Methane Energy of Activation. Transition State

Chapter 2

Methane Energy of Activation. Transition State

Hydrocarbons

• Certain organic compounds contain only two elements, hydrogen and carbon, and hence are known as hydrocarbons. On the basis of structure, hydrocarbons are divided into two main classes, aliphatic and aromatic. Aliphatic hydrocarbons are further divided into families: alkanes, alkenes, alkynes, and their cyclic analogs (cycloalkanes, etc.). We shall take up these families in the order given.

• The simplest member of the alkane family and, indeed, one of the simplest of all organic compounds is methane, CH4 . We shall suedey this single compound at some length, since most of what we learn about it can be carried over with minor modifications to any.

Structure of methane

• As we discussed in the previous chapter each of the four hydrogen atoms is bonded to the carbon atom by a covalent bond, that is, by the sharing of a pair of electrons. When carbon is bonded to four other atoms, its bonding orbitals (sp* orbitals, formed by the mixing of one s and three/? orbitals) are directed to the corners of a tetrahedron (Fig. 2. la). This tetrahedral arrangement is the one that permits the orbitals to be as far apart as possible. For each of these orbitals to overlap most effectively the spherical s orbital of a hydrogen atom, and thus to form the strongest bond, each hydrogen nucleus must be located at a corner of this tetrahedron.

• The tetrahedral structure of methane has been verified by electron diffraction which shows beyond question the arrangement of atoms in such simple molecules. Later on, we shall examine some of the evidence that led chemists to accept this tetrahedral structure long before quantum mechanics or electron diffraction was know.

• We shall ordinarily write methane with a dash to represent each pair of Eltrons shared by carbon and hydrogen (I). To focus our attention on individual electrons, we may sometimes indicate a pair of electrons by a pair of dots (II). Finally, when we wish to consider the actual shape of the molecule, we shall use a simple three-dimensional picture (HI)

Physical properties

• As we discussed in the previous chapter (Sec. 1.18), the unit of such a non-ionic compound, whether solid, liquid, or gas, is the molecule. Because the methane molecule is highly symmetrical, the polarities of the individual carbon-hydrogen bonds cancel out; as a result, the molecule itself is non-polar. Attraction between such non-polar molecules is limited to van der Waals forces; for such small molecules, these attractive forces must be tiny compared with the enormous forces between, say, sodium and chloride ions. It is not surprising, then, that these attractive forces are easily overcome by thermal energy, so that melting, and boiling occur at very low temperatures: m.p -183, b.p. - 101.5. (Compare these values with the corresponding ones for sodium chloride: m.p. 801. b.p. 1413.) As a consequence, methane is a gas at ordinary temperatures. 

Source

• Methane is an end product of the anaerobic ("without air") decay of plants, that is, of the breakdown of certain very complicated molecules. *As such, it is the major constituent (up to 97 ) of natural gas. It is the dangerously/m/ow/? of the coal mine and can be seen as marsh gas bubbling to the surface of swamps. If methane is wanted in very pure form, it can be separated from the other constituents of natural gas (mostly other alkanes) by fractional distillation. Most of it, of course, is consumed as fuel without purification. According to one theory, the origins of life go back to a primitive earth surrounded by an atmosphere of methane, water, ammonia, and hydrogen. Energy radiation from the sun, lightning discharges- -broke these simple molecules into reactive fragments (free radicals, Sec. 2.12); these combined to form larger molecules which eventually yielded the enormously complicated organic compounds that make up living organisms. (Recent detection of organic molecules in space has even led to the speculation that "organic seeds for life could have existed in interstellar clouds/') Evidence that this could have happened was found in 1953 by the Nobel Prize winner Harold C. Urey and his student Stanley Miller at the University of Chicago. They showed that an electric discharge converts a mixture of methane, water, antimonial, and hydrogen into a large number of organic compounds, including amino acids, the building blocks from which proteins, the "stuff of life*' (Chap. 36), are made. (It is perhaps appropriate that we begin this study of organic chemistry with methane and its conversion. into free radicals.) The methane generated in the final decay of a once-living organism may well be the very substance from which in the final analysis the organism was derived. **. . . early to earth, ashes to ashes, dust to dust. . .."

Reactions

• In its chemical properties as in its physical properties, methane sets the pattern for the alkane family (Sec. 3.18). Typically, it reacts only with highly reactive substances or under very vigorous conditions, which, as we shall see, amounts to the same thing. At this point we shall take up only its oxidation: by oxygen, by halogens, and even by. 

Oxidation. Heat of combustion

• Combustion to carbon dioxide and water is characteristic of organic compounds; under special conditions it is used to determine their content of carbon and hydrogen. Combustion of methane is the principal reaction taking place during the.

• Through controlled partial oxidation of methane and the high-temperature catalytic reaction with water, methane is an increasingly important source of products other than heat: of hydrogen, used in the manufacture of ammonia; of mixtures of carbon monoxide and hydrogen, used in the manufacture of methanolized other alcohols; and of acetylene (Sec. 8.5), itself the starting point of large-scale production of many organic compounds. Oxidation by halogens is of particular interest to us partly because we know more about it than the other reactions of methane and, in one way or another, is the topic of discussion throughout the remainder of this chapter.

Chlorination: a substitution reaction

• Under the influence of ultraviolet light or at a temperature of 250-400 a mixture of the two gases, methane and chlorine, reacts vigorously to yield hydrogen chloride and a compound of formula CH3 C1. We say that methane has undergone chlorination, and we call the product, CH3C1, chloromethane or methyl chloride (CH3 = methyl). Chlorination is a typical example of a broad class of organic reactions known as substitution. A chlorine atom has been substituted for a hydrogen atom methane, and the hydrogen atom thus replaced is found combined with a second atom of chlorine.

Control of chlorination

• Chlorination of methane may yield any one of four organic products, depending upon the stage to which the reaction is carried. Can we control this reaction so that methyl chloride is the principal organic product? That is, can we limit the reaction to the first stage, wow chlorination? We might at first expect naively, as it turns out to accomplish this by providing only one mole of chlorine for each mole of methane. But let us see what happens if we do so. At the beginning of the reaction there is only methane for the chlorine to react with, and consequently only the first stage of chlorination takes place. This reaction, however, yields methyl chloride, so that as the reaction proceeds methane disappears and methyl chloride takes its place. As the proportion of methyl chloride grows, it competes with the methane for the available chlorine. By the time the concentration of methyl chloride exceeds that of methane, chlorine is more likely to attack methyl chloride than methane, and the second stage of chlorination becomes more important than the first. A large amount of methylene chloride is formed, which in a similar way is chlorinated to chloroform and this, in turn, is chlorinated to carbon tetrachloride. When we finally work up the reaction product, we find that it is a mixture of all four chlorinated methane's together with some unreacted methane. 

• The reaction may, however, be limited almost entirely to monochlorination if we use a large excess of methane. In this case, even at the very end of the reaction unreacted methane greatly exceeds methyl chloride. Chlorine is more likely to attack methane than methyl chloride, and thus the first stage of chlorination is the principal reaction. Because of the great difference in their boiling points, it is easy to separate the excess methane (b.p. -161.5) from the methyl chloride (b.p. -24) so that the methane can be mixed with more chlorine and put through the process again. While there is a low conversion of methane into methyl chloride in each cycle, the yield of methyl chloride based on the chlorine consumed is quite high. The use of a large excess of one reactant is a common device of the organic chemist when he wishes to limit reaction to only one of a number of reactive sites in the molecule of that reactant.

Reaction with other halogens: halogenation

• Methane reacts with bromine, again at high temperatures or under the influence of ultraviolet light, to yield the corresponding bromomethane's: methyl bromide, methylene bromide, bromoform, and carbon tetrabromide. HBr + CH4 -"^ CH3 Br Methane Methyl Bromide Bromination takes place somewhat less readily than chlorination. Methane does not react with iodine at all. With fluorine it reacts so vigorously that, even in the dark and at room temperature, the reaction must be carefully controlled: the reactants, diluted with an inert gas, are mixed at low pressure. We can, therefore, arrange the halogens in order of reactivity. Reactivity of halogens F2 > C12 > Br2 O I 2) This same order of reactivity holds for the reaction of the halogens with other alkanes and, indeed, with most other organic compounds. The spread of reactivities is so great that only chlorination and bromination proceed at such rates as to be generally useful.

Relative reactivity

• Throughout our study of organic chemistry, we shall constantly be interested in relative reactivities. We shall compare the reactivities of various reagents toward the same organic compound, the reactivities of different organic compounds toward the same reagent, and even the reactivities of different sites in an organic molecule toward the same reagent. It should be understood that when we compare reactivities we compare rates of reaction. When we say that chlorine is more reactive than bromine toward methane, we mean that under the same conditions (same concentration, same temperature, etc.) chlorine reacts with methane faster than does bromine. From another point of view, we mean that the bromine reaction must be carried out under more vigorous conditions (higher concentration or higher temperature) if it is to take place as fast as the chlorine reaction. When we say that methane and iodine do not react at all, we mean that the reaction is too slow to be significant. We shall want to know not only what these relative reactivities are, but also, whenever possible, how to account for them. To see what factors, cause one reaction to be faster than another, we shall take up in more detail this matter of the different reactivities of the halogens toward methane. Before we can do this, however, we must understand a little more about the reaction itself.

Reaction mechanisms

• It is important for us to know not only what happens in a chemical reaction but also how it happens, that is, to know not only the facts but also the theory. For example, we know that methane and chlorine under the influence of heat or light form methyl chloride and hydrogen chloride. Just how is a molecule of methane converted into a molecule of methyl chloride? Does this transformation involve more than one step, and. if so, what are these steps? Just what is the function of heat or light? The answer to questions like these, that is, the detailed, step-by-step description of a chemical reaction, is called a mechanism. It is only a hypothesis; it is advanced to account for the facts. As more facts are discovered, the mechanism must also account for them, or else be modified so that it does account for them; it may even be necessary to discard a mechanism and to propose a new one. It would be difficult to say that a mechanism had ever been proved. If, however, a mechanism accounts satisfactorily for a wide variety of facts; if we make predictions based upon this mechanism and find these predictions borne out; if the mechanism is consistent with mechanisms for other, related reactions; then the mechanism is said to be well established, and it becomes part of the theory of organic chemistry. Why are we interested in the mechanisms of reactions? As an important part of the theory of organic chemistry, they help make up the framework on which we hang the facts we learn. An understanding of mechanisms will help us to see a pattern in the complicated and confusing multitude of organic reactions. We shall find that many apparently unrelated reactions proceed by the same or similar mechanisms, so that most of what we have already learned about one reaction may be applied directly to many new ones. By knowing how a reaction takes place, we can make changes in the experimental conditions not by trial and error, but logically that will improve the yield of the product we want, or that will even alter the course of the reaction completely and give us an entirely different product. As our understanding of reactions grows* so does our power to control. 

Mechanism of chlorination. Free radicals

• It will be worthwhile to examine the mechanism of chlorination of methane in some detail. The same mechanism holds for bromination as well as chlorination, and for other alkanes as well as methane; it even holds for many compounds which, while not alkanes, contain alkane-like portions in their molecules. Closelyrelated mechanisms are involved in oxidation (combustion) and other reactions of alkanes. More important, this mechanism illustrates certain general principles that can be carried over to a wide range of chemical reactions. Finally, by studying the evidence that supports the mechanism, we can learn something of how a chemist finds out what goes on during a chemical reaction. Among the facts that must be accounted for are these: (a) Methane and chlorine do not react in tar dark at room temperature, (b) Reaction takes place readily, however, in the dark at temperatures over 250, or (c) under the influence of ultraviolet light at room temperature, (d) When the reaction is induced by light, many (several thousand) molecules of methyl chloride are obtained for each photon of light that is absorbed by the system, (e) The presence of a small amount of oxygen slows down the reaction for a period of time, after which the reaction proceeds normally; the length of this period depends upon how much oxygen is present. The mechanism that accounts for these facts most satisfactorily, and hence is generally accepted, is shown in the following equation:

•  The chlorine molecule undergoes homolysis (Sec. 1.14): that is, cleavage of the chlorine-chlorine bond takes place in a symmetrical way, so that each atom retains one electron of the pair that formed the covalent bond. This odd electron is not paired as are all the other electrons of the chlorine atom; that is, it does not have a partner of opposite spin (Sec. 1.6). An atom or group of atoms possessing an odd (unpaired) electron is called a free radical. In writing the symbol for a free radical, we generally include a dot to represent the odd electron just as we include a plus or minus sign in the symbol of an ion.

Chain reactions

• The chlorination of methane is an example of a chain reaction, a reaction that involves a series of steps, each of which generates a reactive substance that brings about the next step. While chain reactions may vary widely in their details, they all have certain fundamental characteristics in common.

• First in the chain of reactions is a chain-initiating step, in which energy is absorbed and a reactive particle generated; in the present reaction it is the cleavage of chlorine into atoms (step 1). There are one or more chain-propagating steps, each of which consumes a reactive particle and generates another; here they are the reaction of chlorine atoms with methane (step 2), and of methyl radicals with chlorine (step 3). Finally, there are chain-terminating steps, in which reactive particles are consumed but not generated; in the chlorination of methane these would involve the union of two of the reactive particles, or the capture of one of them by the walls of the reaction vessel. Under one set of conditions, about 10,000 molecules of methyl chloride are formed for every quantum (photon) of light absorbed. Each photon cleaves one chlorine molecule to form two chlorine atoms, each of which starts a chain. On the average, each chain consists of 5000 repetitions of the chain-propagating cycle before it is finally stopped.

Inhibitors

• The CH3OO- radical is much less reactive than the CH3 - radical, and can do little to continue the chain. By combining with a methyl radical, one oxygen molecule breaks a chain, and thus prevents the formation of thousands of molecules of methyl chloride; this, of course, slows down the reaction tremendously. After all the oxygen molecules present have combined with methyl radicals* the reaction is free to proceed at its normal rate. A substance that slows down or stops a reaction even though present in small amount is called an inhibitor. The period of time during which inhibition lasts, and after which the reaction proceeds normally, is called the inhibition period. Inhibition by a relatively small amount of an added material is quite characteristic of chain reactions of any type and is often one of the clues that first leads us to suspect that we are dealing with a chain reaction. It is hard to see how else a few molecules could prevent the reaction of so many. (We shall frequently encounter the use of oxygen to inhibit free-radical reactions.)

Heat of reaction

• In our consideration of the chlorination of methane, we have so far been concerned chiefly with the particles involved molecules and atoms- and the changes that they undergo. As with any reaction, however, it is important to consider also the energy changes involved, since these changes determine to a large extent how fast the reaction will go, and, in fact, whether it will take place at all. By using the values of bond dissociation energies given in Table 1.2 (p. 21), we can calculate the energy changes that take place in a great number of reactions. In the conversion of methane into methyl chloride, two bonds are broken, CH3 -H and CI Cl, consuming 104 f 58, or a total of 162 kcal/mole. At the same time two new bonds are formed, CH3 - Cl and H Cl, liberating 84 f 103, or a total of 187 kcal/mole. The result is the liberation of 25 kcal of heat for every mole of CH3-H + Cl -CI > CH3-C! + H-C1 104 58 84 103. 162 187 A// = -25 kcal methane that is converted into methyl chloride; this is, then, an exothermic reaction. (This calculation, we note, does not depend on our knowing the mechanism of the reaction.) When heat is liberated, the heat content (enthalpy), //, of the molecules themselves must decrease; the change in heat content, A//, is therefore given a negative* sign. (In the case of an endothermic reaction, where heat is absorbed, the increase in heat content of the molecules is indicated by a positive A//.)

Energy of activation

• To see what actually happens during a chemical reaction, let us look more closely at a specific example, the attack of chlorine atoms on methane: Cl- -I- CH3 H > H Cl 4- CH3 - \H = + 1 kcal E^ 4 kcal (104) (103) This reaction is comparatively simple: it occurs in the gas phase, and is thus not complicated by the presence of a solvent; it involves the interaction of single atom and the simplest of organic molecules. Yet from it we can learn certain principles that apply to any reaction. Just what must happen if this reaction is to occur? First of all, a chlorine atom and a methane molecule must collide. Since chemical forces are of extremely short range, a hydrogen-chlorine bond can form only when the atoms are in close contact. Next, to be effective, the collision must provide a certain minimum amount of energy. Formation of the H Cl bond liberates 103 kcal/mole: breaking CHj H bond requires 104 kcal/mole. We might have expected that only 1 kcal/ mole additional energy would be needed for reaction to occur; however, this is not so. Bond-breaking and bond-making evidently are not perfectly synchronized, and the energy liberated by the one process is not completely available for the other. Experiment has shown that if reaction is to occur, an additional 4 kcal/mole of energy must be supplied. The minimum amount of energy that must be provided by a collision for reaction to occur is called the energy of activation, "acl . 

• Its source is the kinetic energy of the moving particles. Most collisions provide less than this minimum quantity and are fruitless, the original particles simply bouncing apart. Only solid collisions between particles one or both of which are moving unusually fast are energetic enough to bring about reaction. In the present example, at 275, only about one collision in 40 is sufficiently energetic. Finally, in addition to being sufficiently energetic, the collisions must occur when the particles are properly oriented. 

• At the instant of collision, the methane molecule must be turned in such a way as to present a hydrogen atom to the full force of the impact. In the present example, only about one collision in eight is properly oriented. In genera], then, a chemical reaction requires collisions of sufficient energy (act) and of Proper orientation. 

• There is an energy of activation for nearly every reaction where bonds are broken, even for exothermic reactions, in which bombmaking liberates more energy than is consumed by bond breaking. The attack of bromine atoms on methane is more highly endothermic, with a A// of +16 kcal. '* Br- + CH3 H > H Br + CH3 - A/f = +16 Kocaj Exact = 18 kcal *< (104) (88) Rr liking the CH3 H bond, as before, requires 104 kcal/mole, of which only 88 kcal is provided by formation of the H Br bond. It is evident that, even if this 88 kcal were completely available for bond-breaking, at least an additional 16 kcal/mole would have to be supplied by the collision. In other words, the act of an endothermic reaction must be at least as large as the A//. As is generally true, the &ct of the present reaction (18 kcal) is actually somewhat larger than the A//.

Progress of reaction: energy changes

• These energy relationships can be seen more clearly in diagrams like Figs. 2.2 and 2.3. Progress of reaction is represented by horizontal movement from reactants on the left to products on the right. Potential energy (that is, all energy except kinetic) at any stage of reaction is indicated by the height of the curve. Let us follow the course of reaction in Fig. 2.2. We start in a potential energy valley with a methane molecule and a chlorine atom. These particles are moving, and hence possess kinetic energy in addition to the potential energy shown. The exact amount of kinetic energy varies with the particular pair of particles, since some move faster than others. They collide, and kinetic energy is converted into potential energy. With this increase in potential energy, reaction begins, and we move up the energy hill. If enough kinetic energy is converted, we reach the top of the hill and start down the far.

• An energy diagram of the sort shown in Figs. 2.2 and 2.3 is particularly useful because it tells us not only about the reaction we are considering, but also about the reverse reaction. Let us move from right to left in Fig. 2.2, for example. We see that the reaction CH3 - + H-C1 > CH3 H + Cl- A# = -1, aot = 3 (103) (104) has an energy of activation of 3 kcal, since in this case we climb the hill from the higher valley. 'Fuscous, of course, an exothermic reaction with a A// of 1 kcal. In the same way we can see from Fig. 2.3 that the reaction CH3. + H Br > CH3^~H + Br A// =-16, ;<* = 2 (88) (104) has an energy of activation of 2 kcal and is exothermic with a A// of - 16 kcal. (We notice that, even though exothermic, these last two reactions have energies of activation.) 

• If there is no hill to climb in going from chlorine atoms to a chlorine molecule, but simply a slope to descend, the cleavage of a chlorine molecule must involve simply the ascent of a slope as shown in Fig. 2.4. The act for the cleavage of a chlorine molecule, then, must equal the A//, that is, 58 kcal. This equality of ^.t and A/7 is believed to hold generally' for reactions in which molecules dissociate into radicals.

Rate of reaction

• A chemical reaction is the result of collisions of sufficient energy and proper orientation. The rate of reaction, therefore, must be the rate at which these effective collisions occur, the number of effective collisions, let us say, that occur during each second within each cc of reaction space. We can then express the rate as the product of three factors.

• At a given* temperature the molecules of a particular compound have an average velocity and hence an average kinetic energy that is characteristic of this system; in fact, the temperature is a measure of this average kinetic energy. But the individual n,) locule's do not all travel with the same velocity, some moving faster than the average and some slower. The distribution of velocities is shown in Fig. 2,5 by the familiar bell-shaped curve that describes the distribution among individuals of so many qualities, for example, height, intelligence, income, or even Jefe expectancy. The number of molecules with a particular velocity is greatest.

• The exact relationship between energy of activation and fraction of collisions with that energy is: e~ E*ct/RT = fraction of collisions with energy greater than where e = 2.7 1 8 (base of natural logarithms) R = 1.986 (gas constant) T = absolute temperature. Using P for the probability factor and Z for the collision frequency, we arrive at the rate equation: rate = This exponential /relationship is important to us in that it indicates that a small difference in act has a large effect on the fraction of sufficiently energetic collisions, and hence on the rate of reaction. For example, at 275, out of every million collisions, 10,000 provide sufficient energy if act = 5 kcal, 100 provide sufficient energy if act = 10 kcal, and only one provides sufficient energy if act = 15 kcal. This means that (all other things being equal) a reaction with acyl = 5 kcal will go 100 times as fast as one with fact = 10 kcal, and 10,000 times as fast as one with apt = 15.

• We have examined the factors that determine rate of reaction. What we have learned may be used in many ways. To speed up a particular reaction, for example, we know that we might raise the temperature, or increase the concentration of reactants, or even (in ways that we shall take up later) lower the Akl . Of immediate interest, however, is the matter of relative reactivities. Let us see, therefore, how our knowledge of reaction rates can help us to account for the fact that one reaction proceeds faster than another, even though conditions for the two reactions are identical.

Relative rates of reaction

• We have seen that the rate of a reaction can be expressed as a product of three factors: rate = collision frequency x energy factor x probability factor Two reactions could proceed at different rates because of differences in any or all these factors. To account for a difference in rate, we must first see in which of these factors the difference lice's.

• As an example, let us compare the reactivities of chlorine and bromine atoms toward methane; that is, let us compare the rates, under the same conditions, of the two reactions: Cl- + CH3-H > H-C1 + CHr A// = +1, act = 4 Br . 4- CH3 H > H-Br + CH3 - A# = +16, ac t = 18.

• As we encounter, again and again, differences in reactivity, we shall in general attribute them to differences in "act ; in many cases we shall be able to account for these differences in act on the basis of differences in molecular structure. // must be understood that we are justified in doing this only when the reactions being compared are so closely related that differences in collision frequency and in probability factor are comparatively insignificant.

Relative reactivities of halogens toward methane

• With this background, let us return to the reaction between methane and the various halogens, and see if we can account for the order of reactivity given before, p2 > C\2 > Br2 > 12 , and in particular for the fact that iodine does not react at all. From the table of bond dissociation energies (Table -1 .2, p. 21) we can calculate for each of the four halogens the AA/ for each of the three steps of halogenation. Since act has been measured for only a. few of these reactions, let us see what tentative conclusions we can reach using only A//.

• (1) X2 > 2X- Atf=+38 +58 +46 +36 (2) X- + CH4 > HX + CHr -32 +1 +16 +33 (3) CH3 - + X2 > CH3X + X- -70 -26 -24 -20.

• For the endothermic reaction of an iodine atom with methane, EMt can be no less than 33 kcal, and is probably somewhat larger. Even for this minimum value of 33 kcal, an iodine atom must collide with an enormous number of methane molecules (1012 or a million million at 275) before reaction is likely to occur. Virtually no iodine atoms last this long, but instead recombine to form iodine molecules; the reaction therefore proceeds at a negligible rate. Iodine atoms are easy to form; it is their inability to abstract hydrogen from methane that prevents iodination from occurring.

• Values of EACt for step (2), we notice, parallel the values ofA//. Since the same bond, CH3 -H, is being broken in every case, the differences in A// reflect differences in bond dissociation energy among the various hydrogen-halogen bonds. Ultimately, it appears, the reactivity of a halogen toward methane depends upon the strength of the bond which that halogen forms with hydrogen. One further point requires clarification. We have said that an "act of 33 kcal is too great for the reaction between iodine atoms and methane to proceed at a significant rate; yet the initial step in each of these halogenations requires an even.

Structure of the methyl radical, sp 2 Hybridization

• We have spent a good part of this chapter discussing the formation and reactions of the methyl free radical CH3 - . Just what is this molecule like? What is its shape? How are the electrons distributed and, in particular, where is the odd electron? These are important questions, for the answers apply not only to this simple radical but to any free radical, however complicated, that we shall encounter. The shape, naturally, underlies the three-dimensional chemistry the stereochemistry of free radicals. The location of the odd electron is intimately involved with the stabilization of free radicals by substituent groups. As we did when we "made" methane (Sec. 1.11), let us start with the electronic configuration of necessary orbitals: three strongly directed sp 2 orbitals which, as we saw before, lie in a plane that includes the carbon nucleus, and are directed to the corners of an equilateral triangle.

Transition state

• Clearly, the concept of act is to be our key to the understanding of chemical reactivity. To make it useful, we need a further concept: transition state, A chemical reaction is presumably a continuous process involving a gradual transition from reactants to products. It has been found extremely helpful, however, to consider the arrangement of atoms at an intermediate stage of reaction as though it were an actual molecule. This intermediate structure is called the transition state; its energy content corresponds to the top of the energy hill .

• The transition state concept is useful for this reason: we can analyze the structure of the transition state very much as though it were a molecule, and attempt to estimate its stability. Any factor that stabilizes the transition state relative to the reactants tends to lower the energy of activation; that is to say, any factor that lowers the top of the energy hill more than it lowers the reactant valley reduces the net height we must climb during reaction. Transition state stability will be the basis whether explicit or implicit of almost every discussion of reactivity in this book. But the transition state is only a fleeting arrangement of atoms which, by its very nature lying at the top of an energy hill cannot be isolated and examined. How can we possibly know anything about its structure? Well, let us take as an example the transition state for the abstraction of hydrogen from methane by a halogen atom, and see where a little thinking will lead us. To start with, we can certainly say this: the carbon-hydrogen bond is stretched but not entirely broken, and the hydrogen-halogen bond has started to form but is not yet complete. This condition could be represented as H f H | 18- 8- H-C-H + -X > H-C--H-X A I A . Reactants Transition state H H-C- + H-X Products where the dashed lines indicate partly broken or partly formed. 

Reactivity and development of the transition state

• For the abstraction of hydrogen from methane by a halogen atom, we have seen that the transition state differs from the reactants and this difference is, of course, what we are looking for chiefly in being like the products. This is generally true for reactions in which free radicals (or, for that matter, carbonium ions or carbanions) are formed.

• But just how much does this particular transition state resemble the products? How far have bond-breaking and bond-making gone? How flat has the methyl group become, and to what extent does it carry the odd electron?

• Surprisingly, we can answer even questions like these, at least in a relative way. In a set of similar reactions, the higher the E&ct , the later the transition state is reached in the reaction process. Of the theoretical considerations underlying this postulate, we shall mention only this: the difference in electronic distribution that we call a difference in structure corresponds to a difference in energy; the greater the difference in structure, the greater the difference in energy. If ,& is high, the transition state differs greatly from the reactants in energy and, presumably, also in electronic structure; if E^ is low, the transition state differs little from the reactants in energy and, presumably, also in electronic structure.

• Practically, this postulate has been found extremely useful in the interpretation of experimental results; among other things, as we shall see, it enables us to account for the relationship between reactivity and selectivity.

• Abstraction of hydrogen by the less reactive bromine atom, in contrast, has a very high act . The transition state is reached only after reaction is well along toward completion and when the carbon-hydrogen bond is more nearly broken. The geometry and electron distribution has begun to approach that of the products, and carbon may well be almost trigonal. The methyl group has developed much free-radical character.

Molecular formula: its fundamental importance

• In this chapter we have been concerned with the structure of methane: the way in which atoms are put together to form a molecule of methane. But first we had to know what kinds of atoms these are and how many of them make up the molecule; we had to know that methane is CH4. Before we can assign a structural formula to a compound, we must first know its molecular formula^ Much of the chapter has been spent in discussing the substitution of chlorine for the hydrogen of methane. But first we had to know that there is substitution, that each step of the reaction yields a product that contains one less hydrogen atom and one more chlorine atom than the reactant; we had to know that CH4 is converted successively into CH3C1, CH2C12, CHC13, and CC14. Before we can discuss the reactions of an organic compound, we must first know the molecular formulas of the products.

• Tet us review a little of what we know about the assigning of a molecular formula to a compound. We must carry out: 

• (a) a qualitative elemental analysis, to find out what kinds of atoms are present in the molecule; 

• (b) a quantitative elemental analysis, to find out the relative numbers of the different kinds of atoms, that is, to determine the empirical formula', 

• (c) a molecular weight determination, which (combined with the empirical formula) shows the actual numbers of the different kinds of atoms, that is, gives us the molecular formula^ Most of this should be familiar to the student from previous courses in chemistry. What we shall concentrate on here will be the application of these principles to organic analysis.

Qualitative elemental analysis

• The presence of carbon or hydrogen in a compound is detected by combustion: heating with copper oxide, which converts carbon into carbon dioxide and hydrogen into water. (Problem: How could each of these products be identified?)

• Covalently bonded halogen, nitrogen, and sulfur must be converted into inorganic ions, which can then be detected in already familiar ways* This conversion is accomplished in either of two ways: (a) through sodium fusion, treatment with hot molten sodium metal; (C,H,X, N,S) + Na -^-> NOx~ -I- Nan- + Naans--Na+ or (b> through Schoninger oxidation by oxygen gas. (C,H,X,N,S) + O2 ^^> Nox- + Na+NO2 - + Nafso- -Na

• By these methods, we could show, for example, that.methane contains carbon and hydrogen, or that methyl chloride contains carbon, hydrogen, and chlorine. Further tests would show the absence of any other element in these compounds, except possibly oxygen, for which there is no simple chemical test; presence or absence of oxygen would be shown by a quantitative analysis. 

Quantitative elemental analysis: carbon, hydrogen, and halogen

• Knowing what elements make up a compound, we must next determine the proportions in which they are present. To do this, we carry out very much the same analysis as before, only this time on a quantitative basis. To find out the relative amounts of carbon and hydrogen in methane, for example, we would completely oxidize a measured amount of methane and weigh the carbon dioxide and water formed. In a quantitative combustion, a weighed sample of the organic compound is passed through a combustion train: a tube packed with copper oxide heated to 600-800, followed by a tube containing a drying agent (usually Dehydrate, magnesium perchlorate) and a tube containing a strong base (usually Azcarate, sodium hydroxide on asbestos). The water formed is absorbed by the drying agent, and the carbon dioxide is absorbed by the base; the increase in weight of each tube gives the weight of product formed. For example, we might find that a sample of methane weighing 9.67 mg produced 26.53 mg of CO2 and 21.56 mg of H2O. Now, only the fraction C/CO2 = 12.01/44.01 of the carbon dioxide is carbon, and only the fraction 2H/H2O = 2.016/18.02 of the water is hydrogen. Therefore wt. C 26.53 x 12.01/44.01 wt. H - 21.56 x 2.016/18.02. 

Empirical formula

• Knowing the percentage composition of a compound, we can now calculate the empirical formula: the simplest formula that shows the relative numbers of the different kinds of atoms in a molecule. For example, in 100 g (taken for convenience) of methane there are 74.9 g of carbon and 24.9 g of hydrogen, according to our quantitative analysis. Dividing each quantity by*the proper atomic weight gives the number of gram-aims of each element.

• Calculate the percentage composition and then the empirical formula for each of the following compounds: (a) Combustion of a 3.02-mg sample of a compound gave 8.86 mg of carbon dioxide and 5.43 mg of water, (b) Combustion of an 8.23-mg sample of a compound gave 9.62 mg of carbon dioxide and 3.94 mg of water. Analysis of a 5.32-mg sample of the same compound by the Carius method gave 13.49 mg of silver chloride. 

Molecular weights. Molecular 

• At this stage we know what kinds of atoms make up the molecule we are studying, and in what ratio they are present. This knowledge is summarized in the empirical formula. But this is not enough. On the basis of just the empirical formula, a molecule of methane, for example, might contain one carbon and four hydrogens, or two carbons and eight hydrogens, or any multiple of CH4. We still have to find the molecular formula: the formula that shows the actual number of each kind of atom in a molecule. To find the molecular formula, we must determine the molecular weight: today, almost certainly by mass spectrometry, which gives an exact value (Sec. 13.2). Ethane, for example, has an empirical formula of CH3. A molecular weight of 30 is found, indicating that, of the possible molecular formulas, C2H6 must be the correct.