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Solutions to Problems 191 forms. Any answer that uses one of the intermediate cations shown above is, by definition, wrong! Look again at the result obtained from D in part (b). Had we used the most stable conformation of D, the cation would be identical to that of A and would necessarily give the same product. Since the observed product is not the same, the intermediate must not be the same, thus the ring flip prior to loss of bromide. Is there any lesson here? It appears that the silver ion-promoted reaction prefers an axial Br. Furthermore, inspection of the reactions in part (b) reveals that in each case an adjacent axial group reacts with the carbocation. So we return to compound B. Examine its structure. There's something funny going on here. If we flip to the alternative all-axial H₅C₆ chair and have silver remove bromide, we get which should ring-close + :OH like a shot to give the same oxacyclopropane product that forms upon treatment of B with base. Since that doesn't happen, let's think of something different. We know that this conformation is poor, over 4 kcal mol⁻¹ higher in energy than the alternative. According to Table 2-1, less than 0.1% (one in a thousand) of the molecules of B will adopt this conformation at any time. So the likelihood is that the reaction of B with silver ion proceeds through the all-equatorial conforma- tion, contrary to the apparent preference of silver for removing axial bromine. But it would still be wrong to simply suppose that silver removes Br from this conformation, because the result would be the same cation that formed from isomer C, which converted to cyclohexenol. In case you hadn't noticed, we've just ruled out all the carbocations as possibilities. What's left? Perhaps silver can't promote removal of equatorial Br to give a cation, but, instead, rearrangement occurs simultaneously with Br loss, bypassing the ring carbocation: + H OH OH O C₆H₅ Br C₆H₅ BrAg -AgBr C₆H₅ + Ag+ 0 CH O which is C₆H₅ C₆H₅ Two points: Notice that the cyclohexane ring bond that migrates is the one located anti to the C-Br bond being broken (this anti business-backside attack in SN2, the E2 transition state, and now this-keeps showing up, doesn't it?). Furthermore, the cation that results places a positive charge on an substituted carbon atom, permitting stabilization by resonance with oxygen's lone pairs and providing the driving force for the rearrangement. 75. (c) 76. (c) 77. (c) 78. (a) P.S. Regarding the question in Section 9-3 of this study guide, the alcohol forms three products. CH₃ + H + CH₃ CH₃ CH₃ + CH₃ CH₃