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Chapter 31 Suggested solutions for Chapter 31 261 used Suggested solution The starting material is SO it has lost nitrogen and gained - one molecule of methanol. We can see the MeO group at 3.5 and the four CH: groups in the ring are still there 8H m at All that is left is a multiplet at 4.2, obviously next to MeO, and a pair of alkene signals at 5.5 and 5.8, coupled with = 17.9 Hz - obviously an One end of the alkene (5.5) is coupled to one proton, the other (5.8) 2. We now have these fragments. ? H ? A ?. ? MeO H ? H H ? These add up to too much! Clearly, the CH attached to OMe and the CH attached to the alkene must be the same and the CH₂ at the other end of the alkene must be one end of the chain of four. We now have a structure, but it doesn't join up. ? H A the - ? OMe ne H H of an This is the test of your belief the dotted ends must be joined up to give the structure of A. Yes, This was the discovery of by this does put an E-alkene in a seven-membered ring, and it is difficult to draw. but you were warned H Angew. Int. Ed. 1980. 19. 1032. If you were that the compound is unstable. The group next to the CHOMe group is diastereotopic and so really on the you might have the coupling constants are different. noticed that a is Z-TM chiral so this compound must be a ? H single diastereoisomer too. though A we don't know which. = ? OMe but a H H of the Problem 5 E-TM Why do these reactions give different alkene geometries? 0 0 OH 0 HCO₂Et base '.9 Purpose of the problem n). Revision of enol chemistry from Chapters 21 and 27 with the question of enol added. the Suggested solution The geometry of the enol product of the first reaction changes easily because of delocalization and tautomerism. It settles down in the more favourable geometry under thermodynamic control and the Z-alkene is favoured because of hydrogen bonding. H H. 0 OH 0 OH 0 0 H CHO 0