7.8 The Driving Force for Chemical Reactions

So far, our focus on elementary steps has been on how they occur, using curved arrows to account for bonds breaking and bonds forming. But why should these elementary steps occur in the first place? That is, what is their driving force?

The driving force for a reaction reflects the extent to which the reaction favors products over reactants, and that tendency increases with increasing stability of the products relative to the reactants. Thus, to understand a reaction’s driving force, it is a matter of identifying and evaluating the factors that help stabilize the products, destabilize the reactants, or both.

Although there are various factors that dictate the stability of a species, we can often make reasonable predictions about a reaction’s driving force by examining just two factors. One is charge stability, which, as we saw in Chapter 6, can vary dramatically from one species to another. The other is total bond energy. As we saw in Chapter 1, each covalent bond provides stability to a molecule, and the amount of stabilization depends heavily on the type of bond that is formed—that is, which atoms are bound together, and whether it is by a single, double, or triple bond. (Review the tables of bond energies in Tables 1-2 and 1-3 on p. 10.)

A reaction’s driving force generally increases with:

●  Greater charge stabilization in the products relative to the reactants.

●  Greater total bond energy in the products relative to the reactants.

With this in mind, we can see that a coordination step such as the one in Equation 7-28 is unambiguously driven toward products.

A chemical reaction compares the charge and stability of a compound before and after coordination. The reaction shows bromine anion carrying four lone pairs of electron reacting with a compound having a central carbon atom, further carrying a positive charge surrounded by three methyl groups linked by a single bond each. A curved arrow points from the bromine anion toward the central carbon atom. Two bullet points above the structure read, �Two formal charges; less stability.� It is followed by a rightward arrow denoting coordination to show a condensed structural formula of a compound highlighting an addition of a bromine atom carrying three lone pairs of electrons to the central carbon atom carrying three methyl groups. Two bullet points above the structure read, �New covalent bond; greater stability.�

Charge stability heavily favors products because there are two formal charges in the reactants but no formal charges in the products. Furthermore, total bond energy favors products because, during the course of the reaction, one covalent bond is formed, giving carbon an octet, and no bonds are broken.

Charge stability and bond energy do not always work in the same direction. Consider the proton transfer step in Equation 7-29.

A chemical reaction represents a change in bond energy and charge stability during proton transfer. The chemical reaction shows a nitrogen atom of an amino group carrying two lone pairs of electrons and a negative charge reacting with hydrogen chloride, with chlorine carrying three lone pairs of electrons. A curved arrow points from a nitrogen atom toward the hydrogen atom of hydrogen chloride. Another curved arrow points from a single bond between hydrogen and chlorine toward chlorine. The text above the reactants read, single bond between hydrogen and chlorine is broken and is stronger than the single bond between hydrogen and nitrogen. It is followed by a rightward arrow denoting proton transfer to show an addition of a hydrogen atom to the nitrogen atom by a single bond, and the release of chlorine anion carrying four lone pairs of electrons. A text reads, �The negative charge is better accommodated on chlorine than nitrogen.� The value of equilibrium constant is equal to 1 times 10 to the power of 43.

In this case, a negative charge appears in both the reactants and the products. In the reactants, the negative charge is on N, whereas it is on Cl in the products. Because Cl is significantly larger (i.e., lower in the periodic table) than N, the negative charge is better accommodated on Cl than on N, so we say that charge stability favors products. Bond energy, however, favors the reactants because the H—Cl bond that appears on the reactant side (431 kJ/mol; 103 kcal/mol) is stronger than the H—N bond that appears on the product side (389 kJ/mol; 93 kcal/mol). Despite the disagreement, notice that the products are heavily favored; HCl is a much stronger acid than NH3, giving rise to a very large Keq of 1 × 1043.

YOUR TURN 7.14
SHOW ANSWERS

Consult Appendix A to verify that HCl is a stronger acid than NH3. Write the pKa values of the two acids below. Do these pKa values agree with the fact that the reaction in Equation 7-29 heavily favors products?

pKa of HCl ________________ pKa of NH3 ________________

HCl pKa =–7; NH3 pKa = 36. HCl is a stronger acid than NH3 (it has a more negative pKa) and the reaction favors the product side in Equation 7-29 (away from the stronger acid). This is confirmed by the large, positive equilibrium constant (Keq = 1043).

The previous example leads to the following general rule:

When charge stability and bond energy favor opposite sides of a chemical reaction, charge stability usually wins.

This idea is very useful when considering the E2 step in Equation 7-30.

A chemical reaction represents the formation of bonds and charge stability during E2 step. The reaction shows the oxygen atom of a hydroxyl group carrying two lone pairs of electrons and a negative charge; two CH2 group linked by a single bond with the left group linked to a hydrogen atom by a single bond while the right group is linked to a chlorine ion carrying three lone pairs of electrons by a single bond. The bond between CH2 and hydrogen; and between CH2 and chlorine are labeled sigma bonds formed. A curved arrow points from an oxygen atom of hydroxyl group toward hydrogen atom while another curved arrow points from a single bond between CH2 and hydrogen toward single bond between two CH 2 groups. A third curved arrow is drawn, pointing from a single bond between CH2 and chlorine toward chlorine. It is followed by a rightward arrow denoting E2, to show an addition of a hydrogen atom to the hydroxyl group by a single bond, with the oxygen atom carrying two lone pairs of electrons. The single bond is labeled, sigma bond formed. It also shows the formation of a double bond between two CH2 groups and labeled pi-bond formed; and release of a chlorine anion carrying four lone pairs of electrons, labeled the negative charge is better stabilized on chlorine than oxygen.

Charge stability favors the products because the negative charge is better accommodated on Cl than on O. Bond energy, however, favors the reactants because, on the product side, a σ and a π bond are formed, while two σ bonds are broken. Effectively, there is a net gain of one σ bond and a net loss of one π bond and, as we learned in Chapter 3, a σ bond is generally stronger. Experimentally, we know that the product side of the reaction is heavily favored, in agreement with the prediction obtained using charge stability.

Solved Problem 7.19

Determine which side of this carbocation rearrangement is favored.

A chemical reaction represents choice for favorable carbocation rearrangement. The reaction shows a closed six-ring structure with two vacant single bonds at carbon-1 and 2, with a positive charge at carbon-2. A curved arrow from a single bond at carbon-1 terminates at carbon-2 of the closed ring. It is followed by a rightward arrow to show a closed six-ring structure with a positive charge at carbon-1 and two vacant single bonds at carbon-2.

Think
SHOW SECTION

On the two sides of the reaction, is there a difference in charge stability? Is there a difference in total bond energy?

Solve
SHOW SECTION

The positive charge is on a tertiary carbon in the reactant and is on a secondary carbon in the product. Because a tertiary carbocation is more stable (Section 6.6e), charge stability favors the reactant side. Total bond energy is not a significant factor in this reaction because a C—C σ bond is broken in the reactant and another one is formed in the product. Overall, then, this reaction will favor the reactant side.

problem 7.20 Determine which side of each of the following carbocation rearrangements is favored.

Two chemical reactions to represent the choice for favorable site of a carbocation rearrangement. The first chemical reaction shows a closed six-ring structure with alternate double bonds and single bonds, and a side chain of four carbon atoms emerging at carbon-2 of the closed ring. A vacant single bond emerges from carbon-2 of the side chain carrying a positive charge, marked at carbon-3. A curved arrow points from a vacant single bond on the side chain toward carbon-3 carrying a positive charge. The resultant shows an exchange of the charge and a single bond between carbon-2 and 3 of the side chain. The second chemical reaction shows a closed six-ring structure with a hydrogen atom linked to carbon-1 by a single bond. It also shows a vacant single bond at carbon-1 and a positive charge at carbon-2. A curved arrow points from a hydrogen atom toward the carbon atom carrying positive charge. The resultant shows a shift of positive charge to carbon-1 and shift of the hydrogen atom to carbon-2.

Caution must be taken when using charge stability and total bond energy to predict the driving force for an elementary step. One reason is that a reaction that is product favored is not guaranteed to occur at a rate that is practical. For example, both charge stability and total bond energy heavily favor the products of the SN2 reaction in Equation 7-31, but essentially no reaction occurs.

A chemical reaction shows a SN 2 chemical reaction not favored as product stability that does not occur at a practical rate. Its shows a chlorine anion carrying four lone pairs of electrons reacting with a propane carrying a double bond between C1 and C2 and a bromine carrying three lone pairs of electrons at C3. A curved arrow points from chlorine to C2 while another curved arrow points toward bromine from a single bond between C2 and C3. It is followed by a rightward arrow, labeled SN 2 marked with a red cross to show a replacement of bromine with chlorine carrying three lone pairs of electrons and a release of bromine anion carrying four lone pairs of electrons labeled, the negative charge is more stable on bromine than chlorine.

Understanding the reason for this requires reaction kinetics, a topic discussed in Chapter 9.

A second reason to be cautious is that other factors can come into play. Consider, for example, the heterolysis step in Equation 7-32, which is the reverse of the coordination step in Equation 7-28.

A chemical reaction represents hydrolysis resulting in the breakage of bonds to form two charges. It shows a condensed structural formula of a compound having a central carbon atom surrounded by three methyl groups and a bromine ion carrying three lone pairs of electrons linked by a single bond each in a tetrahedral arrangement. A curved arrow from a bond between carbon and bromine, labeled bond broken, is shown to point toward bromine. It is followed by a rightward arrow denoting heterolysis to show a release of bromine anion carrying four lone pairs of electrons and a central carbon carrying a positive charge, surrounded by three methyl groups linked by a single bond, each in a trigonal planar arrangement. The two charges are labeled two formal charges.

It might seem that this step shouldn’t take place at all because both charge stability and total bond energy heavily disfavor the products. However, heterolysis steps such as this one often appear in multistep mechanisms. As it turns out, factors that influence the driving force include effects from the solvent, as well as entropy—topics that we discuss in Chapter 9.

By and large, charge stability and total bond energy are the dominant contributors to the driving force of elementary steps in most situations. These are the ones you should focus on the most. We will discuss other factors in the context of the reactions in which they are relevant.