ORGANIC CHEMISTRY 307 CHAPTER 3 LECTURE NOTES R. Boikess II. Principles of Organic Reactions 1. Chemical reactions are the result of bond breaking and bond making. a. Most (but not all) bond making and bond breaking tends to be associated with a functional group. b. Two types of bond breaking (oversimplified) i. Heterolytic, most common. Shared pair winds up with one of the atoms. This atom will therefore have an electron excess (and often a negative charge) and the other one will have an electron deficiency (and often a positive charge). Example HCl or ROH. The more EN atom gets the pair. These reactions tend to take place in solution. ii. Homolytic, less common, but simpler. The two atoms of the bond split the shared pair, each one gets its electron back. No charges generated. Example HCl and Cl 2. These reactions tend to take place in the gas phase or in nonpolar solvents. So complications due to solvation and other intermolecular interactions tend to be minimized. c. Two types of bond making (oversimplified) i. Heterolytic, Much more important. One of the atoms going to form the bond comes with an available electron pair; the other comes with an electron deficiency or incomplete octet that permits it to accept the pair. Example NH 3 and H + or AlCl 3 and Cl -. This bond forming reaction is a Lewis Acid-Lewis Base reaction. ii. Homolytic, each atom brings an electron (note that each atom must therefore have an odd number of valence shell electrons. 2. Understand and systematize reactions by keeping track of the electrons that are involved in bond breaking and making. We 1
show what happens to them using curved arrows, in more or less the same way we use curved arrows to work out contributing resonance structures. (V&S Section 1.5). Being able to use curved arrows to show reaction mechanisms and contributing structures is one of the key skills you must master in order to succeed in this course. An arrow with one hook tracks one electron, with two hooks tracks an electron pair. The direction of the arrow is always the one in which the electrons are moving. Discuss what the arrows mean in detail. Simple examples: homolytic cleavage of a Cl-Cl bond. Heterolytic cleavage of H-Cl bond and formation of an O-H bond with water in a typical acid base reaction. Because this is an essential skill to be mastered you should expend substantial time and effort in practicing it. 3. More about understanding reactions. Kinetics and Thermodynamics, (Review from Gen Chem) Key Concepts a. Free Energy criterion for spontaneity and stability i. ΔG = ΔH TΔS discuss H (standard heat of formation and bond energies) and S (microstates, disorder) ii. ΔGº = -RTlnK b. Kinetics: time and reactivity 2
i. rate laws and mechanistic information, rate determining step ii. Arrhenius and barriers c. Tie kinetics and thermo together with a simple graphical picture called a reaction coordinate diagram. Very useful (See Figure 2.1 in Vollhardt) even if not very precise. i. y-axis, choice of what to plot, we can do free energy (usually, but not always approximated by enthalpy) or potential energy ii. x-axis reaction coordinate iii. Examples: one step exothermic and endothermic. Two steps to show an intermediate. iv Review the material on reaction profiles in HP Sections 13.7-13.10. Being able to draw and interpret hese kinds of diagrams is another important skill. d. Transition state theory. Discuss meaning and importance. i. Analysis in terms of ΔH # (partial bond making and breaking) and ΔS # (relative disorganization) III Reactions of Alkanes. A. Homolytic Reactions Because they do not have functional groups, alkanes almost always react through homolytic mechanisms. Such reactions begin with a homolytic cleavage of a bond. Homolytic cleavage of a bond forms odd electron species called free radicals. Free radicals are very reactive, relatively unstable species, and once formed tend to lead to a variety of products. Rule: High reactivity = low selectivity. (You may think of various real life analogies for this rule). Thus these kinds of reactions are generally (with a number of exceptions) not a good way to prepare a particular product. Homolytic cleavage of a bond requires the input of energy. The energy is generally provided in the form of heat. For substances such as Cl 2 or Br 2, that are colored, the energy can also be provided as light of a wavelength that can be absorbed by the substance. It is better to use light when possible because it can be much more targeted, giving energy only to a species that 3
can absorb that wavelength of light. Since alkanes are always colorless, visible light cannot be used to cleave alkane bonds. B. Bond Strength The energy (actually heat) required to break a bond is equal in magnitude to the energy released when the bond formed in the first place. This energy is called the bond dissociation energy, DH (D for dissociation, Hº for standard enthalpy change). The measurement of DH is a very good way to measure bond strength. So even though free radical reactions are usually not good preparative reactions, they are of interest because they provide very good information about bond strength and structure. When we make these measurements we find a very clear correlation between bond strength and structure in alkanes. The more H s that are bonded to a given carbon, the stronger is each of those C-H bonds. So the C-H bond in methane, CH 4, (4 H s/c) is stronger than the C-H bond in ethane, CH 3 CH 3, (3 H s/c). In propane or other n-alkanes there are two types of C-H bonds: one to the C s at the ends of the chain that have 3 H s attached and one to C s in the middle of the chain that have two H s attached. As our rule predicts, the C-H bonds on the CH 2 groups are weaker than those of the CH 3 groups. Finally, when there is only one H on a C, as on C-2 of isobutane the bond is still weaker. So the trend in C-H bond strength is CH 4 > 1 > 2 > 3. C. Radical Stability How do we explain this trend? Consider isobutane, which has 1, 2, and 3 C-H bonds. When we cleave any one of these bonds we start with isobutane, but form a different free radical in each case. (In other words in each case we have the same starting material but a different product Therefore the relative DH values must reflect the relative stabilities of the free radicals. The more stable the free radical, the easier it is to break a C-H bond to form it; that is the lower is DH. So the order of stability of free radicals is 3 > 2 > 1 >. CH 3. [Figure 3.1] How do we explain this order of stability? Why is a free radical relatively unstable? Because it is electron deficient; it has an incomplete octet. How do we stabilize an electron deficient species? By feeding electron density 4
toward the center of electron deficiency from elsewhere in the species thereby distributing the electron deficiency over a larger area. The way in which electron density is fed toward the C with the incomplete octet (the one with the dot) is called hyperconjugation. We picture the C with the dot as sp 2 hybridized with the single electron in an unhybridized p- orbital.[figure 3.2] This unhybridized p-orbital can overlap with the orbitals of σ-bonds to adjacent C atoms. The more adjacent C s the more hyperconjugation and the more the free radical is stabilized. [Figure 3.3] This trend in radical stability also controls the way in which C-C bonds (which are weaker than C-H bonds) break. When an alkane is heated, the C- C bonds will cleave before the C-H bonds because the C-C bonds are weaker. As a result heating alkanes gives mixtures of products with different carbon skeletons. Such reactions are not useful if we want to make a specific compound or a few specific compounds. They can be useful in petroleum chemistry, however. Remember the two important generalities we have presented. 1. High Reactivity = Low Selectivity and vice versa 2. Alkyl groups feed electrons toward electron deficient centers. D. Halogenation of Alkanes In organic chemistry, we are most interested in reactions that can be used to prepare particular compounds of interest or that proceed in relatively predictable ways. As we have mentioned the formation of free radicals generally requires vigorous conditions and once formed the radicals themselves are highly reactive and thus not very selective. Therefore free radical reactions generally are not useful or instructive. But there are exceptions. We shall focus on the free radical chlorination and bromination of alkanes, which is instructive and sometimes useful. 1. Mechanism (the step by step description of how a reaction takes place) Let s look at how these reactions proceed and what we can use the reactions for. An understanding of the mechanism will help us understand how the reactions can be used. They proceed through a three part mechanism. 5
a. Initiation. A free radical reaction requires free radicals, species with an odd number of electrons in order to proceed. One of the most common ways to get free radicals is via a homolytic bond cleavage. So energy input is required. Both Cl 2 and Br 2 are colored, which means they absorb visible light. So we can use light (symbol hυ ) under relatively mild conditions to add energy and form free radicals. Of course we can always use heat instead, but light is more efficient. The step in which free radicals are formed is called initiation. Every reaction that we will encounter that proceeds by a free radical mechanism must have an initiation step. Cl 2 2Cl. Notice that the odd electron of a free radical is always shown and is represented by a dot. b. Propagation Once formed, a highly reactive free radical seeks to form a bond with anything that can bring an electron. What is most available are the H atoms bonded to the carbons of the alkane. We can represent the alkane as RH and the reaction as RH + Cl. R. + H-Cl In and of itself this reaction doesn t get us anyplace. It is not very favorable energetically, nor is it very unfavorable. We make an H-Cl bond and break a C-H bond. These bonds are of comparable strength and the two free radicals are of comparable instability. Table 3.1 lists the bond energies that allow you to make these assessments for many reactions. Refer to it as needed. (Remember ΔH = bonds broken bonds made) But the alkyl radical now seeks a species that can bring an electron. If it finds a Cl 2 then the following reaction takes place R. + Cl-Cl RCl + Cl. This reaction is relatively exothermic because an R-Cl bond (made) is much stronger than a Cl-Cl bond (broken). It has two key features. The 6
product of the reaction is actually a new compound. The alkane has been transformed into an alkyl halide, The other product of the reaction is the Cl., which started this two step sequence. This two step sequence called propagation will keep repeating as long as there are Cl 2 and C-H bonds remaining. Unless some other reaction can take place to get rid of the free radicals. It is because of this repeat that the reaction is called a free radical chain mechanism. Notice that the sum of the two propagation steps is always the overall stoichiometry of the reaction: RH + Cl 2 RCl + HCl c. Termination A reaction that destroys free radicals is called a termination reaction because it ends the chain. Any reaction between two free radicals is a termination. For example R. + Cl. or R. + R. or Cl. + Cl. are all termination reactions. They are all very rapid and very exothermic, yet we are still able to carry out free radical reactions where there are millions of propagations before a termination takes place. The reason is because the concentration of a free radical is always very low. It is so reactive that it grabs the first reasonably favorable thing it can find rather than waiting around for something even better (another free radical). Again we see the high reactivity-low selectivity rule. Here is an animated mechanism of the chlorination of methane. You can also view it on the book s web site (bcs.whfreeman.com/vollhardtschore5e). 2. Scope and Utility Questions: a. What about the other halogens? A bond energy analysis of the propagation steps helps rationalize the behavior of the other halogens. (You can find these bond energies in Table 3.1) i. With F 2, the first propagation step is very exothermic because the H-F bond is a very strong one. As a result reactions of alkanes with F 2 are very vigorous, even explosive, not very selective, and not very useful 7
unless selectivity is not an issue and we take special care to control conditions. ii. With I 2 the first propagation step is quite endothermic (why?) and therefore a chain mechanism cannot set up. iii. With Br 2 the first step is endothermic, but not so endothermic as to prevent chains from setting up. Because Br. is not so reactive in the first propagation step, bromination of alkanes is more selective than chlorination under corresponding conditions. b. What can we expect and how can we control selectivity? There are two issues: i. the number of halogens in the product. For example in the simplest case the reaction of CH 4 with Cl 2 there are four possible products, methyl chloride, CH 3 Cl, methylene chloride, CH 2 Cl 2, chloroform, CHCl 3, and carbon tetrachloride, CCl 4. To some extent we can control the product by adjusting the relative amounts of alkane and halogen. A large excess of alkane tends to favor the monohalogenated product. A large excess of halogen tends to give the perhalogenated product. It is better to make the partially halogenated products by some other method, unless, as in this case the starting materials are very cheap. Note that in compounds with only one type of H such as methane, or ethane, or neopentane or the unsubstituted cycloalkanes this is the only issue. ii. the number of different monohalogenated products that form from alkanes with more than one kind of hydrogen. For example propane can form n-propyl chloride and isopropyl chloride. For a given alkane, there are two factors that determine the mixture of monohalogenated products under a given set of conditions. One is the relative reactivity of the different hydrogens. Remember those that form 8
the more stable radicals have weaker bonds and are also more reactive (more easily abstracted by the X..) The other factor is the number of each type of H, the statistics. For example the two kinds of H in propane are in the ratio of 6:2. So if they had the same reactivity (which they do not) the 1 H would be 3 times more likely to react. The exact quantitative relative reactivities of the H s depend on several factors in addition to whether they are 1, 2, or 3. First, as we have already seen the selectivity is greater for Br 2 than for Cl 2. Second the reaction conditions have an important effect on relative reactivity. Milder conditions (lower temperature) enhance selectivity. Therefore we can find the relevant relative reactivities only by measuring them under the given conditions. A rough rule of thumb is that for Cl 2 at room temperature (25 C) the reactivity ratio of 3 :2 :1 hydrogens is 5:4:1. We must be given the relative reactivities, but of course we can determine the statistics by just looking at the structure. Given the relative reactivities we can calculate the relative yields of the various products. This is a simple calculation that you must be able to do. Se Exercise 3.7 in V&S. A simple formula is # of H s x relative reactivity/total # of H s = yield. (can be converted to percent by multiplying by 100%) E. Interpretation of Thermochemical Relationships Energy diagrams such as those in Figures 3.1 and 3.12 are very useful in helping us to understand and analyze relative stabilities as well as the relationship between structure and stability. Later on we will use this approach to help us understand resonance energy and strain energy. So you should learn how to construct and interpret these kinds of diagrams. a. They are not really graphs, only the y-axis has meaning. It shows increasing energy (or enthalpy). Remember that high energy or enthalpy is bad and low is good. So lower on the page is more stable and higher on the page is less stable 9
b. Ideally, we want to compare processes that either start or finish with the same substances. i. For example in comparing radical stabilities in Fig 3.1 we start with the same alkane and just abstract different H atoms. ii. In comparing isomeric alkane stabilities using heats of combustion in Figure 3.12, we finish with the same amounts of CO 2 and H 2 O c. When looking at endothermic reactions, the larger the energy change, the less stable the product as in Figure 3.1 when the starting material is the same. Or the more stable the starting material when the product is the same. d. When looking at exothermic processes, the larger the energy change (absolute value), the less stable the starting material as in Figure 3.12 when the product is the same. Or the more stable the product when the starting material is the same. e. As we will see, we can also use these kinds of diagrams to assess the energy contribution of such things as strain, which is destabilizing or resonance, which is stabilizing. The approach is to compare a measured value of some property of a substance, such as heat of combustion, with a calculated value of the property, using just bond energies. (HP 9.10) Remember bond energies are given as positive numbers so ΔH of reaction = bonds broken bonds made. In the case of a strained substance, the measured value of the enthalpy of an exothermic process will be more negative than the calculated value. The difference between the two values is called the strain energy. In the case of resonance the measured value of the enthalpy of an exothermic process will be less negative than the calculated value. The difference between the two values is called the resonance energy. 10