Ch. 5 - Alkenes: Structure, Nomenclature, and an Introduction to Reactivity • Thermodynamics and Kinetics

Bruice8th EditionOrganic ChemistryISBN: 9780135213711Not the one you use?Change textbook

Bruice 8th Edition

Ch. 5 - Alkenes: Structure, Nomenclature, and an Introduction to Reactivity • Thermodynamics and Kinetics

Problem 20a

Chapter 6, Problem 20a

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

a. For a reaction with ∆H° = -12 kcal/mol and ∆S° = 0.01 kcal mol^-1 K^-1, calculate the ∆G° and the equilibrium constant at:

1. 30 °C and 2. 150 °C.

b. How does ∆G° change as T increases?

c. How does K_eq change as T increases?

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Textbook Question

For each of the reactions in Problem 15, indicate which reactant is the nucleophile and which is the electrophile.

1615

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Textbook Question

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

1311

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Textbook Question

Why is the percentage of molecules with the substituent in an equatorial position greater for isopropylcyclohexane than for fluorocyclohexane?

1103

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Textbook Question

The ∆G° for conversion of “axial” fluorocyclohexane to “equatorial” fluorocyclohexane at 25 °C is -0.25kcal/mol. Calculate the percentage of fluorocyclohexane molecules that have the fluoro substituent in an equatorial position at equilibrium.

1151

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Textbook Question

a. For which reaction in each set will ∆S° be more significant?

b. For which reaction will ∆S° be positive?

1. A ⇌ B or A + B ⇌ C

2. A ⇌ B + C or A + B ⇌ C + D

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