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Chapter 4: Problem 68

The energy barrier for rotation about the \(\mathrm{C}-\mathrm{C}\) bond in ethane is about \(3 \mathrm{kcal}\), which suggests that the energy required to bring one pair of hydrogen into an eclipsed arrangement is 1 kcal. Calculate how many kilocalories the planar form and extreme boat form of cyclohexane are likely to be unstable relative to the chair form on account of \(\mathrm{H}-\mathrm{H}\) eclipsing and flagpole interactions.

Short Answer

Expert verified
The planar form of cyclohexane is likely to be unstable by \(3 \mathrm{kcal}\) and the boat form by \(6 \mathrm{kcal}\) relative to the chair form when considering H-H eclipsing and flagpole interactions.

Step by step solution

01

Identify the number of H-H eclipsing interactions in each conformation

Assuming the planar conformation for cyclohexane, the H-H eclipsing interactions would be for every adjacent hydrogens with bond stability of 1 kcal. In a hexagon, there are a total of 6 hydrogen atoms, but since each H-atom participates in 2 interaction, therefore, there would be 3 distinct H-H eclipsing interactions in planar cyclohexane. For the chair conformation, there are no H-H eclipsing interactions since the atoms are staggered, making it more stable than the planar conformation. In the boat conformation, there are 4 H-H eclipsing interactions between hydrogen atoms 1 - 2, 2 - 3, 4 - 5, and 5 - 6 and 2 flagpole interactions between hydrogen atoms 1 - 4 and 3 - 6.
02

Calculate the instability of the planar and boat forms relative to the chair form

As the chair form doesn't have any H-H eclipsing and flagpole interactions, we simply need to calculate the energy difference between the other 2 forms and chair form. Since we are given an energy barrier of 1 kcal for each H-H eclipsing interaction: In the planar form, there are 3 H-H eclipsing interactions giving a total of \(3\times 1\mathrm{kcal} = 3 \mathrm{kcal}\) of interaction energy. In the boat form, there are 4 H-H eclipsing interactions and 2 flagpole interactions giving a total of \(4 \times 1 \mathrm{kcal} + 2 \times 1 \mathrm{kcal} = 6 \mathrm{kcal}\) of interaction energy.
03

Present the results

In conclusion, the planar form of cyclohexane is likely to be unstable by \(3 \mathrm{kcal}\) and the boat form by \(6 \mathrm{kcal}\) relative to the chair form when considering H-H eclipsing and flagpole interactions.

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Most popular questions from this chapter

Write structural formulas for all of the possible cistrans isomers of the following compounds: a. \(1,2,3\) -trimethylcyclopropane b. 1,3 -dichlorocyclopentane c. \(1,1,3\) -trimethylcyclohexane d. (3-methylcyclobuty1)-3-methylcyclobutane

Would you expect cis- or trans-1, 2-dimethylcyclopropane to be the more stable? Explain.

(a) trans-1, 2-Dimethylcyclohexane exists about \(99 \%\) in the diequatorial conformation; trans-1, 2-Dibromocyclohexane on the other hand, exists about equally in diequatorial and diaxial conformations. Furthermore, the fraction of the diaxial conformation decreases with increasing polarity of the solvent. How do you account for the contrast between the dimethyl and dibromo compounds? (b) If trans-3-cis-4-dibromo-tert-butylcyclohexane is subjected to prolonged heating, it is converted into an equilibrium mixture (about \(50: 50\) ) of itself and a diastereomer. What is the diastereomer likely to be? How do you account for the approximately equal stability of these two diastereomers? [Here, and in (c), consider the more stable conformation of each diastereomer to be the one with an equatorial tert-buty1 group]. (c) There, are two more diastereomeric 3,4 -dibromo tert-butylcyclohexanes. What are they? How do you account for the fact that neither is present to an appreciable extent in the equilibrium mixture?

Which of the following compounds are resolvable, and which are non-resolvable? Which are truly meso compounds? Use models as well as drawings. (d) tran-1, 2-cyclohexanediol (a) cis-1,2-cyclohexanediol (b) trans-1,2-cyclohexanediol (e) cis-1,4-cyclohexanediol (c) cis-1,3-cyclohexanediol (f) trans-1,4-cyclohexanediol

There are two isomers of decalin. One is of higher energy and the other is more stable. Which is the stabler form? Why?
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