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Ch. 5 - Chemical Reaction Analysis: Thermodynamics and Kinetics
Mullins - Organic Chemistry: A Learner Centered Approach 1st Edition
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Tutti i libri di testoMullins 1st EditionCh. 5 - Chemical Reaction Analysis: Thermodynamics and KineticsProblema 60
Capitolo 4, Problema 60

Keto–enol tautomerism is a reaction we discuss in detail in Chapter 19. Estimate the equilibrium constant of this reaction (BDE for C―C π bond = 65 kcal/mol ; for C―O π bond = 85 kcal/mol).
Keto-enol tautomerism reaction showing the equilibrium between a carbonyl compound and its enol form.

Guida verificata passo dopo passo
1
Understand the concept of keto-enol tautomerism: This is a chemical equilibrium between a ketone (or aldehyde) and its enol form. The equilibrium constant depends on the relative stability of the keto and enol forms, which can be estimated using bond dissociation energies (BDEs).
Identify the relevant bond dissociation energies (BDEs): The problem provides the BDE for the C=C π bond (65 kcal/mol) and the C=O π bond (85 kcal/mol). These values will help us compare the stability of the keto and enol forms.
Calculate the energy difference between the keto and enol forms: The keto form has a C=O π bond, while the enol form has a C=C π bond. The energy difference can be calculated as ΔE = BDE(C=O) - BDE(C=C). Use MathML to represent this calculation: ΔE=BDE(C=O)-BDE(C=C)
Relate the energy difference to the equilibrium constant: The equilibrium constant (K_eq) can be estimated using the relationship ΔG = -RT ln(K_eq), where ΔG is the Gibbs free energy change (approximately equal to ΔE in this case). Rearrange to solve for K_eq: Keq=e-ΔE/RT
Substitute values into the equation: Use the calculated ΔE, the gas constant R (approximately 1.987 cal/(mol·K)), and the temperature T (assume room temperature, 298 K) to estimate K_eq. Ensure units are consistent when performing the calculation.

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Keto-Enol Tautomerism

Keto-enol tautomerism is a chemical equilibrium between a keto form (a carbonyl compound) and its corresponding enol form (an alcohol with a double bond). This process involves the migration of a hydrogen atom and the shift of a double bond, which can significantly influence the reactivity and stability of organic compounds. Understanding this equilibrium is crucial for predicting the behavior of compounds in various chemical reactions.
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Tautomerization Mechanisms

Bond Dissociation Energy (BDE)

Bond dissociation energy (BDE) is the energy required to break a specific bond in a molecule, resulting in the formation of free radicals. In the context of keto-enol tautomerism, the BDE values for the C-C π bond and the C-O π bond are essential for estimating the stability of the keto and enol forms. Higher BDE values indicate stronger bonds, which can affect the position of the equilibrium between the two tautomers.
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How to calculate enthalpy using bond dissociation energies.

Equilibrium Constant (K)

The equilibrium constant (K) quantifies the ratio of the concentrations of products to reactants at equilibrium for a reversible reaction. In the case of keto-enol tautomerism, K can be calculated using the BDE values, reflecting the relative stability of the keto and enol forms. A larger K value indicates a greater concentration of products (usually the more stable form), while a smaller K suggests that the reactants are favored.
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The relationship between equilibrium constant and pKa.
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Reaction (c), on the other hand, is favored (∆G° < 0). Identify the bonds formed and broken and explain this result in light of (a) and (b).

(c)

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The hydrogenation of alkenes is a reaction we study in Chapter 9.

(b) Is this reaction favored or disfavored in terms of entropy?

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Parts (a)–(d) of this assessment assist in the development of what will become a common theme in organic reactions and should be worked in order. [Think carefully about how each question relates to the others.]

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(b) Mechanistically, the reaction occurs as shown below. Why is this reaction favored? Based on the stability of the anions, estimate Keq.

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Calculate Keq for the following acid–base reaction.

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Reactions (a) and (b) are disfavored overall (∆G° > 0), yet they are favored based on ∆H°. Identify the bonds formed and broken for (a) and (b).

(b)

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