Table of contents

1. A Review of General Chemistry5h 5m
2. Molecular Representations1h 14m
3. Acids and Bases2h 46m
4. Alkanes and Cycloalkanes4h 17m
5. Chirality3h 39m
6. Thermodynamics and Kinetics1h 22m
7. Substitution Reactions1h 48m
8. Elimination Reactions2h 30m
9. Alkenes and Alkynes2h 9m
10. Addition Reactions3h 19m
11. Radical Reactions1h 58m
12. Alcohols, Ethers, Epoxides and Thiols2h 42m
13. Alcohols and Carbonyl Compounds2h 17m
14. Synthetic Techniques1h 26m
15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
16. Conjugated Systems6h 13m
17. Ultraviolet Spectroscopy51m
18. Aromaticity2h 34m
19. Reactions of Aromatics: EAS and Beyond5h 1m
20. Phenols55m
21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
22. Carboxylic Acid Derivatives: NAS2h 51m
23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
24. Enolate Chemistry: Reactions at the Alpha-Carbon1h 53m
25. Condensation Chemistry2h 9m
26. Amines1h 43m
27. Heterocycles2h 0m
28. Carbohydrates5h 53m
29. Amino Acids4h 20m
30. Peptides and Proteins2h 42m
31. Catalysis in Organic Reactions1h 30m
32. Lipids 2h 50m
33. The Organic Chemistry of Metabolic Pathways2h 52m
34. Nucleic Acids1h 32m
35. Transition Metals6h 14m
36. Synthetic Polymers1h 49m

6. Thermodynamics and Kinetics

Gibbs Free Energy

Organic Chemistry

6. Thermodynamics and Kinetics

Gibbs Free Energy

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7:59 minutes

Problem 20a

Textbook Question

Calculate the percentage of isopropylcyclohexane molecules that have the isopropyl substituent in an equatorial position at equilibrium. (Its ∆G° value at 25 °C is -2.1 kcal/mol.)

Verified step by step guidance

Step 1: Understand the problem. The question asks for the percentage of isopropylcyclohexane molecules with the isopropyl group in the equatorial position at equilibrium. This involves using the Gibbs free energy change (∆G°) to calculate the equilibrium constant (K) and then determining the percentage distribution between equatorial and axial positions.

Step 2: Recall the relationship between ∆G° and the equilibrium constant (K). The formula is:

∆G° = -RT ln(K)

, where R is the gas constant (1.987 cal/(mol·K)) and T is the temperature in Kelvin (25 °C = 298 K). Rearrange the formula to solve for K:

K = e

^-∆G°/(RT).

Step 3: Substitute the given values into the formula. Use ∆G° = -2.1 kcal/mol (convert to cal: -2100 cal/mol), R = 1.987 cal/(mol·K), and T = 298 K. Calculate K using the formula:

K = e

^{-(-2100)/(1.987 × 298)}. This will give the equilibrium constant, which represents the ratio of molecules in the equatorial position to those in the axial position.

Step 4: Use the equilibrium constant (K) to determine the percentage of molecules in the equatorial position. The ratio of equatorial to axial molecules is given by K, and the total number of molecules is the sum of equatorial and axial molecules. The percentage of equatorial molecules can be calculated using the formula:

% Equatorial = (K / (1 + K)) × 100

Step 5: Interpret the result. Once the percentage is calculated, it represents the proportion of isopropylcyclohexane molecules with the isopropyl group in the equatorial position at equilibrium. This is the more stable conformation due to steric hindrance being minimized in the equatorial position.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Equatorial vs. Axial Position

In cyclohexane derivatives, substituents can occupy equatorial or axial positions. Equatorial positions are generally more stable due to reduced steric strain and torsional strain, allowing larger groups to be farther from other atoms. Understanding the preference for equatorial positioning is crucial for predicting the stability of isopropylcyclohexane.

Gibbs Free Energy (∆G°)

Gibbs Free Energy (∆G°) indicates the spontaneity of a reaction or process at standard conditions. A negative ∆G° value, such as -2.1 kcal/mol, suggests that the formation of the equatorial isopropylcyclohexane is favored at equilibrium. This concept is essential for calculating the distribution of conformers in the equilibrium state.

Equilibrium Constant (K)

The equilibrium constant (K) relates the concentrations of reactants and products at equilibrium. It can be derived from the Gibbs Free Energy change using the equation ∆G° = -RT ln(K). By calculating K from the given ∆G° value, one can determine the ratio of equatorial to axial isopropylcyclohexane molecules, which is key to answering the question.

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Related Practice

Textbook Question

For the following acid–base reactions studied in Assessment 5.25, draw a likely transition state. Be sure to indicate in your drawing the degree to which bonds are broken or formed.

(a)

Textbook Question

For the following acid–base reactions studied in Assessment 5.25, draw a likely transition state. Be sure to indicate in your drawing the degree to which bonds are broken or formed.

Textbook Question

For the following acid–base reactions studied in Assessment 5.25, draw a likely transition state. Be sure to indicate in your drawing the degree to which bonds are broken or formed.

(b)

Textbook Question

Calculate ∆G° for the conversion of “axial” methylcyclohexane to “equatorial” methylcyclohexane at 25 °C.

Textbook Question

The relative rate of reaction for the cis alkene (E) is given in Table 22.2. What do you expect the relative rate of reaction for the trans alkene to be?

Textbook Question

Explain why the half-life (the time it takes for one-half of the compound to be metabolized) of Xylocaine is longer than that of Novocaine.

Textbook Question

In a reaction in which reactant A is in equilibrium with product B at 25 °C, what relative amounts of A and B are present at equilibrium if ∆G° at 25 °C is

a. 2.72 kcal/mol?

b. 0.65 kcal/mol?

Textbook Question

In a reaction in which reactant A is in equilibrium with product B at 25 °C, what relative amounts of A and B are present at equilibrium if ∆G° at 25 °C is

c. -2.72 kcal/mol?

d. -0.65 kcal/mol?

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