北京化工大学：《有机化学》课程教学资源（双语习题与答案）Chapter 03 Reactions of Alkanes（Bond-Dissociation Energies, Radical Halogenation, and Relative Reactivity）.pdf_大学文库

3 Reactions of Alkanes: Bond-Dissociation Energies, Radical Halogenation, and Relative Reactivity Discussing reactions of alkanes at the start of an organic chemistry course allows us to learn to work with several concepts that will be useful later. These include the idea of a general reaction mechanism that describes how any member of an entire class of compounds is likely to behave under certain reaction conditions. We also see the relationship between the energy concepts introduced in Chapter 2 and reactions that require more than one step. Alkanes do not contain any functional groups: They are made up of nonpolar COC and COH bonds, and nothing else. Therefore, alkanes are essentially unreactive toward ionic or polar materials; indeed, alkanes are just about the least reactive of all compound classes. Therefore, their chemistry is limited to processes that can lead to cleavage of nonpolar bonds. Thus, the only reasonable way for an alkane bond to cleave is homolytically, leaving one electron with each of the formerly bonded atoms: . This kind of bond cleavage is difficult and only occurs at high temperatures or in the presence of certain especially reactive species like halogen atoms. This chapter covers three major ways alkane bonds are cleaved: pyrolysis (high temperature), halogenation (by halogen atoms), and combustion (high temperature and oxygen). You will note a strong emphasis on discussions involving bond energies. This should not surprise you, because bond cleavage requires an input of energy. The mechanism presents the reaction in terms of a step-by-step, bond-by-bond analysis that is helpful for spotting trends and analogies. Outline of the Chapter 33-1 Strength of Alkane Bonds: Radicals Exactly what does it take to cleave bonds in alkanes? 33-2 Alkyl Radicals and Hyperconjugation The nature of the species obtained upon alkane bond cleavage. 33-3 Conversion of Petroleum: Pyrolysis A practical example. 33-4 Chlorination of Methane: The Radical Chain Mechanism Radical substitution of chlorine for hydrogen in methane: mechanism and energetics. 33-5 Other Radical Halogenations of Methane Similarities and differences. 33-6 Chlorination of Higher Alkanes What happens when substitution for different hydrogens in an alkane can give different products? CC C H C or C C H 38 1559T_ch03_38-54 10/22/05 2:59 Page 38

Keys to the Chapter • 39 33-7 Selectivity with Other Halogens Energetic comparisons of the reactions involving F2 and Br2. 33-8 Synthetic Aspects More practical considerations. 33-9 Synthetic Chlorine Compounds and the Stratospheric Ozone Layer Halogens in the “real world.” 3-10 Combustion and the Relative Stability of Alkanes Included is a detailed introduction to the evaluation of the energetics associated with a chemical reaction. Keys to the Chapter 3-1. Strength of Alkane Bonds: Radicals A minor but annoying point of confusion is often encountered when one discusses bond strengths. A bond’s strength, or more properly, bond-dissociation energy (DH°), is defined as the energy released when a bond forms or, equivalently, the energy input required to break a bond: Aj Bj On AOB H° DH° Energy is released. AOB On Aj Bj H° DH° Energy is put in. Inspection of these two equations shows that the bonded molecule AOB is more stableOlower in energy contentOthan the separated atoms A and B by an amount equal to DH°. When the linkages in a molecule are strong (high DH°), the molecule is usually relatively low in energy content (e.g., stable). As long as you remember that DH° is the energy that has to be put in to break a bond, you won’t fall into the common trap of associating large DH° values with high-energy species. Large DH° values imply low energy, strongly bonded, stable species. The tables and figures in this section should further help you develop a comfortable understanding of the meaning of DH° values, in preparation for their use later on. 3-2. Alkyl Radicals and Hyperconjugation Homolytic cleavage of any bond in an alkane generates radicals: species with a single unpaired electron where an attached group used to be. The section illustrates four such examples: methyl, jCH3; ethyl, jCH2CH3; isopropyl, jCH(CH3)2; and tert-butyl, jC(CH3)3. Several points are made in the section. First, radical carbons are sp2 hybridized (planar), not sp3 hybridized (tetrahedral, as in alkanes). Why should this be? A partial reason goes back to basic electrostatics. The shape of a species will be that which minimizes repulsion between electrons around a central atom (remember valence shell electron pair repulsion, or VSEPR?). In ammonia, NH3, the four electron pairs around N are best accommodated by a pyramidal shape based on sp3 hybridization: Repulsion between the lone pair and the electrons in the NOH bonds is important in causing this geometry to be preferred. Reducing the number of nonbonded electrons from two to one as in methyl radical, jCH3, changes the situation. Now, electron repulsion between the pairs in the COH bonds dominates, a situation leading to sp2 hybridization and trigonal planar geometry, which allows the COH bonding electrons to spread out far away from one another. The second main point in the chapter is the stabilization of a radical center by the presence of alkyl groups attached to the radical carbon. So, tert-butyl radical is more stable than isopropyl, which is better than ethyl; methyl radical is the least stable. Hyperconjugation is one concept often used to explain this stabilization. Physically, and electrostatically, the radical carbon can be viewed as somewhat electron deficient (7 valence electrons instead of an octet). Hyperconjugation provides a means for bonds in neighboring alkyl groups to “lend” a little electron density to the radical center, thereby making it feel a little less electron-poor. In doing so, the 1559T_ch03_38-54 10/22/05 2:59 Page 39

40 • Chapter 3 REACTIONS OF ALKANES alkyl groups effectively take some of the electron deficiency onto themselves, spreading it out, or “delocalizing” it. Delocalization of an electron deficiency or an electron excess over more than just the atom on which it is nominally located is often an energetically favorable, stabilizing process. It effectively allows the “problem” to be diluted over a larger area, rather than being the concentrated burden of a single atom. The ability of alkyl groups to stabilize electron-deficient centers like radicals is often taken to imply that they are better donors of electrons than hydrogen atoms. Alkyl groups are therefore referred to as electron donating. 3-3. Conversion of Petroleum: Pyrolysis The practical, “real-world” aspects of bond cleavage and radical formation are explored. Pyrolysis is a process that often gives mixtures of many products, and methods have been developed (mostly within the petroleum industry) to control this reaction somewhat. We will frequently explore the issue of reaction control: the modification of conditions under which a chemical transformation is carried out in order to give a desired molecule as a major or exclusive product. 3-4. Chlorination of Methane: The Radical Chain Mechanism In this section the reaction of methane with chlorine molecules is discussed. The process CH4 Cl2 On HCl CH3Cl is important because it converts a nonfunctionalized molecule (an alkane) into a molecule containing a functional group (a haloalkane). Once the functional group is present, many more kinds of chemical reactions will become possible. This section also presents the mechanism of this reaction in full detail. Pay close attention not only to the steps of the reaction (initiation, propagation, and termination) but also to the finer details relating H°, Ea, and transition state structure for each step. Although some of the terminology introduced here is appropriate only for radical mechanisms and not for the majority of reactions to come later, the type of information that the mechanism contains is critical to an understanding of how and why organic reactions occur. Take some time in this section to study each reaction step. What are its energetic circumstances, under what conditions does it occur, what role does it play in the overall process? Try to establish a feeling for the species involved as “stable” or “unstable,” “reactive” or “unreactive,” relatively speaking. Reaction mechanisms are intended to allow one to make sense out of organic chemistry. Give this one the time to do that for you. Be sure that you understand the procedure for calculating the H° value for a chemical reaction from the DH° values of the bonds taking part in the transformation. The general formula is To illustrate with a reaction different from those in the text, let us calculate H° for the process C2H6 H2 n 2 CH4. Using the data from Tables 3-1 and 3-2 in the text, Comment: This is a “hydrocracking” process that, although exothermic, requires very high temperatures to occur (cleavage of COC bond is necessary). reaction [90 104] [2(105)] 16 kcal mol1 kcal mol1 DH 90 104 105 Note 2 methane C—H bonds CH3 CH3 H H 2 CH3 H H reaction DH (bonds broken) Energy input DH (bonds formed) Energy output H 1559T_ch03_38-54 10/22/05 2:59 Page 40

Keys to the Chapter • 41 Note on energetics: The overall enthalpy of a radical chain reaction is the sum of the H° values for only the propagation steps. If we “sum up” the species in these steps, we see that the free atoms and radicals “cancel out,” leaving only the molecular species of the overall reaction: Propagation step 1 H° 2 kcal mol1 Propagation step 2 H° 27 kcal mol1 Sum: H° 25 kcal mol1 Removing the Clj and jCH3 that appear on both sides of the equation leaves just the molecules of the overall process. What about the initiation and termination steps and their H°’s? They are separate. Their H° values are not a part of the enthalpy change as it is defined for the stoichiometric reaction. When we measure the heat of a radical reaction experimentally, the value we obtain will not be precisely equal to H° for the propagation steps alone; initiation and termination steps are occurring, too, and their H°’s will introduce an error. This deviation will usually be small, however, because initiation and termination steps occur only infrequently relative to the propagations, and because the H°’s for the endothermic initiation are for the most part canceled out by those of the exothermic termination processes. 3-5. Other Radical Halogenations of Methane One of the best features of organic chemistry is the fact that one mechanism can hold for many individual reactions. Thus, the same types of steps that occur in the chlorination of methane are followed in its reactions with the other halogens. The similarities are qualitative, however. Differences in energetics are significant, and as a result the reaction-coordinate diagrams differ in appearance in a way that will become important as we continue through the chapter. 3-6. Chlorination of Higher Alkanes The same mechanism also applies qualitatively to chlorination of other alkanes. The only difference is in the nature of the COH bonds available in the alkane to be broken. They are generally less strong than those in methane, following a DH° order of CH4 1° 2° 3°. (Note: 1° primary, 2° secondary, and 3° tertiary. These are commonly used symbols.) The weakest (3°) are the most readily broken; thus, alkanes with different types of COH bonds display a built-in selectivity of 3° 2° 1° in their reactions with chlorine. This section describes this selectivity quantitatively, illustrating how both reactivity differences and statistical factors combine to produce the observed ratios of products in several representative systems. 3-7. Selectivity with Other Halogens An extension of the previous sections. The most significant point is that reactivity and selectivity in radical halogenations are inversely related. Simply put, the more reactive halogens are less picky and show less preference for 3° vs. 2° vs. 1° COH bonds relative to less reactive halogens. The reason lies in the different activation energies associated with the COH bond-breaking step. The values for fluorine are all very small, and very similar to one another. Fluorine thus reacts very rapidly with any COH bond in a molecule. The reactions for bromine have large activation energies, with significant differences associated with the different types of COH bonds present. The result is that bromine is much, much slower than fluorine to react with any alkane, and bromine is much more discriminating (selective) in its reactions, too, greatly preferring 3° over 2° or 1° COH bonds. The contrasts between reaction-coordinate diagrams for the two halogens provide a pictorial representation of these differences. 3-8. Synthetic Aspects Synthesis is one primary function associated with organic chemistry. In synthesis, we strive to produce a desired product in good yield with high selectivity, to minimize the effort required to separate this material from side products. In particular, a reaction that gives rise to a hard-to-separate mixture of many components is synthetically useless. In this chapter we have seen a large number of possible permutations of a single reaction: alkane halogenation. Not all of the examples shown are equally useful synthetically. The best ones start with CH4 Cl CH3 Cl2 HCl CH3 CH3Cl Cl CH3 Cl2 CH3Cl Cl CH4 Cl HCl CH3 1559T_ch03_38-54 10/22/05 2:59 Page 41

42 • Chapter 3 REACTIONS OF ALKANES an alkane in which all hydrogens are chemically indistinguishable (methane, ethane, neopentane), because they can produce only one monohalogenation product. In the case of most alkanes, synthetic utility will be determined by the number of different types of hydrogens present and whether the desired product derives from substitution of a more reactive or a less reactive hydrogen in the molecule. In isobutane, for example, there are one 3° and nine 1° hydrogens. If we desire to halogenate at the 3° center, the natural selectivity of bromine makes it the obvious halogen to choose. If we desire to halogenate a 1° carbon, a less selective, more reactive halogen would allow us to take best advantage of the statistical factor of nine possible 1° hydrogens available to be replaced in each molecule. Thus, 3-10. Combustion and Relative Stability In order to obtain thermodynamic information experimentally, several methods may be used. The measurement of equilibrium constants gives energy differences between species. Directly measuring the heat of a reaction accomplishes the same thing. When the reaction is combustion of a hydrocarbon, the result is a measure of the energy content of the compound relative to that of the product molecules, CO2 and H2O. Such data allow comparisons to be made between related compounds, which in turn reveal factors influencing the relative stabilities of different structures. Solutions to Problems 13. This problem is really a reminder of material from the previous chapter. For shorthand purposes, we use the symbols 1° primary, 2° secondary, and 3° tertiary. (a) (b) (c) (d) 14. (a) Remember: Identify radicals as 1°, 2°, or 3° by the radical carbon. None of the other carbons matter. Two hyperconjugation pictures may be drawn for 1-methylpropyl radical, one with two 1-Methylpropyl (sec-butyl; see Table 2-4) Secondary (2), more stable CH3CH2CHCH3 Butyl radical Primary (1), less stable CH3CH2CH2CH2 CH3 1 2 2 3 CH2 CH2 CH2 CH2 H C As you will see in Chapter 4, most “ring” compounds can be treated just like molecules without rings. CH3 1 1 2 3 CH3CH2CH2CH2CH3 1 2 1 CH3CH2CH2CH3 1 2 1 CH3 CH CH3 Br2 CH3 CH3 C CH3 Major product CH3 Br CH3 CH CH3 F2 F CH3 CH2 CH CH3 Major product CH3 1559T_ch03_38-54 10/22/05 2:59 Page 42

Solutions to Problems • 45 18. (a) Two: 1-halobutane and 2-halobutane (b) Three (see solution to Problem 19) (c) Four (see solution to Problem 19) (d) Four: 19. (a) (i) CH3CH2CH2CH2CH2Cl (1-chloropentane), CH3CH2CH2CHClCH3 (2-chloropentane), and CH3CH2CHClCH2CH3 (3-chloropentane). (ii) CH3CH2CH(CH3)CH2CH2Cl (1-chloro-3-methylpentane), CH3CH2CH(CH3)CHClCH3 (2-chloro-3-methylpentane), CH3CH2CCl(CH3)CH2CH3 (3-chloro-3-methylpentane), and CH3CH2CH(CH2Cl)CH2CH3 [3-(chloromethyl)pentane]. (b) To answer this question, you first have to count up and identify by type (1°, 2°, or 3°) all the hydrogens whose removal could lead to each one of the products. Then multiply the number of hydrogens you counted by the relative reactivity for that type of hydrogen in a chlorination reaction at 25°C, the conditions for the reaction stated in the problem. This procedure gives you the relative amount of product corresponding to removal of those hydrogens. After you have done this for all the products, you convert these relative amounts into percentage yields by normalizing to 100% (as shown below). (i) 1-Chloropentane is formed by chlorination of any of the six primary hydrogens (each with a relative reactivity 1) on carbons 1 and 5; its relative yield is 6 1 6. 2-Chloropentane is formed by chlorination of any of the four secondary hydrogens (each with a relative reactivity 4) on carbons 2 and 4; its relative yield is 4 4 16. 3-Chloropentane is formed by chlorination of any of the two secondary hydrogens (each with a relative reactivity 4) on carbon 3; its relative yield is 2 4 8. The absolute % yield for each product is calculated as follows: 100% % yield of that product So, Yield of 1-chloropentane 100% 6/(6 16 8) 100% 6/30 20% Yield of 2-chloropentane 100% 16/30 53% Yield of 3-chloropentane 100% 8/30 27% (ii) 1-Chloro-3-methylpentane is formed by chlorination of any of the six primary hydrogens (relative reactivity 1) on carbons 1 and 5; its relative yield is 6 1 6. 2-Chloro-3- methylpentane is formed by chlorination of any of the four secondary hydrogens (relative reactivity 4) on carbons 2 and 4; its relative yield is 4 4 16. 3-Chloro-3-methylpentane is formed by chlorination of the single tertiary hydrogen (relative reactivity 5) on carbon 3; its relative yield is 1 5 5. 3-(Chloromethyl)pentane is formed by chlorination of any of the three primary hydrogens (relative reactivity 1) on the methyl group attached to carbon 3; its relative yield is 1 3 3. Relative yield of the product Sum of relative yields for all products (Halomethyl)- cyclopentane CH2X 1-Halo-1-methylcyclopentane X CH3 1-Halo-2-methylcyclopentane CH3 X 1-Halo-3-methylcyclopentane CH3 X 1559T_ch03_38-54 10/22/05 2:59 Page 45