Chirality in Nature and Chiral Environments Although the different enantiomers of a chiral molecule have the same physical properties, they usually have different biological properties. For example, a change in chirality can affect the biological properties of many drugs, such as fluoxetine, a heavily prescribed medication sold under the trade name Prozac. Racemic fluoxetine is an effective antidepressant but has no activity against migraine. The pure S enantiomer, however, works remarkably well in preventing migraine. Other examples of how chirality affects biological properties are given in the Chapter 5 Chemistry Matters at the end of this chapter. Why do different enantiomers have different biological properties? To have a biological effect, a substance typically must fit into an appropriate receptor that has a complementary shape. But because biological receptors are chiral, only one enantiomer of a chiral substrate can fit, just as only a right hand can fit into a right-handed glove. The mirror-image enantiomer will be a misfit, like a left hand in a right-handed glove. A representation of the interaction between a chiral molecule and a chiral biological receptor is shown in : one enantiomer fits the receptor perfectly, but the other does not. Interaction of a chiral object with a chiral receptor. A left hand interacts with a chiral object, much as a biological receptor interacts with a chiral molecule. (a) One enantiomer fits into the hand perfectly: green thumb , red palm , and gray pinkie finger , with the blue substituent exposed . (b) The other enantiomer, however, can’t fit into the hand. When the green thumb and gray pinkie finger interact appropriately, the palm holds a blue substituent rather than a red one, with the red substituent exposed . The hand-in-glove fit of a chiral substrate into a chiral receptor is relatively straightforward, but it’s less obvious how a prochiral substrate can undergo a selective reaction. Take the reaction of ethanol with NAD + catalyzed by yeast alcohol dehydrogenase. As we saw at the end of , this reaction occurs with exclusive removal of the pro- R hydrogen from ethanol and with addition only to the Re face of the NAD + carbon. We can understand this result by imagining that the chiral enzyme receptor again has three binding sites, as in . When green and gray substituents of a prochiral substrate are held appropriately, however, only one of the two red substituents—say, the pro- S one—is also held while the other, pro- R, substituent is exposed for reaction. We describe the situation by saying that the receptor provides a chiral environment for the substrate. In the absence of a chiral environment, the two red substituents are chemically identical, but in the presence of a chiral environment, they are chemically distinctive ( a). The situation is similar to what happens when you pick up a coffee mug. By itself, the mug has a plane of symmetry and is achiral. When you pick up the mug, however, your hand provides a chiral environment so that one side becomes much more accessible and easier to drink from than the other ( b). (a) When a prochiral molecule is held in a chiral environment, the two seemingly identical substituents are distinguishable. (b) Similarly, when an achiral coffee mug is held in the chiral environment of your hand, it’s much easier to drink from one side than the other because the two sides of the mug are now distinguishable. Chiral Drugs The many hundreds of different pharmaceutical agents approved for use by the U.S. Food and Drug Administration come from many sources. Many drugs are isolated directly from plants or bacteria, and others are made by chemical modification of naturally occurring compounds. An estimated 33%, however, are made entirely in the laboratory and have no relatives in nature. The S enantiomer of ibuprofen soothes the aches and pains of athletic injuries much more effectively than the R enantiomer. (credit: “World Athletic Championships 2007 in Osaka” by Eckhard Pecher/Wikimedia Commons, CC BY 2.5) Those drugs that come from natural sources, either directly or after laboratory modification, are usually chiral and are generally found only as a single enantiomer rather than as a racemate. Penicillin V, for example, an antibiotic isolated from the Penicillium mold, has the 2 S ,5 R ,6 R configuration. Its enantiomer, which does not occur naturally but can be made in the laboratory, has no antibiotic activity. In contrast to drugs from natural sources, drugs that are made entirely in the laboratory are either achiral or, if chiral, are often produced and sold as racemates. Ibuprofen, for example, has one chirality center and is sold commercially under such trade names as Advil, Nuprin, and Motrin as a 50 : 50 mixture of R and S. It turns out, however, that only the S enantiomer is active as an analgesic and anti-inflammatory agent. The R enantiomer of ibuprofen is inactive, although it is slowly converted in the body to the active S form. Not only is it chemically wasteful to synthesize and administer an enantiomer that does not serve the intended purpose, many instances are now known where the presence of the “wrong” enantiomer in a racemic mixture either affects the body’s ability to utilize the “right” enantiomer or has unintended pharmacological effects of its own. The presence of ( R )-ibuprofen in the racemic mixture, for instance, slows the rate at which the S enantiomer takes effect in the body, from 12 minutes to 38 minutes. To get around this problem, pharmaceutical companies attempt to devise methods of enantioselective synthesis, which allow them to prepare only a single enantiomer rather than a racemic mixture. Viable methods have been developed for the preparation of ( S )-ibuprofen, which is marketed in Europe. We’ll look further into enantioselective synthesis in the Chapter 19 Chemistry Matters . absolute configuration achiral Cahn–Ingold–Prelog rules chiral chiral environment chirality center configuration dextrorotatory diastereomer enantiomer epimer levorotatory meso compound optically active prochiral prochirality center pro- R configuration pro- S configuration R configuration racemate Re face resolution S configuration Si face specific rotation, [ α ] D Summary In this chapter, we’ve looked at some of the causes and consequences of molecular handedness—a topic of particular importance in understanding biological chemistry. The subject can be a bit complex but is so important that it’s worthwhile spending time to become familiar with it. An object or molecule that is not superimposable on its mirror image is said to be chiral , meaning “handed.” A chiral molecule is one that does not have a plane of symmetry cutting through it so that one half is a mirror image of the other half. The most common cause of chirality in organic molecules is the presence of a tetrahedral, sp 3 -hybridized carbon atom bonded to four different groups—a so-called chirality center . Chiral compounds can exist as a pair of nonsuperimposable mirror-image stereoisomers called enantiomers . Enantiomers are identical in all physical properties except for the direction in which they rotate plane-polarized light. The stereochemical configuration of a chirality center can be specified as either R ( rectus ) or S ( sinister ) by using the Cahn–Ingold–Prelog rules . First rank the four substituents on the chiral carbon atom, and then orient the molecule so that the lowest-ranked group points directly back. If a curved arrow drawn in the direction of decreasing rank (1 → 2 → 3) for the remaining three groups is clockwise, the chirality center has the R configuration. If the direction is counterclockwise, the chirality center has the S configuration. Some molecules have more than one chirality center. Enantiomers have opposite configuration at all chirality centers, whereas diastereomers have the same configuration in at least one center but opposite configurations at the others. Epimers are diastereomers that differ in configuration at only one chirality center. A compound with n chirality centers can have a maximum of 2 n stereoisomers. A meso compound contains a chirality center but is achiral overall because it has a plane of symmetry. Racemic mixtures, or racemates , are 50 : 50 mixtures of (+) and (−) enantiomers. Racemates and individual diastereomers differ in their physical properties, such as solubility, melting point, and boiling point. A molecule is prochiral if it can be converted from achiral to chiral in a single chemical step. A prochiral sp 2 -hybridized atom has two faces, described as either Re or Si . An sp 3 -hybridized atom is a prochirality center if, by changing one of its attached atoms, a chirality center results. The atom whose replacement leads to an R chirality center is pro- R , and the atom whose replacement leads to an S chirality center is pro- S . Additional Problems Visualizing Chemistry Chirality and Optical Activity Assigning Configuration to Chirality Centers Meso Compounds Prochirality General Problems

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Why This Chapter? Many chemical reactions are like pole vaulters going over the bar. They need a big, initial push of activation energy. (credit: modification of work “UChicago Pole Vault” by Eric Guo/Flickr, CC BY 2.0) All chemical reactions, whether they take place in the laboratory or in living organisms, follow the same “rules.” Reactions in living organisms often look more complex than laboratory reactions because of the size of the biomolecules and the involvement of biological catalysts called enzymes, but the principles governing all chemical reactions are the same. To understand both organic and biological chemistry, it’s necessary to know not just what occurs but also why and how chemical reactions take place. In this chapter, we’ll start with an overview of the fundamental kinds of organic reactions, we’ll see why reactions occur, and we’ll see how reactions can be described. Once this background is out of the way, we’ll then be ready to begin studying the details of organic chemistry in future chapters. When first approached, organic chemistry might seem overwhelming. It’s not so much that any one part is difficult to understand, it’s that there are so many parts: tens of millions of compounds, dozens of functional groups, and an apparently endless number of reactions. With study, though, it becomes evident that there are only a few fundamental ideas that underlie all organic reactions. Far from being a collection of isolated facts, organic chemistry is a beautifully logical subject that is unified by a few broad themes. When these themes are understood, learning organic chemistry becomes much easier and memorization is minimized. The aim of this book is to describe the themes and clarify the patterns that unify organic chemistry in future chapters.

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Kinds of Organic Reactions Organic chemical reactions can be organized broadly in two ways—by what kinds of reactions occur and by how those reactions occur. Let’s look first at the kinds of reactions that take place. There are four general types of organic reactions: additions, eliminations, substitutions, and rearrangements . Addition reactions occur when two reactants add together to form a single product with no atoms “left over.” An example that we’ll be studying soon is the reaction of an alkene, such as ethylene, with HBr to yield an alkyl bromide. Elimination reactions are, in a sense, the opposite of addition reactions. They occur when a single reactant splits into two products, often with the formation of a small molecule such as water or HBr. An example is the acid-catalyzed reaction of an alcohol to yield water and an alkene. Substitution reactions occur when two reactants exchange parts to give two new products. An example is the reaction of an ester such as methyl acetate with water to yield a carboxylic acid plus an alcohol. Similar reactions occur in many biological pathways, including the metabolism of dietary fats. Rearrangement reactions occur when a single reactant undergoes a reorganization of bonds and atoms to yield an isomeric product. An example is the conversion of dihydroxyacetone phosphate into its constitutional isomer glyceraldehyde 3-phosphate, a step in the glycolysis pathway by which carbohydrates are metabolized. Problems

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How Organic Reactions Occur: Mechanisms Having looked at the kinds of reactions that take place, let’s now see how they occur. An overall description of how a reaction occurs is called a reaction mechanism . A mechanism describes in detail exactly what takes place at each stage of a chemical transformation—which bonds are broken and in what order, which bonds are formed and in what order, and what the relative rates are for each step. A complete mechanism must also account for all reactants used and all products formed. All chemical reactions involve bond-breaking and bond-making. When two molecules come together, react, and yield products, specific bonds in the reactant molecules are broken and specific bonds in the product molecules are formed. Fundamentally, there are two ways in which a covalent two-electron bond can break. A bond can break in an electronically unsymmetrical way so that both bonding electrons remain with one product fragment, leaving the other with a vacant orbital, or a bond can break in an electronically symmetrical way so that one electron remains with each product fragment. The unsymmetrical cleavage is said to be heterolytic , and the symmetrical cleavage is said to be homolytic . We’ll develop this point in more detail later, but note for now that the movement of two electrons in the unsymmetrical process is indicated using a full-headed curved arrow ( ), whereas the movement of one electron in the symmetrical process is indicated using a half-headed, or “fishhook,” arrow ( ). Just as there are two ways in which a bond can break, there are two ways in which a covalent two-electron bond can form. A bond can form in an electronically unsymmetrical way if both bonding electrons are donated to the new bond by one reactant, or in a symmetrical way if one electron is donated by each reactant. Processes that involve unsymmetrical bond-breaking and bond-making are called polar reactions . Polar reactions involve species that have an even number of electrons and thus have only electron pairs in their orbitals. Polar processes are by far the more common reaction type in both organic and biological chemistry, and a large part of this book is devoted to their description. Processes that involve symmetrical bond-breaking and bond-making are called radical reactions . A radical , often called a free radical , is a neutral chemical species that contains an odd number of electrons and thus has a single, unpaired electron in one of its orbitals. In addition to polar and radical reactions, there is a third, less commonly encountered process called a pericyclic reaction . Rather than explain pericyclic reactions now, though, we’ll look at them more carefully in Chapter 30.

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Polar Reactions Polar reactions occur because of the electrical attraction between positively polarized and negatively polarized centers on functional groups in molecules. To see how these reactions take place, let’s first recall the discussion of polar covalent bonds in and then look more deeply into the effects of bond polarity on organic molecules. Most organic compounds are electrically neutral; they have no net charge, either positive or negative. We saw in , however, that certain bonds within a molecule, particularly the bonds in functional groups, are polar. Bond polarity is a consequence of an unsymmetrical electron distribution in a bond and is due to the difference in electronegativity of the bonded atoms. Elements such as oxygen, nitrogen, fluorine, and chlorine are more electronegative than carbon, so a carbon atom bonded to one of these atoms has a partial positive charge ( δ +). Metals are less electronegative than carbon, so a carbon atom bonded to a metal has a partial negative charge ( δ −). Electrostatic potential maps of chloromethane and methyllithium illustrate these charge distributions, showing that the carbon atom in chloromethane is electron-poor (blue) while the carbon in methyllithium is electron-rich (red). The polarity patterns of some common functional groups are shown in . Note that carbon is always positively polarized except when bonded to a metal. Polarity Patterns in Some Common Functional Groups Compound type Functional group structure This discussion of bond polarity is oversimplified in that we’ve considered only bonds that are inherently polar due to differences in electronegativity. Polar bonds can also result from the interaction of functional groups with acids or bases. Take an alcohol such as methanol, for example. In neutral methanol, the carbon atom is somewhat electron-poor because the electronegative oxygen attracts the electrons in the C−O bond. On protonation of the methanol oxygen by an acid, however, a full positive charge on oxygen attracts the electrons in the C−O bond much more strongly and makes the carbon much more electron-poor. We’ll see numerous examples throughout this book of reactions that are catalyzed by acids because of the resultant increase in bond polarity upon protonation. Yet a further consideration is the polarizability (as opposed to polarity) of atoms in a molecule. As the electric field around a given atom changes because of changing interactions with solvent or other polar molecules nearby, the electron distribution around that atom also changes. The measure of this response to an external electrical influence is called the polarizability of the atom. Larger atoms with more loosely held electrons are more polarizable, and smaller atoms with fewer, tightly held electrons are less polarizable. Thus, sulfur is more polarizable than oxygen, and iodine is more polarizable than chlorine. The effect of this higher polarizability of sulfur and iodine is that carbon–sulfur and carbon–iodine bonds, although nonpolar according to electronegativity values ( ), nevertheless usually react as if they were polar. What does functional-group polarity mean with respect to chemical reactivity? Because unlike charges attract, the fundamental characteristic of all polar organic reactions is that electron-rich sites react with electron-poor sites. Bonds are made when an electron-rich atom donates a pair of electrons to an electron-poor atom, and bonds are broken when one atom leaves with both electrons from the former bond. As we saw in , the movement of an electron pair during a polar reaction is indicated using a curved, full-headed arrow to show where electrons move when reactant bonds are broken and product bonds are formed during the reaction. In referring to the electron-rich and electron-poor species involved in polar reactions, chemists use the words nucleophile and electrophile . A nucleophile is a substance that is “nucleus-loving.” (Remember that a nucleus is positively charged.) A nucleophile has a negatively polarized, electron-rich atom and can form a bond by donating a pair of electrons to a positively polarized, electron-poor atom. Nucleophiles can be either neutral or negatively charged; ammonia, water, hydroxide ion, and chloride ion are examples. An electrophile , by contrast, is “electron-loving.” An electrophile has a positively polarized, electron-poor atom and can form a bond by accepting a pair of electrons from a nucleophile. Electrophiles can be either neutral or positively charged. Acids (H + donors), alkyl halides, and carbonyl compounds are examples ( ). Some nucleophiles and electrophiles. Electrostatic potential maps identify the nucleophilic (negative) and electrophilic (positive) atoms. Note that neutral compounds can often react either as nucleophiles or as electrophiles, depending on the circumstances. After all, if a compound is neutral yet has an electron- rich nucleophilic site, it must also have a corresponding electron- poor electrophilic site. Water, for instance, acts as an electrophile when it donates H + but acts as a nucleophile when it donates a nonbonding pair of electrons. Similarly, a carbonyl compound acts as an electrophile when it reacts at its positively polarized carbon atom, yet acts as a nucleophile when it reacts at its negatively polarized oxygen atom. If the definitions of nucleophiles and electrophiles sound similar to those given in for Lewis acids and Lewis bases, that’s because there is indeed a correlation. Lewis bases are electron donors and behave as nucleophiles, whereas Lewis acids are electron acceptors and behave as electrophiles. Thus, much of organic chemistry is explainable in terms of acid–base reactions. The main difference is that the words acid and base are used broadly in all fields of chemistry, while the words nucleophile and electrophile are used primarily in organic chemistry when carbon bonding is involved. Identifying Electrophiles and Nucleophiles Which of the following species is likely to behave as a nucleophile and which as an electrophile? (a) NO 2 + (b) CN − (c) CH 3 NH 2 (d) (CH 3 ) 3 S + Strategy A nucleophile has an electron-rich site, either because it is negatively charged or because it has a functional group containing an atom that has a lone pair of electrons. An electrophile has an electron-poor site, either because it is positively charged or because it has a functional group containing an atom that is positively polarized. Solution (a) NO 2 + (nitronium ion) is likely to be an electrophile because it is positively charged. (b) :C≡N − (cyanide ion) is likely to be a nucleophile because it is negatively charged. (c) CH 3 NH 2 (methylamine) might be either a nucleophile or an electrophile, depending on the circumstances. The lone pair of electrons on the nitrogen atom makes methylamine a potential nucleophile, while positively polarized N−H hydrogens make methylamine a potential acid (electrophile). (d) (CH 3 ) 3 S + (trimethylsulfonium ion) is likely to be an electrophile because it is positively charged. Problems

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An Example of a Polar Reaction: Addition of HBr to Ethylene Let’s look at a typical polar process—the addition reaction of an alkene, such as ethylene, with hydrogen bromide. When ethylene is treated with HBr at room temperature, bromoethane is produced. Overall, the reaction can be formulated as The reaction is an example of a polar reaction type known as an electrophilic addition reaction and can be understood using the general ideas discussed in the previous section. Let’s begin by looking at the two reactants. What do we know about ethylene? We know from that a carbon–carbon double bond results from the orbital overlap of two sp 2 -hybridized carbon atoms. The σ part of the double bond results from sp 2 – sp 2 overlap, and the π part results from p – p overlap. What kind of chemical reactivity might we expect from a C═C bond? We know that alkanes, such as ethane, are relatively inert because all valence electrons are tied up in strong, nonpolar, C−C and C−H bonds. Furthermore, the bonding electrons in alkanes are relatively inaccessible to approaching reactants because they are sheltered in σ bonds between nuclei. The electronic situation in alkenes is quite different, however. For one thing, double bonds have a greater electron density than single bonds—four electrons in a double bond versus only two in a single bond. In addition, the electrons in the π bond are accessible to approaching reactants because they are located above and below the plane of the double bond rather than being sheltered between the nuclei ( ). As a result, the double bond is nucleophilic and the chemistry of alkenes is dominated by reactions with electrophiles. A comparison of carbon–carbon single and double bonds. A double bond is both more accessible to approaching reactants than a single bond and more electron-rich (more nucleophilic). An electrostatic potential map of ethylene indicates that the double bond is the region of highest negative charge . What about the second reactant, HBr? As a strong acid, HBr is a powerful proton (H + ) donor and electrophile. Thus, the reaction between HBr and ethylene is a typical electrophile–nucleophile combination, characteristic of all polar reactions. We’ll see more details about alkene electrophilic addition reactions shortly, but for the present we can imagine the reaction as taking place by the pathway shown in . The reaction begins when the alkene nucleophile donates a pair of electrons from its C═C bond to HBr to form a new C−H bond plus Br − , as indicated by the path of the curved arrows in the first step of . One curved arrow begins at the middle of the double bond (the source of the electron pair) and points to the hydrogen atom in HBr (the atom to which a bond will form). This arrow indicates that a new C−H bond forms using electrons from the former C═C bond. Simultaneously, a second curved arrow begins in the middle of the H−Br bond and points to the Br, indicating that the H−Br bond breaks and the electrons remain with the Br atom, giving Br − . MECHANISM The electrophilic addition reaction of ethylene and HBr. The reaction takes place in two steps, both of which involve electrophile–nucleophile interactions. When one of the alkene carbon atoms bonds to the incoming hydrogen, the other carbon atom, having lost its share of the double-bond electrons, now has only six valence electrons and is left with a positive charge. This positively charged species—a carbon-cation, or carbocation —is itself an electrophile that can accept an electron pair from nucleophilic Br − anion in a second step, forming a C−Br bond and yielding the observed addition product. Once again, a curved arrow in shows the electron-pair movement from Br − to the positively charged carbon. The electrophilic addition of HBr to ethylene is only one example of a polar process; there are many others that we’ll study in depth in later chapters. But regardless of the details of individual reactions, all polar reactions take place between an electron-poor site and an electron-rich site and involve the donation of an electron pair from a nucleophile to an electrophile. Problems

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Using Curved Arrows in Polar Reaction Mechanisms It takes practice to use curved arrows properly in reaction mechanisms, but there are a few rules and a few common patterns you should look for that will help you become more proficient: RULE 1 Electrons move from a nucleophilic source (Nu: or Nu: − ) to an electrophilic sink (E or E + ) . The nucleophilic source must have an electron pair available, usually either as a lone pair or in a multiple bond. For example: The electrophilic sink must be able to accept an electron pair, usually because it has either a positively charged atom or a positively polarized atom in a functional group. For example: RULE 2 The nucleophile can be either negatively charged or neutral. If the nucleophile is negatively charged, the atom that donates an electron pair becomes neutral. For example: If the nucleophile is neutral, the atom that donates the electron pair acquires a positive charge. For example: RULE 3 The electrophile can be either positively charged or neutral. If the electrophile is positively charged, the atom bearing that charge becomes neutral after accepting an electron pair. For example: If the electrophile is neutral, the atom that ultimately accepts the electron pair acquires a negative charge. For this to happen, however, the negative charge must be stabilized by being on an electronegative atom such as oxygen, nitrogen, or a halogen. Carbon and hydrogen do not typically stabilize a negative charge. For example: The result of Rules 2 and 3 together is that charge is conserved during the reaction. A negative charge in one of the reactants gives a negative charge in one of the products, and a positive charge in one of the reactants gives a positive charge in one of the products. RULE 4 The octet rule must be followed. That is, no second-row atom can be left with ten electrons (or four for hydrogen). If an electron pair moves to an atom that already has an octet (or two electrons for hydrogen), another electron pair must simultaneously move from that atom to maintain the octet. When two electrons move from the C═C bond of ethylene to the hydrogen atom of H 3 O + , for instance, two electrons must leave that hydrogen. This means that the H−O bond must break and the electrons must stay with the oxygen, giving neutral water. gives another example of drawing curved arrows. Using Curved Arrows in Reaction Mechanisms Add curved arrows to the following polar reaction to show the flow of electrons: Strategy Look at the reaction, and identify the bonding changes that have occurred. In this case, a C−Br bond has broken and a C−C bond has formed. The formation of the C−C bond involves donation of an electron pair from the nucleophilic carbon atom of the reactant on the left to the electrophilic carbon atom of CH 3 Br, so we draw a curved arrow originating from the lone pair on the negatively charged C atom and pointing to the C atom of CH 3 Br. At the same time that the C−C bond forms, the C−Br bond must break so that the octet rule is not violated. We therefore draw a second curved arrow from the C−Br bond to Br. The bromine is now a stable Br − ion. Solution Problems

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Radical Reactions Radical reactions are much less common than polar reactions but are nevertheless important in some industrial processes and biological pathways. We’ll look at them in more detail in Sections 10.2 and 10.3 but will briefly see how they occur at this point. A radical is highly reactive because it contains an atom with an odd number of electrons (usually seven) in its valence shell, rather than a noble-gas octet. The radical can achieve a valence-shell octet in several ways however. For instance, it might abstract an atom and one bonding electron from another reactant, leaving behind a new radical. The net result is a radical substitution reaction. Alternatively, a reactant radical might add to a double bond, taking one electron from the double bond and yielding a new radical. The net result is a radical addition reaction. An example of an industrially useful radical reaction is the reaction of chlorine with methane to yield chloromethane, which is used to manufacture the solvents dichloromethane (CH 2 Cl 2 ) and chloroform (CHCl 3 ). The process begins with irradiation of Cl 2 with ultraviolet light to break the relatively weak Cl − Cl bond of Cl 2 and produce chlorine radicals ( · Cl). Chlorine radicals then react with methane by abstracting a hydrogen atom to give HCl and a methyl radical ( · CH 3 ) that reacts further with Cl 2 to give chloromethane plus a new chlorine radical that cycles back and repeats the first step. Thus, once the sequence has started, it becomes a self-sustaining cycle of repeating steps ( a ) and ( b ), making the overall process a chain reaction. As a biological example of a radical reaction, look at the synthesis of prostaglandins , a large class of molecules found in virtually all body tissues and fluids. A number of pharmaceuticals are based on or derived from prostaglandins, including medicines that induce labor during childbirth, reduce intraocular pressure in glaucoma, control bronchial asthma, and help treat congenital heart defects. Prostaglandin biosynthesis is initiated by the abstraction of a hydrogen atom from arachidonic acid by an iron–oxygen radical, thereby generating a carbon radical in a substitution reaction. Don’t be intimidated by the size of the molecules; focus on the changes that occur in each step. (To help you do that, the unchanged part of the molecule is “ghosted,” with only the reactive part clearly visible.) Following the initial abstraction of a hydrogen atom, the carbon radical then reacts with O 2 to give an oxygen radical, which reacts with a C═C bond within the same molecule in an addition reaction. Several further transformations ultimately yield prostaglandin H 2 . Problems

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Describing a Reaction: Equilibria, Rates, and Energy Changes Every chemical reaction can go in either a forward or reverse direction. Reactants can go forward to products, and products can revert to reactants. As you may remember from your general chemistry course, the position of the resulting chemical equilibrium is expressed by an equation in which K eq , the equilibrium constant, is equal to the product concentrations multiplied together, divided by the reactant concentrations multiplied together, with each concentration raised to the power of its coefficient in the balanced equation. For the generalized reaction a A + b B ⇌ c C + d D we have K eq = [ C ] c [ D ] d [ A ] a [ B ] b The value of the equilibrium constant tells which side of the reaction arrow is energetically favored. If K eq is much larger than 1, then the product concentration term [C] c [D] d is much larger than the reactant concentration term [A] a [B] b , and the reaction proceeds as written from left to right. If K eq is near 1, appreciable amounts of both reactant and product are present at equilibrium. And if K eq is much smaller than 1, the reaction does not take place as written but instead goes in the reverse direction, from right to left. In the reaction of ethylene with HBr, for example, we can write the following equilibrium expression and determine experimentally that the equilibrium constant at room temperature is approximately 7.1 × 10 7 : Because K eq is relatively large, the reaction proceeds as written and more than 99.999 99% of the ethylene is converted into bromoethane. For practical purposes, an equilibrium constant greater than about 10 3 means that the amount of reactant left over will be barely detectable (less than 0.1%). What determines the magnitude of the equilibrium constant? For a reaction to have a favorable equilibrium constant and proceed as written, the energy of the products must be lower than the energy of the reactants. In other words, energy must be released. This situation is analogous to that of a rock poised precariously in a high-energy position near the top of a hill. When it rolls downhill, the rock releases energy until it reaches a more stable, low-energy position at the bottom. The energy change that occurs during a chemical reaction is called the Gibbs free-energy change (Δ G ) , which is equal to the free energy of the products minus the free energy of the reactants: Δ G = G products − G reactants . For a favorable reaction, Δ G has a negative value, meaning that energy is lost by the chemical system and released to the surroundings, usually as heat. Such reactions are said to be exergonic . For an unfavorable reaction, Δ G has a positive value, meaning that energy is absorbed by the chemical system from the surroundings. Such reactions are said to be endergonic . You might also recall from general chemistry that the standard free-energy change for a reaction is denoted as Δ G °, where the superscript ° means that the reaction is carried out under standard conditions, with pure substances in their most stable form at 1 atm pressure and a specified temperature, usually 298 K. For biological reactions, the standard free-energy change is denoted as Δ G °′ and refers to a reaction carried out at pH = 7.0 with solute concentrations of 1.0 M. Because the equilibrium constant, K eq , and the standard free-energy change, Δ G °, both measure whether a reaction is favorable, they are mathematically related by the equation Δ G ∘ = − R T ln K eq or K eq = e − Δ G ∘ / R T where R = 8.314 J/(K ⋅ mol) = 1 .987 cal/(K ⋅ mol) T = Kelvin temperature e = 2.718 ln K eq = natural logarithm of K eq For example, the reaction of ethylene with HBr has K eq = 7.1 × 10 7 , so Δ G ° = −44.8 kJ/mol (−10.7 kcal/mol) at 298 K: K eq = 7.1 × 10 7 and ln K eq = 18.08 Δ G ∘ = − R T ln K eq = − [ 8.314 J/(K ⋅ mol)](298 K)(18 .08) = − 44,800 J/mol= − 44.8 kJ/mol The free-energy change Δ G is made up of two terms, an enthalpy term, Δ H , and a temperature-dependent entropy term, T Δ S . Of the two terms, the enthalpy term is often larger and more dominant. Δ G ∘ = Δ H ∘ − T Δ S ∘ For the reaction of ethylene with HBr at room temperature (298 K), the approximate values are The enthalpy change (Δ H ) , also called the heat of reaction , is a measure of the change in total bonding energy during a reaction. If Δ H is negative, as in the reaction of HBr with ethylene, the products have less energy than the reactants. Thus, the products are more stable and have stronger bonds than the reactants, heat is released, and the reaction is said to be exothermic . If Δ H is positive, the products are less stable and have weaker bonds than the reactants, heat is absorbed, and the reaction is said to be endothermic . For example, if a reaction breaks reactant bonds with a total strength of 380 kJ/mol and forms product bonds with a total strength of 400 kJ/mol, then Δ H for the reaction is 380 kJ/mol - 400 kJ/mol = −20 kJ/mol and the reaction is exothermic. The entropy change (Δ S ) is a measure of the change in the amount of molecular randomness, or freedom of motion, that accompanies a reaction. For example, in an elimination reaction of the type A ⟶ B + C there is more freedom of movement and molecular randomness in the products than in the reactant because one molecule has split into two. Thus, there is a net increase in entropy during the reaction and Δ S has a positive value. On the other hand, for an addition reaction of the type A + B ⟶ C the opposite is true. Because such reactions restrict the freedom of movement of two molecules by joining them together, the product has less randomness than the reactants and Δ S has a negative value. The reaction of ethylene and HBr to yield bromoethane, which has Δ S ° = −0.132 kJ/(K · mol), is an example. describes the thermodynamic terms more fully. Explanation of Thermodynamic Quantities: Δ G ° = Δ H ° − T Δ S ° Term Name Explanation Δ G ° Gibbs free-energy change The energy difference between reactants and products. When Δ G ° is negative, the reaction is exergonic , has a favorable equilibrium constant, and can occur spontaneously. When Δ G ° is positive, the reaction is endergonic , has an unfavorable equilibrium constant, and cannot occur spontaneously. Δ H ° Enthalpy change The heat of reaction, or difference in strength between the bonds broken in a reaction and the bonds formed. When Δ H ° is negative, the reaction releases heat and is exothermic . When Δ H ° is positive, the reaction absorbs heat and is endothermic . Δ S ° Entropy change The change in molecular randomness during a reaction. When Δ S ° is negative, randomness decreases. When Δ S ° is positive, randomness increases. Knowing the value of K eq for a reaction is useful, but it’s important to realize its limitations. An equilibrium constant tells only the position of the equilibrium, or how much product is theoretically possible. It doesn’t tell the rate of reaction, or how fast the equilibrium is established. Some reactions are extremely slow even though they have favorable equilibrium constants. Gasoline is stable at room temperature, for instance, because the rate of its reaction with oxygen is slow at 298 K. Only at higher temperatures, such as contact with a lighted match, does gasoline react rapidly with oxygen and undergo complete conversion to the equilibrium products water and carbon dioxide. Rates (how fast a reaction occurs) and equilibria (how much a reaction occurs) are entirely different. Rate → Is the reaction fast or slow? Equilibrium → In what direction does the reaction proceed? Problems

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Describing a Reaction: Bond Dissociation Energies We’ve just seen that heat is released (negative Δ H ) when a bond is formed because the products are more stable and have stronger bonds than the reactants. Conversely, heat is absorbed (positive Δ H ) when a bond is broken because the products are less stable and have weaker bonds than the reactants. The amount of energy needed to break a given bond to produce two radical fragments when the molecule is in the gas phase at 25 °C is a quantity called the bond strength , or bond dissociation energy ( D ) . Each specific bond has its own characteristic strength, and extensive tables of such data are available. For example, a C−H bond in methane has a bond dissociation energy D = 439.3 kJ/mol (105.0 kcal/mol), meaning that 439.3 kJ/mol must be added to break a C−H bond of methane to give the two radical fragments ·CH 3 and ·H. Conversely, 439.3 kJ/mol of energy is released when a methyl radical and a hydrogen atom combine to form methane. lists some other bond strengths. Some Bond Dissociation Energies, D Bond D (kJ/mol) Bond D (kJ/mol) Bond D (kJ/mol) H ― H 436 (CH 3 ) 2 CH ― H 410 C 2 H 5 ― CH 3 370 H ― F 570 (CH 3 ) 2 CH ― Cl 354 (CH 3 ) 2 CH ― CH 3 369 H ― Cl 431 (CH 3 ) 2 CH ― Br 299 (CH 3 ) 3 C ― CH 3 363 H ― Br 366 (CH 3 ) 3 C ― H 400 H 2 C═CH ― CH 3 426 H ― I 298 (CH 3 ) 3 C ― Cl 352 H 2 C═CHCH 2 ― CH 3 318 Cl ― Cl 242 (CH 3 ) 3 C ― Br 293 H 2 C═CH 2 728 Br ― Br 194 (CH 3 ) 3 C ― I 227 427 I ― I 152 H 2 C═CH ― H 464 325 CH 3 ― H 439 H 2 C═CH ― Cl 396 374 CH 3 ― Cl 350 H 2 C═CHCH 2 ― H 369 HO ― H 497 CH 3 ― Br 294 H 2 C═CHCH 2 ― Cl 298 HO ― OH 211 CH 3 ― I 239 472 CH 3 O ― H 440 CH 3 ― OH 385 400 CH 3 S ― H 366 CH 3 ― NH 2 386 375 C 2 H 5 O ― H 441 C 2 H 5 ― H 421 300 352 C 2 H 5 ― Cl 352 336 CH 3 CH 2 O ― CH 3 355 C 2 H 5 ― Br 293 464 NH 2 ― H 450 C 2 H 5 ― I 233 HC≡C ― H 558 H ― CN 528 C 2 H 5 ― OH 391 CH 3 ― CH 3 377 Think again about the connection between bond strengths and chemical reactivity. In an exothermic reaction, more heat is released than is absorbed. But because making bonds in the products releases heat and breaking bonds in the reactants absorbs heat, the bonds in the products must be stronger than the bonds in the reactants. In other words, exothermic reactions are favored by products with strong bonds and by reactants with weak, easily broken bonds. Sometimes, particularly in biochemistry, reactive substances that undergo highly exothermic reactions, such as ATP (adenosine triphosphate), are referred to as “energy-rich” or “high-energy” compounds. Such a label doesn’t mean that ATP is special or different from other compounds, it only means that ATP has relatively weak bonds that require a relatively small amount of heat to break, thus leading to a larger release of heat when a strong new bond forms in a reaction. When a typical organic phosphate such as glycerol 3-phosphate reacts with water, for instance, only 9 kJ/mol of heat is released (Δ H °′ = −9 kJ/mol), but when ATP reacts with water, 30 kJ/mol of heat is released (Δ H °′ = −30 kJ/mol). The difference between the two reactions is due to the fact that the bond broken in ATP is substantially weaker than the bond broken in glycerol 3-phosphate. We’ll see the metabolic importance of this reaction in later chapters.

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Describing a Reaction: Energy Diagrams and Transition States For a reaction to take place, reactant molecules must collide and reorganization of atoms and bonds must occur. Let’s again look at the addition reaction of HBr and ethylene. As the reaction proceeds, ethylene and HBr approach each other, the ethylene π bond and the H−Br bond break, a new C−H bond forms in step 1 and a new C−Br bond forms in step 2 . To depict graphically the energy changes that occur during a reaction, chemists use energy diagrams, such as that in . The vertical axis of the diagram represents the total energy of all reactants, and the horizontal axis, called the reaction coordinate , represents the progress of the reaction from beginning to end. Let’s see how the addition of HBr to ethylene can be described in an energy diagram. An energy diagram for the first step in the reaction of ethylene with HBr. The energy difference between reactants and the transition state, Δ G ‡ , defines the reaction rate. The energy difference between reactants and carbocation product, Δ G °, defines the position of the equilibrium. At the beginning of the reaction, ethylene and HBr have the total amount of energy indicated by the reactant level on the left side of the diagram in . As the two reactants collide and reaction commences, their electron clouds repel each other, causing the energy level to rise. If the collision has occurred with enough force and proper orientation, however, the reactants continue to approach each other despite the rising repulsion until the new C−H bond starts to form. At some point, a structure of maximum energy is reached, a structure called the transition state . The transition state represents the highest-energy structure involved in this step of the reaction. It is unstable and can’t be isolated, but we can imagine it to be an activated complex of the two reactants in which both the C═C π bond and H−Br bond are partially broken and the new C−H bond is partially formed ( ). A hypothetical transition-state structure for the first step of the reaction of ethylene with HBr. The C═C π bond and H−Br bond are just beginning to break, and the C−H bond is just beginning to form. The energy difference between reactants and the transition state is called the activation energy, Δ G ‡ , and determines how rapidly the reaction occurs at a given temperature. (The double-dagger superscript, ‡ , always refers to the transition state.) A high activation energy results in a slow reaction because few collisions occur with enough energy for the reactants to reach the transition state. A low activation energy results in a rapid reaction because almost all collisions occur with enough energy for the reactants to reach the transition state. As an analogy, you might think of reactants that need enough energy to climb the activation barrier to the transition state as hikers who need enough energy to climb to the top of a mountain pass. If the pass is a high one, the hikers need a lot of energy and surmount the barrier with difficulty. If the pass is low, the hikers need less energy and reach the top easily. As a rough generalization, many organic reactions have activation energies in the range 40 to 150 kJ/mol (10–35 kcal/mol). The reaction of ethylene with HBr, for example, has an activation energy of approximately 140 kJ/mol (34 kcal/mol). Reactions with activation energies less than 80 kJ/mol take place at or below room temperature, while reactions with higher activation energies normally require a higher temperature to give the reactants enough energy to climb the activation barrier. Once the transition state is reached, the reaction can either continue on to give the carbocation product or revert back to reactants. When reversion to reactants occurs, the transition-state structure comes apart and an amount of free energy corresponding to −Δ G ‡ is released. When the reaction continues on to give the carbocation, the new C−H bond forms fully and an amount of energy is released corresponding to the difference between the transition state and carbocation product. The net energy change for the step, ∆ G °, is represented in the diagram as the difference in level between reactant and product. Since the carbocation is higher in energy than the starting alkene, the step is endergonic, has a positive value of ∆ G °, and absorbs energy. Not all energy diagrams are like that shown for the reaction of ethylene and HBr. Each reaction has its own energy profile. Some reactions are fast (small Δ G ‡ ) and some are slow (large Δ G ‡ ); some have a negative Δ G °, and some have a positive Δ G °. illustrates some different possibilities. Some hypothetical energy diagrams: (a) a fast exergonic reaction (low Δ G ‡ , negative Δ G °); (b) a slow exergonic reaction (high Δ G ‡ , negative Δ G °); (c) a fast endergonic reaction (small Δ G ‡ , small positive Δ G °); (d) a slow endergonic reaction (high Δ G ‡ , positive Δ G °). Problems

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Describing a Reaction: Intermediates How can we describe the carbocation formed in the first step of the reaction of ethylene with HBr? The carbocation is clearly different from the reactants, yet it isn’t a transition state and it isn’t a final product. We call the carbocation, which exists only transiently during the course of the multistep reaction, a reaction intermediate . As soon as the intermediate is formed in the first step by reaction of ethylene with H + , it reacts further with Br − in a second step to give the final product, bromoethane. This second step has its own activation energy (Δ G ‡ ), its own transition state, and its own energy change (Δ G °). We can picture the second transition state as an activated complex between the electrophilic carbocation intermediate and the nucleophilic bromide anion, in which Br − donates a pair of electrons to the positively charged carbon atom as the new C−Br bond just starts to form. A complete energy diagram for the overall reaction of ethylene with HBr is shown in . In essence, we draw a diagram for each of the individual steps and then join them so that the carbocation product of step 1 is the reactant for step 2. As indicated in , the reaction intermediate lies at an energy minimum between steps. Because the energy level of the intermediate is higher than the level of either the reactant that formed it or the product it yields, the intermediate can’t normally be isolated. It is, however, more stable than its two neighboring transition states. An energy diagram for the reaction of ethylene with HBr. Two separate steps are involved, each with its own activation energy (Δ G ‡ ) and free-energy change (Δ G °). The overall Δ G ‡ for the complete reaction is the energy difference between reactants and the highest transition state (which corresponds to Δ G 1 ‡ in this case), and the overall Δ G ° for the reaction is the energy difference between reactants and final products. Each step in a multistep process can always be considered separately. Each step has its own Δ G ‡ and its own Δ G °. The overall activation energy that controls the rate of the reaction, however, is the energy difference between initial reactants and the highest transition state, regardless of which step it occurs in. The overall Δ G ° of the reaction is the energy difference between reactants and final products. The biological reactions that take place in living organisms have the same energy requirements as reactions that take place in the laboratory and can be described in similar ways. They are, however, constrained by the fact that they must have low enough activation energies to occur at moderate temperatures, and they must release energy in relatively small amounts to avoid overheating the organism. These constraints can be met through the use of large, structurally complex, enzyme catalysts that alter the mechanism of a reaction to a pathway that can proceed through a series of small steps rather than one or two large steps. Thus, an energy diagram for a biological reaction might look like that in . Energy diagrams for a typical, enzyme-catalyzed biological reaction and an uncatalyzed laboratory reaction . The biological reaction involves many steps, each of which has a relatively small activation energy and small energy change. The end result is the same, however. Drawing a Reaction Energy Diagram Sketch an energy diagram for a one-step reaction that is fast and highly exergonic. Strategy A fast reaction has a small Δ G ‡ , and a highly exergonic reaction has a large negative Δ G °. Solution Drawing a Reaction Energy Diagram Sketch an energy diagram for a two-step exergonic reaction whose second step has a higher-energy transition state than its first step. Show Δ G ‡ and Δ G ° for the overall reaction. Strategy A two-step reaction has two transition states and an intermediate between them. The Δ G ‡ for the overall reaction is the energy change between reactants and the highest-energy transition state—the second one in this case. An exergonic reaction has a negative overall Δ G °. Solution Problems

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A Comparison Between Biological Reactions and Laboratory Reactions Beginning in the next chapter, we’ll be seeing a lot of reactions, some that are important in laboratory chemistry yet don’t occur in nature and others that have counterparts in biological pathways. In comparing laboratory reactions with biological reactions, several differences are apparent. For one, laboratory reactions are usually carried out in an organic solvent such as diethyl ether or dichloromethane to dissolve the reactants and bring them into contact, whereas biological reactions occur in the aqueous medium within cells. For another, laboratory reactions often take place over a wide range of temperatures without catalysts, while biological reactions take place at the temperature of the organism and are catalyzed by enzymes. We’ll look at enzymes in more detail in Section 26.10 , but you may already be aware that an enzyme is a large, globular, protein molecule that contains in its structure a protected pocket called its active site . The active site is lined by acidic or basic groups as needed for catalysis and has precisely the right shape to bind and hold a substrate molecule in the orientation necessary for reaction. shows a molecular model of hexokinase, along with an X-ray crystal structure of the glucose substrate and adenosine diphosphate (ADP) bound in the active site. Hexokinase is an enzyme that catalyzes the initial step of glucose metabolism—the transfer of a phosphate group from ATP to glucose, giving glucose 6-phosphate and ADP. The structures of ATP and ADP were shown at the end of Section 6.8 . Note how the hexokinase-catalyzed phosphorylation reaction of glucose is written. It’s common when writing biological equations to show only the structures of the primary reactant and product, while abbreviating the structures of various biological “reagents” and by-products such as ATP and ADP. A curved arrow intersecting the straight reaction arrow indicates that ATP is also a reactant and ADP also a product. Models of hexokinase in space-filling and wire-frame formats, showing the cleft that contains the active site where substrate binding and reaction catalysis occur. At the bottom is an X-ray crystal structure of the enzyme active site, showing the positions of both glucose and ADP as well as a lysine amino acid that acts as a base to deprotonate glucose. Yet a third difference between laboratory and biological reactions is that laboratory reactions are often done using relatively small, simple reagents such as Br 2 , HCl, NaBH 4 , CrO 3 , and so forth, while biological reactions usually involve relatively complex “reagents” called coenzymes . In the hexokinase-catalyzed phosphorylation of glucose just shown, ATP is the coenzyme. As another example, compare the H 2 molecule, a laboratory reagent that adds to a carbon–carbon double bond to yield an alkane, with the reduced nicotinamide adenine dinucleotide (NADH) molecule, a coenzyme that effects an analogous addition of hydrogen to a double bond in many biological pathways. Of all the atoms in the coenzyme, only the one hydrogen atom shown in red is transferred to the double-bond substrate. Don’t be intimidated by the size of the ATP or NADH molecule; most of the structure is there to provide an overall shape for binding to the enzyme and to provide appropriate solubility behavior. When looking at biological molecules, focus on the small part of the molecule where the chemical change takes place. One final difference between laboratory and biological reactions is in their specificity. A catalyst might be used in the laboratory to catalyze the reaction of thousands of different substances, but an enzyme, because it can only bind a specific substrate molecule having a specific shape, will usually catalyze only a specific reaction. It’s this exquisite specificity that makes biological chemistry so remarkable and that makes life possible. summarizes some of the differences between laboratory and biological reactions. A Comparison of Typical Laboratory and Biological Reactions Laboratory reaction Biological reaction Solvent Organic liquid, such as ether Aqueous environment in cells Temperature Wide range; −80 to 150 °C Temperature of organism Catalyst Either none, or very simple Large, complex enzymes needed Reagent size Usually small and simple Relatively complex coenzymes Specificity Little specificity for substrate Very high specificity for substrate A Comparison of Typical Laboratory and Biological Reactions Where Do Drugs Come From? It has been estimated that major pharmaceutical companies in the United States spent some $200 billion on drug research and development in 2020, while government agencies and private foundations spent another $28 billion. What does this money buy? From 1983 to 2022, the money resulted in a total of 1237 new molecular entities (NMEs)—new biologically active chemical substances approved for sale as drugs by the U.S. Food and Drug Administration (FDA). Where do the new drugs come from? According to a study carried out several years ago at the U.S. National Cancer Institute, only about 33% of new drugs are entirely synthetic and completely unrelated to any naturally occurring substance. The remaining 67% take their lead, to a greater or lesser extent, from nature. Vaccines and genetically engineered proteins of biological origin account for 15% of NMEs, but most new drugs come from natural products , a catchall term generally taken to mean small molecules found in bacteria, plants, algae, and other living organisms. Unmodified natural products isolated directly from the producing organism account for 24% of NMEs, while natural products that have been chemically modified in the laboratory account for the remaining 28%. Many years of work go into screening many thousands of substances to identify a single compound that might ultimately gain approval as an NME. But after that single compound has been identified, the work has just begun because it takes an average of 9 to 10 years for a drug to make it through the approval process. First, the safety of the drug in animals must be demonstrated and an economical method of manufacture must be devised. With these preliminaries out of the way, an Investigational New Drug (IND) application is submitted to the FDA for permission to begin testing in humans. Introduced in June, 2006, Gardasil is the first vaccine ever approved for the prevention of cancer. Where do new drugs like this come from? (credit: modification of work “COVIran Barekat vaccine production” by Sadegh Nikgostar/Wikimedia Commons, CC BY 4.0) Human testing takes, or should take, 5 to 7 years and is divided into three phases. Phase I clinical trials are carried out on a small group of healthy volunteers to establish safety and look for side effects. Several months to a year are needed, and only about 70% of drugs pass at this point. Phase II clinical trials next test the drug for 1 to 2 years in several hundred patients with the target disease or condition, looking both for safety and efficacy, and only about 33% of the original group pass. Finally, phase III trials are undertaken on a large sample of patients to document definitively the drug’s safety, dosage, and efficacy. If the drug is one of the 25% of the original group that make it to the end of phase III, all the data are then gathered into a New Drug Application (NDA) and sent to the FDA for review and approval, which can take another 2 years. Ten years have elapsed and at least $500 million has been spent, with only a 20% success rate for the drugs that began testing. Finally, though, the drug will begin to appear in medicine cabinets. The following timeline shows the process. activation energy (Δ G ‡ ) active site addition reaction bond dissociation energy ( D ) carbocation electrophile elimination reaction endergonic endothermic enthalpy change (Δ H ) entropy change (Δ S ) enzyme exergonic exothermic Gibbs free-energy change (Δ G ) heat of reaction nucleophile polar reaction polarizability radical radical reaction reaction coordinate reaction intermediate reaction mechanism rearrangement reaction substitution reaction transition state Summary All chemical reactions, whether in the laboratory or in living organisms, follow the same chemical rules. To understand both organic and biological chemistry, it’s necessary to know not just what occurs but also why and how chemical reactions take place. In this chapter, we’ve taken a brief look at the fundamental kinds of organic reactions, we’ve seen why reactions occur, and we’ve seen how reactions can be described. There are four common kinds of reactions: addition reactions take place when two reactants add together to give a single product; elimination reactions take place when one reactant splits apart to give two products; substitution reactions take place when two reactants exchange parts to give two new products; and rearrangement reactions take place when one reactant undergoes a reorganization of bonds and atoms to give an isomeric product. A full description of how a reaction occurs is called its mechanism . There are two general kinds of mechanisms by which most reactions take place: radical mechanisms and polar mechanisms. Polar reactions, the more common type, occur because of an attractive interaction between a nucleophilic (electron-rich) site in one molecule and an electrophilic (electron-poor) site in another molecule. A bond is formed in a polar reaction when the nucleophile donates an electron pair to the electrophile. This transfer of electrons is indicated by a curved arrow showing the direction of electron travel from the nucleophile to the electrophile. Radical reactions involve species that have an odd number of electrons. A bond is formed when each reactant donates one electron. The energy changes that take place during reactions can be described by considering both rates (how fast the reactions occur) and equilibria (how much the reactions occur). The position of a chemical equilibrium is determined by the value of the free-energy change (Δ G ) for the reaction, where Δ G = Δ H − T Δ S . The enthalpy term (Δ H ) corresponds to the net change in strength of chemical bonds broken and formed during the reaction; the entropy term (Δ S ) corresponds to the change in the amount of molecular randomness during the reaction. Reactions that have negative values of Δ G release energy, are said to be exergonic , and have favorable equilibria. Reactions that have positive values of Δ G absorb energy, are said to be endergonic , and have unfavorable equilibria. A reaction can be described pictorially using an energy diagram that follows the reaction course from reactants through transition state to product. The transition state is an activated complex occurring at the highest-energy point of a reaction. The amount of energy needed by reactants to reach this high point is the activation energy, Δ G ‡ . The higher the activation energy, the slower the reaction. Many reactions take place in more than one step and involve the formation of a reaction intermediate . An intermediate is a species that lies at an energy minimum between steps on the reaction curve and is formed briefly during the course of a reaction. Additional Problems Visualizing Chemistry Energy Diagrams and Reaction Mechanisms Polar Reactions Radical Reactions General Problems

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Why This Chapter? Tomatoes are good for you . Their red color is due to lycopene, which has 13 double bonds. (credit: modification of work “Tomatoes” by Jeremy Keith/Flickr, CC BY 2.0) Carbon–carbon double bonds are present in most organic and biological molecules, so a good understanding of their behavior is needed. In this chapter, we’ll look at some consequences of alkene stereoisomerism and then focus on the broadest and most general class of alkene reactions, the electrophilic addition reaction. Carbon-carbon triple bonds, by contrast, occur much less commonly, so we’ll not spend much time on their chemistry. An alkene , sometimes called an olefin from the German term for oil forming, is a hydrocarbon that contains a carbon–carbon double bond, while an alkyne is a hydrocarbon that contains a carbon-carbon triple bond. Alkenes occur abundantly in nature. Ethylene, for instance, is a plant hormone that induces ripening in fruit, and α -pinene is the major component of turpentine. Lycopene, found in fruits such as watermelon and papaya as well as tomatoes, is an antioxidant with numerous health benefits such as sun protection and cardiovascular protection.

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Industrial Preparation and Use of Alkenes Ethylene and propylene, the simplest alkenes, are the two most important organic chemicals produced industrially. Approximately 220 million tons of ethylene and 138 million tons of propylene are produced worldwide each year for use in the synthesis of polyethylene, polypropylene, ethylene glycol, acetic acid, acetaldehyde, and a host of other substances ( ). Compounds derived industrially from ethylene and propylene . Ethylene, propylene, and butene are synthesized from (C 2 –C 8 ) alkanes by a process called steam cracking at temperatures up to 900 °C. The cracking process is complex, although it undoubtedly involves radical reactions. The high-temperature reaction conditions cause spontaneous breaking of C−C and C−H bonds, with the resultant formation of smaller fragments. We might imagine, for instance, that a molecule of butane splits into two ethyl radicals, each of which then loses a hydrogen atom to generate two molecules of ethylene. Steam cracking is an example of a reaction whose energetics are dominated by entropy (Δ S °) rather than by enthalpy (Δ H °) in the free-energy equation Δ G ° = Δ H ° − T Δ S ° discussed in . Although the bond dissociation energy D for a carbon–carbon single bond is relatively high (about 370 kJ/mol) and cracking is endothermic, the large positive entropy change resulting from the fragmentation of one large molecule into several smaller pieces, together with the high temperature, makes the T Δ S ° term larger than the Δ H ° term, thereby favoring the cracking reaction.

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Calculating the Degree of Unsaturation Because of its double bond, an alkene has fewer hydrogens than an alkane with the same number of carbons—C n H 2 n for an alkene versus C n H 2 n +2 for an alkane—and is therefore referred to as unsaturated . Ethylene, for example, has the formula C 2 H 4 , whereas ethane has the formula C 2 H 6 . In general, each ring or double bond in a molecule corresponds to a loss of two hydrogens from the alkane formula C n H 2 n +2 . Knowing this relationship, it’s possible to work backward from a molecular formula to calculate a molecule’s degree of unsaturation —the number of rings and/or multiple bonds present in the molecule. Let’s assume that we want to find the structure of an unknown hydrocarbon. A molecular weight determination yields a value of 82 amu, which corresponds to a molecular formula of C 6 H 10 . Since the saturated C 6 alkane (hexane) has the formula C 6 H 14 , the unknown compound has two fewer pairs of hydrogens (H 14 − H 10 = H 4 = 2 H 2 ) so its degree of unsaturation is 2. The unknown therefore contains either two double bonds, one ring and one double bond, two rings, or one triple bond. There’s still a long way to go to establish its structure, but the simple calculation has told us a lot about the molecule. Similar calculations can be carried out for compounds containing elements other than just carbon and hydrogen. Organohalogen compounds (C, H, X, where X = F, Cl, Br, or I) A halogen substituent acts as a replacement for hydrogen in an organic molecule, so we can add the number of halogens and hydrogens to arrive at an equivalent hydrocarbon formula from which the degree of unsaturation can be found. For example, the formula C 4 H 6 Br 2 is equivalent to the hydrocarbon formula C 4 H 8 and thus corresponds to one degree of unsaturation. Organooxygen compounds (C, H, O) Oxygen forms two bonds, so it doesn’t affect the formula of an equivalent hydrocarbon and can be ignored when calculating the degree of unsaturation. You can convince yourself of this by seeing what happens when an oxygen atom is inserted into an alkane bond: C−C becomes C−O−C or C−H becomes C−O−H, and there is no change in the number of hydrogen atoms. For example, the formula C 5 H 8 O is equivalent to the hydrocarbon formula C 5 H 8 and thus corresponds to two degrees of unsaturation. Organonitrogen compounds (C, H, N) Nitrogen forms three bonds, so an organonitrogen compound has one more hydrogen than a related hydrocarbon. We therefore subtract the number of nitrogens from the number of hydrogens to arrive at the equivalent hydrocarbon formula. Again, you can convince yourself of this by seeing what happens when a nitrogen atom is inserted into an alkane bond: C−C becomes C−NH−C or C−H becomes C−NH 2 , meaning that one additional hydrogen atom has been added. We must therefore subtract this extra hydrogen atom to arrive at the equivalent hydrocarbon formula. For example, the formula C 5 H 9 N is equivalent to C 5 H 8 and thus has two degrees of unsaturation. To summarize: Add the number of halogens to the number of hydrogens. Ignore the number of oxygens . Subtract the number of nitrogens from the number of hydrogens. Problems

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Naming Alkenes Alkenes are named using a series of rules similar to those for alkanes ( ), with the suffix -ene used instead of -ane to identify the functional group. There are three steps to this process. STEP 1 Name the parent hydrocarbon. Find the longest carbon chain containing the double bond, and name the compound accordingly, using the suffix -ene: STEP 2 Number the carbon atoms in the chain. Begin at the end nearer the double bond or, if the double bond is equidistant from the two ends, begin at the end nearer the first branch point. This rule ensures that the double-bond carbons receive the lowest possible numbers. STEP 3 Write the full name. Number the substituents according to their positions in the chain, and list them alphabetically. Indicate the position of the double bond by giving the number of the first alkene carbon and placing that number directly before the parent name. If more than one double bond is present, indicate the position of each and use one of the suffixes -diene, -triene, and so on. We should also note that IUPAC changed their naming recommendations in 1993 to place the locant indicating the position of the double bond immediately before the - ene suffix rather than before the parent name: but-2-ene rather than 2-butene, for instance. This change has not been widely accepted by the chemical community in the United States, however, so we’ll stay with the older but more commonly used names. Be aware, though, that you may occasionally encounter the newer system. Cycloalkenes are named similarly, but because there is no chain end to begin from, we number the cycloalkene so that the double bond is between C1 and C2 and the first substituent has as low a number as possible. It’s not necessary to indicate the position of the double bond in the name because it’s always between C1 and C2. As with open-chain alkenes, the newer but not yet widely accepted naming rules place the locant immediately before the suffix in a cyclic alkene. For historical reasons, there are a few alkenes whose names are firmly entrenched in common usage but don’t conform to the rules. For example, the alkene derived from ethane should be called ethene, but the name ethylene has been used for so long that it is accepted by IUPAC. lists several other common names that are often used and are recognized by IUPAC. Note also that a =CH 2 substituent is called a methylene group , a H 2 C═CH– substituent is called a vinyl group , and a H 2 C═ CHCH 2 – substituent is called an allyl group . Common Names of Some Alkenes Compound Systematic name Common name H 2 C═ CH 2 Ethene Ethylene CH 3 CH═ CH 2 Propene Propylene 2-Methylpropene Isobutylene 2-Methyl-1,3-butadiene Isoprene Problems

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Cis–Trans Isomerism in Alkenes We saw in the chapter on Structure and Bonding that the carbon–carbon double bond can be described in two ways. In valence bond language ( ), the carbons are sp 2 -hybridized and have three equivalent hybrid orbitals that lie in a plane at angles of 120° to one another. The carbons form a σ bond by a head-on overlap of sp 2 orbitals and form a π bond by sideways overlap of unhybridized p orbitals oriented perpendicular to the sp 2 plane, as shown in . In molecular orbital language ( ), interaction between the p orbitals leads to one bonding and one antibonding π molecular orbital. The π bonding MO has no node between nuclei and results from a combination of p orbital lobes with the same algebraic sign. The π antibonding MO has a node between nuclei and results from a combination of lobes with different algebraic signs, as shown in . Although essentially free rotation around single bonds is possible ( ), the same is not true of double bonds. For rotation to occur around a double bond, the π bond must break and re-form ( ). Thus, the barrier to double-bond rotation must be at least as great as the strength of the π bond itself, an estimated 350 kJ/mol (84 kcal/mol). Recall that the barrier to bond rotation in ethane is only 12 kJ/mol. The π bond must break for rotation to take place around a carbon–carbon double bond . The lack of rotation around carbon–carbon double bonds is of more than just theoretical interest; it also has chemical consequences. Imagine the situation for a disubstituted alkene such as 2-butene. (Disubstituted means that two substituents other than hydrogen are bonded to the double-bond carbons.) The two methyl groups in 2-butene can either be on the same side of the double bond or on opposite sides, a situation similar to that in disubstituted cycloalkanes ( ). Since bond rotation can’t occur, the two 2-butenes can’t spontaneously interconvert; they are different, isolable compounds. As with disubstituted cycloalkanes, we call such compounds cis–trans stereoisomers. The compound with substituents on the same side of the double bond is called cis -2-butene, and the isomer with substituents on opposite sides is trans -2-butene ( ). Cis and trans isomers of 2-butene . The cis isomer has the two methyl groups on the same side of the double bond, and the trans isomer has methyl groups on opposite sides. Cis–trans isomerism is not limited to disubstituted alkenes. It can occur whenever both double-bond carbons are attached to two different groups. If one of the double-bond carbons is attached to two identical groups, however, cis–trans isomerism is not possible ( ). The requirement for cis–trans isomerism in alkenes . Compounds that have one of their carbons bonded to two identical groups can’t exist as cis–trans isomers. Only when both carbons are bonded to two different groups is cis–trans isomerism possible. Problems

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Alkene Stereochemistry and the E , Z Designation The cis–trans naming system used in the previous section works only with disubstituted alkenes—compounds that have two substituents other than hydrogen on the double bond. With trisubstituted and tetrasubstituted double bonds, a more general method is needed for describing double-bond geometry. ( Trisubstituted means three substituents other than hydrogen on the double bond; tetrasubstituted means four substituents other than hydrogen.) The method used for describing alkene stereochemistry is called the E,Z system and employs the same Cahn–Ingold–Prelog sequence rules given in for specifying the configuration of a chirality center. Let’s briefly review the sequence rules and then see how they’re used to specify double-bond geometry. For a more thorough review, reread . RULE 1 Considering each of the double-bond carbons separately, look at the two substituents attached and rank them according to the atomic number of the first atom in each (8 for oxygen, 6 for carbon, 1 for hydrogen, and so forth). An atom with higher atomic number ranks higher than an atom with lower atomic number. RULE 2 If a decision can’t be reached by ranking the first atoms in the two substituents, look at the second, third, or fourth atoms away from the double-bond until the first difference is found. RULE 3 Multiple-bonded atoms are equivalent to the same number of single-bonded atoms. Once the two groups attached to each double-bonded carbon have been ranked as either higher or lower, look at the entire molecule. If the higher-ranked groups on each carbon are on the same side of the double bond, the alkene is said to have a Z configuration , for the German zusammen, meaning “together.” If the higher-ranked groups are on opposite sides, the alkene has an E configuration , for the German entgegen, meaning “opposite.” (For a simple way to remember which is which, note that the groups are on “ze zame zide” in the Z isomer.) As an example, look at the following two isomers of 2-chloro-2-butene. Because chlorine has a higher atomic number than carbon, a −Cl substituent is ranked higher than a −CH 3 group. Methyl is ranked higher than hydrogen, however, so isomer (a) is designated E because the higher-ranked groups are on opposite sides of the double bond. Isomer (b) has a Z configuration because its higher-ranked groups are on ze zame zide of the double bond. For further practice, work through each of the following examples to convince yourself that the assignments are correct: Assigning E and Z Configurations to Alkenes Assign E or Z configuration to the double bond in the following compound: Strategy Look at the two substituents connected to each double-bonded carbon, and determine their ranking using the Cahn–Ingold–Prelog rules. Then, check whether the two higher-ranked groups are on the same or opposite sides of the double bond. Solution The left-hand carbon has −H and −CH 3 substituents, of which −CH 3 ranks higher by sequence rule 1. The right-hand carbon has −CH(CH 3 ) 2 and −CH 2 OH substituents, which are equivalent by rule 1. By rule 2, however, −CH 2 OH ranks higher than −CH(CH 3 ) 2 because the substituent −CH 2 OH has an oxygen as its highest second atom, but −CH(CH 3 ) 2 has a carbon as its highest second atom. The two higher-ranked groups are on the same side of the double bond, so we assign a Z configuration. Problems

Alkene Stereochemistry and theE,ZDesignation

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Stability of Alkenes Although the cis–trans interconversion of alkene isomers does not occur spontaneously, it can often be made to happen by treating the alkene with a strong acid catalyst. If we interconvert cis -2-butene with trans -2-butene and allow them to reach equilibrium, we find that they aren’t of equal stability. The trans isomer is more stable than the cis isomer by 2.8 kJ/mol (0.66 kcal/mol) at room temperature, corresponding to a 76 : 24 ratio. Cis alkenes are less stable than their trans isomers because of steric strain between the two larger substituents on the same side of the double bond. This is the same kind of steric interference that we saw previously in the axial conformation of methylcyclohexane ( ). Although it’s sometimes possible to find relative stabilities of alkene isomers by establishing a cis–trans equilibrium through treatment with strong acid, a more general method is to take advantage of the fact that alkenes undergo a hydrogenation reaction to give the corresponding alkane when treated with H 2 gas in the presence of a catalyst such as palladium or platinum. Energy diagrams for the hydrogenation reactions of cis - and trans -2-butene are shown in . Because cis -2-butene is less stable than trans -2-butene by 2.8 kJ/mol, the energy diagram shows the cis alkene at a higher energy level. After reaction, however, both curves are at the same energy level (butane). It therefore follows that Δ G ° for reaction of the cis isomer must be larger than Δ G ° for reaction of the trans isomer by 2.8 kJ/mol. In other words, more energy is released in the hydrogenation of the cis isomer than the trans isomer because the cis isomer has more energy to begin with. Energy diagrams for hydrogenation of cis - and trans -2-butene . The cis isomer is higher in energy than the trans isomer by about 2.8 kJ/mol and therefore releases more energy when hydrogenated. If we were to measure the so-called heats of hydrogenation (Δ H ° hydrog ) for two double-bond isomers and find their difference, we could determine the relative stabilities of cis and trans isomers without having to measure an equilibrium position. cis -2-Butene, for instance, has Δ H ° hydrog = −119 kJ/mol(−28.3 kcal/mol), while trans -2-butene has Δ H ° hydrog = −115 kJ/mol(−27.4 kcal/mol)—a difference of 4 kJ/mol. The 4 kJ/mol energy difference between the 2-butene isomers calculated from heats of hydrogenation agrees reasonably well with the 2.8 kJ/mol energy difference calculated from equilibrium data, but the values aren’t exactly the same for two reasons. First, there is probably some experimental error, because heats of hydrogenation are difficult to measure accurately. Second, heats of reaction and equilibrium constants don’t measure exactly the same thing. Heats of reaction measure enthalpy changes, Δ H °, whereas equilibrium constants measure free-energy changes, Δ G °, so we might expect a slight difference between the two. Heats of Hydrogenation of Some Alkenes Substitution Alkene Δ H ° hydrog (kJ/mol) (kcal/mol) Ethylene H 2 C═ CH 2 −136 −32.6 Monosubstituted CH 3 CH═ CH 2 −125 −29.9 Disubstituted CH 3 CH═ CHCH 3 (cis) CH 3 CH═ CHCH 3 (trans) (CH 3 ) 2 C═ CH 2 −119 −115 −118 −28.3 −27.4 −28.2 Trisubstituted (CH 3 ) 2 C═ CHCH 3 −112 −26.7 Tetrasubstituted (CH 3 ) 2 C═ C(CH 3 ) 2 −110 −26.4 lists some representative data for the hydrogenation of different alkenes and shows that alkenes become more stable with increasing substitution. That is, alkenes follow the stability order: The stability order of substituted alkenes is due to a combination of two factors. One is a stabilizing interaction between the C═C π orbital and adjacent C−H σ bonds on substituents. In valence-bond language, the interaction is called hyperconjugation . In a molecular orbital description, there is a bonding MO that extends over the four-atom C═C−C−H grouping, as shown in . The more substituents present on the double bond, the more hyperconjugation occurs and the more stable the alkene. Hyperconjugation is a stabilizing interaction between the C═C π orbital and adjacent C−H σ bonds on substituents . The more substituents there are, the greater the stabilization of the alkene. A second factor that contributes to alkene stability involves bond strengths. A bond between a sp 2 carbon and a sp 3 carbon is somewhat stronger than a bond between two sp 3 carbons. Thus, in comparing 1-butene and 2-butene, the monosubstituted isomer has one sp 3 – sp 3 bond and one sp 3 – sp 2 bond, while the disubstituted isomer has two sp 3 – sp 2 bonds. More highly substituted alkenes always have a higher ratio of sp 3 – sp 2 bonds to sp 3 – sp 3 bonds than less highly substituted alkenes and are therefore more stable. Problems

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Electrophilic Addition Reactions of Alkenes Before beginning a detailed discussion of alkene reactions, let’s review briefly some conclusions from the previous chapter. We said in that alkenes behave as nucleophiles (Lewis bases) in polar reactions, donating a pair of electrons from their electron-rich C═C bond to an electrophile (Lewis acid). For example, reaction of 2-methylpropene with HBr yields 2-bromo-2-methylpropane. A careful study of this and similar reactions by the British chemist Christopher Ingold and others in the 1930s led to the generally accepted mechanism shown in for an electrophilic addition reaction . MECHANISM Mechanism of the electrophilic addition of HBr to 2-methylpropene . The reaction occurs in two steps, protonation and bromide addition, and involves a carbocation intermediate. The reaction begins with an attack on the hydrogen of the electrophile HBr by the electrons of the nucleophilic π bond. Two electrons from the π bond form a new σ bond between the entering hydrogen and an alkene carbon, as shown by the curved arrow at the top of . The resulting carbocation intermediate is itself an electrophile, which can accept an electron pair from nucleophilic Br − ion to form a C−Br bond and yield a neutral addition product. An energy diagram for the overall electrophilic addition reaction ( ) has two peaks (transition states) separated by a valley (carbocation intermediate). The energy level of the intermediate is higher than that of the starting alkene, but the reaction as a whole is exergonic (negative Δ G °). The first step, protonation of the alkene to yield the intermediate cation, is relatively slow. But once the cation intermediate is formed, it rapidly reacts to yield the final alkyl bromide product. The relative rates of the two steps are indicated in by the fact that Δ G 1 ‡ is larger than Δ G 2 ‡ . Energy diagram for the two-step electrophilic addition of HBr to 2-methylpropene . The first step is slower than the second step. Electrophilic addition to alkenes is successful not only with HBr but with HCl, HI, and H 2 O as well. Note that HI is usually generated in the reaction mixture by treating potassium iodide with phosphoric acid and that a strong acid catalyst is needed for the addition of water. Writing Organic Reactions This is a good place to mention that the equations for organic reactions are sometimes written in different ways to emphasize different points. In describing a laboratory process, for instance, the reaction of 2-methylpropene with HCl might be written in the format A + B ⟶ C to emphasize that both reactants are equally important for the purposes of the discussion. The solvent and notes about other reaction conditions such as temperature are written either above or below the reaction arrow. Alternatively, we might write the same reaction in a format to emphasize that 2-methylpropene is the reactant whose chemistry is of greater interest. The second reactant, HCl, is placed above the reaction arrow together with notes about solvent and reaction conditions. In describing a biological process, the reaction is almost always written to show only the structures of the primary reactant and product, while abbreviating the structures of various biological “reagents” and by-products with a curved arrow that intersects the straight reaction arrow. As discussed in , the reaction of glucose with ATP to give glucose 6-phosphate plus ADP would then be written as

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Orientation of Electrophilic Additions: Markovnikov’s Rule Look carefully at the electrophilic addition reactions shown in the previous section. In each case, an unsymmetrically substituted alkene has given a single addition product rather than the mixture that might be expected. For example, 2-methylpropene might react with HCl to give both 2-chloro-2-methylpropane and 1-chloro-2-methylpropane, but it doesn’t. It gives only 2-chloro-2-methylpropane as the sole product. Similarly, it’s invariably the case in biological alkene addition reactions that only a single product is formed. We say that such reactions are regiospecific ( ree -jee-oh-specific) when only one of two possible orientations of an addition occurs. After looking at the results of many such reactions, the Russian chemist Vladimir Markovnikov proposed in 1869 what has become known as: Markovnikov’s rule In the addition of HX to an alkene, the H attaches to the carbon with fewer alkyl substituents and the X attaches to the carbon with more alkyl substituents. When both double-bonded carbon atoms have the same degree of substitution, a mixture of addition products results. Because carbocations are involved as intermediates in these electrophilic addition reactions, Markovnikov’s rule can be restated in the following way: Markovnikov’s rule restated In the addition of HX to an alkene, the more highly substituted carbocation is formed as the intermediate rather than the less highly substituted one. For example, addition of H + to 2-methylpropene yields the intermediate tertiary carbocation rather than the alternative primary carbocation, and addition to 1-methylcyclohexene yields a tertiary cation rather than a secondary one. Why should this be? Predicting the Product of an Electrophilic Addition Reaction What product would you expect from reaction of HCl with 1-ethylcyclopentene? Strategy When solving a problem that asks you to predict a reaction product, begin by looking at the functional group(s) in the reactants and deciding what kind of reaction is likely to occur. In the present instance, the reactant is an alkene that will probably undergo an electrophilic addition reaction with HCl. Next, recall what you know about electrophilic addition reactions to predict the product. You know that electrophilic addition reactions follow Markovnikov’s rule, so H + will add to the double-bond carbon that has one alkyl group (C2 on the ring) and the Cl will add to the double-bond carbon that has two alkyl groups (C1 on the ring). Solution The expected product is 1-chloro-1-ethylcyclopentane. Synthesizing a Specific Compound What alkene would you start with to prepare the following alkyl halide? There may be more than one possibility. Strategy When solving a problem that asks how to prepare a given product, always work backward. Look at the product, identify the functional group(s) it contains, and ask yourself, “How can I prepare that functional group?” In the present instance, the product is a tertiary alkyl chloride, which can be prepared by reaction of an alkene with HCl. The carbon atom bearing the −Cl atom in the product must be one of the double-bond carbons in the reactant. Draw and evaluate all possibilities. Solution There are three possibilities, all of which could give the desired product according to Markovnikov’s rule. Problems

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Carbocation Structure and Stability To understand why Markovnikov’s rule works, we need to learn more about the structure and stability of carbocations and about the general nature of reactions and transition states. The first point to explore involves structure. A great deal of experimental evidence has shown that carbocations are planar. The trivalent carbon is sp 2 -hybridized, and the three substituents are oriented toward the corners of an equilateral triangle, as indicated in . Because there are only six valence electrons on carbon and all six are used in the three σ bonds, the p orbital extending above and below the plane is unoccupied. The structure of a carbocation . The trivalent carbon is sp 2 -hybridized and has a vacant p orbital perpendicular to the plane of the carbon and three attached groups. The second point to explore involves carbocation stability. 2-Methylpropene might react with H + to form a carbocation having three alkyl substituents (a tertiary ion, 3°), or it might react to form a carbocation having one alkyl substituent (a primary ion, 1°). Since the tertiary alkyl chloride, 2-chloro-2-methylpropane, is the only product observed, formation of the tertiary cation is evidently favored over formation of the primary cation. Thermodynamic measurements show that, indeed, the stability of carbocations increases with increasing substitution so that the stability order is tertiary > secondary > primary > methyl. One way of determining carbocation stabilities is to measure the amount of energy required to form a carbocation by dissociation of the corresponding alkyl halide, R − X → R + + :X − . As shown in , tertiary alkyl halides dissociate to give carbocations more easily than secondary or primary ones. Thus, trisubstituted carbocations are more stable than disubstituted ones, which are more stable than monosubstituted ones. The data in are taken from measurements made in the gas phase, but a similar stability order is found for carbocations in solution. The dissociation enthalpies are much lower in solution because polar solvents can stabilize the ions, but the order of carbocation stability remains the same. A plot of dissociation enthalpy versus substitution pattern for the gas-phase dissociation of alkyl chlorides to yield carbocations . More highly substituted alkyl halides dissociate more easily than less highly substituted ones. Why are more highly substituted carbocations more stable than less highly substituted ones? There are at least two reasons. Part of the answer has to do with inductive effects, and part has to do with hyperconjugation. Inductive effects, discussed in in connection with polar covalent bonds, result from the shifting of electrons in a σ bond in response to the electronegativity of nearby atoms. In the present instance, electrons from a relatively larger and more polarizable alkyl group can shift toward a neighboring positive charge more easily than the electron from a hydrogen. Thus, the more alkyl groups attached to the positively charged carbon, the more electron density shifts toward the charge and the more inductive stabilization of the cation occurs ( ). A comparison of inductive stabilization for methyl, primary, secondary, and tertiary carbocations . The more alkyl groups that are bonded to the positively charged carbon, the more electron density shifts toward the charge, making the charged carbon less electron-poor (blue in electrostatic potential maps). Hyperconjugation, discussed in in connection with the stabilities of substituted alkenes, is the stabilizing interaction between a p orbital and properly oriented C−H σ bonds on neighboring carbons that are roughly parallel to the p orbital. The more alkyl groups there are on the carbocation, the more possibilities there are for hyperconjugation and the more stable the carbocation. shows the molecular orbital for the ethyl carbocation, CH 3 CH 2 + , and indicates the difference between the C−H bond perpendicular to the cation p orbital and the two C−H bonds more parallel to the cation p orbital. Only these roughly parallel C−H bonds are oriented properly to take part in hyperconjugation. Stabilization of the ethyl carbocation, CH 3 CH 2 + , through hyperconjugation . Interaction of neighboring C−H σ bonds with the vacant p orbital stabilizes the cation and lowers its energy. The molecular orbital shows that only the two C−H bonds more parallel to the cation p orbital are oriented properly. The C−H bond perpendicular to the cation p orbital can’t take part properly. Problems

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