Chapter 4: Problem 70
Would you expect cis- or trans-1, 2-dimethylcyclopropane to be the more stable? Explain.
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These are the key concepts you need to understand to accurately answer the question.
Chapter 4: Problem 70
Would you expect cis- or trans-1, 2-dimethylcyclopropane to be the more stable? Explain.
These are the key concepts you need to understand to accurately answer the question.
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Get started for free(a) How many pairs of eclipsed hydrogen are present in planar cyclobutane? (b) In puckered cyclobutane ?
(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?
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.
There are two isomers of decalin. One is of higher energy and the other is more stable. Which is the stabler form? Why?
Investigate the thermodynamic feasibility of the following propagation steps for opening the rings of cycloalkanes with \(\mathrm{n}=2-6 \mathrm{by}\) a free-radical mechanism. (1) \(\left(\mathrm{CH}_{2}\right)_{\mathrm{n}}+\mathrm{Br}^{-} \rightarrow \mathrm{BrCH}_{2}\left(\mathrm{CH}_{2}+\mathrm{n}-2 \mathrm{CH}_{2}\right.\) (2) \(\mathrm{BrCH}_{2}\left(\mathrm{CH}_{2}+\mathrm{n}-2 \mathrm{CH}_{2}+\mathrm{Br}_{2} \rightarrow\left(\mathrm{CH}_{2}\right)_{\mathrm{n}-2}\left(\mathrm{CH}_{2} \mathrm{Br}\right)_{2}+\mathrm{Br}^{-}\right.\) Use 83 kcal for the bond-dissociation energy of a normal C-C bond and \(68 \mathrm{kcal}\) for the bond-dissociation energy of a C-Br bond. (An easy way to solve a problem of this type is to first calculate AH of each step for cyclohexane where there is no strain, and then make suitable corrections for the strain which is present for small values of n.) Refer to the following table for your answer. Table 1. Total Strain of Cycloalkanes $$ \begin{array}{|c|c|c|} \hline \begin{array}{c} \text { Cycloalkane } \\ \left(\mathrm{CH}_{2}\right)_{\mathrm{n}} \end{array} & \mathrm{n} & \begin{array}{c} \text { Total strain } \\ (\mathrm{kcal} / \text { moles }) \end{array} \\ \hline \text { ethylene } & 2 & 22.4 \\ \hline \text { cyclopropane } & 3 & 27.6 \\ \hline \text { cyclobutane } & 4 & 26.4 \\ \hline \text { cyclopentane } & 5 & 6.5 \\ \hline \text { cyclohexane } & 6 & 0.0 \\ \hline \end{array} $$
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