Solution - 5.6 Diastereomers
Note carefully the difference between enantiomers and diastereomers: enantiomers have opposite configurations at all chirality centers, whereas diastereomers have opposite configurations at some (one or more) chirality centers but the same configuration at others. A full description of the four stereoisomers of threonine is given in TABLE 5.2. Of the four, only the $2 S, 3 R$ isomer, $[\alpha]_{\mathrm{D}}=-28.3$, occurs naturally in plants and animals and is an essential nutrient for humans. This result is typical: most biological molecules are chiral, and usually only one stereoisomer is found in nature.

| TABLE 5.2 Relationships among the Four Stereoisomers of
 Threonine |
| :--- | :--- |
| Stereoisomer Enantiomer Diastereomer
 $2 R, 3 R$ $2 S, 3 S$ $2 R, 3 S$ and $2 S, 3 R$
 $2 S, 3 S$ $2 R, 3 R$ $2 R, 3 S$ and $2 S, 3 R$ |

TABLE 5.2 Relationships among the Four Stereoisomers of
Threonine

| Stereoisomer | Enantiomer | Diastereomer |
| :--- | :--- | :--- |
| $2 R, 3 S$ | $2 S, 3 R$ | $2 R, 3 R$ and $2 S, 3 S$ |
| $2 S, 3 R$ | $2 R, 3 S$ | $2 R, 3 R$ and $2 S, 3 S$ |

In the special case where two diastereomers differ at only one chirality center but are the same at all others, we say that the compounds are epimers. Cholestanol and coprostanol, for instance, are both found in human feces, and both have nine chirality centers. Eight of the nine are identical, but the one at C5 is different. Thus, cholestanol and coprostanol are epimeric at C5.

Cholestanol

Coprostanol


Solution - 5.6 Diastereomers
Cholestanol

Coprostanol

Epimers
PROBLEM One of the following molecules (a)-(d) is D-erythrose 4-phosphate, an intermediate in the Calvin
5-13 photosynthetic cycle by which plants incorporate $\mathrm{CO}_{2}$ into carbohydrates. If D-erythrose 4-phosphate has $R$ stereochemistry at both chirality centers, which of the structures is it? Which of the remaining three structures is the enantiomer of D-erythrose 4-phosphate, and which are diastereomers?
(a)

(b)

(c)

(d)

PROBLEM How many chirality centers does morphine have? How many stereoisomers of morphine are 5-14 possible in principle?


Morphine -
PROBLEM Assign $R$ or $S$ configuration to each chirality center in the following molecular model of the amino
5-15 acid isoleucine (blue $=\mathrm{N}$ ):


Morphine - 5.7 Meso Compounds
Let's look at another example (Section 5.4) of a compound with more than one chirality center: the tartaric acid used by Pasteur. The four stereoisomers can be drawn as follows:

2R,3R

2S,3S

The $2 R, 3 R$ and $2 S, 3 S$ structures are nonsuperimposable mirror images and therefore represent a pair of enantiomers. A close look at the $2 R, 3 S$ and $2 S, 3 R$ structures, however, shows that they are superimposable, and thus identical, as can be seen by rotating one structure $180^{\circ}$.

Identical
The $2 R, 3 S$ and $2 S, 3 R$ structures are identical because the molecule has a plane of symmetry and is therefore achiral. The symmetry plane cuts through the C2-C3 bond, making one half of the molecule a mirror image of the other half (FIGURE 5.12). Because of the plane of symmetry, the molecule is achiral, despite the fact that it has two chirality centers. Such compounds, which are achiral, yet contain chirality centers, are called meso compounds (me-zo). Thus, tartaric acid exists in three stereoisomeric forms: two enantiomers and one meso form.

FIGURE 5.12 A symmetry plane through the C2-C3 bond of meso-tartaric acid makes the molecule achiral.
Some physical properties of the three stereoisomers are listed in TABLE 5.3. The (+)- and (-)-tartaric acids have identical melting points, solubilities, and densities, but they differ in the sign of their rotation of planepolarized light. The meso isomer, by contrast, is diastereomeric with the (+) and (-) forms. It has no mirrorimage relationship to (+)- and (-)-tartaric acids, is a different compound altogether, and thus has different
physical properties.
TABLE 5.3 Some Properties of the Stereoisomers of Tartaric Acid Some Properties of the Stereoisomers of Tartaric Acid


Morphine - 5.7 Meso Compounds
| Stereoisomer | Melting point $\left({ }^{\circ} \mathrm{C}\right)$ | $[\alpha]_{\mathrm{D}}$ | Density $\left(\mathrm{g} / \mathrm{cm}^{3}\right)$ | Solubility at $20{ }^{\circ} \mathrm{C}$ <br> $(\mathrm{g} / 100 \mathrm{~mL} \mathrm{H}$ |
| :--- | :--- | ---: | :--- | :--- |
| $(+)$ | $168-170$ | +12 | 1.7598 | 139.0 |
| $(-)$ | $168-170$ | -12 | 1.7598 | 139.0 |
| Meso | $146-148$ | 0 | 1.6660 | 125.0 |


Distinguishing Chiral Compounds from Meso Compounds -
Does cis-1,2-dimethylcyclobutane have any chirality centers? Is it chiral?


Strategy -
To see whether a chirality center is present, look for a carbon atom bonded to four different groups. To see whether the molecule is chiral, look for the presence or absence of a symmetry plane. Not all molecules with chirality centers are chiral overall-meso compounds are an exception.


Solution -
A look at the structure of cis-1,2-dimethylcyclobutane shows that both methyl-bearing ring carbons (C1 and C2) are chirality centers. Overall, though, the compound is achiral because there is a symmetry plane bisecting the ring between C 1 and C 2 . Thus, the molecule is a meso compound.

PROBLEM Which of the following structures represent meso compounds?
5-16
(a)

(b)

(c)

(d)

PROBLEM Which of the following have a meso form? (Recall that the -ol suffix refers to an alcohol, ROH.)
5-17
(a) 2,3-Butanediol
(b) 2,3-Pentanediol
(c) 2,4-Pentanediol

PROBLEM Does the following structure represent a meso compound? If so, indicate the symmetry plane.
5-18


Solution - 5.8 Racemic Mixtures and the Resolution of Enantiomers
To end this discussion of stereoisomerism, let's return for a last look at Pasteur's pioneering work, described in Section 5.4. Pasteur took an optically inactive tartaric acid salt and found that he could crystallize from it two optically active forms having what we would now call $2 R, 3 R$ and $2 S, 3 S$ configurations. But what was the optically inactive form he started with? It couldn't have been meso-tartaric acid, because meso-tartaric acid is a different chemical compound and can't interconvert with the two chiral enantiomers without breaking and re-forming chemical bonds.

The answer is that Pasteur started with a $50: 50$ mixture of the two chiral tartaric acid enantiomers. Such a mixture is called a racemate (rass-uh-mate), or racemic mixture, and is denoted by either the symbol $( \pm)$ or the prefix $d, l$ to indicate an equal mixture of dextrorotatory and levorotatory forms. Racemates show no optical rotation because the (+) rotation from one enantiomer exactly cancels the ( - ) rotation from the other. Through good luck, Pasteur was able to separate, or resolve, racemic tartaric acid into its (+) and (-) enantiomers. Unfortunately, the fractional crystallization technique he used doesn't work for most racemates, so other methods are needed.

The most common method for resolving the racemate of a chiral carboxylic acid $\left(\mathrm{RCO}_{2} \mathrm{H}\right)$ is to carry out an acidbase reaction between the acid and an amine base $\left(\mathrm{RNH}_{2}\right)$ to yield an ammonium salt:


Solution - 5.8 Racemic Mixtures and the Resolution of Enantiomers
To understand how this method of resolution works, let's see what happens when a racemic mixture of chiral acids, such as (+)- and (-)-lactic acids, reacts with an achiral amine base, such as methylamine, $\mathrm{CH}_{3} \mathrm{NH}_{2}$. The situation is analogous to what happens when left and right hands (chiral) pick up a ball (achiral). Both left and right hands pick up the ball equally well, and the products-ball in right hand versus ball in left hand-are mirror images. In the same way, both (+)- and (-)-lactic acid react with methylamine equally well, and the product is a racemic mixture of the two enantiomers methylammonium (+)-lactate and methylammonium (-)-lactate (FIGURE 5.13).
(R)

(S)

Racemic lactic acid
(50\% R, 50\% S)

Racemic ammonium salt
(50\% R, 50\% S)

FIGURE 5.13 Reaction of racemic lactic acid with achiral methylamine gives a racemic mixture of ammonium salts.
Now let's see what happens when the racemic mixture of (+)- and (-)-lactic acids reacts with a single enantiomer of a chiral amine base, such as ( $R$ )-1-phenylethylamine. The situation is analogous to what happens when left and right hands (chiral) put on a right-handed glove (also chiral). Left and right hands don't put on the righthanded glove in the same way, so the products-right hand in right glove versus left hand in right glove-are not mirror images; they're similar but different.


Solution - 5.8 Racemic Mixtures and the Resolution of Enantiomers
In the same way, (+)- and (-)-lactic acids react with ( $R$ )-1-phenylethylamine to give two different products (FIGURE 5.14). ( $R$ )-Lactic acid reacts with ( $R$ )-1-phenylethylamine to give the $R, R$ salt, and ( $S$ )-lactic acid reacts with the $R$ amine to give the $S, R$ salt. The two salts are diastereomers, not enantiomers. They have different chemical and physical properties, and it may therefore be possible to separate them by crystallization or some other means. Once separated, acidification of the two diastereomeric salts with a strong acid makes it possible to isolate the two pure enantiomers of lactic acid and to recover the chiral amine for reuse.

FIGURE 5.14 Reaction of racemic lactic acid with ( $R$ )-1-phenylethylamine yields a mixture of diastereomeric ammonium salts, which have different properties and can be separated.


Predicting the Chirality of a Reaction Product -
We'll see in Section 21.3 that carboxylic acids $\left(\mathrm{RCO}_{2} \mathrm{H}\right)$ react with alcohols ( $\mathrm{R}^{\prime} \mathrm{OH}$ ) to form esters ( $\mathrm{RCO}_{2} \mathrm{R}^{\prime}$ ).

Suppose that ( $\pm$ )-lactic acid reacts with $\mathrm{CH}_{3} \mathrm{OH}$ to form the ester, methyl lactate. What stereochemistry would you expect the product(s) to have? What is the relationship of the products?


Solution -
Reaction of a racemic acid with an achiral alcohol such as methanol yields a racemic mixture of mirror-image (enantiomeric) products.

PROBLEM Suppose that acetic acid $\left(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right)$ reacts with (S)-2-butanol to form an ester (see Worked Example 5-19 5.6). What stereochemistry would you expect the product(s) to have? What is the relationship of the products?

PROBLEM What stereoisomers would result from reaction of ( $\pm$ )-lactic acid with ( $S$ )-1-phenylethylamine, and 5-20 what is the relationship between them?


Solution - 5.9 A Review of Isomerism
As noted on several previous occasions, isomers are compounds with the same chemical formula but different structures. We've seen several kinds of isomers in the past few chapters, and it's a good idea at this point to see how they relate to one another (FIGURE 5.15).

FIGURE 5.15 A summary of the different kinds of isomers.
There are two fundamental types of isomers, both of which we've now encountered: constitutional isomers and stereoisomers.

Constitutional isomers (Section 3.2) are compounds whose atoms are connected differently. Among the kinds of constitutional isomers we've seen are skeletal, functional, and positional isomers.

| Different carbon skeletons | <smiles>CC(C)C</smiles> | and | $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ |
| :---: | :---: | :---: | :---: |
|  | 2-Methylpropane |  | Butane |
| Different functional groups | $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$ | and | $\mathrm{CH}_{3} \mathrm{OCH}_{3}$ |
|  | Ethyl alcohol |  | Dimethyl ether |
| Different position of functional groups | <smiles>CC(C)N</smiles> | and | $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}$ |
|  | Isopropylamine |  | Propylamine |

Stereoisomers (Section 4.2) are compounds whose atoms are connected in the same order but with a different spatial arrangement. Among the kinds of stereoisomers we've seen are enantiomers, diastereomers, and cis-trans isomers of cycloalkanes. Actually, cis-trans isomers are just a subclass of diastereomers because they are non-mirror-image stereoisomers:

```
Enantiomers (nonsuperimposable mirror-image stereoisomers)
```

```
Diastereomers
(nonsuperimposable
non-mirror-image
stereoisomers)
```

Configurational diastereomers


Solution - 5.9 A Review of Isomerism
```
Diastereomers
(nonsuperimposable
non-mirror-image
stereoisomers)
```

Configurational diastereomers

Cis-trans diastereomers (substituents on same side or opposite side of double bond or ring)

(R)-Lactic acid

(2R,3R)-2-Amino-3hydroxybutanoic acid

trans-1,3-Dimethylcyclopentane

(S)-Lactic acid

(2R,3S)-2-Amino-3hydroxybutanoic acid

cis-1,3-Dimethylcyclopentane

PROBLEM What kinds of isomers are the following pairs?
5-21 (a) (S)-5-Chloro-2-hexene and chlorocyclohexane
(b) $(2 R, 3 R)$-Dibromopentane and $(2 S, 3 R)$-dibromopentane


Solution - 5.10 Chirality at Nitrogen, Phosphorus, and Sulfur
As noted previously, the most common cause of chirality in a molecule is the presence of four different substituents bonded to a tetrahedral atom. Although that atom is usually carbon, it doesn't necessarily have to be. Nitrogen, phosphorus, and sulfur atoms are all commonly encountered in organic molecules, and can all be chirality centers. We know, for instance, that trivalent nitrogen is tetrahedral, with its lone pair of electrons acting as the fourth "substituent" (Section 1.10). Is trivalent nitrogen chiral? Does a compound such as ethylmethylamine exist as a pair of enantiomers?

The answer is both yes and no. Yes in principle, but no in practice. It turns out that most trivalent nitrogen compounds undergo a rapid umbrella-like inversion that interconverts enantiomers, so we can't isolate individual enantiomers except in special cases.

A similar situation occurs in trivalent phosphorus compounds, called phosphines, but the inversion at phosphorus is substantially slower than inversion at nitrogen, so stable chiral phosphines can be isolated. ( $R$ )and ( $S$ )-methylpropylphenylphosphine, for instance, are configurationally stable for several hours at $100{ }^{\circ} \mathrm{C}$. We'll see the importance of phosphine chirality in Section 26.7 in connection with the synthesis of chiral amino acids.

(R)-Methylpropylphenylphosphine (configurationally stable)


Solution - 5.10 Chirality at Nitrogen, Phosphorus, and Sulfur
(R)-Methylpropylphenylphosphine (configurationally stable)

Divalent sulfur compounds are achiral, but trivalent sulfur compounds called sulfonium salts $\left(\mathrm{R}_{3} \mathrm{~S}^{+}\right)$can be chiral. Like phosphines, sulfonium salts undergo relatively slow inversion, so chiral sulfonium salts are configurationally stable and can be isolated. Perhaps the best known example is the coenzyme $S$-adenosylmethionine, the so-called biological methyl donor, which is involved in many metabolic pathways as a source of $\mathrm{CH}_{3}$ groups. (The " $S$ " in the name $S$-adenosylmethionine stands for sulfur and means that the adenosyl group is attached to the sulfur atom of the amino acid methionine.) The molecule has $S$ stereochemistry at sulfur and is configurationally stable for several days at room temperature. Its $R$ enantiomer is also known but is not biologically active.

(S)-S-Adenosylmethionine

Adenosine


Solution - 5.11 Prochirality
Closely related to the concept of chirality, and particularly important in biological chemistry, is the notion of prochirality. A molecule is said to be prochiral if it can be converted from achiral to chiral in a single chemical step. For instance, an unsymmetrical ketone like 2-butanone is prochiral because it can be converted to the chiral alcohol 2-butanol by the addition of hydrogen, as we'll see in Section 17.4.

Which enantiomer of 2-butanol is produced depends on which face of the planar carbonyl group undergoes reaction. To distinguish between the possibilities, we use the stereochemical descriptors Re and Si. Rank the
three groups attached to the trigonal, $s p^{2}$-hybridized carbon, and imagine curved arrows from the highest to second-highest to third-highest ranked substituents. The face on which the arrows curve clockwise is designated the Re face (similar to $R$ ), and the face on which the arrows curve counterclockwise is designated the $\boldsymbol{S i}$ face (similar to $S$ ). In this example, addition of hydrogen from the Re face gives ( $S$ )-2-butane, and addition from the Si face gives ( $R$ )-2-butane.

In addition to compounds with planar, $s p^{2}$-hybridized atoms, compounds with tetrahedral, $s p^{3}$-hybridized atoms can also be prochiral. An $s p^{3}$-hybridized atom is said to be a prochirality center if, by changing one of its attached groups, it becomes a chirality center. The $-\mathrm{CH}_{2} \mathrm{OH}$ carbon atom of ethanol, for instance, is a prochirality center because changing one of its attached -H atoms converts it into a chirality center.


Solution - 5.11 Prochirality
Ethanol
To distinguish between the two identical atoms (or groups of atoms) on a prochirality center, we imagine a change that will raise the ranking of one atom over the other without affecting its rank with respect to other attached groups. On the $-\mathrm{CH}_{2} \mathrm{OH}$ carbon of ethanol, for instance, we might imagine replacing one of the ${ }^{1} \mathrm{H}$ atoms (protium) by ${ }^{2} \mathrm{H}$ (deuterium). The newly introduced ${ }^{2} \mathrm{H}$ atom ranks higher than the remaining ${ }^{1} \mathrm{H}$ atom, but it remains lower than other groups attached to the carbon. Of the two identical atoms in the original compound, the atom whose replacement leads to an $R$ chirality center is said to be pro- $\boldsymbol{R}$ and the atom whose replacement leads to an $S$ chirality center is pro-S.

A large number of biological reactions involve prochiral compounds. One of the steps in the citric acid cycle by which food is metabolized, for instance, is the addition of $\mathrm{H}_{2} \mathrm{O}$ to fumarate to give malate. Addition of -OH occurs on the Si face of a fumarate carbon and gives $(S)$-malate as product.

As another example, studies with deuterium-labeled substrates have shown that the reaction of ethanol with the coenzyme nicotinamide adenine dinucleotide $\left(\mathrm{NAD}^{+}\right)$, catalyzed by yeast alcohol dehydrogenase, occurs with exclusive removal of the pro- $R$ hydrogen from ethanol and with addition only to the Re face of $\mathrm{NAD}^{+}$.

Ethanol
NAD ${ }^{+}$
Acetaldehyde
NADH
Determining the stereochemistry of reactions at prochirality centers is a powerful method for studying detailed mechanisms in biochemical reactions. As just one example, the conversion of citrate to cis-aconitate in the citric acid cycle has been shown to occur with loss of a pro- $R$ hydrogen, implying that the OH and H groups leave from opposite sides of the molecule.


Solution - 5.11 Prochirality
PROBLEM Identify the indicated hydrogens in the following molecules as pro- $R$ or pro- $S$ :

5-22 (a)

(S)-Glyceraldehyde
(b)

Phenylalanine

PROBLEM Identify the indicated faces of carbon atoms in the following molecules as Re or Si :

5-23 (a)

Hydroxyacetone
(b)

Crotyl alcohol

PROBLEM The lactic acid that builds up in tired muscles is formed from pyruvate. If the reaction occurs with
5-24 addition of hydrogen to the Re face of pyruvate, what is the stereochemistry of the product?

PROBLEM The aconitase-catalyzed addition of water to cis-aconitate in the citric acid cycle occurs with the
5-25 following stereochemistry. Does the addition of the OH group occur on the Re or Si face of the substrate? What about the addition of the H? Do the H and OH groups add from the same side of the double bond or from opposite sides?

cis-Aconitate
(2R,3S)-Isocitrate


Solution - 5.12 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.

(S)-Fluoxetine (prevents migraine)

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 FIGURE 5.16: one enantiomer fits the receptor perfectly, but the other does not.

FIGURE 5.16 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 $\mathrm{NAD}^{+}$catalyzed by yeast alcohol dehydrogenase. As we saw at the end of Section 5.11, this reaction occurs with exclusive removal of the pro- $R$ hydrogen from ethanol and with addition only to the Re face of the $\mathrm{NAD}^{+}$carbon.


Solution - 5.12 Chirality in Nature and Chiral Environments
We can understand this result by imagining that the chiral enzyme receptor again has three binding sites, as in

FIGURE 5.16. 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 (FIGURE 5.17a). 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 (FIGURE 5.17b).

FIGURE 5.17 (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.

FIGURE 5.18 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.

Penicillin V ( $2 S, 5 R, 6 R$ configuration)
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.

(S)-Ibuprofen (an active analgesic agent)


Chiral Drugs -
(S)-Ibuprofen (an active analgesic agent)

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.


Key Terms -
- 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, $[\alpha]_{D}$


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, $s p^{3}$-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 $\boldsymbol{R}$ (rectus) or $\boldsymbol{S}$ (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 $(1 \rightarrow 2 \rightarrow 3)$ 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.


Summary -
A molecule is prochiral if it can be converted from achiral to chiral in a single chemical step. A prochiral $s p^{2}$-hybridized atom has two faces, described as either $\mathbf{R e}$ or $\mathbf{S i}$. An $s p^{3}$-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- $\boldsymbol{R}$, and the atom whose replacement leads to an $S$ chirality center is pro- $\boldsymbol{S}$.


Visualizing Chemistry -
PROBLEM Which of the following structures are identical? (Green $=\mathrm{Cl}$.)
5-26
(a)

(b)

(c)

(d)

PROBLEM Assign $R$ or $S$ configurations to the chirality centers in the following molecules (blue $=\mathrm{N}$ ):

5-27 (a)

Serine
(b)

Adrenaline

PROBLEM Which, if any, of the following structures represent meso compounds? (Blue $=\mathrm{N}$, green $=\mathrm{Cl}$.)
(a)

(b)

(c)

PROBLEM Assign $R$ or $S$ configuration to each chirality center in pseudoephedrine, an over-the-counter 5-29 decongestant found in cold remedies (blue $=\mathrm{N}$ ).

PROBLEM Orient each of the following drawings so that the lowest-ranked group is toward the rear, and then 5-30 assign $R$ or $S$ configuration:
(a)

(b)

(c)

Chirality and Optical Activity
PROBLEM Which of the following objects are chiral?
5-31
(a) A basketball
(b) A fork
(c) A wine glass
(d) A golf club
(e) A spiral staircase
(f) A snowflake

PROBLEM Which of the following compounds are chiral? Draw them, and label the chirality centers.
5-32 (a) 2,4-Dimethylheptane (b) 5-Ethyl-3,3-dimethylheptane (c) cis-1,4-Dichlorocyclohexane
PROBLEM Draw chiral molecules that meet the following descriptions:
5-33 (a) A chloroalkane, $\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{Cl}$
(b) An alcohol, $\mathrm{C}_{6} \mathrm{H}_{14} \mathrm{O}$
(c) An alkene, $\mathrm{C}_{6} \mathrm{H}_{12}$
(d) An alkane, $\mathrm{C}_{8} \mathrm{H}_{18}$


Visualizing Chemistry -
PROBLEM Eight alcohols have the formula $\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}$. Draw them. Which are chiral?
5-34
PROBLEM Draw compounds that fit the following descriptions:
5-35 (a) A chiral alcohol with four carbons (b) A chiral carboxylic acid with the formula $\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{2}$
(c) A compound with two chirality centers
(d) A chiral aldehyde with the formula $\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{BrO}$

PROBLEM Erythronolide B is the biological precursor of erythromycin, a broad-spectrum antibiotic. How 5-36 many chirality centers does erythronolide B have? Identify them.

Erythronolide B


Assigning Configuration to Chirality Centers -
PROBLEM Which of the following pairs of structures represent the same enantiomer, and which represent
5-37 different enantiomers?
(a)

(b)

(c)

PROBLEM What is the relationship between the specific rotations of ( $2 R, 3 R$ )-dichloropentane and ( $2 S, 3 S$ )-
5-38 dichloropentane? Between $(2 R, 3 S)$-dichloropentane and $(2 R, 3 R)$-dichloropentane?
PROBLEM What is the stereochemical configuration of the enantiomer of ( $2 S, 4 R$ )-2,4-octanediol?
5-39
PROBLEM What are the stereochemical configurations of the two diastereomers of ( $2 S, 4 R$ )-2,4-octanediol? (A
5-40 diol is a compound with two - OH groups.)
PROBLEM Orient each of the following drawings so that the lowest-ranked group is toward the rear, and then
5-41 assign $R$ or $S$ configuration:
(a)

(b)

(c)


Assigning Configuration to Chirality Centers -
(b)

(c)

PROBLEM Assign Cahn-Ingold-Prelog rankings to the following sets of substituents:
5-42 (a)
(a) $-\mathrm{CH}=\mathrm{CH}_{2},-\mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2},-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3},-\mathrm{CH}_{2} \mathrm{CH}_{3}$
(b) $-\mathrm{C} \equiv \mathrm{CH},-\mathrm{CH}=\mathrm{CH}_{2},-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}$,
(c) $-\mathrm{CO}_{2} \mathrm{CH}_{3},-\mathrm{COCH}_{3},-\mathrm{CH}_{2} \mathrm{OCH}_{3},-\mathrm{CH}_{2} \mathrm{CH}_{3}$
(d) $-\mathrm{C} \equiv \mathrm{N},-\mathrm{CH}_{2} \mathrm{Br},-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br},-\mathrm{Br}$

PROBLEM Assign $R$ or $S$ configurations to each chirality center in the following molecules:

5-43 (a)

(b)

(c)

PROBLEM Assign $R$ or $S$ configuration to each chirality center in the following molecules:
5-44 (a)

(b)

(c)

PROBLEM Assign $R$ or $S$ configuration to each chirality center in the following biological molecules:

5-45 (a)

Biotin
(b)

Prostaglandin $\mathrm{E}_{\mathbf{1}}$

PROBLEM Draw tetrahedral representations of the following molecules:
5-46 (a) (S)-2-Chlorobutane
(b) (R)-3-Chloro-1-pentene $\left[\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCH}(\mathrm{Cl}) \mathrm{CH}_{2} \mathrm{CH}_{3}\right]$

PROBLEM Assign $R$ or $S$ configuration to each chirality center in the following molecules:
5-47 (a)

(b)


Assigning Configuration to Chirality Centers -
PROBLEM Assign $R$ or $S$ configuration to each chirality center in the following molecules:
5-47 (a)

(b)

PROBLEM Assign $R$ or $S$ configurations to the chirality centers in ascorbic acid (vitamin C).
5-48


Ascorbic acid -
PROBLEM Assign $R$ or $S$ stereochemistry to the chirality centers in the following Newman projections:
5-49 (a)

(b)

PROBLEM Xylose is a common sugar found in many types of wood, including maple and cherry. Because it
5-50 is much less prone to cause tooth decay than sucrose, xylose has been used in candy and chewing gum. Assign $R$ or $S$ configurations to the chirality centers in xylose.

(+)-Xylose


Meso Compounds -
PROBLEM Draw examples of the following:
5-51 (a) A meso compound with the formula $\mathrm{C}_{8} \mathrm{H}_{18}$ (b) A meso compound with the formula $\mathrm{C}_{9} \mathrm{H}_{20}$
(c) A compound with two chirality centers, one $R$ and the other $S$

PROBLEM Draw the meso form of each of the following molecules, and indicate the plane of symmetry in each: 5-52
(a)

(b)

(c)

PROBLEM Draw the structure of a meso compound that has five carbons and three chirality centers. 5-53

PROBLEM Ribose, an essential part of ribonucleic acid (RNA), has the following structure:
5-54

Ribose
(a) How many chirality centers does ribose have? Identify them.
(b) How many stereoisomers of ribose are there?
(c) Draw the structure of the enantiomer of ribose.
(d) Draw the structure of a diastereomer of ribose.

PROBLEM On reaction with hydrogen gas in the presence of a platinum catalyst, ribose (Problem 5-54) is 5-55 converted into ribitol. Is ribitol optically active or inactive? Explain.

Ribitol

Prochirality
PROBLEM Identify the indicated hydrogens in the following molecules as pro- $R$ or pro- $S$ :

-56 (a)

Malic acid
(b)

Methionine
(c)

Cysteine

PROBLEM Identify the indicated faces in the following molecules as Re or Si :

5-57 (a)

Pyruvate
(b)

Crotonate

PROBLEM One of the steps in fat metabolism is the hydration of crotonate to yield 3-hydroxybutyrate. The
5-58 reaction occurs by addition of -OH to the Si face at C 3 , followed by protonation at C 2 , also from the Si face. Draw the product of the reaction, showing the stereochemistry of each step.

Crotonate

3-Hydroxybutyrate


Meso Compounds -
Crotonate

3-Hydroxybutyrate

PROBLEM The dehydration of citrate to yield cis-aconitate, a step in the citric acid cycle, involves the pro- $R$ 5-59 "arm" of citrate rather than the pro- $S$ arm. Which of the following two products is formed?

PROBLEM The first step in the metabolism of glycerol, formed by digestion of fats, is phosphorylation of the
5-60 pro- $R-\mathrm{CH}_{2} \mathrm{OH}$ group by reaction with adenosine triphosphate (ATP) to give the corresponding glycerol phosphate plus adenosine diphosphate (ADP). Show the stereochemistry of the product.

Glycerol
Glycerol phosphate
PROBLEM One of the steps in fatty-acid biosynthesis is the dehydration of ( $R$ )-3-hydroxybutyryl ACP to give
5-61 trans-crotonyl ACP. Does the reaction remove the pro- $R$ or the pro- $S$ hydrogen from C2?

(R)-3-Hydroxybutyryl ACP

trans-Crotonyl ACP


General Problems -
PROBLEM Draw all possible stereoisomers of 1,2-cyclobutanedicarboxylic acid, and indicate the
5-62 interrelationships. Which, if any, are optically active? Do the same for 1,3-cyclobutanedicarboxylic acid.

PROBLEM Draw tetrahedral representations of the two enantiomers of the amino acid cysteine,
5-63 $\mathrm{HSCH}_{2} \mathrm{CH}\left(\mathrm{NH}_{2}\right) \mathrm{CO}_{2} \mathrm{H}$, and identify each as $R$ or $S$.
PROBLEM The naturally occurring form of the amino acid cysteine (Problem 5-63) has the $R$ configuration
5-64 at its chirality center. On treatment with a mild oxidizing agent, two cysteines join to give cystine, a disulfide. Assuming that the chirality center is not affected by the reaction, is cystine optically active? Explain.

Cysteine
Cystine
PROBLEM Draw tetrahedral representations of the following molecules:
5-65 (a) The $2 S, 3 R$ enantiomer of 2,3-dibromopentane
(b) The meso form of 3,5-heptanediol

PROBLEM Assign $R$ or $S$ configurations to the chiral centers in cephalexin, trade-named Keflex, the most
5-66 widely prescribed antibiotic in the United States.


Cephalexin -
PROBLEM Chloramphenicol, a powerful antibiotic isolated in 1947 from the Streptomyces venezuelae
5-67 bacterium, is active against a broad spectrum of bacterial infections and is particularly valuable against typhoid fever. Assign $R$ or $S$ configurations to the chirality centers in chloramphenicol.


Chloramphenicol -
PROBLEM Allenes are compounds with adjacent carbon-carbon double bonds. Many allenes are chiral, even
5-68 though they don't contain chirality centers. Mycomycin, for example, a naturally occurring antibiotic isolated from the bacterium Nocardia acidophilus, is chiral and has $[\alpha]_{D}=-130$. Explain why mycomycin is chiral.
$\mathrm{HC} \equiv \mathrm{C}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}=\mathrm{C}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}$
Mycomycin
PROBLEM Long before chiral allenes were known (Problem 5-68), the resolution of 5-69 4-methylcyclohexylideneacetic acid into two enantiomers had been carried out. Why is it chiral? What geometric similarity does it have to allenes?

4-Methylcyclohexylideneacetic acid
PROBLEM ( $S$ )-1-Chloro-2-methylbutane undergoes light-induced reaction with $\mathrm{Cl}_{2}$ to yield a mixture of 5-70 products, among which are 1,4-dichloro-2-methylbutane and 1,2-dichloro-2-methylbutane.
(a) Write the reaction, showing the correct stereochemistry of the reactant.
(b) One of the two products is optically active, but the other is optically inactive. Which is which?


Chloramphenicol -
PROBLEM How many stereoisomers of 2,4-dibromo-3-chloropentane are there? Draw them, and indicate
5-71 which are optically active.
PROBLEM Draw both cis- and trans-1,4-dimethylcyclohexane in their more stable chair conformations.
5-72 (a) How many stereoisomers are there of cis-1,4-dimethylcyclohexane, and how many of trans-1,4-dimethylcyclohexane?
(b) Are any of the structures chiral?
(c) What are the stereochemical relationships among the various stereoisomers of 1,4-dimethylcyclohexane?

PROBLEM Draw both cis- and trans-1,3-dimethylcyclohexane in their more stable chair conformations.
5-73 (a) How many stereoisomers are there of cis-1,3-dimethylcyclohexane, and how many of trans-1,3-dimethylcyclohexane?
(b) Are any of the structures chiral?
(c) What are the stereochemical relationships among the various stereoisomers of 1,3-dimethylcyclohexane?

PROBLEM cis-1,2-Dimethylcyclohexane is optically inactive even though it has two chirality centers. Explain. 5-74

PROBLEM We'll see in Chapter 11 that alkyl halides react with hydrosulfide ion ( $\mathrm{HS}^{-}$) to give a product whose
5-75 stereochemistry is inverted from that of the reactant.
Draw the reaction of (S)-2-bromobutane with $\mathrm{HS}^{-}$ion to yield 2-butanethiol, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{SH}) \mathrm{CH}_{3}$. Is the stereochemistry of the product $R$ or $S$ ?


Chloramphenicol -
An alkyl
bromide
PROBLEM Ketones react with sodium acetylide (the sodium salt of acetylene, $\mathrm{Na}^{+-}: \mathrm{C} \equiv \mathrm{CH}$ ) to give alcohols.
5-76 For example, the reaction of sodium acetylide with 2-butanone yields 3-methyl-1-pentyn-3-ol:

(a) Is the product chiral?
(b) Assuming that the reaction takes place with equal likelihood from both Re and Si faces of the carbonyl group, is the product optically active? Explain.

PROBLEM Imagine that a reaction similar to that in Problem 5-76 is carried out between sodium acetylide and
5-77 (R)-2-phenylpropanal to yield 4-phenyl-1-pentyn-3-ol:

( $R$ )-2-Phenylpropanal
4-Phenyl-1-pentyn-3-ol
(a) Is the product chiral?
(b) Draw both major and minor reaction products, assuming that the reaction takes place preferentially from the Re face of the carbonyl group. Is the product mixture optically active? Explain.


An Overview of Organic Reactions -
FIGURE 6.1 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)


CHAPTER CONTENTS - 6.1 Kinds of Organic Reactions
6.2 How Organic Reactions Occur: Mechanisms
6.3 Polar Reactions
6.4 An Example of a Polar Reaction: Addition of HBr to Ethylene
6.5 Using Curved Arrows in Polar Reaction Mechanisms
6.6 Radical Reactions
6.7 Describing a Reaction: Equilibria, Rates, and Energy Changes
6.8 Describing a Reaction: Bond Dissociation Energies
6.9 Describing a Reaction: Energy Diagrams and Transition States
6.10 Describing a Reaction: Intermediates
6.11 A Comparison Between Biological Reactions and Laboratory Reactions

WHY THIS CHAPTER? 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.


CHAPTER CONTENTS - 6.1 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.


This one -
reactant...

Ethylene (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.

PROBLEM Classify each of the following reactions as an addition, elimination, substitution, or rearrangement:
6-1 (a) $\mathrm{CH}_{3} \mathrm{Br}+\mathrm{KOH} \longrightarrow \mathrm{CH}_{3} \mathrm{OH}+\mathrm{KBr}$
(b) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br} \longrightarrow \mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}+\mathrm{HBr}$
(c) $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}+\mathrm{H}_{2} \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{3}$


This one - 6.2 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 ( $\curvearrowright$ ), whereas the movement of one electron in the symmetrical process is indicated using a half-headed, or "fishhook," arrow ( $\AA$ ).

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.


This one - 6.2 How Organic Reactions Occur: Mechanisms
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.


This one - 6.3 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 Section 2.1 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 Section 2.1, 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 ( $\delta+$ ). Metals are less electronegative than carbon, so a carbon atom bonded to a metal has a partial negative charge ( $\delta-$ ). 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).

Chloromethane

Methyllithium

The polarity patterns of some common functional groups are shown in TABLE 6.1. Note that carbon is always positively polarized except when bonded to a metal.

TABLE 6.1 Polarity Patterns in Some Common Functional
Groups

| Compound type | Functional group structure |
| :--- | :--- |
| Alcohol | $\_{\mathrm{C}}^{\mathrm{C+}-\mathrm{O-}}$ |
| Alkene | C= |

Symmetrical, nonpolar


This one - 6.3 Polar Reactions
Symmetrical, nonpolar

| Alkyl halide | <smiles>[X]C(C)(C)C</smiles> |
| :---: | :---: |
| Amine | $\begin{aligned} & \backslash \delta+\delta- \\ & -\mathrm{C}-\mathrm{NH}_{2} \end{aligned}$ |
| Ether | <smiles></smiles> |
| Thiol | <smiles>CC(C)(C)[SnH3]</smiles> |
| Nitrile | $-{ }^{\delta+} \equiv{ }^{\delta-} \mathrm{N}$ |
| Grignard reagent | <smiles></smiles> |
| Alkyllithium | <smiles>CC(C)(C)[Te]</smiles> |

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 $\mathrm{C}-\mathrm{O}$ 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.


This one - 6.3 Polar Reactions
Methanol-weakly
electron-poor carbon
Protonated methanol-
strongly electron-poor carbon
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 (FIGURE 2.3), 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 Section 2.11, 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.


This one - 6.3 Polar Reactions
As we saw in Section 2.11, 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 ( $\mathrm{H}^{+}$donors), alkyl halides, and carbonyl compounds are examples (FIGURE 6.2).

FIGURE 6.2 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 $\mathrm{H}^{+}$ 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.


This one - 6.3 Polar Reactions
If the definitions of nucleophiles and electrophiles sound similar to those given in Section 2.11 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) $\mathrm{NO}_{2}{ }^{+}$
(b) $\mathrm{CN}^{-}$
(c) $\mathrm{CH}_{3} \mathrm{NH}_{2}$
(d) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~S}^{+}$


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) $\mathrm{NO}_{2}{ }^{+}$(nitronium ion) is likely to be an electrophile because it is positively charged.
(b) : $\mathrm{C} \equiv \mathrm{N}^{-}$(cyanide ion) is likely to be a nucleophile because it is negatively charged.
(c) $\mathrm{CH}_{3} \mathrm{NH}_{2}$ (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 $\mathrm{N}-\mathrm{H}$ hydrogens make methylamine a potential acid (electrophile).
(d) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~S}^{+}$(trimethylsulfonium ion) is likely to be an electrophile because it is positively charged.

PROBLEM Which of the following species are likely to be nucleophiles and which electrophiles? Which might 6-2 be both?
(a) $\mathrm{CH}_{3} \mathrm{Cl}$ (b) $\mathrm{CH}_{3} \mathrm{~S}^{-}$
(c)

(d)

PROBLEM An electrostatic potential map of boron trifluoride is shown. Is $\mathrm{BF}_{3}$ likely to be a nucleophile or an 6-3 electrophile? Draw a Lewis structure for $\mathrm{BF}_{3}$, and explain your answer.


Solution - 6.4 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 Section 1.8 that a carbon-carbon double bond results from the orbital overlap of two $s p^{2}$-hybridized carbon atoms. The $\sigma$ part of the double bond results from $s p^{2}-s p^{2}$ overlap, and the $\pi$ part results from $p-p$ overlap.

What kind of chemical reactivity might we expect from a $\mathrm{C}=\mathrm{C}$ bond? We know that alkanes, such as ethane, are relatively inert because all valence electrons are tied up in strong, nonpolar, $\mathrm{C}-\mathrm{C}$ and $\mathrm{C}-\mathrm{H}$ bonds. Furthermore, the bonding electrons in alkanes are relatively inaccessible to approaching reactants because they are sheltered in $\sigma$ 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 $\pi$ 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 (FIGURE 6.3). As a result, the double bond is nucleophilic and the chemistry of alkenes is dominated by reactions with electrophiles.

> Carbon-carbon $\sigma$ bond: stronger; less accessible bonding electrons

Carbon-carbon $\pi$ bond: weaker; more accessible electrons

FIGURE 6.3 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.


Solution - 6.4 An Example of a Polar Reaction: Addition of HBr to Ethylene
What about the second reactant, HBr ? As a strong acid, HBr is a powerful proton $\left(\mathrm{H}^{+}\right)$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 FIGURE 6.4. The reaction begins when the alkene nucleophile donates a pair of electrons from its $\mathrm{C}=\mathrm{C}$ bond to HBr to form a new $\mathrm{C}-\mathrm{H}$ bond plus $\mathrm{Br}^{-}$, as indicated by the path of the curved arrows in the first step of FIGURE 6.4. 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 $\mathrm{C}-\mathrm{H}$ bond forms using electrons from the former $\mathrm{C}=\mathrm{C}$ bond. Simultaneously, a second curved arrow begins in the middle of the $\mathrm{H}-\mathrm{Br}$ bond and points to the Br , indicating that the $\mathrm{H}-\mathrm{Br}$ bond breaks and the electrons remain with the Br atom, giving $\mathrm{Br}^{-}$.


FIGURE 6.4 MECHANISM -
The electrophilic addition reaction of ethylene and HBr . The reaction takes place in two steps, both of which involve electrophile-nucleophile interactions.
(1) A hydrogen atom on the electrophile HBr is attacked by $\pi$ electrons from the nucleophilic double bond, forming a new $\mathrm{C}-\mathrm{H}$ bond. This leaves the other carbon atom with a + charge and a vacant $p$ orbital. Simultaneously, two electrons from the $\mathrm{H}-\mathrm{Br}$ bond move onto bromine, giving bromide anion.

2 Bromide ion donates an electron pair to the positively charged carbon atom, forming a $\mathrm{C}-\mathrm{Br}$ bond and yielding the neutral addition product.

Ethylene
(1)

Bromoethane

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 $\mathrm{Br}^{-}$anion in a second step, forming a $\mathrm{C}-\mathrm{Br}$ bond and yielding the observed addition product. Once again, a curved arrow in FIGURE 6.4 shows the electron-pair movement from $\mathrm{Br}^{-}$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.

PROBLEM What product would you expect from reaction of cyclohexene with HBr ? With HCl ?

PROBLEM Reaction of HBr with 2-methylpropene yields 2-bromo-2-methylpropane. What is the structure of 6-5 the carbocation formed during the reaction? Show the mechanism of the reaction.


FIGURE 6.4 MECHANISM - 6.5 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 $\mathrm{Nu}^{-}{ }^{-}$) to an electrophilic sink ( $\mathbf{E}$ or $\mathbf{E}^{+}$). 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:

Electrons usually flow to one of these electrophiles.

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 $\mathrm{C}=\mathrm{C}$ bond of ethylene to the hydrogen atom of $\mathrm{H}_{3} \mathrm{O}^{+}$, for instance, two electrons must leave that hydrogen. This means that the $\mathrm{H}-\mathrm{O}$ bond must break and the electrons must stay with the oxygen, giving neutral water.

Worked Example 6.2 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 $\mathrm{C}-\mathrm{Br}$ bond has broken and a $\mathrm{C}-\mathrm{C}$ bond has formed. The formation of the $\mathrm{C}-\mathrm{C}$ bond involves donation of an electron pair from the nucleophilic carbon atom of the reactant on the left to the electrophilic carbon atom of $\mathrm{CH}_{3} \mathrm{Br}$, so we draw a curved arrow originating from the lone pair on the negatively charged C atom and pointing to the C atom of $\mathrm{CH}_{3} \mathrm{Br}$. At the same time that the $\mathrm{C}-\mathrm{C}$ bond forms, the $\mathrm{C}-\mathrm{Br}$ bond must break so that the octet rule is not violated. We therefore draw a second curved arrow from the $\mathrm{C}-\mathrm{Br}$ bond to Br . The bromine is now a stable $\mathrm{Br}^{-}$ion.


Solution -
PROBLEM Add curved arrows to the following polar reactions to indicate the flow of electrons in each: 6-6
(a)

(b)

(c)

PROBLEM Predict the products of the following polar reaction, a step in the citric acid cycle for food
6-7 metabolism, by interpreting the flow of electrons indicated by the curved arrows:


Solution - 6.6 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 $\mathbf{1 0 . 2}$ and $\mathbf{1 0 . 3}$ 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 $\left(\mathrm{CH}_{2} \mathrm{Cl}_{2}\right)$ and chloroform $\left(\mathrm{CHCl}_{3}\right)$. The process begins with irradiation of $\mathrm{Cl}_{2}$ with ultraviolet light to break the relatively weak $\mathrm{Cl}-\mathrm{Cl}$ bond of $\mathrm{Cl}_{2}$ and produce chlorine radicals $(\cdot \mathrm{Cl})$.

$$
: \ddot{C}: \ddot{C l}: \ddot{\mathrm{C}}: \quad \xrightarrow{\text { Light }} 2: \ddot{\mathrm{c}} \cdot
$$

Chlorine radicals then react with methane by abstracting a hydrogen atom to give HCl and a methyl radical $\left(\cdot \mathrm{CH}_{3}\right)$ that reacts further with $\mathrm{Cl}_{2}$ 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.


Solution - 6.6 Radical Reactions
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 $\mathrm{O}_{2}$ to give an oxygen radical, which reacts with a $\mathrm{C}=\mathrm{C}$ bond within the same molecule in an addition reaction. Several further transformations ultimately yield prostaglandin $\mathrm{H}_{2}$.

PROBLEM Radical chlorination of alkanes is not generally useful because mixtures of products often result 6-8 when more than one kind of $\mathrm{C}-\mathrm{H}$ bond is present in the substrate. Draw and name all monochloro substitution products $\mathrm{C}_{6} \mathrm{H}_{13} \mathrm{Cl}$ you might obtain by reaction of 2-methylpentane with $\mathrm{Cl}_{2}$.

PROBLEM Using curved fishhook arrows, propose a mechanism for the formation of the cyclopentane ring of 6-9 prostaglandin $\mathrm{H}_{2}$.


Solution - 6.7 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 $K_{\text {eq }}$, 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

$$
a \mathrm{~A}+b \mathrm{~B} \rightleftharpoons c \mathrm{C}+d \mathrm{D}
$$

we have

$$
K_{\mathrm{eq}}=\frac{[\mathrm{C}]^{c}[\mathrm{D}]^{d}}{[\mathrm{~A}]^{a}[\mathrm{~B}]^{b}}
$$

The value of the equilibrium constant tells which side of the reaction arrow is energetically favored. If $K_{\text {eq }}$ is much larger than 1, then the product concentration term [C] ${ }^{c}[\mathrm{D}]^{d}$ is much larger than the reactant concentration term $[\mathrm{A}]^{a}[\mathrm{~B}]^{b}$, and the reaction proceeds as written from left to right. If $K_{\text {eq }}$ is near 1, appreciable amounts of both reactant and product are present at equilibrium. And if $K_{\text {eq }}$ 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 $7.1 \times 10^{7}$ :


Solution - 6.7 Describing a Reaction: Equilibria, Rates, and Energy Changes
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 $7.1 \times 10^{7}$ :

$$
\begin{aligned}
& \mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}+\mathrm{HBr} \rightleftarrows \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br} \\
& K_{\text {eq }}=\frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}\right]}{\left\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}\right}=7.1 \times 10^{7}
\end{aligned}
$$

Because $K_{\text {eq }}$ is relatively large, the reaction proceeds as written and more than $99.99999 \%$ 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 highenergy 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.


Solution - 6.7 Describing a Reaction: Equilibria, Rates, and Energy Changes
The energy change that occurs during a chemical reaction is called the Gibbs free-energy change ( $\boldsymbol{\Delta} \boldsymbol{G} \boldsymbol{G}$ ), which is equal to the free energy of the products minus the free energy of the reactants: $\Delta G=G_{\text {products }}-G_{\text {reactants }}$. For a favorable reaction, $\Delta G$ 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, $\Delta G$ 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 $\Delta G^{\circ}$, where the superscript ${ }^{\circ}$ 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 $\Delta G^{\circ \prime}$ and refers to a reaction carried out at
$\mathrm{pH}=7.0$ with solute concentrations of 1.0 M .

Because the equilibrium constant, $K_{\text {eq }}$, and the standard free-energy change, $\Delta G^{\circ}$, both measure whether a reaction is favorable, they are mathematically related by the equation

$$
\Delta G^{\circ}=-R T \ln K_{\mathrm{eq}} \quad \text { or } \quad K_{\mathrm{eq}}=e^{-\Delta G^{\circ} / R T}
$$

where

$$
\begin{aligned}
& R=8.314 \mathrm{~J} /(\mathrm{K} \cdot \mathrm{~mol})=1.987 \mathrm{cal} /(\mathrm{K} \cdot \mathrm{~mol}) \\
& T=\text { Kelvin temperature } \\
& e=2.718
\end{aligned}
$$


Solution - 6.7 Describing a Reaction: Equilibria, Rates, and Energy Changes
$\ln K_{\text {eq }}=$ natural logarithm of $K_{\text {eq }}$
For example, the reaction of ethylene with HBr has $K_{\mathrm{eq}}=7.1 \times 10^{7}$, so $\Delta G^{\circ}=-44.8 \mathrm{~kJ} / \mathrm{mol}(-10.7 \mathrm{kcal} / \mathrm{mol})$ at 298 K:

$$
\begin{aligned}
K_{\mathrm{eq}} & =7.1 \times 10^{7} \quad \text { and } \quad \ln K_{\mathrm{eq}}=18.08 \\
\Delta G^{\circ} & =-R T \ln K_{\mathrm{eq}}=-8.314 \mathrm{~J} /(\mathrm{K} \cdot \mathrm{~mol})(18.08) \\
& =-44,800 \mathrm{~J} / \mathrm{mol}=-44.8 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
$$

The free-energy change $\Delta G$ is made up of two terms, an enthalpy term, $\Delta H$, and a temperature-dependent entropy term, $T \Delta S$. Of the two terms, the enthalpy term is often larger and more dominant.

$$
\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}
$$

For the reaction of ethylene with HBr at room temperature ( 298 K ), the approximate values are


Solution - 6.7 Describing a Reaction: Equilibria, Rates, and Energy Changes
$$
\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}
$$

For the reaction of ethylene with HBr at room temperature ( 298 K ), the approximate values are

$$
\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}+\mathrm{HBr} \rightleftarrows \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br} \quad\left\{\begin{array}{l}
\Delta G^{\circ}=-44.8 \mathrm{~kJ} / \mathrm{mol} \\
\Delta H^{\circ}=-84.1 \mathrm{~kJ} / \mathrm{mol} \\
\Delta S^{\circ}=-0.132 \mathrm{~kJ} /(\mathrm{K} \cdot \mathrm{~mol}) \\
K_{\mathrm{eq}}=7.1 \times 10^{7}
\end{array}\right.
$$

The enthalpy change $(\boldsymbol{\Delta H})$, also called the heat of reaction, is a measure of the change in total bonding energy during a reaction. If $\Delta H$ 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 $\Delta H$ 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 $380 \mathrm{~kJ} / \mathrm{mol}$ and forms product bonds with a total strength of 400 $\mathrm{kJ} / \mathrm{mol}$, then $\Delta H$ for the reaction is $400 \mathrm{~kJ} / \mathrm{mol}-380 \mathrm{~kJ} / \mathrm{mol}=-20 \mathrm{~kJ} / \mathrm{mol}$ and the reaction is exothermic.


Solution - 6.7 Describing a Reaction: Equilibria, Rates, and Energy Changes
The entropy change ( $\boldsymbol{\Delta} \boldsymbol{S}$ ) 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

$$
\mathbf{A} \longrightarrow \mathbf{B}+\mathbf{C}
$$

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 $\Delta S$ has a positive value.

On the other hand, for an addition reaction of the type

$$
\mathbf{A}+\mathbf{B} \longrightarrow \mathbf{C}
$$

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 $\Delta S$ has a negative value. The reaction of ethylene and HBr to yield bromoethane, which has $\Delta S^{\circ}=-0.132 \mathrm{~kJ} /(\mathrm{K} \cdot \mathrm{mol})$, is an example. TABLE 6.2 describes the thermodynamic terms more fully.

TABLE 6.2 Explanation of Thermodynamic Quantities: $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$


Solution - 6.7 Describing a Reaction: Equilibria, Rates, and Energy Changes
TABLE 6.2 Explanation of Thermodynamic Quantities: $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$

| Term | Name | Explanation |
| :--- | :--- | :--- |
| $\Delta G^{\circ}$ | Gibbs
 free-
 energy
 change | The energy difference between reactants and products. When $\Delta G^{\circ}$ is negative, the reaction
 is exergonic, has a favorable equilibrium constant, and can occur spontaneously. When $\Delta G^{\circ}$
 is positive, the reaction is endergonic, has an unfavorable equilibrium constant, and cannot
 occur spontaneously. |
| $\Delta \boldsymbol{H}^{\circ}$ | Enthalpy
 change | The heat of reaction, or difference in strength between the bonds broken in a reaction and
 the bonds formed. When $\Delta H^{\circ}$ is negative, the reaction releases heat and is exothermic.
 When $\Delta H^{\circ}$ is positive, the reaction absorbs heat and is endothermic. |
| $\Delta \boldsymbol{S}^{\circ}$ | Entropy
 change | The change in molecular randomness during a reaction. When $\Delta S^{\circ}$ is negative, randomness
 decreases. When $\Delta S^{\circ}$ is positive, randomness increases. |

Knowing the value of $K_{\text {eq }}$ 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.


Equilibrium $\rightarrow$ In what direction does the reaction proceed? -
PROBLEM Which reaction is more energetically favored, one with $\Delta G^{\circ}=-44 \mathrm{~kJ} / \mathrm{mol}$ or one with $\Delta G^{\circ}=+44 \mathrm{~kJ} /$ 6-10 mol?

PROBLEM Which reaction is more exergonic, one with $K_{\text {eq }}=1000$ or one with $K_{\text {eq }}=0.001$ ? 6-11


Equilibrium $\rightarrow$ In what direction does the reaction proceed? - 6.8 Describing a Reaction: Bond Dissociation Energies
We've just seen that heat is released (negative $\Delta H$ ) when a bond is formed because the products are more stable and have stronger bonds than the reactants. Conversely, heat is absorbed (positive $\Delta H$ ) 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 $25^{\circ} \mathrm{C}$ is a quantity called the bond strength, or bond dissociation energy ( $D$ ).

$$
\mathrm{A}: \mathrm{B} \xrightarrow[\text { energy }]{\text { Bond dissociation }} \mathrm{A} \cdot+\cdot \mathrm{B}
$$

Each specific bond has its own characteristic strength, and extensive tables of such data are available. For example, a $\mathrm{C}-\mathrm{H}$ bond in methane has a bond dissociation energy $D=439.3 \mathrm{~kJ} / \mathrm{mol}(105.0 \mathrm{kcal} / \mathrm{mol})$, meaning that $439.3 \mathrm{~kJ} / \mathrm{mol}$ must be added to break a $\mathrm{C}-\mathrm{H}$ bond of methane to give the two radical fragments $\cdot \mathrm{CH}_{3}$ and $\cdot$ H. Conversely, $439.3 \mathrm{~kJ} / \mathrm{mol}$ of energy is released when a methyl radical and a hydrogen atom combine to form methane. TABLE 6.3 lists some other bond strengths.

TABLE 6.3 Some Bond Dissociation Energies, $\boldsymbol{D}$


Equilibrium $\rightarrow$ In what direction does the reaction proceed? - 6.8 Describing a Reaction: Bond Dissociation Energies
| Bond | D (kJ/mol) | Bond | D (kJ/mol) | Bond | D (kJ/mol) |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathrm{H}-\mathrm{H}$ | 436 | $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{H}$ | 410 | $\mathrm{C}_{2} \mathrm{H}_{5}-\mathrm{CH}_{3}$ | 370 |
| H-F | 570 | $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{CI}$ | 354 | $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{CH}_{3}$ | 369 |
| H-CI | 431 | $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{Br}$ | 299 | $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{3}$ | 363 |
| $\mathrm{H}-\mathrm{Br}$ | 366 | $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{H}$ | 400 | $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}_{3}$ | 426 |
| $\mathrm{H}-\mathrm{I}$ | 298 | $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CI}$ | 352 | $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCH}_{2}-\mathrm{CH}_{3}$ | 318 |
| $\mathrm{CI}-\mathrm{CI}$ | 242 | $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Br}$ | 293 | $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}$ | 728 |


Equilibrium $\rightarrow$ In what direction does the reaction proceed? - 6.8 Describing a Reaction: Bond Dissociation Energies
| $\mathrm{Br}-\mathrm{Br}$ | 194 | $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{I}$ | 227 | <smiles>Cc1ccccc1</smiles> | 427 |
| I-I | 152 | $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{H}$ | 464 | <smiles>CCc1ccccc1</smiles> | 325 |
| $\mathrm{CH}_{3}-\mathrm{H}$ | 439 | $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CI}$ | 396 | <smiles>CC=O</smiles> | 374 |
| $\mathrm{CH}_{3}-\mathrm{CI}$ | 350 | $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCH}_{2}-\mathrm{H}$ | 369 | $\mathrm{HO}-\mathrm{H}$ | 497 |
| $\mathrm{CH}_{3}-\mathrm{Br}$ | 294 | $\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCH}_{2}-\mathrm{CI}$ | 298 | $\mathrm{HO}-\mathrm{OH}$ | 211 |
| $\mathrm{CH}_{3}-\mathrm{I}$ | 239 | <smiles>c1ccccc1</smiles> | 472 | $\mathrm{CH}_{3} \mathrm{O}-\mathrm{H}$ | 440 |
| $\mathrm{CH}_{3}-\mathrm{OH}$ | 385 | <smiles>Clc1ccccc1</smiles> | 400 | $\mathrm{CH}_{3} \mathrm{~S}-\mathrm{H}$ | 366 |
| $\mathrm{CH}_{3}-\mathrm{NH}_{2}$ | 386 | <smiles>Cc1ccccc1</smiles> | 375 | $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}-\mathrm{H}$ | 441 |


Equilibrium $\rightarrow$ In what direction does the reaction proceed? - 6.8 Describing a Reaction: Bond Dissociation Energies
| $\mathrm{C}_{2} \mathrm{H}_{5}-\mathrm{H}$ | 421 | <smiles>ClCc1ccccc1</smiles> | 300 | <smiles>CC(C)=O</smiles> | 352 |
| $\mathrm{C}_{2} \mathrm{H}_{5}-\mathrm{CI}$ | 352 | <smiles>Brc1ccccc1</smiles> | 336 | $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}-\mathrm{CH}_{3}$ | 355 |


Equilibrium $\rightarrow$ In what direction does the reaction proceed? - 6.8 Describing a Reaction: Bond Dissociation Energies
TABLE 6.3 Some Bond Dissociation Energies, $\boldsymbol{D}$

| Bond | $\boldsymbol{D}(\mathrm{kJ} / \mathrm{mol})$ | Bond | $\boldsymbol{D}(\mathrm{kJ} / \mathrm{mol})$ | Bond | $\boldsymbol{D}(\mathrm{kJ} / \mathrm{mol})$ |
| :--- | :--- | :--- | :--- | :--- | :--- |
| $\mathrm{C}_{2} \mathrm{H}_{5}-\mathrm{Br}$ | 293 |  | 464 | $\mathrm{NH}_{2}-\mathrm{H}$ | 450 |
| $\mathrm{C}_{2} \mathrm{H}_{5}-\mathrm{I}$ | 233 | $\mathrm{HC} \equiv \mathrm{C}-\mathrm{H}$ | 558 | $\mathrm{H}-\mathrm{CN}$ | 528 |
| $\mathrm{C}_{2} \mathrm{H}_{5}-\mathrm{OH}$ | 391 | $\mathrm{CH}_{3}-\mathrm{CH}_{3}$ | 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.


Equilibrium $\rightarrow$ In what direction does the reaction proceed? - 6.8 Describing a Reaction: Bond Dissociation Energies
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 $9 \mathrm{~kJ} / \mathrm{mol}$ of heat is released ( $\Delta H^{\circ}=-9 \mathrm{~kJ} / \mathrm{mol}$ ), but when ATP reacts with water, $30 \mathrm{~kJ} / \mathrm{mol}$ of heat is released $\left(\Delta H^{\circ \prime}=-30 \mathrm{~kJ} / \mathrm{mol}\right)$. 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.
$\Delta H^{0^{\prime}}=-9 \mathrm{~kJ} / \mathrm{mol}$

Adenosine triphosphate (ATP)
Adenosine diphosphate (ADP)


Equilibrium $\rightarrow$ In what direction does the reaction proceed? - 6.9 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 $\pi$ bond and the $\mathrm{H}-\mathrm{Br}$ bond break, a new $\mathrm{C}-\mathrm{H}$ bond forms in step 1 and a new $\mathrm{C}-\mathrm{Br}$ bond forms in step 2.

To depict graphically the energy changes that occur during a reaction, chemists use energy diagrams, such as that in FIGURE 6.5. 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.

FIGURE 6.5 An energy diagram for the first step in the reaction of ethylene with HBr. The energy difference between reactants and the transition state, $\Delta G^{\ddagger}$, defines the reaction rate. The energy difference between reactants and carbocation product, $\Delta G^{\circ}$, 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 FIGURE 6.5. 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 $\mathrm{C}-\mathrm{H}$ bond starts to form. At some point, a structure of maximum energy is reached, a structure called the transition state.


Equilibrium $\rightarrow$ In what direction does the reaction proceed? - 6.9 Describing a Reaction: Energy Diagrams and Transition States
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 $\mathrm{C}=\mathrm{C} \pi$ bond and $\mathrm{H}-\mathrm{Br}$ bond are partially broken and the new $\mathrm{C}-\mathrm{H}$ bond is partially formed (FIGURE 6.6).

FIGURE 6.6 A hypothetical transition-state structure for the first step of the reaction of ethylene with HBr . The $\mathrm{C}=\mathrm{C} \pi$ bond and $\mathrm{H}-\mathrm{Br}$ bond are just beginning to break, and the $\mathrm{C}-\mathrm{H}$ bond is just beginning to form.

The energy difference between reactants and the transition state is called the activation energy, $\boldsymbol{\Delta} \boldsymbol{G}^{*}$, and determines how rapidly the reaction occurs at a given temperature. (The double-dagger superscript, ${ }^{\ddagger}$, 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.


Equilibrium $\rightarrow$ In what direction does the reaction proceed? - 6.9 Describing a Reaction: Energy Diagrams and Transition States
As a rough generalization, many organic reactions have activation energies in the range 40 to $150 \mathrm{~kJ} / \mathrm{mol}$ ( $10-35 \mathrm{kcal} / \mathrm{mol}$ ). The reaction of ethylene with HBr , for example, has an activation energy of approximately $140 \mathrm{~kJ} / \mathrm{mol}(34 \mathrm{kcal} / \mathrm{mol})$. Reactions with activation energies less than $80 \mathrm{~kJ} / \mathrm{mol}$ 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 $-\Delta G^{\ddagger}$ is released. When the reaction continues on to give the carbocation, the new $\mathrm{C}-\mathrm{H}$ 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, $\Delta G^{\circ}$, 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 $\Delta G^{\circ}$, 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 $\Delta G^{\ddagger}$ ) and some are slow (large $\Delta G^{\dagger}$ ); some have a negative $\Delta G^{\circ}$, and some have a positive $\Delta G^{\circ}$. FIGURE 6.7 illustrates some different possibilities.


Equilibrium $\rightarrow$ In what direction does the reaction proceed? - 6.9 Describing a Reaction: Energy Diagrams and Transition States
FIGURE 6.7 Some hypothetical energy diagrams: (a) a fast exergonic reaction (low $\Delta G^{\ddagger}$, negative $\Delta G^{\circ}$ ); (b) a slow exergonic reaction (high $\Delta G^{\ddagger}$, negative $\Delta G^{\circ}$ ); (c) a fast endergonic reaction (small $\Delta G^{\ddagger}$, small positive $\Delta G^{\circ}$ ); (d) a slow endergonic reaction (high $\Delta G^{\ddagger}$, positive $\Delta G^{\circ}$ ).
PROBLEM Which reaction is faster, one with $\Delta G^{\ddagger}=+45 \mathrm{~kJ} / \mathrm{mol}$ or one with $\Delta G^{\ddagger}=+70 \mathrm{~kJ} / \mathrm{mol}$ ?


Equilibrium $\rightarrow$ In what direction does the reaction proceed? - 6.10 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 $\mathrm{H}^{+}$, it reacts further with $\mathrm{Br}^{-}$in a second step to give the final product, bromoethane. This second step has its own activation energy $\left(\Delta G^{\ddagger}\right)$, its own transition state, and its own energy change ( $\Delta G^{\circ}$ ). We can picture the second transition state as an activated complex between the electrophilic carbocation intermediate and the nucleophilic bromide anion, in which $\mathrm{Br}^{-}$donates a pair of electrons to the positively charged carbon atom as the new $\mathrm{C}-\mathrm{Br}$ bond just starts to form.

A complete energy diagram for the overall reaction of ethylene with HBr is shown in FIGURE 6.8. 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 FIGURE 6.8, 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.


Equilibrium $\rightarrow$ In what direction does the reaction proceed? - 6.10 Describing a Reaction: Intermediates
FIGURE 6.8 An energy diagram for the reaction of ethylene with $\mathbf{H B r}$. Two separate steps are involved, each with its own activation energy $\left(\Delta G^{\ddagger}\right)$ and free-energy change ( $\Delta G^{\circ}$ ). The overall $\Delta G^{\ddagger}$ for the complete reaction is the energy difference between reactants and the highest transition state (which corresponds to $\Delta G 1^{\ddagger}$ in this case), and the overall $\Delta G^{\circ}$ 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 $\Delta G^{\ddagger}$ and its own $\Delta G^{\circ}$. 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 $\Delta G^{\circ}$ 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 FIGURE 6.9.

FIGURE 6.9 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 $\Delta G^{\ddagger}$, and a highly exergonic reaction has a large negative $\Delta G^{\circ}$.


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 $\Delta G^{\ddagger}$ and $\Delta G^{\circ}$ for the overall reaction.


Strategy -
A two-step reaction has two transition states and an intermediate between them. The $\Delta G^{\ddagger}$ 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 $\Delta G^{\circ}$.