{2} \\mathrm{~S}_{3}(S)$ | -143.1 | -140.6 | 200.4 |\n| boron |  |  |  |\n| B(s) | 0 | 0 | 5.86 |\n| $\\mathrm{B}(\\mathrm{g})$ | 565.0 | 521.0 | 153.4 |\n| $\\mathrm{B}_{2} \\mathrm{O}_{3}(S)$ | -1273.5 | -1194.3 | 53.97 |\n| $\\mathrm{B}_{2} \\mathrm{H}_{6}(\\mathrm{~g})$ | 36.4 | 87.6 | 232.1 |\n| $\\mathrm{H}_{3} \\mathrm{BO}_{3}(\\mathrm{~s})$ | -1094.33 | -968.92 | 88.83 |\n| $\\mathrm{BF}_{3}(\\mathrm{~g})$ | -1136.0 | -1119.4 | 254.4 |\n| $\\mathrm{BCl}_{3}(\\mathrm{~g})$ | -403.8 | -388.7 | 290.1 |"}
{"id": 5248, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| $\\mathrm{BCl}_{3}(\\mathrm{~g})$ | -403.8 | -388.7 | 290.1 |\n| $\\mathrm{B}_{3} \\mathrm{~N}_{3} \\mathrm{H}_{6}(\\mathrm{l})$ | -540.99 | -392.79 | 199.58 |\n| $\\mathrm{HBO}_{2}(s)$ | -794.25 | -723.41 | 37.66 |\n| bromine |  |  |  |\n| $\\mathrm{Br}_{2}(1)$ | 0 | 0 | 152.23 |\n| $\\mathrm{Br}_{2}(\\mathrm{~g})$ | 30.91 | 3.142 | 245.5 |\n| $\\operatorname{Br}(\\mathrm{g})$ | 111.88 | 82.429 | 175.0 |\n| $\\mathrm{Br}^{-}(\\mathrm{aq})$ | -120.9 | -102.82 | 80.71 |\n| $\\mathrm{BrF}_{3}(\\mathrm{~g})$ | -255.60 | -229.45 | 292.42 |\n| $\\operatorname{HBr}(\\mathrm{g})$ | -36.3 | -53.43 | 198.7 |\n| cadmium |  |  |  |\n| $\\mathrm{Cd}(s)$ | 0 | 0 | 51.76 |\n| $\\mathrm{Cd}(\\mathrm{g})$ | 112.01 | 77.41 | 167.75 |\n| $\\mathrm{Cd}^{2+}(\\mathrm{aq})$ | -75.90 | -77.61 | -73.2 |"}
{"id": 5249, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \nTABLE G1"}
{"id": 5250, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{CdO}(s)$ | -258.2 | -228.4 | 54.8 |\n| $\\mathrm{CdCl}_{2}(s)$ | -391.5 | -343.9 | 115.3 |\n| $\\mathrm{CdSO}_{4}(\\mathrm{~S})$ | -933.3 | -822.7 | 123.0 |\n| $\\operatorname{CdS}(s)$ | -161.9 | -156.5 | 64.9 |\n| calcium |  |  |  |\n| $\\mathrm{Ca}(s)$ | 0 | 0 | 41.6 |\n| $\\mathrm{Ca}(\\mathrm{g})$ | 178.2 | 144.3 | 154.88 |\n| $\\mathrm{Ca}^{2+}(\\mathrm{aq})$ | -542.96 | -553.04 | -55.2 |\n| $\\mathrm{CaO}(s)$ | -634.9 | -603.3 | 38.1 |\n| $\\mathrm{Ca}(\\mathrm{OH})_{2}(\\mathrm{~s})$ | -985.2 | -897.5 | 83.4 |\n| $\\mathrm{CaSO}_{4}(\\mathrm{~s})$ | -1434.5 | -1322.0 | 106.5 |\n| $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(s)$ | -2022.63 | -1797.45 | 194.14 |"}
{"id": 5251, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(s)$ | -2022.63 | -1797.45 | 194.14 |\n| $\\mathrm{CaCO}_{3}(s)$ (calcite) | -1220.0 | -1081.4 | 110.0 |\n| $\\mathrm{CaSO}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}(s)$ | -1752.68 | -1555.19 | 184.10 |\n| carbon |  |  |  |\n| C(s) (graphite) | 0 | 0 | 5.740 |\n| $\\mathrm{C}(s)$ (diamond) | 1.89 | 2.90 | 2.38 |\n| $\\mathrm{C}(\\mathrm{g})$ | 716.681 | 671.2 | 158.1 |\n| $\\mathrm{CO}(\\mathrm{g})$ | -110.52 | -137.15 | 197.7 |\n| $\\mathrm{CO}_{2}(\\mathrm{~g})$ | -393.51 | -394.36 | 213.8 |\n| $\\mathrm{CO}_{3}{ }^{2-}(\\mathrm{aq})$ | -677.1 | -527.8 | -56.9 |\n| $\\mathrm{CH}_{4}(\\mathrm{~g})$ | -74.6 | -50.5 | 186.3 |\n| $\\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{I})$ | -239.2 | -166.6 | 126.8 |\n| $\\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{g})$ | -201.0 | -162.3 | 239.9 |"}
{"id": 5252, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \nTABLE G1"}
{"id": 5253, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{CCl}_{4}(\\mathrm{I})$ | -128.2 | -62.5 | 214.4 |\n| $\\mathrm{CCl}_{4}(\\mathrm{~g})$ | -95.7 | -58.2 | 309.7 |\n| $\\mathrm{CHCl}_{3}(\\mathrm{I})$ | -134.1 | -73.7 | 201.7 |\n| $\\mathrm{CHCl}_{3}(\\mathrm{~g})$ | -103.14 | -70.34 | 295.71 |\n| $\\mathrm{CS}_{2}($ I) | 89.70 | 65.27 | 151.34 |\n| $\\mathrm{CS}_{2}(\\mathrm{~g})$ | 116.9 | 66.8 | 238.0 |\n| $\\mathrm{C}_{2} \\mathrm{H}_{2}(\\mathrm{~g})$ | 227.4 | 209.2 | 200.9 |\n| $\\mathrm{C}_{2} \\mathrm{H}_{4}(\\mathrm{~g})$ | 52.4 | 68.4 | 219.3 |\n| $\\mathrm{C}_{2} \\mathrm{H}_{6}(\\mathrm{~g})$ | -84.0 | -32.0 | 229.2 |\n| $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(\\mathrm{I})$ | -484.3 | -389.9 | 159.8 |"}
{"id": 5254, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(\\mathrm{I})$ | -484.3 | -389.9 | 159.8 |\n| $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(\\mathrm{g})$ | -434.84 | -376.69 | 282.50 |\n| $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(\\mathrm{I})$ | -277.6 | -174.8 | 160.7 |\n| $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(\\mathrm{g})$ | -234.8 | -167.9 | 281.6 |\n| $\\mathrm{HCO}_{3}{ }^{-}(\\mathrm{aq})$ | -691.11 | -587.06 | 95 |\n| $\\mathrm{C}_{3} \\mathrm{H}_{8}(\\mathrm{~g})$ | -103.8 | -23.4 | 270.3 |\n| $\\mathrm{C}_{6} \\mathrm{H}_{6}(\\mathrm{~g})$ | 82.927 | 129.66 | 269.2 |\n| $\\mathrm{C}_{6} \\mathrm{H}_{6}(\\mathrm{I})$ | 49.1 | 124.50 | 173.4 |\n| $\\mathrm{CH}_{2} \\mathrm{Cl}_{2}($ I | -124.2 | -63.2 | 177.8 |\n| $\\mathrm{CH}_{2} \\mathrm{Cl}_{2}(\\mathrm{~g})$ | -95.4 | -65.90 | 270.2 |\n| $\\mathrm{CH}_{3} \\mathrm{Cl}(\\mathrm{g})$ | -81.9 | -60.2 | 234.6 |\n| $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{Cl}(\\mathrm{I})$ | -136.52 | -59.31 | 190.79 |"}
{"id": 5255, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{Cl}(\\mathrm{I})$ | -136.52 | -59.31 | 190.79 |\n| $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{Cl}(\\mathrm{g})$ | -112.17 | -60.39 | 276.00 |\n| $\\mathrm{C}_{2} \\mathrm{~N}_{2}(\\mathrm{~g})$ | 308.98 | 297.36 | 241.90 |\n| HCN( $)$ | 108.9 | 125.0 | 112.8 |"}
{"id": 5256, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \nTABLE G1"}
{"id": 5257, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{HCN}(\\mathrm{g})$ | 135.5 | 124.7 | 201.8 |\n| cesium |  |  |  |\n| $\\mathrm{Cs}^{+}(\\mathrm{aq})$ | -248 | -282.0 | 133 |\n| chlorine |  |  |  |\n| $\\mathrm{Cl}_{2}(\\mathrm{~g})$ | 0 | 0 | 223.1 |\n| $\\mathrm{Cl}(\\mathrm{g})$ | 121.3 | 105.70 | 165.2 |\n| $\\mathrm{Cl}^{-}(a q)$ | -167.2 | -131.2 | 56.5 |\n| $\\mathrm{ClF}(\\mathrm{g})$ | -54.48 | -55.94 | 217.78 |\n| $\\mathrm{ClF}_{3}(\\mathrm{~g})$ | -158.99 | -118.83 | 281.50 |\n| $\\mathrm{Cl}_{2} \\mathrm{O}(\\mathrm{g})$ | 80.3 | 97.9 | 266.2 |\n| $\\mathrm{Cl}_{2} \\mathrm{O}_{7}(\\mathrm{I})$ | 238.1 | - | - |\n| $\\mathrm{Cl}_{2} \\mathrm{O}_{7}(\\mathrm{~g})$ | 272.0 | - | - |\n| $\\mathrm{HCl}(\\mathrm{g})$ | -92.307 | -95.299 | 186.9 |\n| $\\mathrm{HClO}_{4}($ l | -40.58 | - | - |"}
{"id": 5258, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| $\\mathrm{HClO}_{4}($ l | -40.58 | - | - |\n| chromium |  |  |  |\n| $\\mathrm{Cr}(\\mathrm{s})$ | 0 | 0 | 23.77 |\n| $\\mathrm{Cr}(\\mathrm{g})$ | 396.6 | 351.8 | 174.50 |\n| $\\mathrm{CrO}_{4}{ }^{2-}(\\mathrm{aq})$ | -881.2 | -727.8 | 50.21 |\n| $\\mathrm{Cr}_{2} \\mathrm{O}_{7}{ }^{2-}(\\mathrm{aq})$ | -1490.3 | -1301.1 | 261.9 |\n| $\\mathrm{Cr}_{2} \\mathrm{O}_{3}(S)$ | -1139.7 | -1058.1 | 81.2 |\n| $\\mathrm{CrO}_{3}(S)$ | -589.5 | - | - |\n| $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}(\\mathrm{~S})$ | -1806.7 | - | - |\n| cobalt |  |  |  |\n| $\\mathrm{Co}(\\mathrm{S})$ | 0 | 0 | 30.0 |"}
{"id": 5259, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \nTABLE G1\n\n| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{Co}^{2+}(\\mathrm{aq})$ | -67.4 | -51.5 | -155 |\n| $\\mathrm{Co}^{3+}(\\mathrm{aq})$ | 92 | 134 | -305.0 |\n| $\\mathrm{CoO}(s)$ | -237.9 | -214.2 | 52.97 |\n| $\\mathrm{Co}_{3} \\mathrm{O}_{4}(s)$ | -910.02 | -794.98 | 114.22 |\n| $\\mathrm{Co}_{\\left(\\mathrm{NO}_{3}\\right)_{2}(s)}$ | -420.5 | - | - |\n\ncopper"}
{"id": 5260, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \ncopper\n\n| $\\mathrm{Cu}(s)$ | 0 | 0 | 33.15 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{Cu}(g)$ | 338.32 | 298.58 | 166.38 |\n| $\\mathrm{Cu}^{+}(a q)$ | 51.9 | 50.2 | -26 |\n| $\\mathrm{Cu}^{2+}(a q)$ | 64.77 | 65.49 | -99.6 |\n| $\\mathrm{CuO}(s)$ | -157.3 | -129.7 | 42.63 |\n| $\\mathrm{Cu}_{2} \\mathrm{O}(s)$ | -168.6 | -146.0 | 93.14 |\n| $\\mathrm{CuS}^{(s)}$ | -53.1 | -53.6 | 66.5 |\n| $\\mathrm{Cu}_{2} \\mathrm{~S}(s)$ | -79.5 | -86.2 | 120.9 |\n| $\\mathrm{CuSO}_{4}(s)$ | -771.36 | -662.2 | 109.2 |\n| $\\mathrm{Cu}^{\\left(\\mathrm{NO}_{3}\\right)_{2}(s)}$ | -302.9 | - | - |\n\nfluorine\n\n| $\\mathrm{F}_{2}(g)$ | 0 | 0 | 202.8 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{~F}(g)$ | 79.4 | 62.3 | 158.8 |\n| $\\mathrm{~F}^{-}(\\mathrm{aq})$ | -332.6 | -278.8 | -13.8 |\n| $\\mathrm{~F}_{2} \\mathrm{O}(g)$ | 24.7 | 41.9 | 247.43 |\n| $\\mathrm{HF}(g)$ | -273.3 | -275.4 | 173.8 |\n\nhydrogen\n\n| $\\mathrm{H}_{2}(g)$ | 0 | 0 | 130.7 |\n| :--- | :--- | :--- | :--- |\n\nTABLE G1"}
{"id": 5261, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{H}(\\mathrm{g})$ | 217.97 | 203.26 | 114.7 |\n| $\\mathrm{H}^{+}(a q)$ | 0 | 0 | 0 |\n| $\\mathrm{OH}^{-}(a q)$ | -230.0 | -157.2 | -10.75 |\n| $\\mathrm{H}_{3} \\mathrm{O}^{+}(a q)$ | -285.8 |  | 69.91 |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})$ | -285.83 | -237.1 | 70.0 |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ | -241.82 | -228.59 | 188.8 |\n| $\\mathrm{H}_{2} \\mathrm{O}_{2}($ I) | -187.78 | -120.35 | 109.6 |\n| $\\mathrm{H}_{2} \\mathrm{O}_{2}(\\mathrm{~g})$ | -136.3 | -105.6 | 232.7 |\n| $\\mathrm{HF}(\\mathrm{g})$ | -273.3 | -275.4 | 173.8 |\n| $\\mathrm{HCl}(\\mathrm{g})$ | -92.307 | -95.299 | 186.9 |\n| $\\operatorname{HBr}(\\mathrm{g})$ | -36.3 | -53.43 | 198.7 |\n| $\\mathrm{HI}(\\mathrm{g})$ | 26.48 | 1.70 | 206.59 |"}
{"id": 5262, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| $\\mathrm{HI}(\\mathrm{g})$ | 26.48 | 1.70 | 206.59 |\n| $\\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{~g})$ | -20.6 | -33.4 | 205.8 |\n| $\\mathrm{H}_{2} \\mathrm{Se}(\\mathrm{g})$ | 29.7 | 15.9 | 219.0 |\n| $\\mathrm{HNO}_{3}$ | -206.64 | - | - |\n| iodine |  |  |  |\n| $\\mathrm{I}_{2}(\\mathrm{~s})$ | 0 | 0 | 116.14 |\n| $\\mathrm{I}_{2}(\\mathrm{~g})$ | 62.438 | 19.3 | 260.7 |\n| $\\mathrm{I}(\\mathrm{g})$ | 106.84 | 70.2 | 180.8 |\n| $\\mathrm{I}^{-}(a q)$ | -55.19 | -51.57 | 11.13 |\n| $\\mathrm{IF}(\\mathrm{g})$ | 95.65 | -118.49 | 236.06 |\n| $\\mathrm{ICl}(\\mathrm{g})$ | 17.78 | -5.44 | 247.44 |\n| $\\operatorname{IBr}(\\mathrm{g})$ | 40.84 | 3.72 | 258.66 |\n| $\\mathrm{IF}_{7}(\\mathrm{~g})$ | -943.91 | -818.39 | 346.44 |"}
{"id": 5263, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \nTABLE G1"}
{"id": 5264, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{HI}(\\mathrm{g})$ | 26.48 | 1.70 | 206.59 |\n| iron |  |  |  |\n| $\\mathrm{Fe}(s)$ | 0 | 0 | 27.3 |\n| $\\mathrm{Fe}(\\mathrm{g})$ | 416.3 | 370.7 | 180.5 |\n| $\\mathrm{Fe}^{2+}(\\mathrm{aq})$ | -89.1 | -78.90 | -137.7 |\n| $\\mathrm{Fe}^{3+}(\\mathrm{aq})$ | -48.5 | -4.7 | -315.9 |\n| $\\mathrm{Fe}_{2} \\mathrm{O}_{3}(S)$ | -824.2 | -742.2 | 87.40 |\n| $\\mathrm{Fe}_{3} \\mathrm{O}_{4}(S)$ | -1118.4 | -1015.4 | 146.4 |\n| $\\mathrm{Fe}(\\mathrm{CO})_{5}(\\mathrm{l})$ | -774.04 | -705.42 | 338.07 |\n| $\\mathrm{Fe}(\\mathrm{CO})_{5}(\\mathrm{~g})$ | -733.87 | -697.26 | 445.18 |\n| FeCl $2(S)$ | -341.79 | -302.30 | 117.95 |\n| $\\mathrm{FeCl}_{3}(S)$ | -399.49 | -334.00 | 142.3 |\n| $\\mathrm{FeO}(s)$ | -272.0 | -255.2 | 60.75 |"}
{"id": 5265, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| $\\mathrm{FeO}(s)$ | -272.0 | -255.2 | 60.75 |\n| $\\mathrm{Fe}(\\mathrm{OH})_{2}(\\mathrm{~s})$ | -569.0 | -486.5 | 88. |\n| $\\mathrm{Fe}(\\mathrm{OH})_{3}(\\mathrm{~S})$ | -823.0 | -696.5 | 106.7 |\n| $\\mathrm{FeS}(s)$ | -100.0 | -100.4 | 60.29 |\n| $\\mathrm{Fe}_{3} \\mathrm{C}(s)$ | 25.10 | 20.08 | 104.60 |\n| lead |  |  |  |\n| $\\mathrm{Pb}(\\mathrm{s})$ | 0 | 0 | 64.81 |\n| $\\mathrm{Pb}(\\mathrm{g})$ | 195.2 | 162. | 175.4 |\n| $\\mathrm{Pb}^{2+}(\\mathrm{aq})$ | -1.7 | -24.43 | 10.5 |\n| $\\mathrm{PbO}(s)$ (yellow) | -217.32 | -187.89 | 68.70 |\n| $\\mathrm{PbO}(s)$ (red) | -218.99 | -188.93 | 66.5 |\n| $\\mathrm{Pb}(\\mathrm{OH})_{2}(\\mathrm{~s})$ | -515.9 | - | - |"}
{"id": 5266, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \nTABLE G1\n\n| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{PbS}(s)$ | -100.4 | -98.7 | 91.2 |\n| $\\mathrm{Pb}\\left(\\mathrm{NO}_{3}\\right)_{2}(\\mathrm{~s})$ | -451.9 | - | - |\n| $\\mathrm{PbO}_{2}(S)$ | -277.4 | -217.3 | 68.6 |\n| $\\mathrm{PbCl}_{2}(S)$ | -359.4 | -314.1 | 136.0 |\n| lithium |  |  |  |\n| Li(s) | 0 | 0 | 29.1 |\n| $\\mathrm{Li}(\\mathrm{g})$ | 159.3 | 126.6 | 138.8 |\n| $\\mathrm{Li}^{+}(a q)$ | -278.5 | -293.3 | 13.4 |\n| $\\mathrm{LiH}(s)$ | -90.5 | -68.3 | 20.0 |\n| Li(OH)(S) | -487.5 | -441.5 | 42.8 |\n| $\\operatorname{LiF}(s)$ | -616.0 | -587.5 | 35.7 |\n| $\\mathrm{Li}_{2} \\mathrm{CO}_{3}(s)$ | -1216.04 | -1132.19 | 90.17 |\n\nmagnesium\n\n| $\\operatorname{Mg}^{2+}(a q)$ | -466.9 | -454.8 | -138.1 |\n| :--- | :--- | :--- | :--- |\n\nmanganese"}
{"id": 5267, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| $\\operatorname{Mg}^{2+}(a q)$ | -466.9 | -454.8 | -138.1 |\n| :--- | :--- | :--- | :--- |\n\nmanganese\n\n| $\\mathrm{Mn}(s)$ | 0 | 0 | 32.0 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{Mn}(g)$ | 280.7 | 238.5 | 173.7 |\n| $\\mathrm{Mn}^{2+}(a q)$ | -220.8 | -228.1 | -73.6 |\n| $\\mathrm{MnO}(s)$ | -385.2 | -362.9 | 59.71 |\n| $\\mathrm{MnO}_{2}(s)$ | -520.03 | -465.1 | 53.05 |\n| $\\mathrm{Mn}_{2} \\mathrm{O}_{3}(s)$ | -958.97 | -881.15 | 110.46 |\n| $\\mathrm{Mn}_{3} \\mathrm{O}_{4}(s)$ | -541.4 | -1283.23 | 155.64 |\n| $\\mathrm{MnO}_{4}{ }^{-}(\\mathrm{aq})$ | -653.0 | -500.7 | 59 |\n| $\\mathrm{MnO}_{4}{ }^{2-}(\\mathrm{aq})$ |  | 191.2 |  |\n\nTABLE G1\n\n| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :--- | :--- | :--- | :--- |"}
{"id": 5268, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| $\\mathrm{Hg}(1)$ | 0 | 0 | 75.9 |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{Hg}(\\mathrm{g})$ | 61.4 | 31.8 | 175.0 |\n| $\\mathrm{Hg}^{2+}(\\mathrm{aq})$ |  | 164.8 |  |\n| $\\mathrm{Hg}^{2+}(\\mathrm{aq})$ | 172.4 | 153.9 | 84.5 |\n| $\\mathrm{HgO}(\\mathrm{s})(\\mathrm{red})$ | -90.83 | -58.5 | 70.29 |\n| $\\mathrm{HgO}(s)$ (yellow) | -90.46 | -58.43 | 71.13 |\n| $\\mathrm{HgCl}_{2}(\\mathrm{~s})$ | -224.3 | -178.6 | 146.0 |\n| $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}(s)$ | -265.4 | -210.7 | 191.6 |\n| $\\mathrm{HgS}(\\mathrm{s})(\\mathrm{red})$ | -58.16 | -50.6 | 82.4 |\n| $\\operatorname{HgS}(s)$ (black) | -53.56 | -47.70 | 88.28 |\n| $\\mathrm{HgSO}_{4}(\\mathrm{~S})$ | -707.51 | -594.13 | 0.00 |\n\nnickel\n\n| $\\mathrm{Ni}^{2+}(a q)$ | -64.0 | -46.4 | -159 |\n| :--- | :--- | :--- | :--- |\n\nnitrogen"}
{"id": 5269, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \nnickel\n\n| $\\mathrm{Ni}^{2+}(a q)$ | -64.0 | -46.4 | -159 |\n| :--- | :--- | :--- | :--- |\n\nnitrogen\n\n| $\\mathrm{N}_{2}(g)$ | 0 | 0 | 191.6 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{~N}(g)$ | 472.704 | 455.5 | 153.3 |\n| $\\mathrm{NO}(g)$ | 90.25 | 87.6 | 210.8 |\n| $\\mathrm{NO}_{2}(g)$ | 33.2 | 51.30 | 240.1 |\n| $\\mathrm{~N}_{2} \\mathrm{O}(g)$ | 81.6 | 103.7 | 220.0 |\n| $\\mathrm{~N}_{2} \\mathrm{O}_{3}(g)$ | 83.72 | 139.41 | 312.17 |\n| $\\mathrm{NO}_{3}-(a q)$ | -205.0 | -108.7 | 146.4 |\n| $\\mathrm{~N}_{2} \\mathrm{O}_{4}(g)$ | 11.1 | 99.8 | 304.4 |\n| $\\mathrm{~N}_{2} \\mathrm{O}_{5}(g)$ | 11.3 | 115.1 | 355.7 |\n\nTABLE G1"}
{"id": 5270, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{NH}_{3}(\\mathrm{~g})$ | -45.9 | -16.5 | 192.8 |\n| $\\mathrm{NH}_{4}{ }^{+}(\\mathrm{aq})$ | -132.5 | -79.31 | 113.4 |\n| $\\mathrm{N}_{2} \\mathrm{H}_{4}(\\mathrm{l})$ | 50.63 | 149.43 | 121.21 |\n| $\\mathrm{N}_{2} \\mathrm{H}_{4}(\\mathrm{~g})$ | 95.4 | 159.4 | 238.5 |\n| $\\mathrm{NH}_{4} \\mathrm{NO}_{3}(S)$ | -365.56 | -183.87 | 151.08 |\n| $\\mathrm{NH}_{4} \\mathrm{Cl}(\\mathrm{s})$ | -314.43 | -202.87 | 94.6 |\n| $\\mathrm{NH}_{4} \\mathrm{Br}(\\mathrm{s})$ | -270.8 | -175.2 | 113.0 |\n| $\\mathrm{NH}_{4} \\mathrm{I}(\\mathrm{s})$ | -201.4 | -112.5 | 117.0 |\n| $\\mathrm{NH}_{4} \\mathrm{NO}_{2}(s)$ | -256.5 | - | - |\n| $\\mathrm{HNO}_{3}(\\mathrm{I})$ | -174.1 | -80.7 | 155.6 |\n| $\\mathrm{HNO}_{3}(\\mathrm{~g})$ | -133.9 | -73.5 | 266.9 |"}
{"id": 5271, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| $\\mathrm{HNO}_{3}(\\mathrm{~g})$ | -133.9 | -73.5 | 266.9 |\n| $\\mathrm{HNO}_{3}(\\mathrm{aq})$ | -207.4 | -110.5 | 146 |"}
{"id": 5272, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \noxygen\n\n| $\\mathrm{O}_{2}(g)$ | 0 | 0 | 205.2 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{O}(g)$ | 249.17 | 231.7 | 161.1 |\n| $\\mathrm{O}_{3}(g)$ | 142.7 | 163.2 | 238.9 |\n\nphosphorus\n\n| $\\mathrm{P}_{4}(s)$ | 0 | 0 | 164.4 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{P}_{4}(g)$ | 58.91 | 24.4 | 280.0 |\n| $\\mathrm{P}(g)$ | 314.64 | 278.25 | 163.19 |\n| $\\mathrm{PH}_{3}(g)$ | 5.4 | 13.5 | 210.2 |\n| $\\mathrm{PCl}_{3}(g)$ | -287.0 | -267.8 | 311.78 |\n| $\\mathrm{PCl}_{5}(g)$ | -374.9 | -305.0 | 364.4 |\n| $\\mathrm{P}_{4} \\mathrm{O}_{6}(s)$ | -1640.1 | - | - |\n\nTABLE G1"}
{"id": 5273, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{P}_{4} \\mathrm{O}_{10}(s)$ | -2984.0 | -2697.0 | 228.86 |\n| $\\mathrm{PO}_{4}{ }^{3-}(\\mathrm{aq})$ | -1277 | -1019 | -222 |\n| $\\mathrm{HPO}_{3}(\\mathrm{~s})$ | -948.5 | - | - |\n| $\\mathrm{HPO}_{4}{ }^{2-}(\\mathrm{aq})$ | -1292.1 | -1089.3 | -33 |\n| $\\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{2-}(\\mathrm{aq})$ | -1296.3 | -1130.4 | 90.4 |\n| $\\mathrm{H}_{3} \\mathrm{PO}_{2}(s)$ | -604.6 | - | - |\n| $\\mathrm{H}_{3} \\mathrm{PO}_{3}(s)$ | -964.4 | - | - |\n| $\\mathrm{H}_{3} \\mathrm{PO}_{4}(S)$ | -1279.0 | -1119.1 | 110.50 |\n| $\\mathrm{H}_{3} \\mathrm{PO}_{4}(\\mathrm{l})$ | -1266.9 | -1124.3 | 110.5 |\n| $\\mathrm{H}_{4} \\mathrm{P}_{2} \\mathrm{O}_{7}(S)$ | -2241.0 | - | - |\n| $\\mathrm{POCl}_{3}(\\mathrm{l})$ | -597.1 | -520.8 | 222.5 |"}
{"id": 5274, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| $\\mathrm{POCl}_{3}(\\mathrm{l})$ | -597.1 | -520.8 | 222.5 |\n| $\\mathrm{POCl}_{3}(\\mathrm{~g})$ | -558.5 | -512.9 | 325.5 |\n| potassium |  |  |  |\n| K(s) | 0 | 0 | 64.7 |\n| K(g) | 89.0 | 60.5 | 160.3 |\n| $\\mathrm{K}^{+}(a q)$ | -252.4 | -283.3 | 102.5 |\n| KF( $s$ ) | -576.27 | -537.75 | 66.57 |\n| $\\mathrm{KCl}(\\mathrm{s})$ | -436.5 | -408.5 | 82.6 |\n| rubidium |  |  |  |\n| $\\mathrm{Rb}^{+}(\\mathrm{aq})$ | -246 | -282.2 | 124 |\n| silicon |  |  |  |\n| Si(s) | 0 | 0 | 18.8 |\n| $\\mathrm{Si}(\\mathrm{g})$ | 450.0 | 405.5 | 168.0 |\n| $\\mathrm{SiO}_{2}(\\mathrm{~S})$ | -910.7 | -856.3 | 41.5 |"}
{"id": 5275, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \nTABLE G1"}
{"id": 5276, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{SiH}_{4}(\\mathrm{~g})$ | 34.3 | 56.9 | 204.6 |\n| $\\mathrm{H}_{2} \\mathrm{SiO}_{3}(S)$ | -1188.67 | -1092.44 | 133.89 |\n| $\\mathrm{H}_{4} \\mathrm{SiO}_{4}(\\mathrm{~S})$ | -1481.14 | -1333.02 | 192.46 |\n| $\\mathrm{SiF}_{4}(\\mathrm{~g})$ | -1615.0 | -1572.8 | 282.8 |\n| $\\mathrm{SiCl}_{4}(I)$ | -687.0 | -619.8 | 239.7 |\n| $\\mathrm{SiCl}_{4}(\\mathrm{~g})$ | -662.75 | -622.58 | 330.62 |\n| SiC(s, beta cubic) | -73.22 | -70.71 | 16.61 |\n| SiC(s, alpha hexagonal) | -71.55 | -69.04 | 16.48 |\n| silver |  |  |  |\n| $\\mathrm{Ag}(s)$ | 0 | 0 | 42.55 |\n| $\\mathrm{Ag}(\\mathrm{g})$ | 284.9 | 246.0 | 172.89 |\n| $\\mathrm{Ag}^{+}(\\mathrm{aq})$ | 105.6 | 77.11 | 72.68 |\n| $\\mathrm{Ag}_{2} \\mathrm{O}(s)$ | -31.05 | -11.20 | 121.3 |"}
{"id": 5277, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| $\\mathrm{Ag}_{2} \\mathrm{O}(s)$ | -31.05 | -11.20 | 121.3 |\n| $\\mathrm{AgCl}(\\mathrm{s})$ | -127.0 | -109.8 | 96.3 |\n| $\\mathrm{Ag}_{2} \\mathrm{~S}(\\mathrm{~S})$ | -32.6 | -40.7 | 144.0 |\n| sodium |  |  |  |\n| $\\mathrm{Na}(S)$ | 0 | 0 | 51.3 |\n| $\\mathrm{Na}(\\mathrm{g})$ | 107.5 | 77.0 | 153.7 |\n| $\\mathrm{Na}^{+}(\\mathrm{aq})$ | -240.1 | -261.9 | 59 |\n| $\\mathrm{Na}_{2} \\mathrm{O}(s)$ | -414.2 | -375.5 | 75.1 |\n| $\\mathrm{NaCl}(\\mathrm{s})$ | -411.2 | -384.1 | 72.1 |\n| strontium |  |  |  |\n| $\\mathrm{Sr}^{2+}(\\mathrm{aq})$ | -545.8 | -557.3 | -32.6 |\n| sulfur |  |  |  |"}
{"id": 5278, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \nTABLE G1"}
{"id": 5279, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{S}_{8}(s)$ (rhombic) | 0 | 0 | 256.8 |\n| $\\mathrm{S}(\\mathrm{g})$ | 278.81 | 238.25 | 167.82 |\n| $S^{2-}(a q)$ | 41.8 | 83.7 | 22 |\n| $\\mathrm{SO}_{2}(\\mathrm{~g})$ | -296.83 | -300.1 | 248.2 |\n| $\\mathrm{SO}_{3}(\\mathrm{~g})$ | -395.72 | -371.06 | 256.76 |\n| $\\mathrm{SO}_{4}{ }^{2-}(\\mathrm{aq})$ | -909.3 | -744.5 | 20.1 |\n| $\\mathrm{S}_{2} \\mathrm{O}_{3}{ }^{2-}(\\mathrm{aq})$ | -648.5 | -522.5 | 67 |\n| $\\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{~g})$ | -20.6 | -33.4 | 205.8 |\n| $\\mathrm{HS}^{-}(\\mathrm{aq})$ | -17.7 | 12.6 | 61.1 |\n| $\\mathrm{H}_{2} \\mathrm{SO}_{4}(\\mathrm{I})$ | -813.989 | -690.00 | 156.90 |\n| $\\mathrm{HSO}_{4}{ }^{2-}(\\mathrm{aq})$ | -885.75 | -752.87 | 126.9 |"}
{"id": 5280, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| $\\mathrm{HSO}_{4}{ }^{2-}(\\mathrm{aq})$ | -885.75 | -752.87 | 126.9 |\n| $\\mathrm{H}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{7}(\\mathrm{~S})$ | -1273.6 | - | - |\n| $\\mathrm{SF}_{4}(\\mathrm{~g})$ | -728.43 | -684.84 | 291.12 |\n| $\\mathrm{SF}_{6}(\\mathrm{~g})$ | -1220.5 | -1116.5 | 291.5 |\n| $\\mathrm{SCl}_{2}($ I) | -50 | - | - |\n| $\\mathrm{SCl}_{2}(\\mathrm{~g})$ | -19.7 | - | - |\n| $\\mathrm{S}_{2} \\mathrm{Cl}_{2}(\\mathrm{I})$ | -59.4 | - | - |\n| $\\mathrm{S}_{2} \\mathrm{Cl}_{2}(\\mathrm{~g})$ | -19.50 | -29.25 | 319.45 |\n| $\\mathrm{SOCl}_{2}(\\mathrm{~g})$ | -212.55 | -198.32 | 309.66 |\n| $\\mathrm{SOCl}_{2}(\\mathrm{I})$ | -245.6 | - | - |\n| $\\mathrm{SO}_{2} \\mathrm{Cl}_{2}($ I | -394.1 | - | - |\n| $\\mathrm{SO}_{2} \\mathrm{Cl}_{2}(\\mathrm{~g})$ | -354.80 | -310.45 | 311.83 |\n| tin |  |  |  |\n| $\\mathrm{Sn}(\\mathrm{s})$ | 0 | 0 | 51.2 |"}
{"id": 5281, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \nTABLE G1\n\n| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}{ }^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| $\\operatorname{Sn}(\\mathrm{g})$ | 301.2 | 266.2 | 168.5 |\n| $\\mathrm{SnO}(\\mathrm{s})$ | -285.8 | -256.9 | 56.5 |\n| $\\mathrm{SnO}_{2}(\\mathrm{~S})$ | -577.6 | -515.8 | 49.0 |\n| $\\mathrm{SnCl}_{4}(\\mathrm{I})$ | -511.3 | -440.1 | 258.6 |\n| $\\mathrm{SnCl}_{4}(\\mathrm{~g})$ | -471.5 | -432.2 | 365.8 |\n| titanium |  |  |  |\n| $\\mathrm{Ti}(s)$ | 0 | 0 | 30.7 |\n| $\\mathrm{Ti}(\\mathrm{g})$ | 473.0 | 428.4 | 180.3 |\n| $\\mathrm{TiO}_{2}(s)$ | -944.0 | -888.8 | 50.6 |\n| $\\mathrm{TiCl}_{4}($ ) | -804.2 | -737.2 | 252.4 |\n| $\\mathrm{TiCl}_{4}(\\mathrm{~g})$ | -763.2 | -726.3 | 353.2 |\n\ntungsten"}
{"id": 5282, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \ntungsten\n\n| $\\mathrm{W}(s)$ | 0 | 0 | 32.6 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{~W}(g)$ | 849.4 | 807.1 | 174.0 |\n| $\\mathrm{WO}_{3}(s)$ | -842.9 | -764.0 | 75.9 |\n\nzinc\n\n| $\\mathrm{Zn}(s)$ | 0 | 0 | 41.6 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{Zn}(g)$ | 130.73 | 95.14 | 160.98 |\n| $\\mathrm{Zn}^{2+}(a q)$ | -153.9 | -147.1 | -112.1 |\n| $\\mathrm{ZnO}(s)$ | -350.5 | -320.5 | 43.7 |\n| $\\mathrm{ZnCl}_{2}(s)$ | -415.1 | -369.43 | 111.5 |\n| $\\mathrm{ZnS}^{(s)}$ | -206.0 | -201.3 | 57.7 |\n| $\\mathrm{ZnSO}_{4}(s)$ | -982.8 | -871.5 | 110.5 |\n| $\\mathrm{ZnCO}_{3}(s)$ | -812.78 | -731.57 | 82.42 |\n\nTABLE G1\n\n| Substance | $\\Delta H_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $\\Delta G_{\\mathrm{f}}^{\\circ}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{\\circ}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :--- | :--- | :--- | :--- |\n\ncomplexes"}
{"id": 5283, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{4}\\left(\\mathrm{NO}_{2}\\right)_{2}\\right] \\mathrm{NO}_{3}, \\mathrm{cis}$ | -898.7 | - | - |\n| :---: | :---: | :---: | :---: |\n| $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{4}\\left(\\mathrm{NO}_{2}\\right)_{2}\\right] \\mathrm{NO}_{3}$, trans | -896.2 | - | - |\n| $\\mathrm{NH}_{4}\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{2}\\left(\\mathrm{NO}_{2}\\right)_{4}\\right]$ | -837.6 | - | - |\n| $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{2}\\left(\\mathrm{NO}_{2}\\right)_{4}\\right]_{3}$ | -2733.0 | - | - |\n| $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{4} \\mathrm{Cl}_{2}\\right] \\mathrm{Cl}$, cis | -874.9 | - | - |\n| [ $\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{4} \\mathrm{Cl}_{2} \\mathrm{Cl}$, trans | -877.4 | - | - |\n| $\\left[\\mathrm{Co}(\\mathrm{en})_{2}\\left(\\mathrm{NO}_{2}\\right)_{2}\\right] \\mathrm{NO}_{3}$, cis | -689.5 | - | - |\n| [Co(en) $2_{2} \\mathrm{Cl}_{2}$ ]Cl, cis | -681.2 | - | - |\n| [Co(en) $)_{2} \\mathrm{Cl}_{2}$ ]Cl, trans | -677.4 | - | - |"}
{"id": 5284, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| [Co(en) $)_{2} \\mathrm{Cl}_{2}$ ]Cl, trans | -677.4 | - | - |\n| $\\left[\\mathrm{Co}(\\mathrm{en})_{3}\\right]\\left(\\mathrm{ClO}_{4}\\right)_{3}$ | -762.7 | - | - |\n| [Co(en) ${ }_{3} \\mathrm{Br}_{2}$ | -595.8 | - | - |\n| [Co(en) ${ }_{3} \\mathrm{I}_{2}$ | -475.3 | - | - |\n| [Co(en) $3_{3} \\mathrm{I}_{3}$ | -519.2 | - | - |\n| $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]\\left(\\mathrm{ClO}_{4}\\right)_{3}$ | -1034.7 | -221.1 | 615 |\n| $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{5} \\mathrm{NO}_{2}\\right]\\left(\\mathrm{NO}_{3}\\right)_{2}$ | -1088.7 | -412.9 | 331 |\n| $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]\\left(\\mathrm{NO}_{3}\\right)_{3}$ | -1282.0 | -524.5 | 448 |\n| $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{5} \\mathrm{Cl]Cl}{ }_{2}\\right.$ | -1017.1 | -582.5 | 366.1 |\n| $\\left[\\mathrm{Pt}\\left(\\mathrm{NH}_{3}\\right)_{4}\\right] \\mathrm{Cl}_{2}$ | -725.5 | - | - |\n| $\\left[\\mathrm{Ni}^{\\left.\\left(\\mathrm{NH}_{3}\\right)_{6}\\right] \\mathrm{Cl}_{2}}\\right.$ | -994.1 | - | - |"}
{"id": 5285, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \n| $\\left[\\mathrm{Ni}^{\\left.\\left(\\mathrm{NH}_{3}\\right)_{6}\\right] \\mathrm{Cl}_{2}}\\right.$ | -994.1 | - | - |\n| $\\left[\\mathrm{Ni}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right] \\mathrm{Br}_{2}$ | -923.8 | - | - |\n| $\\left[\\mathrm{Ni}\\left(\\mathrm{NH}_{3}\\right)_{6} \\mathrm{II}_{2}\\right.$ | -808.3 | - | - |"}
{"id": 5286, "contents": "2268. Standard Thermodynamic Properties for Selected Substances - \nTABLE G1"}
{"id": 5287, "contents": "2270. Ionization Constants of Weak Acids - \nIonization Constants of Weak Acids\n\n| Acid | Formula | $K_{a}$ at $25^{\\circ} \\mathrm{C}$ | Lewis Structure |\n| :---: | :---: | :---: | :---: |\n| acetic | $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathbf{H}$ | $1.8 \\times 10^{-5}$ | <smiles>CC([O])=[O+]</smiles> |\n| arsenic | $\\mathbf{H}_{3} \\mathrm{AsO}_{4}$ | $5.5 \\times 10^{-3}$ | <smiles></smiles> |\n|  | $\\mathrm{H}_{2} \\mathrm{AsO}_{4}{ }^{-}$ | $1.7 \\times 10^{-7}$ |  |\n|  | $\\mathrm{HAsO}_{4}{ }^{2-}$ | $3.0 \\times 10^{-12}$ |  |\n| arsenous | $\\mathbf{H}_{3} \\mathrm{AsO}_{3}$ | $5.1 \\times 10^{-10}$ | <smiles>[O]OAsO</smiles> |\n| boric | $\\mathbf{H}_{3} \\mathrm{BO}_{3}$ | $5.4 \\times 10^{-10}$ | <smiles>OB(O)O</smiles> |\n| carbonic | $\\mathbf{H}_{2} \\mathrm{CO}_{3}$ | $4.3 \\times 10^{-7}$ | <smiles>O=C(O)O</smiles> |\n|  | $\\mathrm{HCO}_{3}{ }^{-}$ | $4.7 \\times 10^{-11}$ |  |\n| cyanic | HCNO | $2 \\times 10^{-4}$ | $\\ddot{\\mathrm{N}}=\\mathrm{C}=\\dot{0} \\text {. }$ |\n| formic | $\\mathrm{HCO}_{2} \\mathbf{H}$ | $1.8 \\times 10^{-4}$ | <smiles>OC=[Os]</smiles> |\n\nTABLE H1"}
{"id": 5288, "contents": "2270. Ionization Constants of Weak Acids - \n| Acid | Formula | $K_{a}$ at $25^{\\circ} \\mathrm{C}$ | Lewis Structure |\n| :---: | :---: | :---: | :---: |\n| hydrazoic | HN3 | $2.5 \\times 10^{-5}$ |  |\n| hydrocyanic | HCN | $4.9 \\times 10^{-10}$ |  |\n| hydrofluoric | HF | $6.4 \\times 10^{-4}$ |  |\n| hydrogen peroxide | $\\mathbf{H}_{2} \\mathrm{O}_{2}$ | $2.4 \\times 10^{-12}$ | $\\mathrm{H}-\\ddot{\\mathrm{O}}-\\ddot{\\mathrm{O}}-\\mathrm{H}$ |\n| hydrogen selenide | $\\mathrm{H}_{2} \\mathrm{Se}$ | $1.29 \\times 10^{-4}$ |  |\n|  | HSe ${ }^{-}$ | $1 \\times 10^{-12}$ |  |\n| hydrogen sulfate ion | $\\mathrm{HSO}_{4}{ }^{-}$ | $1.2 \\times 10^{-2}$ | <smiles>[O]S(=[Se])(=[Se])O[Te]</smiles> |\n| hydrogen sulfide | $\\mathrm{H}_{2} \\mathrm{~S}$ | $8.9 \\times 10^{-8}$ |  |\n|  | HS ${ }^{-}$ | $1.0 \\times 10^{-19}$ |  |\n| hydrogen telluride | $\\mathbf{H}_{\\mathbf{2}} \\mathbf{T e}$ | $2.3 \\times 10^{-3}$ |  |\n|  | HTe ${ }^{-}$ | $1.6 \\times 10^{-11}$ |  |\n| hypobromous | $\\mathbf{H B r O}$ | $2.8 \\times 10^{-9}$ |  |\n| hypochlorous | HClO | $2.9 \\times 10^{-8}$ |  |"}
{"id": 5289, "contents": "2270. Ionization Constants of Weak Acids - \n| hypochlorous | HClO | $2.9 \\times 10^{-8}$ |  |\n| nitrous | $\\mathbf{H N O} 2$ | $4.6 \\times 10^{-4}$ | <smiles>[O]N[O+]</smiles> |\n| oxalic | $\\mathbf{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ | $6.0 \\times 10^{-2}$ |  |\n|  | $\\mathrm{HC}_{2} \\mathrm{O}_{4}{ }^{-}$ | $6.1 \\times 10^{-5}$ |  |\n| phosphoric | $\\mathbf{H}_{3} \\mathrm{PO}_{4}$ | $7.5 \\times 10^{-3}$ | <smiles>O=[O+]P(=O)([O-])O</smiles> |\n|  | $\\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{-}$ | $6.2 \\times 10^{-8}$ |  |\n|  | $\\mathrm{HPO}_{4}{ }^{2-}$ | $4.2 \\times 10^{-13}$ |  |"}
{"id": 5290, "contents": "2271. TABLE H1 - \n| Acid | Formula | $K_{a}$ at $25^{\\circ} \\mathrm{C}$ | Lewis Structure |\n| :---: | :---: | :---: | :---: |\n| phosphorous | $\\mathbf{H}_{3} \\mathrm{PO}_{3}$ | $5 \\times 10^{-2}$ | <smiles>[O]PH([Se])[Se]</smiles> |\n|  | $\\mathrm{H}_{2} \\mathrm{PO}_{3}{ }^{-}$ | $2.0 \\times 10^{-7}$ |  |\n| sulfurous | $\\mathbf{H}_{2} \\mathrm{SO}_{3}$ | $1.6 \\times 10^{-2}$ | <smiles>[O]S([O])([O])[Os]</smiles> |\n|  | $\\mathrm{HSO}_{3}{ }^{-}$ | $6.4 \\times 10^{-8}$ |  |"}
{"id": 5291, "contents": "2272. TABLE H1 - \n1126 H \u2022 Ionization Constants of Weak Acids"}
{"id": 5292, "contents": "2274. Ionization Constants of Weak Bases - \nIonization Constants of Weak Bases\n\n| Base | Lewis Structure | $K_{\\text {b }}$ at $25{ }^{\\circ} \\mathrm{C}$ |\n| :---: | :---: | :---: |\n| ammonia | <smiles>[NH]</smiles> | $1.8 \\times 10^{-5}$ |\n| dimethylamine | <smiles>[1H]N+([2H])NC</smiles> | $5.9 \\times 10^{-4}$ |\n| methylamine | <smiles>C=[NH+]</smiles> | $4.4 \\times 10^{-4}$ |\n| phenylamine (aniline) | <smiles>c1ccccc1</smiles> | $4.3 \\times 10^{-10}$ |\n| trimethylamine | <smiles>CN1CCC1</smiles> | $6.3 \\times 10^{-5}$ |\n\nTABLE I1\n\n1128 I \u2022 Ionization Constants of Weak Bases"}
{"id": 5293, "contents": "2276. Solubility Products - \nSolubility Products\n\n| Substance | $K_{\\text {sp }}$ at $25^{\\circ} \\mathrm{C}$ |\n| :---: | :---: |\n| aluminum |  |\n| $\\mathrm{Al}(\\mathrm{OH})_{3}$ | $2 \\times 10^{-32}$ |\n| barium |  |\n| $\\mathrm{BaCO}_{3}$ | $1.6 \\times 10^{-9}$ |\n| $\\mathrm{BaC}_{2} \\mathrm{O}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$ | $1.1 \\times 10^{-7}$ |\n| $\\mathrm{BaSO}_{4}$ | $2.3 \\times 10^{-8}$ |\n| $\\mathrm{BaCrO}_{4}$ | $8.5 \\times 10^{-11}$ |\n| $\\mathrm{BaF}_{2}$ | $2.4 \\times 10^{-5}$ |\n| $\\mathrm{Ba}(\\mathrm{OH})_{2} \\cdot 8 \\mathrm{H}_{2} \\mathrm{O}$ | $5.0 \\times 10^{-3}$ |\n| $\\mathrm{Ba}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$ | $6 \\times 10^{-39}$ |\n| $\\mathrm{Ba}_{3}\\left(\\mathrm{AsO}_{4}\\right)_{2}$ | $1.1 \\times 10^{-13}$ |\n| bismuth |  |\n| BiO(OH) | $4 \\times 10^{-10}$ |\n| BiOCl | $1.8 \\times 10^{-31}$ |\n| $\\mathrm{Bi}_{2} \\mathrm{~S}_{3}$ | $1 \\times 10^{-97}$ |\n| cadmium |  |\n| $\\mathrm{Cd}(\\mathrm{OH})_{2}$ | $5.9 \\times 10^{-15}$ |\n| CdS | $1.0 \\times 10^{-28}$ |\n| $\\mathrm{CdCO}_{3}$ | $5.2 \\times 10^{-12}$ |\n\nTABLE J1"}
{"id": 5294, "contents": "2276. Solubility Products - \n| Substance | $K_{\\text {sp }}$ at $25^{\\circ} \\mathrm{C}$ |\n| :---: | :---: |\n| calcium |  |\n| $\\mathrm{Ca}(\\mathrm{OH})_{2}$ | $1.3 \\times 10^{-6}$ |\n| $\\mathrm{CaCO}_{3}$ | $8.7 \\times 10^{-9}$ |\n| $\\mathrm{CaSO} 4 \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$ | $6.1 \\times 10^{-5}$ |\n| $\\mathrm{CaC}_{2} \\mathrm{O}_{4} \\cdot \\mathrm{H}_{2} \\mathrm{O}$ | $1.96 \\times 10^{-8}$ |\n| $\\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$ | $1.3 \\times 10^{-32}$ |\n| $\\mathrm{CaHPO}_{4}$ | $7 \\times 10^{-7}$ |\n| $\\mathrm{CaF}_{2}$ | $4.0 \\times 10^{-11}$ |\n| chromium |  |\n| $\\mathrm{Cr}(\\mathrm{OH})_{3}$ | $6.7 \\times 10^{-31}$ |\n| cobalt |  |\n| $\\mathrm{Co}(\\mathrm{OH})_{2}$ | $2.5 \\times 10^{-16}$ |\n| $\\operatorname{CoS}(\\alpha)$ | $5 \\times 10^{-22}$ |\n| $\\operatorname{CoS}(\\beta)$ | $3 \\times 10^{-26}$ |\n| $\\mathrm{CoCO}_{3}$ | $1.4 \\times 10^{-13}$ |\n| $\\mathrm{Co}(\\mathrm{OH})_{3}$ | $2.5 \\times 10^{-43}$ |\n| copper |  |\n| CuCl | $1.2 \\times 10^{-6}$ |\n| CuBr | $6.27 \\times 10^{-9}$ |\n| CuI | $1.27 \\times 10^{-12}$ |\n| CuSCN | $1.6 \\times 10^{-11}$ |"}
{"id": 5295, "contents": "2276. Solubility Products - \n| CuI | $1.27 \\times 10^{-12}$ |\n| CuSCN | $1.6 \\times 10^{-11}$ |\n| $\\mathrm{Cu}_{2} \\mathrm{~S}$ | $2.5 \\times 10^{-48}$ |\n| $\\mathrm{Cu}(\\mathrm{OH})_{2}$ | $2.2 \\times 10^{-20}$ |\n| CuS | $8.5 \\times 10^{-45}$ |\n| TABLE J1 |  |"}
{"id": 5296, "contents": "2276. Solubility Products - \n| Substance | $K_{\\text {sp }}$ at $25^{\\circ} \\mathrm{C}$ |\n| :---: | :---: |\n| $\\mathrm{CuCO}_{3}$ | $2.5 \\times 10^{-10}$ |\n| iron |  |\n| $\\mathrm{Fe}(\\mathrm{OH})_{2}$ | $1.8 \\times 10^{-15}$ |\n| $\\mathrm{FeCO}_{3}$ | $2.1 \\times 10^{-11}$ |\n| FeS | $3.7 \\times 10^{-19}$ |\n| $\\mathrm{Fe}(\\mathrm{OH})_{3}$ | $4 \\times 10^{-38}$ |\n| lead |  |\n| $\\mathrm{Pb}(\\mathrm{OH})_{2}$ | $1.2 \\times 10^{-15}$ |\n| $\\mathrm{PbF}_{2}$ | $4 \\times 10^{-8}$ |\n| $\\mathrm{PbCl}_{2}$ | $1.6 \\times 10^{-5}$ |\n| $\\mathrm{PbBr}_{2}$ | $4.6 \\times 10^{-6}$ |\n| $\\mathrm{PbI}_{2}$ | $1.4 \\times 10^{-8}$ |\n| $\\mathrm{PbCO}_{3}$ | $1.5 \\times 10^{-15}$ |\n| PbS | $7 \\times 10^{-29}$ |\n| $\\mathrm{PbCrO}_{4}$ | $2 \\times 10^{-16}$ |\n| $\\mathrm{PbSO}_{4}$ | $1.3 \\times 10^{-8}$ |\n| $\\mathrm{Pb}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$ | $1 \\times 10^{-54}$ |\n| magnesium |  |\n| $\\mathrm{Mg}(\\mathrm{OH})_{2}$ | $8.9 \\times 10^{-12}$ |\n| $\\mathrm{MgCO}_{3} \\cdot 3 \\mathrm{H}_{2} \\mathrm{O}$ | ca $1 \\times 10^{-5}$ |"}
{"id": 5297, "contents": "2276. Solubility Products - \n| $\\mathrm{MgCO}_{3} \\cdot 3 \\mathrm{H}_{2} \\mathrm{O}$ | ca $1 \\times 10^{-5}$ |\n| $\\mathrm{MgNH}_{4} \\mathrm{PO}_{4}$ | $3 \\times 10^{-13}$ |\n| $\\mathrm{MgF}_{2}$ | $6.4 \\times 10^{-9}$ |\n| $\\mathrm{MgC}_{2} \\mathrm{O}_{4}$ | $7 \\times 10^{-7}$ |\n| manganese |  |"}
{"id": 5298, "contents": "2276. Solubility Products - \n| Substance | $K_{\\text {sp }}$ at $25^{\\circ} \\mathrm{C}$ |\n| :---: | :---: |\n| $\\mathrm{Mn}(\\mathrm{OH})_{2}$ | $2 \\times 10^{-13}$ |\n| $\\mathrm{MnCO}_{3}$ | $8.8 \\times 10^{-11}$ |\n| MnS | $2.3 \\times 10^{-13}$ |\n| mercury |  |\n| $\\mathrm{Hg}_{2} \\mathrm{O} \\cdot \\mathrm{H}_{2} \\mathrm{O}$ | $3.6 \\times 10^{-26}$ |\n| $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$ | $1.1 \\times 10^{-18}$ |\n| $\\mathrm{Hg}_{2} \\mathrm{Br}_{2}$ | $1.3 \\times 10^{-22}$ |\n| $\\mathrm{Hg}_{2} \\mathrm{I}_{2}$ | $4.5 \\times 10^{-29}$ |\n| $\\mathrm{Hg}_{2} \\mathrm{CO}_{3}$ | $9 \\times 10^{-15}$ |\n| $\\mathrm{Hg}_{2} \\mathrm{SO}_{4}$ | $7.4 \\times 10^{-7}$ |\n| $\\mathrm{Hg}_{2} \\mathrm{~S}$ | $1.0 \\times 10^{-47}$ |\n| $\\mathrm{Hg}_{2} \\mathrm{CrO}_{4}$ | $2 \\times 10^{-9}$ |\n| HgS | $1.6 \\times 10^{-54}$ |\n| nickel |  |\n| $\\mathrm{Ni}(\\mathrm{OH})_{2}$ | $1.6 \\times 10^{-16}$ |\n| $\\mathrm{NiCO}_{3}$ | $1.4 \\times 10^{-7}$ |\n| NiS( $\\alpha$ ) | $4 \\times 10^{-20}$ |\n| NiS( $\\beta$ ) | $1.3 \\times 10^{-25}$ |\n| potassium |  |"}
{"id": 5299, "contents": "2276. Solubility Products - \n| NiS( $\\alpha$ ) | $4 \\times 10^{-20}$ |\n| NiS( $\\beta$ ) | $1.3 \\times 10^{-25}$ |\n| potassium |  |\n| $\\mathrm{KClO}_{4}$ | $1.05 \\times 10^{-2}$ |\n| $\\mathrm{K}_{2} \\mathrm{PtCl}_{6}$ | $7.48 \\times 10^{-6}$ |\n| $\\mathrm{KHC}_{4} \\mathrm{H}_{4} \\mathrm{O}_{6}$ | $3 \\times 10^{-4}$ |\n| silver |  |\n| ${ }_{2}^{1} \\mathrm{Ag}_{2} \\mathrm{O}\\left(\\mathrm{Ag}^{+}+\\mathrm{OH}^{-}\\right)$ | $2 \\times 10^{-8}$ |"}
{"id": 5300, "contents": "2276. Solubility Products - \nTABLE J1"}
{"id": 5301, "contents": "2276. Solubility Products - \n| Substance | $K_{\\text {sp }}$ at $25^{\\circ} \\mathrm{C}$ |\n| :---: | :---: |\n| AgCl | $1.6 \\times 10^{-10}$ |\n| AgBr | $5.0 \\times 10^{-13}$ |\n| AgI | $1.5 \\times 10^{-16}$ |\n| AgCN | $1.2 \\times 10^{-16}$ |\n| AgSCN | $1.0 \\times 10^{-12}$ |\n| $\\mathrm{Ag}_{2} \\mathrm{~S}$ | $1.6 \\times 10^{-49}$ |\n| $\\mathrm{Ag}_{2} \\mathrm{CO}_{3}$ | $8.1 \\times 10^{-12}$ |\n| $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}$ | $9.0 \\times 10^{-12}$ |\n| $\\mathrm{Ag}_{4} \\mathrm{Fe}(\\mathrm{CN})_{6}$ | $1.55 \\times 10^{-41}$ |\n| $\\mathrm{Ag}_{2} \\mathrm{SO}_{4}$ | $1.2 \\times 10^{-5}$ |\n| $\\mathrm{Ag}_{3} \\mathrm{PO}_{4}$ | $1.8 \\times 10^{-18}$ |\n| strontium |  |\n| $\\mathrm{Sr}(\\mathrm{OH})_{2} \\cdot 8 \\mathrm{H}_{2} \\mathrm{O}$ | $3.2 \\times 10^{-4}$ |\n| $\\mathrm{SrCO}_{3}$ | $7 \\times 10^{-10}$ |\n| $\\mathrm{SrCrO}_{4}$ | $3.6 \\times 10^{-5}$ |\n| $\\mathrm{SrSO}_{4}$ | $3.2 \\times 10^{-7}$ |\n| $\\mathrm{SrC}_{2} \\mathrm{O}_{4} \\cdot \\mathrm{H}_{2} \\mathrm{O}$ | $4 \\times 10^{-7}$ |\n| thallium |  |\n| TlCl | $1.7 \\times 10^{-4}$ |"}
{"id": 5302, "contents": "2276. Solubility Products - \n| thallium |  |\n| TlCl | $1.7 \\times 10^{-4}$ |\n| TISCN | $1.6 \\times 10^{-4}$ |\n| $\\mathrm{Tl}_{2} \\mathrm{~S}$ | $6 \\times 10^{-22}$ |\n| $\\mathrm{Tl}(\\mathrm{OH})_{3}$ | $6.3 \\times 10^{-46}$ |\n| tin |  |\n| $\\mathrm{Sn}(\\mathrm{OH})_{2}$ | $3 \\times 10^{-27}$ |\n| TABLE J1 |  |"}
{"id": 5303, "contents": "2276. Solubility Products - \n| Substance | $K_{\\text {sp }}$ at $25^{\\circ} \\mathrm{C}$ |\n| :--- | :--- |\n| SnS | $1 \\times 10^{-26}$ |\n| $\\mathrm{Sn}(\\mathrm{OH})_{4}$ | $1.0 \\times 10^{-57}$ |\n| zinc |  |\n| $\\mathrm{ZnCO}_{3}$ | $2 \\times 10^{-10}$ |\n\nTABLE J1"}
{"id": 5304, "contents": "2277. APPENDIX K - \nFormation Constants for Complex Ions\n\nFormation Constants for Complex Ions"}
{"id": 5305, "contents": "2277. APPENDIX K - \n| Equilibrium | $K_{f}$ |\n| :---: | :---: |\n| $\\mathrm{Al}^{3+}+6 \\mathrm{~F}^{-} \\rightleftharpoons\\left[\\mathrm{AlF}_{6}\\right]^{3-}$ | $7 \\times 10^{19}$ |\n| $\\mathrm{Cd}^{2+}+4 \\mathrm{NH}_{3} \\rightleftharpoons\\left[\\mathrm{Cd}\\left(\\mathrm{NH}_{3}\\right)_{4}\\right]^{2+}$ | $1.3 \\times 10^{7}$ |\n| $\\mathrm{Cd}^{2+}+4 \\mathrm{CN}^{-} \\rightleftharpoons\\left[\\mathrm{Cd}(\\mathrm{CN})_{4}\\right]^{2-}$ | $3 \\times 10^{18}$ |\n| $\\mathrm{Co}^{2+}+6 \\mathrm{NH}_{3} \\rightleftharpoons\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]^{2+}$ | $1.3 \\times 10^{5}$ |\n| $\\mathrm{Co}^{3+}+6 \\mathrm{NH}_{3} \\rightleftharpoons\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]^{3+}$ | $2.3 \\times 10^{33}$ |\n| $\\mathrm{Cu}^{+}+2 \\mathrm{CN} \\rightleftharpoons\\left[\\mathrm{Cu}(\\mathrm{CN})_{2}\\right]^{-}$ | $1.0 \\times 10^{16}$ |\n| $\\mathrm{Cu}^{2+}+4 \\mathrm{NH}_{3} \\rightleftharpoons\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{4}\\right]^{2+}$ | $1.7 \\times 10^{13}$ |"}
{"id": 5306, "contents": "2277. APPENDIX K - \n| $\\mathrm{Fe}^{2+}+6 \\mathrm{CN}^{-} \\rightleftharpoons\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{4-}$ | $1.5 \\times 10^{35}$ |\n| $\\mathrm{Fe}^{3+}+6 \\mathrm{CN}^{-} \\rightleftharpoons\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{3-}$ | $2 \\times 10^{43}$ |\n| $\\mathrm{Fe}^{3+}+6 \\mathrm{SCN}^{-} \\rightleftharpoons\\left[\\mathrm{Fe}(\\mathrm{SCN})_{6}\\right]^{3-}$ | $3.2 \\times 10^{3}$ |\n| $\\mathrm{Hg}^{2+}+4 \\mathrm{Cl}^{-} \\rightleftharpoons\\left[\\mathrm{HgCl}_{4}\\right]^{2-}$ | $1.1 \\times 10^{16}$ |\n| $\\mathrm{Ni}^{2+}+6 \\mathrm{NH}_{3} \\rightleftharpoons\\left[\\mathrm{Ni}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]^{2+}$ | $2.0 \\times 10^{8}$ |\n| $\\mathrm{Ag}^{+}+2 \\mathrm{Cl}^{-} \\rightleftharpoons\\left[\\mathrm{AgCl}_{2}\\right]^{-}$ | $1.8 \\times 10^{5}$ |\n| $\\mathrm{Ag}^{+}+2 \\mathrm{CN}^{-} \\rightleftharpoons\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]^{-}$ | $1 \\times 10^{21}$ |\n| $\\mathrm{Ag}^{+}+2 \\mathrm{NH}_{3} \\rightleftharpoons\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}$ | $1.7 \\times 10^{7}$ |"}
{"id": 5307, "contents": "2277. APPENDIX K - \n| $\\mathrm{Zn}^{2+}+4 \\mathrm{CN}^{-} \\rightleftharpoons\\left[\\mathrm{Zn}(\\mathrm{CN})_{4}\\right]^{2-}$ | $2.1 \\times 10^{19}$ |\n| $\\mathrm{Zn}^{2+}+4 \\mathrm{OH}^{-} \\rightleftharpoons\\left[\\mathrm{Zn}(\\mathrm{OH})_{4}\\right]^{2-}$ | $2 \\times 10^{15}$ |"}
{"id": 5308, "contents": "2277. APPENDIX K - \nTABLE K1\n\n| Equilibrium | $K_{\\mathrm{f}}$ |\n| :--- | :--- |\n| $\\mathrm{Fe}^{3+}+\\mathrm{SCN}^{-} \\rightleftharpoons[\\mathrm{Fe}(\\mathrm{SCN})]^{2+}$ | $8.9 \\times 10^{2}$ |\n| $\\mathrm{Ag}^{+}+4 \\mathrm{SCN}^{-} \\rightleftharpoons\\left[\\mathrm{Ag}(\\mathrm{SCN})_{4}\\right]^{3-}$ | $1.2 \\times 10^{10}$ |\n| $\\mathrm{~Pb}^{2+}+4 \\mathrm{I}^{-} \\rightleftharpoons\\left[\\mathrm{PbI}_{4}\\right]^{2-}$ | $3.0 \\times 10^{4}$ |\n| $\\mathrm{Pt}^{2+}+4 \\mathrm{Cl}^{-} \\rightleftharpoons\\left[\\mathrm{PtCl}_{4}\\right]^{2-}$ | $1 \\times 10^{16}$ |\n| $\\mathrm{Cu}^{2+}+4 \\mathrm{CN} \\rightleftharpoons\\left[\\mathrm{Cu}(\\mathrm{CN})_{4}\\right]^{2-}$ | $1.0 \\times 10^{25}$ |\n| $\\mathrm{Co}^{2+}+4 \\mathrm{SCN}^{-} \\rightleftharpoons\\left[\\mathrm{Co}(\\mathrm{SCN})_{4}\\right]^{2-}$ | $1 \\times 10^{3}$ |\n\nTABLE K1"}
{"id": 5309, "contents": "2279. Standard Electrode (Half-Cell) Potentials - \nStandard Electrode (Half-Cell) Potentials"}
{"id": 5310, "contents": "2279. Standard Electrode (Half-Cell) Potentials - \n| Half-Reaction | $E^{\\circ}(\\mathrm{V})$ |\n| :---: | :---: |\n| $\\mathrm{Ag}^{+}+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Ag}$ | +0.7996 |\n| $\\mathrm{AgCl}+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Ag}+\\mathrm{Cl}^{-}$ | +0.22233 |\n| $\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]^{-}+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Ag}+2 \\mathrm{CN}^{-}$ | -0.31 |\n| $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Ag}+\\mathrm{CrO}_{4}{ }^{2-}$ | +0.45 |\n| $\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Ag}+2 \\mathrm{NH}_{3}$ | +0.373 |\n| $\\left[\\mathrm{Ag}\\left(\\mathrm{S}_{2} \\mathrm{O}_{3}\\right)_{2}\\right]^{3+}+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Ag}+2 \\mathrm{~S}_{2} \\mathrm{O}_{3}{ }^{2-}$ | +0.017 |\n| $\\left[\\mathrm{AlF}_{6}\\right]^{3-}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Al}+6 \\mathrm{~F}^{-}$ | -2.07 |\n| $\\mathrm{Al}^{3+}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Al}$ | -1.662 |\n| $\\mathrm{Am}^{3+}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Am}$ | -2.048 |\n| $\\mathrm{Au}^{3+}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Au}$ | +1.498 |"}
{"id": 5311, "contents": "2279. Standard Electrode (Half-Cell) Potentials - \n| $\\mathrm{Au}^{3+}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Au}$ | +1.498 |\n| $\\mathrm{Au}^{+}+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Au}$ | +1.692 |\n| $\\mathrm{Ba}^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Ba}$ | -2.912 |\n| $\\mathrm{Be}^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Be}$ | -1.847 |\n| $\\mathrm{Br}_{2}(a q)+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Br}^{-}$ | +1.0873 |\n| $\\mathrm{Ca}^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Ca}$ | -2.868 |\n| $\\mathrm{Ce}^{3}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Ce}$ | -2.483 |\n| $\\mathrm{Ce}^{4+}+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Ce}^{3+}$ | +1.61 |\n| $\\mathrm{Cd}^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Cd}$ | -0.4030 |\n| $\\left[\\mathrm{Cd}(\\mathrm{CN})_{4}\\right]^{2-}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Cd}+4 \\mathrm{CN}^{-}$ | -1.09 |"}
{"id": 5312, "contents": "2279. Standard Electrode (Half-Cell) Potentials - \nTABLE L1"}
{"id": 5313, "contents": "2279. Standard Electrode (Half-Cell) Potentials - \n| Half-Reaction | $E^{0}(\\mathrm{~V})$ |\n| :---: | :---: |\n| $\\left[\\mathrm{Cd}\\left(\\mathrm{NH}_{3}\\right)_{4}\\right]^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Cd}+4 \\mathrm{NH}_{3}$ | -0.61 |\n| $\\mathrm{CdS}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Cd}+\\mathrm{S}^{2-}$ | -1.17 |\n| $\\mathrm{Cl}_{2}+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Cl}^{-}$ | +1.35827 |\n| $\\mathrm{ClO}_{4}^{-}+\\mathrm{H}_{2} \\mathrm{O}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{ClO}_{3}^{-}+2 \\mathrm{OH}^{-}$ | +0.36 |\n| $\\mathrm{ClO}_{3}^{-}+\\mathrm{H}_{2} \\mathrm{O}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{ClO}_{2}^{-}+2 \\mathrm{OH}^{-}$ | +0.33 |\n| $\\mathrm{ClO}_{2}^{-}+\\mathrm{H}_{2} \\mathrm{O}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{ClO}^{-}+2 \\mathrm{OH}^{-}$ | +0.66 |\n| $\\mathrm{ClO}^{-}+\\mathrm{H}_{2} \\mathrm{O}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Cl}^{-}+2 \\mathrm{OH}^{-}$ | +0.89 |\n| $\\mathrm{ClO}_{4}^{-}+2 \\mathrm{H}_{3} \\mathrm{O}^{+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{ClO}_{3}^{-}+3 \\mathrm{H}_{2} \\mathrm{O}$ | +1.189 |"}
{"id": 5314, "contents": "2279. Standard Electrode (Half-Cell) Potentials - \n| $\\mathrm{ClO}_{3}^{-}+3 \\mathrm{H}_{3} \\mathrm{O}^{+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{HClO}_{2}+4 \\mathrm{H}_{2} \\mathrm{O}$ | +1.21 |\n| $\\mathrm{HClO}+\\mathrm{H}_{3} \\mathrm{O}^{+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Cl}^{-}+2 \\mathrm{H}_{2} \\mathrm{O}$ | +1.482 |\n| $\\mathrm{HClO}+\\mathrm{H}_{3} \\mathrm{O}^{+}+\\mathrm{e}^{-} \\longrightarrow \\frac{1}{2} \\mathrm{Cl}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}$ | +1.611 |\n| $\\mathrm{HClO}_{2}+2 \\mathrm{H}_{3} \\mathrm{O}^{+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{HClO}+3 \\mathrm{H}_{2} \\mathrm{O}$ | +1.628 |\n| $\\mathrm{Co}^{3+}+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Co}^{2+}\\left(2 \\mathrm{~mol} / / \\mathrm{H}_{2} \\mathrm{SO}_{4}\\right)$ | +1.83 |\n| $\\mathrm{Co}^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Co}$ | -0.28 |\n| $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]^{3+}+\\mathrm{e}^{-} \\longrightarrow\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]^{2+}$ | +0.1 |\n| $\\mathrm{Co}(\\mathrm{OH})_{3}+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Co}(\\mathrm{OH})_{2}+\\mathrm{OH}^{-}$ | +0.17 |"}
{"id": 5315, "contents": "2279. Standard Electrode (Half-Cell) Potentials - \n| $\\mathrm{Co}(\\mathrm{OH})_{3}+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Co}(\\mathrm{OH})_{2}+\\mathrm{OH}^{-}$ | +0.17 |\n| $\\mathrm{Cr}^{3}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Cr}$ | -0.744 |\n| $\\mathrm{Cr}^{3+}+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Cr}^{2+}$ | -0.407 |\n| $\\mathrm{Cr}^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Cr}$ | -0.913 |\n| $\\left[\\mathrm{Cu}(\\mathrm{CN})_{2}\\right]^{-}+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Cu}+2 \\mathrm{CN}^{-}$ | -0.43 |\n| $\\mathrm{CrO}_{4}{ }^{2-}+4 \\mathrm{H}_{2} \\mathrm{O}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Cr}(\\mathrm{OH})_{3}+5 \\mathrm{OH}^{-}$ | -0.13 |\n| $\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+14 \\mathrm{H}_{3} \\mathrm{O}^{+}+6 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Cr}^{3+}+21 \\mathrm{H}_{2} \\mathrm{O}$ | +1.232 |\n| $\\left[\\mathrm{Cr}(\\mathrm{OH})_{4}\\right]^{-}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Cr}+4 \\mathrm{OH}^{-}$ | -1.2 |"}
{"id": 5316, "contents": "2279. Standard Electrode (Half-Cell) Potentials - \nTABLE L1"}
{"id": 5317, "contents": "2279. Standard Electrode (Half-Cell) Potentials - \n| Half-Reaction | $E^{\\circ}(\\mathrm{V})$ |\n| :---: | :---: |\n| $\\mathrm{Cr}(\\mathrm{OH})_{3}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Cr}+3 \\mathrm{OH}^{-}$ | -1.48 |\n| $\\mathrm{Cu}^{2+}+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Cu}^{+}$ | +0.153 |\n| $\\mathrm{Cu}^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Cu}$ | +0.34 |\n| $\\mathrm{Cu}^{+}+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Cu}$ | $+0.521$ |\n| $\\mathrm{F}_{2}+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{~F}^{-}$ | +2.866 |\n| $\\mathrm{Fe}^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Fe}$ | -0.447 |\n| $\\mathrm{Fe}^{3+}+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Fe}^{2+}$ | +0.771 |\n| $\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{3-}+\\mathrm{e}^{-} \\longrightarrow\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{4-}$ | +0.36 |\n| $\\mathrm{Fe}(\\mathrm{OH})_{2}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Fe}+2 \\mathrm{OH}^{-}$ | -0.88 |\n| $\\mathrm{FeS}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Fe}+\\mathrm{S}^{2-}$ | -1.01 |\n| $\\mathrm{Ga}^{3+}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Ga}$ | -0.549 |\n| $\\mathrm{Gd}^{3+}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Gd}$ | -2.279 |"}
{"id": 5318, "contents": "2279. Standard Electrode (Half-Cell) Potentials - \n| $\\mathrm{Gd}^{3+}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Gd}$ | -2.279 |\n| $\\frac{1}{2} \\mathrm{H}_{2}+\\mathrm{e}^{-} \\longrightarrow \\mathrm{H}^{-}$ | -2.23 |\n| $2 \\mathrm{H}_{2} \\mathrm{O}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{H}_{2}+2 \\mathrm{OH}^{-}$ | -0.8277 |\n| $\\mathrm{H}_{2} \\mathrm{O}_{2}+2 \\mathrm{H}_{3} \\mathrm{O}^{+}+2 \\mathrm{e}^{-} \\longrightarrow 4 \\mathrm{H}_{2} \\mathrm{O}$ | +1.776 |\n| $2 \\mathrm{H}_{3} \\mathrm{O}^{+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{H}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}$ | 0.00 |\n| $\\mathrm{HO}_{2}^{-}+\\mathrm{H}_{2} \\mathrm{O}+2 \\mathrm{e}^{-} \\longrightarrow 3 \\mathrm{OH}^{-}$ | +0.878 |\n| $\\mathrm{Hf}^{4+}+4 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Hf}$ | -1.55 |\n| $\\mathrm{Hg}^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Hg}$ | +0.851 |\n| $2 \\mathrm{Hg}^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Hg}_{2}{ }^{2+}$ | +0.92 |\n| $\\mathrm{Hg}_{2}{ }^{2+}+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Hg}$ | +0.7973 |\n| $\\left[\\mathrm{HgBr}_{4}\\right]^{2-}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Hg}+4 \\mathrm{Br}^{-}$ | +0.21 |"}
{"id": 5319, "contents": "2279. Standard Electrode (Half-Cell) Potentials - \n| $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Hg}+2 \\mathrm{Cl}^{-}$ | +0.26808 |"}
{"id": 5320, "contents": "2279. Standard Electrode (Half-Cell) Potentials - \nTABLE L1"}
{"id": 5321, "contents": "2279. Standard Electrode (Half-Cell) Potentials - \n| Half-Reaction | $E^{\\circ}(\\mathrm{V})$ |\n| :---: | :---: |\n| $\\left[\\mathrm{Hg}(\\mathrm{CN})_{4}\\right]^{2-}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Hg}+4 \\mathrm{CN}^{-}$ | -0.37 |\n| $\\left[\\mathrm{HgI}_{4}\\right]^{2-}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Hg}+4 \\mathrm{I}^{-}$ | -0.04 |\n| $\\mathrm{HgS}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Hg}+\\mathrm{S}^{2-}$ | -0.70 |\n| $\\mathrm{I}_{2}+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{I}^{-}$ | +0.5355 |\n| $\\mathrm{In}^{3+}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{In}$ | -0.3382 |\n| $\\mathrm{K}^{+}+\\mathrm{e}^{-} \\longrightarrow \\mathrm{K}$ | -2.931 |\n| $\\mathrm{La}^{3+}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{La}$ | -2.52 |\n| $\\mathrm{Li}^{+}+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Li}$ | -3.04 |\n| $\\mathrm{Lu}^{3+}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Lu}$ | $-2.28$ |\n| $\\mathrm{Mg}^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Mg}$ | $-2.372$ |\n| $\\mathrm{Mn}^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Mn}$ | -1.185 |"}
{"id": 5322, "contents": "2279. Standard Electrode (Half-Cell) Potentials - \n| $\\mathrm{Mn}^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Mn}$ | -1.185 |\n| $\\mathrm{MnO}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Mn}(\\mathrm{OH})_{2}+2 \\mathrm{OH}^{-}$ | -0.05 |\n| $\\mathrm{MnO}_{4}^{-}+2 \\mathrm{H}_{2} \\mathrm{O}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{MnO}_{2}+4 \\mathrm{OH}^{-}$ | +0.558 |\n| $\\mathrm{MnO}_{2}+4 \\mathrm{H}^{+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Mn}^{2+}+2 \\mathrm{H}_{2} \\mathrm{O}$ | +1.23 |\n| $\\mathrm{MnO}_{4}^{-}+8 \\mathrm{H}^{+}+5 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Mn}^{2+}+4 \\mathrm{H}_{2} \\mathrm{O}$ | +1.507 |\n| $\\mathrm{Na}^{+}+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Na}$ | $-2.71$ |\n| $\\mathrm{Nd}^{3+}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Nd}$ | -2.323 |\n| $\\mathrm{Ni}^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Ni}$ | -0.257 |\n| $\\left[\\mathrm{Ni}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Ni}+6 \\mathrm{NH}_{3}$ | -0.49 |"}
{"id": 5323, "contents": "2279. Standard Electrode (Half-Cell) Potentials - \n| $\\mathrm{NiO}_{2}+4 \\mathrm{H}^{+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Ni}^{2+}+2 \\mathrm{H}_{2} \\mathrm{O}$ | +1.593 |\n| $\\mathrm{NiO}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Ni}(\\mathrm{OH})_{2}+2 \\mathrm{OH}^{-}$ | +0.49 |\n| $\\mathrm{NiS}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Ni}+\\mathrm{S}^{2-}$ | +0.76 |\n| $\\mathrm{NO}_{3}{ }^{-}+4 \\mathrm{H}^{+}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{NO}+2 \\mathrm{H}_{2} \\mathrm{O}$ | +0.957 |\n| $\\mathrm{NO}_{3}{ }^{-}+3 \\mathrm{H}^{+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{HNO}_{2}+\\mathrm{H}_{2} \\mathrm{O}$ | +0.92 |"}
{"id": 5324, "contents": "2279. Standard Electrode (Half-Cell) Potentials - \nTABLE L1"}
{"id": 5325, "contents": "2279. Standard Electrode (Half-Cell) Potentials - \n| Half-Reaction | $E^{\\circ}(\\mathrm{V})$ |\n| :---: | :---: |\n| $\\mathrm{NO}_{3}{ }^{-}+\\mathrm{H}_{2} \\mathrm{O}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{NO}_{2}{ }^{-}+2 \\mathrm{OH}^{-}$ | +0.10 |\n| $\\mathrm{Np}^{3+}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Np}$ | -1.856 |\n| $\\mathrm{O}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}+4 \\mathrm{e}^{-} \\longrightarrow 4 \\mathrm{OH}^{-}$ | +0.401 |\n| $\\mathrm{O}_{2}+2 \\mathrm{H}^{+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}_{2}$ | +0.695 |\n| $\\mathrm{O}_{2}+4 \\mathrm{H}^{+}+4 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{H}_{2} \\mathrm{O}$ | +1.229 |\n| $\\mathrm{Pb}^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Pb}$ | -0.1262 |\n| $\\mathrm{PbO}_{2}+\\mathrm{SO}_{4}{ }^{2-}+4 \\mathrm{H}^{+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{PbSO}_{4}+2 \\mathrm{H}_{2} \\mathrm{O}$ | +1.69 |\n| $\\mathrm{PbS}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Pb}+\\mathrm{S}^{2-}$ | -0.95 |\n| $\\mathrm{PbSO}_{4}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Pb}+\\mathrm{SO}_{4}{ }^{2-}$ | -0.3505 |"}
{"id": 5326, "contents": "2279. Standard Electrode (Half-Cell) Potentials - \n| $\\mathrm{PbSO}_{4}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Pb}+\\mathrm{SO}_{4}{ }^{2-}$ | -0.3505 |\n| $\\mathrm{Pd}^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Pd}$ | +0.987 |\n| $\\left[\\mathrm{PdCl}_{4}\\right]^{2-}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Pd}+4 \\mathrm{Cl}^{-}$ | +0.591 |\n| $\\mathrm{Pt}^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Pt}$ | +1.20 |\n| $\\left[\\mathrm{PtBr}_{4}\\right]^{2-}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Pt}+4 \\mathrm{Br}^{-}$ | +0.58 |\n| $\\left[\\mathrm{PtCl}_{4}\\right]^{2-}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Pt}+4 \\mathrm{Cl}^{-}$ | +0.755 |\n| $\\left[\\mathrm{PtCl}_{6}\\right]^{2-}+2 \\mathrm{e}^{-} \\longrightarrow\\left[\\mathrm{PtCl}_{4}\\right]^{2-}+2 \\mathrm{Cl}^{-}$ | +0.68 |\n| $\\mathrm{Pu}^{3}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Pu}$ | -2.03 |\n| $\\mathrm{Ra}^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Ra}$ | -2.92 |\n| $\\mathrm{Rb}^{+}+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Rb}$ | -2.98 |\n| $\\left[\\mathrm{RhCl}_{6}\\right]^{3-}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Rh}+6 \\mathrm{Cl}^{-}$ | +0.44 |"}
{"id": 5327, "contents": "2279. Standard Electrode (Half-Cell) Potentials - \n| $\\left[\\mathrm{RhCl}_{6}\\right]^{3-}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Rh}+6 \\mathrm{Cl}^{-}$ | +0.44 |\n| $\\mathrm{S}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{S}^{2-}$ | -0.47627 |\n| $\\mathrm{S}+2 \\mathrm{H}^{+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{H}_{2} \\mathrm{~S}$ | +0.142 |\n| $\\mathrm{Sc}^{3+}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Sc}$ | -2.09 |\n| $\\mathrm{Se}+2 \\mathrm{H}^{+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{H}_{2} \\mathrm{Se}$ | -0.399 |\n| $\\left[\\mathrm{SiF}_{6}\\right]^{2-}+4 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Si}+6 \\mathrm{~F}^{-}$ | -1.2 |"}
{"id": 5328, "contents": "2279. Standard Electrode (Half-Cell) Potentials - \nTABLE L1"}
{"id": 5329, "contents": "2279. Standard Electrode (Half-Cell) Potentials - \n| Half-Reaction | $E^{\\circ}(\\mathrm{V})$ |\n| :---: | :---: |\n| $\\mathrm{SiO}_{3}{ }^{2-}+3 \\mathrm{H}_{2} \\mathrm{O}+4 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Si}+6 \\mathrm{OH}^{-}$ | -1.697 |\n| $\\mathrm{SiO}_{2}+4 \\mathrm{H}^{+}+4 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Si}+2 \\mathrm{H}_{2} \\mathrm{O}$ | -0.86 |\n| $\\mathrm{Sm}^{3+}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Sm}$ | -2.304 |\n| $\\mathrm{Sn}^{4+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Sn}^{2+}$ | +0.151 |\n| $\\mathrm{Sn}^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Sn}$ | -0.1375 |\n| $\\left[\\mathrm{SnF}_{6}\\right]^{2-}+4 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Sn}+6 \\mathrm{~F}^{-}$ | -0.25 |\n| $\\mathrm{SnS}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Sn}+\\mathrm{S}^{2-}$ | -0.94 |\n| $\\mathrm{Sr}^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Sr}$ | -2.89 |\n| $\\mathrm{TeO}_{2}+4 \\mathrm{H}^{+}+4 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Te}+2 \\mathrm{H}_{2} \\mathrm{O}$ | +0.593 |\n| $\\mathrm{Th}^{4+}+4 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Th}$ | -1.90 |\n| $\\mathrm{Ti}^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Ti}$ | -1.630 |"}
{"id": 5330, "contents": "2279. Standard Electrode (Half-Cell) Potentials - \n| $\\mathrm{Ti}^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Ti}$ | -1.630 |\n| $\\mathrm{U}^{3+}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{U}$ | -1.79 |\n| $\\mathrm{V}^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{V}$ | -1.19 |\n| $\\mathrm{Y}^{3+}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Y}$ | -2.37 |\n| $\\mathrm{Zn}^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Zn}$ | -0.7618 |\n| $\\left[\\mathrm{Zn}(\\mathrm{CN})_{4}\\right]^{2-}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Zn}+4 \\mathrm{CN}^{-}$ | -1.26 |\n| $\\left[\\mathrm{Zn}\\left(\\mathrm{NH}_{3}\\right)_{4}\\right]^{2+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Zn}+4 \\mathrm{NH}_{3}$ | -1.04 |\n| $\\mathrm{Zn}(\\mathrm{OH})_{2}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Zn}+2 \\mathrm{OH}^{-}$ | -1.245 |\n| $\\left[\\mathrm{Zn}(\\mathrm{OH})_{4}\\right]^{2}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Zn}+4 \\mathrm{OH}^{-}$ | -1.199 |\n| $\\mathrm{ZnS}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Zn}+\\mathrm{S}^{2-}$ | -1.40 |\n| $\\mathrm{Zr}^{4}+4 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Zr}$ | -1.539 |"}
{"id": 5331, "contents": "2281. Half-Lives for Several Radioactive Isotopes - \nHalf-Lives for Several Radioactive Isotopes"}
{"id": 5332, "contents": "2281. Half-Lives for Several Radioactive Isotopes - \n| Isotope | Half-Life ${ }^{1}$ | Type of Emission ${ }^{2}$ | Isotope | Half-Life ${ }^{3}$ | Type of Emission ${ }^{4}$ |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| ${ }_{6}^{14} \\mathrm{C}$ | 5730 y | $\\left(\\beta^{-}\\right)$ | ${ }_{83}^{210} \\mathrm{Bi}$ | 5.01 d | $\\left(\\beta^{-}\\right)$ |\n| ${ }_{7}^{13} \\mathrm{~N}$ | 9.97 m | $\\left(\\beta^{+}\\right)$ | ${ }_{83}^{212} \\mathrm{Bi}$ | 60.55 m | ( $\\alpha$ or $\\beta^{-}$) |\n| ${ }_{9}^{15} \\mathrm{~F}$ | $4.1 \\times 10^{-22} \\mathrm{~s}$ | (p) | ${ }_{84}^{210} \\mathrm{Po}$ | 138.4 d | ( $)$ |\n| ${ }_{11}^{24} \\mathrm{Na}$ | 15.00 h | $\\left(\\beta^{-}\\right)$ | ${ }_{84}^{212} \\mathrm{Po}$ | $3 \\times 10^{-7} \\mathrm{~s}$ | ( $)$ |\n| ${ }_{15}^{32} \\mathrm{P}$ | 14.29 d | $\\left(\\beta^{-}\\right)$ | ${ }_{84}^{216} \\mathrm{Po}$ | 0.15 s | ( $)$ |\n| ${ }_{19}^{40} \\mathrm{~K}$ | $1.27 \\times 10^{9} \\mathrm{y}$ | ( $\\beta$ or E.C.) | ${ }_{84}^{218} \\mathrm{Po}$ | 3.05 m | ( $\\alpha$ ) |"}
{"id": 5333, "contents": "2281. Half-Lives for Several Radioactive Isotopes - \n| ${ }_{26}^{49} \\mathrm{Fe}$ | 0.08 s | $\\left(\\beta^{+}\\right)$ | ${ }_{85}^{215} \\mathrm{At}$ | $1.0 \\times 10^{-4} \\mathrm{~s}$ | ( $)$ |\n| ${ }_{26}^{60} \\mathrm{Fe}$ | $2.6 \\times 10^{6} \\mathrm{y}$ | $\\left(\\beta^{-}\\right)$ | ${ }_{85}^{218} \\mathrm{At}$ | 1.6 s | ( $)$ |\n| ${ }_{27}^{60} \\mathrm{Co}$ | 5.27 y | $\\left(\\beta^{-}\\right)$ | ${ }_{86}^{220} \\mathrm{Rn}$ | 55.6 s | ( $\\alpha$ ) |\n| ${ }_{37}^{87} \\mathrm{Rb}$ | $4.7 \\times 10^{10} \\mathrm{y}$ | $\\left(\\beta^{-}\\right)$ | ${ }_{86}^{222} \\mathrm{Rn}$ | 3.82 d | ( $\\alpha$ ) |\n| ${ }_{38}^{90} \\mathrm{Sr}$ | 29 y | $\\left(\\beta^{-}\\right)$ | ${ }_{88}^{224} \\mathrm{Ra}$ | 3.66 d | ( $)$ |\n| ${ }_{49}^{115} \\mathrm{In}$ | $5.1 \\times 10^{15} \\mathrm{y}$ | $\\left(\\beta^{-}\\right)$ | ${ }_{88}^{226} \\mathrm{Ra}$ | 1600 y | ( $\\alpha$ ) |\n| ${ }_{53}^{131} \\mathrm{I}$ | 8.040 d | $\\left(\\beta^{-}\\right)$ | ${ }_{88}^{228} \\mathrm{Ra}$ | 5.75 y | $\\left(\\beta^{-}\\right)$ |"}
{"id": 5334, "contents": "2281. Half-Lives for Several Radioactive Isotopes - \n| ${ }_{58}^{142} \\mathrm{Ce}$ | $5 \\times 10^{15} \\mathrm{y}$ | ( $\\alpha$ ) | ${ }_{89}^{228} \\mathrm{Ac}$ | 6.13 h | $\\left(\\beta^{-}\\right)$ |\n| ${ }_{81}^{208} \\mathrm{Tl}$ | 3.07 m | $\\left(\\beta^{-}\\right)$ | ${ }_{90}^{228} \\mathrm{Th}$ | 1.913 y | ( $\\alpha$ ) |\n| ${ }_{82}^{210} \\mathrm{~Pb}$ | 22.3 y | $\\left(\\beta^{-}\\right)$ | ${ }_{90}^{232} \\mathrm{Th}$ | $1.4 \\times 10^{10} \\mathrm{y}$ | ( $\\alpha$ ) |\n| ${ }_{82}^{212} \\mathrm{~Pb}$ | 10.6 h | $\\left(\\beta^{-}\\right)$ | ${ }_{90}^{233} \\mathrm{Th}$ | 22 m | $\\left(\\beta^{-}\\right)$ |\n| ${ }_{21}^{214} \\mathrm{~Pb}$ | 26.8 m | $\\left(\\beta^{-}\\right)$ | ${ }_{90}^{234} \\mathrm{Th}$ | 24.10 d | $\\left(\\beta^{-}\\right)$ |"}
{"id": 5335, "contents": "2281. Half-Lives for Several Radioactive Isotopes - \n[^18]2 E.C. = electron capture, S.F. = Spontaneous fission\n$3 \\mathrm{y}=$ years, $\\mathrm{d}=$ days, $\\mathrm{h}=$ hours, $\\mathrm{m}=$ minutes, $\\mathrm{s}=$ seconds"}
{"id": 5336, "contents": "2281. Half-Lives for Several Radioactive Isotopes - \n| Isotope | Half-Life ${ }^{1}$ | Type of Emission ${ }^{2}$ | Isotope | Half-Life ${ }^{3}$ | Type of Emission ${ }^{4}$ |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| ${ }_{83}^{206} \\mathrm{Bi}$ | 6.243 d | (E.C.) | ${ }_{91}^{233} \\mathrm{~Pa}$ | 27 d | $\\left(\\beta^{-}\\right)$ |\n| ${ }_{92}^{233} \\mathrm{U}$ | $1.59 \\times 10^{5} \\mathrm{y}$ | ( $\\alpha$ ) | ${ }_{96}^{242} \\mathrm{Cm}$ | 162.8 d | ( $\\alpha$ ) |\n| ${ }_{92}^{234} \\mathrm{U}$ | $2.45 \\times 10^{5} \\mathrm{y}$ | ( $\\alpha$ ) | ${ }_{97}^{243} \\mathrm{Bk}$ | 4.5 h | ( $\\alpha$ or E.C.) |\n| ${ }_{92}^{235} \\mathrm{U}$ | $7.03 \\times 10^{8} \\mathrm{y}$ | ( $\\alpha$ ) | ${ }_{99}^{253} \\mathrm{Es}$ | 20.47 d | ( $\\alpha$ ) |\n| ${ }_{92}^{238} \\mathrm{U}$ | $4.47 \\times 10^{9} \\mathrm{y}$ | ( $\\alpha$ ) | ${ }_{100}^{254} \\mathrm{Fm}$ | 3.24 h | ( $\\alpha$ or S.F.) |\n| ${ }_{92}^{239} \\mathrm{U}$ | 23.54 m | $\\left(\\beta^{-}\\right)$ | ${ }_{100}^{255} \\mathrm{Fm}$ | 20.1 h | ( $\\alpha$ ) |"}
{"id": 5337, "contents": "2281. Half-Lives for Several Radioactive Isotopes - \n| ${ }_{93}^{239} \\mathrm{~Np}$ | 2.3 d | $\\left(\\beta^{-}\\right)$ | ${ }_{101}^{256} \\mathrm{Md}$ | 76 m | ( $\\alpha$ or E.C.) |\n| ${ }_{94}^{239} \\mathrm{Pu}$ | $2.407 \\times 10^{4} \\mathrm{y}$ | ( $\\alpha$ ) | ${ }_{102}^{254} \\mathrm{No}$ | 55 s | ( $\\alpha$ ) |\n| ${ }_{94}^{239} \\mathrm{Pu}$ | $6.54 \\times 10^{3} \\mathrm{y}$ | ( $\\alpha$ ) | ${ }_{103}^{257} \\mathrm{Lr}$ | 0.65 s | ( $\\alpha$ ) |\n| ${ }_{94}^{241} \\mathrm{Pu}$ | 14.4 y | ( $\\alpha$ or $\\beta^{-}$) | ${ }_{105}^{260} \\mathrm{Ha}$ | 1.5 s | ( $\\alpha$ or S.F.) |\n| ${ }_{95}^{241} \\mathrm{Am}$ | 432.2 y | ( $\\alpha$ ) | ${ }_{106}^{263} \\mathrm{Sg}$ | 0.8 s | ( $\\alpha$ or S.F.) |"}
{"id": 5338, "contents": "2281. Half-Lives for Several Radioactive Isotopes - \nTABLE M1"}
{"id": 5339, "contents": "2283. Chapter 1 - \n1. Place a glass of water outside. It will freeze if the temperature is below $0^{\\circ} \\mathrm{C}$.\n2. (a) law (states a consistently observed phenomenon, can be used for prediction); (b) theory (a widely accepted explanation of the behavior of matter); (c) hypothesis (a tentative explanation, can be investigated by experimentation)\n3. (a) symbolic, microscopic; (b) macroscopic; (c) symbolic, macroscopic; (d) microscopic\n4. Macroscopic. The heat required is determined from macroscopic properties.\n5. Liquids can change their shape (flow); solids can't. Gases can undergo large volume changes as pressure changes; liquids do not. Gases flow and change volume; solids do not.\n6. The mixture can have a variety of compositions; a pure substance has a definite composition. Both have the same composition from point to point.\n7. Molecules of elements contain only one type of atom; molecules of compounds contain two or more types of atoms. They are similar in that both are comprised of two or more atoms chemically bonded together.\n8. Answers will vary. Sample answer: Gatorade contains water, sugar, dextrose, citric acid, salt, sodium chloride, monopotassium phosphate, and sucrose acetate isobutyrate.\n9. (a) element; (b) element; (c) compound; (d) mixture; (e) compound; (f) compound; (g) compound; (h) mixture\n10. In each case, a molecule consists of two or more combined atoms. They differ in that the types of atoms change from one substance to the next.\n11. Gasoline (a mixture of compounds), oxygen, and to a lesser extent, nitrogen are consumed. Carbon dioxide and water are the principal products. Carbon monoxide and nitrogen oxides are produced in lesser amounts.\n12. (a) Increased as it would have combined with oxygen in the air thus increasing the amount of matter and therefore the mass. (b) 0.9 g\n13. (a) 200.0 g ; (b) The mass of the container and contents would decrease as carbon dioxide is a gaseous product and would leave the container. (c) 102.3 g"}
{"id": 5340, "contents": "2283. Chapter 1 - \n13. (a) 200.0 g ; (b) The mass of the container and contents would decrease as carbon dioxide is a gaseous product and would leave the container. (c) 102.3 g\n14. (a) physical; (b) chemical; (c) chemical; (d) physical; (e) physical\n15. physical\n16. The value of an extensive property depends upon the amount of matter being considered, whereas the value of an intensive property is the same regardless of the amount of matter being considered.\n17. Being extensive properties, both mass and volume are directly proportional to the amount of substance under study. Dividing one extensive property by another will in effect \"cancel\" this dependence on amount, yielding a ratio that is independent of amount (an intensive property).\n18. about a yard\n19. (a) kilograms; (b) meters; (c) meters/second; (d) kilograms/cubic meter; (e) kelvin; (f) square meters; (g) cubic meters\n20. (a) centi-, $\\times 10^{-2}$; (b) deci-, $\\times 10^{-1}$; (c) Giga-, $\\times 10^{9}$; (d) kilo-, $\\times 10^{3}$; (e) milli-, $\\times 10^{-3}$; (f) nano-, $\\times 10^{-9}$; (g) pico-, $\\times 10^{-12}$; (h) tera-, $\\times 10^{12}$"}
{"id": 5341, "contents": "2283. Chapter 1 - \n21. (a) $\\mathrm{m}=18.58 \\mathrm{~g}, \\mathrm{~V}=5.7 \\mathrm{~mL}$. (b) $\\mathrm{d}=3.3 \\mathrm{~g} / \\mathrm{mL}$ (c) dioptase (copper cyclosilicate, $\\mathrm{d}=3.28-3.31 \\mathrm{~g} / \\mathrm{mL}$ ); malachite (basic copper carbonate, $\\mathrm{d}=3.25-4.10 \\mathrm{~g} / \\mathrm{mL}$ ); Paraiba tourmaline (sodium lithium boron silicate with copper, $\\mathrm{d}=2.82-3.32 \\mathrm{~g} / \\mathrm{mL}$ )\n22. (a) displaced water volume $=2.8 \\mathrm{~mL}$; (b) displaced water mass $=2.8 \\mathrm{~g}$; (c) The block mass is 2.76 g , essentially equal to the mass of displaced water $(2.8 \\mathrm{~g})$ and consistent with Archimedes' principle of buoyancy.\n23. (a) $7.04 \\times 10^{2}$; (b) $3.344 \\times 10^{-2}$; (c) $5.479 \\times 10^{2}$; (d) $2.2086 \\times 10^{4}$; (e) $1.00000 \\times 10^{3}$; (f) $6.51 \\times 10^{-8}$; (g) $7.157 \\times 10^{-3}$\n24. (a) exact; (b) exact; (c) uncertain; (d) exact; (e) uncertain; (f) uncertain\n25. (a) two; (b) three; (c) five; (d) four; (e) six; (f) two; (g) five\n26. (a) 0.44 ; (b) 9.0 ; (c) 27 ; (d) 140 ; (e) $1.5 \\times 10^{-3}$; (f) 0.44"}
{"id": 5342, "contents": "2283. Chapter 1 - \n27. (a) $2.15 \\times 10^{5}$; (b) $4.2 \\times 10^{6}$; (c) 2.08 ; (d) 0.19 ; (e) 27,440 ; (f) 43.0\n28. (a) Archer X; (b) Archer W; (c) Archer Y\n29. (a) $\\frac{1.0936 \\mathrm{yd}}{1 \\mathrm{~m}}$; (b) $\\frac{0.94635 \\mathrm{~L}}{1 \\mathrm{qt}}$; (c) $\\frac{2.2046 \\mathrm{lb}}{1 \\mathrm{~kg}}$\n30. $\\frac{2.0 \\mathrm{~L}}{67.6 \\mathrm{fl} \\mathrm{oz}}=\\frac{0.030 \\mathrm{~L}}{1 \\mathrm{fl} \\mathrm{oz}}$"}
{"id": 5343, "contents": "2283. Chapter 1 - \nOnly two significant figures are justified.\n61. $68-71 \\mathrm{~cm} ; 400-450 \\mathrm{~g}$\n63. 355 mL\n65. $8 \\times 10^{-4} \\mathrm{~cm}$\n67. yes; weight $=89.4 \\mathrm{~kg}$\n69. $5.0 \\times 10^{-3} \\mathrm{~mL}$\n71. (a) $1.3 \\times 10^{-4} \\mathrm{~kg}$; (b) $2.32 \\times 10^{8} \\mathrm{~kg}$; (c) $5.23 \\times 10^{-12} \\mathrm{~m}$; (d) $8.63 \\times 10^{-5} \\mathrm{~kg}$; (e) $3.76 \\times 10^{-1} \\mathrm{~m}$; (f) $5.4 \\times 10^{-5}$ m ; (g) $1 \\times 10^{12} \\mathrm{~s}$; (h) $2.7 \\times 10^{-11} \\mathrm{~s}$; (i) $1.5 \\times 10^{-4} \\mathrm{~K}$\n73. 45.4 L\n75. $1.0160 \\times 10^{3} \\mathrm{~kg}$\n77. (a) 394 ft ; (b) 5.9634 km ; (c) $6.0 \\times 10^{2}$; (d) 2.64 L ; (e) $5.1 \\times 10^{18} \\mathrm{~kg}$; (f) 14.5 kg ; (g) 324 mg\n79. $0.46 \\mathrm{~m} ; 1.5 \\mathrm{ft} / \\mathrm{cubit}$\n81. Yes, the acid's volume is 123 mL .\n83. 62.6 in (about 5 ft 3 in .) and 101 lb"}
{"id": 5344, "contents": "2283. Chapter 1 - \n81. Yes, the acid's volume is 123 mL .\n83. 62.6 in (about 5 ft 3 in .) and 101 lb\n85. (a) $3.81 \\mathrm{~cm} \\times 8.89 \\mathrm{~cm} \\times 2.44 \\mathrm{~m}$; (b) 40.6 cm\n87. $2.70 \\mathrm{~g} / \\mathrm{cm}^{3}$\n89. (a) 81.6 g ; (b) 17.6 g\n91. (a) 5.1 mL ; (b) 37 L\n93. $5371{ }^{\\circ} \\mathrm{F}, 3239 \\mathrm{~K}$\n95. $-23^{\\circ} \\mathrm{C}, 250 \\mathrm{~K}$\n97. $-33.4^{\\circ} \\mathrm{C}, 239.8 \\mathrm{~K}$\n99. $113{ }^{\\circ} \\mathrm{F}$"}
{"id": 5345, "contents": "2284. Chapter 2 - \n1. The starting materials consist of one green sphere and two purple spheres. The products consist of two green spheres and two purple spheres. This violates Dalton's postulate that that atoms are not created during a chemical change, but are merely redistributed.\n2. This statement violates Dalton's fourth postulate: In a given compound, the numbers of atoms of each type (and thus also the percentage) always have the same ratio.\n3. Dalton originally thought that all atoms of a particular element had identical properties, including mass. Thus, the concept of isotopes, in which an element has different masses, was a violation of the original idea. To account for the existence of isotopes, the second postulate of his atomic theory was modified to state that atoms of the same element must have identical chemical properties.\n4. Both are subatomic particles that reside in an atom's nucleus. Both have approximately the same mass. Protons are positively charged, whereas neutrons are uncharged.\n5. (a) The Rutherford atom has a small, positively charged nucleus, so most $\\alpha$ particles will pass through empty space far from the nucleus and be undeflected. Those $\\alpha$ particles that pass near the nucleus will be deflected from their paths due to positive-positive repulsion. The more directly toward the nucleus the $\\alpha$ particles are headed, the larger the deflection angle will be. (b) Higher-energy $\\alpha$ particles that pass near the nucleus will still undergo deflection, but the faster they travel, the less the expected angle of deflection.\n(c) If the nucleus is smaller, the positive charge is smaller and the expected deflections are smaller-both in terms of how closely the $\\alpha$ particles pass by the nucleus undeflected and the angle of deflection. If the nucleus is larger, the positive charge is larger and the expected deflections are larger-more $\\alpha$ particles will be deflected, and the deflection angles will be larger. (d) The paths followed by the $\\alpha$ particles match the predictions from (a), (b), and (c).\n6. (a) ${ }^{133} \\mathrm{Cs}^{+}$; (b) ${ }^{127} \\mathrm{I}^{-}$; (c) ${ }^{31} \\mathrm{P}^{3-}$; (d) ${ }^{57} \\mathrm{Co}^{3+}$"}
{"id": 5346, "contents": "2284. Chapter 2 - \n7. (a) Carbon-12, ${ }^{12} \\mathrm{C}$; (b) This atom contains six protons and six neutrons. There are six electrons in a neutral ${ }^{12} \\mathrm{C}$ atom. The net charge of such a neutral atom is zero, and the mass number is 12 . (c) The preceding answers are correct. (d) The atom will be stable since C-12 is a stable isotope of carbon. (e) The preceding answer is correct. Other answers for this exercise are possible if a different element of isotope is chosen.\n8. (a) Lithium- 6 contains three protons, three neutrons, and three electrons. The isotope symbol is ${ }^{6} \\mathrm{Li}$ or ${ }_{3}^{6} \\mathrm{Li}$.\n(b) ${ }^{6} \\mathrm{Li}^{+}$or ${ }_{3}^{6} \\mathrm{Li}^{+}$\n9. (a) Iron, 26 protons, 24 electrons, and 32 neutrons; (b) iodine, 53 protons, 54 electrons, and 74 neutrons\n10. (a) 3 protons, 3 electrons, 4 neutrons; (b) 52 protons, 52 electrons, 73 neutrons; (c) 47 protons, 47 electrons, 62 neutrons; (d) 7 protons, 7 electrons, 8 neutrons; (e) 15 protons, 15 electrons, 16 neutrons"}
{"id": 5347, "contents": "2284. Chapter 2 - \n11. Let us use neon as an example. Since there are three isotopes, there is no way to be sure to accurately predict the abundances to make the total of 20.18 amu average atomic mass. Let us guess that the abundances are $9 \\% \\mathrm{Ne}-22,91 \\% \\mathrm{Ne}-20$, and only a trace of $\\mathrm{Ne}-21$. The average mass would be 20.18 amu . Checking the nature's mix of isotopes shows that the abundances are $90.48 \\% \\mathrm{Ne}-20,9.25 \\% \\mathrm{Ne}-22$, and $0.27 \\% \\mathrm{Ne}-21$, so our guessed amounts have to be slightly adjusted.\n12. 79.90 amu\n13. Turkey source: $20.3 \\%$ (of 10.0129 amu isotope); US source: $19.1 \\%$ (of 10.0129 amu isotope)\n14. The symbol for the element oxygen, $O$, represents both the element and one atom of oxygen. A molecule of oxygen, $\\mathrm{O}_{2}$, contains two oxygen atoms; the subscript 2 in the formula must be used to distinguish the diatomic molecule from two single oxygen atoms.\n15. (a) molecular $\\mathrm{CO}_{2}$, empirical $\\mathrm{CO}_{2}$; (b) molecular $\\mathrm{C}_{2} \\mathrm{H}_{2}$, empirical CH ; (c) molecular $\\mathrm{C}_{2} \\mathrm{H}_{4}$, empirical $\\mathrm{CH}_{2}$; (d) molecular $\\mathrm{H}_{2} \\mathrm{SO}_{4}$, empirical $\\mathrm{H}_{2} \\mathrm{SO}_{4}$\n16. (a) $\\mathrm{C}_{4} \\mathrm{H}_{5} \\mathrm{~N}_{2} \\mathrm{O}$; (b) $\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}$; (c) HO ; (d) $\\mathrm{CH}_{2} \\mathrm{O}$; (e) $\\mathrm{C}_{3} \\mathrm{H}_{4} \\mathrm{O}_{3}$"}
{"id": 5348, "contents": "2284. Chapter 2 - \n17. (a) $\\mathrm{CH}_{2} \\mathrm{O}$; (b) $\\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}$\n18. (a) ethanol"}
{"id": 5349, "contents": "2284. Chapter 2 - \n(b) methoxymethane, more commonly known as dimethyl ether"}
{"id": 5350, "contents": "2284. Chapter 2 - \n(c) These molecules have the same chemical composition (types and number of atoms) but different chemical structures. They are structural isomers.\n19. Use the molecular formula to find the molar mass; to obtain the number of moles, divide the mass of compound by the molar mass of the compound expressed in grams.\n20. Formic acid. Its formula has twice as many oxygen atoms as the other two compounds (one each). Therefore, 0.60 mol of formic acid would be equivalent to 1.20 mol of a compound containing a single oxygen atom.\n21. The two masses have the same numerical value, but the units are different: The molecular mass is the mass of 1 molecule while the molar mass is the mass of $6.022 \\times 10^{23}$ molecules.\n22. (a) $256.48 \\mathrm{~g} / \\mathrm{mol}$; (b) $72.150 \\mathrm{~g} \\mathrm{~mol}^{-1}$; (c) $378.103 \\mathrm{~g} \\mathrm{~mol}^{-1}$; (d) $58.080 \\mathrm{~g} \\mathrm{~mol}^{-1}$; (e) $180.158 \\mathrm{~g} \\mathrm{~mol}^{-1}$\n23. (a) $197.382 \\mathrm{~g} \\mathrm{~mol}^{-1}$; (b) $257.163 \\mathrm{~g} \\mathrm{~mol}^{-1}$; (c) $194.193 \\mathrm{~g} \\mathrm{~mol}^{-1}$; (d) $60.056 \\mathrm{~g} \\mathrm{~mol}^{-1}$; (e) $306.464 \\mathrm{~g} \\mathrm{~mol}^{-1}$\n24. (a) 0.819 g ;\n(b) 307 g ;\n(c) 0.23 g ;\n(d) $1.235 \\times 10^{6} \\mathrm{~g}(1235 \\mathrm{~kg})$;\n(e) 765 g\n25. (a) 99.41 g ;\n(b) 2.27 g ;\n(c) 3.5 g ;"}
{"id": 5351, "contents": "2284. Chapter 2 - \n(e) 765 g\n25. (a) 99.41 g ;\n(b) 2.27 g ;\n(c) 3.5 g ;\n(d) 222 kg ;\n(e) 160.1 g\n26. (a) 9.60 g ; (b) 19.2 g ; (c) 28.8 g\n27. zirconium: $2.038 \\times 10^{23}$ atoms; 30.87 g; silicon: $2.038 \\times 10^{23}$ atoms; 9.504 g ; oxygen: $8.151 \\times 10^{23}$ atoms; 21.66 g\n28. $\\mathrm{AlPO}_{4}: 1.000 \\mathrm{~mol}$ or 26.98 g Al\n$\\mathrm{Al}_{2} \\mathrm{Cl}_{6}: 1.994 \\mathrm{~mol}$ or 53.74 g Al\n$\\mathrm{Al}_{2} \\mathrm{~S}_{3}: 3.00 \\mathrm{~mol}$ or 80.94 g Al\nThe $\\mathrm{Al}_{2} \\mathrm{~S}_{3}$ sample thus contains the greatest mass of Al.\n29. $3.113 \\times 10^{25} \\mathrm{C}$ atoms\n30. 0.865 servings, or about 1 serving.\n31. $20.0 \\mathrm{~g} \\mathrm{H}_{2} \\mathrm{O}$ represents the least number of molecules since it has the least number of moles."}
{"id": 5352, "contents": "2285. Chapter 3 - \n1. The spectrum consists of colored lines, at least one of which (probably the brightest) is red.\n2. 3.15 m\n3. $3.233 \\times 10^{-19} \\mathrm{~J} ; 2.018 \\mathrm{eV}$\n4. $\\nu=4.568 \\times 10^{14} \\mathrm{~s}^{-1} ; \\lambda=656.3 \\mathrm{~nm}$; Energy $\\mathrm{mol}^{-1}=1.823 \\times 10^{5} \\mathrm{~J} \\mathrm{~mol}^{-1}$; red\n5. (a) $\\lambda=8.69 \\times 10^{-7} \\mathrm{~m} ; E=2.29 \\times 10^{-19} \\mathrm{~J}$; (b) $\\lambda=4.59 \\times 10^{-7} \\mathrm{~m} ; E=4.33 \\times 10^{-19} \\mathrm{~J}$; The color of (a) is red; (b) is blue.\n6. $E=9.502 \\times 10^{-15} \\mathrm{~J} ; \\nu=1.434 \\times 10^{19} \\mathrm{~s}^{-1}$\n7. Red: $660 \\mathrm{~nm} ; 4.54 \\times 10^{14} \\mathrm{~Hz} ; 3.01 \\times 10^{-19} \\mathrm{~J}$. Green: $520 \\mathrm{~nm} ; 5.77 \\times 10^{14} \\mathrm{~Hz} ; 3.82 \\times 10^{-19} \\mathrm{~J}$. Blue: 440 nm ; $6.81 \\times 10^{14} \\mathrm{~Hz} ; 4.51 \\times 10^{-19} \\mathrm{~J}$. Somewhat different numbers are also possible.\n8. $5.49 \\times 10^{14} \\mathrm{~s}^{-1}$; no\n9. Quantized energy means that the electrons can possess only certain discrete energy values; values between those quantized values are not permitted.\n10. 2.856 eV"}
{"id": 5353, "contents": "2285. Chapter 3 - \n9. Quantized energy means that the electrons can possess only certain discrete energy values; values between those quantized values are not permitted.\n10. 2.856 eV\n11. $-8.716 \\times 10^{-18} \\mathrm{~J}$\n12. $-3.405 \\times 10^{-20} \\mathrm{~J}$\n13. $33.9 \\AA$\n14. $1.471 \\times 10^{-17} \\mathrm{~J}$\n15. Both involve a relatively heavy nucleus with electrons moving around it, although strictly speaking, the Bohr model works only for one-electron atoms or ions. According to classical mechanics, the Rutherford model predicts a miniature \"solar system\" with electrons moving about the nucleus in circular or elliptical orbits that are confined to planes. If the requirements of classical electromagnetic theory that electrons in such orbits would emit electromagnetic radiation are ignored, such atoms would be stable, having constant energy and angular momentum, but would not emit any visible light (contrary to observation). If classical electromagnetic theory is applied, then the Rutherford atom would emit electromagnetic radiation of continually increasing frequency (contrary to the observed discrete spectra), thereby losing energy until the atom collapsed in an absurdly short time (contrary to the observed long-term stability of atoms). The Bohr model retains the classical mechanics view of circular orbits confined to planes having constant energy and angular momentum, but restricts these to quantized values dependent on a single quantum number, $n$. The orbiting electron in Bohr's model is assumed not to emit any electromagnetic radiation while moving about the nucleus in its stationary orbits, but the atom can emit or absorb electromagnetic radiation when the electron changes from one orbit to another. Because of the quantized orbits, such \"quantum jumps\" will produce discrete spectra, in agreement with observations."}
{"id": 5354, "contents": "2285. Chapter 3 - \n16. Both models have a central positively charged nucleus with electrons moving about the nucleus in accordance with the Coulomb electrostatic potential. The Bohr model assumes that the electrons move in circular orbits that have quantized energies, angular momentum, and radii that are specified by a single quantum number, $n=1,2,3, \\ldots$, but this quantization is an ad hoc assumption made by Bohr to incorporate quantization into an essentially classical mechanics description of the atom. Bohr also assumed that electrons orbiting the nucleus normally do not emit or absorb electromagnetic radiation, but do so when the electron switches to a different orbit. In the quantum mechanical model, the electrons\ndo not move in precise orbits (such orbits violate the Heisenberg uncertainty principle) and, instead, a probabilistic interpretation of the electron's position at any given instant is used, with a mathematical function $\\psi$ called a wavefunction that can be used to determine the electron's spatial probability distribution. These wavefunctions, or orbitals, are three-dimensional stationary waves that can be specified by three quantum numbers that arise naturally from their underlying mathematics (no ad hoc assumptions required): the principal quantum number, $n$ (the same one used by Bohr), which specifies shells such that orbitals having the same $n$ all have the same energy and approximately the same spatial extent; the angular momentum quantum number 1 , which is a measure of the orbital's angular momentum and corresponds to the orbitals' general shapes, as well as specifying subshells such that orbitals having the same $l$ (and $n$ ) all have the same energy; and the orientation quantum number $m$, which is a measure of the $z$ component of the angular momentum and corresponds to the orientations of the orbitals. The Bohr model gives the same expression for the energy as the quantum mechanical expression and, hence, both properly account for hydrogen's discrete spectrum (an example of getting the right answers for the wrong reasons, something that many chemistry students can sympathize with), but gives the wrong expression for the angular momentum (Bohr orbits necessarily all have non-zero angular momentum, but some quantum orbitals [ $s$ orbitals] can have zero angular momentum)."}
{"id": 5355, "contents": "2285. Chapter 3 - \n17. $n$ determines the general range for the value of energy and the probable distances that the electron can be from the nucleus. $l$ determines the shape of the orbital. $m_{1}$ determines the orientation of the orbitals of the same $l$ value with respect to one another. $m_{s}$ determines the spin of an electron.\n18. (a) $2 p$; (b) $4 d$; (c) $6 s$\n19. (a) $3 d$; (b) 1 s ; (c) $4 f$\n20."}
{"id": 5356, "contents": "2285. Chapter 3 - \n41. (a) x. 2, y. 2, z. 2; (b) x. 1, y. 3, z. 0 ; (c) x. $400 \\frac{1}{2}$, y. $210 \\frac{1}{2}$, z. $320 \\frac{1}{2}$; (d) x. 1, y. 2 , z. 3 ; (e) $x . l=0, m_{l}=0$, y. $l=1, m_{l}=-1,0$, or +1, z. $l=2, m_{l}=-2,-1,0,+1,+2$\n43. 12\n45.\n\n| $n$ | $I$ | $m_{I}$ | $s$ |\n| :---: | :---: | :---: | :---: |\n| 4 | 0 | 0 | $+\\frac{1}{2}$ |\n| 4 | 0 | 0 | $-\\frac{1}{2}$ |\n| 4 | 1 | -1 | $+\\frac{1}{2}$ |\n| 4 | 1 | 0 | $+\\frac{1}{2}$ |\n| 4 | 1 | +1 | $+\\frac{1}{2}$ |\n| 4 | 1 | -1 | $-\\frac{1}{2}$ |"}
{"id": 5357, "contents": "2285. Chapter 3 - \n47. For example, $\\mathrm{Na}^{+}: 1 s^{2} 2 s^{2} 2 p^{6}$; $\\mathrm{Ca}^{2+}: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$; $\\mathrm{Sn}^{2+}: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{10} 5 s^{2}$; $\\mathrm{F}^{-}$: $1 s^{2} 2 s^{2} 2 p^{6} ; \\mathrm{O}^{2-}: 1 s^{2} 2 s^{2} 2 p^{6} ; \\mathrm{Cl}^{-}: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$.\n48. (a) $1 s^{2} 2 s^{2} 2 p^{3}$; (b) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{2}$; (c) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{6}$; (d)\n```\n$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{2} 4 d^{10} 5 p^{4}$; (e) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{2} 4 d^{10} 5 p^{6} 6 s^{2} 4 f^{9}$\n```"}
{"id": 5358, "contents": "2285. Chapter 3 - \n51. The charge on the ion.\n52. (a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)"}
{"id": 5359, "contents": "2285. Chapter 3 - \n53. Zr\n54. $\\mathrm{Rb}^{+}, \\mathrm{Se}^{2-}$\n55. Although both (b) and (c) are correct, (e) encompasses both and is the best answer.\n56. K\n57. $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} 4 p^{6} 5 s^{2} 4 d^{10} 5 p^{6} 6 s^{2} 4 f^{14} 5 d^{10}$\n58. Co has 27 protons, 27 electrons, and 33 neutrons: $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{7}$. I has 53 protons, 53 electrons, and 78 neutrons: $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{10} 5 s^{2} 5 p^{5}$.\n59. Cl\n60. O\n61. $\\mathrm{Rb}<\\mathrm{Li}<\\mathrm{N}<\\mathrm{F}$\n62. 15 ( 5 A )\n63. $\\mathrm{Mg}<\\mathrm{Ca}<\\mathrm{Rb}<\\mathrm{Cs}$\n64. $\\mathrm{Si}^{++}<\\mathrm{Al}^{3+}<\\mathrm{Ca}^{2+}<\\mathrm{K}^{+}$\n65. $\\mathrm{Se}, \\mathrm{As}^{-}$\n66. $\\mathrm{Mg}^{2+}<\\mathrm{K}^{+}<\\mathrm{Br}^{-}<\\mathrm{As}^{3-}$\n67. $\\mathrm{O}, \\mathrm{IE}_{1}$\n68. Ra\n69. (a) metal, inner transition metal; (b) nonmetal, representative element; (c) metal, representative element;"}
{"id": 5360, "contents": "2285. Chapter 3 - \n67. $\\mathrm{O}, \\mathrm{IE}_{1}$\n68. Ra\n69. (a) metal, inner transition metal; (b) nonmetal, representative element; (c) metal, representative element;\n(d) nonmetal, representative element; (e) metal, transition metal; (f) metal, inner transition metal; (g) metal, transition metal; (h) nonmetal, representative element; (i) nonmetal, representative element; (j) metal, representative element\n70. (a) He ; (b) Be ; (c) Li; (d) O\n71. (a) krypton, Kr ; (b) calcium, Ca ; (c) fluorine, F ; (d) tellurium, Te\n72. (a) ${ }_{11}^{23} \\mathrm{Na}$; (b) ${ }_{54}^{129} \\mathrm{Xe}$; (c) ${ }_{33}^{73} \\mathrm{As}$; (d) ${ }_{88}^{226} \\mathrm{Ra}$\n73. Ionic: $\\mathrm{KCl}, \\mathrm{MgCl}_{2}$; Covalent: $\\mathrm{NCl}_{3}, \\mathrm{ICl}, \\mathrm{PCl}_{5}, \\mathrm{CCl}_{4}$\n74. (a) covalent; (b) ionic, $\\mathrm{Ba}^{2+}, \\mathrm{O}^{2-}$; (c) ionic, $\\mathrm{NH}_{4}{ }^{+}, \\mathrm{CO}_{3}{ }^{2-}$; (d) ionic, $\\mathrm{Sr}^{2+}, \\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{-}$; (e) covalent; (f) ionic, $\\mathrm{Na}^{+}, \\mathrm{O}^{2-}$"}
{"id": 5361, "contents": "2285. Chapter 3 - \n75. (a) CaS ; (b) $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{4}$; (c) $\\mathrm{AlBr}_{3}$; (d) $\\mathrm{Na}_{2} \\mathrm{HPO}_{4}$; (e) $\\mathrm{Mg}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$"}
{"id": 5362, "contents": "2286. Chapter 4 - \n1. The protons in the nucleus do not change during normal chemical reactions. Only the outer electrons move. Positive charges form when electrons are lost.\n2. $\\mathrm{P}, \\mathrm{I}, \\mathrm{Cl}$, and O would form anions because they are nonmetals. Mg , $\\mathrm{In}, \\mathrm{Cs}, \\mathrm{Pb}$, and Co would form cations because they are metals.\n3. (a) $\\mathrm{P}^{3-}$; (b) $\\mathrm{Mg}^{2+}$; (c) $\\mathrm{Al}^{3+}$; (d) $\\mathrm{O}^{2-}$; (e) $\\mathrm{Cl}^{-}$; (f) $\\mathrm{Cs}^{+}$\n4. (a) $[\\mathrm{Ar}] 4 s^{2} 3 d^{10} 4 p^{6}$; (b) $[\\mathrm{Kr}] 4 d^{10} 5 s^{2} 5 p^{6}$ (c) $1 s^{2}$ (d) $[\\mathrm{Kr}] 4 d^{10}$; (e) $[\\mathrm{He}] 2 s^{2} 2 p^{6}$; (f) $[\\mathrm{Ar}] 3 d^{10}$; (g) $1 s^{2}$ (h) $[\\mathrm{He}] 2 s^{2} 2 p^{6}$ (i) $[\\mathrm{Kr}] 4 d^{10} 5 s^{2}(\\mathrm{j})[\\mathrm{Ar}] 3 d^{7}(\\mathrm{k})[\\mathrm{Ar}] 3 d^{6}$, (l) $[\\mathrm{Ar}] 3 d^{10} 4 s^{2}$"}
{"id": 5363, "contents": "2286. Chapter 4 - \n5. (a) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1}$; $\\mathrm{Al}^{3+}$ : $1 s^{2} 2 s^{2} 2 p^{6}$; (b) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{5}$; $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6}$; (c) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 5 s^{2}$; $\\mathrm{Sr}^{2+}: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6}$; (d) $1 s^{2} 2 s^{1}$; $\\mathrm{Li}^{+}: 1 s^{2}$; (e) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{3} ; 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6}$; (f) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{4} ; 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$"}
{"id": 5364, "contents": "2286. Chapter 4 - \n6. NaCl consists of discrete ions arranged in a crystal lattice, not covalently bonded molecules.\n7. ionic: (b), (d), (e), (g), and (i); covalent: (a), (c), (f), (h), (j), and (k)\n8. (a) Cl; (b) O; (c) O; (d) S; (e) N; (f) P; (g) N\n9. (a) H, C, N, O, F; (b) H, I, Br, Cl, F; (c) H, P, S, O, F; (d) Na, Al, H, P, O; (e) Ba, H, As, N, O\n10. $\\mathrm{N}, \\mathrm{O}, \\mathrm{F}$, and Cl\n11. (a) HF; (b) CO; (c) OH ; (d) PCl ; (e) NH ; (f) PO ; (g) CN\n12. (a) cesium chloride; (b) barium oxide; (c) potassium sulfide; (d) beryllium chloride; (e) hydrogen bromide;\n(f) aluminum fluoride\n13. (a) RbBr ; (b) MgSe ; (c) $\\mathrm{Na}_{2} \\mathrm{O}$; (d) $\\mathrm{CaCl}_{2}$; (e) HF ; (f) GaP ; (g) $\\mathrm{AlBr}_{3}$; (h) $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{4}$\n14. (a) $\\mathrm{ClO}_{2}$; (b) $\\mathrm{N}_{2} \\mathrm{O}_{4}$; (c) $\\mathrm{K}_{3} \\mathrm{P}$; (d) $\\mathrm{Ag}_{2} \\mathrm{~S}$; (e) $\\mathrm{AIF}_{3} \\cdot 3 \\mathrm{H}_{2} \\mathrm{O}$; (f) $\\mathrm{SiO}_{2}$"}
{"id": 5365, "contents": "2286. Chapter 4 - \n15. (a) chromium(III) oxide; (b) iron(II) chloride; (c) chromium(VI) oxide; (d) titanium(IV) chloride; (e) cobalt(II) chloride hexahydrate; (f) molybdenum(IV) sulfide\n16. (a) $\\mathrm{K}_{3} \\mathrm{PO}_{4}$; (b) $\\mathrm{CuSO}_{4}$; (c) $\\mathrm{CaCl}_{2}$; (d) $\\mathrm{TiO}_{2}$; (e) $\\mathrm{NH}_{4} \\mathrm{NO}_{3}$; (f) $\\mathrm{NaHSO}_{4}$\n17. (a) manganese(IV) oxide; (b) mercury(I) chloride; (c) iron(III) nitrate; (d) titanium(IV) chloride; (e) copper(II) bromide\n18. (a) eight electrons:"}
{"id": 5366, "contents": "2286. Chapter 4 - \n$$\n: \\ddot{A s}:^{3-}\n$$\n\n(b) eight electrons:\n\n$$\n: \\ddot{:}\n$$\n\n(c) no electrons $\\mathrm{Be}^{2+}$\n(d) eight electrons:\n\n(e) no electrons $\\mathrm{Ga}^{3+}$\n(f) no electrons $\\mathrm{Li}^{+}$\n(g) eight electrons:\n\n$: \\ddot{N}:$\n36. (a)\n\n$$\n\\mathrm{Mg}^{2+} \\quad: \\ddot{\\mathrm{S}} \\mathbf{:}^{2-}\n$$\n\n(b)\n\n$$\n\\mathrm{Al}^{3+} \\quad \\because \\ddot{\\mathrm{O}}{ }^{2-}\n$$\n\n(c)\n\n$$\n\\mathrm{Ga}^{3+} \\quad: \\ddot{\\mathrm{c}} \\overline{:} \\bar{\\square}\n$$\n\n(d)\n\n$$\n\\mathrm{K}^{+} \\quad \\ddot{\\ddot{\\mathrm{O}}^{2-}}\n$$\n\n(e)\n\n$$\n\\mathrm{Li}^{+} \\quad: \\ddot{\\mathrm{N}}^{3-}\n$$\n\n(f)\n\n$$\n\\mathrm{K}^{+} \\quad: \\ddot{\\mathrm{F}} \\bar{\\square}\n$$\n\n38.\n\n: $\\mathrm{P} \\equiv \\mathrm{P}$ :\n40. (a)\n$: \\ddot{o}=\\ddot{0}:$\n\nIn this case, the Lewis structure is inadequate to depict the fact that experimental studies have shown two unpaired electrons in each oxygen molecule.\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n(f)\n\n(g)\n\n(h)\n\n(i)\n$\\mathrm{H}-\\mathrm{C} \\equiv \\mathrm{C}-\\mathrm{H}$\n(j)\n$: \\ddot{\\mathrm{Cl}}-\\mathrm{C} \\equiv \\mathrm{N}:$\n(k)\n$C \\equiv C^{2+}$\n42. (a) $\\mathrm{SeF}_{6}$ :\n\n(b) $\\mathrm{XeF}_{4}$ :\n\n(c) $\\mathrm{SeCl}_{3}{ }^{+}$:"}
{"id": 5367, "contents": "2286. Chapter 4 - \n(b) $\\mathrm{XeF}_{4}$ :\n\n(c) $\\mathrm{SeCl}_{3}{ }^{+}$:\n\n(d) $\\mathrm{Cl}_{2} \\mathrm{BBCl}_{2}$ :\n\n44. Two valence electrons per Pb atom are transferred to Cl atoms; the resulting $\\mathrm{Pb}^{2+}$ ion has a $6 s^{2}$ valence shell configuration. Two of the valence electrons in the HCl molecule are shared, and the other six are located on the Cl atom as lone pairs of electrons.\n46.\n\n\n48.\n\n50. (a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n52.\n\n54. Each bond includes a sharing of electrons between atoms. Two electrons are shared in a single bond; four electrons are shared in a double bond; and six electrons are shared in a triple bond.\n56. (a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)\n\n58.\n\n60. (a)\n\n$$\n\\ddot{O}=c=\\ddot{o}\n$$\n\n(b)\n\n$$\n: C \\equiv O:\n$$\n\nCO has the strongest carbon-oxygen bond because there is a triple bond joining C and $\\mathrm{O} . \\mathrm{CO}_{2}$ has double bonds.\n62. (a) H: $0, \\mathrm{Cl}: 0$; (b) $\\mathrm{C}: 0, \\mathrm{~F}: 0$; (c) $\\mathrm{P}: 0, \\mathrm{Cl} 0$; (d) $\\mathrm{P}: 0, \\mathrm{~F}: 0$\n64. Cl in $\\mathrm{Cl}_{2}: 0 ; \\mathrm{Cl}$ in $\\mathrm{BeCl}_{2}: 0 ; \\mathrm{Cl}$ in $\\mathrm{ClF}_{5}: 0$\n66. (a)\n\n(b)\n\n(c)\n\n(d)\n\n68. HOCl\n70. The structure that gives zero formal charges is consistent with the actual structure:\n\n72. $\\mathrm{NF}_{3}$;\n\n74."}
{"id": 5368, "contents": "2286. Chapter 4 - \n75. The placement of the two sets of unpaired electrons in water forces the bonds to assume a tetrahedral arrangement, and the resulting HOH molecule is bent. The HBeH molecule (in which Be has only two electrons to bond with the two electrons from the hydrogens) must have the electron pairs as far from one another as possible and is therefore linear.\n77. Space must be provided for each pair of electrons whether they are in a bond or are present as lone pairs. Electron-pair geometry considers the placement of all electrons. Molecular structure considers only the bonding-pair geometry.\n79. As long as the polar bonds are compensated (for example. two identical atoms are found directly across the central atom from one another), the molecule can be nonpolar.\n81. (a) Both the electron geometry and the molecular structure are octahedral. (b) Both the electron geometry and the molecular structure are trigonal bipyramid. (c) Both the electron geometry and the molecular structure are linear. (d) Both the electron geometry and the molecular structure are trigonal planar.\n83. (a) electron-pair geometry: octahedral, molecular structure: square pyramidal; (b) electron-pair geometry: tetrahedral, molecular structure: bent; (c) electron-pair geometry: octahedral, molecular structure: square planar; (d) electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal; (e) electron-pair geometry: trigonal bypyramidal, molecular structure: seesaw; (f) electron-pair geometry: tetrahedral, molecular structure: bent (109 ${ }^{\\circ}$ )\n85. (a) electron-pair geometry: trigonal planar, molecular structure: bent ( $120^{\\circ}$ ); (b) electron-pair geometry: linear, molecular structure: linear; (c) electron-pair geometry: trigonal planar, molecular structure: trigonal planar; (d) electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal; (e) electron-pair geometry: tetrahedral, molecular structure: tetrahedral; (f) electron-pair geometry: trigonal bipyramidal, molecular structure: seesaw; (g) electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal"}
{"id": 5369, "contents": "2286. Chapter 4 - \n87. All of these molecules and ions contain polar bonds. Only $\\mathrm{ClF}_{5}, \\mathrm{ClO}_{2}{ }^{-}, \\mathrm{PCl}_{3}, \\mathrm{SeF}_{4}$, and $\\mathrm{PH}_{2}{ }^{-}$have dipole moments.\n89. $\\mathrm{SeS}_{2}, \\mathrm{CCl}_{2} \\mathrm{~F}_{2}, \\mathrm{PCl}_{3}$, and ClNO all have dipole moments.\n91. $P$\n93. nonpolar\n95. (a) tetrahedral; (b) trigonal pyramidal; (c) bent (109$) ; ~(d) ~ t r i g o n a l ~ p l a n a r ; ~(e) ~ b e n t ~(109 \u3147) ; ~(f) ~ b e n t ~(109 \u3147) ; ~$\n(g) $\\mathrm{CH}_{3} \\mathrm{CCH}$ tetrahedral, $\\mathrm{CH}_{3} \\mathrm{CCH}$ linear; (h) tetrahedral; (i) $\\mathrm{H}_{2} \\mathrm{CCCH}_{2}$ linear; $\\mathrm{H}_{2} \\mathrm{CCCH}_{2}$ trigonal planar\n97.\n$B-A$ - $\\quad$ CO $_{2}$, linear\n$B-\\dot{A} \\_{B}^{\\bullet} \\mathrm{H}_{2} \\mathrm{O}$, bent with an approximately $109^{\\circ}$ angle\n$B-\\ddot{A} \\mathrm{SO}_{2}$, bent with an approximately $120^{\\circ}$ angle\n99. (a)"}
{"id": 5370, "contents": "2286. Chapter 4 - \n(b)\n\n```\n#}=c=\\ddot{S\n```\n\n(c)"}
{"id": 5371, "contents": "2287. : $\\mathrm{C} \\equiv \\mathrm{s}$ : - \n(d) $\\mathrm{CS}_{3}{ }^{2-}$ includes three regions of electron density (all are bonds with no lone pairs); the shape is trigonal planar; $\\mathrm{CS}_{2}$ has only two regions of electron density (all bonds with no lone pairs); the shape is linear\n101. The Lewis structure is made from three units, but the atoms must be rearranged:\n\n103. The molecular dipole points away from the hydrogen atoms.\n105. The structures are very similar. In the model mode, each electron group occupies the same amount of space, so the bond angle is shown as $109.5^{\\circ}$. In the \"real\" mode, the lone pairs are larger, causing the hydrogens to be compressed. This leads to the smaller angle of $104.5^{\\circ}$."}
{"id": 5372, "contents": "2288. Chapter 5 - \n1. Similarities: Both types of bonds result from overlap of atomic orbitals on adjacent atoms and contain a maximum of two electrons. Differences: $\\sigma$ bonds are stronger and result from end-to-end overlap and all single bonds are $\\sigma$ bonds; $\\pi$ bonds between the same two atoms are weaker because they result from side-by-side overlap, and multiple bonds contain one or more $\\pi$ bonds (in addition to a $\\sigma$ bond).\n2. Bonding: One $\\sigma$ bond and one $\\pi$ bond. The $s$ orbitals are filled and do not overlap. The $p$ orbitals overlap along the axis to form a $\\sigma$ bond and side-by-side to form the $\\pi$ bond.\n\n3. No, two of the $p$ orbitals (one on each N ) will be oriented end-to-end and will form a $\\sigma$ bond.\n\n4. Hybridization is introduced to explain the geometry of bonding orbitals in valance bond theory.\n5. There are no $d$ orbitals in the valence shell of carbon.\n6. trigonal planar, $s p^{2}$; trigonal pyramidal (one lone pair on A ) $s p^{3}$; T -shaped (two lone pairs on $\\mathrm{A} s p^{3} d$, or (three lone pairs on A) $s p^{3} d^{2}$\n7. (a) Each $S$ has a bent $\\left(109^{\\circ}\\right)$ geometry, $s p^{3}$\n\n(b) Bent ( $120^{\\circ}$ ), $s p^{2}$\n\n(c) Trigonal planar, $s p^{2}$\n\n(d) Tetrahedral, $s p^{3}$\n\n8. (a) $\\mathrm{XeF}_{2}$\n(b)\n\n(c) linear (d) $s p^{3} d$\n9. (a)"}
{"id": 5373, "contents": "2288. Chapter 5 - \n8. (a) $\\mathrm{XeF}_{2}$\n(b)\n\n(c) linear (d) $s p^{3} d$\n9. (a)\n\n\n(b) P atoms, trigonal pyramidal; S atoms, bent, with two lone pairs; Cl atoms, trigonal pyramidal; (c) Hybridization about $\\mathrm{P}, \\mathrm{S}$, and Cl is, in all cases, $s p^{3}$; (d) Oxidation states $\\mathrm{P}+1, \\mathrm{~S}-1 \\frac{1}{3}, \\mathrm{Cl}+5, \\mathrm{O}-2$. Formal charges: P 0; S 0; Cl +2: O-1\n10.\n\n\n\n\nPhosphorus and nitrogen can form $s p^{3}$ hybrids to form three bonds and hold one lone pair in $\\mathrm{PF}_{3}$ and $\\mathrm{NF}_{3}$, respectively. However, nitrogen has no valence $d$ orbitals, so it cannot form a set of $s p^{3} d$ hybrid orbitals to bind five fluorine atoms in $\\mathrm{NF}_{5}$. Phosphorus has $d$ orbitals and can bind five fluorine atoms with $s p^{3} d$ hybrid orbitals in $\\mathrm{PF}_{5}$.\n21. A triple bond consists of one $\\sigma$ bond and two $\\pi$ bonds. A $\\sigma$ bond is stronger than a $\\pi$ bond due to greater overlap.\n23. (a)"}
{"id": 5374, "contents": "2288. Chapter 5 - \n(b) The terminal carbon atom uses $s p^{3}$ hybrid orbitals, while the central carbon atom is $s p$ hybridized. (c) Each of the two $\\pi$ bonds is formed by overlap of a $2 p$ orbital on carbon and a nitrogen $2 p$ orbital.\n25. (a) $s p^{2}$; (b) $s p$; (c) $s p^{2}$; (d) $s p^{3}$; (e) $s p^{3}$; (f) $s p^{3} d$; (g) $s p^{3}$\n27. (a) $s p^{2}$, delocalized; (b) $s p$, localized; (c) $s p^{2}$, delocalized; (d) $s p^{3}$, delocalized\n29.\n\nOrbitals in an isolated $C$ atom\n\n\nOrbitals in the $s p$ hybridized C in $\\mathrm{CO}_{2}$"}
{"id": 5375, "contents": "2288. Chapter 5 - \nEach of the four electrons is in a separate orbital and overlaps with an electron on an oxygen atom.\n31. (a) Similarities: Both are bonding orbitals that can contain a maximum of two electrons. Differences: $\\sigma$\norbitals are end-to-end combinations of atomic orbitals, whereas $\\pi$ orbitals are formed by side-by-side overlap of orbitals. (b) Similarities: Both are quantum-mechanical constructs that represent the probability of finding the electron about the atom or the molecule. Differences: $\\psi$ for an atomic orbital describes the behavior of only one electron at a time based on the atom. For a molecule, $\\psi$ represents a mathematical combination of atomic orbitals. (c) Similarities: Both are orbitals that can contain two electrons. Differences: Bonding orbitals result in holding two or more atoms together. Antibonding orbitals have the effect of destabilizing any bonding that has occurred.\n33. An odd number of electrons can never be paired, regardless of the arrangement of the molecular orbitals. It will always be paramagnetic.\n35. Bonding orbitals have electron density in close proximity to more than one nucleus. The interaction between the bonding positively charged nuclei and negatively charged electrons stabilizes the system.\n37. The pairing of the two bonding electrons lowers the energy of the system relative to the energy of the nonbonded electrons."}
{"id": 5376, "contents": "2288. Chapter 5 - \n37. The pairing of the two bonding electrons lowers the energy of the system relative to the energy of the nonbonded electrons.\n39. (a) $\\mathrm{H}_{2}$ bond order $=1, \\mathrm{H}_{2}{ }^{+}$bond order $=0.5, \\mathrm{H}_{2}{ }^{-}$bond order $=0.5$, strongest bond is $\\mathrm{H}_{2}$; (b) $\\mathrm{O}_{2}$ bond order $=2, \\mathrm{O}_{2}{ }^{2+}$ bond order $=3 ; \\mathrm{O}_{2}{ }^{2-}$ bond order $=1$, strongest bond is $\\mathrm{O}_{2}{ }^{2+}$; (c) $\\mathrm{Li}_{2}$ bond order $=1, \\mathrm{Be}_{2}{ }^{+}$ bond order $=0.5, \\mathrm{Be}_{2}$ bond order $=0$, strongest bond is $\\mathrm{Li}_{2}$; (d) $\\mathrm{F}_{2}$ bond order $=1, \\mathrm{~F}_{2}{ }^{+}$bond order $=1.5$, $\\mathrm{F}_{2}{ }^{-}$bond order $=0.5$, strongest bond is $\\mathrm{F}_{2}{ }^{+}$; (e) $\\mathrm{N}_{2}$ bond order $=3, \\mathrm{~N}_{2}{ }^{+}$bond order $=2.5, \\mathrm{~N}_{2}{ }^{-}$bond order $=2.5$, strongest bond is $\\mathrm{N}_{2}$\n41. (a) $\\mathrm{H}_{2}$; (b) $\\mathrm{N}_{2}$; (c) O ; (d) $\\mathrm{C}_{2}$; (e) $\\mathrm{B}_{2}$\n43. Yes, fluorine is a smaller atom than Li, so atoms in the $2 s$ orbital are closer to the nucleus and more stable.\n45. $2+$"}
{"id": 5377, "contents": "2288. Chapter 5 - \n43. Yes, fluorine is a smaller atom than Li, so atoms in the $2 s$ orbital are closer to the nucleus and more stable.\n45. $2+$\n47. $\\mathrm{N}_{2}$ has s-p mixing, so the $\\pi$ orbitals are the last filled in $\\mathrm{N}_{2}{ }^{2+} . \\mathrm{O}_{2}$ does not have s-p mixing, so the $\\sigma_{p}$ orbital fills before the $\\pi$ orbitals."}
{"id": 5378, "contents": "2289. Chapter 6 - \n1. (a) 12.01 amu ; (b) 12.01 amu ; (c) 144.12 amu ; (d) 60.05 amu\n2. (a) 123.896 amu ; (b) 18.015 amu ; (c) 164.086 amu ; (d) 60.052 amu ; (e) 342.297 amu\n3. (a) 56.107 amu ;\n(b) 54.091 amu ;\n(c) 199.9976 amu ;\n(d) 97.9950 amu\n4. (a) $\\% \\mathrm{~N}=82.24 \\%$\n\\% H = 17.76\\%;\n(b) $\\% \\mathrm{Na}=29.08 \\%$\n\\% $\\mathrm{S}=40.56 \\%$\n\\% O = 30.36\\%;\n(c) $\\% \\mathrm{Ca}^{2+}=38.76 \\%$\n5. $\\% \\mathrm{NH}_{3}=38.2 \\%$\n6. (a) $\\mathrm{CS}_{2}$\n(b) $\\mathrm{CH}_{2} \\mathrm{O}$\n7. $\\mathrm{C}_{6} \\mathrm{H}_{6}$\n8. $\\mathrm{Mg}_{3} \\mathrm{Si}_{2} \\mathrm{H}_{3} \\mathrm{O}_{8}$ (empirical formula), $\\mathrm{Mg}_{6} \\mathrm{Si}_{4} \\mathrm{H}_{6} \\mathrm{O}_{16}$ (molecular formula)\n9. $\\mathrm{C}_{15} \\mathrm{H}_{15} \\mathrm{~N}_{3}$\n10. We need to know the number of moles of sulfuric acid dissolved in the solution and the volume of the solution.\n11. (a) 0.679 M ; (b) 1.00 M ; (c) 0.06998 M ; (d) 1.75 M ; (e) 0.070 M ; (f) 6.6 M"}
{"id": 5379, "contents": "2289. Chapter 6 - \n12. (a) determine the number of moles of glucose in 0.500 L of solution; determine the molar mass of glucose; determine the mass of glucose from the number of moles and its molar mass; (b) 27 g\n13. (a) $37.0 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{SO}_{4}, 3.63 \\times 10^{3} \\mathrm{~g} \\mathrm{H}_{2} \\mathrm{SO}_{4}$; (b) $3.8 \\times 10^{-7} \\mathrm{~mol} \\mathrm{NaCN}, 1.9 \\times 10^{-5} \\mathrm{~g} \\mathrm{NaCN}$; (c) $73.2 \\mathrm{~mol} \\mathrm{H} \\mathrm{H}_{2} \\mathrm{CO}$, $2.20 \\mathrm{~kg} \\mathrm{H}_{2} \\mathrm{CO}$; (d) $5.9 \\times 10^{-7} \\mathrm{~mol} \\mathrm{FeSO}_{4}, 8.9 \\times 10^{-5} \\mathrm{~g} \\mathrm{FeSO}_{4}$\n14. (a) Determine the molar mass of $\\mathrm{KMnO}_{4}$; determine the number of moles of $\\mathrm{KMnO}_{4}$ in the solution; from the number of moles and the volume of solution, determine the molarity; (b) $1.15 \\times 10^{-3} \\mathrm{M}$\n15. (a) $5.04 \\times 10^{-3} \\mathrm{M}$; (b) 0.499 M ; (c) 9.92 M ; (d) $1.1 \\times 10^{-3} \\mathrm{M}$\n16. $0.025 M$\n17. 0.5000 L\n18. 1.9 mL\n19. (a) $0.125 M$; (b) $0.04888 M$; (c) $0.206 M$; (d) $0.0056 M$\n20. 11.9 M\n21. 1.6 L"}
{"id": 5380, "contents": "2289. Chapter 6 - \n20. 11.9 M\n21. 1.6 L\n22. (a) The dilution equation can be used, appropriately modified to accommodate mass-based concentration units: $\\%$ mass $_{1} \\times$ mass $_{1}=\\%$ mass $_{2} \\times$ mass $_{2}$. This equation can be rearranged to isolate mass ${ }_{1}$ and the given quantities substituted into this equation. (b) 58.8 g\n23. 114 g\n24. $1.75 \\times 10^{-3} \\mathrm{M}$\n25. $95 \\mathrm{mg} / \\mathrm{dL}$\n26. $2.38 \\times 10^{-4} \\mathrm{~mol}$\n27. 0.29 mol"}
{"id": 5381, "contents": "2290. Chapter 7 - \n1. An equation is balanced when the same number of each element is represented on the reactant and product sides. Equations must be balanced to accurately reflect the law of conservation of matter.\n2. (a) $\\mathrm{PCl}_{5}(s)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{POCl}_{3}(l)+2 \\mathrm{HCl}(a q)$; (b)\n$3 \\mathrm{Cu}(s)+8 \\mathrm{HNO}_{3}(a q) \\longrightarrow 3 \\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+4 \\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{NO}(g) ;(\\mathrm{c}) \\mathrm{H}_{2}(g)+\\mathrm{I}_{2}(s) \\longrightarrow 2 \\mathrm{HI}(s) ;(\\mathrm{d})$\n$4 \\mathrm{Fe}(s)+3 \\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{Fe}_{2} \\mathrm{O}_{3}(s)$; (e) $2 \\mathrm{Na}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{NaOH}(a q)+\\mathrm{H}_{2}(g)$; (f)\n$\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}(s) \\longrightarrow \\mathrm{Cr}_{2} \\mathrm{O}_{3}(s)+\\mathrm{N}_{2}(g)+4 \\mathrm{H}_{2} \\mathrm{O}(g) ;(\\mathrm{g}) \\mathrm{P}_{4}(s)+6 \\mathrm{Cl}_{2}(g) \\longrightarrow 4 \\mathrm{PCl}_{3}(l) ;(\\mathrm{h})$\n$\\mathrm{PtCl}_{4}(s) \\longrightarrow \\mathrm{Pt}(s)+2 \\mathrm{Cl}_{2}(g)$"}
{"id": 5382, "contents": "2290. Chapter 7 - \n$\\mathrm{PtCl}_{4}(s) \\longrightarrow \\mathrm{Pt}(s)+2 \\mathrm{Cl}_{2}(g)$\n3. (a) $\\mathrm{CaCO}_{3}(s) \\longrightarrow \\mathrm{CaO}(s)+\\mathrm{CO}_{2}(g)$; (b) $2 \\mathrm{C}_{4} \\mathrm{H}_{10}(g)+13 \\mathrm{O}_{2}(g) \\longrightarrow 8 \\mathrm{CO}_{2}(g)+10 \\mathrm{H}_{2} \\mathrm{O}(g)$; (c)\n$\\mathrm{MgCl}_{2}(a q)+2 \\mathrm{NaOH}(a q) \\longrightarrow \\mathrm{Mg}(\\mathrm{OH})_{2}(s)+2 \\mathrm{NaCl}(a q) ;(\\mathrm{d}) 2 \\mathrm{H}_{2} \\mathrm{O}(g)+2 \\mathrm{Na}(s) \\longrightarrow 2 \\mathrm{NaOH}(s)+\\mathrm{H}_{2}(g)$\n4. (a) $\\mathrm{Ba}\\left(\\mathrm{NO}_{3}\\right)_{2}, \\mathrm{KClO}_{3}$; (b) $2 \\mathrm{KClO}_{3}(s) \\longrightarrow 2 \\mathrm{KCl}(s)+3 \\mathrm{O}_{2}(g)$; (c)\n$2 \\mathrm{Ba}\\left(\\mathrm{NO}_{3}\\right)_{2}(s) \\longrightarrow 2 \\mathrm{BaO}(s)+2 \\mathrm{~N}_{2}(g)+5 \\mathrm{O}_{2}(g)$; (d) $2 \\mathrm{Mg}(s)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{MgO}(s) ;$\n$4 \\mathrm{Al}(s)+3 \\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{Al}_{2} \\mathrm{O}_{3}(s) ; 4 \\mathrm{Fe}(s)+3 \\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{Fe}_{2} \\mathrm{O}_{3}(s)$"}
{"id": 5383, "contents": "2290. Chapter 7 - \n5. (a) $4 \\mathrm{HF}(a q)+\\mathrm{SiO}_{2}(s) \\longrightarrow \\mathrm{SiF}_{4}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$; (b) complete ionic equation:\n$2 \\mathrm{Na}^{+}(a q)+2 \\mathrm{~F}^{-}(a q)+\\mathrm{Ca}^{2+}(a q)+2 \\mathrm{Cl}^{-}(a q) \\longrightarrow \\mathrm{CaF}_{2}(s)+2 \\mathrm{Na}^{+}(a q)+2 \\mathrm{Cl}^{-}(a q)$, net ionic equation: $2 \\mathrm{~F}^{-}(a q)+\\mathrm{Ca}^{2+}(a q) \\longrightarrow \\mathrm{CaF}_{2}(s)$\n6. (a)\n$2 \\mathrm{~K}^{+}(a q)+\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}(a q)+\\mathrm{Ba}^{2+}(a q)+2 \\mathrm{OH}^{-}(a q) \\longrightarrow 2 \\mathrm{~K}^{+}(a q)+2 \\mathrm{OH}^{-}(a q)+\\mathrm{BaC}_{2} \\mathrm{O}_{4}(s) \\quad$ (complete)\n$\\mathrm{Ba}^{2+}(a q)+\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}(a q) \\longrightarrow \\mathrm{BaC}_{2} \\mathrm{O}_{4}(s) \\quad$ (net)\n(b)\n$\\mathrm{Pb}^{2+}(a q)+2 \\mathrm{NO}_{3}{ }^{-}(a q)+2 \\mathrm{H}^{+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q) \\longrightarrow \\mathrm{PbSO}_{4}(s)+2 \\mathrm{H}^{+}(a q)+2 \\mathrm{NO}_{3}{ }^{-}(a q) \\quad$ (complete)"}
{"id": 5384, "contents": "2290. Chapter 7 - \n$\\mathrm{Pb}^{2+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q) \\longrightarrow \\mathrm{PbSO}_{4}(s) \\quad$ (net)\n(c) $\\mathrm{CaCO}_{3}(s)+2 \\mathrm{H}^{+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q) \\longrightarrow \\mathrm{CaSO}_{4}(s)+\\mathrm{CO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l) \\quad$ (complete)\n$\\mathrm{CaCO}_{3}(s)+2 \\mathrm{H}^{+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q) \\longrightarrow \\mathrm{CaSO}_{4}(s)+\\mathrm{CO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l) \\quad$ (net)\n7. (a) oxidation-reduction (addition); (b) acid-base (neutralization); (c) oxidation-reduction (combustion)\n8. It is an oxidation-reduction reaction because the oxidation state of the silver changes during the reaction.\n9. (a) $\\mathrm{H}+1, \\mathrm{P}+5, \\mathrm{O}-2$; (b) $\\mathrm{Al}+3, \\mathrm{H}+1, \\mathrm{O}-2$; (c) $\\mathrm{Se}+4, \\mathrm{O}-2$; (d) $\\mathrm{K}+1, \\mathrm{~N}+3, \\mathrm{O}-2$; (e) In $+3, \\mathrm{~S}-2$; (f) $\\mathrm{P}+3, \\mathrm{O}-2$\n10. (a) acid-base; (b) oxidation-reduction: Na is oxidized, $\\mathrm{H}^{+}$is reduced; (c) oxidation-reduction: Mg is oxidized, $\\mathrm{Cl}_{2}$ is reduced; (d) acid-base; (e) oxidation-reduction: $\\mathrm{P}^{3-}$ is oxidized, $\\mathrm{O}_{2}$ is reduced; (f) acid-base"}
{"id": 5385, "contents": "2290. Chapter 7 - \n11. (a) $2 \\mathrm{HCl}(g)+\\mathrm{Ca}(\\mathrm{OH})_{2}(s) \\longrightarrow \\mathrm{CaCl}_{2}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$; (b)\n$\\mathrm{Sr}(\\mathrm{OH})_{2}(a q)+2 \\mathrm{HNO}_{3}(a q) \\longrightarrow \\mathrm{Sr}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n12. (a) $2 \\mathrm{Al}(s)+3 \\mathrm{~F}_{2}(g) \\longrightarrow 2 \\mathrm{AlF}_{3}(s)$; (b) $2 \\mathrm{Al}(s)+3 \\mathrm{CuBr}_{2}(a q) \\longrightarrow 3 \\mathrm{Cu}(s)+2 \\mathrm{AlBr}_{3}(a q)$; (c)\n$\\mathrm{P}_{4}(s)+5 \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{P}_{4} \\mathrm{O}_{10}(s) ;(\\mathrm{d}) \\mathrm{Ca}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Ca}(\\mathrm{OH})_{2}(a q)+\\mathrm{H}_{2}(g)$\n13. (a) $\\mathrm{Mg}(\\mathrm{OH})_{2}(s)+2 \\mathrm{HClO}_{4}(a q) \\longrightarrow \\mathrm{Mg}^{2+}(a q)+2 \\mathrm{ClO}_{4}{ }^{-}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$; (b)\n$\\mathrm{SO}_{3}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{HSO}_{4}{ }^{-}(a q)$, (a solution of $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ ); (c)"}
{"id": 5386, "contents": "2290. Chapter 7 - \n$\\mathrm{SrO}(s)+\\mathrm{H}_{2} \\mathrm{SO}_{4}(l) \\longrightarrow \\mathrm{SrSO}_{4}(s)+\\mathrm{H}_{2} \\mathrm{O}$\n14. $\\mathrm{H}_{2}(g)+\\mathrm{F}_{2}(g) \\longrightarrow 2 \\mathrm{HF}(g)$\n15. $2 \\mathrm{NaBr}(a q)+\\mathrm{Cl}_{2}(g) \\longrightarrow 2 \\mathrm{NaCl}(a q)+\\mathrm{Br}_{2}(l)$\n16. $2 \\mathrm{LiOH}(a q)+\\mathrm{CO}_{2}(g) \\longrightarrow \\mathrm{Li}_{2} \\mathrm{CO}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)$\n17. (a) $\\mathrm{Ca}(\\mathrm{OH})_{2}(s)+\\mathrm{H}_{2} \\mathrm{~S}(g) \\longrightarrow \\mathrm{CaS}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$; (b)\n$\\mathrm{Na}_{2} \\mathrm{CO}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{~S}(g) \\longrightarrow \\mathrm{Na}_{2} \\mathrm{~S}(a q)+\\mathrm{CO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l)$\n18. (a) step 1: $\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\longrightarrow 2 \\mathrm{NH}_{3}(g)$, step 2 :\n$\\mathrm{NH}_{3}(g)+\\mathrm{HNO}_{3}(a q) \\longrightarrow \\mathrm{NH}_{4} \\mathrm{NO}_{3}(a q) \\longrightarrow \\mathrm{NH}_{4} \\mathrm{NO}_{3}(s)$ (after drying); (b)\n$\\mathrm{H}_{2}(g)+\\mathrm{Br}_{2}(l) \\longrightarrow 2 \\mathrm{HBr}(g)$; (c) $\\mathrm{Zn}(s)+\\mathrm{S}(s) \\longrightarrow \\mathrm{ZnS}(s)$ and"}
{"id": 5387, "contents": "2290. Chapter 7 - \n$\\mathrm{ZnS}(s)+2 \\mathrm{HCl}(a q) \\longrightarrow \\mathrm{ZnCl}_{2}(a q)+\\mathrm{H}_{2} \\mathrm{~S}(g)$\n19. (a) $\\mathrm{Sn}^{4+}(a q)+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Sn}^{2+}(a q)$, (b) $\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}(a q)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Ag}(s)+2 \\mathrm{NH}_{3}(a q)$; (c)\n$\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}(s)+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Hg}(l)+2 \\mathrm{Cl}^{-}(a q) ;(\\mathrm{d}) 2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{O}_{2}(g)+4 \\mathrm{H}^{+}(a q)+4 \\mathrm{e}^{-}$; (e)\n$6 \\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{IO}_{3}^{-}(a q)+10 \\mathrm{e}^{-} \\longrightarrow \\mathrm{I}_{2}(s)+12 \\mathrm{OH}^{-}(a q) ;(\\mathrm{f})$\n$\\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{SO}_{3}{ }^{2-}(a q) \\longrightarrow \\mathrm{SO}_{4}{ }^{2-}(a q)+2 \\mathrm{H}^{+}(a q)+2 \\mathrm{e}^{-}$; (g)\n$8 \\mathrm{H}^{+}(a q)+\\mathrm{MnO}_{4}^{-}(a q)+5 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Mn}^{2+}(a q)+4 \\mathrm{H}_{2} \\mathrm{O}(l) ;(\\mathrm{h})$"}
{"id": 5388, "contents": "2290. Chapter 7 - \n$\\mathrm{Cl}^{-}(a q)+6 \\mathrm{OH}^{-}(a q) \\longrightarrow \\mathrm{ClO}_{3}^{-}(a q)+3 \\mathrm{H}_{2} \\mathrm{O}(l)+6 \\mathrm{e}^{-}$\n20. (a) $\\mathrm{Sn}^{2+}(a q)+2 \\mathrm{Cu}^{2+}(a q) \\longrightarrow \\mathrm{Sn}^{4+}(a q)+2 \\mathrm{Cu}^{+}(a q)$; (b)\n$\\mathrm{H}_{2} \\mathrm{~S}(g)+\\mathrm{Hg}_{2}{ }^{2+}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{Hg}(l)+\\mathrm{S}(s)+2 \\mathrm{H}_{3} \\mathrm{O}^{+}(a q) ;$ (c)\n$5 \\mathrm{CN}^{-}(a q)+2 \\mathrm{ClO}_{2}(a q)+3 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 5 \\mathrm{CNO}^{-}(a q)+2 \\mathrm{Cl}^{-}(a q)+2 \\mathrm{H}_{3} \\mathrm{O}^{+}(a q) ;$ (d)\n$\\mathrm{Fe}^{2+}(a q)+\\mathrm{Ce}^{4+}(a q) \\longrightarrow \\mathrm{Fe}^{3+}(a q)+\\mathrm{Ce}^{3+}(a q)$; (e)\n$2 \\mathrm{HBrO}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+2 \\mathrm{Br}^{-}(a q)+\\mathrm{O}_{2}(g)$"}
{"id": 5389, "contents": "2290. Chapter 7 - \n21. (a) $2 \\mathrm{MnO}_{4}{ }^{-}(a q)+3 \\mathrm{NO}_{2}{ }^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{MnO}_{2}(s)+3 \\mathrm{NO}_{3}{ }^{-}(a q)+2 \\mathrm{OH}^{-}(a q)$; (b)\n$3 \\mathrm{MnO}_{4}{ }^{2-}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{MnO}_{4}^{-}(a q)+4 \\mathrm{OH}^{-}(a q)+\\mathrm{MnO}_{2}(s)$ (in base); (c)\n$\\mathrm{Br}_{2}(l)+\\mathrm{SO}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 4 \\mathrm{H}^{+}(a q)+2 \\mathrm{Br}^{-}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q)$\n22. (a) $0.435 \\mathrm{~mol} \\mathrm{Na}, 0.217 \\mathrm{~mol} \\mathrm{Cl}_{2}, 15.4 \\mathrm{~g} \\mathrm{Cl}_{2}$; (b) $0.005780 \\mathrm{~mol} \\mathrm{HgO}, 2.890 \\times 10^{-3} \\mathrm{~mol} \\mathrm{O}_{2}, 9.248 \\times 10^{-2} \\mathrm{~g} \\mathrm{O}_{2}$;"}
{"id": 5390, "contents": "2290. Chapter 7 - \n(c) $8.00 \\mathrm{~mol} \\mathrm{NaNO}_{3}, 6.8 \\times 10^{2} \\mathrm{~g} \\mathrm{NaNO}_{3}$; (d) $1665 \\mathrm{~mol} \\mathrm{CO}_{2}, 73.3 \\mathrm{~kg} \\mathrm{CO}_{2}$; (e) $18.86 \\mathrm{~mol} \\mathrm{CuO}, 2.330 \\mathrm{~kg} \\mathrm{CuCO} 3$;\n(f) $0.4580 \\mathrm{~mol} \\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{Br}_{2}, 86.05 \\mathrm{~g} \\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{Br}_{2}$\n23. (a) $0.0686 \\mathrm{~mol} \\mathrm{Mg}, 1.67 \\mathrm{~g} \\mathrm{Mg}$; (b) $2.701 \\times 10^{-3} \\mathrm{~mol} \\mathrm{O}_{2}, 0.08644 \\mathrm{~g} \\mathrm{O}_{2}$; (c) $6.43 \\mathrm{~mol} \\mathrm{MgCO}_{3}, 542 \\mathrm{~g} \\mathrm{MgCO}_{3}$ (d) $768 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{O}, 13.8 \\mathrm{~kg} \\mathrm{H}_{2} \\mathrm{O}$; (e) $16.31 \\mathrm{~mol} \\mathrm{BaO}_{2}, 2762 \\mathrm{~g} \\mathrm{BaO}_{2}$; (f) $0.207 \\mathrm{~mol} \\mathrm{C}_{2} \\mathrm{H}_{4}, 5.81 \\mathrm{~g} \\mathrm{C}_{2} \\mathrm{H}_{4}$"}
{"id": 5391, "contents": "2290. Chapter 7 - \n24. (a) volume HCl solution $\\longrightarrow \\mathrm{mol} \\mathrm{HCl} \\longrightarrow \\mathrm{mol} \\mathrm{GaCl}_{3}$; (b) $1.25 \\mathrm{~mol} \\mathrm{GaCl}_{3}, 2.2 \\times 10^{2} \\mathrm{~g} \\mathrm{GaCl}_{3}$\n25. (a) $5.337 \\times 10^{22}$ molecules; (b) $10.41 \\mathrm{~g} \\mathrm{Zn}(\\mathrm{CN})_{2}$\n26. $\\mathrm{SiO}_{2}+3 \\mathrm{C} \\longrightarrow \\mathrm{SiC}+2 \\mathrm{CO}, 4.50 \\mathrm{~kg} \\mathrm{SiO}$\n27. $5.00 \\times 10^{3} \\mathrm{~kg}$\n28. $1.28 \\times 10^{5} \\mathrm{~g} \\mathrm{CO}_{2}$\n29. 161.4 mL KI solution\n30. $176 \\mathrm{~g} \\mathrm{TiO}_{2}$\n31. The limiting reactant is Cl 2 .\n32. Percent yield $=31 \\%$\n33. $\\mathrm{g} \\mathrm{CCl}_{4} \\longrightarrow \\mathrm{~mol} \\mathrm{CCl}_{4} \\longrightarrow \\operatorname{mol~CCl}_{2} \\mathrm{~F}_{2} \\longrightarrow \\mathrm{~g} \\mathrm{CCl}_{2} \\mathrm{~F}_{2}$, percent yield $=48.3 \\%$\n34. percent yield $=91.3 \\%$\n35. Convert mass of ethanol to moles of ethanol; relate the moles of ethanol to the moles of ether produced using the stoichiometry of the balanced equation. Convert moles of ether to grams; divide the actual grams of ether (determined through the density) by the theoretical mass to determine the percent yield; 87.6\\%"}
{"id": 5392, "contents": "2290. Chapter 7 - \n36. The conversion needed is $\\mathrm{mol} \\mathrm{Cr} \\longrightarrow \\mathrm{mol} \\mathrm{H}_{3} \\mathrm{PO}_{4}$. Then compare the amount of Cr to the amount of acid present. Cr is the limiting reactant.\n37. $\\mathrm{Na}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ is the limiting reactant. percent yield $=86.56 \\%$\n38. Only four molecules can be made.\n39. This amount cannot be weighted by ordinary balances and is worthless.\n40. $3.4 \\times 10^{-3} \\mathrm{MH}_{2} \\mathrm{SO}_{4}$\n41. $9.6 \\times 10^{-3} \\mathrm{MCl}^{-}$\n42. $22.4 \\%$\n43. The empirical formula is $\\mathrm{BH}_{3}$. The molecular formula is $\\mathrm{B}_{2} \\mathrm{H}_{6}$.\n44. 49.6 mL\n45. 13.64 mL\n46. 0.0122 M\n47. 34.99 mL KOH\n48. The empirical formula is $\\mathrm{WCl}_{4}$."}
{"id": 5393, "contents": "2291. Chapter 8 - \n1. The cutting edge of a knife that has been sharpened has a smaller surface area than a dull knife. Since pressure is force per unit area, a sharp knife will exert a higher pressure with the same amount of force and cut through material more effectively.\n2. Lying down distributes your weight over a larger surface area, exerting less pressure on the ice compared to standing up. If you exert less pressure, you are less likely to break through thin ice.\n3. $0.809 \\mathrm{~atm} ; 82.0 \\mathrm{kPa}$\n4. $2.2 \\times 10^{2} \\mathrm{kPa}$\n5. Earth: $14.7 \\mathrm{lb} \\mathrm{in}{ }^{-2}$; Venus: $1.30 \\times 10^{3} \\mathrm{lb} \\mathrm{in}^{-2}$\n6. (a) 101.5 kPa ; (b) 51 torr drop\n7. (a) 264 torr; (b) $35,200 \\mathrm{~Pa}$; (c) 0.352 bar\n8. (a) 623 mm Hg ; (b) 0.820 atm ; (c) 83.1 kPa\n9. With a closed-end manometer, no change would be observed, since the vaporized liquid would contribute equal, opposing pressures in both arms of the manometer tube. However, with an open-ended manometer, a higher pressure reading of the gas would be obtained than expected, since $P_{\\text {gas }}=P_{\\mathrm{atm}}+P_{\\mathrm{vol}}$ liquid.\n10. As the bubbles rise, the pressure decreases, so their volume increases as suggested by Boyle's law.\n11. (a) The number of particles in the gas increases as the volume increases. (b) temperature, pressure\n12. The curve would be farther to the right and higher up, but the same basic shape.\n13. About 12.5 L\n14. $3.40 \\times 10^{3}$ torr\n15. 12.1 L\n16. 217 L\n17. $8.190 \\times 10^{-2} \\mathrm{~mol} ; 5.553 \\mathrm{~g}$"}
{"id": 5394, "contents": "2291. Chapter 8 - \n15. 12.1 L\n16. 217 L\n17. $8.190 \\times 10^{-2} \\mathrm{~mol} ; 5.553 \\mathrm{~g}$\n18. (a) $7.24 \\times 10^{-2} \\mathrm{~g}$; (b) 23.1 g ; (c) $1.5 \\times 10^{-4} \\mathrm{~g}$\n19. 5561 L\n20. 46.4 g\n21. For a gas exhibiting ideal behavior:"}
{"id": 5395, "contents": "2291. Chapter 8 - \n22. (a) $1.85 \\mathrm{~L} \\mathrm{CCl}_{2} \\mathrm{~F}_{2}$; (b) $4.66 \\mathrm{~L} \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{~F}$\n23. 0.644 atm\n24. The pressure decreases by a factor of 3 .\n25. $4.64 \\mathrm{~g} \\mathrm{~L}^{-1}$\n26. 38.8 g\n27. $72.0 \\mathrm{~g} \\mathrm{~mol}^{-1}$\n28. $88.1 \\mathrm{~g} \\mathrm{~mol}^{-1} ; \\mathrm{PF}_{3}$\n29. 141 atm\n30. $\\mathrm{CH}_{4}: 276 \\mathrm{kPa} ; \\mathrm{C}_{2} \\mathrm{H}_{6}: 27 \\mathrm{kPa} ; \\mathrm{C}_{3} \\mathrm{H}_{8}: 3.4 \\mathrm{kPa}$\n31. Yes\n32. 740 torr\n33. (a) Determine the moles of HgO that decompose; using the chemical equation, determine the moles of $\\mathrm{O}_{2}$ produced by decomposition of this amount of HgO ; and determine the volume of $\\mathrm{O}_{2}$ from the moles of $\\mathrm{O}_{2}$, temperature, and pressure. (b) 0.308 L\n34. (a) Determine the molar mass of $\\mathrm{CCl}_{2} \\mathrm{~F}_{2}$. From the balanced equation, calculate the moles of $\\mathrm{H}_{2}$ needed for the complete reaction. From the ideal gas law, convert moles of $\\mathrm{H}_{2}$ into volume. (b) $3.72 \\times 10^{3} \\mathrm{~L}$\n35. (a) Balance the equation. Determine the grams of $\\mathrm{CO}_{2}$ produced and the number of moles. From the ideal gas law, determine the volume of gas. (b) $7.43 \\times 10^{5} \\mathrm{~L}$\n36. 42.00 L"}
{"id": 5396, "contents": "2291. Chapter 8 - \n36. 42.00 L\n37. (a) 18.0 L ; (b) 0.533 atm\n38. $10.57 \\mathrm{~L} \\mathrm{O}_{2}$\n39. $5.40 \\times 10^{5} \\mathrm{~L}$\n40. $\\mathrm{XeF}_{4}$\n41. 4.2 hours\n42. Effusion can be defined as the process by which a gas escapes through a pinhole into a vacuum. Graham's law states that with a mixture of two gases A and B: $\\left(\\frac{\\text { rate } A}{\\text { rate } B}\\right)=\\left(\\frac{\\text { molar mass of } B}{\\text { molar mass of } A}\\right)^{1 / 2}$. Both A and B are in the same container at the same temperature, and therefore will have the same kinetic energy:\n$\\mathrm{KE}_{\\mathrm{A}}=\\mathrm{KE}_{\\mathrm{B}} \\mathrm{KE}=\\frac{1}{2} m v^{2}$\nTherefore, $\\frac{1}{2} m_{\\mathrm{A}} v_{\\mathrm{A}}^{2}=\\frac{1}{2} m_{\\mathrm{B}} v_{\\mathrm{B}}^{2}$\n$\\frac{v_{\\mathrm{A}}^{2}}{v_{\\mathrm{B}}^{2}}=\\frac{m_{\\mathrm{B}}}{m_{\\mathrm{A}}}$\n$\\left(\\frac{v_{\\mathrm{A}}^{2}}{v_{\\mathrm{B}}^{2}}\\right)^{1 / 2}=\\left(\\frac{m_{\\mathrm{B}}}{m_{\\mathrm{A}}}\\right)^{1 / 2}$\n$\\frac{v_{\\mathrm{A}}}{v_{\\mathrm{B}}}=\\left(\\frac{m_{\\mathrm{B}}}{m_{\\mathrm{A}}}\\right)^{1 / 2}$\n43. $\\mathrm{F}_{2}, \\mathrm{~N}_{2} \\mathrm{O}, \\mathrm{Cl}_{2}, \\mathrm{H}_{2} \\mathrm{~S}$"}
{"id": 5397, "contents": "2291. Chapter 8 - \n43. $\\mathrm{F}_{2}, \\mathrm{~N}_{2} \\mathrm{O}, \\mathrm{Cl}_{2}, \\mathrm{H}_{2} \\mathrm{~S}$\n44. $1.4 ; 1.2$\n45. 51.7 cm\n46. Yes. At any given instant, there are a range of values of molecular speeds in a sample of gas. Any single molecule can speed up or slow down as it collides with other molecules. The average speed of all the molecules is constant at constant temperature.\n47. $\\mathrm{H}_{2} \\mathrm{O}$. Cooling slows the speeds of the He atoms, causing them to behave as though they were heavier.\n48. (a) The number of collisions per unit area of the container wall is constant. (b) The average kinetic energy doubles. (c) The root mean square speed increases to $\\sqrt{2}$ times its initial value; $u_{\\text {rms }}$ is proportional to $\\sqrt{\\mathrm{KE}_{\\text {avg }}}$.\n49. (a) equal; (b) less than; (c) $29.48 \\mathrm{~g} \\mathrm{~mol}^{-1}$; (d) $1.0966 \\mathrm{~g} \\mathrm{~L}^{-1}$; (e) $0.129 \\mathrm{~g} / \\mathrm{L}$; (f) $4.01 \\times 10^{5} \\mathrm{~g}$; net lifting capacity $=384 \\mathrm{lb}$; (g) 270 L ; (h) $39.1 \\mathrm{~kJ} \\mathrm{~min}^{-1}$\n50. Gases C, E, and F\n51. The gas behavior most like an ideal gas will occur under the conditions in (b). Molecules have high speeds and move through greater distances between collision; they also have shorter contact times and interactions are less likely. Deviations occur with the conditions described in (a) and (c). Under conditions of (a), some gases may liquefy. Under conditions of (c), most gases will liquefy.\n52. $\\mathrm{SF}_{6}$"}
{"id": 5398, "contents": "2291. Chapter 8 - \n52. $\\mathrm{SF}_{6}$\n53. (a) A straight horizontal line at 1.0; (b) When real gases are at low pressures and high temperatures, they behave close enough to ideal gases that they are approximated as such; however, in some cases, we see that at a high pressure and temperature, the ideal gas approximation breaks down and is significantly different from the pressure calculated by the ideal gas equation. (c) The greater the compressibility, the more the volume matters. At low pressures, the correction factor for intermolecular attractions is more significant, and the effect of the volume of the gas molecules on Z would be a small lowering compressibility. At higher pressures, the effect of the volume of the gas molecules themselves on Z would increase compressibility (see Figure 8.35). (d) Once again, at low pressures, the effect of intermolecular attractions on Z would be more important than the correction factor for the volume of the gas molecules themselves, though perhaps still small. At higher pressures and low temperatures, the effect of intermolecular attractions would be larger. See Figure 8.35. (e) Low temperatures"}
{"id": 5399, "contents": "2292. Chapter 9 - \n1. The temperature of 1 gram of burning wood is approximately the same for both a match and a bonfire. This is an intensive property and depends on the material (wood). However, the overall amount of produced heat depends on the amount of material; this is an extensive property. The amount of wood in a bonfire is much greater than that in a match; the total amount of produced heat is also much greater, which is why we can sit around a bonfire to stay warm, but a match would not provide enough heat to keep us from getting cold.\n2. Heat capacity refers to the heat required to raise the temperature of the mass of the substance 1 degree; specific heat refers to the heat required to raise the temperature of 1 gram of the substance 1 degree. Thus, heat capacity is an extensive property, and specific heat is an intensive one.\n3. (a) $47.6 \\mathrm{~J} /{ }^{\\circ} \\mathrm{C} ; 11.38 \\mathrm{cal}{ }^{\\circ} \\mathrm{C}^{-1}$; (b) $407 \\mathrm{~J} /{ }^{\\circ} \\mathrm{C}$; $97.3 \\mathrm{cal}{ }^{\\circ} \\mathrm{C}^{-1}$\n4. $1310 \\mathrm{~J} ; 313 \\mathrm{cal}$\n5. $7.15{ }^{\\circ} \\mathrm{C}$\n6. (a) $0.390 \\mathrm{~J} / \\mathrm{g}{ }^{\\circ} \\mathrm{C}$; (b) Copper is a likely candidate.\n7. We assume that the density of water is $1.0 \\mathrm{~g} / \\mathrm{cm}^{3}(1 \\mathrm{~g} / \\mathrm{mL})$ and that it takes as much energy to keep the water at $85^{\\circ} \\mathrm{F}$ as to heat it from $72^{\\circ} \\mathrm{F}$ to $85^{\\circ} \\mathrm{F}$. We also assume that only the water is going to be heated. Energy required $=7.47 \\mathrm{kWh}$"}
{"id": 5400, "contents": "2292. Chapter 9 - \n8. lesser; more heat would be lost to the coffee cup and the environment and so $\\Delta T$ for the water would be lesser and the calculated $q$ would be lesser\n9. greater, since taking the calorimeter's heat capacity into account will compensate for the thermal energy transferred to the solution from the calorimeter; this approach includes the calorimeter itself, along with the solution, as \"surroundings\": $q_{\\mathrm{rxn}}=-\\left(q_{\\text {solution }}+q_{\\text {calorimeter }}\\right)$; since both $q_{\\text {solution }}$ and $q_{\\text {calorimeter }}$ are negative, including the latter term $\\left(q_{\\mathrm{rxn}}\\right)$ will yield a greater value for the heat of the dissolution\n10. The temperature of the coffee will drop 1 degree.\n11. $5.7 \\times 10^{2} \\mathrm{~kJ}$\n12. $38.5^{\\circ} \\mathrm{C}$\n13. -2.2 kJ ; The heat produced shows that the reaction is exothermic.\n14. 1.4 kJ\n15. 22.6. Since the mass and the heat capacity of the solution is approximately equal to that of the water, the two-fold increase in the amount of water leads to a two-fold decrease of the temperature change.\n16. 11.7 kJ\n17. $30 \\%$\n18. 0.24 g\n19. $1.4 \\times 10^{2}$ Calories\n20. The enthalpy change of the indicated reaction is for exactly 1 mol HCL and 1 mol NaOH ; the heat in the example is produced by 0.0500 mol HCl and 0.0500 mol NaOH .\n21. $25 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n22. $81 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n23. 5204.4 kJ\n24. $1.83 \\times 10^{-2} \\mathrm{~mol}$\n25. $-802 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$"}
{"id": 5401, "contents": "2292. Chapter 9 - \n24. $1.83 \\times 10^{-2} \\mathrm{~mol}$\n25. $-802 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n26. $15.5 \\mathrm{~kJ} /{ }^{\\circ} \\mathrm{C}$\n27. 7.43 g\n28. Yes.\n29. 459.6 kJ\n30. $-494 \\mathrm{~kJ} / \\mathrm{mol}$\n31. $44.01 \\mathrm{~kJ} / \\mathrm{mol}$\n32. -394 kJ\n33. 265 kJ\n34. $90.3 \\mathrm{~kJ} / \\mathrm{mol}$\n35. (a) $-1615.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$; (b) $-484.3 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$; (c) 164.2 kJ ; (d) -232.1 kJ\n36. $-54.04 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n37. $-2660 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n38. -66.4 kJ\n39. -122.8 kJ\n40. 3.7 kg\n41. On the assumption that the best rocket fuel is the one that gives off the most heat, $\\mathrm{B}_{2} \\mathrm{H}_{6}$ is the prime candidate.\n42. -88.2 kJ\n43. (a) $\\mathrm{C}_{3} \\mathrm{H}_{8}(g)+5 \\mathrm{O}_{2}(g) \\longrightarrow 3 \\mathrm{CO}_{2}(g)+4 \\mathrm{H}_{2} \\mathrm{O}(l)$; (b) 1570 L air; (c) $-104.5 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$; (d) $75.4^{\\circ} \\mathrm{C}$\n44. (a) -114 kJ ;\n(b) 30 kJ ;\n(c) -1055 kJ"}
{"id": 5402, "contents": "2292. Chapter 9 - \n44. (a) -114 kJ ;\n(b) 30 kJ ;\n(c) -1055 kJ\n45. The specific average bond distance is the distance with the lowest energy. At distances less than the bond distance, the positive charges on the two nuclei repel each other, and the overall energy increases.\n46. The greater bond energy is in the figure on the left. It is the more stable form.\n47."}
{"id": 5403, "contents": "2292. Chapter 9 - \n$$\n\\begin{array}{lc}\n\\mathrm{HCl}(g) \\longrightarrow \\frac{1}{2} \\mathrm{H}_{2}(g)+\\frac{1}{2} \\mathrm{Cl}_{2}(g) & \\Delta H_{1}^{\\circ}=-\\Delta H_{\\mathrm{f}[\\mathrm{HCl}(g)]}^{\\circ} \\\\\n\\frac{1}{2} \\mathrm{H}_{2}(g) \\longrightarrow \\mathrm{H}(g) & \\Delta H_{2}^{\\circ}=\\Delta H_{\\mathrm{f}[\\mathrm{H}(g)]}^{\\circ} \\\\\n\\frac{1}{2} \\mathrm{Cl}_{2}(g) \\longrightarrow \\mathrm{Cl}(g) & \\Delta H_{3}^{\\circ}=\\Delta H_{\\mathrm{f}[\\mathrm{Cl}(g)]}^{\\circ} \\\\\n\\hline \\mathrm{HCl}(g) \\longrightarrow \\mathrm{H}(g)+\\mathrm{Cl}(g) & \\Delta H_{298}^{\\circ}=\\Delta H_{1}^{\\circ}+\\Delta H_{2}^{\\circ}+\\Delta H_{3}^{\\circ} \\\\\nD_{\\mathrm{HCl}}^{\\circ}=\\Delta H_{298}^{\\circ}=\\Delta H_{\\mathrm{f}[\\mathrm{HCl}(g)]}^{\\circ}+\\Delta H_{\\mathrm{f}[\\mathrm{H}(g)]}^{\\circ}+\\Delta H_{\\mathrm{f}[\\mathrm{Cl}(g)]}^{\\circ} \\\\\n& =-(-92.307 \\mathrm{~kJ})+217.97 \\mathrm{~kJ}+121.3 \\mathrm{~kJ} \\\\\n& =431.6 \\mathrm{~kJ}\n\\end{array}\n$$\n\n96. The $\\mathrm{S}-\\mathrm{F}$ bond in $\\mathrm{SF}_{4}$ is stronger.\n97."}
{"id": 5404, "contents": "2292. Chapter 9 - \n96. The $\\mathrm{S}-\\mathrm{F}$ bond in $\\mathrm{SF}_{4}$ is stronger.\n97.\n\n\n\nThe $\\mathrm{C}-\\mathrm{C}$ single bonds are longest.\n100. (a) When two electrons are removed from the valence shell, the Ca radius loses the outermost energy level and reverts to the lower $n=3$ level, which is much smaller in radius. (b) The +2 charge on calcium pulls the oxygen much closer compared with K , thereby increasing the lattice energy relative to a less charged ion. (c) Removal of the $4 s$ electron in Ca requires more energy than removal of the $4 s$ electron in $K$ because of the stronger attraction of the nucleus and the extra energy required to break the pairing of the electrons. The second ionization energy for K requires that an electron be removed from a lower energy level, where the attraction is much stronger from the nucleus for the electron. In addition, energy is required to unpair two electrons in a full orbital. For Ca , the second ionization potential requires removing only a lone electron in the exposed outer energy level. (d) In Al, the removed electron is relatively unprotected and unpaired in a $p$ orbital. The higher energy for Mg mainly reflects the unpairing of the $2 s$ electron.\n102. (d)\n104. $4008 \\mathrm{~kJ} / \\mathrm{mol}$; both ions in MgO have twice the charge of the ions in LiF; the bond length is very similar and both have the same structure; a quadrupling of the energy is expected based on the equation for lattice energy\n106. (a) $\\mathrm{Na}_{2} \\mathrm{O}$; $\\mathrm{Na}^{+}$has a smaller radius than $\\mathrm{K}^{+}$; (b) BaS ; Ba has a larger charge than K ; (c) BaS ; Ba and S have larger charges; (d) BaS; S has a larger charge\n108. (e)"}
{"id": 5405, "contents": "2293. Chapter 10 - \n1. Liquids and solids are similar in that they are matter composed of atoms, ions, or molecules. They are incompressible and have similar densities that are both much larger than those of gases. They are different in that liquids have no fixed shape, and solids are rigid.\n2. They are similar in that the atoms or molecules are free to move from one position to another. They differ in that the particles of a liquid are confined to the shape of the vessel in which they are placed. In contrast,\na gas will expand without limit to fill the space into which it is placed.\n3. All atoms and molecules will condense into a liquid or solid in which the attractive forces exceed the kinetic energy of the molecules, at sufficiently low temperature.\n4. (a) Dispersion forces occur as an atom develops a temporary dipole moment when its electrons are distributed asymmetrically about the nucleus. This structure is more prevalent in large atoms such as argon or radon. A second atom can then be distorted by the appearance of the dipole in the first atom. The electrons of the second atom are attracted toward the positive end of the first atom, which sets up a dipole in the second atom. The net result is rapidly fluctuating, temporary dipoles that attract one another (e.g., Ar). (b) A dipole-dipole attraction is a force that results from an electrostatic attraction of the positive end of one polar molecule for the negative end of another polar molecule (e.g., ICI molecules attract one another by dipole-dipole interaction). (c) Hydrogen bonds form whenever a hydrogen atom is bonded to one of the more electronegative atoms, such as a fluorine, oxygen, or nitrogen atom. The electrostatic attraction between the partially positive hydrogen atom in one molecule and the partially negative atom in another molecule gives rise to a strong dipole-dipole interaction called a hydrogen bond (e.g., HF...HF ).\n5. The London forces typically increase as the number of electrons increase."}
{"id": 5406, "contents": "2293. Chapter 10 - \n5. The London forces typically increase as the number of electrons increase.\n6. (a) $\\mathrm{SiH}_{4}<\\mathrm{HCl}<\\mathrm{H}_{2} \\mathrm{O}$; (b) $\\mathrm{F}_{2}<\\mathrm{Cl}_{2}<\\mathrm{Br}_{2}$; (c) $\\mathrm{CH}_{4}<\\mathrm{C}_{2} \\mathrm{H}_{6}<\\mathrm{C}_{3} \\mathrm{H}_{8}$; (d) $\\mathrm{N}_{2}<\\mathrm{O}_{2}<\\mathrm{NO}$\n7. Only rather small dipole-dipole interactions from C - H bonds are available to hold $n$-butane in the liquid state. Chloroethane, however, has rather large dipole interactions because of the $\\mathrm{Cl}-\\mathrm{C}$ bond; the interaction, therefore, is stronger, leading to a higher boiling point.\n8. $-85^{\\circ} \\mathrm{C}$. Water has stronger hydrogen bonds, so it melts at a higher temperature.\n9. The hydrogen bond between two hydrogen fluoride molecules is stronger than that between two water molecules because the electronegativity of $F$ is greater than that of $O$. Consequently, the partial negative charge on F is greater than that on O . The hydrogen bond between the partially positive H and the larger partially negative F will be stronger than that formed between H and O .\n10. H -bonding is the principle IMF holding the protein strands together. The H -bonding is between the $\\mathrm{N}-\\mathrm{H}$ and $\\mathrm{C}=\\mathrm{O}$.\n11. (a) hydrogen bonding, dipole-dipole attraction, and dispersion forces; (b) dispersion forces; (c) dipoledipole attraction and dispersion forces\n12. The water molecules have strong intermolecular forces of hydrogen bonding. The water molecules are thus attracted strongly to one another and exhibit a relatively large surface tension, forming a type of \"skin\" at its surface. This skin can support a bug or paper clip if gently placed on the water."}
{"id": 5407, "contents": "2293. Chapter 10 - \n13. Temperature has an effect on intermolecular forces: The higher the temperature, the greater the kinetic energies of the molecules and the greater the extent to which their intermolecular forces are overcome, and so the more fluid (less viscous) the liquid. The lower the temperature, the less the intermolecular forces are overcome, and so the less viscous the liquid.\n14. (a) As the water reaches higher temperatures, the increased kinetic energies of its molecules are more effective in overcoming hydrogen bonding, and so its surface tension decreases. Surface tension and intermolecular forces are directly related. (b) The same trend in viscosity is seen as in surface tension, and for the same reason.\n15. $1.7 \\times 10^{-4} \\mathrm{~m}$\n16. The heat is absorbed by the ice, providing the energy required to partially overcome intermolecular attractive forces in the solid and causing a phase transition to liquid water. The solution remains at $0^{\\circ} \\mathrm{C}$ until all the ice is melted. Only the amount of water existing as ice changes until the ice disappears. Then the temperature of the water can rise.\n17. We can see the amount of liquid in an open container decrease and we can smell the vapor of some liquids.\n18. The vapor pressure of a liquid decreases as the strength of its intermolecular forces increases.\n19. As the temperature increases, the average kinetic energy of the molecules of gasoline increases and so a greater fraction of molecules have sufficient energy to escape from the liquid than at lower temperatures.\n20. They are equal when the pressure of gas above the liquid is exactly 1 atm.\n21. approximately $95^{\\circ} \\mathrm{C}$\n22. (a) At 5000 feet, the atmospheric pressure is lower than at sea level, and water will therefore boil at a lower temperature. This lower temperature will cause the physical and chemical changes involved in cooking\nthe egg to proceed more slowly, and a longer time is required to fully cook the egg. (b) As long as the air surrounding the body contains less water vapor than the maximum that air can hold at that temperature, perspiration will evaporate, thereby cooling the body by removing the heat of vaporization required to vaporize the water."}
{"id": 5408, "contents": "2293. Chapter 10 - \n23. Dispersion forces increase with molecular mass or size. As the number of atoms composing the molecules in this homologous series increases, so does the extent of intermolecular attraction via dispersion forces and, consequently, the energy required to overcome these forces and vaporize the liquids.\n24. The boiling point of $\\mathrm{CS}_{2}$ is higher than that of $\\mathrm{CO}_{2}$ partially because of the higher molecular weight of $\\mathrm{CS}_{2}$; consequently, the attractive forces are stronger in $\\mathrm{CS}_{2}$. It would be expected, therefore, that the heat of vaporization would be greater than that of $9.8 \\mathrm{~kJ} / \\mathrm{mol}$ for $\\mathrm{CO}_{2}$. A value of $28 \\mathrm{~kJ} / \\mathrm{mol}$ would seem reasonable. A value of $-8.4 \\mathrm{~kJ} / \\mathrm{mol}$ would indicate a release of energy upon vaporization, which is clearly implausible.\n25. The thermal energy (heat) needed to evaporate the liquid is removed from the skin.\n26. 1130 kJ\n27. (a) 13.0 kJ ; (b) It is likely that the heat of vaporization will have a larger magnitude since in the case of vaporization the intermolecular interactions have to be completely overcome, while melting weakens or destroys only some of them.\n28. At low pressures and $0.005^{\\circ} \\mathrm{C}$, the water is a gas. As the pressure increases to 4.6 torr, the water becomes a solid; as the pressure increases still more, it becomes a liquid. At $40^{\\circ} \\mathrm{C}$, water at low pressure is a vapor; at pressures higher than about 75 torr, it converts into a liquid. At $-40^{\\circ} \\mathrm{C}$, water goes from a gas to a solid as the pressure increases above very low values.\n29. (a) gas; (b) gas; (c) gas; (d) gas; (e) solid; (f) gas\n30."}
{"id": 5409, "contents": "2293. Chapter 10 - \n61. Yes, ice will sublime, although it may take it several days. Ice has a small vapor pressure, and some ice molecules form gas and escape from the ice crystals. As time passes, more and more solid converts to gas until eventually the clothes are dry.\n63. (a)\n\n(b)\n\n(c)"}
{"id": 5410, "contents": "2293. Chapter 10 - \n(e) liquid phase (f) sublimation\n65. (e) molecular crystals\n67. Ice has a crystalline structure stabilized by hydrogen bonding. These intermolecular forces are of comparable strength and thus require the same amount of energy to overcome. As a result, ice melts at a single temperature and not over a range of temperatures. The various, very large molecules that compose butter experience varied van der Waals attractions of various strengths that are overcome at various temperatures, and so the melting process occurs over a wide temperature range.\n69. (a) ionic; (b) covalent network; (c) molecular; (d) metallic; (e) covalent network; (f) molecular; (g)\nmolecular; (h) ionic; (i) ionic\n71. $\\mathrm{X}=$ ionic; $\\mathrm{Y}=$ metallic; $\\mathrm{Z}=$ covalent network\n73. (b) metallic solid\n75. The structure of this low-temperature form of iron (below $910^{\\circ} \\mathrm{C}$ ) is body-centered cubic. There is oneeighth atom at each of the eight corners of the cube and one atom in the center of the cube.\n77. eight\n79. 12\n81. (a) $1.370 \\AA$; (b) $19.26 \\mathrm{~g} / \\mathrm{cm}$\n83. (a) $2.176 \\AA$; (b) $3.595 \\mathrm{~g} / \\mathrm{cm}^{3}$\n85. The crystal structure of Si shows that it is less tightly packed (coordination number 4) in the solid than Al (coordination number 12).\n87. In a closest-packed array, two tetrahedral holes exist for each anion. If only half the tetrahedral holes are occupied, the numbers of anions and cations are equal. The formula for cadmium sulfide is CdS.\n89. $\\mathrm{Co}_{3} \\mathrm{O}_{4}$"}
{"id": 5411, "contents": "2293. Chapter 10 - \n89. $\\mathrm{Co}_{3} \\mathrm{O}_{4}$\n91. In a simple cubic array, only one cubic hole can be occupied be a cation for each anion in the array. The ratio of thallium to iodide must be 1:1; therefore, the formula for thallium is Tli.\n93. $59.95 \\%$; The oxidation number of titanium is +4 .\n95. Both ions are close in size: $\\mathrm{Mg}, 0.65 ; \\mathrm{Li}, 0.60$. This similarity allows the two to interchange rather easily. The difference in charge is generally compensated by the switch of $\\mathrm{Si}^{4+}$ for $\\mathrm{Al}^{3+}$.\n97. $\\mathrm{Mn}_{2} \\mathrm{O}_{3}$\n99. $1.48 \\AA$\n101. $2.874 \\AA$\n103. $20.2^{\\circ}$\n105. $1.74 \\times 10^{4} \\mathrm{eV}$"}
{"id": 5412, "contents": "2294. Chapter 11 - \n1. A solution can vary in composition, while a compound cannot vary in composition. Solutions are homogeneous at the molecular level, while other mixtures are heterogeneous.\n2. (a) The process is endothermic as the solution is consuming heat. (b) Attraction between the $\\mathrm{K}^{+}$and $\\mathrm{NO}_{3}{ }^{-}$ ions is stronger than between the ions and water molecules (the ion-ion interactions have a lower, more negative energy). Therefore, the dissolution process increases the energy of the molecular interactions, and it consumes the thermal energy of the solution to make up for the difference. (c) No, an ideal solution is formed with no appreciable heat release or consumption.\n3. (a) ion-dipole forces; (b) dipole-dipole forces; (c) dispersion forces; (d) dispersion forces; (e) hydrogen bonding\n4. Heat is released when the total intermolecular forces (IMFs) between the solute and solvent molecules are stronger than the total IMFs in the pure solute and in the pure solvent: Breaking weaker IMFs and forming stronger IMFs releases heat. Heat is absorbed when the total IMFs in the solution are weaker than the total of those in the pure solute and in the pure solvent: Breaking stronger IMFs and forming weaker IMFs absorbs heat.\n5. Crystals of NaCl dissolve in water, a polar liquid with a very large dipole moment, and the individual ions become strongly solvated. Hexane is a nonpolar liquid with a dipole moment of zero and, therefore, does not significantly interact with the ions of the NaCl crystals.\n6. (a) $\\mathrm{Fe}\\left(\\mathrm{NO}_{3}\\right)_{3}$ is a strong electrolyte, thus it should completely dissociate into $\\mathrm{Fe}^{3+}$ and $\\mathrm{NO}_{3}{ }^{-}$ions. Therefore, (z) best represents the solution. (b) $\\mathrm{Fe}\\left(\\mathrm{NO}_{3}\\right)_{3}(s) \\longrightarrow \\mathrm{Fe}^{3+}(a q)+3 \\mathrm{NO}_{3}{ }^{-}(a q)$"}
{"id": 5413, "contents": "2294. Chapter 11 - \n7. (a) high conductivity (solute is an ionic compound that will dissociate when dissolved); (b) high conductivity (solute is a strong acid and will ionize completely when dissolved); (c) nonconductive (solute is a covalent compound, neither acid nor base, unreactive towards water); (d) low conductivity (solute is a weak base and will partially ionize when dissolved)\n8. (a) ion-dipole; (b) hydrogen bonds; (c) dispersion forces; (d) dipole-dipole attractions; (e) dispersion forces\n9. The solubility of solids usually decreases upon cooling a solution, while the solubility of gases usually decreases upon heating.\n10. $40 \\%$\n11. 2.8 g\n12. 2.9 atm\n13. 102 L HCl\n14. The strength of the bonds between like molecules is stronger than the strength between unlike molecules. Therefore, some regions will exist in which the water molecules will exclude oil molecules and other regions will exist in which oil molecules will exclude water molecules, forming a heterogeneous region.\n15. Both form homogeneous solutions; their boiling point elevations are the same, as are their lowering of vapor pressures. Osmotic pressure and the lowering of the freezing point are also the same for both solutions.\n16. (a) Find number of moles of $\\mathrm{HNO}_{3}$ and $\\mathrm{H}_{2} \\mathrm{O}$ in 100 g of the solution. Find the mole fractions for the components. (b) The mole fraction of $\\mathrm{HNO}_{3}$ is 0.378 . The mole fraction of $\\mathrm{H}_{2} \\mathrm{O}$ is 0.622 ."}
{"id": 5414, "contents": "2294. Chapter 11 - \n17. (a) $X_{\\mathrm{Na}_{2} \\mathrm{CO}_{3}}=0.0119 ; X_{\\mathrm{H}_{2} \\mathrm{O}}=0.988$; (b) $X_{\\mathrm{NH}_{4} \\mathrm{NO}_{3}}=0.0928 ; X_{\\mathrm{H}_{2} \\mathrm{O}}=0.907$; (c) $X_{\\mathrm{Cl}_{2}}=0.192$; $X_{\\mathrm{CH}_{2} \\mathrm{CI}_{2}}=0.808 ;$ (d) $X_{\\mathrm{C}_{5} \\mathrm{H}_{9} \\mathrm{~N}}=0.00426 ; X_{\\mathrm{CHCl}_{3}}=0.997$\n18. In a 1 M solution, the mole is contained in exactly 1 L of solution. In a 1 m solution, the mole is contained in exactly 1 kg of solvent.\n19. (a) Determine the molar mass of $\\mathrm{HNO}_{3}$. Determine the number of moles of acid in the solution. From the number of moles and the mass of solvent, determine the molality. (b) 33.7 m\n20. (a) $6.70 \\times 10^{-1} \\mathrm{~m}$; (b) 5.67 m ; (c) 2.8 m ; (d) 0.0358 m\n21. 1.08 m\n22. (a) Determine the molar mass of sucrose; determine the number of moles of sucrose in the solution; convert the mass of solvent to units of kilograms; from the number of moles and the mass of solvent, determine the molality; determine the difference between the boiling point of water and the boiling point of the solution; determine the new boiling point. (b) $100.5^{\\circ} \\mathrm{C}$"}
{"id": 5415, "contents": "2294. Chapter 11 - \n23. (a) Determine the molar mass of sucrose; determine the number of moles of sucrose in the solution; convert the mass of solvent to units of kilograms; from the number of moles and the mass of solvent, determine the molality; determine the difference between the freezing temperature of water and the freezing temperature of the solution; determine the new freezing temperature. (b) $-1.8^{\\circ} \\mathrm{C}$\n24. (a) Determine the molar mass of $\\mathrm{Ca}\\left(\\mathrm{NO}_{3}\\right)_{2}$; determine the number of moles of $\\mathrm{Ca}\\left(\\mathrm{NO}_{3}\\right)_{2}$ in the solution; determine the number of moles of ions in the solution; determine the molarity of ions, then the osmotic pressure. (b) 2.67 atm\n25. (a) Determine the molal concentration from the change in boiling point and $K_{\\mathrm{b}}$; determine the moles of solute in the solution from the molal concentration and mass of solvent; determine the molar mass from the number of moles and the mass of solute. (b) $2.1 \\times 10^{2} \\mathrm{~g} \\mathrm{~mol}^{-1}$\n26. No. Pure benzene freezes at $5.5^{\\circ} \\mathrm{C}$, and so the observed freezing point of this solution is depressed by $\\Delta T_{\\mathrm{f}}$ $=5.5-0.4=5.1^{\\circ} \\mathrm{C}$. The value computed, assuming no ionization of HCl , is $\\Delta T_{\\mathrm{f}}=(1.0 \\mathrm{~m})\\left(5.14{ }^{\\circ} \\mathrm{C} / \\mathrm{m}\\right)=5.1$ ${ }^{\\circ} \\mathrm{C}$. Agreement of these values supports the assumption that HCl is not ionized.\n27. $144 \\mathrm{~g} \\mathrm{~mol}^{-1}$\n28. $0.870^{\\circ} \\mathrm{C}$\n29. $\\mathrm{S}_{8}$"}
{"id": 5416, "contents": "2294. Chapter 11 - \n27. $144 \\mathrm{~g} \\mathrm{~mol}^{-1}$\n28. $0.870^{\\circ} \\mathrm{C}$\n29. $\\mathrm{S}_{8}$\n30. $1.39 \\times 10^{4} \\mathrm{~g} \\mathrm{~mol}^{-1}$\n31. 54 g\n32. $100.26^{\\circ} \\mathrm{C}$\n33. (a) $X_{\\mathrm{CH}_{3} \\mathrm{OH}}=0.590 ; X_{\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}}=0.410$; (b) Vapor pressures are: $\\mathrm{CH}_{3} \\mathrm{OH}$ : 55 torr; $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$ : 18 torr; (c) $\\mathrm{CH}_{3} \\mathrm{OH}: 0.75 ; \\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}: 0.25$\n34. The ions and compounds present in the water in the beef lower the freezing point of the beef below $-1^{\\circ} \\mathrm{C}$.\n35. $\\Delta \\mathrm{bp}=K_{\\mathrm{b}} m=\\left(1.20^{\\circ} \\mathrm{C} / m\\right)\\left(\\frac{9.41 \\mathrm{~g} \\times \\frac{1 \\mathrm{~mol} \\mathrm{Hg} \\mathrm{Cl}_{2}}{271.496 \\mathrm{~g}}}{0.03275 \\mathrm{~kg}}\\right)=1.27^{\\circ} \\mathrm{C}$"}
{"id": 5417, "contents": "2294. Chapter 11 - \nThe observed change equals the theoretical change; therefore, no dissociation occurs.\n71.\n\n| Colloidal System | Dispersed Phase | Dispersion Medium |\n| :--- | :--- | :--- |\n| starch dispersion | starch | water |\n| smoke | solid particles | air |\n| fog | water | air |\n| pearl | water | calcium carbonate $\\left(\\mathrm{CaCO}_{3}\\right)$ |\n| whipped cream | air | cream |\n| floating soap | air | soap |\n| jelly | fruit juice | pectin gel |\n| milk | butterfat | water |\n| ruby | chromium(III) oxide $\\left(\\mathrm{Cr}_{2} \\mathrm{O}_{3}\\right)$ | aluminum oxide $\\left(\\mathrm{Al}_{2} \\mathrm{O}_{3}\\right)$ |\n\n73. Colloidal dispersions consist of particles that are much bigger than the solutes of typical solutions. Colloidal particles are either very large molecules or aggregates of smaller species that usually are big enough to scatter light. Colloids are homogeneous on a macroscopic (visual) scale, while solutions are homogeneous on a microscopic (molecular) scale.\n74. If they are placed in an electrolytic cell, dispersed particles will move toward the electrode that carries a charge opposite to their own charge. At this electrode, the charged particles will be neutralized and will coagulate as a precipitate."}
{"id": 5418, "contents": "2295. Chapter 12 - \n1. A reaction has a natural tendency to occur and takes place without the continual input of energy from an external source.\n2. (a) spontaneous; (b) nonspontaneous; (c) spontaneous; (d) nonspontaneous; (e) spontaneous; (f) spontaneous\n3. Although the oxidation of plastics is spontaneous, the rate of oxidation is very slow. Plastics are therefore kinetically stable and do not decompose appreciably even over relatively long periods of time.\n4. There are four initial microstates and four final microstates.\n$\\Delta S=k \\ln \\frac{W_{\\mathrm{f}}}{W_{\\mathrm{i}}}=1.38 \\times 10^{-23} \\mathrm{~J} / \\mathrm{K} \\times \\ln \\frac{4}{4}=0$\n5. The probability for all the particles to be on one side is $\\frac{1}{32}$. This probability is noticeably lower than the $\\frac{1}{8}$ result for the four-particle system. The conclusion we can make is that the probability for all the particles to stay in only one part of the system will decrease rapidly as the number of particles increases, and, for instance, the probability for all molecules of gas to gather in only one side of a room at room temperature and pressure is negligible since the number of gas molecules in the room is very large.\n6. There is only one initial state. For the final state, the energy can be contained in pairs A-C, A-D, B-C, or B-D. Thus, there are four final possible states.\n$\\Delta S=k \\ln \\left(\\frac{W_{\\mathrm{f}}}{W_{\\mathrm{i}}}\\right)=1.38 \\times 10^{-23} \\mathrm{~J} / \\mathrm{K} \\times \\ln \\left(\\frac{4}{1}\\right)=1.91 \\times 10^{-23} \\mathrm{~J} / \\mathrm{K}$"}
{"id": 5419, "contents": "2295. Chapter 12 - \n7. The masses of these molecules would suggest the opposite trend in their entropies. The observed trend is a result of the more significant variation of entropy with a physical state. At room temperature, $\\mathrm{I}_{2}$ is a solid, $\\mathrm{Br}_{2}$ is a liquid, and $\\mathrm{Cl}_{2}$ is a gas.\n8. (a) $\\mathrm{C}_{3} \\mathrm{H}_{7} \\mathrm{OH}(I)$ as it is a larger molecule (more complex and more massive), and so more microstates\ndescribing its motions are available at any given temperature. (b) $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(g)$ as it is in the gaseous state. (c) $2 \\mathrm{H}(\\mathrm{g})$, since entropy is an extensive property, and so two H atoms (or two moles of H atoms) possess twice as much entropy as one atom (or one mole of atoms).\n9. (a) Negative. The relatively ordered solid precipitating decreases the number of mobile ions in solution.\n(b) Negative. There is a net loss of three moles of gas from reactants to products. (c) Positive. There is a net increase of seven moles of gas from reactants to products.\n10. $\\mathrm{C}_{6} \\mathrm{H}_{6}(l)+7.5 \\mathrm{O}_{2}(g) \\longrightarrow 3 \\mathrm{H}_{2} \\mathrm{O}(g)+6 \\mathrm{CO}_{2}(g)$"}
{"id": 5420, "contents": "2295. Chapter 12 - \nThere are 7.5 moles of gas initially, and $3+6=9$ moles of gas in the end. Therefore, it is likely that the entropy increases as a result of this reaction, and $\\Delta S$ is positive.\n21. (a) $107 \\mathrm{~J} / \\mathrm{K}$; (b) $-86.4 \\mathrm{~J} / \\mathrm{K}$; (c) $133.2 \\mathrm{~J} / \\mathrm{K}$; (d) $118.8 \\mathrm{~J} / \\mathrm{K}$; (e) $-326.6 \\mathrm{~J} / \\mathrm{K}$; (f) $-171.9 \\mathrm{~J} / \\mathrm{K}$; (g) $-7.2 \\mathrm{~J} / \\mathrm{K}$\n23. $100.6 \\mathrm{~J} / \\mathrm{K}$\n25. (a) $-198.1 \\mathrm{~J} / \\mathrm{K}$; (b) $-348.9 \\mathrm{~J} / \\mathrm{K}$\n27. As $\\Delta S_{\\text {univ }}<0$ at each of these temperatures, melting is not spontaneous at either of them. The given values for entropy and enthalpy are for NaCl at 298 K . It is assumed that these do not change significantly at the higher temperatures used in the problem.\n29. (a) $2.86 \\mathrm{~J} / \\mathrm{K}$; (b) $24.8 \\mathrm{~J} / \\mathrm{K}$; (c) $-113.2 \\mathrm{~J} / \\mathrm{K}$; (d) $-24.7 \\mathrm{~J} / \\mathrm{K}$; (e) $15.5 \\mathrm{~J} / \\mathrm{K}$; (f) $290.0 \\mathrm{~J} / \\mathrm{K}$\n31. The reaction is nonspontaneous at room temperature."}
{"id": 5421, "contents": "2295. Chapter 12 - \nAbove $400 \\mathrm{~K}, \\Delta G$ will become negative, and the reaction will become spontaneous.\n33. (a) 465.1 kJ nonspontaneous; (b) -106.86 kJ spontaneous; (c) -291.9 kJ spontaneous; (d) -83.4 kJ spontaneous; (e) -406.7 kJ spontaneous; (f) -154.3 kJ spontaneous\n35. (a) The standard free energy of formation is $-1124.3 \\mathrm{~kJ} / \\mathrm{mol}$. (b) The calculation agrees with the value in Appendix G because free energy is a state function (just like the enthalpy and entropy), so its change depends only on the initial and final states, not the path between them.\n37. (a) The reaction is nonspontaneous; (b) Above $566^{\\circ} \\mathrm{C}$ the process is spontaneous.\n39. (a) $1.5 \\times 10^{2} \\mathrm{~kJ}$; (b) -21.9 kJ ; (c) -5.34 kJ ; (d) -0.383 kJ ; (e) 18 kJ ; (f) 71 kJ\n41. (a) $K=41$; (b) $K=0.053$; (c) $K=6.9 \\times 10^{13}$; (d) $K=1.9$; (e) $K=0.04$\n42. (a) 22.1 kJ ;\n(b) $98.9 \\mathrm{~kJ} / \\mathrm{mol}$\n44. $90 \\mathrm{~kJ} / \\mathrm{mol}$"}
{"id": 5422, "contents": "2295. Chapter 12 - \n42. (a) 22.1 kJ ;\n(b) $98.9 \\mathrm{~kJ} / \\mathrm{mol}$\n44. $90 \\mathrm{~kJ} / \\mathrm{mol}$\n46. (a) Under standard thermodynamic conditions, the evaporation is nonspontaneous; (b) $K_{p}=0.031$; (c) The evaporation of water is spontaneous; (d) $P_{\\mathrm{H}_{2} \\mathrm{O}}$ must always be less than $K_{p}$ or less than 0.031 atm .0 .031 atm represents air saturated with water vapor at $25^{\\circ} \\mathrm{C}$, or $100 \\%$ humidity.\n48. (a) Nonspontaneous as $\\Delta G^{\\circ}>0$; (b) $\\Delta G^{\\circ}=\\Delta G^{\\circ}+R T \\ln Q$,\n$\\Delta G=1.7 \\times 10^{3}+\\left(8.314 \\times 310 \\times \\ln \\frac{28}{120}\\right)=-2.1 \\mathrm{~kJ}$. The forward reaction to produce F6P is spontaneous under these conditions.\n50. $\\Delta G$ is negative as the process is spontaneous. $\\Delta H$ is positive as with the solution becoming cold, the dissolving must be endothermic. $\\Delta S$ must be positive as this drives the process, and it is expected for the dissolution of any soluble ionic compound.\n52. (a) Increasing $P_{\\mathrm{O}_{2}}$ will shift the equilibrium toward the products, which increases the value of $K . \\Delta G^{\\circ}$ therefore becomes more negative.\n(b) Increasing $P_{\\mathrm{O}_{2}}$ will shift the equilibrium toward the products, which increases the value of $K . \\Delta G^{\\circ}$ therefore becomes more negative.\n(c) Increasing $P_{\\mathrm{O}_{2}}$ will shift the equilibrium the reactants, which decreases the value of $K . \\Delta G^{\\circ}$ therefore becomes more positive."}
{"id": 5423, "contents": "2296. Chapter 13 - \n1. The reaction can proceed in both the forward and reverse directions.\n2. When a system has reached equilibrium, no further changes in the reactant and product concentrations occur; the forward and reverse reactions continue to proceed, but at equal rates.\n3. Not necessarily. A system at equilibrium is characterized by constant reactant and product concentrations, but the values of the reactant and product concentrations themselves need not be equal.\n4. Equilibrium cannot be established between the liquid and the gas phase if the top is removed from the\nbottle because the system is not closed; one of the components of the equilibrium, the $\\mathrm{Br}_{2}$ vapor, would escape from the bottle until all liquid disappeared. Thus, more liquid would evaporate than can condense back from the gas phase to the liquid phase.\n5. (a) $K_{c}=\\left[\\mathrm{Ag}^{+}\\right]\\left[\\mathrm{Cl}^{-}\\right]<1$. AgCl is insoluble; thus, the concentrations of ions are much less than 1 M ; (b) $K_{c}=\\frac{1}{\\left[\\mathrm{~Pb}^{2+}\\right]\\left[\\mathrm{Cl}^{-}\\right]^{2}}>1$ because $\\mathrm{PbCl}_{2}$ is insoluble and formation of the solid will reduce the concentration of ions to a low level ( $<1 M$ ).\n6. Since $K_{c}=\\frac{\\left[\\mathrm{C}_{6} \\mathrm{H}_{6}\\right]}{\\left[\\mathrm{C}_{2} \\mathrm{H}_{2}\\right]^{3}}$, a value of $K_{c} \\approx 10$ means that $\\mathrm{C}_{6} \\mathrm{H}_{6}$ predominates over $\\mathrm{C}_{2} \\mathrm{H}_{2}$. In such a case, the reaction would be commercially feasible if the rate to equilibrium is suitable.\n7. $K_{c}>1$"}
{"id": 5424, "contents": "2296. Chapter 13 - \n7. $K_{c}>1$\n8. (a) $Q_{c}=\\frac{\\left\\mathrm{CH}_{3} \\mathrm{Cl}\\right}{\\left[\\mathrm{CH}_{4}\\right]\\left[\\mathrm{Cl}_{2}\\right]}$; (b) $Q_{c}=\\frac{[\\mathrm{NO}]^{2}}{\\left[\\mathrm{~N}_{2}\\right]\\left[\\mathrm{O}_{2}\\right]}$; (c) $Q_{c}=\\frac{\\left[\\mathrm{SO}_{3}\\right]^{2}}{\\left[\\mathrm{SO}_{2}\\right]^{2}\\left[\\mathrm{O}_{2}\\right]}$; (d) $Q_{c}=\\left[\\mathrm{SO}_{2}\\right]$; (e) $Q_{c}=\\frac{1}{\\left[\\mathrm{P}_{4}\\right]\\left[\\mathrm{O}_{2}\\right]^{5}}$; (f) $Q_{c}=\\frac{[\\mathrm{Br}]^{2}}{\\left[\\mathrm{Br}_{2}\\right]} ;$ (g) $Q_{c}=\\frac{\\left[\\mathrm{CO}_{2}\\right]}{\\left[\\mathrm{CH}_{4}\\right]\\left[\\mathrm{O}_{2}\\right]^{2}}$; (h) $Q_{c}=\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]^{5}$\n9. (a) $Q_{\\mathrm{c}} 25$ proceeds left; (b) $Q_{P} 0.22$ proceeds right; (c) $Q_{c}$ undefined proceeds left; (d) $Q_{P} 1.00$ proceeds right; (e) $Q_{P} 0$ proceeds right; (f) $Q_{c} 4$ proceeds left\n10. The system will shift toward the reactants to reach equilibrium.\n11. (a) homogenous; (b) homogenous; (c) homogenous; (d) heterogeneous; (e) heterogeneous; (f) homogenous; (g) heterogeneous; (h) heterogeneous\n12. This situation occurs in (a) and (b)."}
{"id": 5425, "contents": "2296. Chapter 13 - \n12. This situation occurs in (a) and (b).\n13. (a) $K_{P}=1.6 \\times 10^{-4}$; (b) $K_{P}=50.2$; (c) $K_{c}=5.34 \\times 10^{-39}$; (d) $K_{c}=4.60 \\times 10^{-3}$\n14. $K_{P}=P_{\\mathrm{H}_{2} \\mathrm{O}}=0.042$.\n15. $Q_{c}=\\frac{\\left[\\mathrm{NH}_{4}{ }^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]}{\\left[\\mathrm{NH}_{3}\\right]}$\n16. The amount of $\\mathrm{CaCO}_{3}$ must be so small that $\\mathrm{P}_{\\mathrm{CO}_{2}}$ is less than $K_{P}$ when the $\\mathrm{CaCO}_{3}$ has completely decomposed. In other words, the starting amount of $\\mathrm{CaCO}_{3}$ cannot completely generate the full $P_{\\mathrm{CO}_{2}}$ required for equilibrium.\n17. The change in enthalpy may be used. If the reaction is exothermic, the heat produced can be thought of as a product. If the reaction is endothermic the heat added can be thought of as a reactant. Additional heat would shift an exothermic reaction back to the reactants but would shift an endothermic reaction to the products. Cooling an exothermic reaction causes the reaction to shift toward the product side; cooling an endothermic reaction would cause it to shift to the reactants' side.\n18. No, it is not at equilibrium. Because the system is not confined, products continuously escape from the region of the flame; reactants are also added continuously from the burner and surrounding atmosphere.\n19. Add $\\mathrm{N}_{2}$; add $\\mathrm{H}_{2}$; decrease the container volume; heat the mixture."}
{"id": 5426, "contents": "2296. Chapter 13 - \n19. Add $\\mathrm{N}_{2}$; add $\\mathrm{H}_{2}$; decrease the container volume; heat the mixture.\n20. (a) $T$ increase $=$ shift right, $V$ decrease $=$ shift left; (b) $T$ increase $=$ shift right, $V=$ no effect; (c) $T$ increase $=$ shift left, $V$ decrease $=$ shift left; (d) $T$ increase $=$ shift left, $V$ decrease $=$ shift right.\n21. (a) $K_{c}=\\frac{\\left[\\mathrm{CH}_{3} \\mathrm{OH}\\right]}{\\left[\\mathrm{H}_{2}\\right]^{2}[\\mathrm{CO}]}$; (b) $\\left[\\mathrm{H}_{2}\\right]$ increases, $[\\mathrm{CO}]$ decreases, $\\left[\\mathrm{CH}_{3} \\mathrm{OH}\\right]$ increases; (c), $\\left[\\mathrm{H}_{2}\\right]$ increases, $[\\mathrm{CO}]$ decreases, $\\left[\\mathrm{CH}_{3} \\mathrm{OH}\\right]$ decreases; (d), $\\left[\\mathrm{H}_{2}\\right]$ increases, $[\\mathrm{CO}]$ increases, $\\left[\\mathrm{CH}_{3} \\mathrm{OH}\\right]$ increases; (e), $\\left[\\mathrm{H}_{2}\\right]$ increases, [CO] increases, $\\left[\\mathrm{CH}_{3} \\mathrm{OH}\\right]$ decreases"}
{"id": 5427, "contents": "2296. Chapter 13 - \n22. (a) $K_{c}=\\frac{[\\mathrm{CO}]\\left[\\mathrm{H}_{2}\\right]}{\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]}$; (b) $\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]$ no change, $[\\mathrm{CO}]$ no change, $\\left[\\mathrm{H}_{2}\\right]$ no change; (c) $\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]$ decreases, $[\\mathrm{CO}]$ decreases, $\\left[\\mathrm{H}_{2}\\right]$ decreases; (d) $\\left[\\mathrm{H}_{2} \\mathrm{O}\\right.$ ] increases, [CO] increases, [ $\\mathrm{H}_{2}$ ] decreases; (f) [ $\\mathrm{H}_{2} \\mathrm{O}$ ] decreases, [CO] increases, $\\left[\\mathrm{H}_{2}\\right]$ increases. In (b), (c), (d), and (e), the mass of carbon will change, but its concentration (activity) will not change.\n23. Only (b)\n24. Add NaCl or some other salt that produces $\\mathrm{Cl}^{-}$to the solution. Cooling the solution forces the equilibrium to the right, precipitating more $\\mathrm{AgCl}(s)$.\n25. Though the solution is saturated, the dynamic nature of the solubility equilibrium means the opposing processes of solid dissolution and precipitation continue to occur (just at equal rates, meaning the dissolved ion concentrations and the amount of undissolved solid remain constant). The radioactive $\\mathrm{Ag}^{+}$\nions detected in the solution phase come from dissolution of the added solid, and their presence is countered by precipitation of nonradioactive $\\mathrm{Ag}^{+}$.\n26. $K_{c}=\\frac{[\\mathrm{C}]^{2}}{\\mathrm{~A}^{2}}$. $[\\mathrm{A}]=0.1 \\mathrm{M},[\\mathrm{B}]=0.1 \\mathrm{M},[\\mathrm{C}]=1 \\mathrm{M}$; and $[\\mathrm{A}]=0.01,[\\mathrm{~B}]=0.250,[\\mathrm{C}]=0.791$."}
{"id": 5428, "contents": "2296. Chapter 13 - \n27. $K_{c}=6.00 \\times 10^{-2}$\n28. $K_{c}=0.50$\n29. $K_{P}=1.9 \\times 10^{3}$\n30. $K_{P}=3.06$\n31. (a) $-2 x,+2 x$; (b) $+\\frac{4}{3} x,-\\frac{2}{3} x,-2 x$; (c) $-2 x, 3 x$; (d) $+x,-x,-3 x$; (e) $+x$; (f) $-\\frac{1}{4} x$.\n32. Activities of pure crystalline solids equal 1 and are constant; however, the mass of Ni does change.\n33. $\\left[\\mathrm{NH}_{3}\\right]=9.1 \\times 10^{-2} \\mathrm{M}$\n34. $P_{\\mathrm{BrCl}}=4.9 \\times 10^{-2} \\mathrm{~atm}$\n35. $[\\mathrm{CO}]=2.04 \\times 10^{-4} \\mathrm{M}$\n36. $P_{\\mathrm{H}_{2} \\mathrm{O}}=3.64 \\times 10^{-3} \\mathrm{~atm}$\n37. Calculate $Q$ based on the calculated concentrations and see if it is equal to $K_{c}$. Because $Q$ does equal 4.32, the system must be at equilibrium.\n38. (a) $\\left[\\mathrm{NO}_{2}\\right]=1.17 \\times 10^{-3} \\mathrm{M}$\n$\\left[\\mathrm{N}_{2} \\mathrm{O}_{4}\\right]=0.128 \\mathrm{M}$\n(b) The assumption that $x$ is negligibly small compared to 0.129 is confirmed by comparing the initial concentration of the $\\mathrm{N}_{2} \\mathrm{O}_{4}$ to its concentration at equilibrium (they differ by just 1 in the least significant digit's place).\n39. (a) $\\left[\\mathrm{H}_{2} \\mathrm{~S}\\right]=0.810 \\mathrm{~atm}$"}
{"id": 5429, "contents": "2296. Chapter 13 - \n39. (a) $\\left[\\mathrm{H}_{2} \\mathrm{~S}\\right]=0.810 \\mathrm{~atm}$\n$\\left[\\mathrm{H}_{2}\\right]=0.014 \\mathrm{~atm}$\n$\\left[\\mathrm{S}_{2}\\right]=0.0072 \\mathrm{~atm}$\n(b) The assumption that $2 x$ is negligibly small compared to 0.824 is confirmed by comparing the initial concentration of the $\\mathrm{H}_{2} \\mathrm{~S}$ to its concentration at equilibrium ( 0.824 atm versus 0.810 atm , a difference of less than $2 \\%$ ).\n40. 507 g\n41. 330 g\n42. (a) 0.33 mol .\n(b) $\\left[\\mathrm{CO}_{2}=0.50 \\mathrm{M}\\right.$. Added $\\mathrm{H}_{2}$ forms some water as a result of a shift to the left after $\\mathrm{H}_{2}$ is added.\n43. (a) $K_{c}=\\frac{\\left[\\mathrm{NH}_{3}\\right]^{4}\\left[\\mathrm{O}_{2}\\right]^{7}}{\\left[\\mathrm{NO}_{2}\\right]^{4}\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]^{6}}$. (b) $\\left[\\mathrm{NH}_{3}\\right]$ must increase for $Q_{c}$ to reach $K_{c}$. (c) The increase in system volume would lower the partial pressures of all reactants (including $\\mathrm{NO}_{2}$ ). (d) $P_{\\mathrm{O}_{2}}=49$ torr\n44. $P_{\\mathrm{N}_{2} \\mathrm{O}_{3}}=1.90 \\mathrm{~atm}$ and $P_{\\mathrm{NO}}=P_{\\mathrm{NO}_{2}}=1.90 \\mathrm{~atm}$"}
{"id": 5430, "contents": "2296. Chapter 13 - \n45. In each of the following, the value of $\\Delta G$ is not given at the temperature of the reaction. Therefore, we must calculate $\\Delta G$ from the values $\\Delta H^{\\circ}$ and $\\Delta S$ and then calculate $\\Delta G$ from the relation $\\Delta G=\\Delta H^{\\circ}-T \\Delta S^{\\circ}$.\n(a) $K=1.07 \\times 10^{-13}$;\n(b) $K=2.42 \\times 10^{-3}$;\n(c) $K=2.73 \\times 10^{4}$;\n(d) $K=0.229$;\n(e) $K=16.1$\n46. The standard free energy change is $\\Delta G_{298}^{\\circ}=-R T \\ln K=4.84 \\mathrm{~kJ} / \\mathrm{mol}$. When reactants and products are in their standard states ( 1 bar or 1 atm ), $Q=1$. As the reaction proceeds toward equilibrium, the reaction shifts left (the amount of products drops while the amount of reactants increases): $Q<1$, and $\\Delta G_{298}$ becomes less positive as it approaches zero. At equilibrium, $Q=K$, and $\\Delta G=0$.\n47. The reaction will be spontaneous at temperatures greater than 287 K .\n48. $K=5.35 \\times 10^{15}$"}
{"id": 5431, "contents": "2296. Chapter 13 - \nThe process is exothermic.\n98. $1.0 \\times 10^{-8} \\mathrm{~atm}$. This is the maximum pressure of the gases under the stated conditions.\n100. $x=1.29 \\times 10^{-5} \\mathrm{~atm}=P_{\\mathrm{O}_{2}}$\n102. -0.16 kJ"}
{"id": 5432, "contents": "2297. Chapter 14 - \n1. One example for $\\mathrm{NH}_{3}$ as a conjugate acid: $\\mathrm{NH}_{2}{ }^{-}+\\mathrm{H}^{+} \\longrightarrow \\mathrm{NH}_{3}$; as a conjugate base:\n$\\mathrm{NH}_{4}{ }^{+}(a q)+\\mathrm{OH}^{-}(a q) \\longrightarrow \\mathrm{NH}_{3}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)$\n2. (a) $\\mathrm{H}_{3} \\mathrm{O}^{+}(a q) \\longrightarrow \\mathrm{H}^{+}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)$; (b) $\\mathrm{HCl}(\\mathrm{aq}) \\longrightarrow \\mathrm{H}^{+}(a q)+\\mathrm{Cl}^{-}(a q)$; (c)\n$\\mathrm{NH}_{3}(a q) \\longrightarrow \\mathrm{H}^{+}(a q)+\\mathrm{NH}_{2}^{-}(a q) ;$ (d) $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}(a q) \\longrightarrow \\mathrm{H}^{+}(a q)+\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}(a q)$; (e)\n$\\mathrm{NH}_{4}{ }^{+}(a q) \\longrightarrow \\mathrm{H}^{+}(a q)+\\mathrm{NH}_{3}(a q) ;$ (f) $\\mathrm{HSO}_{4}{ }^{-}(a q) \\longrightarrow \\mathrm{H}^{+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q)$\n3. (a) $\\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{H}^{+}(a q) \\longrightarrow \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)$; (b) $\\mathrm{OH}^{-}(a q)+\\mathrm{H}^{+}(a q) \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}(l)$; (c)"}
{"id": 5433, "contents": "2297. Chapter 14 - \n$\\mathrm{NH}_{3}(a q)+\\mathrm{H}^{+}(a q) \\longrightarrow \\mathrm{NH}_{4}^{+}(a q) ;$ (d) $\\mathrm{CN}^{-}(a q)+\\mathrm{H}^{+}(a q) \\longrightarrow \\mathrm{HCN}(a q)$; (e)\n$\\mathrm{S}^{2-}(a q)+\\mathrm{H}^{+}(a q) \\longrightarrow \\mathrm{HS}^{-}(a q) ;$ (f) $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}(a q)+\\mathrm{H}^{+}(a q) \\longrightarrow \\mathrm{H}_{3} \\mathrm{PO}_{4}(a q)$\n4. (a) $\\mathrm{H}_{2} \\mathrm{O}, \\mathrm{O}^{2-}$; (b) $\\mathrm{H}_{3} \\mathrm{O}^{+}, \\mathrm{OH}^{-}$; (c) $\\mathrm{H}_{2} \\mathrm{CO}_{3}, \\mathrm{CO}_{3}{ }^{2-}$; (d) $\\mathrm{NH}_{4}^{+}, \\mathrm{NH}_{2}^{-}$; (e) $\\mathrm{H}_{2} \\mathrm{SO}_{4}, \\mathrm{SO}_{4}{ }^{2-}$; (f) $\\mathrm{H}_{3} \\mathrm{O}_{2}{ }^{+}, \\mathrm{HO}_{2}^{-}$; (g) $\\mathrm{H}_{2} \\mathrm{~S} ; \\mathrm{S}^{2-}$; (h) $\\mathrm{H}_{6} \\mathrm{~N}_{2}{ }^{2+}, \\mathrm{H}_{4} \\mathrm{~N}_{2}$"}
{"id": 5434, "contents": "2297. Chapter 14 - \n5. The labels are $\\mathrm{Br} \\varnothing$ nsted-Lowry acid $=\\mathrm{BA}$; its conjugate base $=\\mathrm{CB}$; Br\u00f8nsted-Lowry base $=\\mathrm{BB}$; its conjugate acid = CA. (a) $\\mathrm{HNO}_{3}(\\mathrm{BA}), \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{BB}), \\mathrm{H}_{3} \\mathrm{O}^{+}(\\mathrm{CA}), \\mathrm{NO}_{3}{ }^{-}(\\mathrm{CB}) ;$ (b) $\\mathrm{CN}^{-}(\\mathrm{BB}), \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{BA}), \\mathrm{HCN}(\\mathrm{CA}), \\mathrm{OH}^{-}(\\mathrm{CB})$; (c) $\\mathrm{H}_{2} \\mathrm{SO}_{4}(\\mathrm{BA}), \\mathrm{Cl}^{-}(\\mathrm{BB}), \\mathrm{HCl}(\\mathrm{CA}), \\mathrm{HSO}_{4}{ }^{-}(\\mathrm{CB})$; (d) $\\mathrm{HSO}_{4}^{-}(\\mathrm{BA}), \\mathrm{OH}^{-}(\\mathrm{BB}), \\mathrm{SO}_{4}{ }^{2-}(\\mathrm{CB}), \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{CA}) ;$ (e) $\\mathrm{O}^{2-}(\\mathrm{BB})$, $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{BA}) \\mathrm{OH}^{-}(\\mathrm{CB}$ and CA$)$; (f) $\\left[\\mathrm{Cu}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{3}(\\mathrm{OH})\\right]^{+}(\\mathrm{BB}),\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{3+}(\\mathrm{BA}),\\left[\\mathrm{Cu}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{4}\\right]^{2+}(\\mathrm{CA}),\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2}"}
{"id": 5435, "contents": "2297. Chapter 14 - \n\\mathrm{O}\\right)_{4}\\right]^{2+}(\\mathrm{CA}),\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{5}(\\mathrm{OH})\\right]^{2+}(\\mathrm{CB})$; (g) $\\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{BA}), \\mathrm{NH}_{2}{ }^{-}(\\mathrm{BB}), \\mathrm{HS}^{-}(\\mathrm{CB}), \\mathrm{NH}_{3}(\\mathrm{CA})$"}
{"id": 5436, "contents": "2297. Chapter 14 - \n6. Amphiprotic species may either gain or lose a proton in a chemical reaction, thus acting as a base or an acid. An example is $\\mathrm{H}_{2} \\mathrm{O}$. As an acid: $\\mathrm{H}_{2} \\mathrm{O}(a q)+\\mathrm{NH}_{3}(a q) \\rightleftharpoons \\mathrm{NH}_{4}+(a q)+\\mathrm{OH}^{-}(a q)$. As a base: $\\mathrm{H}_{2} \\mathrm{O}(a q)+\\mathrm{HCl}(a q) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{Cl}^{-}(a q)$\n7. amphiprotic: (a) $\\mathrm{NH}_{3}+\\mathrm{H}_{3} \\mathrm{O}^{+} \\longrightarrow \\mathrm{NH}_{4} \\mathrm{OH}+\\mathrm{H}_{2} \\mathrm{O}, \\mathrm{NH}_{3}+\\mathrm{OCH}_{3}-\\longrightarrow \\mathrm{NH}_{2}^{-}+\\mathrm{CH}_{3} \\mathrm{OH}$; (b) $\\mathrm{HPO}_{4}{ }^{2-}+\\mathrm{OH}^{-} \\longrightarrow \\mathrm{PO}_{4}{ }^{3-}+\\mathrm{H}_{2} \\mathrm{O}, \\mathrm{HPO}_{4}{ }^{2-}+\\mathrm{HClO}_{4} \\longrightarrow \\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{-}+\\mathrm{ClO}_{4}{ }^{-}$; not amphiprotic: (c) $\\mathrm{Br}^{-}$; (d) $\\mathrm{NH}_{4}{ }^{+}$; (e) $\\mathrm{AsO}_{4}{ }^{3-}$"}
{"id": 5437, "contents": "2297. Chapter 14 - \n8. In a neutral solution $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=\\left[\\mathrm{OH}^{-}\\right]$. At $40^{\\circ} \\mathrm{C},\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=\\left[\\mathrm{OH}^{-}\\right]=\\left(2.910 \\times 10^{-14}\\right)^{1 / 2}=1.7 \\times 10^{-7}$.\n9. $x=3.051 \\times 10^{-7} M=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=\\left[\\mathrm{OH}^{-}\\right] ; \\mathrm{pH}=-\\log 3.051 \\times 10^{-7}=-(-6.5156)=6.5156 ; \\mathrm{pOH}=\\mathrm{pH}=6.5156$\n10. (a) $\\mathrm{pH}=3.587 ; \\mathrm{pOH}=10.413$; (b) $\\mathrm{pOH}=0.68 ; \\mathrm{pH}=13.32$; (c) $\\mathrm{pOH}=3.85 ; \\mathrm{pH}=10.15$; (d) $\\mathrm{pOH}=-0.40$; pH $=14.4$\n11. $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=3.0 \\times 10^{-7} \\mathrm{M} ;\\left[\\mathrm{OH}^{-}\\right]=3.3 \\times 10^{-8} \\mathrm{M}$\n12. $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=1 \\times 10^{-2} \\mathrm{M} ;\\left[\\mathrm{OH}^{-}\\right]=1 \\times 10^{-12} \\mathrm{M}$\n13. $\\left[\\mathrm{OH}^{-}\\right]=3.1 \\times 10^{-12} \\mathrm{M}$"}
{"id": 5438, "contents": "2297. Chapter 14 - \n13. $\\left[\\mathrm{OH}^{-}\\right]=3.1 \\times 10^{-12} \\mathrm{M}$\n14. The salt ionizes in solution, but the anion slightly reacts with water to form the weak acid. This reaction also forms $\\mathrm{OH}^{-}$, which causes the solution to be basic.\n15. $\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]>\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right]>\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right] \\approx\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}\\right]>\\left[\\mathrm{OH}^{-}\\right]$\n16. The oxidation state of the sulfur in $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ is greater than the oxidation state of the sulfur in $\\mathrm{H}_{2} \\mathrm{SO}_{3}$.\n17. $\\mathrm{Mg}(\\mathrm{OH})_{2}(s)+2 \\mathrm{HCl}(a q) \\longrightarrow \\mathrm{Mg}^{2+}(a q)+2 \\mathrm{Cl}^{-}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$"}
{"id": 5439, "contents": "2297. Chapter 14 - \nBB BA CB CA\n35. $K_{\\mathrm{a}}=2.3 \\times 10^{-11}$\n37. The stronger base or stronger acid is the one with the larger $K_{\\mathrm{b}}$ or $K_{\\mathrm{a}}$, respectively. In these two examples, they are $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NH}$ and $\\mathrm{CH}_{3} \\mathrm{NH}_{3}{ }^{+}$.\n39. triethylamine\n41. (a) $\\mathrm{HSO}_{4}{ }^{-}$; higher electronegativity of the central ion. (b) $\\mathrm{H}_{2} \\mathrm{O}$; $\\mathrm{NH}_{3}$ is a base and water is neutral, or decide on the basis of $K_{\\mathrm{a}}$ values. (c) $\\mathrm{HI} ; \\mathrm{PH}_{3}$ is weaker than $\\mathrm{HCl} ; \\mathrm{HCl}$ is weaker than HI . Thus, $\\mathrm{PH}_{3}$ is weaker than HI . (d) $\\mathrm{PH}_{3}$; in binary compounds of hydrogen with nonmetals, the acidity increases for the element lower in a group. (e) HBr ; in a period, the acidity increases from left to right; in a group, it increases from top to bottom. Br is to the left and below S , so HBr is the stronger acid."}
{"id": 5440, "contents": "2297. Chapter 14 - \n43. (a) $\\mathrm{NaHSeO}_{3}<\\mathrm{NaHSO}_{3}<\\mathrm{NaHSO}_{4}$; in polyoxy acids, the more electronegative central element-S, in this case-forms the stronger acid. The larger number of oxygen atoms on the central atom (giving it a higher oxidation state) also creates a greater release of hydrogen atoms, resulting in a stronger acid. As a salt, the acidity increases in the same manner. (b) $\\mathrm{ClO}_{2}{ }^{-}<\\mathrm{BrO}_{2}{ }^{-}<\\mathrm{IO}_{2}{ }^{-}$; the basicity of the anions in a series of acids will be the opposite of the acidity in their oxyacids. The acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br , and I is the least electronegative of the three. (c) $\\mathrm{HOI}<\\mathrm{HOBr}<\\mathrm{HOCl}$; in a series of the same form of oxyacids, the acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br , and I is the least"}
{"id": 5441, "contents": "2297. Chapter 14 - \nelectronegative of the three. (d) $\\mathrm{HOCl}<\\mathrm{HOClO}<\\mathrm{HOClO}_{2}<\\mathrm{HOClO}_{3}$; in a series of oxyacids of the same central element, the acidity increases as the number of oxygen atoms increases (or as the oxidation state of the central atom increases). (e) $\\mathrm{HTe}^{-}<\\mathrm{HS}^{-} \\ll \\mathrm{PH}_{2}^{-}<\\mathrm{NH}_{2}^{-} ; \\mathrm{PH}_{2}^{-}$and $\\mathrm{NH}_{2}-$ are anions of weak bases, so they act as strong bases toward $\\mathrm{H}^{+}$. $\\mathrm{HTe}^{-}$and $\\mathrm{HS}^{-}$are anions of weak acids, so they have less basic character. In a periodic group, the more electronegative element has the more basic anion. (f) $\\mathrm{BrO}_{4}{ }^{-}<\\mathrm{BrO}_{3}{ }^{-}<\\mathrm{BrO}_{2}{ }^{-}<\\mathrm{BrO}^{-}$; with a larger number of oxygen atoms (that is, as the oxidation state of the central ion increases), the corresponding acid becomes more acidic and the anion consequently less basic.\n45. $\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]>\\left[\\mathrm{C}_{6} \\mathrm{H}_{4} \\mathrm{OH}\\left(\\mathrm{CO}_{2} \\mathrm{H}\\right)\\right]>\\left[\\mathrm{H}^{+}\\right] 0>\\left[\\mathrm{C}_{6} \\mathrm{H}_{4} \\mathrm{OH}\\left(\\mathrm{CO}_{2}\\right)^{-}\\right] \\gg\\left[\\mathrm{C}_{6} \\mathrm{H}_{4} \\mathrm{O}\\left(\\mathrm{CO}_{2} \\mathrm{H}\\right)^{-}\\right]>\\left[\\mathrm{OH}^{-}\\right]$"}
{"id": 5442, "contents": "2297. Chapter 14 - \n47. 1. Assume that the change in initial concentration of the acid as the equilibrium is established can be neglected, so this concentration can be assumed constant and equal to the initial value of the total acid concentration. 2. Assume we can neglect the contribution of water to the equilibrium concentration of $\\mathrm{H}_{3} \\mathrm{O}^{+}$.\n48. (b) The addition of HCl\n50. (a) Adding HCl will add $\\mathrm{H}_{3} \\mathrm{O}^{+}$ions, which will then react with the $\\mathrm{OH}^{-}$ions, lowering their concentration. The equilibrium will shift to the right, increasing the concentration of $\\mathrm{HO}_{2}$, and decreasing the concentration of $\\mathrm{NO}_{2}{ }^{-}$ions. (b) Adding $\\mathrm{HNO}_{2}$ increases the concentration of $\\mathrm{HNO}_{2}$ and shifts the equilibrium to the left, increasing the concentration of $\\mathrm{NO}_{2}{ }^{-}$ions and decreasing the concentration of $\\mathrm{OH}^{-}$ions. (c) Adding NaOH adds $\\mathrm{OH}^{-}$ions, which shifts the equilibrium to the left, increasing the concentration of $\\mathrm{NO}_{2}{ }^{-}$ions and decreasing the concentrations of $\\mathrm{HNO}_{2}$. (d) Adding NaCl has no effect on the concentrations of the ions. (e) Adding $\\mathrm{KNO}_{2}$ adds $\\mathrm{NO}_{2}{ }^{-}$ions and shifts the equilibrium to the right, increasing the $\\mathrm{HNO}_{2}$ and $\\mathrm{OH}^{-}$ion concentrations."}
{"id": 5443, "contents": "2297. Chapter 14 - \n52. This is a case in which the solution contains a mixture of acids of different ionization strengths. In solution, the $\\mathrm{HCO}_{2} \\mathrm{H}$ exists primarily as $\\mathrm{HCO}_{2} \\mathrm{H}$ molecules because the ionization of the weak acid is suppressed by the strong acid. Therefore, the $\\mathrm{HCO}_{2} \\mathrm{H}$ contributes a negligible amount of hydronium ions to the solution. The stronger acid, HCl , is the dominant producer of hydronium ions because it is completely ionized. In such a solution, the stronger acid determines the concentration of hydronium ions, and the ionization of the weaker acid is fixed by the $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]$produced by the stronger acid.\n54. (a) $K_{\\mathrm{b}}=1.8 \\times 10^{-5}$; (b) $K_{\\mathrm{a}}=4.5 \\times 10^{-4}$; (c) $K_{\\mathrm{b}}=6.4 \\times 10^{-5}$; (d) $K_{\\mathrm{a}}=5.6 \\times 10^{-10}$\n56. $K_{\\mathrm{a}}=1.2 \\times 10^{-2}$\n58. (a) $K_{\\mathrm{b}}=4.3 \\times 10^{-12}$ (b) $K_{\\mathrm{a}}=1.6 \\times 10^{10}$ (c) $K_{\\mathrm{b}}=5.9 \\times 10^{8}$ (d) $K_{\\mathrm{b}}=4.2 \\times 10^{-3}$ (e) $K_{\\mathrm{b}}=2.3 \\times 10^{5}$ (f) $K_{\\mathrm{b}}=6.3 \\times 10^{-13}$"}
{"id": 5444, "contents": "2297. Chapter 14 - \n60. (a) $\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{ClO}^{-}\\right]}{[\\mathrm{HClO}]}=\\frac{(x)(x)}{(0.0092-x)} \\approx \\frac{(x)(x)}{0.0092}=2.9 \\times 10^{-8}$"}
{"id": 5445, "contents": "2297. Chapter 14 - \nSolving for $x$ gives $1.63 \\times 10^{-5} \\mathrm{M}$. This value is less than $5 \\%$ of 0.0092 , so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:\n$\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=\\left[\\mathrm{ClO}^{-}\\right]=1.6 \\times 10^{-5} \\mathrm{M}$\n$\\left[\\mathrm{HClO}^{-}\\right]=0.0092 \\mathrm{M}$\n$\\left[\\mathrm{OH}^{-}\\right]=6.1 \\times 10^{-10} \\mathrm{M}$;\n(b) $\\frac{\\left[\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{3}{ }^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]}{\\left[\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{2}\\right]}=\\frac{(x)(x)}{(0.0784-x)} \\approx \\frac{(x)(x)}{0.0784}=4.3 \\times 10^{-10}$"}
{"id": 5446, "contents": "2297. Chapter 14 - \nSolving for $x$ gives $5.81 \\times 10^{-6} \\mathrm{M}$. This value is less than $5 \\%$ of 0.0784 , so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:\n$\\left[\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{3}{ }^{+}\\right]=\\left[\\mathrm{OH}^{-}\\right]=5.8 \\times 10^{-6} \\mathrm{M}$\n$\\left[\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{2}\\right]=0.0784 \\mathrm{M}$\n$\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=1.7 \\times 10^{-9} \\mathrm{M}$;\n(c) $\\frac{\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]\\left[\\mathrm{CN}^{-}\\right]}{[\\mathrm{HCN}]}=\\frac{(x)(x)}{(0.0810-x)} \\approx \\frac{(x)(x)}{0.0810}=4.9 \\times 10^{-10}$"}
{"id": 5447, "contents": "2297. Chapter 14 - \nSolving for $x$ gives $6.30 \\times 10^{-6} \\mathrm{M}$. This value is less than $5 \\%$ of 0.0810 , so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:\n$\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=\\left[\\mathrm{CN}^{-}\\right]=6.3 \\times 10^{-6} \\mathrm{M}$\n$[\\mathrm{HCN}]=0.0810 \\mathrm{M}$\n$\\left[\\mathrm{OH}^{-}\\right]=1.6 \\times 10^{-9} \\mathrm{M}$;\n(d) $\\frac{\\left[\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{NH}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]}{\\left[\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{~N}\\right]}=\\frac{(x)(x)}{(0.11-x)} \\approx \\frac{(x)(x)}{0.11}=6.3 \\times 10^{-5}$"}
{"id": 5448, "contents": "2297. Chapter 14 - \nSolving for $x$ gives $2.63 \\times 10^{-3} \\mathrm{M}$. This value is less than $5 \\%$ of 0.11 , so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:\n$\\left[\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{NH}^{+}\\right]=\\left[\\mathrm{OH}^{-}\\right]=2.6 \\times 10^{-3} \\mathrm{M}$\n$\\left[\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{~N}\\right]=0.11 \\mathrm{M}$\n$\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=3.8 \\times 10^{-12} \\mathrm{M}$;\n(e) $\\frac{\\left[\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{5}(\\mathrm{OH})^{+}\\right]\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]}{\\left[\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{2+}\\right]}=\\frac{(x)(x)}{(0.120-x)} \\approx \\frac{(x)(x)}{0.120}=1.6 \\times 10^{-7}$"}
{"id": 5449, "contents": "2297. Chapter 14 - \nSolving for $x$ gives $1.39 \\times 10^{-4} \\mathrm{M}$. This value is less than $5 \\%$ of 0.120 , so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:\n$\\left[\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{5}(\\mathrm{OH})^{+}\\right]=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=1.4 \\times 10^{-4} \\mathrm{M}$\n$\\left[\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{2+}\\right]=0.120 \\mathrm{M}$\n$\\left[\\mathrm{OH}^{-}\\right]=7.2 \\times 10^{-11} \\mathrm{M}$\n62. $\\mathrm{pH}=2.41$\n64. $\\left[\\mathrm{C}_{10} \\mathrm{H}_{14} \\mathrm{~N}_{2}\\right]=0.049 \\mathrm{M} ;\\left[\\mathrm{C}_{10} \\mathrm{H}_{14} \\mathrm{~N}_{2} \\mathrm{H}^{+}\\right]=1.9 \\times 10^{-4} \\mathrm{M} ;\\left[\\mathrm{C}_{10} \\mathrm{H}_{14} \\mathrm{~N}_{2} \\mathrm{H}_{2}{ }^{2+}\\right]=1.4 \\times 10^{-11} \\mathrm{M} ;\\left[\\mathrm{OH}^{-}\\right]=1.9 \\times 10^{-4}$ $M ;\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=5.3 \\times 10^{-11} \\mathrm{M}$\n66. $K_{\\mathrm{a}}=1.2 \\times 10^{-2}$\n68. $K_{\\mathrm{b}}=1.77 \\times 10^{-5}$\n70. (a) acidic; (b) basic; (c) acidic; (d) neutral"}
{"id": 5450, "contents": "2297. Chapter 14 - \n68. $K_{\\mathrm{b}}=1.77 \\times 10^{-5}$\n70. (a) acidic; (b) basic; (c) acidic; (d) neutral\n72. $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]$and $\\left[\\mathrm{HCO}_{3}{ }^{-}\\right]$are practically equal\n74. $\\left[\\mathrm{C}_{6} \\mathrm{H}_{4}\\left(\\mathrm{CO}_{2} \\mathrm{H}\\right)_{2}\\right] 7.2 \\times 10^{-3} \\mathrm{M},\\left[\\mathrm{C}_{6} \\mathrm{H}_{4}\\left(\\mathrm{CO}_{2} \\mathrm{H}\\right)\\left(\\mathrm{CO}_{2}\\right)^{-}\\right]=\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right] 2.8 \\times 10^{-3} \\mathrm{M},\\left[\\mathrm{C}_{6} \\mathrm{H}_{4}\\left(\\mathrm{CO}_{2}\\right)_{2}{ }^{2-}\\right] 3.9 \\times 10^{-6} \\mathrm{M}$, $\\left[\\mathrm{OH}^{-}\\right] 3.6 \\times 10^{-12} \\mathrm{M}$\n76. (a) $K_{\\mathrm{a} 2}=1.5 \\times 10^{-11}$;\n(b) $K_{\\mathrm{b}}=4.3 \\times 10^{-12}$;\n(c) $\\frac{\\left[\\mathrm{Te}^{2-}\\right]\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]}{\\left[\\mathrm{HTe}^{-}\\right]}=\\frac{(x)(0.0141+x)}{(0.0141-x)} \\approx \\frac{(x)(0.0141)}{0.0141}=1.5 \\times 10^{-11}$"}
{"id": 5451, "contents": "2297. Chapter 14 - \nSolving for $x$ gives $1.5 \\times 10^{-11} M$. Therefore, compared with $0.014 M$, this value is negligible $(1.1 \\times$ $10^{-7} \\%$ ).\n78. Excess $\\mathrm{H}_{3} \\mathrm{O}^{+}$is removed primarily by the reaction: $\\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{-}(a q) \\longrightarrow \\mathrm{H}_{3} \\mathrm{PO}_{4}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)$ Excess base is removed by the reaction: $\\mathrm{OH}^{-}(a q)+\\mathrm{H}_{3} \\mathrm{PO}_{4}(a q) \\longrightarrow \\mathrm{H}_{2} \\mathrm{PO}_{4}{ }^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)$\n80. $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]=1.5 \\times 10^{-4} \\mathrm{M}$\n82. $\\left[\\mathrm{OH}^{-}\\right]=4.2 \\times 10^{-4} \\mathrm{M}$"}
{"id": 5452, "contents": "2297. Chapter 14 - \n84. (a) The added HCl will increase the concentration of $\\mathrm{H}_{3} \\mathrm{O}^{+}$slightly, which will react with $\\mathrm{CH}_{3} \\mathrm{CO}_{2}^{-}$and produce $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$ in the process. Thus, $\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}\\right]$decreases and $\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right]$ increases. (b) The added $\\mathrm{KCH}_{3} \\mathrm{CO}_{2}$ will increase the concentration of $\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}\\right]$which will react with $\\mathrm{H}_{3} \\mathrm{O}^{+}$and produce $\\mathrm{CH}_{3} \\mathrm{CO}_{2}$ H in the process. Thus, $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]$decreases slightly and $\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right]$ increases. (c) The added NaCl will have no effect on the concentration of the ions. (d) The added KOH will produce $\\mathrm{OH}^{-}$ions, which will react with the $\\mathrm{H}_{3} \\mathrm{O}^{+}$, thus reducing $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]$. Some additional $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$ will dissociate, producing $\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}\\right]$ions in the process. Thus, $\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right]$ decreases slightly and $\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}\\right]$increases. (e) The added $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$ will increase its concentration, causing more of it to dissociate and producing more $\\left[\\mathrm{CH}_{3}"}
{"id": 5453, "contents": "2297. Chapter 14 - \n(e) The added $\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}$ will increase its concentration, causing more of it to dissociate and producing more $\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}\\right]$and $\\mathrm{H}_{3} \\mathrm{O}^{+}$in the process. Thus, $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]$increases slightly and $\\left[\\mathrm{CH}_{3} \\mathrm{CO}_{2}{ }^{-}\\right]$increases."}
{"id": 5454, "contents": "2297. Chapter 14 - \n86. $\\mathrm{pH}=8.95$\n88. $37 \\mathrm{~g}(0.27 \\mathrm{~mol})$\n90. (a) $\\mathrm{pH}=5.222$; (b) The solution is acidic. (c) $\\mathrm{pH}=5.220$\n92. At the equivalence point in the titration of a weak base with a strong acid, the resulting solution is slightly acidic due to the presence of the conjugate acid. Thus, pick an indicator that changes color in the acidic range and brackets the pH at the equivalence point. Methyl orange is a good example.\n94.\n(a) $\\mathrm{pH}=2.50$; (b)\n(b) $\\mathrm{pH}=4.01$;\n(c) $\\mathrm{pH}=5.60$;\n; (d) $\\mathrm{pH}=8.35$; (e) $\\mathrm{pH}=11.08$"}
{"id": 5455, "contents": "2298. Chapter 15 - \n1. $(\\mathrm{a})$\n\n$$\n\\mathrm{AgI}(s) \\rightleftharpoons \\mathrm{Ag}^{+}(a q)+\\mathrm{I}^{-}(a q)\n$$\n\n$x \\quad \\underline{x}$\n(b)\n$\\mathrm{CaCO}_{3}(s) \\rightleftharpoons \\mathrm{Ca}^{2+}(a q)+\\mathrm{CO}_{3}{ }^{2-}(a q)$\n$\\underline{x} \\quad x$\n(c)\n$\\begin{array}{ll}\\mathrm{Mg}(\\mathrm{OH})_{2}(s) \\rightleftharpoons \\mathrm{Mg}^{2+}(a q) & +2 \\mathrm{OH}^{-}(a q) \\\\ x & \\underline{2 x}\\end{array}$\n(d)\n$\\mathrm{Mg}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}(s) \\rightleftharpoons 3 \\mathrm{Mg}^{2+}(a q)+2 \\mathrm{PO}_{4}{ }^{3-}(a q)$\n\n$$\n\\underline{x} \\quad \\frac{2}{3} x\n$$"}
{"id": 5456, "contents": "2298. Chapter 15 - \n(e)\n$\\mathrm{Ca}_{5}\\left(\\mathrm{PO}_{4}\\right)_{3} \\mathrm{OH}(s) \\rightleftharpoons 5 \\mathrm{Ca}^{2+}(a q)+3 \\mathrm{PO}_{4}{ }^{3-}(a q)+\\mathrm{OH}^{-}(a q)$\n$\\underline{5 x} \\quad \\underline{3 x} \\quad x$\n3. There is no change. A solid has an activity of 1 whether there is a little or a lot.\n5. The solubility of silver bromide at the new temperature must be known. Normally the solubility increases and some of the solid silver bromide will dissolve.\n7. $\\mathrm{CaF}_{2}, \\mathrm{MnCO}_{3}$, and ZnS\n9. (a) $\\mathrm{LaF}_{3}(s) \\rightleftharpoons \\mathrm{La}^{3+}(a q)+3 \\mathrm{~F}^{-}(a q) \\quad K_{\\mathrm{sp}}=\\left[\\mathrm{La}^{3+}\\right]\\left[\\mathrm{F}^{-}\\right]^{3}$;\n(b) $\\mathrm{CaCO}_{3}(s) \\rightleftharpoons \\mathrm{Ca}^{2+}(a q)+\\mathrm{CO}_{3}{ }^{2-}(a q) \\quad K_{\\text {sp }}=\\left[\\mathrm{Ca}^{2+}\\right]\\left[\\mathrm{CO}_{3}{ }^{2-}\\right]$;\n(c) $\\mathrm{Ag}_{2} \\mathrm{SO}_{4}(s) \\rightleftharpoons 2 \\mathrm{Ag}^{+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q) \\quad K_{\\mathrm{sp}}=\\left[\\mathrm{Ag}^{+}\\right]^{2}\\left[\\mathrm{SO}_{4}{ }^{2-}\\right]$;"}
{"id": 5457, "contents": "2298. Chapter 15 - \n(d) $\\mathrm{Pb}(\\mathrm{OH})_{2}(s) \\rightleftharpoons \\mathrm{Pb}^{2+}(a q)+2 \\mathrm{OH}^{-}(a q) \\quad K_{\\mathrm{sp}}=\\left[\\mathrm{Pb}^{2+}\\right]\\left[\\mathrm{OH}^{-}\\right]^{2}$\n11. (a) $1.77 \\times 10^{-7}$; (b) $1.6 \\times 10^{-6}$; (c) $2.2 \\times 10^{-9}$; (d) $7.91 \\times 10^{-22}$\n13. (a) $2 \\times 10^{-2} \\mathrm{M}$; (b) $1.5 \\times 10^{-3} \\mathrm{M}$; (c) $2.27 \\times 10^{-9} \\mathrm{M}$; (d) $2.2 \\times 10^{-10} \\mathrm{M}$\n15. (a) $6.4 \\times 10^{-9} M=\\left[\\mathrm{Ag}^{+}\\right],\\left[\\mathrm{Cl}^{-}\\right]=0.025 M$. Check: $\\frac{6.4 \\times 10^{-9} M}{0.025 M} \\times 100 \\%=2.6 \\times 10^{-5} \\%$,an insignificant change;\n(b) $2.2 \\times 10^{-5} M=\\left[\\mathrm{Ca}^{2+}\\right],\\left[\\mathrm{F}^{-}\\right]=0.0013 M$. Check: $\\frac{2.26 \\times 10^{-5} M}{0.00133 M} \\times 100 \\%=1.70 \\%$. This value is less than $5 \\%$ and can be ignored."}
{"id": 5458, "contents": "2298. Chapter 15 - \n(c) $0.2238 M=\\left[\\mathrm{SO}_{4}^{2-}\\right] ;\\left[\\mathrm{Ag}^{+}\\right]=7.4 \\times 10^{-3} M$. Check: $\\frac{3.7 \\times 10^{-3}}{0.2238} \\times 100 \\%=1.64 \\times 10^{-2}$; the condition is satisfied.\n(d) $\\left[\\mathrm{OH}^{-}\\right]=2.8 \\times 10^{-3} M$; $5.7 \\times 10^{-12} M=\\left[\\mathrm{Zn}^{2+}\\right]$. Check: $\\frac{5.7 \\times 10^{-12}}{2.8 \\times 10^{-3}} \\times 100 \\%=2.0 \\times 10^{-7} \\% ; x$ is less than $5 \\%$ of $\\left[\\mathrm{OH}^{-}\\right]$and is, therefore, negligible.\n17. (a) $\\left[\\mathrm{Cl}^{-}\\right]=7.6 \\times 10^{-3} \\mathrm{M}$"}
{"id": 5459, "contents": "2298. Chapter 15 - \nCheck: $\\frac{7.6 \\times 10^{-3}}{0.025} \\times 100 \\%=30 \\%$\nThis value is too large to drop $x$. Therefore solve by using the quadratic equation:\n$\\left[\\mathrm{Ti}^{+}\\right]=3.1 \\times 10^{-2} \\mathrm{M}$\n$\\left[\\mathrm{Cl}^{-}\\right]=6.1 \\times 10^{-3}$\n(b) $\\left[\\mathrm{Ba}^{2+}\\right]=7.7 \\times 10^{-4} \\mathrm{M}$"}
{"id": 5460, "contents": "2298. Chapter 15 - \nCheck: $\\frac{7.7 \\times 10^{-4}}{0.0313} \\times 100 \\%=2.4 \\%$\nTherefore, the condition is satisfied.\n$\\left[\\mathrm{Ba}^{2+}\\right]=7.7 \\times 10^{-4} \\mathrm{M}$\n$\\left[\\mathrm{F}^{-}\\right]=0.0321 \\mathrm{M}$;\n(c) $\\mathrm{Mg}\\left(\\mathrm{NO}_{3}\\right)_{2}=0.02444 \\mathrm{M}$\n$\\left[\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right]=2.9 \\times 10^{-5}$\nCheck: $\\frac{2.9 \\times 10^{-5}}{0.02444} \\times 100 \\%=0.12 \\%$\nThe condition is satisfied; the above value is less than $5 \\%$.\n$\\left[\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right]=2.9 \\times 10^{-5} \\mathrm{M}$\n$\\left[\\mathrm{Mg}^{2+}\\right]=0.0244 \\mathrm{M}$\n(d) $\\left[\\mathrm{OH}^{-}\\right]=0.0501 \\mathrm{M}$\n$\\left[\\mathrm{Ca}^{2+}\\right]=3.15 \\times 10^{-3}$\nCheck: $\\frac{3.15 \\times 10^{-3}}{0.050} \\times 100 \\%=6.28 \\%$\nThis value is greater than $5 \\%$, so a more exact method, such as successive approximations, must be used.\n$\\left[\\mathrm{Ca}^{2+}\\right]=2.8 \\times 10^{-3} \\mathrm{M}$\n$\\left[\\mathrm{OH}^{-}\\right]=0.053 \\times 10^{-2} \\mathrm{M}$\n19. The changes in concentration are greater than $5 \\%$ and thus exceed the maximum value for disregarding the change."}
{"id": 5461, "contents": "2298. Chapter 15 - \n19. The changes in concentration are greater than $5 \\%$ and thus exceed the maximum value for disregarding the change.\n21. $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$ is the most soluble Ca salt in $\\mathrm{mol} / \\mathrm{L}$, and it is also the most soluble Ca salt in $\\mathrm{g} / \\mathrm{L}$.\n23. $4.8 \\times 10^{-3} M=\\left[\\mathrm{SO}_{4}{ }^{2-}\\right]=\\left[\\mathrm{Ca}^{2+}\\right]$; Since this concentration is higher than $2.60 \\times 10^{-3} M$, \"gyp\" water does not meet the standards.\n25. Mass $\\left(\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}\\right)=0.72 \\mathrm{~g} / \\mathrm{L}$"}
{"id": 5462, "contents": "2298. Chapter 15 - \n25. Mass $\\left(\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}\\right)=0.72 \\mathrm{~g} / \\mathrm{L}$\n27. (a) $\\left[\\mathrm{Ag}^{+}\\right]=\\left[\\mathrm{I}^{-}\\right]=1.3 \\times 10^{-5} \\mathrm{M}$; (b) $\\left[\\mathrm{Ag}^{+}\\right]=2.88 \\times 10^{-2} \\mathrm{M},\\left[\\mathrm{SO}_{4}{ }^{2-}\\right]=1.44 \\times 10^{-2} \\mathrm{M}$; (c) $\\left[\\mathrm{Mn}^{2+}\\right]=3.7 \\times 10^{-5} \\mathrm{M}$, $\\left[\\mathrm{OH}^{-}\\right]=7.4 \\times 10^{-5} \\mathrm{M}$; (d) $\\left[\\mathrm{Sr}^{2+}\\right]=4.3 \\times 10^{-2} \\mathrm{M},\\left[\\mathrm{OH}^{-}\\right]=8.6 \\times 10^{-2} \\mathrm{M}$; (e) $\\left[\\mathrm{Mg}^{2+}\\right]=1.3 \\times 10^{-4} \\mathrm{M},\\left[\\mathrm{OH}^{-}\\right]=2.6 \\times$ $10^{-4} \\mathrm{M}$.\n29. (a) $1.45 \\times 10^{-4}$; (b) $8.2 \\times 10^{-55}$; (c) $1.35 \\times 10^{-4}$; (d) $1.18 \\times 10^{-5}$; (e) $1.08 \\times 10^{-10}$\n31. (a) $\\mathrm{CaCO}_{3}$ does precipitate. (b) The compound does not precipitate. (c) The compound does not precipitate.\n(d) The compound precipitates.\n33. $3.03 \\times 10^{-7} \\mathrm{M}$\n35. $9.2 \\times 10^{-13} \\mathrm{M}$"}
{"id": 5463, "contents": "2298. Chapter 15 - \n(d) The compound precipitates.\n33. $3.03 \\times 10^{-7} \\mathrm{M}$\n35. $9.2 \\times 10^{-13} \\mathrm{M}$\n37. $\\left[\\mathrm{Ag}^{+}\\right]=1.8 \\times 10^{-3} \\mathrm{M}$\n39. $6.3 \\times 10^{-4}$\n41. (a) 2.25 L ; (b) $7.2 \\times 10^{-7} \\mathrm{~g}$\n43. $100 \\%$ of it is dissolved\n45. (a) $\\mathrm{Hg}_{2}{ }^{2+}$ and $\\mathrm{Cu}^{2+}$ : Add $\\mathrm{SO}_{4}{ }^{2-}$. (b) $\\mathrm{SO}_{4}{ }^{2-}$ and $\\mathrm{Cl}^{-}$: Add $\\mathrm{Ba}^{2+}$. (c) $\\mathrm{Hg}^{2+}$ and $\\mathrm{Co}^{2+}$ : Add $\\mathrm{S}^{2-}$. (d) $\\mathrm{Zn}^{2+}$ and $\\mathrm{Sr}^{2+}$ : Add $\\mathrm{OH}^{-}$until $\\left[\\mathrm{OH}^{-}\\right]=0.050 \\mathrm{M}$. (e) $\\mathrm{Ba}^{2+}$ and $\\mathrm{Mg}^{2+}$ : Add $\\mathrm{SO}_{4}{ }^{2-}$. (f) $\\mathrm{CO}_{3}{ }^{2-}$ and $\\mathrm{OH}^{-}$: Add $\\mathrm{Ba}^{2+}$.\n47. AgI will precipitate first.\n49. $1.5 \\times 10^{-12} \\mathrm{M}$\n51. 3.99 kg\n53. (a) $3.1 \\times 10^{-11}$; (b) $\\left[\\mathrm{Cu}^{2+}\\right]=2.6 \\times 10^{-3} ;\\left[\\mathrm{IO}_{3}{ }^{-}\\right]=5.3 \\times 10^{-3}$"}
{"id": 5464, "contents": "2298. Chapter 15 - \n55. $1.8 \\times 10^{-5} \\mathrm{~g} \\mathrm{~Pb}(\\mathrm{OH})_{2}$\n57. $\\mathrm{Mg}(\\mathrm{OH})_{2}(s) \\rightleftharpoons \\mathrm{Mg}^{2+}+2 \\mathrm{OH}^{-} \\quad K_{\\mathrm{sp}}=\\left[\\mathrm{Mg}^{2+}\\right]\\left[\\mathrm{OH}^{-}\\right]^{2}$ $1.23 \\times 10^{-3} \\mathrm{~g} \\mathrm{Mg}(\\mathrm{OH})_{2}$\n59. $\\mathrm{MnCO}_{3}$ will form first since it has the smallest $K_{\\text {sp }}$ value among these homologous compounds and is therefore the least soluble. $\\mathrm{MgCO}_{3} \\cdot 3 \\mathrm{H}_{2} \\mathrm{O}$ will be the last to precipitate since it has the largest K _sp value and is the most soluble. $K_{\\mathrm{sp}}$ value.\n62. when the amount of solid is so small that a saturated solution is not produced\n64. $1.8 \\times 10^{-5} \\mathrm{M}$\n66. $5 \\times 10^{23}$\n68."}
{"id": 5465, "contents": "2298. Chapter 15 - \n|  | $\\left[\\mathrm{Cd}(\\mathrm{CN})_{4}{ }^{2}\\right]$ |  | $[\\mathrm{CN}]$ |\n| :---: | :---: | :---: | :---: |$]\\left[\\mathrm{Cd}^{2+}\\right]$\n\n$$\n\\left[\\mathrm{Cd}^{2+}\\right]=9.5 \\times 10^{-5} \\mathrm{M} ;\\left[\\mathrm{CN}^{-}\\right]=3.8 \\times 10^{-4} \\mathrm{M}\n$$\n\n70. $\\left[\\mathrm{Co}^{3+}\\right]=3.0 \\times 10^{-6} \\mathrm{M} ;\\left[\\mathrm{NH}_{3}\\right]=1.8 \\times 10^{-5} \\mathrm{M}$\n71. 1.3 g\n72. 0.79 g\n73. (a)\n\n(b)\n\n(c)\n\n(d)\n\n\nAcid Base\n(e)\n\n\nAcid Base\n78. (a)\n\n$$\n\\mathrm{HCl}(g)+\\mathrm{PH}_{3}(g) \\longrightarrow\\left[\\mathrm{PH}_{4}\\right]^{+}+[\\because \\ddot{\\mathrm{Cl}}:]^{-}\n$$\n\n\n(b) $\\mathrm{H}_{3} \\mathrm{O}^{+}+\\mathrm{CH}_{3}{ }^{-} \\longrightarrow \\mathrm{CH}_{4}+\\mathrm{H}_{2} \\mathrm{O}$\n\n(c) $\\mathrm{CaO}+\\mathrm{SO}_{3} \\longrightarrow \\mathrm{CaSO} 4$\n\n(d) $\\mathrm{NH}_{4}{ }^{+}+\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{O}^{-} \\longrightarrow \\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}+\\mathrm{NH}_{3}$"}
{"id": 5466, "contents": "2298. Chapter 15 - \n80. 0.0281 g\n82. $\\mathrm{HNO}_{3}(l)+\\mathrm{HF}(l) \\longrightarrow \\mathrm{H}_{2} \\mathrm{NO}_{3}{ }^{+}+\\mathrm{F}^{-} ; \\mathrm{HF}(l)+\\mathrm{BF}_{3}(g) \\longrightarrow \\mathrm{H}^{+}+\\mathrm{BF}_{4}$\n84. (a) $\\mathrm{H}_{3} \\mathrm{BO}_{3}+\\mathrm{H}_{2} \\mathrm{O} \\longrightarrow \\mathrm{H}_{4} \\mathrm{BO}_{4}^{-}+\\mathrm{H}^{+}$; (b) The electronic and molecular shapes are the same-both tetrahedral. (c) The tetrahedral structure is consistent with $s p^{3}$ hybridization.\n86. $0.014 M$\n88. $7.2 \\times 10^{-15} \\mathrm{M}$\n90. $4.4 \\times 10^{-22} \\mathrm{M}$\n93. $\\left[\\mathrm{OH}^{-}\\right]=4.5 \\times 10^{-6} ;\\left[\\mathrm{Al}^{3+}\\right]=2 \\times 10^{-16}$ (molar solubility)\n95. $\\left[\\mathrm{SO}_{4}{ }^{2-}\\right]=0.049 \\mathrm{M} ;\\left[\\mathrm{Ba}^{2+}\\right]=4.7 \\times 10^{-7}$ (molar solubility)\n97. $\\left[\\mathrm{OH}^{-}\\right]=7.6 \\times 10^{-3} \\mathrm{M} ;\\left[\\mathrm{Pb}^{2+}\\right]=2.1 \\times 10^{-11}$ (molar solubility)\n99. 7.66\n101. (a) $K_{\\mathrm{sp}}=\\left[\\mathrm{Mg}^{2+}\\right]\\left[\\mathrm{F}^{-}\\right]^{2}=\\left(1.21 \\times 10^{-3}\\right)\\left(2 \\times 1.21 \\times 10^{-3}\\right)^{2}=7.09 \\times 10^{-9}$"}
{"id": 5467, "contents": "2298. Chapter 15 - \n(b) $7.09 \\times 10^{-7} \\mathrm{M}$\n(c) Determine the concentration of $\\mathrm{Mg}^{2+}$ and $\\mathrm{F}^{-}$that will be present in the final volume. Compare the value of the ion product $\\left[\\mathrm{Mg}^{2+}\\right]\\left[\\mathrm{F}^{-}\\right]^{2}$ with $K_{\\mathrm{sp}}$. If this value is larger than $K_{\\mathrm{sp}}$, precipitation will occur.\n$0.1000 \\mathrm{~L} \\times 3.00 \\times 10^{-3} \\mathrm{M} \\mathrm{Mg}\\left(\\mathrm{NO}_{3}\\right)_{2}=0.3000 \\mathrm{~L} \\times M \\mathrm{Mg}\\left(\\mathrm{NO}_{3}\\right)_{2}$\n$M \\mathrm{Mg}\\left(\\mathrm{NO}_{3}\\right)_{2}=1.00 \\times 10^{-3} \\mathrm{M}$\n$0.2000 \\mathrm{~L} \\times 2.00 \\times 10^{-3} M \\mathrm{NaF}=0.3000 \\mathrm{~L} \\times M \\mathrm{NaF}$\n$M \\mathrm{NaF}=1.33 \\times 10^{-3} \\mathrm{M}$\nion product $=\\left(1.00 \\times 10^{-3}\\right)\\left(1.33 \\times 10^{-3}\\right)^{2}=1.77 \\times 10^{-9}$ This value is smaller than $K_{\\mathrm{sp}}$, so no precipitation will occur.\n(d) $\\mathrm{MgF}_{2}$ is less soluble at $27^{\\circ} \\mathrm{C}$ than at $18{ }^{\\circ} \\mathrm{C}$. Because added heat acts like an added reagent, when it appears on the product side, the Le Ch\u00e2telier's principle states that the equilibrium will shift to the reactants' side to counter the stress. Consequently, less reagent will dissolve. This situation is found in our case. Therefore, the reaction is exothermic."}
{"id": 5468, "contents": "2298. Chapter 15 - \n103. $\\mathrm{BaF}_{2}, \\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$, ZnS ; each is a salt of a weak acid, and the $\\left[\\mathrm{H}_{3} \\mathrm{O}^{+}\\right]$from perchloric acid reduces the equilibrium concentration of the anion, thereby increasing the concentration of the cations\n105. Effect on amount of solid $\\mathrm{CaHPO}_{4},\\left[\\mathrm{Ca}^{2+}\\right],\\left[\\mathrm{OH}^{-}\\right]$: (a) increase, increase, decrease; (b) decrease, increase, decrease; (c) no effect, no effect, no effect; (d) decrease, increase, decrease; (e) increase, no effect, no effect"}
{"id": 5469, "contents": "2299. Chapter 16 - \n1. (a) reduction; (b) oxidation; (c) oxidation; (d) reduction\n2. (a) $\\mathrm{F}_{2}+\\mathrm{Ca} \\longrightarrow 2 \\mathrm{~F}^{-}+\\mathrm{Ca}^{2+}$; (b) $\\mathrm{Cl}_{2}+2 \\mathrm{Li} \\longrightarrow 2 \\mathrm{Li}^{+}+2 \\mathrm{Cl}^{-}$; (c) $3 \\mathrm{Br}_{2}+2 \\mathrm{Fe} \\longrightarrow 2 \\mathrm{Fe}^{3+}+6 \\mathrm{Br}^{-}$; (d) $\\mathrm{MnO}_{4}-+4 \\mathrm{H}^{+}+3 \\mathrm{Ag} \\longrightarrow 3 \\mathrm{Ag}^{+}+\\mathrm{MnO}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}$\n3. Oxidized: (a) $\\mathrm{Sn}^{2+}$; (b) Hg ; (c) Al; reduced: (a) $\\mathrm{H}_{2} \\mathrm{O}_{2}$; (b) $\\mathrm{PbO}_{2}$; (c) $\\mathrm{Cr}_{2} \\mathrm{O}_{7}{ }^{2-}$; oxidizing agent: (a) $\\mathrm{H}_{2} \\mathrm{O}_{2}$; (b) $\\mathrm{PbO}_{2}$; (c) $\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}$; reducing agent: (a) $\\mathrm{Sn}^{2+}$; (b) Hg ; (c) Al\n4. Oxidized = reducing agent: (a) $\\mathrm{SO}_{3}{ }^{2-}$; (b) $\\mathrm{Mn}(\\mathrm{OH})_{2}$; (c) $\\mathrm{H}_{2}$; (d) Al; reduced = oxidizing agent: (a) $\\mathrm{Cu}(\\mathrm{OH})_{2}$; (b) $\\mathrm{O}_{2}$; (c) $\\mathrm{NO}_{3}^{-}$; (d) $\\mathrm{CrO}_{4}{ }^{2-}$"}
{"id": 5470, "contents": "2299. Chapter 16 - \n5. In basic solution, $\\left[\\mathrm{OH}^{-}\\right]>1 \\times 10^{-7} M>\\left[\\mathrm{H}^{+}\\right]$. Hydrogen ion cannot appear as a reactant because its concentration is essentially zero. If it were produced, it would instantly react with the excess hydroxide ion to produce water. Thus, hydrogen ion should not appear as a reactant or product in basic solution.\n6. (a) $\\operatorname{Mg}(s)\\left|\\operatorname{Mg}^{2+}(a q) \\| \\mathrm{Ni}^{2+}(a q)\\right| \\mathrm{Ni}(s)$; (b) $\\mathrm{Cu}(s)\\left|\\mathrm{Cu}^{2+}(a q) \\| \\mathrm{Ag}^{+}(a q)\\right| \\mathrm{Ag}(s)$; (c)\n$\\operatorname{Mn}(s)\\left|\\mathrm{Mn}^{2+}(a q)\\left\\|\\mathrm{Sn}^{2+}(a q)|\\operatorname{Sn}(s) ;(\\mathrm{d}) \\operatorname{Pt}(s)| \\mathrm{Cu}^{+}(a q), \\mathrm{Cu}^{2+}(a q)\\right\\| \\mathrm{Au}^{3+}(a q)\\right| \\mathrm{Au}(s)$\n7. (a) $\\mathrm{Mg}(s)+\\mathrm{Cu}^{2+}(a q) \\longrightarrow \\mathrm{Mg}^{2+}(a q)+\\mathrm{Cu}(s)$; (b) $2 \\mathrm{Ag}^{+}(a q)+\\mathrm{Ni}(s) \\longrightarrow \\mathrm{Ni}^{2+}(a q)+2 \\mathrm{Ag}(s)$"}
{"id": 5471, "contents": "2299. Chapter 16 - \n8. Species oxidized = reducing agent: (a) $\\mathrm{Al}(\\mathrm{s})$; (b) $\\mathrm{NO}(g)$; (c) $\\mathrm{Mg}(s)$; and (d) $\\mathrm{MnO}_{2}(s)$; Species reduced $=$ oxidizing agent: (a) $\\mathrm{Zr}^{4+}(a q)$; (b) $\\mathrm{Ag}^{+}(a q)$; (c) $\\mathrm{SiO}_{3}{ }^{2-}$ (aq); and (d) $\\mathrm{ClO}_{3}{ }^{-}(a q)$\n9. Without the salt bridge, the circuit would be open (or broken) and no current could flow. With a salt bridge, each half-cell remains electrically neutral and current can flow through the circuit.\n10. Active electrodes participate in the oxidation-reduction reaction. Since metals form cations, the electrode would lose mass if metal atoms in the electrode were to oxidize and go into solution. Oxidation occurs at the anode.\n11. (a) +2.115 V (spontaneous); (b) +0.4626 V (spontaneous); (c) +1.0589 V (spontaneous); (d) +0.727 V (spontaneous)\n12. $3 \\mathrm{Cu}(s)+2 \\mathrm{Au}^{3+}(a q) \\longrightarrow 3 \\mathrm{Cu}^{2+}(a q)+2 \\mathrm{Au}(s)$; +1.16 V ; spontaneous\n13. $3 \\mathrm{Cd}(s)+2 \\mathrm{Al}^{3+}(a q) \\longrightarrow 3 \\mathrm{Cd}^{2+}(a q)+2 \\mathrm{Al}(s)$; -1.259 V ; nonspontaneous\n14. (a) $0 \\mathrm{~kJ} / \\mathrm{mol}$; (b) $-83.7 \\mathrm{~kJ} / \\mathrm{mol}$; (c) $+235.3 \\mathrm{~kJ} / \\mathrm{mol}$"}
{"id": 5472, "contents": "2299. Chapter 16 - \n15. (a) standard cell potential: 1.50 V , spontaneous; cell potential under stated conditions: 1.43 V , spontaneous; (b) standard cell potential: 1.405 V , spontaneous; cell potential under stated conditions: 1.423 V, spontaneous; (c) standard cell potential: -2.749 V, nonspontaneous; cell potential under stated conditions: -2.757 V , nonspontaneous\n16. (a) $1.7 \\times 10^{-10}$; (b) $2.6 \\times 10^{-21}$; (c) $4.693 \\times 10^{21}$; (d) $1.0 \\times 10^{-14}$\n17. (a) anode: $\\mathrm{Cu}(s) \\longrightarrow \\mathrm{Cu}^{2+}(a q)+2 \\mathrm{e}^{-}$\ncathode: $2 \\times\\left(\\mathrm{Ag}^{+}(a q)+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Ag}(s)\\right)$\n$E_{\\text {anode }}^{\\circ}=0.34 \\mathrm{~V}$"}
{"id": 5473, "contents": "2299. Chapter 16 - \n$E^{\\circ}=0.7996 \\mathrm{~V}$$\\quad$; (b) $3.5 \\times 10^{15}$; (c) $5.6 \\times 10^{-9}$\n\nM\n34. Batteries are self-contained and have a limited supply of reagents to expend before going dead."}
{"id": 5474, "contents": "2299. Chapter 16 - \nAlternatively, battery reaction byproducts accumulate and interfere with the reaction. Because a fuel cell is constantly resupplied with reactants and products are expelled, it can continue to function as long as reagents are supplied.\n36. $E_{\\text {cell }}$, as described in the Nernst equation, has a term that is directly proportional to temperature. At low temperatures, this term is decreased, resulting in a lower cell voltage provided by the battery to the device-the same effect as a battery running dead.\n38. Mg and Zn\n40. Both examples involve cathodic protection. The (sacrificial) anode is the metal that corrodes (oxidizes or reacts). In the case of iron ( -0.447 V ) and zinc ( -0.7618 V ), zinc has a more negative standard reduction potential and so serves as the anode. In the case of iron and copper ( 0.34 V ), iron has the smaller standard reduction potential and so corrodes (serves as the anode).\n42. While the reduction potential of lithium would make it capable of protecting the other metals, this high potential is also indicative of how reactive lithium is; it would have a spontaneous reaction with most substances. This means that the lithium would react quickly with other substances, even those that would not oxidize the metal it is attempting to protect. Reactivity like this means the sacrificial anode would be depleted rapidly and need to be replaced frequently. (Optional additional reason: fire hazard in the presence of water.)"}
{"id": 5475, "contents": "2299. Chapter 16 - \n46. (a) mass $\\mathrm{Ca}=69.1 \\mathrm{~g}$; (b) $\\begin{aligned} & \\text { mass } \\mathrm{Li}=23.9 \\mathrm{~g} \\\\ & \\text { mass } \\mathrm{H}_{2}=3.48 \\mathrm{~g}\\end{aligned}$; (c) $\\begin{aligned} & \\text { mass } \\mathrm{Al}=31.0 \\mathrm{~g} \\\\ & \\text { mass } \\mathrm{Cl}_{2}=122 \\mathrm{~g}\\end{aligned}$; (d) $\\begin{aligned} & \\text { mass } \\mathrm{Cr}=59.8 \\mathrm{~g} \\\\ & \\operatorname{mass} \\mathrm{Br}_{2}=276 \\mathrm{~g}\\end{aligned}$\n48. 0.79 L"}
{"id": 5476, "contents": "2300. Chapter 17 - \n1. The instantaneous rate is the rate of a reaction at any particular point in time, a period of time that is so short that the concentrations of reactants and products change by a negligible amount. The initial rate is the instantaneous rate of reaction as it starts (as product just begins to form). Average rate is the average of the instantaneous rates over a time period.\n2. rate $=+\\frac{1}{2} \\frac{\\Delta\\left[\\mathrm{CIF}_{3}\\right]}{\\Delta t}=-\\frac{\\Delta\\left[\\mathrm{Cl}_{2}\\right]}{\\Delta t}=-\\frac{1}{3} \\frac{\\Delta\\left[\\mathrm{~F}_{2}\\right]}{\\Delta t}$\n3. (a) average rate, $0-10 \\mathrm{~s}=0.0375 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~s}^{-1}$; average rate, $10-20 \\mathrm{~s}=0.0265 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~s}^{-1}$; (b) instantaneous rate, $15 \\mathrm{~s}=0.023 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~s}^{-1}$; (c) average rate for $B$ formation $=0.0188 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~s}^{-1}$; instantaneous rate for B formation $=0.012 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~s}^{-1}$\n4. Higher molarity increases the rate of the reaction. Higher temperature increases the rate of the reaction. Smaller pieces of magnesium metal will react more rapidly than larger pieces because more reactive surface exists.\n5. (a) Depending on the angle selected, the atom may take a long time to collide with the molecule and, when a collision does occur, it may not result in the breaking of the bond and the forming of the other. (b) Particles of reactant must come into contact with each other before they can react."}
{"id": 5477, "contents": "2300. Chapter 17 - \n6. (a) very slow; (b) As the temperature is increased, the reaction proceeds at a faster rate. The amount of reactants decreases, and the amount of products increases. After a while, there is a roughly equal amount of $B C, A B$, and $C$ in the mixture and a slight excess of $A$.\n7. (a) 2 ; (b) 1\n8. (a) The process reduces the rate by a factor of 4 . (b) Since CO does not appear in the rate law, the rate is not affected.\n9. $4.3 \\times 10^{-5} \\mathrm{~mol} / \\mathrm{L} / \\mathrm{s}$\n10. $7.9 \\times 10^{-13} \\mathrm{~mol} / \\mathrm{L} /$ year\n11. rate $=k$; $k=2.0 \\times 10^{-2} \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~h}^{-1}$ (about $0.9 \\mathrm{~g} \\mathrm{~L}^{-1} \\mathrm{~h}^{-1}$ for the average male); The reaction is zero order.\n12. rate $=k[\\mathrm{NOCl}]^{2} ; k=8.0 \\times 10^{-8} \\mathrm{~L} / \\mathrm{mol} / \\mathrm{h}$; second order\n13. rate $=k[\\mathrm{NO}]^{2}\\left[\\mathrm{Cl}_{2}\\right] ; k=9.1 \\mathrm{~L}^{2} \\mathrm{~mol}^{-2} \\mathrm{~h}^{-1}$; second order in NO; first order in $\\mathrm{Cl}_{2}$\n14. (a) The rate law is second order in $A$ and is written as rate $=k[A]^{2}$. (b) $k=7.88 \\times 10^{-3} \\mathrm{~L} \\mathrm{~mol}^{-1} \\mathrm{~s}^{-1}$\n15. (a) $2.5 \\times 10^{-4} \\mathrm{~mol} / \\mathrm{L} / \\mathrm{min}$"}
{"id": 5478, "contents": "2300. Chapter 17 - \n15. (a) $2.5 \\times 10^{-4} \\mathrm{~mol} / \\mathrm{L} / \\mathrm{min}$\n16. rate $=k\\left[\\mathrm{I}^{-}\\right]\\left[\\mathrm{OCl}^{-}\\right] ; k=6.1 \\times 10^{-2} \\mathrm{~L} \\mathrm{~mol}^{-1} \\mathrm{~s}^{-1}$\n17. Plotting a graph of $\\ln \\left[\\mathrm{SO}_{2} \\mathrm{Cl}_{2}\\right]$ versus $t$ reveals a linear trend; therefore we know this is a first-order reaction:"}
{"id": 5479, "contents": "2300. Chapter 17 - \n$k=2.20 \\times 10^{-5} \\mathrm{~s}^{-1}$\n18.\n\n\n\nThe plot is nicely linear, so the reaction is second order. $k=50.1 \\mathrm{~L} \\mathrm{~mol}^{-1} \\mathrm{~h}^{-1}$\n36. 14.3 d\n38. $8.3 \\times 10^{7} \\mathrm{~s}$\n40. 0.826 s\n42. The reaction is first order. $k=1.0 \\times 10^{7} \\mathrm{~L} \\mathrm{~mol}^{-1} \\mathrm{~min}^{-1}$\n44. $1.16 \\times 10^{3} \\mathrm{~s} ; 20 \\%$ remains\n46. 252 days\n48.\n\n| $[A]_{0}(M)$ | $k \\times 10^{3}\\left(\\mathrm{~s}^{-1}\\right)$ |\n| :--- | :--- |\n| 4.88 | 2.45 |\n| 3.52 | 2.51 |\n| 2.29 | 2.53 |\n| 1.81 | 2.58 |\n| 5.33 | 2.36 |\n| 4.05 | 2.47 |\n| 2.95 | 2.48 |\n| 1.72 | 2.43 |"}
{"id": 5480, "contents": "2300. Chapter 17 - \n50. The reactants either may be moving too slowly to have enough kinetic energy to exceed the activation energy for the reaction, or the orientation of the molecules when they collide may prevent the reaction from occurring.\n51. The activation energy is the minimum amount of energy necessary to form the activated complex in a reaction. It is usually expressed as the energy necessary to form one mole of activated complex.\n52. After finding $k$ at several different temperatures, a plot of $\\ln k$ versus $\\frac{1}{T}$, gives a straight line with the slope $\\frac{-E_{\\mathrm{a}}}{R}$ from which $E_{\\mathrm{a}}$ may be determined.\n53. (a) 4-times faster (b) 128-times faster\n54. $3.9 \\times 10^{15} \\mathrm{~s}^{-1}$\n55. $43.0 \\mathrm{~kJ} / \\mathrm{mol}$\n56. $177 \\mathrm{~kJ} / \\mathrm{mol}$\n57. $E_{\\mathrm{a}}=108 \\mathrm{~kJ} ; A=2.0 \\times 10^{8} \\mathrm{~s}^{-1} ; k=3.2 \\times 10^{-10} \\mathrm{~s}^{-1}$; (b) $1.81 \\times 10^{8} \\mathrm{~h}$ or $7.6 \\times 10^{6}$ day; (c) Assuming that the reaction is irreversible simplifies the calculation because we do not have to account for any reactant that, having been converted to product, returns to the original state.\n58. The $A$ atom has enough energy to react with $B C$; however, the different angles at which it bounces off of $B C$ without reacting indicate that the orientation of the molecule is an important part of the reaction kinetics and determines whether a reaction will occur.\n59. No. In general, for the overall reaction, we cannot predict the effect of changing the concentration without knowing the rate law. Yes. If the reaction is an elementary reaction, then doubling the concentration of $A$ doubles the rate.\n60. Rate $=kA^{2}$; Rate $=k[A]^{3}$"}
{"id": 5481, "contents": "2300. Chapter 17 - \n60. Rate $=kA^{2}$; Rate $=k[A]^{3}$\n61. (a) Rate $_{1}=k\\left[\\mathrm{O}_{3}\\right]$; (b) Rate $_{2}=k\\left\\mathrm{O}_{3}\\right$; (c) Rate ${ }_{3}=k\\mathrm{ClO}$; (d) Rate ${ }_{2}=k\\left\\mathrm{O}_{3}\\right$; (e) Rate ${ }_{3}=k\\left\\mathrm{NO}_{2}\\right$\n62. (a) Doubling $\\left[\\mathrm{H}_{2}\\right]$ doubles the rate. $\\left[\\mathrm{H}_{2}\\right]$ must enter the rate law to the first power. Doubling [NO] increases the rate by a factor of 4 . [ NO ] must enter the rate law to the second power. (b) Rate $=k[\\mathrm{NO}]^{2}\\left[\\mathrm{H}_{2}\\right]$; (c) $k=5.0$ $\\times 10^{3} \\mathrm{~mol}^{-2} \\mathrm{~L}^{-2} \\mathrm{~min}^{-1}$; (d) $0.0050 \\mathrm{~mol} / \\mathrm{L}$; (e) Step II is the rate-determining step. If step I gives $\\mathrm{N}_{2} \\mathrm{O}_{2}$ in adequate amount, steps 1 and 2 combine to give $2 \\mathrm{NO}+\\mathrm{H}_{2} \\longrightarrow \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{N}_{2} \\mathrm{O}$. This reaction corresponds to the observed rate law. Combine steps 1 and 2 with step 3 , which occurs by supposition in a rapid fashion, to give the appropriate stoichiometry.\n63. The general mode of action for a catalyst is to provide a mechanism by which the reactants can unite more\nreadily by taking a path with a lower reaction energy. The rates of both the forward and the reverse reactions are increased, leading to a faster achievement of equilibrium."}
{"id": 5482, "contents": "2300. Chapter 17 - \nreadily by taking a path with a lower reaction energy. The rates of both the forward and the reverse reactions are increased, leading to a faster achievement of equilibrium.\n64. (a) Chlorine atoms are a catalyst because they react in the second step but are regenerated in the third step. Thus, they are not used up, which is a characteristic of catalysts. (b) NO is a catalyst for the same reason as in part (a).\n65. no changes.\n66. The lowering of the transition state energy indicates the effect of a catalyst. (a) B; (b) B\n67. The energy needed to go from the initial state to the transition state is (a) 10 kJ ; (b) 10 kJ\n68. Both diagrams describe two-step, exothermic reactions, but with different changes in enthalpy, suggesting the diagrams depict two different overall reactions."}
{"id": 5483, "contents": "2301. Chapter 18 - \n1. The alkali metals all have a single $s$ electron in their outermost shell. In contrast, the alkaline earth metals have a completed $s$ subshell in their outermost shell. In general, the alkali metals react faster and are more reactive than the corresponding alkaline earth metals in the same period.\n2."}
{"id": 5484, "contents": "2301. Chapter 18 - \n$\\mathrm{Na}+\\mathrm{I}_{2} \\longrightarrow 2 \\mathrm{NaI}$\n$2 \\mathrm{Na}+\\mathrm{Se} \\longrightarrow \\mathrm{Na}_{2} \\mathrm{Se}$\n$2 \\mathrm{Na}+\\mathrm{O}_{2} \\longrightarrow \\mathrm{Na}_{2} \\mathrm{O}_{2}$\n$\\mathrm{Sr}+\\mathrm{I}_{2} \\longrightarrow \\mathrm{SrI}_{2}$\n$\\mathrm{Sr}+\\mathrm{Se} \\longrightarrow \\mathrm{SrSe}$\n$2 \\mathrm{Sr}+\\mathrm{O}_{2} \\longrightarrow 2 \\mathrm{SrO}$\n$2 \\mathrm{Al}+3 \\mathrm{I}_{2} \\longrightarrow 2 \\mathrm{AlI}_{3}$\n$2 \\mathrm{Al}+3 \\mathrm{Se} \\longrightarrow \\mathrm{Al}_{2} \\mathrm{Se}_{3}$\n$4 \\mathrm{Al}+3 \\mathrm{O}_{2} \\longrightarrow 2 \\mathrm{Al}_{2} \\mathrm{O}_{3}$"}
{"id": 5485, "contents": "2301. Chapter 18 - \n$4 \\mathrm{Al}+3 \\mathrm{O}_{2} \\longrightarrow 2 \\mathrm{Al}_{2} \\mathrm{O}_{3}$\n5. The possible ways of distinguishing between the two include infrared spectroscopy by comparison of known compounds, a flame test that gives the characteristic yellow color for sodium (strontium has a red flame), or comparison of their solubilities in water. At $20^{\\circ} \\mathrm{C}, \\mathrm{NaCl}$ dissolves to the extent of $\\frac{35.7 \\mathrm{~g}}{100 \\mathrm{~mL}}$ compared with $\\frac{53.8 \\mathrm{~g}}{100 \\mathrm{~mL}}$ for $\\mathrm{SrCl}_{2}$. Heating to $100^{\\circ} \\mathrm{C}$ provides an easy test, since the solubility of NaCl is $\\frac{39.12 \\mathrm{~g}}{100 \\mathrm{~mL}}$, but that of $\\mathrm{SrCl}_{2}$ is $\\frac{100.8 \\mathrm{~g}}{100 \\mathrm{~mL}}$. Density determination on a solid is sometimes difficult, but there is enough difference ( $2.165 \\mathrm{~g} / \\mathrm{mL} \\mathrm{NaCl}$ and $3.052 \\mathrm{~g} / \\mathrm{mL} \\mathrm{SrCl}_{2}$ ) that this method would be viable and perhaps the easiest and least expensive test to perform.\n7. (a) $2 \\mathrm{Sr}(s)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{SrO}(s)$; (b) $\\mathrm{Sr}(s)+2 \\mathrm{HBr}(g) \\longrightarrow \\mathrm{SrBr}_{2}(s)+\\mathrm{H}_{2}(g)$; (c)\n$\\mathrm{Sr}(s)+\\mathrm{H}_{2}(g) \\longrightarrow \\mathrm{SrH}_{2}(s) ;(\\mathrm{d}) 6 \\mathrm{Sr}(s)+\\mathrm{P}_{4}(s) \\longrightarrow 2 \\mathrm{Sr}_{3} \\mathrm{P}_{2}(s)$; (e)"}
{"id": 5486, "contents": "2301. Chapter 18 - \n$\\mathrm{Sr}(s)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Sr}(\\mathrm{OH})_{2}(a q)+\\mathrm{H}_{2}(g)$\n9. 11 lb\n11. Yes, tin reacts with hydrochloric acid to produce hydrogen gas.\n13. In $\\mathrm{PbCl}_{2}$, the bonding is ionic, as indicated by its melting point of $501^{\\circ} \\mathrm{C}$. In $\\mathrm{PbCl}_{4}$, the bonding is covalent, as evidenced by it being an unstable liquid at room temperature."}
{"id": 5487, "contents": "2301. Chapter 18 - \n$$\n\\begin{aligned}\n& \\text { countercurrent } \\\\\n& \\text { fractionating }\n\\end{aligned}\n$$\n\n15. $2 \\mathrm{CsCl}(l)+\\mathrm{Ca}(g) \\xrightarrow{\\text { tower }} 2 \\mathrm{Cs}(g)+\\mathrm{CaCl}_{2}(l)$\n16. Cathode (reduction): $2 \\mathrm{Li}^{+}+2 \\mathrm{e}^{-} \\longrightarrow 2 \\mathrm{Li}(l)$; Anode (oxidation): $2 \\mathrm{Cl}^{-} \\longrightarrow \\mathrm{Cl}_{2}(g)+2 \\mathrm{e}^{-}$; Overall reaction: $2 \\mathrm{Li}^{+}+2 \\mathrm{Cl}^{-} \\longrightarrow 2 \\mathrm{Li}(l)+\\mathrm{Cl}_{2}(g)$\n17. $0.5035 \\mathrm{~g} \\mathrm{H}_{2}$\n18. Despite its reactivity, magnesium can be used in construction even when the magnesium is going to come in contact with a flame because a protective oxide coating is formed, preventing gross oxidation. Only if the metal is finely subdivided or present in a thin sheet will a high-intensity flame cause its rapid burning.\n19. Extract from ore: $\\mathrm{AlO}(\\mathrm{OH})(s)+\\mathrm{NaOH}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow \\mathrm{Na}\\left\\mathrm{Al}(\\mathrm{OH})_{4}\\right$ Recover: $2 \\mathrm{Na}\\left\\mathrm{Al}(\\mathrm{OH})_{4}\\right+\\mathrm{H}_{2} \\mathrm{SO}_{4}(a q) \\longrightarrow 2 \\mathrm{Al}(\\mathrm{OH})_{3}(s)+\\mathrm{Na}_{2} \\mathrm{SO}_{4}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$"}
{"id": 5488, "contents": "2301. Chapter 18 - \nSinter: $2 \\mathrm{Al}(\\mathrm{OH})_{3}(s) \\longrightarrow \\mathrm{Al}_{2} \\mathrm{O}_{3}(s)+3 \\mathrm{H}_{2} \\mathrm{O}(g)$\nDissolve in $\\mathrm{Na}_{3} \\mathrm{AlF}_{6}(\\mathrm{I})$ and electrolyze: $\\mathrm{Al}^{3+}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Al}(s)$\n25. $25.83 \\%$\n27. 39 kg\n29. (a) $\\mathrm{H}_{3} \\mathrm{BPH}_{3}$ :\n\n(b) $\\mathrm{BF}_{4}^{-}$:\n\n(c) $\\mathrm{BBr}_{3}$ :\n\n(d) $\\mathrm{B}\\left(\\mathrm{CH}_{3}\\right)_{3}$ :\n\n(e) $\\mathrm{B}(\\mathrm{OH})_{3}$ :"}
{"id": 5489, "contents": "2301. Chapter 18 - \n31. $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{2} 3 d^{0}$.\n33. (a) $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{SiH}$ : $s p^{3}$ bonding about Si ; the structure is tetrahedral; (b) $\\mathrm{SiO}_{4}{ }^{4-}: s p^{3}$ bonding about Si ; the structure is tetrahedral; (c) $\\mathrm{Si}_{2} \\mathrm{H}_{6}: s p^{3}$ bonding about each Si ; the structure is linear along the $\\mathrm{Si}-\\mathrm{Si}$ bond; (d) $\\mathrm{Si}(\\mathrm{OH})_{4}: s p^{3}$ bonding about Si ; the structure is tetrahedral; (e) $\\mathrm{SiF}_{6}{ }^{2-}: s p^{3} d^{2}$ bonding about Si; the structure is octahedral\n35. (a) nonpolar; (b) nonpolar; (c) polar; (d) nonpolar; (e) polar\n37. (a) tellurium dioxide or tellurium(IV) oxide; (b) antimony(III) sulfide; (c) germanium(IV) fluoride; (d) silane or silicon(IV) hydride; (e) germanium(IV) hydride\n39. Boron has only $s$ and $p$ orbitals available, which can accommodate a maximum of four electron pairs. Unlike silicon, no $d$ orbitals are available in boron.\n41. (a) $\\Delta H^{\\circ}=87 \\mathrm{~kJ} ; \\Delta G^{\\circ}=44 \\mathrm{~kJ}$; (b) $\\Delta H^{\\circ}=-109.9 \\mathrm{~kJ} ; \\Delta G^{\\circ}=-154.7 \\mathrm{~kJ}$; (c) $\\Delta H^{\\circ}=-510 \\mathrm{~kJ} ; \\Delta G^{\\circ}=-601.5 \\mathrm{~kJ}$"}
{"id": 5490, "contents": "2301. Chapter 18 - \n43. A mild solution of hydrofluoric acid would dissolve the silicate and would not harm the diamond.\n45. In the $N_{2}$ molecule, the nitrogen atoms have an $\\sigma$ bond and two $\\pi$ bonds holding the two atoms together. The presence of three strong bonds makes $\\mathrm{N}_{2}$ a very stable molecule. Phosphorus is a third-period element, and as such, does not form $\\pi$ bonds efficiently; therefore, it must fulfill its bonding requirement by forming three $\\sigma$ bonds.\n47. (a) $\\mathrm{H}=1+, \\mathrm{C}=2+$, and $\\mathrm{N}=3-$; (b) $\\mathrm{O}=2+$ and $\\mathrm{F}=1-$; (c) $\\mathrm{As}=3+$ and $\\mathrm{Cl}=1-$\n49. $\\mathrm{S}<\\mathrm{Cl}<\\mathrm{O}<\\mathrm{F}$\n51. The electronegativity of the nonmetals is greater than that of hydrogen. Thus, the negative charge is better represented on the nonmetal, which has the greater tendency to attract electrons in the bond to itself.\n53. Hydrogen has only one orbital with which to bond to other atoms. Consequently, only one two-electron bond can form.\n55. $0.43 \\mathrm{~g} \\mathrm{H}_{2}$\n57. (a) $\\mathrm{Ca}(\\mathrm{OH})_{2}(a q)+\\mathrm{CO}_{2}(g) \\longrightarrow \\mathrm{CaCO}_{3}(s)+\\mathrm{H}_{2} \\mathrm{O}(l)$; (b) $\\mathrm{CaO}(s)+\\mathrm{SO}_{2}(g) \\longrightarrow \\mathrm{CaSO}_{3}(s)$;\n(c) $2 \\mathrm{NaHCO}_{3}(s)+\\mathrm{NaH}_{2} \\mathrm{PO}_{4}(a q) \\longrightarrow \\mathrm{Na}_{3} \\mathrm{PO}_{4}(a q)+2 \\mathrm{CO}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n59. (a) $\\mathrm{NH}^{2-}$ :"}
{"id": 5491, "contents": "2301. Chapter 18 - \n(b) $\\mathrm{N}_{2} \\mathrm{~F}_{4}$ :\n\n(c) $\\mathrm{NH}_{2}{ }^{-}$:\n\n(d) $\\mathrm{NF}_{3}$ :\n\n(e) $\\mathrm{N}_{3}{ }^{-}$:\n\n61. Ammonia acts as a Br\u00f8nsted base because it readily accepts protons and as a Lewis base in that it has an electron pair to donate.\nBr\u00f8nsted base: $\\mathrm{NH}_{3}+\\mathrm{H}_{3} \\mathrm{O}^{+} \\longrightarrow \\mathrm{NH}_{4}^{+}+\\mathrm{H}_{2} \\mathrm{O}$\nLewis base: $2 \\mathrm{NH}_{3}+\\mathrm{Ag}^{+} \\longrightarrow\\left[\\mathrm{H}_{3} \\mathrm{~N}-\\mathrm{Ag}-\\mathrm{NH}_{3}\\right]^{+}$\n63. (a) $\\mathrm{NO}_{2}$ :\n\n\nNitrogen is $s p^{2}$ hybridized. The molecule has a bent geometry with an ONO bond angle of approximately $120^{\\circ}$.\n(b) $\\mathrm{NO}_{2}{ }^{-}$:\n\n\nNitrogen is $s p^{2}$ hybridized. The molecule has a bent geometry with an ONO bond angle slightly less than $120^{\\circ}$.\n(c) $\\mathrm{NO}_{2}{ }^{+}$:\n\n\nNitrogen is $s p$ hybridized. The molecule has a linear geometry with an ONO bond angle of $180^{\\circ}$.\n65. Nitrogen cannot form a $\\mathrm{NF}_{5}$ molecule because it does not have $d$ orbitals to bond with the additional two fluorine atoms.\n67. (a)\n\n(b)\n\n(c)\n\n(d)\n\n(e)"}
{"id": 5492, "contents": "2301. Chapter 18 - \n(b)\n\n(c)\n\n(d)\n\n(e)\n\n69. (a) $\\mathrm{P}_{4}(s)+4 \\mathrm{Al}(s) \\longrightarrow 4 \\mathrm{AlP}(s)$; (b) $\\mathrm{P}_{4}(s)+12 \\mathrm{Na}(s) \\longrightarrow 4 \\mathrm{Na}_{3} \\mathrm{P}(s)$; (c) $\\mathrm{P}_{4}(s)+10 \\mathrm{~F}_{2}(g) \\longrightarrow 4 \\mathrm{PF}_{5}(l)$; (d) $\\mathrm{P}_{4}(s)+6 \\mathrm{Cl}_{2}(g) \\longrightarrow 4 \\mathrm{PCl}_{3}(l)$ or $\\mathrm{P}_{4}(s)+10 \\mathrm{Cl}_{2}(g) \\longrightarrow 4 \\mathrm{PCl}_{5}(l)$; (e) $\\mathrm{P}_{4}(s)+3 \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{P}_{4} \\mathrm{O}_{6}(s)$ or $\\mathrm{P}_{4}(s)+5 \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{P}_{4} \\mathrm{O}_{10}(s) ;(\\mathrm{f}) \\mathrm{P}_{4} \\mathrm{O}_{6}(s)+2 \\mathrm{O}_{2}(g) \\longrightarrow \\mathrm{P}_{4} \\mathrm{O}_{10}(s)$\n71. 291 mL\n73. 28 tons\n75. (a)\n\n\nTetrahedral\n(b)\n\n\nTrigonal bipyramid\n(c)\n\n\nOctahedral\n(d)"}
{"id": 5493, "contents": "2301. Chapter 18 - \n77. (a) $\\mathrm{P}=3+$; (b) $\\mathrm{P}=5+$; (c) $\\mathrm{P}=3+$; (d) $\\mathrm{P}=5+$; (e) $\\mathrm{P}=3-$; (f) $\\mathrm{P}=5+$\n79. $\\mathrm{FrO}_{2}$\n81. (a) $2 \\mathrm{Zn}(s)+\\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{ZnO}(s)$; (b) $\\mathrm{ZnCO}_{3}(s) \\longrightarrow \\mathrm{ZnO}(s)+\\mathrm{CO}_{2}(g)$; (c)\n$\\mathrm{ZnCO}_{3}(s)+2 \\mathrm{CH}_{3} \\mathrm{COOH}(a q) \\longrightarrow \\mathrm{Zn}\\left(\\mathrm{CH}_{3} \\mathrm{COO}\\right)_{2}(a q)+\\mathrm{CO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l)$; (d)\n$\\mathrm{Zn}(s)+2 \\mathrm{HBr}(a q) \\longrightarrow \\mathrm{ZnBr}_{2}(a q)+\\mathrm{H}_{2}(g)$\n83. $\\mathrm{Al}(\\mathrm{OH})_{3}(s)+3 \\mathrm{H}^{+}(a q) \\longrightarrow \\mathrm{Al}^{3+}+3 \\mathrm{H}_{2} \\mathrm{O}(l) ; \\mathrm{Al}(\\mathrm{OH})_{3}(s)+\\mathrm{OH}^{-} \\longrightarrow\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]^{-}(a q)$"}
{"id": 5494, "contents": "2301. Chapter 18 - \n85. (a) $\\mathrm{Na}_{2} \\mathrm{O}(s)+\\mathrm{H}_{2} \\mathrm{O}(l) \\longrightarrow 2 \\mathrm{NaOH}(a q)$; (b) $\\mathrm{Cs}_{2} \\mathrm{CO}_{3}(s)+2 \\mathrm{HF}(a q) \\longrightarrow 2 \\mathrm{CsF}(a q)+\\mathrm{CO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(l)$;\n(c) $\\mathrm{Al}_{2} \\mathrm{O}_{3}(s)+6 \\mathrm{HClO}_{4}(a q) \\longrightarrow 2 \\mathrm{Al}\\left(\\mathrm{ClO}_{4}\\right)_{3}(a q)+3 \\mathrm{H}_{2} \\mathrm{O}(l)$; (d)\n$\\mathrm{Na}_{2} \\mathrm{CO}_{3}(a q)+\\mathrm{Ba}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q) \\longrightarrow 2 \\mathrm{NaNO}_{3}(a q)+\\mathrm{BaCO}_{3}(s) ;(\\mathrm{e})$\n$\\mathrm{TiCl}_{4}(l)+4 \\mathrm{Na}(s) \\longrightarrow \\mathrm{Ti}(s)+4 \\mathrm{NaCl}(s)$\n87. $\\mathrm{HClO}_{4}$ is the stronger acid because, in a series of oxyacids with similar formulas, the higher the electronegativity of the central atom, the stronger is the attraction of the central atom for the electrons of the oxygen(s). The stronger attraction of the oxygen electron results in a stronger attraction of oxygen for the electrons in the O-H bond, making the hydrogen more easily released. The weaker this bond, the stronger the acid.\n89. As $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ and $\\mathrm{H}_{2} \\mathrm{SeO}_{4}$ are both oxyacids and their central atoms both have the same oxidation number, the acid strength depends on the relative electronegativity of the central atom. As sulfur is more electronegative than selenium, $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ is the stronger acid."}
{"id": 5495, "contents": "2301. Chapter 18 - \n91. $\\mathrm{SO}_{2}, s p^{2} 4+; \\mathrm{SO}_{3}, s p^{2}, 6+; \\mathrm{H}_{2} \\mathrm{SO}_{4}, s p^{3}, 6+$\n93. $\\mathrm{SF}_{6}: \\mathrm{S}=6+; \\mathrm{SO}_{2} \\mathrm{~F}_{2}: \\mathrm{S}=6+$; $\\mathrm{KHS}: \\mathrm{S}=2-$\n95. Sulfur is able to form double bonds only at high temperatures (substantially endothermic conditions), which is not the case for oxygen.\n97. There are many possible answers including: $\\mathrm{Cu}(s)+2 \\mathrm{H}_{2} \\mathrm{SO}_{4}(l) \\longrightarrow \\mathrm{CuSO}_{4}(a q)+\\mathrm{SO}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$ and $\\mathrm{C}(s)+2 \\mathrm{H}_{2} \\mathrm{SO}_{4}(l) \\longrightarrow \\mathrm{CO}_{2}(g)+2 \\mathrm{SO}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n99. $5.1 \\times 10^{4} \\mathrm{~g}$\n101. $\\mathrm{SnCl}_{4}$ is not a salt because it is covalently bonded. A salt must have ionic bonds.\n103. In oxyacids with similar formulas, the acid strength increases as the electronegativity of the central atom increases. $\\mathrm{HClO}_{3}$ is stronger than $\\mathrm{HBrO}_{3} ; \\mathrm{Cl}$ is more electronegative than Br .\n105. (a)"}
{"id": 5496, "contents": "2301. Chapter 18 - \nSquare pyramidal\n(b)\n\n\nLinear\n(c)\n\n\nTrigonal bipyramidal\n(d)\n\n\nSeesaw\n(e)\n\n\nT-shaped\n107. (a) bromine trifluoride; (b) sodium bromate; (c) phosphorus pentabromide; (d) sodium perchlorate; (e) potassium hypochlorite\n109. (a) I: 7+; (b) I: 7+; (c) Cl: 4+; (d) I: 3+; Cl: 1-; (e) F: 0\n111. (a) $s p^{3} d$ hybridized; (b) $s p^{3} d^{2}$ hybridized; (c) $s p^{3}$ hybridized; (d) $s p^{3}$ hybridized; (e) $s p^{3} d^{2}$ hybridized;\n113. (a) nonpolar; (b) nonpolar; (c) polar; (d) nonpolar; (e) polar\n115. The empirical formula is $\\mathrm{XeF}_{6}$, and the balanced reactions are:\n\n$$\n\\begin{aligned}\n& \\mathrm{Xe}(g)+3 \\mathrm{~F}_{2}(g) \\xrightarrow{\\Delta} \\mathrm{XeF}_{6}(s) \\\\\n& \\mathrm{XeF}_{6}(s)+3 \\mathrm{H}_{2}(g) \\longrightarrow 6 \\mathrm{HF}(g)+\\mathrm{Xe}(g)\n\\end{aligned}\n$$"}
{"id": 5497, "contents": "2302. Chapter 19 - \n1. (a) $\\mathrm{Sc}:[\\mathrm{Ar}] 4 s^{2} 3 d^{1}$; (b) $\\mathrm{Ti}:[\\mathrm{Ar}] 4 s^{2} 3 d^{2}$; (c) $\\mathrm{Cr}:[\\mathrm{Ar}] 4 s^{1} 3 d^{5}$; (d) $\\mathrm{Fe}:[\\mathrm{Ar}] 4 s^{2} 3 d^{6}$; (e) $\\mathrm{Ru}:[\\mathrm{Kr}] 5 s^{2} 4 d^{6}$\n2. (a) La: $[\\mathrm{Xe}] 6 s^{2} 5 d^{1}, \\mathrm{La}^{3+}$ : $[\\mathrm{Xe}]$; (b) $\\mathrm{Sm}:[\\mathrm{Xe}] 6 s^{2} 4 f^{6}, \\mathrm{Sm}^{3+}$ : $[\\mathrm{Xe}] 4 f^{5}$; (c) $\\mathrm{Lu}:[\\mathrm{Xe}] 6 s^{2} 4 f^{14} 5 d^{1}, \\mathrm{Lu}^{3+}$ : $[\\mathrm{Xe}] 4 f^{14}$\n3. Al is used because it is the strongest reducing agent and the only option listed that can provide sufficient driving force to convert La (III) into La.\n4. Mo\n5. The $\\mathrm{CaSiO}_{3}$ slag is less dense than the molten iron, so it can easily be separated. Also, the floating slag\nlayer creates a barrier that prevents the molten iron from exposure to $\\mathrm{O}_{2}$, which would oxidize the Fe back to $\\mathrm{Fe}_{2} \\mathrm{O}_{3}$.\n6. $2.57 \\%$\n7. 0.167 V\n8. $E^{\\circ}=-0.6 \\mathrm{~V}, E^{\\circ}$ is negative so this reduction is not spontaneous. $E^{\\circ}=+1.1 \\mathrm{~V}$"}
{"id": 5498, "contents": "2302. Chapter 19 - \n8. $E^{\\circ}=-0.6 \\mathrm{~V}, E^{\\circ}$ is negative so this reduction is not spontaneous. $E^{\\circ}=+1.1 \\mathrm{~V}$\n9. (a) $\\mathrm{Fe}(s)+2 \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q) \\longrightarrow \\mathrm{Fe}^{2+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q)+\\mathrm{H}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$; (b)\n$\\mathrm{FeCl}_{3}(a q)+3 \\mathrm{Na}^{+}(a q)+3 \\mathrm{OH}^{-}(a q) \\longrightarrow \\mathrm{Fe}(\\mathrm{OH})_{3}(s)+3 \\mathrm{Na}^{+}(a q)+3 \\mathrm{Cl}^{+}(a q) ;$ (c)\n$\\mathrm{Mn}(\\mathrm{OH})_{2}(s)+2 \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+2 \\mathrm{Br}^{-}(a q) \\longrightarrow \\mathrm{Mn}^{2+}(a q)+2 \\mathrm{Br}^{-}(a q)+4 \\mathrm{H}_{2} \\mathrm{O}(l) ;$ (d)\n$4 \\mathrm{Cr}(s)+3 \\mathrm{O}_{2}(g) \\longrightarrow 2 \\mathrm{Cr}_{2} \\mathrm{O}_{3}(s) ;(\\mathrm{e}) \\mathrm{Mn}_{2} \\mathrm{O}_{3}(s)+6 \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+6 \\mathrm{Cl}^{-}(a q) \\longrightarrow 2 \\mathrm{MnCl}_{3}(s)+9 \\mathrm{H}_{2} \\mathrm{O}(l)$;\n(f) $\\mathrm{Ti}(s)+x s \\mathrm{~F}_{2}(g) \\longrightarrow \\mathrm{TiF}_{4}(g)$\n10. (a)"}
{"id": 5499, "contents": "2302. Chapter 19 - \n(f) $\\mathrm{Ti}(s)+x s \\mathrm{~F}_{2}(g) \\longrightarrow \\mathrm{TiF}_{4}(g)$\n10. (a)\n$\\mathrm{Cr}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}(a q)+2 \\mathrm{Zn}(s)+2 \\mathrm{H}_{3} \\mathrm{O}^{+}(a q) \\longrightarrow 2 \\mathrm{Zn}^{2+}(a q)+\\mathrm{H}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{Cr}^{2+}(a q)+3 \\mathrm{SO}_{4}{ }^{2-}(a q) ;$\n(b) $4 \\mathrm{TiCl}_{3}(s)+\\mathrm{CrO}_{4}{ }^{2-}(a q)+8 \\mathrm{H}^{+}(a q) \\longrightarrow 4 \\mathrm{Ti}^{4+}(a q)+\\mathrm{Cr}(s)+4 \\mathrm{H}_{2} \\mathrm{O}(l)+12 \\mathrm{Cl}^{-}(a q)$; (c) In acid solution between pH 2 and $\\mathrm{pH} 6, \\mathrm{CrO}_{4}{ }^{2-}$ forms $\\mathrm{HCrO}_{4}{ }^{-}$, which is in equilibrium with dichromate ion. The reaction is $2 \\mathrm{HCrO}_{4}{ }^{-}(a q) \\longrightarrow \\mathrm{Cr}_{2} \\mathrm{O}_{7}{ }^{2-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)$. At other acidic pHs, the reaction is $3 \\mathrm{Cr}^{2+}(a q)+\\mathrm{CrO}_{4}{ }^{2-}(a q)+8 \\mathrm{H}_{3} \\mathrm{O}^{+}(a q) \\longrightarrow 4 \\mathrm{Cr}^{3+}(a q)+12 \\mathrm{H}_{2} \\mathrm{O}(l)$; (d)"}
{"id": 5500, "contents": "2302. Chapter 19 - \n$8 \\mathrm{CrO}_{3}(s)+9 \\mathrm{Mn}(s) \\xrightarrow{\\Delta} 4 \\mathrm{Cr}_{2} \\mathrm{O}_{3}(s)+3 \\mathrm{Mn}_{3} \\mathrm{O}_{4}(s) ;(\\mathrm{e})$\n$\\mathrm{CrO}(s)+2 \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+2 \\mathrm{NO}_{3}^{-}(a q) \\longrightarrow \\mathrm{Cr}^{2+}(a q)+2 \\mathrm{NO}_{3}^{-}(a q)+3 \\mathrm{H}_{2} \\mathrm{O}(l) ;$ (f)\n$\\mathrm{CrCl}_{3}(s)+3 \\mathrm{NaOH}(a q) \\longrightarrow \\mathrm{Cr}(\\mathrm{OH})_{3}(s)+3 \\mathrm{Na}^{+}(a q)+3 \\mathrm{Cl}^{-}(a q)$\n11. (a) $3 \\mathrm{Fe}(s)+4 \\mathrm{H}_{2} \\mathrm{O}(g) \\longrightarrow \\mathrm{Fe}_{3} \\mathrm{O}_{4}(s)+4 \\mathrm{H}_{2}(g)$; (b)\n$3 \\mathrm{NaOH}(a q)+\\mathrm{Fe}\\left(\\mathrm{NO}_{3}\\right)_{3}(a q) \\xrightarrow{\\mathrm{H}_{2} \\mathrm{O}} \\mathrm{Fe}(\\mathrm{OH})_{3}(s)+3 \\mathrm{Na}^{+}(a q)+3 \\mathrm{NO}_{3}-(a q) ;$ (c)\n$\\mathrm{MnO}^{4-}+5 \\mathrm{Fe}^{2+}+8 \\mathrm{H}^{+} \\longrightarrow \\mathrm{Mn}^{2+}+5 \\mathrm{Fe}_{3}+4 \\mathrm{H}_{2} \\mathrm{O}$; (d)"}
{"id": 5501, "contents": "2302. Chapter 19 - \n$\\mathrm{Fe}(s)+2 \\mathrm{H}_{3} \\mathrm{O}^{+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q) \\longrightarrow \\mathrm{Fe}^{2+}(a q)+\\mathrm{SO}_{4}{ }^{2-}(a q)+\\mathrm{H}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)$; (e)\n$4 \\mathrm{Fe}^{2+}(a q)+\\mathrm{O}_{2}(g)+4 \\mathrm{HNO}_{3}(a q) \\longrightarrow 4 \\mathrm{Fe}^{3+}(a q)+2 \\mathrm{H}_{2} \\mathrm{O}(l)+4 \\mathrm{NO}_{3}{ }^{-}(a q) ;(\\mathrm{f})$\n$\\mathrm{FeCO}_{3}(s)+2 \\mathrm{HClO}_{4}(a q) \\longrightarrow \\mathrm{Fe}\\left(\\mathrm{ClO}_{4}\\right)_{2}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{CO}_{2}(g) ;(\\mathrm{g}) 3 \\mathrm{Fe}(s)+2 \\mathrm{O}_{2}(g) \\xrightarrow{\\Delta} \\mathrm{Fe}_{3} \\mathrm{O}_{4}(s)$\n12. $\\mathrm{As} \\mathrm{CN}^{-}$is added,\n$\\mathrm{Ag}^{+}(a q)+\\mathrm{CN}^{-}(a q) \\longrightarrow \\mathrm{AgCN}(s)$\nAs more $\\mathrm{CN}^{-}$is added,\n$\\mathrm{Ag}^{+}(a q)+2 \\mathrm{CN}^{-}(a q) \\longrightarrow\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]^{-}(a q)$\n$\\mathrm{AgCN}(s)+\\mathrm{CN}^{-}(a q) \\longrightarrow\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]^{-}(a q)$"}
{"id": 5502, "contents": "2302. Chapter 19 - \n$\\mathrm{AgCN}(s)+\\mathrm{CN}^{-}(a q) \\longrightarrow\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]^{-}(a q)$\n13. (a) $\\mathrm{Sc}^{3+}$; (b) $\\mathrm{Ti}^{4+}$; (c) $\\mathrm{V}^{5++}$; (d) $\\mathrm{Cr}^{6+}$; (e) $\\mathrm{Mn}^{4+}$; (f) $\\mathrm{Fe}^{2+}$ and $\\mathrm{Fe}^{3+}$; (g) $\\mathrm{Co}^{2+}$ and $\\mathrm{Co}^{3+}$; (h) $\\mathrm{Ni}^{2+}$; (i) $\\mathrm{Cu}^{+}$\n14. (a) $4,\\left[\\mathrm{Zn}(\\mathrm{OH})_{4}\\right]^{2-}$; (b) $6,\\left[\\mathrm{Pd}(\\mathrm{CN})_{6}\\right]^{2-}$; (c) $2,\\left[\\mathrm{AuCl}_{2}\\right]^{-}$; (d) $4,\\left[\\mathrm{Pt}\\left(\\mathrm{NH}_{3}\\right)_{2} \\mathrm{Cl}_{2}\\right]$; (e) $6, \\mathrm{~K}\\left[\\mathrm{Cr}\\left(\\mathrm{NH}_{3}\\right)_{2} \\mathrm{Cl}_{4}\\right]$; (f) 6 , $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]\\left[\\mathrm{Cr}(\\mathrm{CN})_{6}\\right] ;(\\mathrm{g}) 6,\\left[\\mathrm{Co}(\\mathrm{en})_{2} \\mathrm{Br}_{2}\\right] \\mathrm{NO}_{3}$\n15. (a) $\\left[\\mathrm{Pt}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{2} \\mathrm{Br}_{2}\\right]$ :"}
{"id": 5503, "contents": "2302. Chapter 19 - \ncis\n\ntrans\n(b) $\\left[\\mathrm{Pt}\\left(\\mathrm{NH}_{3}\\right)(\\mathrm{py})(\\mathrm{Cl})(\\mathrm{Br})\\right]$ :\n\n\n\n(c) $\\left[\\mathrm{Zn}\\left(\\mathrm{NH}_{3}\\right)_{3} \\mathrm{Cl}\\right]^{+}$:\n\n(d) $\\left[\\mathrm{Pt}\\left(\\mathrm{NH}_{3}\\right)_{3} \\mathrm{Cl}\\right]^{+}$:\n\n(e) $\\left[\\mathrm{Ni}^{\\left.\\left(\\mathrm{H}_{2} \\mathrm{O}\\right){ }_{4} \\mathrm{Cl}_{2}\\right] \\text { : }}\\right.$\n\n(f) $\\left[\\mathrm{Co}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}\\right)_{2} \\mathrm{Cl}_{2}\\right]^{3-}$ :\n\n\n16. (a) tricarbonatocobaltate(III) ion; (b) tetraaminecopper(II) ion; (c) tetraaminedibromocobalt(III) sulfate; (d) tetraamineplatinum(II) tetrachloroplatinate(II); (e) tris-(ethylenediamine)chromium(III) nitrate; (f) diaminedibromopalladium(II); (g) potassium pentachlorocuprate(II); (h) diaminedichlorozinc(II)\n17. (a) none; (b) none; (c) The two Cl ligands can be cis or trans. When they are cis, there will also be an optical isomer.\n18.\n\n\n\n\n\n\n\n37."}
{"id": 5504, "contents": "2302. Chapter 19 - \n37.\n\n39. $\\left[\\mathrm{Co}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right] \\mathrm{Cl}_{2}$ with three unpaired electrons.\n41. (a) 4 ; (b) 2 ; (c) 1 ; (d) 5 ; (e) 0\n43. (a) $\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{4-}$; (b) $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]^{3+}$; (c) $\\left[\\mathrm{Mn}(\\mathrm{CN})_{6}\\right]^{4-}$\n45. The complex does not have any unpaired electrons. The complex does not have any geometric isomers, but the mirror image is nonsuperimposable, so it has an optical isomer.\n47. No. $\\mathrm{Au}^{+}$has a complete $5 d$ sublevel."}
{"id": 5505, "contents": "2303. Chapter 20 - \n1. (a) sodium-24; (b) aluminum-29; (c) krypton-73; (d) iridium-194\n2. (a) ${ }_{14}^{34} \\mathrm{Si}$; (b) ${ }_{15}^{36} \\mathrm{P}$; (c) ${ }_{25}^{57} \\mathrm{Mn}$; (d) ${ }_{56}^{121} \\mathrm{Ba}$\n3. (a) ${ }_{25}^{45} \\mathrm{Mn}^{+1}$; (b) ${ }_{45}^{69} \\mathrm{Rh}^{+2}$; (c) ${ }_{53}^{142} \\mathrm{I}^{-1}$; (d) ${ }_{97}^{243} \\mathrm{Bk}$\n4. Nuclear reactions usually change one type of nucleus into another; chemical changes rearrange atoms. Nuclear reactions involve much larger energies than chemical reactions and have measureable mass changes.\n5. (a), (b), (c), (d), and (e)\n6. (a) A nucleon is any particle contained in the nucleus of the atom, so it can refer to protons and neutrons.\n(b) An $\\alpha$ particle is one product of natural radioactivity and is the nucleus of a helium atom. (c) A $\\beta$ particle is a product of natural radioactivity and is a high-speed electron. (d) A positron is a particle with the same mass as an electron but with a positive charge. (e) Gamma rays compose electromagnetic radiation of high energy and short wavelength. (f) Nuclide is a term used when referring to a single type of nucleus. (g) The mass number is the sum of the number of protons and the number of neutrons in an element. (h) The atomic number is the number of protons in the nucleus of an element."}
{"id": 5506, "contents": "2303. Chapter 20 - \n7. (a) ${ }_{13}^{27} \\mathrm{Al}+{ }_{2}^{4} \\mathrm{He} \\longrightarrow{ }_{15}^{30} \\mathrm{P}+{ }_{0}^{1} \\mathrm{n}$; (b) ${ }_{94}^{239} \\mathrm{Pu}+{ }_{2}^{4} \\mathrm{He} \\longrightarrow{ }_{96}^{242} \\mathrm{Cm}+{ }_{0}^{1} \\mathrm{n}$; (c) ${ }_{7}^{14} \\mathrm{~N}+{ }_{2}^{4} \\mathrm{He} \\longrightarrow{ }_{8}^{17} \\mathrm{O}+{ }_{1}^{1} \\mathrm{H}$; (d) ${ }_{92}^{235} \\mathrm{U} \\longrightarrow{ }_{37}^{96} \\mathrm{Rb}+{ }_{55}^{135} \\mathrm{Cs}+4{ }_{0}^{1} \\mathrm{n}$\n8. (a) ${ }_{7}^{14} \\mathrm{~N}+{ }_{2}^{4} \\mathrm{He} \\longrightarrow{ }_{8}^{17} \\mathrm{O}+{ }_{1}^{1} \\mathrm{H}$; (b) ${ }_{7}^{14} \\mathrm{C}+{ }_{0}^{1} \\mathrm{n} \\longrightarrow{ }_{6}^{14} \\mathrm{C}+{ }_{1}^{1} \\mathrm{H}$; (c) ${ }_{90}^{232} \\mathrm{Th}+{ }_{0}^{1} \\mathrm{n} \\longrightarrow{ }_{90}^{233} \\mathrm{Th}$; (d) ${ }_{92}^{238} \\mathrm{U}+{ }_{1}^{2} \\mathrm{H} \\longrightarrow{ }_{92}^{239} \\mathrm{U}+{ }_{1}^{1} \\mathrm{H}$"}
{"id": 5507, "contents": "2303. Chapter 20 - \n9. (a) 148.8 MeV per atom; (b) $7.808 \\mathrm{MeV} /$ nucleon\n10. $\\alpha$ (helium nuclei), $\\beta$ (electrons), $\\beta^{+}$(positrons), and $\\eta$ (neutrons) may be emitted from a radioactive element, all of which are particles; $\\gamma$ rays also may be emitted.\n11. (a) conversion of a neutron to a proton: ${ }_{0}^{1} \\mathrm{n} \\longrightarrow{ }_{1}^{1} \\mathrm{p}+{ }_{+1}^{0} \\mathrm{e}$; (b) conversion of a proton to a neutron; the positron has the same mass as an electron and the same magnitude of positive charge as the electron has negative charge; when the n :p ratio of a nucleus is too low, a proton is converted into a neutron with the emission of a positron: ${ }_{1}^{1} \\mathrm{p} \\longrightarrow{ }_{0}^{1} \\mathrm{n}+{ }_{+1}^{0} \\mathrm{e}$; (c) In a proton-rich nucleus, an inner atomic electron can be absorbed. In simplest form, this changes a proton into a neutron: ${ }_{1}^{1} \\mathrm{p}+{ }_{-1}^{0} \\mathrm{e} \\longrightarrow{ }_{0}^{1} \\mathrm{p}$\n12. The electron pulled into the nucleus was most likely found in the $1 s$ orbital. As an electron falls from a higher energy level to replace it, the difference in the energy of the replacement electron in its two energy levels is given off as an X-ray."}
{"id": 5508, "contents": "2303. Chapter 20 - \n13. Manganese-51 is most likely to decay by positron emission. The n:p ratio for $\\mathrm{Cr}-53$ is $\\frac{29}{24}=1.21$; for $\\mathrm{Mn}-51$, it is $\\frac{26}{25}=1.04$; for $\\mathrm{Fe}-59$, it is $\\frac{33}{26}=1.27$. Positron decay occurs when the n :p ratio is low. Mn-51 has the lowest n:p ratio and therefore is most likely to decay by positron emission. Besides, ${ }_{24}^{53} \\mathrm{Cr}$ is a stable\nisotope, and ${ }_{26}^{59} \\mathrm{Fe}$ decays by beta emission.\n14. (a) $\\beta$ decay; (b) $\\alpha$ decay; (c) positron emission; (d) $\\beta$ decay; (e) $\\alpha$ decay\n15. ${ }_{92}^{238} \\mathrm{U} \\longrightarrow{ }_{90}^{234} \\mathrm{Th}+{ }_{2}^{4} \\mathrm{He} ;{ }_{90}^{234} \\mathrm{Th} \\longrightarrow{ }_{91}^{234} \\mathrm{~Pa}+{ }_{-1}^{0} \\mathrm{e} ;{ }_{91}^{234} \\mathrm{~Pa} \\longrightarrow{ }_{92}^{234} \\mathrm{U}+{ }_{-1}^{0} \\mathrm{e} ;{ }_{92}^{234} \\mathrm{U} \\longrightarrow{ }_{90}^{230} \\mathrm{Th}+{ }_{2}^{4} \\mathrm{He}$"}
{"id": 5509, "contents": "2303. Chapter 20 - \n${ }_{90}^{230} \\mathrm{Th} \\longrightarrow{ }_{88}^{226} \\mathrm{Ra}+{ }_{2}^{4} \\mathrm{He}{ }_{88}^{226} \\mathrm{Ra} \\longrightarrow{ }_{86}^{22} \\mathrm{Rn}+{ }_{2}^{4} \\mathrm{He} ;{ }_{86}^{22} \\mathrm{Rn} \\longrightarrow{ }_{84}^{218} \\mathrm{Po}+{ }_{2}^{4} \\mathrm{He}$\n16. Half-life is the time required for half the atoms in a sample to decay. Example (answers may vary): For C-14, the half-life is 5770 years. A 10-g sample of C-14 would contain 5 g of C-14 after 5770 years; a $0.20-\\mathrm{g}$ sample of $\\mathrm{C}-14$ would contain 0.10 g after 5770 years.\n17. $\\left(\\frac{1}{2}\\right)^{0.04}=0.973$ or $97.3 \\%$\n18. $2 \\times 10^{3} \\mathrm{y}$\n19. $0.12 \\mathrm{~h}^{-1}$\n20. (a) 3.8 billion years; (b) The rock would be younger than the age calculated in part (a). If Sr was originally in the rock, the amount produced by radioactive decay would equal the present amount minus the initial amount. As this amount would be smaller than the amount used to calculate the age of the rock and the age is proportional to the amount of Sr , the rock would be younger.\n21. $c=0$; This shows that no Pu-239 could remain since the formation of the earth. Consequently, the plutonium now present could not have been formed with the uranium.\n22. 17.5 MeV"}
{"id": 5510, "contents": "2303. Chapter 20 - \n21. $c=0$; This shows that no Pu-239 could remain since the formation of the earth. Consequently, the plutonium now present could not have been formed with the uranium.\n22. 17.5 MeV\n23. (a) ${ }_{83}^{212} \\mathrm{Bi} \\longrightarrow{ }_{84}^{212} \\mathrm{Po}+{ }_{-1}^{0} \\mathrm{e}$; (b) ${ }_{5}^{8} \\mathrm{~B} \\longrightarrow{ }_{4}^{8} \\mathrm{Be}+{ }_{-1}^{0} \\mathrm{e}$; (c) ${ }_{92}^{238} \\mathrm{U}+{ }_{0}^{1} \\mathrm{n} \\longrightarrow{ }_{93}^{239} \\mathrm{~Np}+{ }_{-1}^{0} \\mathrm{~Np}$, ${ }_{93}^{239} \\mathrm{~Np} \\longrightarrow{ }_{94}^{239} \\mathrm{Pu}+{ }_{-1}^{0} \\mathrm{e}$; (d) ${ }_{38}^{90} \\mathrm{Sr} \\longrightarrow{ }_{39}^{90} \\mathrm{Y}+{ }_{-1}^{0} \\mathrm{e}$"}
{"id": 5511, "contents": "2303. Chapter 20 - \n24. (a) ${ }_{95}^{241} \\mathrm{Am}+{ }_{2}^{4} \\mathrm{He} \\longrightarrow{ }_{97}^{244} \\mathrm{Bk}+{ }_{0}^{1} \\mathrm{n}$; (b) ${ }_{94}^{239} \\mathrm{Pu}+15{ }_{0}^{1} \\mathrm{n} \\longrightarrow{ }_{100}^{254} \\mathrm{Fm}+6{ }_{-1}^{0} \\mathrm{e}$; (c) ${ }_{98}^{250} \\mathrm{Cf}+{ }_{5}^{11} \\mathrm{~B} \\longrightarrow{ }_{103}^{257} \\mathrm{Lr}+4{ }_{0}^{1} \\mathrm{n}$; (d) ${ }_{98}^{249} \\mathrm{Cf}+{ }_{7}^{15} \\mathrm{~N} \\longrightarrow{ }_{105}^{260} \\mathrm{Db}+4{ }_{0}^{1} \\mathrm{n}$\n25. Two nuclei must collide for fusion to occur. High temperatures are required to give the nuclei enough kinetic energy to overcome the very strong repulsion resulting from their positive charges.\n26. A nuclear reactor consists of the following:\n27. A nuclear fuel. A fissionable isotope must be present in large enough quantities to sustain a controlled chain reaction. The radioactive isotope is contained in tubes called fuel rods.\n28. A moderator. A moderator slows neutrons produced by nuclear reactions so that they can be absorbed by the fuel and cause additional nuclear reactions.\n29. A coolant. The coolant carries heat from the fission reaction to an external boiler and turbine where it is transformed into electricity.\n30. A control system. The control system consists of control rods placed between fuel rods to absorb neutrons and is used to adjust the number of neutrons and keep the rate of the chain reaction at a safe level.\n31. A shield and containment system. The function of this component is to protect workers from radiation produced by the nuclear reactions and to withstand the high pressures resulting from hightemperature reactions."}
{"id": 5512, "contents": "2303. Chapter 20 - \n31. A shield and containment system. The function of this component is to protect workers from radiation produced by the nuclear reactions and to withstand the high pressures resulting from hightemperature reactions.\n32. The fission of uranium generates heat, which is carried to an external steam generator (boiler). The resulting steam turns a turbine that powers an electrical generator.\n33. Introduction of either radioactive $\\mathrm{Ag}^{+}$or radioactive $\\mathrm{Cl}^{-}$into the solution containing the stated reaction, with subsequent time given for equilibration, will produce a radioactive precipitate that was originally devoid of radiation.\n34. (a) ${ }_{53}^{133} \\mathrm{I} \\longrightarrow{ }_{54}^{133} \\mathrm{Xe}+{ }_{-1}^{0} \\mathrm{e}$; (b) 37.6 days\n35. Alpha particles can be stopped by very thin shielding but have much stronger ionizing potential than beta particles, X -rays, and $\\gamma$-rays. When inhaled, there is no protective skin covering the cells of the lungs, making it possible to damage the DNA in those cells and cause cancer.\n36. (a) $7.64 \\times 10^{9} \\mathrm{~Bq}$; (b) $2.06 \\times 10^{-2} \\mathrm{Ci}$"}
{"id": 5513, "contents": "2304. Chapter 21 - \n1. There are several sets of answers; one is:\n(a) $\\mathrm{C}_{5} \\mathrm{H}_{12}$\n\n(b) $\\mathrm{C}_{5} \\mathrm{H}_{10}$\n\n(c) $\\mathrm{C}_{5} \\mathrm{H}_{8}$\n\n2. Both reactions result in bromine being incorporated into the structure of the product. The difference is the way in which that incorporation takes place. In the saturated hydrocarbon, an existing $\\mathrm{C}-\\mathrm{H}$ bond is broken, and a bond between the C and the Br can then be formed. In the unsaturated hydrocarbon, the only bond broken in the hydrocarbon is the $\\pi$ bond whose electrons can be used to form a bond to one of the bromine atoms in $\\mathrm{Br}_{2}$ (the electrons from the $\\mathrm{Br}-\\mathrm{Br}$ bond form the other $\\mathrm{C}-\\mathrm{Br}$ bond on the other carbon that was part of the $\\pi$ bond in the starting unsaturated hydrocarbon).\n3. Unbranched alkanes have free rotation about the $\\mathrm{C}-\\mathrm{C}$ bonds, yielding all orientations of the substituents about these bonds equivalent, interchangeable by rotation. In the unbranched alkenes, the inability to rotate about the $\\mathrm{C}=\\mathrm{C}$ bond results in fixed (unchanging) substituent orientations, thus permitting different isomers. Since these concepts pertain to phenomena at the molecular level, this explanation involves the microscopic domain.\n4. They are the same compound because each is a saturated hydrocarbon containing an unbranched chain of six carbon atoms.\n5. (a) $\\mathrm{C}_{6} \\mathrm{H}_{14}$\n\n(b) $\\mathrm{C}_{6} \\mathrm{H}_{14}$\n\n(c) $\\mathrm{C}_{6} \\mathrm{H}_{12}$\n\n(d) $\\mathrm{C}_{6} \\mathrm{H}_{12}$\n\n(e) $\\mathrm{C}_{6} \\mathrm{H}_{10}$\n\n(f) $\\mathrm{C}_{6} \\mathrm{H}_{10}$"}
{"id": 5514, "contents": "2304. Chapter 21 - \n(e) $\\mathrm{C}_{6} \\mathrm{H}_{10}$\n\n(f) $\\mathrm{C}_{6} \\mathrm{H}_{10}$\n\n6. (a) 2,2-dibromobutane; (b) 2-chloro-2-methylpropane; (c) 2-methylbutane; (d) 1-butyne; (e) 4-fluoro-4-methyl-1-octyne; (f) trans-1-chloropropene; (g) 4-methyl-1-pentene\n7.\n\n\nn-butane\n\n\n2-methylpropane\n15.\n\ncis-\n\ntrans-\n17. (a) 2,2,4-trimethylpentane; (b) 2,2,3-trimethylpentane, 2,3,4-trimethylpentane, and 2,3,3-trimethylpentane:\n\n\n\n19.\n\n\n1-chlorobutane\n\n\n2-chlorobutane\n\n\n2-chloro-2-methylpropane\n\n\n1-chloro-2-methylpropane (1-chloro-2-methylpropane)\n21. In the following, the carbon backbone and the appropriate number of hydrogen atoms are shown in condensed form:\n\n\n\n\n\n\n\n\n23.\n\n\n\nIn acetylene, the bonding uses $s p$ hybrids on carbon atoms and $s$ orbitals on hydrogen atoms. In benzene, the carbon atoms are $s p^{2}$ hybridized.\n25. (a) $\\mathrm{CH} \\equiv \\mathrm{CCH}_{2} \\mathrm{CH}_{3}+2 \\mathrm{I}_{2} \\longrightarrow \\mathrm{CHI}_{2} \\mathrm{CI}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$\n\n(b) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}+8 \\mathrm{O}_{2} \\longrightarrow 5 \\mathrm{CO}_{2}+6 \\mathrm{H}_{2} \\mathrm{O}$"}
{"id": 5515, "contents": "2304. Chapter 21 - \n27. 65.2 g\n29. $9.328 \\times 10^{2} \\mathrm{~kg}$\n31. (a) ethyl alcohol, ethanol: $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$; (b) methyl alcohol, methanol: $\\mathrm{CH}_{3} \\mathrm{OH}$; (c) ethylene glycol, ethanediol: $\\mathrm{HOCH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}$; (d) isopropyl alcohol, 2-propanol: $\\mathrm{CH}_{3} \\mathrm{CH}(\\mathrm{OH}) \\mathrm{CH}_{3}$; (e) glycerine, 1,2,3-trihydroxypropane: $\\mathrm{HOCH}_{2} \\mathrm{CH}(\\mathrm{OH}) \\mathrm{CH}_{2} \\mathrm{OH}$\n33. (a) 1-ethoxybutane, butyl ethyl ether; (b) 1-ethoxypropane, ethyl propyl ether; (c) 1-methoxypropane, methyl propyl ether\n35. $\\mathrm{HOCH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}$, two alcohol groups; $\\mathrm{CH}_{3} \\mathrm{OCH}_{2} \\mathrm{OH}$, ether and alcohol groups\n37. (a)\n\n(b) $4.593 \\times 10^{2} \\mathrm{~L}$\n39. (a) $\\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{CHCH}_{3}+\\mathrm{H}_{2} \\mathrm{O} \\longrightarrow \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}(\\mathrm{OH}) \\mathrm{CH}_{3}$\n\n(b) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH} \\longrightarrow \\mathrm{CH}_{2}=\\mathrm{CH}_{2}+\\mathrm{H}_{2} \\mathrm{O}$\n\n41. (a)\n\n(b)\n\n(c)"}
{"id": 5516, "contents": "2304. Chapter 21 - \n41. (a)\n\n(b)\n\n(c)\n\n43. A ketone contains a group bonded to two additional carbon atoms; thus, a minimum of three carbon atoms are needed.\n45. Since they are both carboxylic acids, they each contain the -COOH functional group and its characteristics. The difference is the hydrocarbon chain in a saturated fatty acid contains no double or triple bonds, whereas the hydrocarbon chain in an unsaturated fatty acid contains one or more multiple bonds.\n47. (a) $\\mathrm{CH}_{3} \\mathrm{CH}(\\mathrm{OH}) \\mathrm{CH}_{3}$ : all carbons are tetrahedral; (b) $\\mathrm{CH}_{3} \\mathrm{COCH}_{3}$ : the end carbons are tetrahedral and the central carbon is trigonal planar; (c) $\\mathrm{CH}_{3} \\mathrm{OCH}_{3}$ : all are tetrahedral; (d) $\\mathrm{CH}_{3} \\mathrm{COOH}$ : the methyl carbon is tetrahedral and the acid carbon is trigonal planar; (e) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}\\left(\\mathrm{CH}_{3}\\right) \\mathrm{CHCH}_{2}$ : all are tetrahedral except the right-most two carbons, which are trigonal planar\n49.\n\n51. (a) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}+\\mathrm{CH}_{3} \\mathrm{C}(\\mathrm{O}) \\mathrm{OH} \\longrightarrow \\mathrm{CH}_{3} \\mathrm{C}(\\mathrm{O}) \\mathrm{OCH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{3}+\\mathrm{H}_{2} \\mathrm{O}$ :"}
{"id": 5517, "contents": "2304. Chapter 21 - \n(b) $2 \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COOH}+\\mathrm{CaCO}_{3} \\longrightarrow\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COO}\\right)_{2} \\mathrm{Ca}+\\mathrm{CO}_{2}+\\mathrm{H}_{2} \\mathrm{O}$ :\n\n53.\n\n\nCompound A\n\n\nCompound $B$\n\n\nCompound C\n55. Trimethyl amine: trigonal pyramidal, $s p^{3}$; trimethyl ammonium ion: tetrahedral, $s p^{3}$\n57.\n\n\npyridine,\ntrigonal planar, $s p^{2}$\n(nonlinear or bent)\npyridinium ion, trigonal planar, $s p^{2}$\n(trigonal planar)\n59. $\\mathrm{CH}_{3} \\mathrm{NH}_{2}+\\mathrm{H}_{3} \\mathrm{O}^{+} \\longrightarrow \\mathrm{CH}_{3} \\mathrm{NH}_{3}{ }^{+}+\\mathrm{H}_{2} \\mathrm{O}$\n\n61. $\\mathrm{CH}_{3} \\underline{\\mathrm{C}} \\mathrm{H}=\\mathrm{CHCH}_{3}\\left(s p^{2}\\right)+\\mathrm{Cl} \\longrightarrow \\mathrm{CH}_{3} \\underline{\\mathrm{CH}}(\\mathrm{Cl}) \\mathrm{H}(\\mathrm{Cl}) \\mathrm{CH}_{3}\\left(s p^{3}\\right) ; 2 \\underline{\\mathrm{C}}_{6} \\mathrm{H}_{6}\\left(s p^{2}\\right)+15 \\mathrm{O}_{2} \\longrightarrow 12 \\underline{\\mathrm{CO}}_{2}(s p)+6 \\mathrm{H}_{2} \\mathrm{O}$\n63. The carbon in $\\mathrm{CO}_{3}{ }^{2-}$, initially at $s p^{2}$, changes hybridization to $s p$ in $\\mathrm{CO}_{2}$."}
{"id": 5518, "contents": "2304. Chapter 21 - \nErnest Rutherford, \"The Development of the Theory of Atomic Structure,\" ed. J. A. Ratcliffe, in Background to Modern Science, eds. Joseph Needham and Walter Pagel, (Cambridge, UK: Cambridge University Press, 1938), 61-74. Accessed September 22, 2014, https://ia600508.us.archive.org/3/items/backgroundtomode032734mbp/backgroundtomode032734mbp.pdf.\n\nOmiatek, Donna M., Amanda J. Bressler, Ann-Sofie Cans, Anne M. Andrews, Michael L. Heien, and Andrew G. Ewing. \"The Real Catecholamine Content of Secretory Vesicles in the CNS Revealed by Electrochemical Cytometry.\" Scientific Report 3 (2013): 1447, accessed January 14, 2015, doi:10.1038/srep01447.\n\nPer the IUPAC definition, group 12 elements are not transition metals, though they are often referred to as such. Additional details on this group's elements are provided in a chapter on transition metals and coordination chemistry.\n\nNote that orbitals may sometimes be drawn in an elongated \"balloon\" shape rather than in a more realistic \"plump\" shape in order to make the geometry easier to visualize.\n\nThe IUPAC definition of standard pressure was changed from 1 atm to $1 \\mathrm{bar}(100 \\mathrm{kPa})$ in 1982, but the prior definition remains in use by many literature resources and will be used in this text.\n\n\"Quotations by Joseph-Louis Lagrange,\" last modified February 2006, accessed February 10, 2015, http://www-history.mcs.standrews.ac.uk/Quotations/Lagrange.html\n\nUS Energy Information Administration, Primary Energy Consumption by Source and Sector, 2012, http://www.eia.gov/totalenergy/ data/monthly/pdf/flow/css_2012_energy.pdf. Data derived from US Energy Information Administration, Monthly Energy Review (January 2014)."}
{"id": 5519, "contents": "2304. Chapter 21 - \nFrancis D. Reardon et al. \"The Snellen human calorimeter revisited, re-engineered and upgraded: Design and performance characteristics.\" Medical and Biological Engineering and Computing 8 (2006)721-28, http://link.springer.com/article/10.1007/ s11517-006-0086-5.\n\nThis question is taken from the Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service.\n\nTitration of 25.00 mL of $0.100 \\mathrm{M} \\mathrm{HCl}(0.00250 \\mathrm{~mol}$ of HCI$)$ with 0.100 M NaOH .\n2 Titration of 25.00 mL of $0.100 \\mathrm{MCH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\left(0.00250 \\mathrm{~mol}\\right.$ of $\\left.\\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}\\right)$ with 0.100 M NaOH .\n\nHerrlich, P. \"The Responsibility of the Scientist: What Can History Teach Us About How Scientists Should Handle Research That Has the Potential to Create Harm?\" EMBO Reports 14 (2013): 759-764.\n\n\"The Nobel Prize in Chemistry 1995,\" Nobel Prize.org, accessed February 18, 2015, http://www.nobelprize.org/nobel_prizes/ chemistry/laureates/1995/.\n\nThe \" m \" in Tc-99m stands for \"metastable,\" indicating that this is an unstable, high-energy state of Tc-99. Metastable isotopes emit $\\gamma$ radiation to rid themselves of excess energy and become (more) stable.\n\nThis is the Beilstein database, now available through the Reaxys site (www.elsevier.com/online-tools/reaxys).\n2 Peplow, Mark. \"Organic Synthesis: The Robo-Chemist,\" Nature 512 (2014): 20-2.\n\nPhysical properties for $\\mathrm{C}_{4} \\mathrm{H}_{10}$ and heavier molecules are those of the normal isomer, $n$-butane, $n$-pentane, etc."}
{"id": 5520, "contents": "2304. Chapter 21 - \nPhysical properties for $\\mathrm{C}_{4} \\mathrm{H}_{10}$ and heavier molecules are those of the normal isomer, $n$-butane, $n$-pentane, etc.\n\nSTP indicates a temperature of $0^{\\circ} \\mathrm{C}$ and a pressure of 1 atm .\n\nThe IUPAC adopted new nomenclature guidelines in 2013 that require this number to be placed as an \"infix\" rather than a prefix. For example, the new name for 2-propanol would be propan-2-ol. Widespread adoption of this new nomenclature will take some time, and students are encouraged to be familiar with both the old and new naming protocols.\n\nAcids and bases are commercially available as aqueous solutions. This table lists properties (densities and concentrations) of common acid and base solutions. Nominal values are provided in cases where the manufacturer cites a range of concentrations and densities.\n2 This column contains specific gravity data. In the case of this table, specific gravity is the ratio of density of a substance to the density of pure water at the same conditions. Specific gravity is often cited on commercial labels.\n3 This solution is sometimes called \"ammonium hydroxide,\" although this term is not chemically accurate.\n\n\\mathrm{y}=$ years, $\\mathrm{d}=$ days, $\\mathrm{h}=$ hours, $\\mathrm{m}=$ minutes, $\\mathrm{s}=$ seconds"}