北京化工大学：《有机化学》课程教学资源（习题与答案）Chapter 02 Structure and Reactivity：Acids and Bases, Polar and Nonpolar Molecules.pdf_大学文库

18 2 Structure and Reactivity: Acids and Bases, Polar and Nonpolar Molecules Organic chemistry is largely a study of chemical reactions involving organic molecules. The textbook chapter therefore begins with a review of the principles of kinetics and thermodynamics, which apply to all reactions. Then we proceed to discuss acids and bases, which serves two purposes: It provides us with good examples of thermodynamics as applied to reactions, and it illustrates a process that is actually closely related to most of the reactions of polar organic molecules. After an introduction to functional groups and the classes of organic compounds, the chapter turns to a consideration of the simplest of those classes, the nonpolar alkanes. These sections cover (1) how to name organic molecules (nomenclature), (2) the relation of the physical properties of molecules to their molecular structure, (3) flexibility and shape of molecules (conformation), and (4) applications of kinetics and thermodynamics to changes in molecular shape. In Chapter 3 we cover chemical reactions of alkanes. Outline of the Chapter 2-1 Kinetics and Thermodynamics The energetic factors that govern the transformations of molecules. 2-2 Acids and Bases; Electrophiles and Nucleophiles Reviewing chemistry that you’ve seen before, with an eye to chemistry you will see a lot of in the chapters to come. Also, a first look at the notation used to describe how organic reactions take place. 2-3 Functional Groups The “business ends” of molecules: where reactions are likely to occur. 2-4 Straight-Chain and Branched Alkanes Alkanes of various structures. Isomers. 2-5 Nomenclature The first of a group of rules used to unambiguously name any organic compound. 2-6 Physical Properties A topic that is usually not emphasized very much but that does reveal several useful, generalizable points about molecules. 2-7 and 2-8 Conformations A discussion of the spatial arrangements that are possible for atoms in alkanes and the energy changes associated with their interconversions. 1559T_ch02_18-37 10/22/05 1:11 Page 18

Keys to the Chapter • 19 Keys to the Chapter 2-1. Kinetics and Thermodynamics This section introduces ideas associated with energy changes in organic chemistry. Even though some of the terminology may be somewhat familiar to you from freshman chemistry, a few comments may be useful for orientation purposes. In this course you are going to encounter a lot of discussion concerning the energy content of molecules or other species. This term will refer in general to what is called potential energy in physics: energy that is stored in some way and can potentially be released in some process later on. Discussions involving energy will often refer to the stability or instability of various substances or systems. Energy and stability are related in the following way: A species with high-energy content will tend to want to get rid of some of its energy somehow. So, relatively speaking, high-energy species are generally unstable. Heat and energy are also related, so high-energy species will have a tendency to undergo processes that give off lots of heat. However, that a substance is capable of doing such a thing doesn’t necessarily mean that it will do it quickly. The point here is that the rate of a process is the subject of kinetics, whereas the energetic favorability is one of thermodynamics, and the two are very different. Energetically favorable processes can take place at fast rates, slow rates, or in some cases, hardly seem to take place at all. A wooden match in the presence of air is a good chemical example of the latter. The reactions with oxygen of the compounds in the wood as well as on the head of the match are all extremely energetically favorable (thermodynamics), but nothing perceptible happens at room temperature. Why not? The rate of the reaction is too low: The number of molecules actually reacting with the oxygen at room temperature is so small that nothing seems to be happening at all (kinetics). However, when we strike the match—heat the match head with friction—it starts to burn and continues until the whole thing has burned up. The reaction of most organic molecules with oxygen requires energy input to get started even though it ultimately results in net energy output after the reaction has finished. The reason is as follows: In most reactions, old bonds are broken and new ones are formed, but not exactly simultaneously. Some partial breaking of old bonds has to take place before anything else, and that requires an input of energy. Once this process has started, it can lead to the formation of new bonds, and the release of energy—enough to make more old bonds break plus extra in the form of the flame and heat of burning. This initial energy input is the activation energy of the reaction, and it is a key factor governing kinetics: rates of reactions. This section provides a brief mathematical description of each of the main concepts involved in thermodynamics and kinetics as applied to organic chemistry. The equations are generally fairly straightforward in their application. The problems will give you several chances to use them. 2-2. Acids and Bases; Electrophiles and Nucleophiles The beginning of this text section covers the material you are most likely to have encountered in freshman chemistry: the mechanics of acid-base chemistry. The guiding principle is that such processes are reversible and are governed by thermodynamics. Notions of strong and weak acids and bases are based upon the positions of equilibria of the general sort stronger acid stronger base weaker acid weaker base G° 0, where the thermodynamic driving force favors conversion of the stronger acid and stronger base into the weaker ones. While this concept may be familiar to you, you may not be quite as used to the relative nature of the terms strong and weak. In other words, a compound that acts as the strong acid in one such equation may be the weak partner in another, or may even play the role of a base. After all, the range of known acid strengths covers sixty orders of magnitude, and in organic chemistry we will encounter examples from every part of the scale. Water is the most familiar of substances to show such varied behavior. Water acts as a weak acid (the way we normally think of it): 1559T_ch02_18-37 10/22/05 1:11 Page 19

HCl NaOH NaCl H2O stronger stronger weaker weaker acid base base acid But water may act as the strong acid (and conversely, hydroxide as the weak base!): H2O NaNH2 NaOH NH3 stronger stronger weaker weaker acid base base acid And finally, water acts as a base when it encounters a strong enough acid: HCl H2O Cl H3O stronger stronger weaker weaker acid base base acid In describing acid-base reactions, we define a very simple relationship: the one between an acid and its conjugate base (or, conversely, a base and its conjugate acid). Through this relationship it is possible to estimate the strength of acids and bases that we’ve never seen before by making structural comparisons with species with which we are more familiar. We use the notion that, relatively speaking, strong acids have weak conjugate bases, and vice versa. Through this relationship, we may use an analysis of either component of a conjugate acid-base pair to find the strengths of both, relative to other acids and bases. The most common application is to determine the strength of an acid by evaluating effects that stabilize (and make weaker) its conjugate base: increased size and increased electronegativity of the negatively charged atom, and any effects that disperse negative charge away from the negatively charged atom, such as resonance.* By comparing the degree to which these properties are present in each of a pair of conjugate bases, you can usually tell which of the corresponding conjugate acids is stronger or weaker. This text section also reviews the definitions of Lewis acids and bases and compares them with their analogs in organic chemistry: electrophiles and nucleophiles. The latter are the two terms that we use to describe electron-poor and electron-rich atoms in molecules, respectively. Such atoms possess partial or full electrical charges and, as a result, are places where chemical reactivity is usually high. Many of the functional groups are characterized by the presence of electrophilic or nucleophilic carbon atoms, for example. The analogy between a simple inorganic acid-base reaction and an organic nucleophilic substitution is illustrative of these principles. It also utilizes the “curved arrow” notation that we first presented in Chapter 1 when we discussed the shifting of electron pairs to interconvert resonance forms. Here, however, we use the arrows to show the electron movement that takes place when bonds break or form in the course of a chemical reaction. As your first example of a polar organic reaction, pay close attention to the details here: We will be returning to these principles repeatedly. The curved arrow convention is an especially powerful tool to help you understand how and why chemical reactions of organic compounds take place. The more you understand, the less you will have to memorize. 2-3. Functional Groups One look at the 16 classes of organic compounds in Table 2-3 (and these are only some of the most common ones!) will immediately tell you how complicated organic chemistry can become. At the same time, however, closer inspection reveals features of these categories that can greatly simplify learning in this course. Each 20 • Chapter 2 STRUCTURE AND REACTIVITY * You may have learned in freshman chemistry that the bond strength between H and A in H–A relates inversely to its acid strength. This correlation is not as general as you may have been led to believe: It holds only when the acids being compared are all from the same column of the periodic table, such as the hydrogen halides. It fails, for instance, in the series CH4, NH3, H2O, HF, where the acid strength increases as the bond strength goes up! The reason? Acidity relates to heterolytic bond cleavage to give ions, whereas bond strength relates to homolytic bond cleavage to give uncharged species. The two processes are very different. Differences in atomic electronegativity affect heterolytic bond cleavage (and therefore acidity) much more. 1559T_ch02_18-37 10/22/05 1:11 Page 20

Keys to the Chapter • 21 compound class is characterized by a specific atomic grouping called a functional group. Notice that only nine different elements are represented: C, H, S, N, O, and the four halogens. In fact, 11 of these classes contain only carbon, hydrogen, and oxygen. Knowledge of the characteristics of these atoms and the bonds between them, as we will see, will tell us the properties of the functional groups in which they appear. The functional groups will, in turn, provide the key to understanding the chemistry of all the members of the category. Thus, all members of the “alcohol” class of compounds, for instance, have certain common physical and chemical properties, resulting from the presence of the OH group in all of them. This kind of generalizable, qualitative similarity among compounds in any given class allows organic chemistry to be learned in a structured, organized, and, above all, logical way. Functional groups consist either of polarized bonds, whose atoms can attract other polarized or charged species, thereby leading to reactions, or of multiple (double or triple) bonds that also show reactivity for reasons we’ll explore later. Functional groups are the parts of molecules that most often take part in chemical reactions of those molecules. They are the “centers of reactivity” of molecules—where the action is. The most fundamental feature of alkanes relates to this concept of functional groups: Alkanes don’t have any. We’ll see the consequences of this in the next chapter. 2-4 and 2-5. Structures and Names for Alkanes There are a lot of organic compounds. Table 2-4 lists the numbers of isomers of just alkanes, and only goes up to 20 carbons, and already over half a million structures are possible! Imagine how many more structures can be manufactured when functional groups are present, or when the molecules get larger. Obviously not all these possible structures exist in nature or have been prepared in laboratories. Nonetheless, over 80 million different compounds are known at present, and nomenclature is the language that allows anyone interested in any of these materials to communicate about them in a clear and sensible way. The text presents brief descriptions of the problems associated with naming compounds before the systematic procedures of the IUPAC were developed. It then goes on to introduce just the rules necessary for naming simple alkanes: molecules containing only carbon and hydrogen atoms and having only single bonds holding the atoms together. Only four rules are needed at this stage: 1. Identify the longest carbon chain (the parent chain) in the molecule and name it. 2. Name all groups attached to this chain as substituents. 3. Number the carbon atoms of the parent chain from the end that gives the one containing the first substituents the lowest possible number. 4. Assemble the name, using the proper format. Although examples are given in the text and there are lots of problems for you to practice on, here are four additional worked-out examples to further clarify some fine points of the procedure. Example 1. Analysis: The longest chain contains seven carbons, so this is a heptane. However, there are two ways to identify a seven-carbon chain (see numbering). Which one is the parent? The rules specify that, in case of a tie for longest chain, the one with the most substituents is chosen as the parent. The seven-carbon chain labeled “a” CH3CH2CH2CHCH2CH2CH3 CHCH2CH3 CH3 Name: 7 6 5 4 3 2 1 3 2 1 Chain “b” (Proper parent stem) Improper numbering Chain “a” Proper numbering 1559T_ch02_18-37 10/22/05 1:11 Page 21

has one substituent (a sec-butyl group on carbon 4). The seven-carbon chain labeled “b” has two substituents (a methyl on carbon 3 and a propyl on carbon 4), so it wins. The molecule is called 3-methyl-4-propylheptane. Example 2. Analysis: The main chain here is unambiguous and 14 carbons long—the parent is tetradecane. Which is the correct numbering direction, however? Most of the groups are close to the right-hand end and will have low numbers if we number right-to-left. But that is not the criterion for determining which way to number the chain. The rule says to number in the direction that gives the carbon containing the first substituent the lowest possible number. If we number from right-to-left, the first substituted carbon is C3; if left-to-right, it is C2. So, left-to-right is correct, and the molecule’s name is 6,10-diethyl-2,8,8,10,11,12-hexamethyltetradecane. Even though the name that comes from numbering the other way has mostly low numbers, (5,9-diethyl-3,4,5,7,7,13- hexamethyltetradecane), it is wrong—its lowest number is a “3,” and the correct name’s lowest number is a “2.” Example 3. Analysis: A hexane. Numbering left-to-right gives 2,3,4,4,5-pentamethylhexane; right-to-left gives 2,3,3,4,5- pentamethylhexane. The choice is made by comparing substituent numbers from lowest to highest. The name with the lower number at the first point of difference is the winner. So 2,3,3,4,5 is preferred over 2,3,4,4,5. Example 4. Analysis: A nonane with a complicated substituent on carbon 5. Rule 3 illustrates what to do. Number the substituent carbons from the point of attachment to the main chain, outward along the substituent’s longest chain. The substituent has three carbons, so it has a name based on propyl. Then add appropriate numbers and names for groups attached to the substituent chain. So, 1,1,2-trimethylpropyl is the complete name of the substituent. Now, attach the substituent’s name to the name of the main chain to get the name of the whole molecule: 5-(1,1,2-trimethylpropyl)nonane. Note punctuation. It’s not hard, but it does take some careful analysis. CH3 CH CH3 CH3 CH3 CH3 CH3CH3 Name: CH C CH CH3CHCH2CH2CH2CHCH2CCH2C CH3 CH3 CH3 CH3 CH3 CH3 CH3CH2 CH2 CH3 Name: CHCHCH2CH3 1 2 3 1 2 22 • Chapter 2 STRUCTURE AND REACTIVITY 1559T_ch02_18-37 10/22/05 1:11 Page 22

Keys to the Chapter • 23 The notes above refer to the systematic nomenclature method as it is currently used. Please note, however, that there are many nonsystematic names in common use that are holdovers from the olden days and are still used for convenience or by force of habit. A number of compounds whose systematic names are very complicated have been given names that are well understood by people in the business but may seem random to the uninitiated. Several of these are mentioned in the text. One more example provides perspective in this area. Illustrated below is a compound that we eat every day. By the end of this course, you could sit down with the handbook of IUPAC rules and come up with the name 1-[3,4-dihydroxy-2,5-bis(hydroxymethyl)oxacyclopent-2-oxy]-3,4,5-trihydroxy-6-(hydroxymethyl)oxacyclohexane (and even this is only partially complete, lacking certain indicators that distinguish it from other known isomers!). Fortunately, the name used by general consent for this molecule, which is none other than ordinary table sugar, is a lot shorter: sucrose. See, even chemists use common sense sometimes. Fear not, odds are you will never, ever, have to give an IUPAC name to a molecule like this. I never did, at least until I had to write this study guide. 2-6. Physical Properties Every time we encounter a new class of compounds, we will briefly discuss common “physical properties” of members of that compound class. These will include general comments on the nature of the compound under ordinary conditions (e.g., diethylamine, colorless liquid, smells like something died, or, 2-hydroperoxy-2-isopropoxypropane, colorless crystalline solid, blows up like an A-bomb if you look at it cross-eyed). The purpose of these comments is to give you a feeling for what these materials are really like (as well as alerting you to the fact that some organic molecules may not be your friends). For the record, alkanes are colorless gases or liquids, with rather light odors, or white, waxy solids (candle wax is mainly alkanes). More specific discussion will focus on relationships between molecular structure and physical properties for the class of compounds as a whole. In this chapter a brief summary of the kinds of forces that attract molecules to each other is presented. Alkanes, lacking charged atoms or highly polarized bonds, do not exhibit either ionic or dipolar forces. As nonpolar molecules, alkane molecules are attracted to each other by only the rather weak London forces. These can be understood fairly simply. In even a totally unpolarized bond, the electrons are always moving. Even though the average location of the electron pair is exactly half-way between the atoms, at any particular moment in time, the electrons may be closer to one atom or the other: During these moments, the bond is polarized. Because this polarization is not permanent, the partial charges associated with it are only transient, or fleeting in nature, thus the name fleeting dipoles. When two nonpolar molecules are close to each other and a bond in one of them exhibits a fleeting dipole, the electrons in a nearby bond of the other molecule will be pushed away from the fleeting dipole’s “” end and attracted toward its “” end. The positions and movements of all the electrons are said to be “correlated”: A: A A vs. vs. A : A :A “Fleeting dipole” “Fleeting dipole” Average situation 0 0 HO HO O O O OH OH OH H H H CH2OH CH2OH CH2OH 1559T_ch02_18-37 10/22/05 1:11 Page 23

The result will be a new dipole in the second molecule’s bond, “induced” by the original fleeting dipole in the first molecule. As the diagram shows, the polarizations that result lead to an attractive force between the molecules—the so-called London forces. Even though the dipoles involved have only transient existence and all the bonds are nonpolar, it turns out that the odds always favor the presence of some fleeting dipoles in a molecule, and the net result is this weak, but real, London attraction. Because of the weakness of this attraction, alkanes exhibit relatively low melting points and boiling points relative to those of more polar or charged molecules. The nonpolar nature of alkanes results in other physical consequences, such as rather limited ability to serve as solvents for polar compounds (remember “like dissolves like” from freshman chemistry?). Lack of polarized bonds also very much limits the chemistry that alkanes can display. This subject will be taken up in the next chapter. 2-7 and 2-8. Conformations Although we generally draw pictures of molecules in a single geometrical representation, the fact is that no molecule has a single rigid geometry. The electrons in bonds can be viewed as an elastic glue holding the atoms together. The bonds are therefore somewhat flexible and are subject to some degree of bending or stretching. So, even in the simplest molecules like H2, the atoms are capable of some degree of movement with respect to one another. In more complicated molecules, additional forms of internal motion become possible. The conformations of ethane and larger alkanes are a result of rotation about carbon–carbon single bonds, a relatively easy process. This section describes the energetics associated with this rotation and the names associated with the various shapes of the molecules as this rotation occurs. Newman projections provide an “endon” view of these conformations: At this point you should take a look at a set of molecular models so that you can become familiar with these conformations in three dimensions. Conformational energetics can be summarized for alkanes as follows: 1. Eclipsed is 2.9 kcal mol1 higher in energy (less stable) than staggered for ethane. 2. Each CH3–H eclipsing is 0.3 kcal mol1 worse than an H–H eclipsing (relative to corresponding changes in staggered conformation energies). 3. Each CH3–CH3 eclipsing is 2.0 kcal mol1 worse than an H–H eclipsing. 4. Each CH3–CH3 gauche is 0.9 kcal mol1 worse than CH3–CH3 anti. With these individual estimates, the graph of energy vs. rotational angle can be readily sketched for simple alkanes. Note: These “energy” values are actually enthalpies (heat content, or H° values). A: A A : A A :A Original fleeting dipole New fleeting dipole, “induced” by the original one Result Repel Attract Molecule 1: Molecule 2: Attract Attract A: A Electrons will move 24 • Chapter 2 STRUCTURE AND REACTIVITY 1559T_ch02_18-37 10/22/05 1:11 Page 24

Solutions to Problems • 25 Solutions to Problems 22. (a) Remember, H° (reaction) H° (bonds broken) H° (bonds formed). (1) To calculate the H° associated with breaking one of the two bonds in the carbon–carbon double bonds, use H° (CPC) as a bond-breaking contribution and H° (COC) as a bondforming contribution: (2) (b) In reaction (1), two molecules combine to make one. This concentrates the energy content of the system into fewer particles, resulting in a large negative value for S (35 entropy units) for reaction (1). If you like, the system becomes more “ordered,” which is saying much the same thing. In reaction (2), two molecules react to make two different molecules. The dispersal of the energy content of the system undergoes no major change, resulting in a S of approximately zero. (c) For (1) at 25°C, G° H° TS° 27 298(35 103 ) 17 kcal mol1 For (1) at 600°C, G° H° TS° 27 873(35 103 ) 4 kcal mol1 For (2) at either 25°C or 600°C, G° H° 10 kcal mol1 because S° 0. Both reactions have negative G° at 25°C, so both are thermodynamically favorable. At 600°C, the S° for reaction (1) has made its G° value positive: The reaction is therefore energetically unfavorable. Reaction (2) is still just as good as it was at 25°C. 23. Don’t identify the acids until you’ve looked to see which species give up protons; many of the species here can act both as acids and bases! The equilibrium lies to the side of the weaker acid/base pair, as indicated by the unequal lengths of the forward and reverse arrows. From the data in Table 2-2, you can identify the stronger acids as the species with the larger Ka or smaller (less positive or more negative) pKa. The equilibrium constant for each reaction is found by dividing Ka for the acid on the left by Ka for the acid on the right. How did I know that? Here’s how. For the following general reaction: HA1 A2 HA2 A1 we have Ka1 [H][A1 ]/[HA1] and Ka2 [H][A2 ]/[HA2], right? So, Ka1 /Ka2 [H][A1 ][HA2]/[HA1][H][A2 ] [HA2][A1 ]/[HA1][A2 ] Keq) (a) H2O HCN H3O CN Keq 1.3 1011 weaker base weaker acid stronger acid stronger base (b) CH3O NH3 CH3OH NH2 Keq 3.1 1020 weaker base weaker acid stronger acid stronger base H break COH break BrOBr form COBr form HOBr 99 46 68 87 10 kcal mol1 H break CPC break BrOBr form COC form 2COBr 146 46 83 2(68) 27 kcal mol1 1559T_ch02_18-37 10/22/05 1:11 Page 25