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?
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Chloramphenicol -

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?


Chloramphenicol -

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


Chloramphenicol -

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 $S$ ?
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0163.jpg?height=126&width=508&top_left_y=189&top_left_x=413)


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:
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0163.jpg?height=218&width=825&top_left_y=539&top_left_x=414)
(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:
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0163.jpg?height=285&width=874&top_left_y=1012&top_left_x=417)
( $R$ )-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 -


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 -


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.


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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0165.jpg?height=240&width=1104&top_left_y=986&top_left_x=538)
- 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 -


This one -

Ethylene (an alkene)


This one -

- 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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0165.jpg?height=228&width=1421&top_left_y=1740&top_left_x=379)
- 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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0165.jpg?height=286&width=1231&top_left_y=2139&top_left_x=474)


This one -

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 ( $\AA$ ).
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0166.jpg?height=185&width=1042&top_left_y=1043&top_left_x=539)


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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0166.jpg?height=190&width=1080&top_left_y=1366&top_left_x=517)


Processes that involve unsymmetrical bond-breaking and bond-making are called polar reactions. Polar reactions involve species that have an even number of electrons and thus have only electron pairs in their orbitals. Polar processes are by far the more common reaction type in both organic and biological chemistry, and a large part of this book is devoted to their description.


Processes that involve symmetrical bond-breaking and bond-making are called radical reactions. A radical, often called a free radical, is a neutral chemical species that contains an odd number of electrons and thus has a single, unpaired electron in one of its orbitals.


In addition to polar and radical reactions, there is a third, less commonly encountered process called a pericyclic reaction. Rather than explain pericyclic reactions now, though, we'll look at them more carefully in Chapter 30.


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.


This one - 6.3 Polar Reactions

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.


This one - 6.3 Polar Reactions

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).
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0167.jpg?height=308&width=246&top_left_y=445&top_left_x=552)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0167.jpg?height=158&width=128&top_left_y=539&top_left_x=847)


This one - 6.3 Polar Reactions

Chloromethane
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0167.jpg?height=245&width=139&top_left_y=458&top_left_x=1123)


This one - 6.3 Polar Reactions

Methyllithium
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0167.jpg?height=302&width=256&top_left_y=451&top_left_x=1306)


This one - 6.3 Polar Reactions

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


This one - 6.3 Polar Reactions

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


This one - 6.3 Polar Reactions

Symmetrical, nonpolar


This one - 6.3 Polar Reactions

| 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 one - 6.3 Polar Reactions


This one - 6.3 Polar Reactions

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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0169.jpg?height=313&width=977&top_left_y=207&top_left_x=571)


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.
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This one - 6.3 Polar Reactions

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.


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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0169.jpg?height=299&width=952&top_left_y=1493&top_left_x=581)


This one - 6.3 Polar Reactions

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).
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0170.jpg?height=945&width=1456&top_left_y=178&top_left_x=329)


This one - 6.3 Polar Reactions

FIGURE 6.2 Some nucleophiles and electrophiles. Electrostatic potential maps identify the nucleophilic (negative) and electrophilic (positive) atoms.


This one - 6.3 Polar Reactions

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.


WORKED EXAMPLE 6.1 -


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.


Solution -


Solution -

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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0171.jpg?height=267&width=373&top_left_y=818&top_left_x=426)


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
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0171.jpg?height=706&width=1337&top_left_y=1357&top_left_x=386)


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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0172.jpg?height=554&width=909&top_left_y=432&top_left_x=578)


> Carbon-carbon $\sigma$ bond: stronger; less accessible bonding electrons
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0172.jpg?height=296&width=392&top_left_y=741&top_left_x=1105)


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.


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.


FIGURE 6.4 MECHANISM -


FIGURE 6.4 MECHANISM -

Bromoethane


FIGURE 6.4 MECHANISM -

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.


FIGURE 6.4 MECHANISM -

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.


FIGURE 6.4 MECHANISM -

PROBLEM What product would you expect from reaction of cyclohexene with HBr ? With HCl ?
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0173.jpg?height=144&width=516&top_left_y=2118&top_left_x=344)


FIGURE 6.4 MECHANISM -

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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0174.jpg?height=218&width=765&top_left_y=192&top_left_x=420)


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:
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0174.jpg?height=150&width=1069&top_left_y=754&top_left_x=528)


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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0174.jpg?height=159&width=753&top_left_y=1002&top_left_x=745)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0174.jpg?height=139&width=229&top_left_y=1009&top_left_x=1493)


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:
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0174.jpg?height=215&width=847&top_left_y=1345&top_left_x=642)

If the nucleophile is neutral, the atom that donates the electron pair acquires a positive charge. For example:
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0174.jpg?height=259&width=809&top_left_y=1605&top_left_x=661)

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:
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0174.jpg?height=243&width=828&top_left_y=1998&top_left_x=643)

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:
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0175.jpg?height=236&width=881&top_left_y=191&top_left_x=622)


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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0175.jpg?height=304&width=1044&top_left_y=908&top_left_x=535)


RULE 4 -

Worked Example 6.2 gives another example of drawing curved arrows.


(8) WORKED EXAMPLE 6.2 -


Using Curved Arrows in Reaction Mechanisms -

Add curved arrows to the following polar reaction to show the flow of electrons:
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0175.jpg?height=182&width=965&top_left_y=1511&top_left_x=580)


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 -


Solution -

PROBLEM Add curved arrows to the following polar reactions to indicate the flow of electrons in each: 6-6
(a)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0176.jpg?height=172&width=772&top_left_y=204&top_left_x=476)
(b)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0176.jpg?height=159&width=884&top_left_y=384&top_left_x=477)
(c)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0176.jpg?height=177&width=719&top_left_y=551&top_left_x=475)


Solution -

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:
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0176.jpg?height=359&width=516&top_left_y=853&top_left_x=411)


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.


Solution - 6.6 Radical Reactions

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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0176.jpg?height=264&width=719&top_left_y=1646&top_left_x=703)


Solution - 6.6 Radical Reactions

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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0176.jpg?height=305&width=730&top_left_y=2013&top_left_x=687)


Solution - 6.6 Radical Reactions

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 $(\cdot \mathrm{Cl})$.


Solution - 6.6 Radical Reactions

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


Solution - 6.6 Radical Reactions

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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0177.jpg?height=180&width=716&top_left_y=485&top_left_x=702)


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.


Solution - 6.6 Radical Reactions

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.)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0177.jpg?height=495&width=1201&top_left_y=1048&top_left_x=462)


Solution - 6.6 Radical Reactions

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 $\mathrm{H}_{2}$.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0177.jpg?height=654&width=1356&top_left_y=1684&top_left_x=382)


Solution - 6.6 Radical Reactions

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 $\mathrm{Cl}_{2}$.


Solution - 6.6 Radical Reactions

PROBLEM Using curved fishhook arrows, propose a mechanism for the formation of the cyclopentane ring of 6-9 prostaglandin $\mathrm{H}_{2}$.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0178.jpg?height=223&width=1240&top_left_y=295&top_left_x=410)


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 $7.1 \times 10^{7}$ :


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


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.


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 .
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0179.jpg?height=416&width=817&top_left_y=253&top_left_x=649)


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


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


$\ln K_{\text {eq }}=$ 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


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


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.


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


| 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.


Rate $\rightarrow$ Is the reaction fast or slow? -


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, $\boldsymbol{D}$


| 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 |
| $\mathrm{Br}-\mathrm{Br}$ | 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 |
| $\mathrm{C}_{2} \mathrm{H}_{5}-\mathrm{H}$ | 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 |


| 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.


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.
$\Delta H^{0^{\prime}}=-9 \mathrm{~kJ} / \mathrm{mol}$
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0182.jpg?height=294&width=1231&top_left_y=1422&top_left_x=336)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0182.jpg?height=481&width=1443&top_left_y=1752&top_left_x=338)
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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0183.jpg?height=332&width=838&top_left_y=178&top_left_x=638)


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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0183.jpg?height=619&width=925&top_left_y=813&top_left_x=600)


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.


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).
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0183.jpg?height=343&width=475&top_left_y=1994&top_left_x=814)

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.


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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0184.jpg?height=1031&width=1259&top_left_y=1249&top_left_x=430)


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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0185.jpg?height=313&width=825&top_left_y=369&top_left_x=647)

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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0185.jpg?height=730&width=995&top_left_y=1294&top_left_x=565)

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.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0186.jpg?height=568&width=971&top_left_y=388&top_left_x=577)

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.


WORKED EXAMPLE 6.3 -


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 $\Delta G^{\circ}$.


Solution -


WORKED EXAMPLE 6.4 -


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 $\Delta G^{\circ}$.


Solution -

PROBLEM Sketch an energy diagram for a two-step reaction in which both steps are exergonic and in which
6-13 the second step has a higher-energy transition state than the first. Label the parts of the diagram corresponding to reactant, product, intermediate, overall $\Delta G^{\ddagger}$, and overall $\Delta G^{\circ}$.


Solution - 6.11 A Comparison Between Biological Reactions and Laboratory Reactions

Beginning in the next chapter, we'll be seeing a lot of reactions, some that are important in laboratory chemistry yet don't occur in nature and others that have counterparts in biological pathways. In comparing laboratory reactions with biological reactions, several differences are apparent. For one, laboratory reactions are usually carried out in an organic solvent such as diethyl ether or dichloromethane to dissolve the reactants and bring them into contact, whereas biological reactions occur in the aqueous medium within cells. For another, laboratory reactions often take place over a wide range of temperatures without catalysts, while biological reactions take place at the temperature of the organism and are catalyzed by enzymes.


We'll look at enzymes in more detail in Section 26.10, but you may already be aware that an enzyme is a large, globular, protein molecule that contains in its structure a protected pocket called its active site. The active site is lined by acidic or basic groups as needed for catalysis and has precisely the right shape to bind and hold a substrate molecule in the orientation necessary for reaction. FIGURE 6.10 shows a molecular model of hexokinase, along with an X-ray crystal structure of the glucose substrate and adenosine diphosphate (ADP) bound in the active site. Hexokinase is an enzyme that catalyzes the initial step of glucose metabolism-the transfer of a phosphate group from ATP to glucose, giving glucose 6-phosphate and ADP. The structures of ATP and ADP were shown at the end of Section 6.8.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0187.jpg?height=326&width=830&top_left_y=1675&top_left_x=645)


Note how the hexokinase-catalyzed phosphorylation reaction of glucose is written. It's common when writing biological equations to show only the structures of the primary reactant and product, while abbreviating the structures of various biological "reagents" and by-products such as ATP and ADP. A curved arrow intersecting the straight reaction arrow indicates that ATP is also a reactant and ADP also a product.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0188.jpg?height=898&width=1235&top_left_y=191&top_left_x=445)


FIGURE 6.10 Models of hexokinase in space-filling and wire-frame formats, showing the cleft that contains the active site where substrate binding and reaction catalysis occur. At the bottom is an X-ray crystal structure of the enzyme active site, showing the positions of both glucose and ADP as well as a lysine amino acid that acts as a base to deprotonate glucose.