Opp5an-07 - ’ s one suggestion try the reaction without...

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Opportunity 5 Answer, Chemistry 302X- 2007 O HO O O O O KOH CH 3 OH O O O O O O O O O O HO O + LiOH reverse aldol (-) (-) rotate (-) Michael deprotonate protonate deprotonate (-) protonate aldol What Kevin Bartlett and your ancestors spotted (being mathematically inclined they can count to six) was that a Cope rearrangement leads to the enol of the final ketone product. In fact, in the terminology of the trade, this is an oxy-Cope rearrangement, discovered by an obscure postdoc in Wisconsin in 1964. Normally oxy-Cope rearrangements require lots of heat, just like normal Copes. But when the 1,5-diene system is attached to an alkoxide, the rearrangement is accelerated by many powers of ten. So, probably what really happens here is that the alcohol is deprotonated and the rapid oxy-Cope then ensues. But how would one ever tell? Cope O HO O HO 1 2 3 4 5 6 O HO = O O
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There is probably more than one answer to the rhetorial question above, but here
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Unformatted text preview: ’ s one suggestion: try the reaction without the carbonyl group in the five-membered ring. The “Fun in Base” mechanism must fail, whereas the Cope could still work. It is not a perfect answer, though - why not? - we have made one of the competitors impossible - under such circumstances, the “second best” process could take over. Far better would be to do the reaction with optically active starting material. The Cope mechanism must produce optically active product whereas the “aldol” mechanism cannot. Easy if you see it. Remember that if you run a reaction in D 2 O/DO – , all alpha and other acidic positions will exchange. You can ’ t just hope that the one you want will exchange and the others not. Similarly, any deuterium in an alpha or other acidic position will wash out in H 2 O/HO – ....
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Opp5an-07 - ’ s one suggestion try the reaction without...

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