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76 Chapter 4 CYCLOALKANES 52. Addition of Cl₂ in each case generates a substituted ICl₂ unit. Then, in the presence of light, these become I-CI groups, which can chlorinate nearby C-H bonds according to the propagation steps shown in the answer to Problem 51(a). The selectivity comes from the fact that in the reaction for (a) the group can reach the H at position 9 most easily, whereas in (b) it most easily reaches the H at position 14. Your models should look something like these: (a) CH₃ (b) CH₃ CH₃ CH₃ 9 14 H O H CH₂ CI O O I 53. (a) Without making a model, this problem may initially make no sense to you at all. Begin by finding the better chair conformation for A. Of the two possibilities, the one with both alkyl groups equatorial is preferable: H₃C O CH₃ Diaxial CH₃ O Diequatorial CH₃ If we digress a moment for a closer look at these structures, we find that the conformations about the external bond connecting the ring to the large substituent at C4 are not optimal. As discussed briefly in the solution to Problem 33 of this chapter, the free rotation available to a simple alkyl group (such as CH₃) causes it to be more sterically demanding than is a comparable molecular fragment contained in a ring. Furthermore, the ring angles in cyclopropanes are only 60°, further shrinking the volume in space taken up by this substituent. As a result, a better conformation for the diequatorial structure is one in which a 120° rotation about the aforementioned bond takes the two methyl groups as far apart as possible and puts the compressed cyclopropane ring on the side of the bond closest to the methyl group at C3. Similar rotation in the diaxial moves the CH₃ group away from the axial hydrogens on the same face of the cyclohexane ring: CH₃ O Diaxial CH₃ CH₃ Diequatorial CH₃ (b) Compound B results from opening of the cyclopropane ring in A. This process generates a (tert-butyl) group in place of the substituent. For purposes of evaluating steric interactions, this new tert-butyl group is effectively much larger, for two